MAT 271 Project: Partial Fractions for certain rational functions

Transcription

1 MAT 7 Project: Partial Fractios for certai ratioal fuctios Prerequisite kowledge: partial fractios from MAT 7, a very good commad of factorig ad complex umbers from Precalculus. To complete this project, studets will eed to lear the cocept of the complex roots of uity ad the basic properties of the complex expoetial fuctio. The mai purpose of this project is to fid the partial fractio decompositios of the fuctios ad x x + where is a arbitrary positive iteger. To facilitate the discussio we will use the followig otatios: For ay fixed positive iteger, let c k = cos πk d k = si πk x k = e πik for all itegers k. By defiitio of the complex expoetial fuctio, c k is the real part of z k, ad d k is the imagiary part: x k = c k + id k. 0 R. Boerer, Arizoa State Uiversity

2 A brief descriptio of the method Recall that the first step towards fidig the partial fractios of a ratioal fuctio is to factor the deomiator ito real liear ad irreducible quadratic factors. We therefore eed to factor the fuctios x ad x +. This is doe by fidig the complex zeros of each polyomial ad the usig the liear factorizatio theorem. The ext step is to idetify the pairs of solutios which are complex cojugates, ad to replace the product of their complex liear factors by irreducible real quadratic factors. The fial step is to evaluate the umerator coefficiets of the partial fractios. A effective techique for doig that is multiplicatio by oe deomiator at a time ad the applyig the rule of l'hôpital. Due to differet symmetries, we will distiguish the cases where is eve ad is odd for each fuctio separately. First we study the fuctio for the case odd. x The complex zeros of the deomiator are the so-called roots of uity, which are precisely the x k for 0 k. By the liear factorizatio theorem, that meas x = (x x k ) k=0 0 R. Boerer, Arizoa State Uiversity

3 For odd, the imagiary roots of uity are symmetric with respect to the real axis. More specifically, x k = x. k We will prove this with a little calculatio: x k = e πi( k) = e πi e πik = e πik = e πık =. xk Sice is odd, exactly oe of the x k, x 0 =, is real ad therefore is its ow symmetric parter. It will require special treatmet: x = (x ) (x x k ) k= We ow rewrite the product to associate the liear factors with cojugate complex zeros with e proved above: x = (x ) (x x k ) (x x k ) k= x = (x ) (x x k ) k= (x ) x k The ext step is to simplify the pairs of factors. If c is a complex umber, the (x c)(x c ) = x (c + c )x + cc It is a basic property of complex umbers that the product of a complex umber with its comple ig this formula to each c = x k, ad sice the x k are o the complex uit circle, that meas (x x k )(x ) x k = x (x k + )x k + 0 R. Boerer, Arizoa State Uiversity

4 Furthermore, whe you add a complex umber to its cojugate, the imagiary parts cacel ad meas x k + x k = c k, ad therefore (x x k )(x ) x k = x c k x +. This meas that we have foud the factorizatio of x ito real liear ad irreducible quadratic factors. For odd positive iteger, x = (x ) (x c k x + ) k= where c k = cos πk. To cofirm this result, ad illustrate its use at the same time, let us simplify the formula for the case = 3. I that case, the product oly has a sigle factor, sice k goes from to. Furthermore, Thus c = cos π 3 =. x 3 = (x )(x + x + ) We kow this to be true because it is othig but the geometric sum formula for N =. N x k k=0 = xn+ x 0 R. Boerer, Arizoa State Uiversity

5 A more iterestig applicatio of our factorizatio formula for x is the case = 5. I that case, the product cotais two factors, because k goes from to. The ecessary values of c k are ad c = cos π 5 c = cos π 5 = cos 36 = = cos 7 =. These two exact values of cosie are related to the golde ratio ad ca be foud by exploitig similar triagles i a isosceles triagle with base agles 7 ad top agle 36. It follows, the, that x 5 = (x ) x 5 x + x x + There is a famous mathematical theorem that implies that a similar formula does ot exist for = 7, 9,, 3, but exists for = 5 ad = 7 (as well as for certai other itegers). Assigmet : Research to discover the ame ad cotet of this theorem, ad to substatiate the meaig of similar. What usolved mathematical problem is related to this theorem? 0 R. Boerer, Arizoa State Uiversity

6 Assigmet : Fid the factorizatio of x ito real liear ad irreducible quadratic factors whe is eve. Illustrate your result by applyig it to the cases =,6,8. Movig back to our mai ivestigatio, havig foud a factorizatio of x ito real liear ad irreducible quadratic factors for odd, we are ow i a positio to fid the partial fractios of x. The form of this partial fractio decompositio must be x A k x + B k = x c k x + k= + C x To determie the ukow coefficiets, we will multiply the equatio by oe of the deomiator factors at a time ad the apply the rule of l'hôpital. Multiplyig by (x ), we get x x A k x + B k = (x ) x c k x + + C k= Both sides of this equatio are cotiuous fuctios that are defied for all real umbers x. Therefore, the limits of both sides as we let x go to must also be equal, if they exist. We kow that the limit exists because the right side is i fact defied ad cotiuous for all real umbers, icludig x =, hece its limit as x goes to is equal to the fuctio value at x =, which is C. Kowig ow that the limit of the left side exists, we ca apply the rule of l'hôpital to compute it: 0 R. Boerer, Arizoa State Uiversity

7 x lim x x = lim x x = We just showed that C =. We ow do this agai for a arbitrary oe of the quadratic factors. Let us multiply the origial equatio by (x c j + ), where j. The x c j + x = (x c j x + ) k= k j A k x + B k x c k x + + C x + (Aj x + B j ) Both sides are cotiuous fuctios for all complex umbers except the roots of uity. The right side is cotiuous eve at x = x j ad at x = x ȷ. It follows that the right sides, ad therefore both sides, have a limit as x goes to x j ad that the two limits are equal to the value of the right side at x j, which is simply A j x j + B j sice x j is a zero of x c j x +. The same is true as x goes to x ȷ. Therefore, x c j x + x c j A j x j + B j = lim x xj x = lim x xj x = x j c j x j Sice x j =, x j = x j, thus A j x j + B j = x j c j x j. The same calculatio ca ow be made for x ȷ ad yields A j x ȷ + B j = x ȷ c j x ȷ. 0 R. Boerer, Arizoa State Uiversity

8 Our job is ow to solve this system of liear equatios i two ukows for the umbers A j ad B j. By subtractig the equatios, we get A j x j x ȷ = x j c j x j x ȷ c j x ȷ We expad the terms o the right side, factor out the commo to get A j x j x ȷ = x j c j x j x ȷ + c j x ȷ We ow rearrage the terms o the right side, A j x j x ȷ = x j x ȷ c j x j + c j x ȷ ad use this groupig to factor: A j x j x ȷ = x j + x ȷ j x ȷ c j (x j x ȷ ) Dividig both sides by the ozero umber x j x, ȷ we obtai A j = x j + x ȷ c j. Fially, we observe that x j + x ȷ = c j, hece A j = c j Havig thus obtaied the value of A j, we substitute it ito the first of the two equatios of the system we are solvig to solve for B j : 0 R. Boerer, Arizoa State Uiversity

9 B j = x j c j x j A j x j = x j c j x j c j x j After factorig out, we get B j = x j c j x j c j x j = x j c j x j = x j x j c j Whe you subtract twice its real part from a complex umber, you get the egative cojugate of that complex umber. Thus B j = x jx ȷ = We have therefore succeeded i fidig the coefficiets of the partial fractios of for odd : x x c k = x x c k x + + k= x To make the formula more presetable, we factor the commo / out of the sum: x = c k x k= x c k x + + for odd ad c k = cos πk. (x ) 0 R. Boerer, Arizoa State Uiversity

10 To illustrate the use of this formula, we try the case = 3 agai. The sigma sum the cotais oly oe term. Agai, c = for this. We get Let s examie the case = 5 ow. With x 3 = 3 x + x x + x + ad c = 5 c = 5 +. we get x 5 = 5 5 x c k x x 5 + x c k x + x + Assigmet : fid the partial fractio decompositio of illustrate its use for the cases =,6,8. x for eve ad Assigmet 5: Demostrate that for eve, x + has the followig factorizatio ito irreducible quadratics: ad its partial fractio expasio is: x + = (x c k+ x + ) k=0 0 R. Boerer, Arizoa State Uiversity

11 x + = c k+x x c k+ x + Verify that the idefiite itegral of these partial fractios is give by c k+x x c k+ x + dx k=0 = c k+ l(x c k+ x + ) d k+ arcta x c k+ + C d k+ ad that as a corollary of this result, we have x + dx = 0 π si π Assigmet 6: develop aalogous formulas for the case whe is odd. 0 R. Boerer, Arizoa State Uiversity