Taylor Series study notes tute0064
video list , update 2015-03-18
Math Constants rank list , EulerMascheroni sum, bound, pendulum
Zeta function: 01, 02, 03, 04, 05, 06, 07, 08, 09, 10, 11, 12, 13, 14, 15, 16, 17
Gamma function: 01, 02, 03, 04, 05, 06, 07, 08, 09, 10, 11, 12
Taylor Series: 01, 02, 03, 04, 05, 06, 07, 08, 09
Why gamma function this way? Γ(s) = ∫[t=0,t=∞]{ts-1*e-t*dt} eq.CG01
Integral Test, Euler's formula, Integration by parts
How to prove eq.CG16? Taylor series expansion equation
Analysis of a Complex Kind. Prof. Petra Bonfert-Taylor




<a name="Taylor_Index">
2015-02-08-21-30 LiuHH access 
https://www.youtube.com/playlist?list=PL68EAB0099AFEAAA5
2015-02-08-23-51 use YTD download 
Taylor Series
•by MrYouMath
•9 videos
•2,037 views
•Last updated on Apr 22, 2014

<a name="TaylorIndex"> 
Taylor Series - 1 - Motivation and Derivation.mp4
http://www.youtube.com/watch?v=ttnl671QBcE
Taylor Series - 1 - Radius of Convergence (Exercise).mp4
http://www.youtube.com/watch?v=QKvC2uQymf4
Taylor Series - 1 - Radius of Convergence by Singularities (Exercise).mp4
http://www.youtube.com/watch?v=JkukU802NiY
Taylor Series - 2 - Geometric Series and the tortoise.mp4
http://www.youtube.com/watch?v=UUqdmgw7QT8
Taylor Series - 3 - Logarithm and alternating harmonic series.mp4
http://www.youtube.com/watch?v=UZQS8D_zNNE
Taylor Series - 4 - Arctan and Pi.mp4
http://www.youtube.com/watch?v=XBKGuIQK1Ek
Taylor Series - 5 - Binomial Functions and Einsteins E=mc².mp4
http://www.youtube.com/watch?v=tZHRWPOXaaA
Taylor Series - 6 - Trigonometric functions and Foucault's pendulum.mp4
http://www.youtube.com/watch?v=ycxR9HJnazo
Taylor Series - 7 - Exponential Function and Euler's Formula.mp4
http://www.youtube.com/watch?v=rXHQUVwfKwE

<a name="Gamma_Indx"> 
2015-01-01-15-18 start 
On 
2014-03-25-20-36 Liu,Hsinhan access
http://www.youtube.com/playlist?list=PL3E4136E122545FBE
find and download next 12 video files. 

Gamma Function 
•by MrYouMath
•12 videos
•5,940 views
•1 hour, 48 minutes
<a name="GammaIndex"> 
Gamma Function - Part 1 - Functional Equation
http://www.youtube.com/watch?v=2iBNo4j3vRo
Gamma Function - Part 2 - Gauss Representation
http://www.youtube.com/watch?v=a_Dlx7TTjkI
Gamma Function - Part 3 - Weierstrass Representation
http://www.youtube.com/watch?v=d9Oz62Ioue0
Gamma Function - Part 4 - Relationship to Sine
http://www.youtube.com/watch?v=W3ulxj3s90U
Gamma Function - Part 5 - Gamma of 0.5 ( one half)
http://www.youtube.com/watch?v=_vwqsJNKY-c
Gamma Function - Part 6 - Stirling's Approximation
http://www.youtube.com/watch?v=MuAb2dnPD0Q
An intuitive derivation of Stirling’s formula
http://www.rowan.edu/open/depts/math/osler/Stirlings_formula.pdf

Gamma Function - Part 7 - Euler Integral I
http://www.youtube.com/watch?v=VF7ud3Al6d8
Gamma Function - Part 8 - Euler Integral II The Sinc-Function
http://www.youtube.com/watch?v=h4wQ_F1s1UI
Gamma Function - Part 9 - Euler Integral III Fresnel Integral
http://www.youtube.com/watch?v=H0kT-EKbUzM
Gamma Function - Part 10 - Beta Function
http://www.youtube.com/watch?v=Korx_G7eySQ
Gamma Function - Part 11 - Legendre Duplication Formula
http://www.youtube.com/watch?v=yu9NeDXalpA
Gamma Function - Part 12 - Relation to Zeta Function
http://www.youtube.com/watch?v=jQAPlsNY_P0
Above red link has complete lecture. Blue not.
Gauss multiplication formula half way, not done!
https://proofwiki.org/wiki/Gauss_Multiplication_Formula
Γ(z) has a simple pole with residue (-1)n/n! 
at z=-n for n = 0, 1, 2, ...
http://www.math.leidenuniv.nl/~evertse/ant13-8.pdf
2015-01-01-15-38 stop

<a name="ZetaIndex"> 
2014-10-29-07-56
https://www.youtube.com/playlist?list=PL32446FDD4DA932C9
2014-10-29-08-24
MrYouMath_17_files_Riemann.htm
"Zeta Function - Part 1 - Convergence" 
http://www.youtube.com/watch?v=ZlYfEqdlhk0
"Zeta Function - Part 2 - Euler Product Representation" 
http://www.youtube.com/watch?v=I3qSCWNXZKg
"Zeta Function - Part 3 - Euler Product (easy)" 
http://www.youtube.com/watch?v=TDdGisWD5OU
<a name="docA007">
"Zeta Function - Part 4 - Infinitude of Prime Numbers" 
http://www.youtube.com/watch?v=SKa7b-3C32A
"Zeta Function - Part 5 - Prime Zeta Function" 
http://www.youtube.com/watch?v=3eN9tQX3JJ4
"Zeta Function - Part 6 - The Prime Counting Function" 
http://www.youtube.com/watch?v=U16_KTTKtb0
"Zeta Function - Part 7 - Zeta of 2 aka The Basel Problem" 
http://www.youtube.com/watch?v=GeKDmoYHiAk
"Zeta Function - Part 8 - Zeta of 2n - Part 1" 
http://www.youtube.com/watch?v=axQqExF7NsU
<a name="docA008">
"Zeta Function - Part 8 - Zeta of 2n - Part 2" 
http://www.youtube.com/watch?v=XHQ0OzqTjd0 9/17
"Zeta Function - Part 8 - Zeta of 2n - Part 3" 
http://www.youtube.com/watch?v=1f24RZfP6m8 10/17
"Zeta Function - Part 9 - Relation to Gamma Function" 
http://www.youtube.com/watch?v=UEZ4ClCdog8 11/17
"Zeta Function - Part 10 - Jacobi Theta Function" 
http://www.youtube.com/watch?v=-GQFljOVZ7I 12/17
"Zeta Function - Part 11 - Riemann Functional Equation I" 
http://www.youtube.com/watch?v=K6L4Ez4ZVZc 13/17
<a name="docA009">
"Zeta Function - Part 12 - Riemann Functional Equation II" 
http://www.youtube.com/watch?v=TnRnlJBecRg 14/17
"Zeta Function - Part 13 - Trivial Zeros of the Zeta Function" 
http://www.youtube.com/watch?v=G-fqe3BkBnE 15/17
"Zeta Function - Part 14 - Riemann Xi Function" 
http://www.youtube.com/watch?v=QfDbF_qlp58 16/17
"Sine Function Product Formula (Hadamard Factorization Theorem)"
https://www.youtube.com/watch?v=h3Hcioh2F9I 17/17
"What is a function? Why 1+2+3+4+5+.... not equals -1/12 = Zeta(-1)" 
http://www.youtube.com/watch?v=wt6ngy6pDws This video is easy. skipped a312190901


<a name="docAi01"> EulerMascheroni sum bound pendulum 2015-02-13-12-20 start Next is study notes of Taylor Series - 1 - Motivation and Derivation http://www.youtube.com/watch?v=ttnl671QBcE Lecturer is MrYouMath. Notes recorder is Liu,Hsinhan. Recorded notes is about 60% match, not exact. Liu,Hsinhan inserted text is blue Italic. LiuHH notes may contain error! <a name="docAi02"> Begin video 1 of 9 Hello and welcome everyone. This is MrYouMath with new series about the Taylor series. I want start by the motivation of the Taylor series and a little bit derivation and go ahead see what you can do with Taylor series with few example on special functions and their Taylor series and so forth. ..... <a name="docAi03"> Where we start is a simple problem. f(x)=a0+a1x+a2x2 --- eq.TS01 eq.TS01 = equation Taylor Series number 01 f(0), f'(0) and f''(0) are known. eq.TS01 is a quadratic equation. We want to find out a0, a1, a2 three coefficients when we only know f(0), f'(0) and f''(0). 2015-02-13-12-47 stop <a name="docAi04"> 2015-02-13-14-00 start First of all we know the value of f(x) at x=0. We know the value of f(x) first derivative at x=0. We know the value of f(x) second derivative at x=0. In eq.TS01 plug in x=0, we get f(0)=a0+a10+a202 --- eq.TS02 which give us f(0)=a0, since f(0) is known and a0 is unknown. eq.TS02 tell us a0=f(0) --- eq.TS03 <a name="docAi05"> Next go ahead, differentiate eq.TS01 get df(x)/dx=d[a0]/dx+d[a1x]/dx+d[a2x2]/dx --- eq.TS04 a0 is constant. Differentiate a constant get zero. 'differentiate' mean take a small variation. But a constant has no variation. Therefore d[constant]/dx=0 --- eq.TS05 In eq.TS04 d[a1x]/dx, a1 is second constant. a1 not vary, then d[a1x]/dx is same as a1*d[x]/dx Here d[x]/dx is dx/dx, numerator and denominator change same amount, result is one. Then d[a1x]/dx=a1*d[x]/dx=a1*1=a1 --- eq.TS06 <a name="docAi06"> In eq.TS04 d[a2x2]/dx, a2 is third constant. constant contribute nothing in differentiation process. Take constant to front , d[a2x2]/dx become a2*d[x2]/dx .

Now d[x2]/dx is d[x*x]/dx. There are two x (x*x) to be differentiated, Calculus differentiation law say d[x*x]/dx={d[x]/dx}*x + x*{d[x]/dx} --- eq.TS07 <a name="docAi07"> Since dx/dx=1, eq.TS07 become eq.TS08 below d[x*x]/dx=1*x + x*1=2x --- eq.TS08

Reader can use similar method find out d[x3]/dx=3*x2 --- eq.TS09 In general d[xn]/dx=n*xn-1 --- eq.TS10 Now back to d[a2x2]/dx its value is d[a2x2]/dx=a2*2x --- eq.TS11 <a name="docAi08"> Substitute eq.TS05, eq.TS06, eq.TS11 into eq.TS04 get df(x)/dx=0+a1+2x*a2 --- eq.TS12 2015-02-13-14-38 here
In eq.TS12 plug in x=0 get f'(0)=0+a1+0*a2 You will find a1=f'(0) --- eq.TS13 f'(0) is given. unknown a1 become known. <a name="docAi09"> Then you differentiate second time d2f(x)/dx2=d[a1]/dx+d[2x*a2]/dx d2f(x)/dx2=0+2*a2 --- eq.TS14 In eq.TS14 set x=0, then realize 2*a2=f''(0). Get a2=f''(0)/[2*1] --- eq.TS15 f''(0) is given. Unknown a2 become known. In eq.TS14 it is actually f''(0)=1*2*a2 Why? We can see from third power term. <a name="docAi10"> If eq.TS01 has third power term like f(x)=a0+a1x+a2x2+a3x3 --- eq.TS16 First differentiate a3x3 become 3*a3x2 second differentiate a3x3 become 3*2*a3x1 third differentiate a3x3 become 3*2*1*a3x0 a3 has coefficient 3! and 3!*a3=f'''(0) x0=1, therefore a3=f'''(0)/3!=f'''(0)/6 --- eq.TS17 Let us re-substitute eq.TS03, eq.TS13 and eq.TS15 and eq.TS17 into eq.TS01 get
<a name="docAi11">
 
f(x)
=
f(0)
+
f '(0)

1!
*x
+
f ''(0)

2!
*x2
+
f '''(0)

3!
*x3
+ ...
--- eq.TS18
Compare eq.TS01 with eq.TS18. In eq.TS01 a0, a1, a2 are unknown.
In eq.TS18, coefficients f(0), f'(0)/1!, f''(0)/2! are known.
width of above equation a402131527
<a name="docAi12">
2015-02-13-15-38 here 
eq.TS18 has an interesting pattern. We can write 
down f(x) equal f(x) evaluated at x=0, plus 
f prime evaluated at x=0 over 1! multiply x, plus 
f double prime evaluated at x=0 over 2! multiply 
x2. (f prime=f'(x)=d[f(x)]/dx 
    f double prime=f''(x)=d2f(x)/dx2)
If you take three factorial, this 
means 3*2*1 you get 6 (eq.TS18 Purple term divide 
by six) We can extend eq.TS18 to higher order. 
Plus x3 plus x4, but all higher order term 
coefficients are zero. (In eq.TS01 a3=a4=...=0)
<a name="docAi13">
You can think of any function you want. Actually 
any continuous, differentiable function. 
Fourier Series handle dis-continuous function.
Taylor series NOT handle dis-continuous function.
You can assume eq.TS18 is a polynomial with 
infinite many terms in it. You can try to find 
these coefficients a0, a1, a2 etc. They will be 
calculated exactly the same way as we did above. 

<a name="docAi14">
Next let us see more complicated situation. 
We do not know function value at x=0, but we 
know function value at x=x0. (x0≠0)
  f(x)=a0+a1(x-x0)+a2(x-x0)2+a3(x-x0)3 ... --- eq.TS19 
Principle stay the same. Set x=x0 eq.TS19 become 
  f(x0)=a0+a1(x0-x0)+a2(x0-x0)2+a3(x0-x0)3 ... --- eq.TS20 
which is 
  f(x0)=a0+0+0+0 ... --- eq.TS21 
In eq.TS21, a0 is unknown and f(x0) is known. 
eq.TS21 tell us that unknown a0 become known f(x0)

<a name="docAi15">
Let us do first derivative of eq.TS19. 
  f'(x)=1*a1+2*a2(x-x0)1+3*a3(x-x0)2 ... --- eq.TS22 
In eq.TS22 set x=x0 get 
  f'(x0)=1*a1+2*a2(x0-x0)1+3*a3(x0-x0)2 ... --- eq.TS23 
eq.TS23 is actually 
  f'(x0)=a1+0+0+0 ... --- eq.TS24 
In eq.TS24, a1 is unknown and f'(x0) is known. 
eq.TS24 say that unknown a1 become known f'(x0)/1!.

<a name="docAi16"> 
Do same thing for second derivative of eq.TS19. 
  f''(x)=2*1*a2(x-x0)0+3*2*a3(x-x0)1 ... --- eq.TS25 
In eq.TS25 set x=x0 get 
  f''(x0)=2*1*a2+3*2*a3(x0-x0)1 ... --- eq.TS26 
eq.TS26 is actually 
  f''(x0)=2*1*a2+0+0+0 ... --- eq.TS27 
In eq.TS27, a2 is unknown and f''(x0) is known. 
eq.TS27 say that unknown a2 become known f''(x0)/2!.
  a2=f''(x0)/2! --- eq.TS28 

<a name="docAi17">
Do third derivative of eq.TS19. 
  f'''(x)=3*2*1*a3+4*3*2*a4(x-x0)1 ... --- eq.TS29 
In eq.TS29 set x=x0 get 
  f'''(x0)=3*2*1*a3+4*3*2*a4(x0-x0) ... --- eq.TS30 
eq.TS30 is actually 
  f'''(x0)=3*2*1*a3+0+0+0 ... --- eq.TS31 
In eq.TS31, a3 is unknown and f'''(x0) is known. 
eq.TS31 say that unknown a3 become known f'''(x0)/3!.
  a3=f'''(x0)/3! --- eq.TS32
2015-02-13-16-31 stop at video 09:22 

<a name="docAi18">
2015-02-13-19-26 start 
If we have n-th derivative, then 
  f(n)(x0)=n*(n-1)*(n-2)*...*3*2*1*an --- eq.TS33 
We can write eq.TS33 in more compact form 
  f(n)(x0)=n!*an --- eq.TS34 
where n! is n factorial. 
  n!=n*(n-1)*(n-2)*...*3*2*1 --- eq.TS35 
Since an is unknown and f(n)(x0) is known 
  an=f(n)(x0)/n! --- eq.TS36 
Where 
  f(n)(x) =define= dnf(x)/(dx)n --- eq.TS37 
f(n)(x0) is f(n)(x) evalue at x=x0

<a name="docAi19">
If you have a differentiable (continuous) function
(discontinuous function excluded) you can write 
  f(x)=a0+a1(x-x0)+a2(x-x0)2+a3(x-x0)3 ... --- eq.TS19 
If f(x) is exponential function ex, to find an
use 
  an=f(n)(x0)/n! --- eq.TS36 
What do we get by doing eq.TS36 ? Write eq.TS19 as 
<a name="docAi20">
 
f(x)
=
f(0)(x0)

0!
+
f(1)(x0)

1!
*(x-x0)
+
f(2)(x0)

2!
*(x-x0)2
+
f(3)(x0)

3!
*(x-x0)3
+
...
--- eq.TS38
 
f(x)
=
n=∞
n=0
f(n)(x0)

n!
*(x-x0)n
           
--- eq.TS39
width of above equation a402132013
<a name="docAi21"> 
2015-02-13-20-20 here
Write eq.TS38 in compact form get eq.TS39 
Zero-th derivative of a function is defined to be 
function itself. Zero-th factorial is defined to 
be one. eq.TS39 is function f(x) Taylor series 
expression at point x0 (at ONE POINT) 
What is important about it is that your function 
should be differentiable. This Taylor series 
method does not work, if your function is not 
differentiable, not continuous. Important thing 
to know is that function f(x) should be infinite 
many times differentiable. 

<a name="docAi22">
Second important point is that since we add 
infinite many terms, one could question himself 
does the answer has a meaningful value? This is 
the topic about convergence. MrYouMath do not 
bother talk here. This series eq.TS39 does not 
always converge for every value of x. For some 
function ex, sin(x), cos(x) you can plug in any 
value for x, eq.TS39 will converge. Some function 
have radius of convergence. The converge x is not 
infinite, but finite. 

<a name="docAi23">
Let us look at Cauchy-Hadamard Theorem. 
 
radius of
convergence
=
1

lim[n→∞]{sup n√(|an|) }
--- eq.TS40
 
radius of
convergence
=
limit
n→∞
|an|

|an+1|
       
--- eq.TS41
width of above equation a402132058
<a name="docAi24">
2015-02-13-20-59 here 
eq.TS41 is easier to calculate than eq.TS40. If you 
remember the quotient criteria for convergence sums 
or sometime called the D'Alembert's test. eq.TS41 is 
easier to calculate. Calculate |an|/|an+1|
eq.TS40 and eq.TS41 get same answer. If eq.TS40 or 
eq.TS41 exist, you get finite radius of convergence.
You can actual have a zero in eq.TS40 or eq.TS41. 
If you have one over zero, that is infinite. Your 
radius of convergence is infinite large. For example 
ex, sin(x), cos(x) (converge everwhere). 

<a name="docAi25">
Finite radius of convergence, for example logarithm 
and geometric series, for example arctan().
Why am I talk about radius of convergence? Because 
it is natural to consider x be a complex variable. 
Generalize Taylor polynomial called power sums for 
complex numbers. Then you have a circle in the 
complex plane. In this circle really express f(x). 
(this circle define radius of convergence)
Outside of this radius, you do not have convergence. 
Radius of convergence is not known in advance. 
If we talk about real numbers, this is just interval 
of convergence. Series will converge between +r and 
-r. 

<a name="docAi26"> 
Where do we need Taylor Series? First of all you can 
use Taylor Series for cracking a very hard problem 
in physics. It is very easy thing to look at a 
differential equation and solve by method of power 
sum. This often work quite good. Another problem you 
can do is if you have a differential equation very 
hard to solve. You can take Taylor Series of one 
of the functions which make this problem hard. Break 
it up and say your x is pretty small so you are 
pretty good at estimate. You can use Taylor Series 
to show the irrationality of e (2.718281828459045...)
You can find series expression for PI, for e for 
every strange number you can think of. 
I hope you have fun watching this video. See you guys.
2015-02-13-21-53 stop 
2015-02-13-23-15 done first proofread
2015-02-14-07-59 done second proofread


<a name="a402272242"> 2015-02-27-22-42 Liu,Hsinhan more activity please see http://freeman2.com/bioge010.htm English http://freeman2.com/biogc012.htm Chinese
<a name="docAi31"> EulerMascheroni sum bound pendulum 2015-02-25-13-33 start Next is study notes of Taylor Series - 1 - Radius of Convergence (Exercise).mp4 http://www.youtube.com/watch?v=QKvC2uQymf4 Lecturer is MrYouMath. Notes recorder is Liu,Hsinhan. Recorded notes is about 60% match, not exact. Liu,Hsinhan inserted text is blue Italic. LiuHH notes may contain error! <a name="docAi32"> Begin video 2 of 9 Hello welcome everyone. This is MrYouMath again. This is more apply video I want go ahead show you how to calculate Radius of Convergence using ratio test. Write down Taylor series or power sum. We will start slowly and try to find out the radius of convergence
<a name="docAi33">
exp(x) =
n=∞
n=0
xn

n!
=
x0

0!
+
x1

1!
+
x2

2!
+
x3

3!
+ ●●●
x100

100!
+ ●●●
--- eq.TS42
an
=
xn

n!
an+1
=
xn+1

(n+1)!
q =
lim
n→∞
an+1

an
--- eq.TS43
q=
lim
n→∞
xn+1

(n+1)!
n!

xn
=
lim
n→∞
x*n!

(n+1)!
--- eq.TS44
q=
lim
n→∞
x*n!

(n+1)n!
=
lim
n→∞
x

n+1
=0
--- eq.TS45
width of above equation a402251442
<a name="docAi34">
2015-02-25-14-43 here 
eq.TS42 is ex Taylor series expansion. 
eq.TS43 define an, define an+1, define q.
an is n_th Taylor series expansion element.
an+1 is n+1_th Taylor series expansion element.
q is the ratio of ratio test.
ratio test use two neighbor elements ratio to 
find convergence or divergence answer. 
<a name="docAi35">
ratio is far away element (n+1_th) divide by 
near element (n_th). If ratio is less than 1 
and great than -1, this Taylor series expansion 
converge. Otherwise diverge. 
To remove negative value factor, take absolute 
value of ratio, convergence requirement is 0<=q<1 .
Ratio test convergence make sure n+1_th element 
magnitude is smaller than n_th element. 
This smallness MUST MAINTAIN TO INFINITY. The 
geometry series eq.cz48 fail ratio test, 
because the an+1 to an ratio limit value is 1.
but, n approach to infinity limit ratio less 
than one is required. a402251517
<a name="docAi36">
n=∞
n=1
1

n
=
1

1
+
1

2
+
1

3
+
1

4
+
1

5
+
1

6
+
1

7
+
1

8
+ ●●●
--- eq.cz48
n=∞
n=1
1

n
1

1
+
1

2
+
[
1

4
+
1

4
]
+
[
1

8
+
1

8
+
1

8
+
1

8
]
+ ●●●
--- eq.cz49
n=∞
n=1
1

n
1 +
1

2
+
[
1

2
]
+
[
1

2
]
+ ●●●
--- eq.cz50
eq.cz49 and eq.cz50 indicate eq.cz48 diverge.
width of above equation a402251511
2015-02-25-15-17 here
<a name="docAi37">
2015-02-25-17-28 start
eq.TS42 is Taylor series of exponential function. 
How does the ratio test work? We have a ratio q .
q can be calculated as the limit for n approach 
to infinity. This is always important. Take the 
absolute value of an+1 over an. See eq.TS43 "q=" 
an and an+1 are defined in eq.TS43 left side. 
Now plug an and an+1 into q definition eq.TS43 
<a name="docAi38">
Result is eq.TS44 left side. From numerator and 
denominator cancel xn get eq.TS44 right side 
equation. This step contain n!/(n+1)! In eq.TS45 
left side the factorial sign is there to confuse 
you a little bit. What is n!? n! is 1*2*3*4*...*n 
(n+1)! is n! even further to (n+1) In eq.TS45 guy 
below is (n+1)!=(n+1)*n!. Two red n! cancel to 1.
We are left with eq.TS45 right limit equation. 
What will happen is we are taking the limit for 
n to infinity. Very important is that this x is 
fixed. You can imagine x arbitrary high number, 
but x is fixed and n increase to infinity. The 
<a name="docAi39">
limit value is zero (trillion/infinity = 0) We 
got a q value that is not depend on x. It does 
not matter which x you choose, the ratio limit 
is always zero. For a convergence conclusion, 
the q value must be less than one. q=0<1 
radius of convergence is equal to infinity.
It does not mean you can plug in x value be 
infinity. It mean you can plug in arbitrary 
large value. This is our first result. 
2015-02-25-18-10 
<a name="docAi40">
2015-02-25-20-11 start 
"n increase to infinity_1" and 
"radius of convergence equal to infinity_2"
Two "infinity" are different. 
infinity_1 describe sequence index n. 
infinity_1 say the term xn/n! at extreme 
           remote "end" (actually no end). 
           for example xtrillion/(trillion!)
<a name="docAi41"> 
infinity_2 describe "my home yard" radius. 
infinity_2 say the "stone" x-value can be 
           picked in "yard" extreme remote 
           location still satisfy convergence 
           criteria.
Although infinity_2=infinity, but x cannot 
equal to infinity. x can be arbitrary large 
and x be a constant, let variable n move to 
infinity_1 such that xn/n! diminish to zero. 

Liu,Hsinhan inserted text is blue Italic.
LiuHH notes may contain error!
<a name="docAi42">
2015-02-25-20-36 here video 06:12/18:23
Whenever you get a limit ratio q, not depend 
on x, and the limit q value is smaller than 1, 
then you have a infinite radius of convergence. 
If you have q, not depend on x, and the limit q 
value is greater than 1, you have zero radius 
of convergence. 

<a name="docAi43">
Let us look at second example. 
-log(1-x) =
n=∞
n=1
xn

n
=
x1

1
+
x2

2
+
x3

3
+ ●●●
x100

100
+ ●●●
--- eq.TS46
an
=
xn

n
an+1
=
xn+1

(n+1)
q =
lim
n→∞
an+1

an
--- eq.TS47
q=
lim
n→∞
xn+1

(n+1)
n

xn
=
lim
n→∞
x*n

(n+1)
--- eq.TS48
q =  
 |x| *
lim
n→∞
n/n

n/n + 1/n
=
|x|*
1
=|x|
<1
--- eq.TS49
In eq.TS49 if write x without absolute sign, that is error. MUST use |x|, not use x.
width of above equation a402252058
2015-02-25-20-59 stop 
<a name="docAi44">
2015-02-26-21-31 start 
Let us look at second sum here. 
eq.TS46 is function equation -log(1-x) . 
Ratio q is defined at eq.TS47 right side. 
Plug an and an+1 into q definition eq.TS47 
get eq.TS48. Simplify to eq.TS49 . x in eq.TS48 
does not depend on n. x is fixed value. Because 
eq.TS48 has absolute value, we take x out of 
limit sign. It is important to take x out as 
absolute value |x| (see eq.TS49). What happen 
to this guy limit[n→∞]|n/(n+1)| ? To make your 
teacher happy, to prove rigorously. 
<a name="docAi45">
Multiple 1/n to both numerator and denominator. 
eq.TS49 numerator is n/n which is one. 
eq.TS49 denominator is n/n + 1/n which 
is 1 + 1/n When n approach to infinity 1/n is 
zero. Result is q=|x|, this q has to be smaller 
than one for convergence. Remember in previous 
case we have ratio q that is not dependent on 
x (see eq.TS45). Now we have a q ratio which is 
dependent on x. What you see is that |x| has to 
be smaller than one. It mean eq.TS46 -log(1-x) 
only get convergence if -1 < x < 1 
Your radius of convergence is one. → r=1 
2015-02-26-22-06 stop at video 10:30/18:23

<a name="docAi46"> 
2015-02-27-11-58 start
I will have a look at another equation. 
n=∞
n=0
2nxn

3nn!
=
n=∞
n=0
(2*x/3)n

n!
=
1

1
+
2*x1

3*1
+
22x2

322!
+
23x3

333!
+ ...
--- eq.TS50
an
=
2nxn

3nn!
;
an+1
=
2n+1xn+1

3n+1(n+1)!
;
q =
lim
n→∞
an+1

an
--- eq.TS51
q=
lim
n→∞
2n+1xn+1

3n+1(n+1)!
3nn!

2nxn
=
lim
n→∞
2*x

3*(n+1)
docAi47

--- eq.TS52
q =  
2*|x|

3
lim
n→∞
1

n + 1
=
2*|x|

3
*0
=
0
<1
--- eq.TS53
width of above equation a402271219
<a name="docAi48">
2015-02-27-12-22 here
Very interesting to know here is that in eq.TS46 
you have a pole at x=1 [log(1-x)=log(0)=-infinity]
In complex analysis you can show this pole are 
somehow constraining your radius of convergence. 
In complex plane you imagine a circle of radius 
one (why one? because a pole at x=one). Then your 
radius of convergence is as large as nearest 
singularity or pole. 
<a name="docAi49">
Now back to example three eq.TS50 . eq.TS50 left 
sum can be written as second sum. Then eq.TS50 
is similar to eq.TS42 eq.TS50 is exp(2*x/3) Its 
radius of convergence is infinite large. I will 
not go through example 3. Because example 3 is 
same as example 1. 
2015-02-27-12-47 here video 13:00/18:23 
Now give you a new example 4. 
<a name="docAi50">
n=∞
n=0
xn

2n
=
1

1
+
x1

21
+
x2

22
+ ●●●
x100

2100
+ ●●●
--- eq.TS54
a
=
x

2
n=∞
n=0
an
=
a0 + a1 + a2 + a3
+ ●●●
+ a100
+ ●●●
--- eq.TS55
n=∞
n=0
an
=
1

1 - a
=
1

1 - (x/2)
;
|a| < 1
  |x/2|<1   |x|<2
docAi51

--- eq.TS56
q =  
lim
n→∞
an+1

an
=
lim
n→∞
xn+1

2n+1
2n

xn
--- eq.TS57
q =  
lim
n→∞
x

2
=
x

2
< 1 ;
|x/2|<1
|x|< 2
r = 2
--- eq.TS58
n disappear
width of above equation a402271323
<a name="docAi52"> 
2015-02-27-13-31 here
Have a look at eq.TS54. Define a=(x/2) then 
eq.TS54 is a geometric series, see eq.TS55. 
eq.TS56 left equality is geometric sum see 
tute0059.htm#docA718 and tute0057.htm#docA108
Very importantly in eq.TS56 require |a|<1. 
That is |x/2|<1 or |x|<2 What is your radius 
of convergence? |x|<2 tell us the radius of 
convergence is two. 
Above is geometric sum consideration.
Below is ratio test consideration.
Let us check the ratio very fast. 
<a name="docAi53">
see eq.TS57 . From eq.TS57 to 
eq.TS58 the variable n disappear. Limit sign 
control n, but n is gone. Limit sign has no 
use and in eq.TS58 drop limit sign. 
eq.TS58 right side indicate |x|<2 and radius 
of convergence is two r=2.
eq.TS56 geometric sum say r=2. 
eq.TS58 ratio test result r=2. 
They are consistent.
<a name="docAi54">
That is actual it. I showed you three very 
simple equation. How to find their radius of 
convergence. Again a little remark on this. 
Please see video graph at 16:35/18:23
http://www.youtube.com/watch?v=QKvC2uQymf4
Imagine in complex plane, in eq.TS54 if x=2 
we have a singularity. This is really from 
origin [(x,y)=(0,0)] a circle of radius two. 
Your function converge everywhere inside the 
circle until it gets the singularity point. 
<a name="docAi55">
You can ask yourself why does this happen? 
A pole show up and Taylor series is not able 
to represent our function? This is pretty 
simple. If you have a look at the Taylor 
series It is easy to construct a polynomial 
and a polynomial do not have a pole or some 
thing like this. Actually you can not represent 
a function with pole using Taylor polynomial. 
Instead of that you can use Laurent series 
or even more complicated series. But Laurent 
series may be the one to go. 
That is actual it. See you guys.
2015-02-27-14-40 stop
2015-02-27-19-33 done first proofread
2015-02-28-16-12 done second proofread


<a name="MathConstants">
EulerMascheroni sum bound pendulum
2015-03-01-09-58 record start
2015-01-19-15-25 Liu,Hsinhan access
http://www.ebyte.it/library/educards/constants/MathConstants.html
save as Euler-Mascheroni_constant_www.ebyte.it_MathConstants.html
Next table is partial of MathConstants.html
Basic math constants 
Zero and One (and i, and ...)  0 and 1 (and √(-1), and ...)  Can anything be more basic than these two ? (or three, or ...) 
π, Archimedes' constant  3.141 592 653 589 793 238 462 643 •••  Circumference of a disc with unit diameter. 
e, Euler number, Napier's constant  2.718 281 828 459 045 235 360 287 •••  Base of natural logarithms. 
γ, Euler-Mascheroni constant  0.577 215 664 901 532 860 606 512 •••  Limit[n→∞]{(1+1/2+1/3+...1/n) - ln(n)} 
√2, Pythagora's constant  1.414 213 562 373 095 048 801 688 •••  Diagonal of a square with unit side. 
Φ, Golden ratio  1.618 033 988 749 894 848 204 586 •••  Φ = (√5 + 1)/2 = 2.cos(π/5). Diagonal of a unit side pentagon. 
φ, Inverse golden ratio (often confused with Φ)  0.618 033 988 749 894 848 204 586 •••  φ = 1/Φ = Φ -1 =(1-φ)/φ or φ = (√5 - 1)/2. 
δs, Silver ratio / mean  2.414 213 562 373 095 048 801 688 •••  δs = 1+√2. 
Physics Constants
http://www.ebyte.it/library/educards/constants/ConstantsOfPhysicsAndMath.html
2015-03-01-10-06 record stop
<a name="EulerMascheroni"> 
EulerMascheroni sum bound pendulum
γ=0.577215664901532860606512090082402431042 ••• 

http://freeman2.com/tut064a1.gif
2015-02-28-18-32 Liu,Hsinhan scan from Calculus textbook 
by Tom M. Apostol Volume 1 second edition page 377
γ=0.577215664901532860606512090082402431042 ••• from 
http://mathworld.wolfram.com/topics/Euler-MascheroniConstant.html
This graph URL: http://freeman2.com/tut064a1.gif 2015-02-28-20-12

<a name="a40301a"> 
2015-03-01-12-37 start 
When write eq.cz48, LiuHH paid attention to 
Euler-Mascheroni Constant
γ=0.577215664901532860606512090082402431042 ••• 
Euler-Mascheroni constant is defined by eq.TS59 below.
<a name="a40301b">
 
γ
=
limit
n→∞
[
k=n
k=1
1

k
log(n)
]
--- eq.TS59
 
lim
n→∞
k=n
k=1
1

k
= ∞
lim
n→∞
log(n)
= ∞
--- eq.TS60
Euler-Mascheroni constant γ = difference of two infinity.
width of above equation a403011249
<a name="a40301c">
2015-03-01-13-12 here 
eq.TS60 left equation is geometric sum to infinity. 
Its value is infinity. 
eq.TS60 right equation is log of infinity. 
Its value is infinity. 
eq.TS59 say Euler-Mascheroni constant γ is the 
difference of two infinities. γ is finite ! 

<a name="a40301d">
See tut064a1.gif 
Euler-Mascheroni constant γ is the 
infinity many red corner piece sum. 

<a name="a40301e">
MathConstants table 
http://www.ebyte.it/library/educards/constants/MathConstants.html 
say 
1st important Math Constant is PI=3.141 592 653 589 793 238 462 643 ••• 
2nd important Math Constant is  e=2.718 281 828 459 045 235 360 287 ••• 
3rd important Math Constant is  γ=0.577 215 664 901 532 860 606 512 ••• 
4th important Math Constant is √2=1.414 213 562 373 095 048 801 688 ••• 
5th important is Golden ratio   Φ=1.618 033 988 749 894 848 204 586 ••• 
•••
2015-03-01-13-28 stop

<a name="a40302a"> 
2015-03-02-00-28 start 
eq.TS59 is equation for Euler-Mascheroni constant γ 
This equation is difference of two infinity valued 
equations. See eq.TS60, it has two infinities. 
Now the question is that how to show eq.TS59 is 
bounded? 

<a name="a40302b">
2015-03-01-21-17 LiuHH access 
http://www.math.ualberta.ca/~isaac/math214/w08/harmonic_numbers.pdf
save as EulerMascheroni_www.math.ualberta.ca-harmonic_numbers.pdf
Key points summarized in next tut064a2.gif

BoundEulerMascheroni 
EulerMascheroni sum bound pendulum
γ=0.577215664901532860606512090082402431042 ••• 

http://freeman2.com/tut064a2.gif
γ=lim[n→∞]{∑[k=1,n](1/k)-log(n)} a403032305 add red eq. 
In tut064a2.gif move infinite many red&blue squares 
from col.2+col.3+col.4 ...→∞ to col.1 Fill column_1 
to full value one. 2015-03-04-16-11 

<a name="a40302c">
In tut064a2.gif y axis stretch longer than x axis.
Column 1 has width=1 , height=1/1 and area =1*1/1
Column 2 has width=1 , height=1/2 and area =1*1/2
Column 3 has width=1 , height=1/3 and area =1*1/3
Column 4 has width=1 , height=1/4 and area =1*1/4
Column 5 has width=1 , height=1/5 and area =1*1/5
.... to infinity 
Each column include grey, blue and red. 
grey+blue+red= one element in eq.TS60 ∑[k=1,n]1/k
grey+blue    = ∫[x=n,n+1]{dx/x} see tut064a1.gif 
one grey stripe. Or tut064a2.gif one grey+blue stripe.

<a name="a40302d">
Column 1 has grey+blue+red three color area=one. 
Column 1 grey area = column 2 grey+blue+red area.
Column 2 grey area = column 3 grey+blue+red area.
etc.
Column 1 red+blue = Column 1 area - Column 2 area
Column 1 red+blue = 1/1 - 1/2  similarly
Column 2 red+blue = 1/2 - 1/3  
Column 3 red+blue = 1/3 - 1/4  
Column 4 red+blue = 1/4 - 1/5  
Column 5 red+blue = 1/5 - 1/6 
etc.
<a name="a40302e">
Infinite many red+blue area is the sum of each 
individual red+blue area.
Infinite many red+blue area = 
[1/1 - 1/2] + [1/2 - 1/3] + [1/3 - 1/4] + ... →∞
In above expression 
 -1/2+1/2 cancel to zero 
 -1/3+1/3 cancel to zero etc to infinity. 
Infinite many red+blue area = 1/1 - 1/∞ = 1-0 = 1
Infinite many red area is our concern because 
infinite many red area = EulerMascheroni constant
<a name="a40302f"> 
From graph tut064a2.gif blue area take away some 
value. Then red area must be smaller than one. 
Here conclude eq.TS59 
EulerMascheroni constant is bounded by 1.
www.math.ualberta.ca/~isaac/math214/w08/harmonic_numbers.pdf
give only bounds 0 < γ < 1 there is no analytic 
expression for 
γ=0.577215664901532860606512090082402431042 ••• 
2015-03-02-01-05 stop


<a name="docAi61"> update 2015-03-03 EulerMascheroni sum bound pendulum 2015-03-02-14-16 start Next is study notes of Taylor Series - 1 - Radius of Convergence by Singularities (Exercise).mp4 http://www.youtube.com/watch?v=JkukU802NiY Lecturer is MrYouMath. Notes recorder is Liu,Hsinhan. Recorded notes is about 60% match, not exact. Liu,Hsinhan inserted text is blue Italic. LiuHH notes may contain error! <a name="docAi62"> Begin video 3 of 9 Hello and welcome everyone. This is MrYouMath again, and today's video we want have a look at a very interesting function. Try to find its radius of convergence without having these guys an here. We do not want to calculate an Because they are very complicated. (an see eq.TS62 )
<a name="docAi63">
x

ex-1
=
n=∞
n=0
Bn

n!
xn
=
B0x0

0!
+
B1x1

1!
+
B2x2

2!
+ ●●●
--- eq.TS61
cn
=
xn

n!
an
=
Bnxn

n!
q =
lim
n→∞
an+1

an
--- eq.TS62
forget q
cn and q are inserted by LiuHH, both are not used in lecture.
width of above equation a403021441
<a name="docAi64">
2015-03-02-14-41 here
We want to find out how large is the radius of 
convergence for eq.TS61 x/[ex-1]. There is a 
very important theorem in complex analysis. 
Please see video for MrYouMath's graph 
http://www.youtube.com/watch?v=JkukU802NiY
You can only do this for complex number. 
Horizontal axis x is real part of complex.
Vertical axis y is imaginary part of complex.
  complex z =x+i*y --- eq.TS63 
  x=real(z)        --- eq.TS64
  y=imag(z)        --- eq.TS65
<a name="docAi65">
This theorem tell you to find radius of convergence 
the only thing we have to do is we have to find 
the singularity that is closest to the origin. 
For example, imagine we have a singularity here 
(from graph, guess a point at x+i*y=2+i*1)
For some reason, our function x/[ex-1] have a 
singularity here. Function value at this point 
just run off to infinity. In order to find 
radius of convergence, what you have to do is 
take the distance between singularity (x,y)=(2,1) 
and the origin (x,y)=(0,0). This distance is 
your radius of convergence. Pretty easy. Now 
<a name="docAi66"> 
if you have multiple singularity, Choose the 
shortest distance between singularity and the 
origin. This shortest distance is your radius 
of convergence. In eq.TS61, we want to find 
radius of convergence for x/[ex-1], but we do 
not know what is Bn, we do not know its Taylor 
series. In order to find singularity for x/[ex-1] 
there are many methods. In order to do that we 
want have a look. You see x/[ex-1] we have a 
singularity at x=0 (ex-1=e0-1=1-1=0) 
<a name="docAi67">
What is 
important is that when x=0, in x/[ex-1] above 
and below both become zero. We can go ahead do 
an analytic continuation. We want find all the 
zeros of denominator [ex-1]. Rewrite [ex-1] 
as below and switch from real x to complex z 
  ez-1=0  --- eq.TS66
eq.TS66 is same as 
  ez=1    --- eq.TS67
<a name="docAi68">
In order to solve eq.TS67, rewrite 1 as below 
  ez=1=ei*(0+2πk)    --- eq.TS68
Why 2πk? because if you are in a 
complex coordinate system, some time called 
Gauss plane. In this plane point (x,y)=(1,0) 
is same if go around (0,0) a 360 degree circle. 
For integer k, you come back to (x,y)=(1,0) 
eq.TS68 is very easy to solve. 
<a name="docAi69">
Just equate two side power and get 
  z=i*(2πk) --- eq.TS69 
remember k can be any number out of ℤ 04:42/09:28
( LiuHH: k can be any integer number=ℤ )
2015-03-02-15-42 here
All the singularity are on the line 
  z=2πik  --- eq.TS70
( π=3.1415926535..., i=√-1 , k=any integer )
<a name="docAi70">
For k=0, we have a singularity at (x,y)=(0,0) 
this is a removable singularity. k=0 is not a 
problem to us. 
For k=1, we have a singularity at (x,y)=(0,2π) 
For k=-1,we have a singularity at (x,y)=(0,-2π) 
We have even more for z=(0,4π) z=(0,6π) etc. 
What you see the closest singularity is these 
guys (x,y)=(0,2π) and (x,y)=(0,-2π) Draw a circle 
<a name="docAi71"> 
This circle has a radius 2π. From this drawing 
without even doing any ratio test or anything 
else, we can see the radius of convergence for 
x/[ex-1] is 2π. All |x|<2π has convergent series 
representation. 
What happen to the boundary circle radius=2π ? 
This is a very hard question. We actually do not 
know convergence for points on the boundary. 
<a name="docAi72">
Sometime you get something useful, sometime not. 
  |x|<2π --- eq.TS71 
is radius of convergence for x/[ex-1]
The function x/[ex-1] would be living outside 
of boundary circle. But there is no promise of 
convergence. 
I hope above lecture showed you how to solve 
a convergence problem. 
<a name="docAi73">
If you are giving a function f(z) and actually 
you do not know anything about this function.
What you know it has a removable singularity 
here [drawing show point at (x,y)=(0.5,0.5)]
f(z) has a non-removable singularity here
[drawing show point at (x,y)=(-1,2)]
To find radius of convergence, you say OK this 
guy (x,y)=(0.5,0.5) is removable and (x,y)=(-1,2) 
is not removable. 
<a name="docAi74">
So your radius of convergence 
is from (x,y)=(0,0) to (x,y)=(-1,2) z=-1+2i 
|z|=√(-1)*(-1)+2*2=√5
radius of convergence r = √5
Without knowing anything about f(z), just know the 
pole location, you find radius of convergence.
Here concludes our lecture about singularity and 
radius of convergence. I hope you have fun If you 
like my video, please give thumb-up. I hope to 
publish new videos as soon as possible. 
See you guys.
2015-03-02-16-24 stop 


<a name="docAj01"> EulerMascheroni sum bound pendulum 2015-03-02-18-03 start Next is study notes of Taylor Series - 2 - Geometric Series and the tortoise.mp4 http://www.youtube.com/watch?v=UUqdmgw7QT8 Lecturer is MrYouMath. Notes recorder is Liu,Hsinhan. Recorded notes is about 60% match, not exact. Liu,Hsinhan inserted text is blue Italic. LiuHH notes may contain error! <a name="docAj02"> Begin video 4 of 9 Hello welcome everyone. It is MrYouMath again. Today we want to have a look at Geometric Series. We solve tortoise paradox. The tortoise paradox is a very old paradox. In which a tortoise first step doing one meter. Then it goes down one half meter, and then the half of one half meter, which is 0.25 meter. And so forth. Actually tortoise is going ahead. But always half distance it traveled before. Now we want to resolve this problem, The key is Geometric Series. And the Taylor series to Geometric Series.
<a name="docAj03">
f(x)
=
n=∞
n=0
xn
=
1 + x1 + x2 + x3 + x4
+ ●●●
+ x100
+ ●●●
--- eq.TS72
S
=
i=n
i=0
xi
=
1 + x1 + x2 + x3 + x4
+ ●●●
+ xn
               
--- eq.TS73
xS
=
i=n
i=0
xi+1
=
x1 + x2 + x3 + x4
+ ●●●
+ xn+1
               
--- eq.TS74
width of above equation a403021822
<a name="docAj04">
2015-03-02-18-23 here 
Let us look at geometric series eq.TS72 You are 
adding all powers of x. Why use power sum like 
eq.TS72? instead of going to the function? 
Because sometime it is very easy to use another 
approach instead of the differentiation approach 
that I showed you in the first video. 
(differentiation approach see eq.TS18, eq.TS38)
<a name="docAj05">
Because the Taylor series has a very interesting 
property. If you find a Taylor series for a function 
then this series is unique. 
Let us have a look at the sum. Instead of going 
for the infinite sum. I will take finite sum. 
A sum end at x=n. Later I will explain why this 
finite sum is important. If n=1000, it is very 
hard to calculate eq.TS73 for thousand terms . 
2015-03-02-18-48 stop 
<a name="docAj06"> 
2015-03-02-20-00 start 
A lot summations, a lot multiplications. I know 
it is fast for computer, it still takes longer 
time then the way we will learn here. The trick 
to solve this problem is to multiply the whole 
equation eq.TS73 with x. Important thing is that 
x is not equal to zero. If x=0, the function 
have value one any way. Multiply the whole 
equation eq.TS73 with x get eq.TS74. 
<a name="docAj07">
eq.TS73 and eq.TS74 look similar. Both contain 
x1, x2 ... xn If we would subtract both equations. 
eq.TS73 - eq.TS74 we could reduce these guys in 
the middle. Let us do it. 
S-xS
=
1
+ x1 + x2 + x3 + x4 + ●●● + xn
-x1-x2-x3-x4-●●●-xn
-xn+1
--- eq.TS75
width of above equation a403022028
<a name="docAj08">
2015-03-02-20-28 here 
Why it is important that the sum is finite? Because 
when working with infinite sum, you have to be very 
careful in order to be able to rearrange this kind 
equation. The key word for that is absolute convergence. 
If your sum is not absolute convergent, you are not 
allowed to rearrange this. Whenever working with 
infinite sum, be very very careful. Instead to work 
with infinite sum, try to break it to finite sum 
then take the sum for limit to infinite. Now 
simplify eq.TS75 to eq.TS76 below 
  S-xS=1-xn+1 --- eq.TS76 
<a name="docAj09">
In eq.TS75 the term 1 stay there (pass to eq.TS76)
because there is no other -1 to cancel this 1. 
In eq.TS75 red terms cancel blue terms. Last term 
is -xn+1. No other term to cancel it. So in eq.TS76 
we are left with -xn+1. We can go ahead factor out 
S in eq.TS76, get eq.TS77 below 
  S(1-x)=1-xn+1 --- eq.TS77 
It is pretty easy. We are not looking for anything 
else than S. We divide eq.TS77 with (1-x) get 
  S=(1-xn+1)/(1-x) --- eq.TS78 
<a name="docAj10">
In this step we are not allowed to plug in x=1. 
In eq.TS78 assume x≠0 and x≠1. 
We are not interested in the finite sum, but 
infinite sum. In eq.TS78 take limit for n to 
infinite large. The magnitude of x must be smaller 
than one. Imagine you have x=2 to power 1000. 
This is a very large number. But if x=1/2 to 
power 1000 this term 1/21000 is nearly zero. 
It is quite important for us to say, if |x| is 
smaller than one. (x=1 is not allowed) then 
limit[n→∞]xn+1 is zero (again require |x|<1) 
In eq.TS78 leave xn+1 away get eq.TS79 below. 
<a name="docAj11">
f(x)
=
1

1-x
=
1 + x1 + x2 + x3 + x4 + ●●● + xn + ●●●
--- eq.TS79
|x| < 1
width of above equation a403022111
<a name="docAj12">
2015-03-02-21-13 here 
In eq.TS78 let n approach to infinity, get eq.TS79. 
We will do a lot funny stuff with eq.TS79. Earlier 
we asked that x≠0. Now in eq.TS79 we plug in x=0 
you will always get one. Now let us look at the 
formula f(x)=1/(1-x) if x=0 , f(0) is one. Then 
x=0 is valid. This is the reason why in eq.TS79 
require only |x|<1 (not require 0<|x|<1)

<a name="docAj13">
Now let us look at the paradox we talk before. 
Let us imagine you have a tortoise. Someone is 
in front of tortoise and tortoise try to catch 
up with someone. But you only catch up half the 
distance between tortoise and someone. For 
example, we start at two meter distance between 
tortoise and opponent. The first step tortoise 
move forward one meter. Second step is half of 
the distance before. This would be half meter. 
<a name="docAj14">
Third step, we are taking half of one half. That 
is a quarter meter or 1/22 
S
=
1 +
1

2
+
1

22
+
1

23
+
●●●
=
1

1 - 1/2
= 2
--- eq.TS80
width of above equation a403022135
<a name="docAj15">
2015-03-02-21-36 here 
Now doing a further step, we add fourth step 1/23
You can imagine we are doing this all the way along. 
You are running as a crazy guy and try to catch up 
with it. The question is that if you ever reach this 
guy? The result is very easy. You will never reach 
the one ahead of you. eq.TS80 infinite sum to two.
eq.TS80 is nothing else than a geometric series. 
1/2 is your x. 1/22 is x squared. 1/23 is x cube 
and so forth. We know if we have geometric sum, 
eq.TS79 give us the answer 1/(1-1/2) = 2 which mean 
tortoise run forever and cannot exceed two meter. 
<a name="docAj16"> 
We are adding steps and steps and steps, and we will 
never reach above the magic value of two. There are 
a lot series, you add value to it, they will have a 
finite total sum. There are some series you add small 
amount to it. But total sum to infinite (diverging) 
For example geometry series eq.cz48 sum to infinite. 
That is it. See you guys and have a nice day. 
2015-03-02-21-56 stop 
2015-03-02-23-57 done  first proofread 3 of 9, 4 of 9.
2015-03-03-09-12 done second proofread 3 of 9, 4 of 9.


<a name="docAj31"> EulerMascheroni sum bound pendulum 2015-03-03-14-37 start Next is study notes of Taylor Series - 3 - Logarithm and alternating harmonic series http://www.youtube.com/watch?v=UZQS8D_zNNE Lecturer is MrYouMath. Notes recorder is Liu,Hsinhan. Recorded notes is about 60% match, not exact. Liu,Hsinhan inserted text is blue Italic. LiuHH notes may contain error! <a name="docAj32"> Begin video 5 of 9 Hello welcome everyone. This is MrYouMath with new video on the Taylor series. Today we want to have a look at logarithm and the alternating harmonic series. Like the video with geometric series. I will not use the normal approach to find Taylor series and calculating all the derivatives. Because there is an easier way and more intuitive understanding of Taylor series of the logarithm. Let us just start off. I hope to show you some application of this logarithm when we calculate alternating harmonic series. We start again the geometric series
<a name="docAj33">
1

1-x
=
1 + x1 + x2 + x3 + x4 + ●●● + xn + ●●●
 ☺  |x| < 1
--- eq.TS81
width of above equation a403031453
<a name="docAj34">
2015-03-03-15-00 here 
eq.TS81 is very strongly related to logarithm. If 
you look at the Taylor series of this guy 1/[1-x] 
and eq.TS81. Very important, the magnitude of x 
should be smaller than 1, |x|<1 . Now we try to 
integrate eq.TS81 get eq.TS82 below.
<a name="docAj35">
1

1-x
dx
  =  
(1 + x1 + x2 + x3 + x4 + ●●● )*dx
 
--- eq.TS82
|x} < 1
-log(1-x)
=
x
+
x2

2
+
x3

3
+
x4

4
+ ●●●
--- eq.TS83
|x} < 1
log(1-x)
=
-x
x2

2
x3

3
x4

4
- ●●●
--- eq.TS84
|x} < 1
log(1+x)
=
+x
x2

2
x3

3
x4

4
+/- ●●●
--- eq.TS85
|x} < 1
width of above equation a403031528
<a name="docAj36"> 
2015-03-03-15-17 here 
Big question at this point is are we allowed to 
integrate eq.TS82 right side to eq.TS83 right side?
Similar question in docAj67 has explicit equation. 
are we allowed to integrate eq.TS94 right side by 
eq.TS95 right side? 2015-03-04-19-50 notes
This is allowed in this case. Because eq.TS82 right 
side is converging uniformly. You can inter-change 
differentiation and integration. 
<a name="docAj37">
Carry out integration in eq.TS82 get eq.TS83. The 
term in eq.TS83 go higher and higher. For example 
you will get x7/7 . We would get an integration 
constant in front of eq.TS83. If you plug in x=0 
you will see constant of integration is zero. So 
I just leave it out. Not comment any more on the 
constant. 
2015-03-03-15-30 stop
<a name="docAj37b">
2015-03-05-10-53 start 
eq.TS83 with integration constant K is next 
-log(1-x) = K + x +x2/2 +x3/3 +x4/4 + ●●● --- eq.TS83K
In eq.TS83K set x=0 get 
-log(1-0) = K + 0 +x2/2 +03/3 +04/4 + ●●● --- eq.TS83L
or
-log(1) = K  --- eq.TS83M
But -log(1) is -0 or 0, therefore 
integration constant K = 0
2015-03-05-10-58 stop 

<a name="docAj38">
2015-03-03-18-25 start 
We multiply eq.TS83 with -1, get eq.TS84. 
Plug -x for x in eq.TS84, get eq.TS85 . 
The odd power in eq.TS84 change sign in eq.TS85. 
The even power in eq.TS84 keep negative in eq.TS85. 
Because square (-x) get positive x*x. Not change 
sign in eq.TS85. 
Now calculate log(1+x) - log(1-x) get eq.TS86 
<a name="docAj39">
log(1+x)-log(1-x)
=
log
1+x

1-x
=
                   
--- eq.TS86
|x} < 1
=
+x
x2

2
x3

3
x4

4
+/- ●●●
[
-x
x2

2
x3

3
x4

4
- ●●●
]
--- eq.TS87
red even
cancel

black odd
double
log
1+x

1-x
=
2*
[
x
+
x3

3
+
x5

5
+
x7

7
+ ●●●
]
--- eq.TS88
|x} < 1
width of above equation a403031856
<a name="docAj40">
2015-03-03-18-57 here
In eq.TS87, red even power terms cancel out. 
In eq.TS87, black odd power terms double. 
The result is eq.TS88. Now why would I want 
to use eq.TS88 instead of eq.TS84 eq.TS85 ?
Because using eq.TS88 you are able to calculate 
even larger values. 
<a name="docAj41"> 
For example, to calculate 
log(8) Set log[(1+x)/(1-x)] equal to log(8) 
solve for (1+x)/(1-x) = 8 get 1+x=8-8x , x=7/9
In log[(1+x)/(1-x)] set x=7/9 get log(8) 
On the other hand, in eq.TS85 set x=7 also 
get eq.TS85: log(1+7)=log(8) BUT eq.TS85 ask 
|x|<1 and x=7 is not allowed. However in 
eq.TS88 set x=7/9 here |7/9|<1 not violate 
the |x|<1 rule. 

In eq.TS85 plug in x=1 (violate |x|<1 rule !)
<a name="docAj42">
alternating harmonic series
log(1+x) = log(2)
=
+1
1

2
1

3
1

4
+/- ●●●
--- eq.TS89
|x} < 1
width of above equation a403031915
<a name="docAj43">
2015-03-03-19-16 here
We get eq.TS89. It is only allowed to plug in |x| 
smaller than one. Here, plug in x=1? I can do this 
(I can violate |x|<1 rule) because eq.TS89 right 
hand side converges. In the second representation 
x=1 it still converges. The rule |x|<1 is most 
conservative. Sometimes violate rule |x|<1 still 
converges. There is a way to show eq.TS89 
convergence by Leibniz criterion 

<a name="docAj44"> Leibniz's Theorem ∑[n=1,∞]{(-1)n+1an} converges if three of following conditions are satisfied. 1. The an are all positive. 2. an > an+1 for every n. 3. an → 0 when n → ∞ textbook by George, B. Thomas Jr. and Ross L. Finney Calculus sixth ed. page 649 2015-03-03-19-35 here

an → 0 when n → ∞ is required for every convergent series.
<a name="docAj45">
H
=
1
+
1

2
+
1

3
+
1

4
●●● = ∞ ≠ -∞ =
limit
x→1
log(1-x)
--- eq.TS90
width of above equation a403031942
<a name="docAj46"> 
2015-03-03-19-43 here
What is important is that if in eq.TS85 you plug 
in x=-1 you get log(1-1) diverge to minus infinity.
It shows even if your sum diverge, it will not 
have same value it should have. We can use x=1 
for log(2) Otherwise we must follow the rule 
|x|<1 .
That is it. See you guys.
2015-03-03-19-53 stop 


<a name="docAj61"> EulerMascheroni sum bound pendulum 2015-03-04-13-26 start Next is study notes of Taylor Series - 4 - Arctan and Pi http://www.youtube.com/watch?v=XBKGuIQK1Ek Lecturer is MrYouMath. Notes recorder is Liu,Hsinhan. Recorded notes is about 60% match, not exact. Liu,Hsinhan inserted text is blue Italic. LiuHH notes may contain error! <a name="docAj62"> Begin video 6 of 9 Hello welcome everyone. It is MrYouMath again. Today is special video about arctan and PI. First of all, we want find a Taylor series for arctan then we want to do is find a series expression for PI. Actually find one for PI/4. <a name="docAj63"> Let us start from geometric series eq.TS91
1

1-x
  =
1 + x1 + x2 + x3 + x4 + x5 + ●●●
--- eq.TS91
|x| < 1
width of above equation a403041337

<a name="docAj64">
2015-03-04-13-37 here
If we plug in -x2 to replace x in eq.TS91 , we get eq.TS92 below.

1

1-(-x2)
  =  
1 + (-x2)1 + (-x2)2 + (-x2)3 + (-x2)4 + (-x2)5 + ●●●
--- eq.TS92
|x| < 1
width of above equation a403041352

<a name="docAj65">
2015-03-04-13-53 here
eq.TS92 simplify to eq.TS93 below.

1

1+x2
  =
1 -x2 + x4 - x6 + x8 - x10 + ●●●
--- eq.TS93
|x| < 1
width of above equation a403041359
<a name="docAj66"> 
2015-03-04-14-00 here 
We integrate eq.TS93 left side 1/(1+x2) get arctan().
eq.TS93 right side is 
2*odd  power (2,6,10 etc) has negative sign and 
2*even power (4,8,12 etc) has positive sign.
integrate eq.TS93 get eq.TS94 below. 
<a name="docAj67">
1

1+x2
dx
=
(1 -x2 + x4 - x6 + x8 - x10 + ●●● )
dx
--- eq.TS94
|x| < 1
From above to below, inter-change the order of
integration and summation, require eq.TS94 series
has uniform convergence property.
arctan(x)
=
1*dx
x2*dx
x4*dx
-/+ ●●●
--- eq.TS95
|x| < 1
width of above equation a403041417
<a name="docAj68">
2015-03-04-14-18 here 
We know the geometric series is uniform convergence, 
we can inter-change the order of integration and 
summation here from eq.TS94 to eq.TS95 
Next integrate eq.TS95 get eq.TS96 below.
<a name="docAj69">
arctan(x)
=
x
x3

3
x5

5
x7

7
x9

9
x11

11
+/- ●●●
--- eq.TS96
|x| < 1
width of above equation a403041427
<a name="docAj70">
2015-03-04-14-32 here 
First point worth your attention is that in eq.TS96 
right side term sign change +/-. This is alternating 
sum. Second point is that in eq.TS96 each term 
denominator number is same as the power number. 
We have only odd number power. eq.TS96 has a 
integration constant. To see constant value, 
in eq.TS96 plug in x=0 get 0=0. Then constant 
must be zero. Again, the magnitude of x should 
be smaller than one. Now we will play a little 
bit with this representation eq.TS96 . I will 
do something normally you should not do. Later 
I will explained why this is allowed. 

<a name="docAj71"> 
First of all, I plug in x=1 to eq.TS96 You know 
magnitude of x should be smaller than one. Using 
x=1 violate the rule |x| < 1 . 
arctan(1)
=
1
13

3
15

5
17

7
19

9
111

11
+/- ●●●
--- eq.TS97
|x| < 1
width of above equation a403041446
<a name="docAj72"> 
2015-03-04-14-46 here 
This series in eq.TS97 goes back to Madhava, Gregary 
and Leibniz. Madhava was the first one, Gregary was 
second. Leibniz re-discovered series in eq.TS97. 
Most European country name after Leibniz, because 
Leibniz was the first one to show that series in 
eq.TS97 is really convergent. We see arctan(1) is 
equal to PI/4 then eq.TS97 is same as eq.TS98 below.
<a name="docAj73">
π

4
=
1
1

3
1

5
1

7
1

9
1

11
+/- ●●●
--- eq.TS98
width of above equation a403041454
<a name="docAj74">
2015-03-04-14-55 here 
Think about tangent() the value you plug in to get 
one. Remember tangent(45 degree) has value one. 
45 degree is PI over four. That is π/4 in eq.TS98 
left side. eq.TS98 right side has an interesting 
sum. The alternating series of inverse of odd 
numbers. eq.TS98 not converge fast, very slow. 
<a name="docAj75">
I explain why eq.TS98 not converge fast. Because 
I told you eq.TS96 require |x| < 1 But now we are 
actually at the limit (using x=1) What happen is 
we have a very slowly convergent series eq.TS98 . 
In order to increase the speed, Leonhard Euler 
used next equation
<a name="docAj76">
arctan(x) + arctan(y)
=
arctan
x + y

1-xy
--- eq.TS99
width of above equation a403041509
<a name="docAj77">
2015-03-04-15-10 here 
eq.TS99 is addition theorem for tangent angle. We 
can break up eq.TS97 into two smaller values, so 
the convergent rate is much faster than eq.TS97 . 
Euler found arctan(1/2) and arctan(1/3) give you 
arctan(1) again 
[in (x + y)/(1-xy) let x=1/2, y=1/3 
 (x + y)/(1-xy) = (1/2 + 1/3)/{1-(1/2)(1/3)}
 (x + y)/(1-xy) = (5/6)/{1- 1/6} =(5/6)/(5/6)=1
 2015-03-04-15-18 
 ]
<a name="docAj78">
You would get faster convergence. Euler worked a 
lot of these. He break up to smaller and smaller 
pieces. The smaller angle value the faster series 
converge. Euler found the famous formula for π/4 
  π/4 = 4*arctan(1/5) - arctan(1/239) --- eq.TSa0
eq.TSa0 is a fast convergent formula. You get very 
good result. 
That is it. See you guys. 
2015-03-04-15-27 stop 
2015-03-04-21-14 done  first proofread 5 of 9, 6 of 9.
2015-03-05-14-14 done second proofread 5 of 9, 6 of 9.


<a name="docAk01"> EulerMascheroni sum bound pendulum 2015-03-05-20-55 start Next is study notes of Taylor Series - 5 - Binomial Functions and Einsteins E=mc² http://www.youtube.com/watch?v=tZHRWPOXaaA Lecturer is MrYouMath. Notes recorder is Liu,Hsinhan. Recorded notes is about 60% match, not exact. Liu,Hsinhan inserted text is blue Italic. LiuHH notes may contain error! <a name="docAk02"> Begin video 7 of 9 Hello welcome everyone. This is MrYouMath again. Today I want to have a look at Binomial Functions and relate them to Einsteins E=mc² and relate to Newtonian physics. Binomial Functions look like this f(x)=(1+x)r --- eq.TSa1 I want find Taylor series for eq.TSa1. In previous video I used geometric series (see eq.TS91) (In previous video not use Taylor series eq.LAf2 Current equation eq.TSa1, it is impossible to use geometric series. If set x=0, geometric series become 1=1. A dull case.) But now I want stick to normal procedure finding Taylor series. Evaluate f(x) at x=0. We will call this Maclaurin series. (see eq.dzA0) <a name="docAk03"> We find first derivative of eq.TSa1 f'(x)=r*(1+x)r-1 ; f'(0)=r --- eq.TSa2 Differentiate eq.TSa1 get eq.TSa2 if power r=2 please see eq.TS07 and eq.TS08 It is easy to generalize power r=other value case. f'(x) evaluated at x=0, r*(1+x)r-1 become r*(1+0)r-1 , 1 to any power is always one. Therefore f'(0)=r*1=r . Taking the second derivative f''(x)=r*(r-1)*(1+x)r-2 ; f''(0)=r*(r-1) --- eq.TSa3 Reasoning is same as first derivative . <a name="docAk04"> Let us do third derivative f'''(x)=r*(r-1)*(r-2)*(1+x)r-3 --- eq.TSa4 f'''(0)=r*(r-1)*(r-2) --- eq.TSa5 Each derivative reduce (1+x) power by one. It is pretty easy to see the n-th derivative evaluate at x=0, see eq.TSa6 below f(n)(0)=r*(r-1)*(r-2)*...*(r-n+1) --- eq.TSa6 eq.TSa6 is general equation for n-th derivative evaluate at x=0. Now let us check n=3. In eq.TSa6 set n=3 get f(3)(0)=r*(r-1)*(r-3+1)=r*(r-1)*(r-2) --- eq.TSa7 This is correct. eq.TSa7 confirm eq.TSa5. If we plug in f(n)(0) to Taylor series definition equation eq.LAf2 below
<a name="TaylorSeriesEq">
Next is eq.LAf2 (too long, "eq.LAf2" is hidden)
Taylor series expansion for f(x) at x=a ; 2015-01-26-18-12 ;
 
f(x)
=
f(a)
+ (x-a)*f '(a)
+
(x-a)2

2!
d2f(x)

(dx)2
x=a
+
(x-a)3

3!
d3f(x)

(dx)3
x=a
...
+
(x-a)n

n!
dnf(x)

(dx)n
x=a
...
--- eq.LAf2
Taylor series expansion for function f(x) at point x=0 is named
Maclaurin series expansion. It is next equation.
f(x)=f(0) +f'(0)*x/1! +f''(0)*x2/2! +f'''(0)*x3/3! + ... +f(n-1)(0)*xn-1/(n-1)! +Rn --- eq.dzA0

where Rn=f(n)(θx)*xn/(n)! ; 0<θ<1
width of above equation a401261836; a403052149

<a name="docAk05">
 
f(x)
=
n=∞
n=0
f(n)(0)

n!
xn
=
f(0)
+ (x-0)*f '(0)
+
(x-0)2

2!
d2f(x)

(dx)2
x=0
+
(x-0)3

3!
d3f(x)

(dx)3
x=0
...
--- eq.TSa8
width of above equation a403052157

<a name="docAk06">
 
f(x)
=
n=∞
n=0
r(r-1)(r-2)...(r-n+1)

n!
xn
--- eq.TSa9
width of above equation a403052203
<a name="docAk07">
2015-03-05-22-04 here
If we write down eq.TSa8 left end equality equation 
as eq.TSa9. It look a little bit complicated. It is 
hard to remember. Simplify eq.TSa9 a little bit. 
This is commonly done by using binomial coefficient. 
2015-03-05-22-10 stop at video 3:40/15:40

<a name="docAk08">
2015-03-06-14-44 start 
I hope you remember eq.TSb0 from common factories. 
(r  n) is called r over n, or n out of r. 
(r  n) = r!/[n!(r-n)!] as shown in eq.TSb0. 
Red term in eq.TSb1 is inserted one. Numerator red 
let r(r-1)(r-2)●●●(r-n+1) extend all the way to *1
Say in another way, red term explain why in eq.TSb1 
r(r-1)(r-2)●●●(r-n+1)/n! equal r!/[n!(r-n)!].
<a name="docAk09">
(
r

n
)
=
r!

n!(r-n)!
=
r(r-1)(r-2)●●●(r-n+1)

n!
       
--- eq.TSb0
=
r(r-1)(r-2)●●●(r-n+1)*(r-n)(r-n-1)*●●●*2*1

n!*(r-n)(r-n-1)*●●●*2*1
=
r!

n!(r-n)!
--- eq.TSb1
width of above equation a403052345
<a name="docAk10">
2015-03-06-14-57 here 
In eq.TSa9, there are infinite many binomial 
coefficient (r  n). Now substitute eq.TSb0 into 
eq.TSa9. Change all r(r-1)(r-2)●●●(r-n+1)/n! to 
simpler r!/[n!(r-n)!]. eq.TSa9 become eq.TSb2 
below.
 
f(x)
=
n=∞
n=0
r(r-1)(r-2)...(r-n+1)

n!
xn
=
n=∞
n=0
(
r

n
)
xn
--- eq.TSb2
|x| < 1
width of above equation a403052353
<a name="docAk11"> 
2015-03-06-15-16 here 
In eq.TSb2, important point is that |x| < 1 You can 
show |x| < 1 by using convergence criteria. I will do 
a very interesting thing. I will plug in -1/2 for r . 
Also use '-x' in place of 'x'. Under this special 
consideration, eq.TSb2 become eq.TSb3 below. 

2015-03-06-15-26 external mouse disappear 
2015-03-06-15-30 turn off computer 
2015-03-06-16-00 turn on  computer 

<a name="docAk12">
2015-03-06-16-47 start
Recall we start from eq.TSa1
  f(x)=(1+x)r --- eq.TSa1
Set r=-1/2 and set x to -x eq.TSb2 become next
 
f(x)
=
(1-x)-1/2
=
n=∞
n=0
(
-1/2

n
)
(-x)n
--- eq.TSb3
|x| < 1
width of above equation a403061650

<a name="docAk13">
Write eq.TSb3 in long form, get eq.TSb4 below
 
1

1-x
=
1
+
-1/2

1!
(-x)
+
(-1/2)(-3/2)

2!
(-x)2
+
(-1/2)(-3/2)(-5/2)

3!
(-x)3
+ ●●●
--- eq.TSb4
|x| < 1
width of above equation a403061703

<a name="docAk14">
eq.TSb4 simplify to eq.TSb5 below.
 
1

1-x
=
1
+
1

2
x
+
3

8
x2
+
5

16
x3
+ ●●●
--- eq.TSb5
|x| < 1
width of above equation a403061713
<a name="docAk15">
2015-03-06-17-14 here
In eq.TSb5, magically, all negative cancel. Now we 
will use eq.TSb5 for Einsteins equation E=mc²
In eq.TSb5, right side term take only first two 
1+x/2 . Drop all higher order terms. Then we get 
<a name="docAk16">
 
1

1-x
1
+
1

2
x
      drop higher order terms.
--- eq.TSb6
|x| < 1
 
E     =     mc²
=    
m0

1 - v²/c²
     
--- eq.TSb7
width of above equation a403061732
<a name="docAk17">
2015-03-06-17-32 here
In eq.TSb5 if |x| is very small, then eq.TSb6 is 
approximately correct. You can even generalize this. 
If eq.TSb5 has 1/∛1-x, then eq.TSb6 is 1/∛1-x1+x/3. 
If eq.TSb5 has 1/∜1-x, then eq.TSb6 is 1/∜1-x1+x/4. 

eq.TSb7 is Einsteins equation E=mc² Many of you 
might know E=mc² Total energy of a particle equivalent 
to particle mass multiply by speed of light c squared. 
<a name="docAk18">
We do not care physical interpretation now. I just do 
a little mathematical stuff. eq.TSb7 say particle mass 
depend on ratio of velocity v to light speed c. We can 
rewrite E=mc² as eq.TSb7 right end expression. Some 
time people write ratio of velocity v to light speed c 
as gamma γ. But I will stick to v/c. 
<a name="docAk19">
eq.TSb6 is true, because |x| is small, say x=1/100
then x² is (1/100)²=1/10000 and x³ is (1/100)³ 
which is 1/1000000. All higher order terms are 
negligible. Man made speed v is much smaller then 
light speed. v/c is small, eq.TSb6 approximation 
is valid.
Apply eq.TSb6 to eq.TSb7 right side term. See what 
happen to this energy term eq.TSb7. We get eq.TSb8
below. 
<a name="docAk20">
 
E   =   mc²
=    
m0

1 - v²/c²
=  
m0
[
1 +
1

2
(
v

c
)
2


+ h.o.t
]
--- eq.TSb8
width of above equation a403061822
<a name="docAk21"> 
2015-03-06-18-25 here 
In eq.TSb8 multiply things out get 
  E = m0c² +m0v²/2 +m0c²*(h.o.t) --- eq.TSb9 
m0 is called motionless mass. If you are not moving 
at all. Second term is m0v²/2 in which numerator c² 
cancel denominator c². We are left with v² plus 
m0c²*(h.o.t) This higher order terms are not in our 
thinking. If you look at m0v²/2 , this terms is just 
kinetic energy for a particle which is moving with 
velocity v. But m0v²/2 is Newtonian physics equation.
<a name="docAk22">
Newtonian kinetic energy is incorporated in Einstein 
equation. Some of you have ask how can that be? Newton 
was so wrong. Actually, Newton was not wrong. Newton 
did just first term. If you include higher order terms 
You will end up with Einsteins equation. This mean 
Einsteins equation is more general than Newtonian 
equation. The higher speed you get, the less 
negligible these higher order terms become. 
That is it. See you guys.
2015-03-06-18-56 stop 
2015-03-06-20-21 done first proofread
2015-03-06-23-25 done second proofread


<a name="docAk31"> EulerMascheroni sum bound pendulum 2015-03-07-15-06 start Next is study notes of Taylor Series - 6 - Trigonometric functions and Foucault's pendulum http://www.youtube.com/watch?v=ycxR9HJnazo Lecturer is MrYouMath. Notes recorder is Liu,Hsinhan. Recorded notes is about 60% match, not exact. Liu,Hsinhan inserted text is blue Italic. LiuHH notes may contain error! <a name="docAk32"> Begin video 8 of 9 Hello welcome everyone. This is MrYouMath again. Today's video I want talk about Trigonometric functions and their Taylor series. Final step I want look at Foucault's pendulum. It is a quite hard differential equation cannot be solved by normal methods. We will solve pendulum equation by using knowledge about Taylor series for sine function.
<a name="docAk33">
Taylor series expansion for f(x) at x=0 ; Below is eq.TSc0
 
f(x)
=
n=∞
n=0
f(n)(0)

n!
xn
=
f(0)
+ (x-0)*f '(0)
+
(x-0)2

2!
d2f(x)

(dx)2
x=0
+
(x-0)3

3!
d3f(x)

(dx)3
x=0
●●●
 
docAk34
Table of Sine
derivative
expression
one more
derivative
derivative
value at x=0
one more d/dx
value at x=0
f(x)
sin(x)
cos(x)
sin(0)=0
cos(0)=1
df(x)/dx
cos(x)
-sin(x)
cos(0)=1
-sin(0)=0
d²f(x)/(dx)²
-sin(x)
-cos(x)
-sin(0)=0
-cos(0)=-1
d³f(x)/(dx)³
-cos(x)
sin(x)
-cos(0)=-1
sin(0)=0
 
sin(x)
=
f(0)
+
f '(0)

1!
x
+
f ''(0)

2!
x2
+
f '''(0)

3!
x3
+
f(4)(0)

4!
x4
      docAk35

●●● --- eq.TSc1
width of above equation a403071617
<a name="docAk36"> 
2015-03-07-16-32 here 
eq.TSc0 is Taylor series expansion for f(x) at x=0
which is Maclaurin series. We need all derivatives 
at x=0. Just plug derivative values in to the 
formula. Let us have a look at our functions and 
derivative expression and value for f(n)(x) at x=0
They are listed in a table shown above. I will do 
sine function and cosine function at the same time, 
because they are pretty much the same. (video 0:55)
<a name="docAk37">
Below is Liu,Hsinhan words. Please verify.
Table f(x) is sin(x). Function cos(x) is derived. 
Table column one write down derivative times. 
Table column two is derivative expression 
Table column three is one more derivative 
Element sin(x) go down one place is cos(x)
Element sin(x) go right one place is cos(x)
Two blue cos(x) are identical. Repeat on purpose. 
Both cos(x) are first derivative of sin(x).
Similarly, two purple -sin(x) are identical.
two gray -cos(x) are identical. Repeat on purpose. 
Table column four is derivative value at x=0. 
Column five is one more derivative value at x=0. 
<a name="docAk38">
In eq.TSc1 we need derivative value at x=0.
Table entry use next formulas 
  d[sin(x)]/dx = cos(x) --- eq.TSc2 
  d[cos(x)]/dx =-sin(x) --- eq.TSc3 
Please find calculus textbook for these elementary 
relations. Based on eq.TSc2 and eq.TSc3 
  d2[sin(x)]/(dx)2 = d[cos(x)]/dx =-sin(x) --- eq.TSc4 
  d3[sin(x)]/(dx)3 = d[-sin(x)]/dx =-cos(x) --- eq.TSc5 
If continue, result repeat itself 
  d4[sin(x)]/(dx)4 = d[-cos(x)]/dx =+sin(x) --- eq.TSc6 
This fourth derivative result +sin(x) is the start 
function. Repeat pattern show up. Repeat formula 
contribute nothing. 
All we need is eq.TSc2, eq.TSc3, eq.TSc4, eq.TSc5. 
Above is Liu,Hsinhan words. 2015-03-07-17-15
<a name="docAk39">
2015-03-07-17-46 start at video 3:15/16:23
Now let us look at eq.TSc1 sin(x), how can you 
write down sine function? eq.TSc1 is nothing else 
than the Taylor series. (video 03:30) Interesting 
part is that f(0)=0 [see table "sin(0)=0"]
eq.TSc1 constant term f(0) cancel. 
Next term f'(0) is one [see table "cos(0)=1"]
Next f''(0) will be zero. [see "-sin(0)=0"]
f(4)(0) will be zero also. [repeat to eq.TSc6
and see table "sin(0)=0"] What we see is we only 
have odd differentiation divided by odd factorial 
are none zero. (all even differentiation terms 
become zero) Important thing is sine Taylor series 
representation is alternating in sign. plus then 
minus repeat. Now rewrite eq.TSc1 as eq.TSc7 below. 
<a name="docAk40">
 
sin(x)
=
f(0)
+
f '(0)

1!
x
+
f ''(0)

2!
x2
+
f '''(0)

3!
x3
+
f(4)(0)

4!
x4
●●●
--- eq.TSc1
 
sin(x)
=
0
+
1

1!
x
+
0

2!
x2
+
-1

3!
x3
+
0

4!
x4
●●●
--- eq.TSc7
 
sin(x)
=
x
x3

3!
+
x5

5!
x7

7!
+
x9

9!
x11

11!
+/- ●●●
--- eq.TSc8
 
sin(x)
=
n=∞
n=0
(-1)n
x2n+1

(2n+1)!
                   
--- eq.TSc9
width of above equation a403071830
<a name="docAk41"> 
2015-03-07-18-30 here video 04:30/16:23 
eq.TSc1 change to eq.TSc7 to eq.TSc8 and compact 
form is eq.TSc9. 
eq.TSc8 is pretty easy to remember. To write in 
short hand, use sum notation as shown in eq.TSc9.
Doing the same procedure for cos(x) we get next.
<a name="docAk42">
 
cos(x)
=
f(0)
+
f '(0)

1!
x
+
f ''(0)

2!
x2
+
f '''(0)

3!
x3
+
f(4)(0)

4!
x4
●●●
--- eq.TSd0
 
cos(x)
=
1
+
0

1!
x
+
-1

2!
x2
+
0

3!
x3
+
1

4!
x4
●●●
--- eq.TSd1
 
cos(x)
=
1
x2

2!
+
x4

4!
x6

6!
+
x8

8!
x10

10!
+/- ●●●
--- eq.TSd2
 
cos(x)
=
n=∞
n=0
(-1)n
x2n

(2n)!
                   
--- eq.TSd3
width of above equation a403071845
<a name="docAk43">
2015-03-07-18-46 here 
eq.TSc3 is Taylor series representation for cos(x). 
We have alternating sum and we have only even power x 
in cos(x). Sine function is an odd function. If in 
sin(x) plug in -x for x , all term in eq.TSc8 change 
sign. eq.TSc8 also contain periodicity for sin(x). 

Now let us look at application. We want look at a 
pendulum. 
2015-03-07-18-56 stop at video 07:17/16:23 
<a name="tut064a3">

http://freeman2.com/tut064a3.jpg

<a name="docAk44">
2015-03-07-20-36 start
Just imagine a roof top hang a rope and a ball 
attach at rope end. Assume swing very small angle. 
Below is Liu,Hsinhan words. Please verify.
Let ball mass be m. Ball has gravitational force 
  Fg=mg --- eq.TSd4 
g is gravitational acceleration near earth surface. 
  g=9.80 meter/sec/sec --- eq.TSd5 
m is mass of the ball. Expressed in kilo-gram.
Fg is gravitational force toward earth center.
<a name="docAk45">
If one stand on high mountain, eq.TSc4 is still 
true, mass of the ball unchanged. But gravitational 
acceleration g on top of mountain reduced to 
9.76 meter/sec/sec (for example). 
Physics quantity Fg, m, g must have both number 
and physics unit. If mass is kilo-gram, length is 
meter, time is second, then force unit is 'newton'.
One newton force push one kilo-gram mass at 
velocity increase rate one meter/sec/sec.
"meter/sec" is velocity unit. When we consider 
"velocity increase rate" the unit is meter/sec 
increase rate (per second) result is acceleration 
and physics unit is "meter/sec/sec"
Above is Liu,Hsinhan words. 2015-03-07-21-01
<a name="docAk46"> 
We know the rope length is L. The angle between 
rope and the vertical dash line is theta θ . 
(this theta near ceiling, above ball) Now let 
us have a look at the system in more detail.
First of all we see there is an angle theta.
(this theta below ball) Gravitational force 
mg has a tangential component 
  Ft=mg*sin(θ) --- eq.TSd6 
('tangential' mean tangent to dash curve path)
mg has a radial component 
  Fr=mg*cos(θ) --- eq.TSd7 
('radial' mean from ball to ceiling point)
<a name="docAk47">
The radial force just pulling on the rope. 
Weight radial force and rope tension force 
same magnitude and opposite in direction. 
Both cancel to zero. Just get more tension 
in the rope, nothing else. The only force 
component help accelerate ball is tangential 
component Ft. Tangential Ft and radial Fr 
are always perpendicular to each other. 
<a name="docAk48">
We can use sin(x) to find tangential force. 
Let us look at Newton's equation. Newton 
states If you have a system and want to find 
state of equilibrium, you have to take all 
of the forces you have and equate them to 
zero. See next eq.TSd8 
  Fa + Ft = 0 --- eq.TSd8 
Now let us look at Fa and Ft in more detail. 
From eq.TSd6, we already know Ft. 
What is Fa? The accelerating force. 
<a name="docAk49">
Re-write eq.TSd8 as eq.TSd9 below 
  m*a + m*g*sin(θ) = 0 --- eq.TSd9 
where 
  Ft=m*g*sin(θ) --- eq.TSd6
and 
  Fa=m*a --- eq.TSe0
Ft is gravity force tangential component
Fa is accelerating force
eq.TSe0 is Newton's second motion law. 

<a name="Newton2nd"> Newton's second motion law. Acceleration 'a' of an object is proportional to the NET force 'Fa' acting on it and acceleration is inversely proportional to the mass 'm'. Fa=m*a --- eq.TSe0

<a name="docAk50"> 'NET' in 'NET force' is very important. In tut064a3.jpg, ball has a radial force Fr, but Fr cancelled by rope tension. radial direction net force is ZERO. According to Newton's second law, ball do not have acceleration along radial direction. Newton's second law include Newton's first law. Given Fa=0, in eq.TSe0 set Fa=0. Mass cannot become zero, eq.TSe0 require acceleration 'a' be zero. Not change velocity? then maintain same velocity forever. That is Newton's first law. 2015-03-07-22-09 here <a name="docAk51"> What is Fa? By Newton's second law Fa is m*a 'a' is acceleration. Because ball go circular motion. We know 'a' is angular acceleration a=L*d2θ/(dt)2 --- eq.TSe1 where L is rope length, θ is rope angle from dash vertical line. Now substitute eq.TSe1 into eq.TSd9 m*a + m*g*sin(θ) = 0 --- eq.TSd9 get <a name="docAk52"> m*L*d2θ/(dt)2 + m*g*sin(θ) = 0 --- eq.TSe2 eq.TSe2 is pendulum system's equation of motion. Solve eq.TSe2 for θ(t) is solve pendulum problem. Compare eq.TSe2 with sin(x) eq.TSc8
 
sin(x)
=
x
x3

3!
+
x5

5!
x7

7!
+
x9

9!
x11

11!
+/-●●● x
--- eq.TSc8
|x|<<1
width of above equation a403072230
<a name="docAk53">
2015-03-07-22-31 here
There is no close formula to solve eq.TSe2 . 
Because sin(x) is highly non-linear. For us to 
understand a little bit about this system. We 
can use Taylor series expansion for sine 
eq.TSc8 . This method is called small angle 
approximation for sine function. We have very 
small value of x |x|<<1. Those higher power 
terms will get even smaller and negligible. 
Only left with x term. We can say sine of 
very small angle is approximately equal to 
the arc length itself. 
<a name="docAk54">
With this approximation in eq.TSe2 
  m*L*d2θ/(dt)2 + m*g*sin(θ) = 0 --- eq.TSe2
we replace sin(θ) with θ and get 
  m*L*d2θ/(dt)2 + m*g*θ = 0 --- eq.TSe3
eq.TSe3 is same as 
 
d2θ(t)

(dt)2
=
g

L
θ(t)
--- eq.TSe4
width of above equation a403072253
<a name="docAk55">
2015-03-07-22-54 here 
In eq.TSe4 angular displacement θ is function of 
time t. We are not allowed to replace sin(θ) with 
θ for all values, only small angle approximation 
case OK. Now look at eq.TSe4. eq.TSe4 is a pretty 
easy harmonic oscillation. We know how to solve 
eq.TSe4. Its solution is 
  θ(t)=θ0*sin(ω*t+φ0) --- eq.TSe5 
φ0 is phase lag. 
2015-03-07-23-07 stop at video 13:00/16:23
<a name="docAk56"> 
2015-03-07-23-42 start
eq.TSe5 is solution of eq.TSe2 equation of motion. 
We differentiate eq.TSe5 once find 
  dθ(t)/(dt)=ω*θ0*cos(ω*t+φ0) --- eq.TSe6 
Differentiate eq.TSe5 twice find 
  d2θ(t)/(dt)2=-ω20*sin(ω*t+φ0) --- eq.TSe7 
Substitute eq.TSe5 into eq.TSe7 get 
  d2θ(t)/(dt)2=-ω2*θ(t) --- eq.TSe8 
Compare eq.TSe8 with eq.TSe4, we get 
  ω2=g/L --- eq.TSe9 
<a name="docAk57">
Take square root, we get 
  ω=√g/L --- eq.TSf0 
Square root has positive answer and negative answer. 
Here negative answer make no sense. We take only 
positive answer. ω is 2*PI*frequency of pendulum. 
Divide √g/L with 2*PI, get pendulum system frequency. 
  f0=ω/(2π)=√g/L/(2π) --- eq.TSf1
2015-03-08-00-00 here 
2015-03-08-10-05 add begin. 
ω is pendulum angular velocity. Radian per second. 
f0 is pendulum frequency. Cycle per second. 
One cycle has 2π radian. Then radian divide by 
2π is cycle. 2015-03-08-10-10 add stop
<a name="docAk58">
Exact solution is harder than this. In eq.TSc8 
sine approximation, instead of taking one term, 
you can take two terms or even more. Then try to 
solve the differential equation eq.TSe2. At least 
easier than solve eq.TSe2 with sin(θ). The second 
approximation we did not talk about drag force. 
Above solution is valid experiment in vacuum where 
no air resistance take place. 

<a name="docAk59">
What this example tell you is that Taylor series is 
not only a mathematical interest, but Taylor series 
give you a lot power in physics. Taylor series help 
you solve hard problem by easier approximation 
method. This is what we got. We use Taylor series 
to get this result. We also find Taylor series for 
sine function and cosine function. Next video we 
will relate Taylor series with exponential function.
Thank you guys for watching my video. If you like 
them, please give thumb up. If you have questions 
please feel free to ask. Please subscribe. 
See you guys. 
2015-03-08-00-18 stop 
2015-03-08-10-25 done first proofread
2015-03-08-20-37 done second proofread

<a name="docAk60">
Principle of Dynamics by Donald T. Greenwood 
1965 page 117 to 120. Section title Simple Pendulum. 
page 117 equation 3-224 begin large angle calculation
2015-03-08-00-22

small amplitude pendulum on rotating earth 
Dynamics Goodman and Warner , page 410-413 
1961, 1963
2015-03-08-00-27


<a name="docAk61"> EulerMascheroni sum bound pendulum 2015-03-09-13-36 start Next is study notes of Taylor Series - 7 - Exponential Function and Euler's Formula http://www.youtube.com/watch?v=rXHQUVwfKwE Lecturer is MrYouMath. Notes recorder is Liu,Hsinhan. Recorded notes is about 60% match, not exact. Liu,Hsinhan inserted text is blue Italic. LiuHH notes may contain error! <a name="docAk62"> Begin video 9 of 9 Hello welcome everyone. This is MrYouMath again. Today's video we want have a look at exponential function, its Taylor series. We want prove Euler's formula, we prove Euler's number e=2.718281828... is not rational. This is a very interesting video. The primary thing we use is Taylor series for exponential function.
<a name="docAk63">
Taylor series expansion for f(x) at x=0 ; Next equation --- eq.TSf2
 
f(x)
=
n=∞
n=0
f(n)(0)

n!
xn
=
f(0)
+ (x-0)*f '(0)
+
(x-0)2

2!
d2f(x)

(dx)2
x=0
+
(x-0)3

3!
d3f(x)

(dx)3
x=0
...
width of above equation a403091355
<a name="docAk64">
2015-03-09-13-55 here 
eq.TSf2 is Taylor series evaluated at x=0.
Taylor series evaluated at x=a see eq.LAf2 
Formula in eq.TSf2 is simple. We sum n from n=0 
to n→∞. All these derivatives evaluated at x=0. 
It is pretty easy and fast. 
If f(x)=exponential function, its derivative is 
again exponential function. 
  f(x)=exp(x) --- eq.TSf3 
It does not matter how often differentiation we 
take, it always equal exp(x) itself.
  f(n)(x)=exp(x) --- eq.TSf4 
<a name="docAk65">
Below is Liu,Hsinhan words. Please verify.
Show d[exp(x)]/dx=exp(x)
2015-03-09-14-31 here
To show 
  d[exp(x)]/dx=exp(x) --- eq.TSf5 
Do as following. Set 
  y(x)=exp(x) --- eq.TSf6 
Take log() to both side of eq.TSf6 get 
  log[y(x)]=log[exp(x)] --- eq.TSf7 
Since log() and exp() are inverse operation, 
  log[exp(M)]=M --- eq.TSf8 
for any M. That is log[exp(x)]=x . 
eq.TSf7 become 
  log[y(x)]=x --- eq.TSf9 
<a name="docAk66"> 
Now differentiate eq.TSf9 with respect x get 
  d{log[y(x)]}/dx=dx/dx --- eq.TSf9 
Write eq.TSf9 in detail steps as next. 
Attention: eq.TSg0 d[y(x)]/d[y(x)] is inserted one.
 
d

dx
log[y(x)]
=
d

dx
log[y(x)]
d[y(x)]

d[y(x)]
=
dlog[y(x)]

d[y(x)]
d[y(x)]

dx
--- eq.TSg0
 
1

y(x)
d[y(x)]

dx
=
dx

dx
=
1
☺ 
d[y(x)]

dx
=
y(x)
☺ 
d[exp(x)]

dx
=
exp(x)
--- eq.TSg1
width of above equation a403091512
<a name="docAk67">
2015-03-09-15-13 here 
From eq.TSg0 to eq.TSg1 
  dlog[y(x)]/d[y(x)] = 1/y(x) --- eq.TSg2 
that is logarithm differentiation 
dx/dx has value one. Move from eq.TSf9 to eq.TSg1 
eq.TSg1 left end equation result to eq.TSg1 middle 
equation. Refer to eq.TSf6, change y(x) to exp(x)
get eq.TSg1 right end red equation. This red 
equation is the target eq.TSf5. Above steps is 
from textbook by George, B. Thomas Jr. and 
Ross L. Finney Calculus sixth ed. page 409
Above is Liu,Hsinhan words. 2015-03-09-15-22 
<a name="docAk68">
If plug in x=0 to eq.TSf4, you get eq.TSg3 below.
  f(n)(0)=exp(0)=1 --- eq.TSg3
eq.TSg3 is e=2.718281828459045... to power zero. 
No matter which number you take, its zero power 
is always one. There is an exception, number zero 
to zero power This is a little problematic. So I 
will not include zero in my statement. Let us just 
use it. f(n)(0)=1 for all n, we get 
<a name="docAk69">
 
exp(x)
=
f(0)
+
f '(0)

1!
x
+
f ''(0)

2!
x2
+
f '''(0)

3!
x3
+
f(4)(0)

4!
x4
●●●
--- eq.TSg4
 
exp(x)
=
1
+
1

1!
x
+
1

2!
x2
+
1

3!
x3
+
1

4!
x4
●●●
--- eq.TSg5
 
exp(x)
=
n=∞
n=0
xn

n!
          |x| < ∞          
--- eq.TSg6
width of above equation a403091554
<a name="docAk70">
2015-03-09-15-55 here 
We get this very nice looking eq.TSg6 for exp(x). 
Very important, the radius of convergence which 
can be shown using the quotient rule or even better 
using Cauchy-Hadamard Theorem, you can show that 
the radius of convergence is infinitely large for 
exp(x). If x is finite, you can take more and more 
terms to the series eq.TSg6 and get a converging 
sum not depending on x. 
Let us look what we can do with eq.TSg6. 
(video 3:15/17:33)
<a name="docAk71">
 
exp(x)
=
1
+
1

1!
x
+
1

2!
x2
+
1

3!
x3
+
1

4!
x4
●●●
--- eq.TSg7
 
exp(1)
=
e=2.718281828459045...
=
1
+
1

1!
+
1

2!
+
1

3!
+
1

4!
...
--- eq.TSg8
width of above equation a403091616
<a name="docAk72">
2015-03-09-16-16 here 
First thing I want show you is a very nice series 
expression for Euler number e. Set x=1, eq.TSg7 
become eq.TSg8. eq.TSg8 is series expression.
eq.TSg8 has pretty fast converging. Because of these 
factorial in denominator. 
Now we want show this number e=2.718281828459045... 
cannot be rational. That is our second step. 
That is to show (video 4:01) e is irrational 
<a name="docAk73">
 
m

n
=
e
=
1
+
1

1!
+
1

2!
+
1

3!
+
1

4!
...
--- eq.TSg9
width of above equation a403091630
<a name="docAk74">
Below is Liu,Hsinhan words. Please verify.
2015-03-09-16-47 here 
We assume this number e IS RATIONAL which can be 
expressed as integer m divide by integer n . 
Assume there is no common factor between m and n. 
e IS RATIONAL assumption is an ERROR assumption on 
purpose. We will show this assumption get a result 
which is contradictory. Write eq.TSg9 in longer 
form as shown in eq.TSh0. In eq.TSh0 multiply n! 
to all terms, get eq.TSh1. 
<a name="docAk75">
eq.TSh1 left side is an integer. Since n! has a 
factor n which cancel denominator n. then m*(n-1)! 
is an integer. eq.TSh1 is an equality equation. 
eq.TSh1 right side MUST be an integer. 
Add color to eq.TSh1 get eq.TSh2 
eq.TSh2 blue terms are integers. Because n! has a 
factor of 2!, n!/2! is an integer. Similarly 
n!/3! ... n!/(n-1)!=n and n!/n!=1 are all integers. 
To maintain an integer equation, 
eq.TSh2 purple terms must sum to an integer. 
The following show that eq.TSh2 purple terms 
sum to a value less than one. This process get 
a contradictory result, and prove that 
e is irrational 
<a name="docAk76">
 
m

n
=
1
+
1

1!
+
1

2!
+
1

3!
+
...
+
1

n!
+
1

(n+1)!
+
1

(n+2)!
+
...
--- eq.TSh0
 
n!m

n
=
n!
+
n!

1!
+
n!

2!
+
n!

3!
+
...
+
n!

n!
+
n!

(n+1)!
+
n!

(n+2)!
+
...
--- eq.TSh1
 
n!*m

n
=
n!
+
n!

1!
+
n!

2!
+
n!

3!
+
...
+
n!

n!
+
n!

(n+1)!
+
n!

(n+2)!
+
...
--- eq.TSh2
width of above equation a403091646
<a name="docAk77">
2015-03-09-17-15 here 
eq.TSh2 first purple term is n!/(n+1)! 
If n=3, n!/(n+1)! = 3!/(3+1)! = 1/4 < 1/3
If n=4, n!/(n+1)! = 4!/(4+1)! = 1/5 < 1/3
Since n sum to infinity 
If n=99, 99!/(99+1)! = 99!/100! = 1/100 < 1/3

2015-03-09-17-20 UPS beep constantly for two minutes 

<a name="docAk78">
eq.TSh2 second purple term is n!/(n+2)! 
If n=3, n!/(n+2)! = 3!/(3+2)! = 1/4/5 < (1/3)2
If n=4, n!/(n+2)! = 4!/(4+2)! = 1/5/6 < (1/3)2
Since n sum to infinity 
If n=99, 99!/(99+2)! = 99!/101! = 1/100/101 < (1/3)2

Similarly, 
eq.TSh2 third purple term is n!/(n+3)! < (1/3)3
eq.TSh2 fourth purple term is n!/(n+4)! < (1/3)4

<a name="docAk79">
Now (1/3)1 + (1/3)2 + (1/3)3 + (1/3)4 + ... 
bound eq.TSh2 ALL purple terms 
This bound form a geometric series with x=1/3
f(x)
=
1

1-x
=
1 + x1 + x2 + x3 + x4 + ●●● + xn + ●●●
--- eq.TSh3
|x| < 1
1

1-x
- 1 =
x1 + x2 + x3 + x4 + ●●● + xn + ●●●
--- eq.TSh4
|x| < 1
width of above equation a403091735
<a name="docAk80">
2015-03-09-17-35 here 
This bound  (1/3)1 + (1/3)2 + (1/3)3 + (1/3)4 + ... 
sum to (see eq.TSh4)
  1/(1-1/3) -1 = 1/(2/3) -1 = 1.5 -1 = 0.5 --- eq.TSh5 
Here, showed that eq.TSh2 purple terms must be 
smaller than upper bound 0.5 and purple terms 
cannot be an integer. But eq.TSh2 left side is 
an integer and eq.TSh2 right side is not an 
integer. This is a contradictory result. The 
initial assumption "number e IS RATIONAL" is 
wrong. Conclude number e IS IRRATIONAL 
2015-03-09-17-48 here at video 12:00/17:33
Above is Liu,Hsinhan words. 
Skipped many MrYouMath lectures, 
Please watch video for MrYouMath lecture.
2015-03-09-17-50

<a name="docAk81"> 
2015-03-10-02-06 start at video 12:18/17:33
Now let us have a look at something else, that is 
also very interesting. 
<a name="docAk82">
 
exp(x)
=
n=∞
n=0
xn

n!
=
1
+
x

1!
+
x2

2!
+
x3

3!
+
x4

4!
+
●●●
--- eq.TSh6
 
exp(ix)
=
n=∞
n=0
(ix)n

n!
=
1
+
ix

1!
+
(ix)2

2!
+
(ix)3

3!
+
(ix)4

4!
+
●●●
--- eq.TSh7
 
i=√-1
☺ 
i0=1
☺ 
i1=i
☺ 
i2=-1
☺ 
i3=-i
☺ 
i4=1
--- eq.TSh8
 
exp(ix)
=
1
+
i*
x

1!
x2

2!
-i*
x3

3!
+
x4

4!
+
●●●
--- eq.TSh9
width of above equation a403100235
<a name="docAk83">
2015-03-10-02-35 here 
You are looking at the Taylor expansion of 
exponential function again. see eq.TSh6. Now I 
am doing something very tricky. In eq.TSh6 
replace 'x' with 'ix'. 'i' is imaginary unit 
  i=√-1 --- eq.TSi0
'i' is the number that you can square and get -1 
out of it. In eq.TSh6 every 'x' plug in 'ix', 
result is eq.TSh7. In eq.TSh8, listed the result 
of 0-th to 4-th power of 'i'. Use eq.TSh8, 
reduce eq.TSh7 to eq.TSh9. 
<a name="docAk84">
In eq.TSh9 there are two 
parts in it. One is numbers without 'i' , these 
are real part of exp(ix). The other part is 
numbers with 'i'. these are imaginary part of 
exp(ix). Let us just re-write eq.TSh9 as next 
 
exp(ix)
=
[
1
x2

2!
+
x4

4!
-●●●
]
+
i*
[
x

1!
x3

3!
+
x5

5!
-●●●
]
--- eq.TSi1
width of above equation a403100301
<a name="docAk85">
2015-03-10-03-02 here 
Compare eq.TSi1 left  square bracket with eq.TSd2
We see  eq.TSi1 left  square bracket is cos(x) 
Compare eq.TSi1 right square bracket with eq.TSc8
We see  eq.TSi1 right square bracket is sin(x). Then
eq.TSi1 is same as next equation 

Euler's formula eq.TSi2 exp(ix)=cos(x)+i*sin(x) --- eq.TSi2

Euler's formula in tute0064.htm is eq.TSi2 Euler's formula in tute0060.htm is eq.Eul1
<a name="EulerEq">
Euler's formula eq.Eul1
Please see Tom M. Apostol prove eq.Eul1
Please see MrYouMath prove eq.Eul1
 
ei*z
=
cos(z)
+
i*sin(z)
--- eq.Eul1
 
e-i*z
=
cos(z)
i*sin(z)
--- eq.Eul2
 
sin(z)
=
eiz - e-iz

2*i
--- eq.Eul3
 
cos(z)
=
eiz + e-iz

2
--- eq.Eul4
In eq.Eul1 change z to -z get eq.Eul2
[eq.Eul1 - eq.Eul2 ]/[2*i] get eq.Eul3
[eq.Eul1 + eq.Eul2 ]/2 get eq.Eul4 ; i=√(-1)
Euler's formula eq.Eul1 is copied from tute0060.htm#EulerEq
width of above equation a312192130;a403100308
<a name="docAk86"> 
2015-03-10-03-20 here is the end of 
MrYouMath Taylor Series lectures. Please watch 
MrYouMath video files for better explanation.
Because Liu,Hsinhan English is poor and 
LiuHH is not qualified to work independently.
Thank MrYouMath for his wonderful lectures. 
Thank you for visiting Liu,Hsinhan study notes 
pages.
2015-03-10-03-24 stop
2015-03-10-14-11 done first proofread
2015-03-10-14-59 done second proofread

<a name="a40318a">
2015-03-18-21-14
update 2015-03-18 change page end 
[[
file name tute0064.htm mean 
TUTor, English, 63 th .htm 
Chinese version is tutc0063.htm 
]]
to
[[
file name tute0064.htm mean 
TUTor, English, 64 th .htm 
Chinese version is tutc0064.htm 
]]



<a name="docA999"> Following is frequently needed strings. √π √p ε α → ∞ negligible θ0 an an+1 Taylor series representation exponential function and Euler's formula Trigonometric functions and Foucault's pendulum geometric series and the tortoise paradox logarithm and alternating harmonic series arctan and PI x2 E=mc² binomial coefficient singularity radius of convergence xn+1 f(x)=a0+a1x+a2x2π Γ(s) ζ(s) criteria √p √<a style='text-decoration:overline;'>p</a> logarithm Taylor series expansion xn ratio test radius of convergence quotient rule OSLER paper eq.LAf1 eq.LAc1 d2log(x)/(dx)2 us-1 (1-u)t-1 Taylor series expansion equation eq.LAf2 d2f(x)/(dx)2 Figure 1 ck2 Γ(n+1/2)=[(2n)!]*√π/[22n*n!] --- eq.LAb4 Figure 1 Gamma function →∞ ∏[i=0,n] ∑[k=1,∞] Γ(s)Γ(1-s) γ=lim[n→∞]{∑[i=1,n](1/i)-log(n)} eγs Γ(s) = ∫[t=0,t=∞]{ts-1*e-t*dt} eq.CG01 awesome trigonometric Pythagoras Cartesian implicity differentiation ┌ │ ┐ ┘ └ | ⇍ ⇎ ⇏ Α Β Γ Δ Ε Ζ Η Θ Ι Κ Λ Μ Ν Ξ Ο Π Ρ ΢ Σ Τ Υ Φ Χ Ψ Ω α β γ δ ε ζ η θ ι κ λ μ ν ξ ο π ρ ς σ τ υ φ χ ψ ω ¬ ~ ∨ ⊻ ∧ → ← ↔ ⇒ ⇐ ⇔ ↑ ↓ ⇓ ⇑ ⇕ ⇖ ⇗ ⇘ ⇙ ≦ ≠ ≧ < = > ± ≡ ≈ ≌ ≒ ∏ ∑ √ ∛ ∜ ∝ → ∞ ⊕ ⊙ ⊗ 〈v,w〉 ∈∀ ∂ ⊥ ∃ ∋ ∆ ∇ ∟ ∠ ∫ ∬ ∭ ∮ ∥ ○ ● ◎  ∧ ∨ ∩ ∪ ∴ ∵ ∶ ∷ ⊂ ⊃ ⊄ ⊅ ⊆ ⊇ ⊿ + - * / 2014-12-09-13-16 add next for hollow ℂ ℍ ℕ ℙ ℚ ℝ ℤ ℀ ℁ ℂ ℃ ℄ ℅ ℆ ℇ ℈ ℉ ℊ ℋ ℌ ℍ ℎ ℏ ℐ ℑ ℒ ℓ ℔ ℕ № ℗ ℘ ℙ ℚ ℛ ℜ ℝ ℞ ℟ ℠ ℡ ™ ℣ ℤ ℥ Ω ℧ ℨ ℩ K Å ℬ ℭ ℮ ℯ ℰ ℱ Ⅎ ℳ ℴ ℵ ℶ ℷ ℸ  ⅍ ⅎ ■□ ▢▣▤▥▦▧▨▩▪▫ × ÷ ° ◦ º ¹ ² ³ ts-1*e-t*dt ●●● rigorously 嚴謹的 negligible 可忽略的






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