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18.034 SOLUTIONS TO PROBLEM SET 5
Due date: Friday, April 2 in lecture. Late work will be accepted only with a medical note or for another Instituteapproved reason. You are strongly encouraged to work with others, but the final writeup should be entirely your own and based on your own understanding.
(1)(10 points) Read Spotlight on Approximating Functions on pp 616–630. Everybody will receive full credit for this problem. If you have any questions about the material, want to discuss the material further, etc., please come and talk with me or Edward during office hours. There is a serious mistake in the proof of Theorem 9, p. 626 (which is easily corrected by using Theorem 13). Can you see what the mistake is?
Remark: This was not to be turned in. There is a typo in Theorem 9 – all instances of C^0 should be C^1. A more serious mistake is that it is only proved that the Fourier series of f converges uniformly to a continuous function, but it is not proved that this continuous function is f. In Theorem 13 it is proved that the Fourier series converges pointwise to f , which finishes the proof that the Fourier series converges uniformly to f.
(2)(10 points) The Riemann zeta function ζ(s) is defined for real numbers s > 1 by the formula, ∞ . n=1^ ns
ζ(s) =
(In fact the Riemann zeta function can be defined as an analytic function for every complex number s except 1, but certainly not by the series above.) The values of ζ(s) are of importance throughout mathematics, and one of the most famous open problems in mathematics is to prove that every root of ζ(s) is of the form 1 + bi for some real number b. 2 On the interval [−π, π], consider the orthonormal sequence Φ = (φn)n∈Z where,
φn(x) = √
2 π
einx^.
Let u be a real number and consider the function fu(x) = e^ ux.
(a)(5 points) Compute the Fourier coefficients,
1
� (^) π �fu, φn� = √ 2 π (^) −π
eux^ e−inxdx.
Solution: If u = 0, then �f 0 , φ 0 � =
2 π and �f 0 , φn� = 0 for n = 0. Thus suppose that u = 0. Then for every n, � (^) π 1 1 1 e(u−in)x^
π �fu , φn� =√ e(u−in)xdx 2 π
2 π (u^ −^ in)^
−π^ −π
Of course e−inπ^ = einπ^ = (−1)n. So the Fourier coefficient is,
(−1)n^ eπu^ − e−πu �fu, φn� = √. 2 π u^ −^ in
(b)(5 points) Apply Plancherel’s theorem,
�f u , fu� = |�fu, φn�|^2 , n∈Z 1
to get an equation that can be used to find a formula,
� (^1) = g(u),
∞
u^2 + n^2 n= where g(u) is some simple expression involving exponentials, etc. (Hint: One formulation of the answer involves the hyperbolic cotangent).
Solution: For u = 0, there is only one nonzero Fourier coefficient, and Plancherel’s theorem simply gives, (^) � π 12 dx = |√ 2 π |^2 = 2π, −π which is clearly true. The interesting case is if u =� 0. Then we have,
1 (eπu^ − e−πu)^2 |�fu, φn�|^2 =. 2 π |u − in |^2
Remember that u − in is a complex number. Therefore |u − in |^2 = u^2 + n^2. Therefore,
|�fu, φn�|^2 =
1 (eπu^ − e−πu)^2
2 π u^2 + n Applying Plancherel’s theorem, � (^) π ∞ |e^ ux |^2 dx = �^ 1 (e
πu (^) − e−πu) 2
−π (^) n=−∞ 2 π u^2 +^ n
First of all, |eux^ |^2 is just e^2 ux. So the integral is, � (^) π e^2 uxdx =
e^2 πu^ − e−^2 πu . −π 2 u As for the sum, factoring common terms gives,
(eπu^ − e−πu)^2 �∞^1
2 π (^) n=−∞u^2 + n
Notice that the term for −n is the same as the term for n. Singling out the term for n = 0, the series reduces to, (^) � �
(eπu^ − e−πu)^2 1 �∞^1 2 π u^2 + 2^ u^2 + n^2. n= So we have the equation,
e^2 πu^ − e−^2 πu^ (eπu^ − e−πu)^2 1 �^ ∞^1 2 u =^2 π u^2 + 2^ u^2 + n^2. n=
Dividing both sides by (eπu^ − e−πu)^2 / 2 π gives,
π e^2 πu^ − e−^2 πu^1 �∞^1 = + 2 u (eπu^ − e−πu)^2 2
u (^) n=1u^2 + n
The term (e^2 πu^ − e−^2 πu)/(eπu^ − e−πu)^2 is the same as (v^2 − w^2 )/(v − w)^2 where v = eπu^ and w = e−πu^. Since the numerator is a difference of squares, this simplifies to (v + w)/(v − w). Making this simplification and solving for the series gives,
�∞^1 π eπu^ + e−πu 1 = (^) πu − e−πu^
2 u^2
n=1^ u^2 +^ n^2 2 u^ e 2
�∞ (^) π^2 n^ � 2 (2n)!
u^2 n^ = eπu^ + e−πu^ = [
(eπu^ − e−πu)]
∞ π^2 m^ amu^2 m n=0 πu m= �∞^ π^2 l^2 l �∞ 2 m = 2 u π^2 m^ amu. l=0 (2l^ + 1)! m=
Multiplying the two series termbyterm gives,
� �^ ∞^ ∞ (^) π2(m+l)am 2(m+l) 2 u. l=0 m=0^ (2l^ + 1)!
Gathering all terms whose exponent is 2n gives,
∞ n am 2 π^2 n^ u^2 n. n=0 m=0^ (2n^ −^2 m^ + 1)!
Since the two power series are convergent and equal, for each n the coefficient of u^2 n^ of each series is equal. This gives,
2
π^2 n^ = 2
n (^) a m (^) π 2 n (^). (2n)! (^) m=0(2n − 2 m + 1)!
Of course 2π^2 n^ cancels from each side of the equation. Plugging in n = 0 gives a 0 = 1. For each n > 0, solving for an gives,
1 n−^1 am an =. (2n)!
m=0(2n^ −^2 m^ + 1)!
This gives a recursive algorithm for finding the numbers an. It is interesting to notice that each an is a fraction (this can be proved by an easy induction argument).
Plugging in the power series for the righthandside of our original equation,
π 1
∞ ( anπ^2 n^ u^2 n) − 2 = 2 ((a 0 + anπ^2 n^ u^2 n) − 1). 2 u πu (^) n=0 2 u 2 u n=
Since a 0 = 1, this cancels and we get,
�∞^1 �∞ (^) π^2 m+2 2 m u^2 + n
2 am+1u^. n=1 m=
On the other hand, for each integer n, we can expand 1/(u^2 + n^2 ) as a geometric series,
1 1 1 = = u^2 + n^2 n^2 1 + (u/n)^2
1
(^1 2) m (−1)m^
u^2 m^ =
∞ 2 2 m (−1)m^2 m+2u^. n (^) m=0 n (^) m=0 n
Plugging this in gives the series,
� 1 � � (^1 2) m = (−1)m^ u.
∞ ∞ ∞
u^2 + n^2 n^2 m+ n=1 n=1 m= 4
We would like to interchange the sums in this last series. This is justified if the series is absolutely convergent, i.e., if the series of absolute values is convergent. Let |u| < 1. Then the series of absolute values is, � �∞^ ∞^1 �∞ (^1) 2 m (^) = 2 m+2 |u| n= n n^2 − | |u^2. n=1 m= Of course this series is dominated by the series,
1 �^ ∞^1 1 − |u|^2 +^2 n=2 n^ −^1
By the integral test, for example, this last series is convergent. Therefore the original series is absolutely convergent which justifies interchanging the sums. This gives,
�∞^ �∞^1 �∞ 2 m (−1)m^2 m+2 u^2 m^ = (−1)mζ(2m + 2)u. m=0 n=1 n m=
Finally, we get the equality of convergent power series,
�∞^2 m �∞ (^) π^2 m+2 2 m (−1)mζ(2m + 2)u = 2
am+1u. m=0 m=
Therefore the coefficient of u^2 m^ on each side of the equation is equal, i.e.,
ζ(2m + 2) = (−1)m^ am+1^ π^2 m+2, 2 or,
ζ(2m) = (−1)m−^1 am^ π^2 m. 2
This is a true expression, and the recursive formula can be readily used to compute any particular value of ζ(2m). The standard convention, however, is to use the Bernoulli numbers Bn rather than an. The relation between the two of them follows from the straightforward computation,
1 1 1 1 ez^ − 1
z
coth(
z 1
∞ (^) a n (^) z 2 n− (^1). 2
z (^) n=1 22 n
This gives the identities,
Bn = (−1)
n− (^1) (2n)! 22 n^ an,
an = (−1)
n− 122 n (2n)!^ Bn In terms of the Bernoulli numbers,
22 m−^1 ζ(2m) = Bmπ^2 m. (2m)!
In particular, each value of ζ(2m) is a rational number times π^2 m.
Using the recursion relation, we have a 0 = 1, a 1 = 1/3, a 2 = − 1 /45 and a 3 = 2/945. This gives, B 1 = 1/6, B 2 = 1/30, and B 3 = 1/42. So the first three even zeta values are,
π^2 π^4 π^6 ζ(2) = 6
, ζ(4) = 90
, ζ(6) =. 945 Many beautiful results involving the Riemann zeta function and Bernoulli numbers can be found at the following URL.
http://mathworld.wolfram.com/RiemannZetaFunction.html 5
In particular, plugging in x = 0, gives,
π 4 �∞^1 F S(f )(0) =. 2
π (^) k=0(2k + 1)^2
Since the periodic extension f�^ is continuous and piecewise smooth, Theorem 10.2.2 implies that the Fourier series converges pointwise to f�^ at every point of R. In particular, for x = 0, the Fourier series converges pointwise to f (0) = 0, i.e.,
π 4 �∞^1 0 =. 2
π (^) k=0(2k + 1)^2
Therefore, � (^1) π^2 (2k + 1)^2
∞
k=
(b) As mentioned above, the periodic extension f�^ of |x| is continuous and piecewise smooth. There fore, by Theorem 10.2.2, the Fourier series F Sf converges pointwise to f�(x) for every x in R.
(5)(5 points) p. 588, Problem 7
Solution: On the interval [−π, π] the normalized Fourier exponential functions are,
(^1) inx φn(x) = √ e. 2 π
These functions are orthonormal. In this case, f (x) =
2 πφ 1 (x). Therefore �f, φn� =
2 π�φ 1 , φn�. This is 0 unless n = 1, and it is
2 π for n = 1. So the Fourier exponential series is simply,
e^ ix = √ 2 πφ ix 1 (x) =^ e^.
(Not a very exciting Fourier series!)
(6)(10 points) p. 588, Problem 17
Solution: The solution is more readable if we use variable names for the constants. Denote by m the mass of the spring, m = 1 kg. Denote by k the spring constant, k = 1. 01 Nm^. Denote by b the
damping constant, b = 0. (^2) s^ N 2. And denote by F 0 the amplitude of the square wave driving force, i.e., F 0 = 1 N.
Denote by x(t) the position of the spring at time t. The differential equation describing the motion is,
mx��(t) + bx�(t) + kx(t) = f (t).
Because f (t) is periodic of period 2π, in the steady state x(t) will be a twice differentiable, periodic function of period 2π. Therefore the Fourier exponential series of x(t) will converge to x(t) (with respect to the interval [−π, π]).
The normalized Fourier exponential functions on [−π, π] are φn(t) = √^1 2 π e
inx (^). Define p(D) to be,
p(D) = mD^2 + bD + k.
Then the Fourier coefficients an(x) = �x, φn� satisfy the equation,
p(in)an(x) = an(f ), p(in) = −(mn^2 − k) + ibn. 7
Because b =� 0, p(in) is nonzero for each n. Therefore we can solve to get,
1 an(x) = an(f ) = q(n)an(f ), p(in)
where q(n) is the function,
1 �^2 � q(n) = − (mn − k) + ibn. (mn^2 − k)^2 + b^2 n^2
It remains to compute the Fourier coefficients an(f ).
For n = 0, 1
� (^) π 1
� (^) π π a 0 (f ) = √ 2 π
f (t)dt = F 0 dt = F 0 2
−π^2 π^0 Let n be different than 0. Then,
an(f ) = √
2 π
0
π F 0 e−inxdx =
iF 0 � e−inx^
� �π
iF 0 ((−1)n^ − 1). n
2 π 0 n
2 π
Therefore an(f ) = 0 if n is even. If n is odd, then
i 2 F 0 an(f ) = −. n
2 π
Substituting this into our equation for an(x) gives,
π a 0 (x) =
F 0
k
for n even,
an(x) = 0,
and for n odd,
2 F 0 1 �^2 � an(x) = q(n)an(f ) = (^2) − k)^2 + b^2 n^2
−bn + i(mn − k). n
2 π (mn
We can simplify this somewhat by defining, √ 2 π F 0 1 An = , 2
|an(x)| = n
(mn^2 − k)^2 + b^2 n^2
and defining, 2 φn = tan−^1 (Im(an(x))/Re(an(x))) = tan−^1
mn − k . bn
Then we have, 2 an(x) = √ Ane−iφn 2 π
Of course A−n = An, and φ−n = −φn. Therefore the Fourier series is,
1 �^ ∞^1 �^ i(2l+1)t (^) + a −(2l+1)e F S[x] = a 0 (f ) √ −i(2l+1)t 2 π
2 π
a 2 l+1(x)e l=
1 ∞^ A 2 l+ = a 0 (f ) √ 2 π
(exp[i((2l + 1)t − φ 2 l+1)] + exp[−i((2l + 1)t − φ 2 l+1)]). l= 8
