Polynomial Equations and Tangents
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1 Polyomial Equatios ad Tagets Jim lowers presetatio to the Mathematical ssociatio of merica MD-DC-V Sectio Meetig 07 pril 9 9:5 am
2 rilliat.org Puzzle Problem appeared i a Facebook post this past witer What is the sum of x, y, ad z? It is arcta( arcta( arcta( ut how do you evaluate that?
3 Example What is ta(arcta(cos( / 9 arcta(cos(8 / 9 arcta(cos(4 / 9?
4 Example What is ta(arcta ta(arcta(cos( / 9 arcta(cos(8 / 9 arcta(cos(4 / 9 (cos( / 9 arcta(cos(8 / 9 arcta(cos(4 / 9 4?
5 Example College Mathematical Joural, 04 November, p. 57
6 Tagets ta(arcta( =, ta(arcta(=, ta(arcta(= This suggests we work with sums of tagets. ta( ta( ta(
7 Tagets ta(arcta( =, ta(arcta(=, ta(arcta(= This suggests we work with sums of tagets. ta( ta( ta( ta( ta(
8 Sum of Tagets High school trigoometry, but we wat ta(++c ta( ta( ta( ta( ta( cos( cos( si( si( cos( cos( cos( cos( cos( cos( si( cos( cos( cos( cos( si( ta( si( si( cos( cos( si( cos( cos( si( cos( si( ta( si( si( cos( cos( cos( si( cos( cos( si( si(
9 Tagets of sums Note that the taget formula is expressed oly with tagets, ot with other trig fuctios This allows us to evaluate ta(++c as ta((++c ad use the taget formula twice.
10 Taget of sum of three agles ta( C ta(( C ta( ta( C ta( ta( C ta( ta( ta( C ta( ta( ta( ta( ta( C ta( ta( ta( ta( ( ta( ta( ta( C ta( ta( (ta( ta( ta( C ta( ta( ta( C ta( ta( ta( C ta( C ta( ta( ta( ta( C ta( ta( C
11 Symmetric Polyomials ta( C ta( ta( ta( C ta( ta( ta( C ta( ta( ta( ta( C ta( ta( C The terms i the formula are symmetric i ta(, ta( ad ta(c They are symmetric polyomials, as are coefficiets of polyomials This suggests fidig a polyomial equatio which has ta(, ta( ad ta(c as roots; let r, r, ad r be ta(, ta( ad ta(c respectively r r ta( C ( r r a c a c b b x ax bx c 0 r r r r r r r r
12 Taget of four agles ta( ta( ta( C ta( D ta( ta( ta( C ta( ta( ta( D ta( ta( C ta( D ta( ta( C ta( D ta( C D (ta( ta( ta( ta( C ta( ta( D ta( ta( C ta( ta( D ta( C ta( D ta( ta( ta( C ta( D a a a a x 4 a x 4 a x a x a 4 0 Note the mius sig i i frot of Lie. That s because a k, where k is odd, is the egative of the sum of products i the term. This seems to cofirm the patter. That suggests a theorem, but first of all some ew otatio.
13 Some otatio Defie Where the sum is take over all j-subsets of roots. If the umber of roots is less tha j, this is defied to be 0. Example Where a, b, are the roots of a polyomial equatio b c a c c b a b c a b a ab b a 0 [] j j k k k j r r r ] [
14 racket ad race Notatio For example, ( a b c a b c ( ab ac bc [] [] [] The coefficiets of a polyomial equatio are (- k [ ] for k s, where the strig of k s correspods to a k. To avoid writig strigs of s, make aother defiitio { } [...] s If the umber of roots <, the {}=0.
15 racket ad brace otatio The for ay polyomial f(x, f ( x x ax a x {} x ( { } x ( { } This is much cleaer tha complicated summatio symbols or log strigs of moomials. Note this is idepedet of the particular root values or eve of the umber of roots (degree of equatio. j Note that for all j, a j ( { j} d also that I umber coefficiets upward istead of dowward i the covetioal form.
16 racket ad race Notatio Whe it is ecessary to iclude the umber of roots, write [ { } p where p is the umber of roots. This is zero if p<k ad p<, respectively. k ] p
17 Polyomial-taget theorem Suppose that the roots of a equatio are The ta( r, ta( r x a x ax a 0 ta( it(( / i ( a i i0 {} {} {5} a a a5 ri it( / i i {} {4} a a4 ( ai i0
18 Lemma j First ote that a j ( { j} That proves the last equality of the theorem. Lemma: Let s i =ta(r i for all i=,,. The { p} { } { p } s s { } { p} Proof. Does a term i {p} + cotai s + or ot? The terms that do have s + i it have p- variables i it alog with s +. This gives {p-} s +. The terms that do ot have s are precisely the terms i {p}. ddig these together gives lie of the lemma. For the secod lie, {} is simply the product of the s s up to. Multiplyig this by s + gives the product up to +.
19 Proof of theorem y iductio. If =, the we get s = s, or ta(r =ta(r, which is true. Suppose theorem true up to. The we apply the lemma termwise (icluded subscripts ta( i {} {} {5} s {} {4} {} {} {5} s {} {4} ({} r ta( {} {} {} i s ({} {} s ({5} {4} s ( ({} {} s ({4} {} s i {5} {4} r Result is the formula for +. r i {} {} {5} ( {} {} {4} ({} {} {4} {5} s s
20 The origial problem Evaluate Evaluate first the taget of this i terms of the tagets of the origial terms Fid equatio that has,, as roots pply formula arcta( arcta( arcta( ta(arcta( {} {} {} ; ta(arcta( ; ta(arcta( a a a ( x ( x ( x 0 x 6 0 ta(x+y+z=0, so that x+y+z = 0 + for some. The idividual agles are all betwee /4 ad /, so their sum caot exceed /. This implies = ad x+y+z= That is the aswer to the origial problem. x ( 6 ( 6 6x 0 0 0
21 Geometric Solutio
22 Example ta(arcta(cos( / 9 arcta(cos(8 / 9 arcta(cos(4 / 9 ta(arcta(cos(/9= cos(/9 Use triple agle formula cos( / 4cos ( / 9 cos( / 9 8cos ( / 9 6cos( / 9 0; x cos( / 9 x x 0 {} {} {} a a a 0 ( 4
23 Example College Mathematical Joural, 04 November, p. 57 Call this quatity U
24 Example College Mathematical Joural, 04 November, p. 57 arcta( 0 5/ {4} {} {} {} 7 4 arcta( 7 arcta( arcta( 8 ta(arcta( ta( ( 7 ( ( 8 ( 4 4 U a a a a U x x x x x x x x
25 Possible aveues for research Newto s Idetities how do they relate to this problem? Somethig similar for sies ad cosies? Two trig fuctios to work with How does the formula relate to the geometric solutio? Lill s method also deals with tagets ad polyomial equatios. How does it relate to this problem?
26 ckowledgemets lfredo Kraus i Yahoo swers obtaied the same solutio as i this presetatio but did ot relate it to polyomial equatios
27 Newto s Idetities i bracket ad brace otatio j0 k j0 j ( { j}[ k j] 0 for k ( j { j}[ k j] ( k ( k{ k} for k <
28 Lill s Method x +6x +x+6=0
29 Triagle Idetity Foud this o Math Stack Exchage: Show ta(+ta(+ta(c = ta(ta(ta(c if a+b+c=80 degrees; e.g. they are the agles of a triagle Solutio: 0 ta( {} {} ta( ta( ta( C QED ta( C 0 {} {} {} ta( ta( ta( C ta( ta( ta( C ta( ta( ta( C 0
30 Example 4? ( arcta( ( arcta( ( arcta( ( ta(arcta(
31 Example 4 ta(arcta( ( arcta( ( arcta( ( arcta( ( / 6
32 Example 4 ta(arcta( ( arcta( ( arcta( ( arcta( ( Oe ca compute by addig/multiplyig these roots together that they solve this equatio: x 4 6x 5x 0 pply the formula: {} {} {} {4} a a a a 4 0 ( 5 ( 6 ( 5 6
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