/**/
function alert4() //9812060000
{
document.write(''
+'Output may contain error, Please verify first.
'
+'Program environment is MSIE 6.0, please use MSIE
'
+''
);
} //function alert4() 9812060002
/**
textbook page 87 line 19 eqn.6.2
convex minorant used three times
create a command to build it any
where.
must call HelloJensen(hbPar)
outside of
..otherwise width wrong. 200912251340 here calling code is next line or or 9902141245 add w1 to HelloJensen() in tute0030.htm 99,02,07,21,45 the use of w1 from 9902072131 to 9902072143 solve infinite loop problem and no need change from uppercase to lowercase. /**/ function HelloJensen(hjPar1) //9812251342 { strJensen1='' +'<a name="Jensen' +hjPar1 +'"> ' +'Jensen Inequality for convex function' +'
' +' 
 page 87 line 19 eqn.6.2 
<a name="docA001"> Index begin Index this file 200906081910 start This file http://freeman2.com/tute0008.htm is Freeman's reading notes. Although Freeman always keep correct view point. But Freeman's capability is limited, plus no one proofread this file. Then you can still find wrong view point. When read, please put question mark as often as possible. If you suspect any view point wrong, please ask a math expert near by. <a name="docA002">,<a name="textbook"> This file is a note for reading inequality book written by Professor J. Michael Steele The CauchySchwarz Master Class ★★★★★ Below use 'textbook' as abbreviation. Freeman also read web pages online, and will indicate the source URL at discussion point. <a name="docA003"> Freeman study mechanical engineering. Engineering mathematics do not teach inequality. Above book is first inequality book. First time read, it was very hard. Although high school time learned a*a + b*b >= 2*a*b But this little knowledge do not help. <a name="docA004"> This file follow textbook chapter section order, but not continuous, Freeman skip those uncertain sections/problems. This file first function is to learn inequality. Second function is to learn how to use html code to write math equations. <a name="docA005"> Index begin Index this file On 200901271008 Freeman accessed the next page http://www.sftw.umac.mo/~fstitl/10mmo/inequality.html save as sftw.umac.mo_text_math_eqn_good.htm Above page is the main reference for html math equation. In order to let reader to build html math equation, previous file tute0007.htm page end has math symbol and internal code. This file third function is to display how to draw curves in web page. The main engine is XYGraph v2.3  Technical Figures Thank you for read Freeman's inequality page. 200906081947 stop <a name="docA006"> 200908231500 start □ Exercise 1.14 solution Exercise 1.14 problem (a) is solved by hint Exercise 1.14 problem (b) is out of LiuHH's reach. What is "cardinality of a set B⊂Z^{3}" ? Without clear understand the definition, LiuHH is unable to solve problem (b) LiuHH major in mechanical engineering. LiuHH is mathematics admirer and outsider. <a name="docA007"> To solve problems in The CauchySchwarz Master Class LiuHH's goal is to solve 50% of the exercise problems. In reality, solve 40% is good enough. Please do not be surprise that LiuHH skip some text section, skip some problems. 200908231512 stop
<a name="ch08c001"> Index begin Index this file 201002231027 ■ Power Mean attention In previous work, we discussed □ Why require p_{1}+p_{2}+...+p_{n}=1 ? eqn.AE20 It relate to Power Mean directly. Power Mean inequality is next<a name="ch08c002">

<a name="ch08c003"> Here, we consider equation physics dimension consistency. p_{k} is probability, which is a pure number, not carry physics dimension. Assume x_{k} is measured distance, eqn.8.10 left side {term} is sum of pure_number*distance^{s} physics dimension is length to s power. Then {term} rise to 1/s power again, the net result is back to distance first power. [ s*(1/s)=1 ] <a name="ch08c004"> eqn.8.10 right side change from s to t. Same conclusion. Therefore eqn.8.10 is distance compare with distance. The math form in right side of eqn.8.10 is marked as M_{t} Left side is M_{s}. When t approach to zero we have M_{0}. <a name="ch08c005"> M_{0} is expressed in eqn.8.2, its right side ∏x_{k}^{pk} was discussed in tute0011.htm#ch02a063 In brief, because SUM{p_{k}}=1, then ∏x_{k}^{pk} still has the dimension of length. M_{0} can be compared with M_{1} M_{1} etc. 201002231106 stop<a name="ch08c007">
<a name="ch08c006"> Index begin Index this file 201002231626 start ■ Exercise 8.2 problem statement textbook page 131 （Harmonic Means and Recognizable Sums） Suppose x_{1},x_{2},...,x_{n} are positive and let S denote their sum. Show that one has the bound

<a name="ch08c008"> In this problem (and many like it) one gets a nice hint from the fact that there is a simple expression for the sum of the denominator on the righthand side. 201002231638 stop <a name="ch08c009"> Index begin Index this file 201002231641 start ■ Exercise 8.2 hint textbook page 258 By the upsidedown HMAM inequality (8.16)<a name="ch08c010"> //201002231736 change x_{j} to a_{j}

page 126
Line 25 eqn.8.16 

page 258
Line 8 eqn.AZ002 
<a name="ch08c012"> 201002231700 here If we set a_{k}=2Sx_{k} eqn.AZ003 then a_{1}+a_{2}+...+a_{n}=2nSS=(2n1)S eqn.AZ004 and the HMAM bound yields<a name="ch08c013">

201002231706 stop <a name="ch08c014"> Index begin Index this file 201002231708 start ■ Exercise 8.2 solution Exercise 8.2 hint already solved this problem. Few thing to observe. <a name="ch08c015"> In eqn.8.16 set p_{1}=p_{2}=...=p_{n}=1/n eqn.AZ006 to get eqn.AZ002. Then use eqn.AZ003 to relate x_{k} and a_{k} and S. Here S=x_{1}+x_{2}+...+x_{n} eqn.AZ007 which is given in Exercise 8.2 problem statement. <a name="ch08c016"> With a_{k}=2Sx_{k} eqn.AZ003 relation, we can express eqn.AZ002 less than side denominator a_{1}+a_{2}+...+a_{n} =(2Sx_{1})+(2Sx_{2})+...+(2Sx_{n}) =2nS  x_{1}x_{2}...x_{n} =2nS  S <a name="ch08c017"> that is a_{1}+a_{2}+...+a_{n}=(2n1)S eqn.AZ008 Put eqn.AZ008 to eqn.AZ002 left side (once) Put eqn.AZ003 to eqn.AZ002 right side (n times) we get eqn.AZ005. From eqn.AZ005 to eqn.AZ001 is just one step. Problem solved. 201002231724 stop
<a name="ch08c018"> Index begin Index this file 201002231837 start ■ Exercise 8.3 problem statement textbook page 132 （Integral Analogs and Homogeneity in ∑） The importance of weight factor. Weight factor not sum to one? All M_{t}, t≠0 are still length^{1.0} BUT GM M_{0} at t=0 has length^{1.1} AM:length^{1.0} compare?! GM:length^{1.1} Power Mean curve jump to infinity at singular GM. explore local <a name="ch08c019"> (a) Show that for all nonnegative sequence {a_{k}: 1≦k≦n} one has

page 132
Line 3 eqn.8.29 
<a name="ch08c020"> and be sure to notice the differences between this bound and the power mean inequality (8.10) with s=1/3 and t=1/2. //201003021650 key difference ! //eqn.8.29 each term has probability //coefficient=1, sum to n, NOT one! <a name="ch08c021"> (b) By analogy with the bound (8.29) one might carelessly guess that for nonnegative f one has an integral bound<a name="ch08c022">

page 132
Line 8 eqn.8.30 
Show by example that the bound (8.30) does not hold in general. <a name="ch08c023"> Index begin Index this file The likehood of an integral analog can often be explained by a heuristic principle which Hardy, Littlewood and Polya (1952, p4) describes as "homogeneity in ∑". The principle suggests that we consider ∑ in a bound such as (8.29) as a formal symbol. <a name="ch08c024"> In this case we see that the left side is "homogeneous of order two in ∑" while the right side is "homogeneous of order three in ∑" The two sides are therefore incompatible and one should not expect any integral analog. <a name="ch08c025"> On the other hand, in Cauchy's inequality and Holder's inequality both sides are homogeneous of order one in ∑. It is therefor natural  even inevitable  that we should have integral analog for these bounds. 201002231904 stop <a name="ch08c026"> Index begin Index this file 201002231916 start ■ Exercise 8.3 hint explore textbook page 258 local Both sides of the bound (8.29) are homogeneous of order one in (a_{1},a_{2}, ...,a_{n}), so we can assume without loss of generality that a_{1}^{1/3}+a_{2}^{1/3}+...+a_{n}^{1/3}=1 eqn.AZ009 <a name="ch08c027"> Given this, we only need to show that a_{1}^{1/2}+a_{2}^{1/2}+...+a_{n}^{1/2}≦1 eqn.AZ010 and this is remarkable easy. By the normalization, we have a_{k}≦1 for all 1≦k≦n eqn.AZ011 so we also have a_{k}^{1/2} ≦ a_{k}^{1/3} for all 1≦k≦n eqn.AZ012 //if a_{k}=0.5, a_{k}^{1/2}=0.70710＜0.79370=a_{k}^{1/3} and we just take the sum to get our target bound. <a name="ch08c028"> One might want to reflect on what made this exercise so much easier than the proof of the power mean inequality (8.10). For part (b), if we take f(x)=x^{6} eqn.AZ013 to minimize arithmetic, then we see that the putative bound (8.30) falsely asserts 1/16 ≦ 1/27. 201002231930 stop <a name="ch08c029"> explore local 201002232010 start ■ Exercise 8.3 discussion Put power mean inequality (8.10) and Exercise 8.3 target eqn.8.29 side by side, easier for comparison.<a name="ch08c030"> Top line is power mean inequality
width of above equation <a name="ch08c031"> Index begin Index this file 201002232028 here Power mean inequality eqn.8.10 read as M_{s}≦M_{t} eqn.AZ014 where 's' and 't' are inner x_{k} power, not outer {term} power. Power mean inequality tell us that, if s＜t then M_{s}≦M_{t}, because 1/3＜1/2, we have M_{1/3}≦M_{1/2} eqn.AZ015 <a name="ch08c032"> Next exam bottom eqn.8.29. If we use N to represent terms in eqn.8.29, we get N_{1/2}≦N_{1/3} eqn.AZ016 Ex.8.3 M_{1/3}≦M_{1/2} eqn.AZ015 PowMean Here again put two simplified equation side by side. PowMean is 1/3＜1/2 then M_{1/3}≦M_{1/2} Ex.8.3 is 1/3＜1/2 then N_{1/2}≦N_{1/3} <a name="ch08c033"> We found that Exercise 8.3 eqn.8.29 s,t order is opposite to that of power mean inequality !! Go back to two equation and exam carefully. We find that power mean inequality x_{k} has company probability sequence p_{k} <a name="ch08c034"> But Ex.8.3 eqn.8.29 a_{k} do not have probability company ! In this case, we can scale a_{k} sequence freely. Also a_{k} sequence not normalized by probability sequence p_{k}. This difference is the cause of oppositeinequality result. <a name="ch08c035"> Who said one do not need a wife? A wife keep everything in good order. A wife keep power mean curve in monotone increase order. 201002232055 stop <a name="ch08c036"> 201002232056 start ■ Exercise 8.3 solution It is your job for change. Liu,Hsinhan miss wife Liling. LiuHH take a break. 201002232057 stop More related work: explore local
<a name="ch08c040"> So, your analytic challenge is to find the value p^{*} such that
width of above equation <a name="ch08c041"> One expects p^{*} to be some well known mean, but which one is it? 201002241416 stop <a name="ch08c042"> Index begin Index this file 201002241419 start ■ Exercise 8.4 hint textbook page 258 We only need to consider p∈[a,b], and in that case we can write<a name="ch08c043">
<a name="ch08c044"> The identity
width of above equation <a name="ch08c045"> tell us that (b－a)/(a＋b) is a weighted mean of (p－a)/a and (b－p)/b, so we always have the bound
<a name="ch08c046"> Moreover, we have strict inequality here, unless (pa)/a=(bp)/b, so, as Polya (1950) observed, the unique minimum of F(p) is attained at p^{*}=2ab/(a+b), which is the harmonic mean of a and b. 201002241443 stop <a name="ch08c047"> 201002241452 the exact answer x is out of question !! x has no position in the answer process !?<a name="drawEx.8.4"> Index begin Index this file 201002241640
<a name="ch08c048"> 201002241856 start ■ Exercise 8.4 solution Why start from eqn.AZ017 ? To understand Exercise 8.4. Above has a small drawing program. <a name="ch08c049"> Input is Box11. Output is Box12. Input need five numbers, they are described in Box11. eqn.8.31 is drawn in figure. eqn.8.31 is rewritten as eqn.AZ020 F(p)=max(x∈[a,b]){px/x} eqn.AZ020 Actually next equation do the job. F(x)=max(x∈[a,b]){px/x} eqn.AZ021 //F(p) means that p is a variable, //x constant. F(x) is just reverse. <a name="ch08c050"> There are two run buttons "Ex.8.4RUN 0" and "Ex.8.4RUN 1" Program use 't' as variable. Both p and x are fixed number. "Ex.8.4RUN 0" draw {pt/t} //x vary "Ex.8.4RUN 1" draw {tx/x} //p vary <a name="ch08c051"> Index begin Index this file "Ex.8.4RUN 0" draw {pt/t} x vary. Curve is nonlinear. "If you guess is p" is constant. "unknown number x" walk through whole domain. (set x as variable t) When x meet p, F(x) is zero. //not F(p) [F(x) is fine, punish money] Key point to observe in this drawing is that maximum F(x) occur at two end points. This is why we start from eqn.AZ017 . <a name="ch08c052"> "Ex.8.4RUN 1" draw {tx/x} p vary. Curve is linear. Same observation as above. (two ends has maximum value) <a name="ch08c053"> With the knowledge that maximum occur at end points. We can goto Exercise 8.4 hint. Exercise 8.4 key equation is eqn.AZ017 It tell us to choose greater one of two end values. <a name="ch08c054"> If two end values are not equal, then the greater one may trap us to pay higher fine. If two end values are equal, that is the best case. Then we require (pa)/a=(bp)/b eqn.AZ022 <a name="ch08c055"> Index begin Index this file solve for p get answer p^{*}=2ab/(a+b) eqn.AZ023 Take reciprocal get 1/p^{*}=(a+b)/(2ab) 1/p^{*}=1/(2b) + 1/(2a) eqn.AZ024 This is wellknown harmonic mean. Attention: "unknown number x" never show up in solving procedure! 201002241922 stop <a name="ch08c056"> Box12 output has pmin=1.6666666666666667 pmin is harmonic mean of a=1 and b=5. If paste 1.6666666666666667 to Box11, change from 2.5 guess value to 1.6666666666666667 guess value and click "Ex.8.4RUN 0" <a name="ch08c057"> output curve two end values are equal height. Maximum fine minimized, this is our answer. 201002241928 stop 201002251906 done proofread 201002251918 done spelling check<a name="ch08c059">
width of above equation <a name="ch08c060"> where D is the region of Real^{n} defined by //●●change
//because if one x_{k}=0, ∏x_{k}=1 is void. 201002261115 width of above equation <a name="ch08c061"> For practice with this characterization of the geometric mean, use it to give another proof that the geometric mean is superadditive, that is show that the formula (8.33) implies the bound (2.31) on page 34. 201002261103 stop <a name="ch08c062"> Index begin Index this file 201002261151 start ■ Exercise 8.5 hint textbook page 259 For all x∈D we have the bound (a_{1}*a_{2}*...*a_{n})^{1/n} //*1=x_{1}*x_{2}*...*x_{n} =(a_{1}x_{1}*a_{2}x_{2}*...*a_{n}x_{n})^{1/n} //GM≦AM ≦{∑[k=1,n](a_{k}x_{k})}/n eqn.14.57 by the AMGM inequality, and we have equality here if and only if a_{k}x_{k} does not depend on k. <a name="ch08c063"> If we take x_{k}=a_{k} / (a_{1}*a_{2}*...*a_{n})^{1/n} //by ByronSchmuland x_{k}=(a_{1}*a_{2}*...*a_{n})^{1/n} / a_{k} eqn.AZ026 then x∈D and the equality holds in the bound (14.57). This is all one needs to justify the identity (8.33) <a name="ch08c064"> Wonder Now to prove the bound (2.31) on page 34, one now just notes
width of above equation since two choices are better than one. <a name="ch08c065"> Incidentally, this type of argument is exploited systematically in Bechenbech and Bellman (1965) where the formula (8.33) is called quasilinear representation of the geometric mean. 201002261226 stop<a name="ch08c066"> <a name="ch02c097">
page 34 line 19 eqn.2.31 width of above equation <a name="ch08c067"> Index begin Index this file 201002261255 start ■ Exercise 8.5 solution AMGM Inequality AMGM Inequality is general for any irrational weights (page 23, eqn.2.9) a_{1}^{p1}a_{2}^{p2}...a_{n}^{pn} ≦ p_{1}a_{1}+p_{2}a_{2}+...+p_{n}a_{n} eqn.2.9 ALERT: p_{1}+p_{2}+...+p_{n}=1 eqn.AE20 <a name="ch08c068"> Use power mean language, AMGM Inequality is nextGM ＝ M_{0} ＝ next line //∑p_{k}＝1 , k=1,2,...,n is required
GM≦AM power mean style expression eqn.AZ028 width of above equation compare above with eqn.8.33
width of above equation <a name="ch08c069"> 201002261326 here eqn.AZ028 has probability coefficients p_{k}. They are allowed to be unequal. If we set p_{k} to be equal, then p_{1}=p_{2}=...=p_{n}. Because eqn.AE20 require p_{k} sum to one, we find p_{1}=p_{2}=...=p_{n}=1/n eqn.AZ029 put these p_{k}=1/n to eqn.AZ028, <a name="ch08c070"> it is almost the same as eqn.8.33. The difference is that eqn.8.33 use 1/n which is equal p_{k} eqn.8.33 use additional ∏x_{k}=1 ∏x_{k}=x_{1}*x_{2}*x_{3}=1 multiplied (extra) ∑p_{k}=p_{1}+p_{2}+p_{3}=1 summed (routine) <a name="ch08c071"> How do we know ∏x_{k}＝1, k=1,2,...,n eqn.AZ030 This requirement is stated in two steps. First step in eqn.8.33 say (x_{1},x_{2},...,x_{n})∈D eqn.AZ031 <a name="ch08c072"> Index begin Index this file Second step in eqn.AZ025 require set D has the following property ∏x_{k}＝1, k=1,2,...,n In eqn.14.57 equality we used 1=x_{1}*x_{2}*...*x_{n} eqn.AZ032 which is ∏x_{k}＝1. Any number multiply by one not change value. <a name="ch08c073"> In eqn.14.57 INequality we used GM≦AM inequality for the sequence a_{k}x_{k}, (consider multiplication of two numbers a_{k}, x_{k} as one number) Insert x_{k} for a_{k} play trick now. Because 1=x_{1}*x_{2}*...*x_{n} eqn.AZ032 we can define x_{k} to be x_{k}=(a_{1}*a_{2}*...*a_{n})^{1/n} / a_{k} eqn.AZ026 <a name="ch08c074"> To see whether eqn.AZ026 satisfy eqn.AZ032, assume n=3 for simpler example, then x_{1}=(a_{1}*a_{2}*a_{3})^{1/3} / a_{1} x_{2}=(a_{1}*a_{2}*a_{3})^{1/3} / a_{2} x_{3}=(a_{1}*a_{2}*a_{3})^{1/3} / a_{3} x_{1}*x_{2}*x_{3}= [(a_{1}*a_{2}*a_{3})^{1/3}]^{3}/[a_{1}a_{2}a_{3}] =[a_{1}a_{2}a_{3}]/[a_{1}a_{2}a_{3}] = 1 eqn.AZ033 Find eqn.AZ026 do satisfy eqn.AZ032. <a name="ch08c075"> eqn.AZ026 tell us that a_{k}*x_{k}=(a_{1}*a_{2}*...*a_{n})^{1/n} eqn.AZ028 which is constant, independent from index k. All a_{k}*x_{k} have same value (a_{1}*a_{2}*...*a_{n})^{1/n} (a_{k}*x_{k}=constant is key point for constant sequence GM=AM, otherwise GM＜AM) The section Equality GM=AM is easy tell us that when all elements are equal (same as constant) then GM≦AM become GM=AM. This specific GM=AM is minimum value of AM. 201002261414 stop <a name="ch08c076"> Index begin Index this file 201002261422 start To prove eqn.2.31 We pay attention to that eqn.8.33 left side (∏ side) do not contain x_{k} eqn.8.33 right side (∑ side) has x_{k} term. <a name="ch08c077"> eqn.2.31 is same as eqn.8.33 left side. Three terms in eqn.2.31 change to three terms in eqn.AZ027 Give reasoning in eqn.AZ027 is easy. <a name="ch08c078"> a_{k}x_{k} and b_{k}x_{k} are allowed to find minimum independently the result is less than or at most equal to a_{k}x_{k} and b_{k}x_{k} tied together for a minimum. eqn.2.31 is proved. 201002261432 stop<a name="ch08c081">
<a name="ch08c082"> and verify in turn that this implies the poignant formula
width of above equation <a name="ch08c083"> Incidentally, the product formula (8.34) for sin(x)/x is known as Viete's identity, and it has been known since 1593. 201002261612 stop <a name="ch08c084"> Index begin Index this file 201002261617 start ■ Exercise 8.6 hint textbook page 259 The halfangle formula for sine gives<a name="ch08c085">
width of above equation <a name="ch08c088"> 201002261642 here and as k→∞ the bracketed term goes to 1 since sin(t)=t+O(t^{3}) eqn.AZ038 as t→0. //sin(t)/t=t/t+O(t^{3})/t=1+O(t^{2})=1 Upon setting x=PI/2 eqn.AZ039 one gets the second formula after computing the successive values of cosine with help from its half angle formula. <a name="ch08c089"> Naor (1998, pp. 139143) gives a full discussion of Viete's formula, including a fascinating geometric proof. 201002261649 stop <a name="ch08c090"> 201002261655 start ■ Exercise 8.6 solution Exercise 8.6 hint already solved this problem, and this is an easy problem. Nothing to discuss here. 201002261656 stop<a name="ch08c095">
width of above equation 201002261822 stop <a name="ch08c096"> Index begin Index this file 201002261828 start ■ Exercise 8.7 hint textbook page 259 Our assumptions give us the bound [f(t_{0}+h)f(t_{0})]/h≧0 eqn.AZ043 for all h∈(0,∆], and now we just let h→0 to prove the first claim. To address the second claim, one first notes by the power mean inequality, or by Jensen's inequality, that one has<a name="ch08c097">
width of above equation <a name="ch08c098"> Since 0=f(1)≦f(t) for 1≦t eqn.AZ045 we also have f'(1)≧0 eqn.AZ046 and this is precisely the bound (8.35) 201002261849 stop<a name="drawEx.8.7"> 201002261938 Box22 equation definition must use 'x' as variable.
<a name="ch08c099"> Index begin Index this file 201003011455 start ■ Exercise 8.7 solution part (a) discussion. Exercise 8.7 find differentiation of inequality type. Given condition is <a name="ch08c100"> [[ a function f that is differentiable at t_{0} and that satisfies the bound f(t_{0})≦f(t) eqn.AZ040 for all t∈[t_{0},t_{0}+∆) and some ∆＞0. ]] <a name="ch08c101"> Please see the drawing drawEx.8.7 and click "Ex.8.7RUN" button. (reading notes use 't' as variable, drawing use 'x' as variable, since function differentiatef1() for d[f(x)]/dx use 'x' as variable. Yield to function definition cause not consistency. 201003011516) <a name="ch08c102"> Default curve is f(x)=(x1)*(x1)+0.5 eqn.AZ047 Default point is x=3.2, which is t_{0} In the problem statement, given condition f(t_{0})≦f(t) eqn.AZ040 tell us that for a positive small number 0≦h≦∆ we have f(t_{0})≦f(t) here t=t_{0}+h＞t_{0} eqn.AZ048 //The use of ∆ says that d[f(t)]/dt＞0 //is a local phenomenon. Few pace more //ahead may not have d[f(t)]/dt＞0; //201003031134 <a name="ch08c103"> In the drawing, red curve right to dot vertical line is t＞t_{0} section. t_{0} is where dot line located. Rewrite given condition eqn.AZ040 as [f(t_{0}+h)f(t_{0})]≧0 eqn.AZ049 divide by positive number h, get [f(t_{0}+h)f(t_{0})]/h≧0 eqn.AZ050 <a name="ch08c104"> We require h approach to zero, eqn.AZ050 is differentiation at point t_{0} exactly. Then given condition eqn.AZ040 give us f'(t_{0})= eqn.AZ051 limit[h→0][f(t_{0}+h)f(t_{0})]/h≧0 //'=' from derivative definition //'≧' from given condition. eqn.AZ051 is same as eqn.AZ041. Problem part (a) solved easily. 201003011532 here <a name="ch08c105"> Index begin Index this file 201003011542 start For part (b) 201003011615 Next copied from tute0030.htm#ch08a045 <a name="ch08c106"> Please compare next two equations
width of above equation //above Jensen Inequality with base function f(x)=x^{r}, r＞1 //below Jensen Inequality with unspecified base function and <a name="ch08c107"> From both equation greater than side, we see directly that f(x_{k})=x_{k}^{r} eqn.AX106 From both equation less than side, we see f(∑px)={∑px}^{r} eqn.AX107 <a name="ch08c108"> <a name="ch08a047"> eqn.AX106 and eqn.AX107 suggest that f(x)=x^{r} eqn.AX108 Jensen Inequality for convex function require that r>1 eqn.AX109 Above copied from tute0030.htm#ch08a047 201003011631 here <a name="ch08c109"> Jensen give us eqn.AZ052, rewrite eqn.AZ052 as same type of eqn.AZ044 next
page 259 Line 29 eqn.AZ053 (eqn.AX105) width of above equation <a name="ch08c110"> Index begin Index this file In eqn.AZ053 if we set r=1, find f(r=1)=∑p_{k}x_{k}  ∑p_{k}x_{k} = 0 eqn.AZ054 This is just one point r=0 condition. What if r=2, r=3? Do we have f(r=2)＞f(r=3) ? (≧0) f(r=2)＜f(r=3) ? (≧0) f(r=2)＝f(r=3) ? (≧0) <a name="ch08c111"> Here we need Power Mean inequality
width of above equation <a name="ch08c112"> eqn.8.10 short writing is M_{s}≦M_{t} for s≦t eqn.AZ055 Now set s=r and t=r+0.1, we have M_{r}≦M_{r+0.1} for r≧1 eqn.AZ056 Here M_{r} has a outer root power 1/r M_{r+0.1} has outer root power 1/(r+0.1) See eqn.8.10. eqn.AZ056 whole equation take r th power, eqn.AZ056 become M_{r}^{r}≦M_{r+0.1}^{r} for r≧1 eqn.AZ057 <a name="ch08c113"> Write in eqn.AZ053 form for f(r) and f(r+0.1) get
page 259 Line 29 eqn.AZ058 (eqn.AX105) width of above equation <a name="ch08c114"> 201003011718 here 0.5*pow(0.2,1.1)+0.5*pow(0.3,1.1) 0.21811921483686045 0.5*pow(0.2,2.1)+0.5*pow(0.3,2.1) 0.056922365225850285 in x^p if x<1 and p>1 greater p get smaller x^p 201003011722 stop <a name="ch08c115"> Index begin Index this file 201003012001 start Above calculation [[ in x^p if x<1 and p>1 greater p get smaller x^p ]] But we require greater p get greater function value, that is monotone increase. Above calculation take only one term into consideration (the red or blue) did not consider the minus_black terms. <a name="ch08c116"> It is easier to verify numerically. Here is a program local http://freeman2.com/powmean2.htm Added new function, it is marked "Jensen". There are two buttons "JensenA" and "JensenB". "JensenA" modify input boxes value before call function powMeanf() modification target at xsequence value range in [0,17]. <a name="ch08c117"> If you have xsequence value range in [0,1] for example, you need give proper setting (input boxes values) then click "JensenB" button. "Jensen" button draw three curves. Red curve is eqn.AZ053 red term (or blue term same thing) Blue curve is eqn.AZ053 black term ignored ''. Black curve is whole eqn.AZ053 . <a name="ch08c118"> First, fill '1' or '1.5' to 10^ box second click "random5#" button, third click "JensenA" button Black curve has zero value at t=1 In Box 3, output, find a line similar to next line 1; 15.558182090181461; 15.558182090181461; 0 Here t=1, both red term and black term has value 15.55818, their difference redblack (eqn.AZ053) is zero. Zero value is required by eqn.AZ054. <a name="ch08c119"> Key point is for t＞1, black curve is monotone increase in limited domain [t0,t0+∆) Above xsequence range in [0,17] <a name="ch08c120"> Index begin Index this file Below xsequence range in [0,1] First, fill '0' to 10^ box second click "random5#" button, third modify the following boxes (use the following value) y min: 1 , y max: 1 fourth click "JensenB" button (alert: do not click "JensenA" ) <a name="ch08c121"> Output curve show that black curve has negative value for t in [0,1] black curve has positive value for t in ( 1, greater_value ) black curve is monotone increase within visible domain [1,3] We need black curve be monotone increase. <a name="ch08c122"> With above observation, (NOT A PROOF) we can answer What if r=2, r=3? Do we have f(r=2)＞f(r=3) ? (≧0) f(r=2)＜f(r=3) ? (≧0) f(r=2)＝f(r=3) ? (≧0) The answer is monotone increase we have f(r=2)＜f(r=3) <a name="ch08c123"> What we have in hand is eqn.AZ053 is monotone increase, f(r=1)=0 (eqn.AZ054) we use part (a) conclusion get f'(1)≧0 eqn.AZ046 follow textbook [[ and this is precisely the bound (8.35) ]] we solved part (b) Do you agree? 201003012041 here <a name="ch08c124"> 201003012049 start Not really done !! How to go from eqn.AZ044 to eqn.8.35 It look like remote. Why log() show up? why not exp()? why not sin()? Puzzle !! <a name="ch08c125"> Index begin Index this file Function g(t) if variable t is in power position like E^t, //E=2.71828... We know if g(t)=exp(t) eqn.AZ059 then d[g(t)]/dt=exp(t) eqn.AZ060 <a name="b2t_power"> We know if h(t)=B^t eqn.AZ061 then d[h(t)]/dt=h(t)*log(B) eqn.AZ062 //if h(t)=B^t were h(t)=E^t, then //log(B) become log(E) which is ONE. <a name="ch08c126"> eqn.AZ044 is waiting to be differentiated. We do as following
width of above equation <a name="ch08c127">
red to red, blue to blue, all NOT contain 't' (B here) width of above equation <a name="ch08c128">
width of above equation <a name="ch08c129"> We need use the condition f'(1)≧0 eqn.AZ046 then eqn.AZ064≧0 and eqn.AZ064 change t to one. After this two steps, eqn.AZ064 is same as eqn.8.35 Now Exercise 8.7 is really solved. 201003012125 stop 201003021856 done proofread 201003031512 done spelling check <a name="ch08c130"> Update 20100316 made following notes about eqn.AZ027 201003162059 copy next from tute0034.htm <a name="ch09b014"> Wonder, textbook page 259, Line 11 whether eqn.AZ027 is better stated as OUTER_min{inner_min[x∈D](∑[k=1,n]a_{k}*x_{k})/n, inner_min[x∈D](∑[k=1,n]b_{k}*x_{k})/n} ≦min[x∈D]{∑[k=1,n](a_{k}+b_{k})*x_{k}}/n eqn.BB009 <a name="ch09b015"> eqn.AZ027 two team two min add together has risk greater than joint team one min !! 201003161306 stop 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 tute0032.htm means TUTor, English, 32nd .htm Chinese series file name is tutc0001.htm This page, Inequality file twenty six. http://freeman2.com/tute0032.htm First Upload 20100225 (Inequality start from tute0007.htm) Thank you for visiting Freeman's page. Freeman 201002251919 ≦ ≠ ≧ ＜＝＞±≡≈≌≒∏∑√∛∜∝ →∞ ⊕⊙ 〈v,w〉 ∈ ∀∂⊥∃∋∆∇∟∠∫∬∭∮∥○●◎ ∧∨∩∪∴∵∶∷⊂⊃⊄⊅⊆⊇⊿＋－＊／ §‰¼½¾ ⅓⅔⅕⅖⅗⅘⅙⅚⅛⅜⅝⅞⅟←↑→↓↔↕↖↗↘↙ ■□ ▢▣▤▥▦▧▨▩▪▫ × ÷ ° º¹²³ ΑΒΓΔΕΖΗΘΙΚΛΜΝΞΟΠΡΣΤΥΦΧΨΩ ΪΫάέήίΰ αβγδεζηθικλμνξοπρςστυφχψω 