﻿ Taylor Series study notes tute0064 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
Taylor Series
•by MrYouMath
•9 videos
•2,037 views
•Last updated on Apr 22, 2014

Taylor Series - 1 - Motivation and Derivation.mp4
Taylor Series - 1 - Radius of Convergence (Exercise).mp4
Taylor Series - 1 - Radius of Convergence by Singularities (Exercise).mp4
Taylor Series - 2 - Geometric Series and the tortoise.mp4
Taylor Series - 3 - Logarithm and alternating harmonic series.mp4
Taylor Series - 4 - Arctan and Pi.mp4
Taylor Series - 5 - Binomial Functions and Einsteins E=mc².mp4
Taylor Series - 6 - Trigonometric functions and Foucault's pendulum.mp4
Taylor Series - 7 - Exponential Function and Euler's Formula.mp4

2015-01-01-15-18 start
On
2014-03-25-20-36 Liu,Hsinhan access
find and download next 12 video files.

Gamma Function
•by MrYouMath
•12 videos
•5,940 views
•1 hour, 48 minutes
Gamma Function - Part 1 - Functional Equation
Gamma Function - Part 2 - Gauss Representation
Gamma Function - Part 3 - Weierstrass Representation
Gamma Function - Part 4 - Relationship to Sine
Gamma Function - Part 5 - Gamma of 0.5 ( one half)
Gamma Function - Part 6 - Stirling's Approximation
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
Gamma Function - Part 8 - Euler Integral II The Sinc-Function
Gamma Function - Part 9 - Euler Integral III Fresnel Integral
Gamma Function - Part 10 - Beta Function
Gamma Function - Part 11 - Legendre Duplication Formula
Gamma Function - Part 12 - Relation to Zeta Function
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

2014-10-29-07-56
2014-10-29-08-24
MrYouMath_17_files_Riemann.htm
"Zeta Function - Part 1 - Convergence"
"Zeta Function - Part 2 - Euler Product Representation"
"Zeta Function - Part 3 - Euler Product (easy)"
<a name="docA007">
"Zeta Function - Part 4 - Infinitude of Prime Numbers"
"Zeta Function - Part 5 - Prime Zeta Function"
"Zeta Function - Part 6 - The Prime Counting Function"
"Zeta Function - Part 7 - Zeta of 2 aka The Basel Problem"
"Zeta Function - Part 8 - Zeta of 2n - Part 1"
<a name="docA008">
"Zeta Function - Part 8 - Zeta of 2n - Part 2"
"Zeta Function - Part 8 - Zeta of 2n - Part 3"
"Zeta Function - Part 9 - Relation to Gamma Function"
"Zeta Function - Part 10 - Jacobi Theta Function"
"Zeta Function - Part 11 - Riemann Functional Equation I"
<a name="docA009">
"Zeta Function - Part 12 - Riemann Functional Equation II"
"Zeta Function - Part 13 - Trivial Zeros of the Zeta Function"
"Zeta Function - Part 14 - Riemann Xi Function"
"Sine Function Product Formula (Hadamard Factorization Theorem)"
"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

EulerMascheroni sum bound pendulum
2015-02-13-12-20 start
Next is study notes of
Taylor Series - 1 - Motivation and Derivation

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

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!.

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"> t064link()
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.

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

EulerMascheroni sum bound pendulum
2015-02-25-13-33 start
Next is study notes of
Taylor Series - 1 - Radius of Convergence
(Exercise).mp4

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!)
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

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
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"> t064link()
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
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"> t064link()
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

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

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

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
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

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

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
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
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
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
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">
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

EulerMascheroni sum bound pendulum
2015-03-02-18-03 start
Next is study notes of
Taylor Series - 2 - Geometric Series and
the tortoise.mp4

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
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.
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.

EulerMascheroni sum bound pendulum
2015-03-03-14-37 start
Next is study notes of
Taylor Series - 3 - Logarithm and alternating
harmonic series

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"> t064link()
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
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.
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"> t064link()
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

EulerMascheroni sum bound pendulum
2015-03-04-13-26 start
Next is study notes of
Taylor Series - 4 - Arctan and Pi

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"> t064link()
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.

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"> t064link()
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.

EulerMascheroni sum bound pendulum
2015-03-05-20-55 start
Next is study notes of
Taylor Series - 5 - Binomial Functions and
Einsteins E=mc²

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">
 ( rn ) = 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 ( rn ) xn
--- eq.TSb2
|x| < 1
width of above equation a403052353
```<a name="docAk11"> t064link()
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 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-x≈1+x/3.
If eq.TSb5 has 1/∜1-x, then eq.TSb6 is 1/∜1-x≈1+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 c² = m0c² [ 1 + 1 2 ( v c ) 2 + h.o.t ]
--- eq.TSb8
width of above equation a403061822
```<a name="docAk21"> t064link()
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

EulerMascheroni sum bound pendulum
2015-03-07-15-06 start
Next is study notes of
Taylor Series - 6 - Trigonometric functions and
Foucault's pendulum

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"> t064link()
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"> t064link()
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
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
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
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
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=-ω2*θ0*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
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
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

EulerMascheroni sum bound pendulum
2015-03-09-13-36 start
Next is study notes of
Taylor Series - 7 - Exponential Function
and Euler's Formula

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
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

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"> t064link()
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
]]

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+a2x2
√π 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
```

Javascript index
http://freeman2.com/jsindex2.htm   local
Save graph code to same folder as htm files.
http://freeman2.com/jsgraph2.js   local

file name tute0064.htm mean
TUTor, English, 64 th .htm
Chinese version is tutc0064.htm

2015-01-30-10-34 save as tute0064.htm

The address of this file is
http://freeman2.com/tute0064.htm