Mathematics 805 Homework 9 Due Friday, April 3, 1 PM. j r. = (et 1)e tx t. e t 1. = t n! = xn 1. = x n 1

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1 Mathematics 85 Homework 9 Due Friday, April 3, PM. As before, let B n be the Bernoulli polynomial of degree n. a Show that B n B n n n. b By using part a, derive a formula epressing in terms of B r. Answer: a We have n Equating coefficients of t n, we get k j B n B n t n et e t t e t te t t n t n B n B n B n B n n j r et e t t e t n n n!. n n tn n n! tn Notice incidentally that there is no problem with n, because B B, so there is no n term on the left-hand side of the initial equation. b We have B r B r r, r so k j j r r k j [ Br j B r j ] B rk B r. r Because B r B r B r, the above formula is sometimes written as B r k B r /r. A quick check: We computed last week that B 3 3, and indeed B B / Obtain the following Fourier epansions for the Bernoulli polynomials on the interval [, ]: a n B n n n n! b B n n cos k k n n,,... sin k k n n,,... Answer: Start with z 3 n cos nz n

2 valid for z. Let z y. Note that cos nz cosny n cos nycos n n cos ny, so we have y y y 3 n cos ny n n 3 cos ny n y y 3 cos ny n valid for y. Now let y, and we have 3 3 B 6 cos n n cos n n cos n n valid for. This is the case n of formula a, which we use to start the induction. Assume now that formula a is true for n. We will show that formula b is also true for n, using termby-term integration. To see that the Fourier series is uniformly convergent, apply the Weierstrass M-test using the comparison series k n. We have B n B n n n n n n n! n B n y dy n sin ky kk n cos ky dy ] k n B n n n! n sin k k n Now we assume that formula b is true for n, and show that formula a is true for n. This is trickier, because B n and cos, so there are constants of integration to be dealt with. Again, we can integrate term-by-term, because we can apply the Weierstrass M-test using the comparison series k n. We have B n B n n B n y dy n n! B n n n n n! B n n n n! B n n cos ky kk n k n sin ky dy ] k n n n! n We need to show that the constant term in parentheses is. Call it C n, so we have n n! B n C n n cos k k n cos k k n

3 The only way that I can see to show that C n is is to integrate this equation from to with respect to. We know that B k for k,,.... We know that cos k for k,,.... Termby-term integration is again justifiable because we have a uniformly convergent series using the Weierstrass M-test. The conclusion is that C n, which means that C n ; hence, establishing the induction Recall that the series zeta-function is defined by a Show that n n! B n n cos k k n n s is convergent for s > and divergent for s. The Riemann ζs ζn n n B n n s s > n,,... b Show that ζ and ζ8 95. Answer: a Take formula a from the previous problem, and substitute : n B n n n B n B n n n B n n ζn cos k k n n ζn kn n b We know that B 6, so ζ6 6 6! As a quick check, we can compute that , while 5 k We also have B 8 3, so ζ ! Again, we can compute that , while 5 k a Show that if α Z, then and deduce that cos α sin α α sin α α α α cos α α α α cos α α b Plugging in and in the above series for cos α and relabeling, prove i csc n n

4 ii cot n Answer: We begin by epanding cos α in a Fourier series from to. Because the function is an even one, we know that the result will only have terms involving cos n. We will work in terms of eponential functions, using the fact that cos ei e i. As usual, we must compute c separately. We have c c k e iα e iα [ e iα e iα iα ] sin α sin α sin α α α. e iα k e iαk [ e iα k iα k e iαk iα k e iα e iα e ik e iα k e iα k eiαk e iαk sinα k iα k iα k α k k sinα k sinα k sin α α k α k α k α k Notice that c k c k. Therefore cos α k sin α α sin α c k e ik α sin α α c k e ik c k e ik k α sin α α k k α α cos k k e ik e ik sin α sin α α sin α α sinα k α k k α sin α α k ] k α sin α α k eik e ik k α sin α α k cos k α cos α cos α α α cos The Fourier series converges to cos α for, because cos α is continuously differentiable. The series also converges at ± because cos α cos α, and the left- and right-hand derivatives eist at ±. Substituting, we have sin α sin α α α k α α k b Take the previous equation and replace α by to get sin k k k α α k csc k k We can also take the Fourier series and substitute, getting sin α cos α α k α α cos k k cot α α α α k α α sin α α k k k α α α k

5 Relabeling α as gives the desired result Evaluate each improper integral. a sin tan b 3 c cot d log Answer: a We have b We have sin tan lim t lim t t t sin tan lim t du lim u u t t ] t sin t lim cos t lim t cos ] t sin cos. 3 lim t 3 lim t ] t ] lim t /3 lim t /3. t c Note that this is not really an improper integral, because lim cot. However, it is a tricky integral to do: cot cot cot cot cot log log z cot z dz log z cot z dz log d We have y tan y dy tan dy cos sin sin cos cos sin log log lim lim t t t t lim t t cot [ logcos y ] cot log log log u u du lim u t log u du lim t [ u log u u ] y cot y dy tan y dy y tan y dy cot tan log cos sin cos sin cot log t lim t t log u du z cot z dz [ log ] t.

Mathematics 324 Riemann Zeta Function August 5, 2005

Mathematics 324 Riemann Zeta Function August 5, 25 In this note we give an introduction to the Riemann zeta function, which connects the ideas of real analysis with the arithmetic of the integers. Define