Question

I only need help with d, e, and f. Can you please explain especially d in detail? Thank you

1. Here you will prove the famous identity that 1 1 1 3+5+7+ Do not use the Ratio Test anywhere in this question. (a) Express h as a power series centered at 0. Determine its interval of convergence. Hint: This should be short. (b) Let S(:) be the Taylor series of arctan x at 0. Without computing S(2), briefly explain why the function arctan x is analytic at 0. (c) Compute S(x) and show its interval of convergence is (-1,1]. (d) Your overly-excited friend proclaims victory: "Since arctan r is analytic at 0 and its Maclaurin series S(2) has interval of convergence [-1,1], this means that S(1) = which proves (+). Hurray!!” Briefly explain why your friend's reasoning is wrong. (e) Instead, you will use a different approach to prove S(1) = . For NE N, let Pn(2) be the Nth Taylor polynomial of arctan(x) at 0. Show VN EN, Vx € (0,1), PAN +3(x) < aretan r < PAN+1(2). (f) Conclude that (*) holds. What a crazy world, eh? Hint: Use two different limits. Be careful to apply them in the correct order.

Answer 1

Answer: Given that d) The derivative of the arc tausgest is Le are tanca) - it's? From the formula for geometric series i+y.yº+ y + --...-i-y if 141</ itzz - 1-C-x2) = 1+(-x) + (-xe2j2+(-x2)+...+(-x)+... = 1-x² + x2 x + 2 8 _ xot... provided that l-x<that is provided 1«l</ So we have that Darc (tauscrs) 1-x*+ x4. x6+48_22.... if ixl</ This is a taylor series arc tansıx) -11 are tancx)) da peper 1214 6+28_%"7 ..) dx utoraken . -,* {") dez (Sc=5*2" da) ( (-3°Sz"de) (1) +C 3.6+(x-top - + x - x +.-) Evaluating at x-o gives 0 - arctan (o): C, so we get (C1-x}+x4.2 +28_x"? ..) da

2n+ arctarcx)= x - x - + -" +... if 12167 The series we have here is Ź (0)" x H Osing the ratio test, we have that ilin izle n:o 2nt! 2nt 3 no Tante no n02n + 3 Teintl/enti i = lim (2n+1) 1x1 2n + 3 =121? = 121² lim anti non .00 26t3 (2n+3) 1x120 t By the ratio test the series converges absolutely if 1x1²<1 and diverges if 13171 At 2 =1, and x=-1 the series is known to be conve. 8 ge. So the radius of convergence is I and the equality is valid for 2€ L-1,1] arc tan(\/u) - 17a - arctanca) i so, a givers value of x with 1x1>1.