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# Real Analysis I MATH 323

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This 2 page Class Notes was uploaded by Maximilian Turcotte on Monday October 5, 2015. The Class Notes belongs to MATH 323 at Colgate University taught by Staff in Fall. Since its upload, it has received 17 views. For similar materials see /class/219593/math-323-colgate-university in Mathematics (M) at Colgate University.

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Date Created: 10/05/15

2 00 1 7139 Theorem 7 7 2 kg 6 k1 Euler s mnemonic Suppose the polynomial pz has roots 11012 an and constant term equal to 1 Then we have m m m pm 1771771 a1 a2 an What is true of polynomials must also be true of power series Wrong If it were right em having no roots must be a constant So because the roots of sin are lm for each positive integer k we have 7 7 lt17 gtlt17 gtlt17ggtlt17 gtlt17 gtlt17 gtw 2 4 2 2 2 17 17p17 17wm Expanding the product on the right side shows that the coefficient of 2 is 7 21k27r2 On the left the coefficient of 2 is 716 Equating these two gives the desired result Proof Consider rst 2k 1 2k71 n n 3 i 1 1 s1 s1n km cos 7x 1 sin 7x cos km 2 2 2 2 i 1 i 1 7 sin km cos 5x 1 sin 5x cos km 2 sin ix cos km 2 Thus using the fact that we have a telescoping sum we get 2n1m m 7 n 2k1m 2k71m s1nf7s1n 7 glts1nf7s1nfgt V L 2 sin cos In k1 De ne 1 V L fnz 5 Ecoskm so that by the last computation we have except at z 2k7r i 2k 1m sin 2 2 m sin 5 Now de ne 2 n k 71 7 1 En mfnzdm7r7 Z 2 4 k1 k Thus But in View of the other way of writing fnz valid on the interval 07139 except for the right endpoint we can also write 7 7 xQ 4n71m E2n7170 msmfdx Using integration by parts with u hz sinm2 2 L if0ltm 7r 1 ifx0 4 71 and dv sinwdz you should check that h 0 0 limj hz to be sure that h is 71 continuous we get 47171 7r 7 i1 hzcoswdm 71 0 2 7 72 4717 1m 7r 2 7 4n 71m EMA 7 7hzcos 1 EA h zcosfdz Now we have 0739 Hw COS 4n 71m h 471 71m cos 2 7r dz lhm dz 2 0 7r Ww 0 and the last expression does not change with 71 So we see that EMA is the fraction times a quantity that is no larger than 1 07 lh mldz and hence as n 7 00 we have that EMA 7 0 Therefore 71392 00 2 7 7 Z 72 4 k1 2k 7 1 But by breaking up the original sum into odd and even terms we also have 1 1 1 1 1 7r2 272777Z77 k1 k2 k1 21 k12k 71 4 k1 k2 8 so we get 3 00 1 7r ZZE E k 1 and the result follows

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