/**/
function alert4() //9812060000
{
document.write(''
+'Output may contain error, Please verify first.
'
+'Program environment is MSIE 6.0, please use MSIE
'
+''
);
} //function alert4() 9812060002
//>
<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="ch12b001"> Index begin Index this file 201005200632 start ■ Define three corner E Newton's inequality is E_{k1}(x)*E_{k+1}(x)≦E_{k}^{2}(x) eqn.12.2 for 0＜k＜n Assume x=[a,b,c,d,e], n=5 E_{k} is averaged e_{k}. 1≦k＜5 E_{k}(x)=e_{k}(x)/bicof(5,k) eqn.BK001 e_{k} is symmetric sum of five numbers a,b,c,d,e in kth power. <a name="ch12b002"> If k=1, then e_{1}(x)=a+b+c+d+e eqn.BJ108 and E_{1}(x)=e_{1}(x)/bicof(5,1) eqn.BJ123 Here bicof(5,1)= 5!/[1!*(51)!] bicof(5,1)= 5 eqn.BK002 Binomial coefficients is defined at eqn.BJ128 <a name="ch12b003"> E_{k}^{2}(x) lower right corner is k E_{k}^{2}(x) upper right corner is p=2 In the following discussion, we will differentiate polynomial several times and reduce its order. Order of polynomial n must be visible. <a name="ch12b004"> Now use third corner, define three corner E as following _{n}E_{k}^{p}(x) eqn.BK003 is averaged symmetric sum E of n numbers x in kth power seat, whole term take pth power. Above is uppercase E. Same thing apply to lowercase e. <a name="ch12b005"> Index begin Index this file Example 1, if x=[a,b,c,d,e]=[1,2,3,4,5] eqn.BK004 then n=5. For second seat k=2, the square root value is _{n}E_{k}^{p}(x)=_{5}E_{2}^{1/2}(x) eqn.BK005 =[(a*b+a*c+b*c+a*d+b*d+c*d+a*e +b*e+c*e+d*e)/bicof(5,2)]^{1/2} =[(1*2+1*3+2*3+1*4+2*4 +3*4+1*5+2*5+3*5+4*5)/10]^{1/2} =2.9154759474226503 201005200720 here <a name="ch12b006"> Example 2, if y=[f,g,h,i]=[6,7,8,9] eqn.BK006 then n=4. For second seat k=2, the square value is _{n}E_{k}^{p}(y)=_{4}E_{2}^{2}(y) eqn.BK007 =[(+f*g+f*h+f*i+g*h+g*i+h*i)/bicof(4,2)]^{2} =[( 6*7+6*8+6*9+7*8+7*9+8*9)/6]^{2} =3117.361111111111 <a name="ch12b007"> y=[f,g,h,i] has n=4 information, we need n value show up, please see push down diagram. E's lower left corner n number help us to analysis. Following use eqn.BK003 expression for clear explanation. 201005200736 stop <a name="ch12b008"> Index begin Index this file 201005201034 start ■ Numerically test inequality To prove Newton's inequality, take two steps. <a name="ch12b009"> First, relate higher polynomial order _{n}E_{k}(x) to lower polynomial order _{n1}E_{k}(x) Second, prove any polynomial high power end Newton's inequality is true. _{n}E_{k}(x) = _{n1}E_{k}(x) = _{n2}E_{k}(x) please see Push down diagram. Move down a column polynomial reduce power n. _{z}E_{k}(x) unchange. <a name="ch12b010"> Numerically test inequality help us understand the problem. If we have x=[a,b,c,d,e]=[1,2,3,4,5] eqn.BK004 in hand then the polynomial which generate symmetric sums for this x is polynomial P5(t) P5(t)=(ta)*(tb)*(tc) *(td)*(te) eqn.BK008 <a name="ch12b011"> In expanded form it is P5(t)= eqn.BK009 (ta)*(tb)*(tc)*(td)*(te) =t*t*t*t*t t*t*t*t*(a+b+c+d+e) +t*t*t*(a*b+a*c+b*c+a*d+b*d +c*d+a*e+b*e+c*e+d*e) t*t*(a*b*c+a*b*d+a*c*d+b*c*d+a*b*e +a*c*e+b*c*e+a*d*e+b*d*e+c*d*e) +t*(a*b*c*d+a*b*c*e+a*b*d*e +a*c*d*e+b*c*d*e) a*b*c*d*e <a name="ch12b012"> In _{n}e_{k}(x) expression: (n=5) P5(t)= //remind: _{5}e_{0}(x)=1 =(ta)*(tb)*(tc)*(td)*(te) =t*t*t*t*t*_{5}e_{0}(x) t*t*t*t*_{5}e_{1}(x) +t*t*t*_{5}e_{2}(x) t*t*_{5}e_{3}(x) +t*_{5}e_{4}(x) _{5}e_{5}(x) eqn.BK010 <a name="ch12b013"> Index begin Index this file In _{n}E_{k}(x) expression: (n=5) P5(t)= eqn.BK011 =(ta)*(tb)*(tc)*(td)*(te) =t*t*t*t*t*_{5}E_{0}(x)*bicof(5,0) t*t*t*t*_{5}E_{1}(x)*bicof(5,1) +t*t*t*_{5}E_{2}(x)*bicof(5,2) t*t*_{5}E_{3}(x)*bicof(5,3) +t*_{5}E_{4}(x)*bicof(5,4) _{5}E_{5}(x)*bicof(5,5) <a name="ch12b014"> For x=[a,b,c,d,e]=[1,2,3,4,5] in partial numerical value A, P5(t)= eqn.BK012 =(ta)*(tb)*(tc)*(td)*(te) =t*t*t*t*t*_{5}E_{0}(x)*1 t*t*t*t*_{5}E_{1}(x)*5 +t*t*t*_{5}E_{2}(x)*10 t*t*_{5}E_{3}(x)*10 //number is +t*_{5}E_{4}(x)*5 //binomial _{5}E_{5}(x)*1 //coefficients <a name="ch12b015"> In partial numerical value B, P5(t)= eqn.BK013 =(ta)*(tb)*(tc)*(td)*(te) =t*t*t*t*t*1 //_{5}E_{0}(x)=1 t*t*t*t*3*5 //_{5}E_{1}(x)=3 +t*t*t*8.5*10 //_{5}E_{2}(x)=8.5 t*t*22.5*10 //_{5}E_{3}(x)=22.5 +t*54.8*5 //_{5}E_{4}(x)=54.8 120 //_{5}E_{5}(x)=120 Red mark worth your attention. <a name="ch12b016"> In complete numerical value, P5(t)= eqn.BK014 =(ta)*(tb)*(tc)*(td)*(te) =t*t*t*t*t t*t*t*t*15 +t*t*t*85 t*t*225 +t*274 120 201005201124 here <a name="ch12b017"> Index begin Index this file Newton's inequality best view is eqn.BK012. Newton said: For sequence [1,2,3,4,5], its symmetric sums has the following relations //dropped n=5 E_{k1}(x)*E_{k+1}(x)≦E_{k}^{2}(x) eqn.12.2 <a name="ch12b018"> Case n=5, complete set Newton's for k=1: E_{0}(x)*E_{2}(x)≦E_{1}^{2}(x) eqn.BK015 for k=2: E_{1}(x)*E_{3}(x)≦E_{2}^{2}(x) eqn.BK016 for k=3: E_{2}(x)*E_{4}(x)≦E_{3}^{2}(x) eqn.BK017 for k=4: E_{3}(x)*E_{5}(x)≦E_{4}^{2}(x) eqn.BK018 <a name="ch12b019"> eqn.BK013 has numerical value for x=[1,2,3,4,5], that is E_{0}(x)=1 eqn.BK019 E_{1}(x)=3 eqn.BK020 E_{2}(x)=8.5 eqn.BK021 E_{3}(x)=22.5 eqn.BK022 E_{4}(x)=54.8 eqn.BK023 E_{5}(x)=120 eqn.BK024 <a name="ch12b020"> Check Newton's inequality as following for k=1: 1*8.5 ≦ 3^{2} for k=2: 3*22.5≦ 8.5^{2} for k=3: 8.5*54.8≦22.5^{2} for k=4: 22.5*120 ≦54.8^{2} <a name="ch12b021"> or for k=1: 8.5 ≦ 9 for k=2: 67.5 ≦ 72.25 for k=3: 465.8 ≦ 506.25 for k=4: 2700.0 ≦3003.04 Newton's inequality is true for above specific case. 201005201153 here <a name="ch12b022"> Newton's inequality eqn.12.2 and Maclaurin's inequality eqn.12.3 are close related. It is easy to prove inequalities at low power end (n=5, k=0,1) and at high power end (n=5, k=4,5). It is hard to prove inequalities at middle power (n=5, k=2,3). <a name="ch12b023"> Can we change hard to prove middle power inequalities to easy to prove high power end inequality? In Newton's case there is an easy way !! <a name="ch12b024"> Index begin Index this file The key point is that if we start from P5(t) and generate P4(t), P3(t), P2(t), P1(t) which are related to P5(t) by differentiation, <a name="ch12b025"> then we have //Prove _{5}E_{k}(x)=_{4}E_{k}(x)=_{3}E_{k}(x) =_{2}E_{k}(x)=_{1}E_{k}(x) eqn.BK025 (_{u}E_{v} end at u=v; eqn.BK025 whole line enter front stage if k=1) this is an unexpected and surprised relation! It is true! <a name="ch12b026"> With this relation in hand, instead of prove _{5}E_{1}(x)*_{5}E_{3}(x)≦_{5}E_{2}^{2}(x) eqn.BK026 we prove _{3}E_{1}(x)*_{3}E_{3}(x)≦_{3}E_{2}^{2}(x) eqn.BK027 Lower right 1,3,2 is E's power. Lower left 5,3=polynomial order In Push down diagram eqn.BK026 is high order middle terms. <a name="ch12b027"> eqn.BK027 is low order end terms. [_{3}E_{3}(x) is end] eqn.BK025 promise that eqn.BK026 and eqn.BK027 have same value. We prove much easy low order end term inequality conclude that high order middle term inequality is true !! <a name="ch12b028"> LiuHH got above understanding on 20100519 after long reading and long thinking. Textbook say so, but no handson calculation it is hard to believe we have such nice/lucky eqn.BK025. 201005201230 stop <a name="ch12b029"> Index begin Index this file 201005201425 start The following illustrate eqn.BK025 from different angle. Start from eqn.BK011. Differentiate P5(t) get P4(t) Differentiate P4(t) get P3(t) etc.<"ch12b030"> Push down diagram P5 push down to P3

<a name="ch12b033"> 201005201531 here Above is symbolic equation, no numerical value tell us column value are the same. Below is numerical equation, let sequence be [a,b,c,d,e]=[1,2,3,4,5] eqn.BK029 Please goto Program Test Newton, Maclaurin inequalities. <a name="ch12b034"> Change from var a=1,b=2.3,c=3.1,d=4.6,e=6.8 var outType=0 //you can select 0,1 to 8 to var a=1,b=2,c=3,d=4,e=5 var outType=2 //change 0 to 2 Then click [Test box3 command, output to box4] <a name="ch12b035"> get the following [[ E1 //length compare with volume wrong 3 E2 //upper case E* removed n,k factor 8.5 E3 //upper case E* use average value 22.5 E4 //but still wrong, here is outType=2 54.8 E5 //select outType=0 for correct answer 120 ]] From above assumed value, we have the following numerical equations.<a name="ch12b036"> Index begin Index this file

<a name="ch12b037"> Now let us look at the red terms from P5(t)= t^{5}－5*3*t^{4}+10*8.5*t^{3}+... eqn.BK031 to P4(t)= t^{4}－4*3*t^{3}+6*8.5*t^{2}+... eqn.BK032 Study how they change? Why keep 8.5 unchanged? From P5(t) to P4(t) is a differentiation process P4(t) = d[P5(t)]/dt eqn.BK033 P4(t) = d[t^{5}－5*3*t^{4}+10*8.5*t^{3}+...]/dt eqn.BK034 = 5*t^{4}－4*5*3*t^{3}+3*10*8.5*t^{2}+... <a name="ch12b038"> Normalize 5*t^{4} to t^{4} P4(t)/5 = t^{4}－4*3*t^{3}+3*2*8.5*t^{2}+... = t^{4}－4*3*t^{3}+ 6*8.5*t^{2}+... (eqn.BK030 dropped '/5') Red multiply by 3, divide by 5 Red multiply by k, divide by n Multiply by k due to d[t^{k}]/dt=k*t^{k1} Divide by n due to normalize P4(t) <a name="ch12b039"> 8.5 (=_{5}E_{3}=_{4}E_{3}) is unchanged Red "+ 6*" is bicof(4,2), just right. All other _{n}E_{k} apply the same reason, then no change. n≧k OK, n＜k is undefined. when n＝k, this term end here. 201005201630 stop <a name="ch12b040"> 201005201850 start Above are observation, mainly numerical check. But numerical equation is not general, can not be considered as proof. We must use n,k symbol equation to prove, because we can plug any value in to n,k, then, it cover all cases. <a name="ch12b041"> Index begin Index this file ■ Prove E_{k}(x_{1},...,x_{n})=E_{k}(y_{1},...,y_{n1}) The following prove first step of total two step proof. That is if we differentiate a nth order polynomial to one order lower, two polynomial kth seat two E_{k} have same value, // eqn.12.6 _{n}E_{k}(x)=_{n1}E_{k}(x) eqn.BK035 See symbolic equation eqn.BK028 or numerical equation eqn.BK030 Numerical observation why unchange ? is here <a name="ch12b042"> Symbolic proof why unchange is lectured in textbook page 180 to 181. Consider polynomial P(t)=(tx_{1})(tx_{2})...(tx_{n})<a name="ch12b043">

eqn.12.5 
<a name="ch12b044"> 201005202023 here Use Q(t) as normalized derivative of P(t). 'normalize' mean that Q(t) highest power coefficient is one, not n.<a name="ch12b045">

<a name="ch12b046"> Index begin Index this file nk come from d[t^{nk}]/dt = nk*t^{nk1} eqn.BK037 n come from normalize Q(t) = {d[P(t)]/dt}/n eqn.BK038 <a name="ch12b047"> eqn.BK036 line one change to line two as following
width of above equation Red cancel result red; Blue cancel result blue. <a name="ch12b048"> Above is coefficient analysis. Not involve the root of Q(t). But E_{k} of Q(t) is defined on roots of Q(t). See Lowercase e_{k}(x) defined and Uppercase E_{k}(x) defined We take second, rootbased route to find E_{k} of Q(t). Problem given {x_{1},x_{2},...,x_{n}} as roots of P(t). <a name="ch12b049"> After differentiate P(t), get Q(t), Q(t) has a set of roots {y_{1},y_{2},...,y_{n1}} Which should be different from {x_{1},x_{2},...,x_{n}}, unless there are repeated root in {x_{1},x_{2},...,x_{n}}. Repeated root not change analysis. Build Q(t) from _{n1}E_{k}(y) get <a name="ch12b050"> Index begin Index this file Q(t)=P'(t)/n= eqn.BK040 =(ty_{1})(ty_{2})...(ty_{n1})
Why write polynomial this way? at eqn.12.5 width of above equation <a name="ch12b051"> eqn.BK041 and eqn.BK036 both express Q(t), they are the same equation from two different path Equate E_{k} from eqn.BK036 and E_{k} from eqn.BK041 get E_{k}(x_{1},x_{2},...,x_{n}) =E_{k}(y_{1},y_{2},...,y_{n1}) eqn.12.6 If use tute0043.htm express method, eqn.12.6 is next _{n}E_{k}(x)=_{n1}E_{k}(y) eqn.BK042 eqn.12.6 is our main result. 201005202138 stop <a name="ch12b052"> Index begin Index this file 201005202319 start ■ Why is it so remarkable? What benefit we get from eqn.12.6? Textbook page 181 explain as following. ===textbook copy start=== Why is it so remarkable? The lefthand side of the identity (12.6) is a function of the n vector x=(x_{1},x_{2},...,x_{n}) while the right side is a function of the n1 vector y=(y_{1},y_{2},...,y_{n1}) <a name="ch12b053"> Thus if we can prove a relation such as 0≦F(E_{0}(y),E_{1}(y),...,E_{n1}(y)) eqn.BK043 for all y∈[a,b]^{n1}. Then it follows that we also have the relation 0≦F(E_{0}(x),E_{1}(x),...,E_{n1}(x)) eqn.BK044 for all x∈[a,b]^{n}. <a name="ch12b054"> That is, any inequality  or identity  which provides a relation between the n1 quantities E_{0}(y),E_{1}(y),...,E_{n1}(y) and which is valid for all values of y∈[a,b]^{n1} extends automatically to a corresponding relation for the n1 quantities E_{0}(x),E_{1}(x),...,E_{n1}(x) which is valid for all x∈[a,b]^{n}. <a name="ch12b055"> This presents a rare but valuable situation where to prove a relation for function of just n1 variables. This observation can be used in an ad hoc way to produce many special identities which <a name="ch12b056"> otherwise would be completely baffling, and it can also be used systematically to provide seamless induction proofs for results such as Newton's inequalities. ===textbook copy stop=== 201005202339 stop <a name="ch12b057"> Index begin Index this file 201005211050 start We had numerical example x=[a,b,c,d,e]=[1,2,3,4,5] eqn.BK004 P5(t)= eqn.BK014 =(ta)*(tb)*(tc)*(td)*(te) =t*t*t*t*t t*t*t*t*15 +t*t*t*85 t*t*225 +t*274 120 <a name="ch12b058"> That is P5(t)=t^{5}－15*t^{4}+85*t^{3}－225*t^{2}+274*t－120 eqn.BK045 P4(t)=t^{4}－12*t^{3}+51*t^{2}－90*t+54.8 eqn.BK046 P4(t) and P5(t) are related by P4(t)=d[P5(t)]/dt eqn.BK047 <a name="ch12b059"> From one view point that P4(t) is created from P5(t), no need go a redundant loop find P4(t) roots y and reassemble P4(t) from y. <a name="ch12b060"> From second view point that _{n1}E_{k}(y) should be generated by P4(t) roots, just like _{n}E_{k}(x) are generated by P5(t) roots x. Textbook equate two E's E_{k}(x_{1},x_{2},...,x_{n}) =E_{k}(y_{1},y_{2},...,y_{n1}) eqn.12.6 <a name="ch12b061"> or in tute0043.htm notation _{n}E_{k}(x)=_{n1}E_{k}(y) eqn.BK042 where _{n1}E_{k}(y) is created by 'redundant' step, that is _{n1}E_{k}(y) is created by P4(t) roots y. <a name="ch12b062"> Index begin Index this file The following verify that P4(t) roots y create _{n1}E_{k}(y) Starting point is P4(t)=t^{4}－12*t^{3}+51*t^{2}－90*t+54.8 eqn.BK046 end point is still eqn.BK046. <a name="ch12b063"> Please goto (local) http://freeman2.com/polyroot.htm#begin0 Click [04] button for 4th order polynomial. Then fill in the following coefficients 54.8 －90 51 －12 1 where 54.8 goto box marked [C00] and in [C04] fill 1. <a name="ch12b064"> Click [fast get roots] button Box 1, answer has [[ Below is answer, complex polynomial root ([1,2] same as 1+2i) 1st root is [1.355567131841731,0] 2nd root is [2.456088843227607,0] 3rd root is [3.543910070325765,0] 4th root is [4.644433954604896,0] Below is answer too, complex number only, no text. 1.355567131841731+0i 2.456088843227607+0i 3.543910070325765+0i 4.644433954604896+0i ]] <a name="ch12b065"> P4(t) roots y has 1.355567131841731 2.456088843227607 3.543910070325765 4.644433954604896 <a name="ch12b066"> Index begin Index this file Next goto(local) http://freeman2.com/complex2.htm#calculator in [Box3, input JS command] fill [[ f=1.355567131841731 g=2.456088843227607 h=3.543910070325765 i=4.644433954604896 e1=f+g+h+i E1=e1/bicof(4,1) e2=f*g+f*h+f*i+g*h+g*i+h*i E2=e2/bicof(4,2) e3=f*g*h+f*g*i+f*h*i+g*h*i E3=e3/bicof(4,3) e4=f*g*h*i E4=e4/bicof(4,4) E1 E2 E3 E4 ]] <a name="ch12b067"> Next click [Test box3 command, output to box4] Box 4 show up next [[ E1 3 E2 8.5 E3 22.500000654128463 E4 54.80000354686018 ]] <a name="ch12b068"> Above is E1,E2,E3,E4 from P4(t) roots y //P4=4th order polynomial Please goto P5(t) E's compare E1,E2,E3,E4 [P4 no E5] Result show that E1,E2,E3,E4 from P4 and from P5 are the same !! <a name="ch12b069"> They should be identical, because we walked a loop. Above numerical experiments confirmed textbook eqn.12.6 201005211146 stop <a name="ch12b070"> Index begin Index this file 201005211520 start ■ Prove high power end Newton Inequality We have eqn.12.6 in hand, for example, a data set of ten numbers x=[x_{1},x_{2},...,x_{10}] eqn.BK048 and ask to prove _{10}E_{3}(x)*_{10}E_{5}(x)≦_{10}E_{4}^{2}(x) eqn.BK049 is true. <a name="ch12b071"> eqn.12.6 tell us that use x to build a tenth order polynomial p10(t) use x as its roots. Differentiate p10(t) to fifth order polynomial p5(t). Then prove next inequality eqn.BK050 <a name="ch12b072"> _{5}E_{3}(x)* _{5}E_{5}(x)≦ _{5}E_{4}^{2}(x) eqn.BK050 If eqn.BK050 is true, then eqn.BK049 is true by eqn.12.6 Key point is red 5 _{10}E_{5} is 10th order polynomial middle term, hard to prove. _{5}E_{5} is 5th order polynomial end term, easy to prove. <a name="ch12b073"> eqn.BK050 is p5(t)'s high power end inequality. This point can be seen from the term _{5}E_{5}(x) for _{u}E_{v}(x), we must have u≧v. If u=v, that is the end, and no more. Because _{v}E_{v}(x) multiply by t^{zero}, _{v}E_{v}(x) is a constant. Next differentiation _{v1}E_{v}(x) is gone. _{2}E_{3} do NOT exist. <a name="ch12b074"> For above reason, we only need to prove high power end Newton Inequality. Newton's Inequality general form is next E_{j1}(x_{1},x_{2},...,x_{n})*E_{j+1}(x_{1},x_{2},...,x_{n})≦ E_{j}^{2}(x_{1},x_{2},...,x_{n}) eqn.12.7 201005211552 here <a name="ch12b075"> If we have one number in hand, that is x=[x_{1}] eqn.BK051 There is nothing to do. <a name="ch12b076"> Index begin Index this file ■ Prove Newton Inequality n=2 Textbook page 182 If we have two data in hand, that is x=[x_{1},x_{2}] eqn.BK052 Use x_{1},x_{2} as two roots, build a polynomial P2(t). P2(t)=(tx_{1})*(tx_{2}) eqn.BK053 =t*tt*(x_{1}+x_{2})+x_{1}*x_{2} =t*t2*t*[(x_{1}+x_{2})/2]+x_{1}*x_{2} <a name="ch12b077"> On the other hand P2(t)=t^{2}*_{2}E_{0}(x_{1},x_{2}) 2*t^{1}*_{2}E_{1}(x_{1},x_{2}) +1*t^{0}*_{2}E_{2}(x_{1},x_{2}) eqn.BK054 <a name="ch12b078"> For the creation of eqn.BK054 please read start from Chapter 12: Symmetric Sums to at least Uppercase E_{k}(x) defined. eqn.BK053 and eqn.BK054 are both expression of P2(t). <a name="ch12b079"> We have _{2}E_{0}(x_{1},x_{2})=1 eqn.BK055 _{2}E_{1}(x_{1},x_{2})=[(x_{1}+x_{2})/2] eqn.BK056 _{2}E_{2}(x_{1},x_{2})=x_{1}*x_{2} eqn.BK057 <a name="ch12b080"> For n=2 simplest case, Newton's inequality is E_{0}(x_{1},x_{2})* E_{2}(x_{1},x_{2}) ≦ E_{1}^{2}(x_{1},x_{2}) eqn.12.8 Above is textbook equation. Below is tute0043.htm equation. _{2}E_{0}(x)* _{2}E_{2}(x)≦ _{2}E_{1}^{2}(x) eqn.BK058 Put eqn.BK055, eqn.BK056 and eqn.BK057 into eqn.BK058, get 1*[x_{1}*x_{2}]?≦?[(x_{1}+x_{2})/2]^{2} eqn.BK059 <a name="ch12b081"> We assume that x_{1}＞0 and x_{2}＞0 so that x_{1}*x_{2}＞0. Now take square root of eqn.BK059 [x_{1}*x_{2}]^{1/2}?≦?(x_{1}+x_{2})/2 eqn.BK060 eqn.BK060 is certainly true. <a name="ch12b082"> [x_{1}*x_{2}]^{1/2} is Geometric Mean and (x_{1}+x_{2})/2 is Arithmetic Mean GM≦AM is a matter of course. (one year ago, LiuHH is inequality stranger. Now LiuHH is sophomore) 201005211631 here Above is n=2 case. <a name="ch12b083"> Index begin Index this file ■ Prove Newton Inequality n=3 Textbook page 182 Below is n=3 case. We have three data in hand. x=[x_{1},x_{2},x_{3}] eqn.BK061 We use x build third degree polynomial P3(t). To save work express P3(t) with _{3}E_{k}(x) directly <a name="ch12b084"> P3(t)=1*t^{3}*_{3}E_{0}(x_{1},x_{2},x_{3}) 3*t^{2}*_{3}E_{1}(x_{1},x_{2},x_{3}) +3*t^{1}*_{3}E_{2}(x_{1},x_{2},x_{3}) 1*t^{0}*_{3}E_{3}(x_{1},x_{2},x_{3}) eqn.BK062 <a name="ch12b085"> eqn.BK062 coefficient are [1,3,+3,1]. Neglect +/ [1,+3,+3,+1] is bicof(3,0)=1 bicof(3,1)=3 bicof(3,2)=3 bicof(3,3)=1 <a name="ch12b086"> Why eqn.BK062 involve binomial coefficients? Please see from Lowercase e_{k}(x) defined. to Uppercase E_{k}(x) defined. <a name="ch12b087"> P3(t) eqn.BK062 contain two Newton's inequalities. They are E_{0}(x_{1},x_{2},x_{3})* E_{2}(x_{1},x_{2},x_{3})≦ E_{1}^{2}(x_{1},x_{2},x_{3}) eqn.12.9 E_{1}(x_{1},x_{2},x_{3})* E_{3}(x_{1},x_{2},x_{3})≦ E_{2}^{2}(x_{1},x_{2},x_{3}) eqn.12.10 Above is textbook equation. Below is tute0043.htm equation. _{3}E_{0}(x)* _{3}E_{2}(x)≦ _{3}E_{1}^{2}(x) eqn.BK063 _{3}E_{1}(x)* _{3}E_{3}(x)≦ _{3}E_{2}^{2}(x) eqn.BK064 <a name="ch12b088"> Index begin Index this file for n=3 case, eqn.BK063 do not have _{3}E_{3}(x). We can push eqn.BK063 to lower order polynomial. Compare eqn.BK063 with eqn.BK058 below _{3}E_{0}(x)* _{3}E_{2}(x)≦ _{3}E_{1}^{2}(x) eqn.BK063 _{2}E_{0}(x)* _{2}E_{2}(x)≦ _{2}E_{1}^{2}(x) eqn.BK058 <a name="ch12b089"> Lower right corner 0,2,1 are the same in both equation. Lower left corner 3,2 tell us that job of P3(t) hand over to P2(t). Base on eqn.12.6 and already proved n=2 case eqn.BK058, we conclude that eqn.BK063 is true. <a name="ch12b090"> eqn.BK064 is different, because it has the term _{3}E_{3}(x). We can not push eqn.BK064 to lower order polynomial. Because _{v1}E_{v}(x) is undefined. _{v1}E_{v}(x) is nonexist. <a name="ch12b091"> Lower left corner subnote 'v1' means polynomial order is v1. Lower right corner subnote 'v' means Symmetric Sum has order v. We have v1 data in hand, we can not create vth multiplication. Same argument in tute0042.htm Five numbers can not have power six. <a name="ch12b092"> We can not push eqn.BK064 to lower order polynomial. We must prove it right here n=3. This part is done in textbook page 183 <a name="ch12b093"> Index begin Index this file We need to find the definitions _{3}E_{1}(x)=(x_{1}+x_{2}+x_{3})/3 eqn.BK065 _{3}E_{2}(x)=(x_{1}*x_{2}+x_{1}*x_{3}+x_{2}*x_{3})/3 eqn.BK066 _{3}E_{3}(x)=x_{1}*x_{2}*x_{3} eqn.BK067 Above three equations are Symmetric Sum of x_{1},x_{2},x_{3} three data set. <a name="ch12b094"> eqn.BK065 and eqn.BK066 divide by three are taking average. Please see one to one Put eqn.BK065 to eqn.BK067 into eqn.BK064. The equation to prove is next<a name="ch12b095">
width of above equation <a name="ch12b096"> To prove high power end Newton Inequality, it is always easier to prove if use the reciprocal of x. In this case we divide whole equation by the nth multiplied term _{n}E_{n}(x)=(x_{1}*x_{2}*...*x_{n}) eqn.BK301 Inserted equation, use number 300+ <a name="ch12b097"> For this reason (divide by x), we require that the observed data x_{1},x_{2},...,x_{n} all be positive. Zero is not allowed. <a name="ch12b098"> Index begin Index this file (x_{1}*x_{2}*x_{3})^{2} is expressed as _{3}E_{3}^{2}(x) Divide eqn.12.11 by (x_{1}*x_{2}*x_{3})^{2} get
width <a name="ch12b099"> Expand eqn.BK068 get
width <a name="ch12b100"> Cancel like term get
width of above equation <a name="ch12b101"> Whether eqn.BK070 is true? From rearrangement inequality view point, eqn.BK070 is true. Because we can assume the order 0＜x_{1}≦x_{2}≦x_{3} eqn.BK071 then reciprocal give us 1/x_{1}≧1/x_{2}≧1/x_{3} eqn.BK072 <a name="ch12b102"> Consider sequence_A=[1/x_{1}, 1/x_{2}, 1/x_{3}] sequence_B=[1/x_{2}, 1/x_{3}, 1/x_{1}] Rearrangement inequality say sequence_A dot product sequence_A is greater than or equal to sequence_A dot product sequence_B <a name="ch12b103"> Index begin Index this file Whether eqn.BK070 is true? From Cauchy's inequality view point, eqn.BK070 is true. Because we can assume the sequences sequence_C=[1/x_{1}, 1/x_{3}, 1/x_{2}] sequence_D=[1/x_{2}, 1/x_{1}, 1/x_{3}] <a name="ch12b104"> sequence_C dot product sequence_D get eqn.BK070 less than side sequence_C dot product sequence_C get square root of eqn.BK070 greater than side. sequence_D dot product sequence_D get square root of eqn.BK070 greater than side. Net result is eqn.BK070. <a name="ch12b105"> Whether eqn.BK070 is true? From AMGM inequality view point, eqn.BK070 is true. Because we can do three times AMGM inequality, then sum them to eqn.BK070. <a name="ch12b106"> That is 1/√(x_{1}x_{1}x_{2}x_{2})≦[1/(x_{1}x_{1})+1/(x_{2}x_{2})]/2 eqn.BK073 1/√(x_{1}x_{1}x_{3}x_{3})≦[1/(x_{1}x_{1})+1/(x_{3}x_{3})]/2 eqn.BK074 1/√(x_{2}x_{2}x_{3}x_{3})≦[1/(x_{2}x_{2})+1/(x_{3}x_{3})]/2 eqn.BK075 Sum eqn.BK073 to eqn.BK075 get eqn.BK070. <a name="ch12b107"> Conclude: For n=3 case, Newton Inequality eqn.12.10 or eqn.12.11 or eqn.BK064 or eqn.BK070 is true. 201005211859 stop <a name="ch12b108"> Index begin Index this file 201005212230 start ■ Prove Newton Inequality n=any Textbook page 183 Newton's inequality in general case, E_{k1}(x)*E_{k+1}(x)≦E_{k}^{2}(x) eqn.12.2 for 0＜k＜n <a name="ch12b109"> x contain hidden condition that is x has n elements, polynomial has degree=n. Equation contain E_{k+1}(x) which can not exceed E_{n}(x) k+1≦n, then k＜n . Equation contain E_{k1}(x) which can not be smaller than E_{0}(x), 0≦k1 then 0＜k . <a name="ch12b110"> If k+1＜n, we have E_{k1}(x)*E_{k+1}(x)≦E_{k}^{2}(x) eqn.12.12 for 1≦k＜n1 eqn.12.12 condition 1≦k＜n1 eqn.12.2 condition 0＜k＜n The inequality in the group k+1＜n can be push down to lower order polynomial's high end inequality. Done proof in low order polynomial. <a name="ch12b111"> If k+1=n, E_{k+1}(x)=E_{n}(x), we have E_{n2}(x)*E_{n}(x)≦E_{n1}^{2}(x) eqn.12.13 this is a boundary inequality. We can not push E_{n}(x) down to lower order polynomial. Because there is no more same E downward. Must prove eqn.12.13 right here. <a name="ch12b112"> Index begin Index this file We write eqn.12.13 in longhand and use x^{^}_{j} remind us that this x_{j} is a missing term. Why missing? The only one not missing any is _{n}E_{n}(x). All elements multiply together. The next one _{n}E_{n1}(x) has one missing in symmetric sum. <a name="ch12b113"> Example, assume x=[a,b,c,d,e] _{5}E_{5}(x)=a*b*c*d*e no one missing. _{5}E_{4}(x)=b*c*d*e+a*c*d*e +a*b*d*e+a*b*c*e +a*b*c*d In which "b*c*d*e" missing an 'a' "a*c*d*e" missing a 'b' etc. <a name="ch12b114"> If we write "b*c*d*e" as "a^{^}*b*c*d*e", a^{^}=pure number 1 at the step whole equation divide by "a*b*c*d*e" , the missing one "x^{^}_{j}" show up at denominator. Use "a^{^}" or use "x^{^}_{j}" help us do bookkeeping right. 201005212322 stop <a name="ch12b115"> 201005220926 start We begin prove eqn.12.13 which is the high power end Newton's inequality. eqn.12.13 use the following symmetric sums<a name="ch12b116"> E_{n}(x)=x_{1}*x_{2}*...*x_{n} eqn.BK076
'－1' in 'n－1' exclude one missing element "x^{^}_{j}". '－2' in 'n－2' exclude two missing "x^{^}_{j}" and "x^{^}_{k}". width <a name="ch12b117"> Binomial coefficient is defined here Substitute E_{n}(x), E_{n1}(x), E_{n2}(x) into eqn.12.13 get
highest order because x_{1}*x_{2}*...*x_{n} no missing element width of above equation <a name="ch12b118"> 201005221028 here For n elements sequence, we build nth order polynomial. There are n1 Newton's inequality. They are <a name="ch12b119"> Index begin Index this file _{n}E_{n2}(x)*_{n}E_{n}(x)≦_{n}E_{n1}^{2}(x) eqn.12.13 _{n}E_{n3}(x)*_{n}E_{n1}(x)≦_{n}E_{n2}^{2}(x) eqn.BK079 _{n}E_{n4}(x)*_{n}E_{n2}(x)≦_{n}E_{n3}^{2}(x) eqn.BK080 ..... //Red are boundary terms <a name="ch12b120"> _{n}E_{2}(x)* _{n}E_{4}(x)≦ _{n}E_{3}^{2}(x) eqn.BK081 _{n}E_{1}(x)* _{n}E_{3}(x)≦ _{n}E_{2}^{2}(x) eqn.BK082 _{n}E_{0}(x)* _{n}E_{2}(x)≦ _{n}E_{1}^{2}(x) eqn.BK083 <a name="ch12b121"> Boundary Newton's inequality eqn.12.13 high power end and eqn.BK083 low power end are easy to prove. Because there are less terms involved. bicof(10,10)=1, bicof(10,9)=10; bicof(10,1)=10, bicof(10,0)=1. <a name="ch12b122"> Middle Newton's inequality has many terms and hard to sort out a form for inequality argument. Please see a hard work example at here and get trouble at here. bicof(10,5)=252, bicof(10,4)=210; after square 252*252=63504 terms <a name="ch12b123"> Index begin Index this file Now we prove boundary Newton's inequality eqn.12.14, it is much easier than the middle Newton's. 201005221109 here <a name="ch12b124"> We assume all data x_{1},x_{2},...,x_{n} are positive numbers, no zero. We divide eqn.12.14 by square of x_{1}*x_{2}*...*x_{n} convert high power (nth power x_{1}*x_{2}*...*x_{n}) to reciprocal of low power 1/x_{1} or 1/(x_{2}*x_{3}) etc. This division let x^{^}_{j} change from invisible one to visible 1/x_{j}. After division we get<a name="ch12b125">
x hat as ONE, next step x^{^}_{j} show up at denominator. width <a name="ch12b126">
Reciprocal of highest order Newton's inequality width of above equation <a name="ch12b127"> 201005221143 here How to prove eqn.12.15? Textbook page 183 explain as following. ===textbook copy start=== <a name="ch12b128"> Index begin Index this file We could now stick with the pattern that worked for H_{3}, but there is a more graceful way to finish which is almost staring us in the face. <a name="ch12b129"> If we adopt the language of symmetric function, the target bound (12.15) may be written more systematically as <a name="ch12b130">E_{0}(1/x_{1},1/x_{2},...,1/x_{n}) //E_{0} defined to be 1 *E_{2}(1/x_{1},1/x_{2},...,1/x_{n}) //E_{2}=∑[1/(x_{i}x_{j})]/bicof(n,2) ≦E_{1}^{2}(1/x_{1},1/x_{2},...,1/x_{n}) //E_{1}=∑[1/x_{k}]/bicof(n,1) Above is eqn.BK085 <a name="ch12b131"> and one now sees that this inequality is covered by the first bound of the group (12.12). Thus the proof of Newton's inequality is complete. ===textbook copy stop=== 201005221154 here <a name="ch12b132"> Index begin Index this file First time read here, LiuHH was very puzzle. Why?! Newton's inequality is designed for x. Now the data become y=[1/x_{1},1/x_{2},...,1/x_{n}] eqn.BK086 How can I use x equation to prove y property? <a name="ch12b133"> After more think, find out why. Newton's inequality is designed for x. That is very true. x and y are different, that is also true. <a name="ch12b134"> We need to see what limitation applied to data x ? Review from the beginning, Newton's inequality require x has n elements, otherwise it is another set Newton. <a name="ch12b135"> Require x has all positive elements, so that division of x_{1}*x_{2}*...*x_{n} is possible. That is all. <a name="ch12b136"> Then y satisfy these two conditions too. So x's Newton inequality apply to y perfect. However, <a name="ch12b137"> we must differentiate nth order polynomial Pn(t) all the way down to P2(t). ['2' in E_{2}(y) demand P2(t)] This is not a problem, as long as the proof is DONE. Problem 12.1 is solved. 201005221213 stop <a name="ch12b138"> Index begin Index this file 201005221601 start ■ Prove Maclaurin Inequality Maclaurin's inequality asserts that [E_{n}(x)]^{1/n} ≦ [E_{n1}(x)]^{1/(n1)} ≦ ... ≦ [E_{2}(x)]^{1/2} ≦ E_{1}(x) eqn.12.3 <a name="ch12b139"> Here we still use eqn.12.6 With the property _{n}E_{k}(x)=_{n1}E_{k}(y) eqn.BK042 where 1≦k＜n we only prove high power end Maclaurin Inequality. <a name="ch12b140"> Middle power Maclaurin Ineq. go to lower order polynomial's high power end, like Newton did. Please see Push down diagram Newton use 3 E, Maclaurin use 2. _{u}E_{v}(x) u=v indicate reach end. Diagram colored for Newton but same reason apply to Maclaurin. <a name="ch12b141"> Let us exam high power end Maclaurin Inequality [E_{n}(x)]^{1/n} ?≦? [E_{n1}(x)]^{1/(n1)} eqn.BK087 The requirement to x is that x has n elements and all of x elements are positive. If we normalize x such that Geometric Mean of x is one x_{1}*x_{2}*...*x_{n}=1 eqn.BK088 <a name="ch12b142"> Index begin Index this file Normalize x still satisfy two requirements. eqn.BK087 raise power to (n1) eqn.BK087 become 1 ?≦? E_{n1}(x) eqn.BK089 Divide eqn.BK089 by x_{1}*x_{2}*...*x_{n}=1, we have<a name="ch12b143">
width <a name="ch12b144"> which reduce to
width of above equation <a name="ch12b145"> eqn.BK091 is AMGM inequality. for the sequence y=[1/x_{1},1/x_{2},...,1/x_{n}] eqn.BK086 which is true. Proof of Maclaurin Inequality is done. 201005221644 stop <a name="ch12b146"> Index begin Index this file 201005221707 start ■ 'high power end'? why it is t^{0}? The term 'high power end' is used frequently in this page. Sometime it may be confuse. Please see polynomial in terms of E's. In one equation, each E_{u}*t^{v} must have u+v=constant. Please see related topic. <a name="ch12b147"> 'high power end' for E, u=n is ' low power end' for t, v=0 . ' low power end' for E, u=0 is 'high power end' for t, v=n . <a name="ch12b148"> In this page, the term 'high power end' is reserved for E, not for t. If you see notes say 'high power end', look for E, do not look at t! <a name="ch12b149"> Polynomial P6(t) differentiate once become P5(t) E_{u}*t^{v} u+v=constant 6. change to E_{u}*t^{w} u+w=constant 5. t^{v} reduce power by one. E_{u} is constant (no t), E_{u} is unchanged. 201005221725 stop <a name="ch12b150"> Index begin Index this file 201005221830 start ■ Equality in Newton and Maclaurin Newton's inequality eqn.12.2. Maclaurin's inequality eqn.12.3 equality condition is x_{1}=x_{2}=...=x_{n} eqn.BK092 <a name="ch12b151"> For n=3 case, Prove Newton's Inequality, used three methods From rearrangement inequality From Cauchy's inequality From AMGM inequality. <a name="ch12b152"> We explain equality condition from three methods too. The basic point is that we are given one data set x=[x_{1},x_{2},...,x_{n}] eqn.BK093 we can not use x_{n+1}. <a name="ch12b153"> Use numerical example for easy illustration. Assume given seq_A1=[1,2,3,4,5] eqn.BK094 <a name="ch12b154"> Index begin Index this file Rearrangement inequality need just one data set. Second seq. is rearranged from first. Define seq_B1=[3,2,4,1,5] eqn.BK095 <a name="ch12b155"> Rearrangement inequality get seq_A1*seq_B1 ≦ seq_A1*seq_A1 eqn.BK096 1*3+2*2+3*4+4*1+5*5 //48≦ ≦1*1+2*2+3*3+4*4+5*5 //55 How can we do rearrangement still get equality? The only answer is that given seq_A2=[2,2,2,2,2] eqn.BK097 All elements are the same. Rearrange please. Any seq_B2 is still [2,2,2,2,2] <a name="ch12b156"> The condition x_{1}=x_{2}=...=x_{n} eqn.BK092 let us get equality in Newton's inequality eqn.12.2. and Maclaurin's inequality eqn.12.3 <a name="ch12b157"> Next see Cauchy's inequality Cauchy require two sequences. If we have seq_A1=[1,2,3,4,5] eqn.BK094 seq_C1=[2,4,6,8,10] eqn.BK098 Two sequence are proportional. Cauchy will give us equality! <a name="ch12b158"> But, we discuss Symmetric Sums '6,8,10' are not given ! seq_C1 is forbidden. Then create second sequence from given sequence is same as rearrangement. Same result eqn.BK092 let Cauchy give us equality. <a name="ch12b159"> Index begin Index this file Third, see AMGM inequality. We applied three times AMGM inequality. From eqn.BK073 to eqn.BK075. Each one equality condition is x_{i}=x_{j} eqn.BK099 Link one by one, all x_{k} are equal. x_{1}=x_{2},x_{2}=x_{3},x_{3}=x_{4} etc. Again we get eqn.BK092 Please see Equality GM=AM is easy 201005221912 stop 201005231820 done first proofread 201005232123 done second proofread 201005232146 done spelling check 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 tute0043.htm means TUTor, English, 43rd .htm Chinese series file name is tutc0001.htm This page, Inequality file thirty seven. http://freeman2.com/tute0043.htm First Upload 20100523 (Inequality start from tute0007.htm) Thank you for visiting Freeman's page. Freeman 201005232147 ≦ ≠ ≧ ＜＝＞±≡≈≌≒∏∑√∛∜∝ →∞ ⊕⊙ 〈v,w〉 ∈ ∀∂⊥∃∋∆∇∟∠∫∬∭∮∥○●◎ ∧∨∩∪∴∵∶∷⊂⊃⊄⊅⊆⊇⊿＋－＊／ §‰¼½¾ ⅓⅔⅕⅖⅗⅘⅙⅚⅛⅜⅝⅞⅟←↑→↓↔↕↖↗↘↙ ■□ ▢▣▤▥▦▧▨▩▪▫ × ÷ ° ◦º¹²³ ⇒ ⇓ ⇔ ΑΒΓΔΕΖΗΘΙΚΛΜΝΞΟΠΡΣΤΥΦΧΨΩ ΪΫάέήίΰ αβγδεζηθικλμνξοπρςστυφχψω 