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Exercises for march 26 and march 27th entries

March 28, 2012

Here are the exercises for the last two posts, Convolution and good kernels and Cesàro summability and Fejér’s theorem.

Exercise 1: Let f be a real-valued function which is 2\pi-periodic and integrable on any finite interval. Show that, for a,b \in \textbf{R},

\displaystyle\int_a^b f(x)dx = \int_{a+2\pi}^{b+2\pi}f(x)dx = \int_{a-2\pi}^{b-2\pi}f(x)dx.

And that

\displaystyle\int_{-\pi}^{\pi} f(x+a)dx = \int_{-\pi}^{\pi} f(x)dx = \int_{-\pi+a}^{\pi+a} f(x)dx.

Solution:

Let g:\textbf{R} \rightarrow \textbf{R} be defined as g(x) = x + 2\pi. Then g'(x) = 1 and by the periodicity of f we have f(x) = f(g(x)). So

\displaystyle\int_a^bf(x)dx = \int_a^b f (g (x)) g'(x)dx = F \circ g(b) - F \circ g(a)

where F is a primitive for f. Indeed, by the chain rule we have \frac{d}{dx}(F \circ g (x)) = f(g(x))g'(x) so we only need to apply the fundamental theorem of calculus. But again by the fundamental theorem of calculus, we have

\displaystyle\int_{a+2\pi}^{b+2\pi}f(x)dx = F(b+2\pi) - F(a+2\pi) = F \circ g(b) - F \circ g (a)

so we may conclude.The same argument gives the second equality of the first point. What this means is that for such a function f, the integral of f over an integral is invariant by a translation of length equal to the period of f.

For the second point, note that with a similar argument as above, we get

\displaystyle\int_{-\pi}^{\pi} f(x+a)dx = \int_{-\pi}^{\pi}f(g(x))g'(x)dx = \int_{g(-\pi)}^{g(\pi)}f(x)dx = \int_{-\pi+a}^{\pi+a}f(x)dx

for g(x) = x+a. As for the second equality, we have

\displaystyle\int_{-\pi+a}^{\pi+a}f(x)dx = \int_{-\pi+a}^{-\pi}f(x)dx + \int_{-\pi}^{\pi}f(x)dx + \int_{\pi}^{\pi+a}f(x)dx.

However, applying what we learned in the first set of equalities of this exercise, we can translate the domain of integration of the first term in the right side by 2\pi and get

\displaystyle\int_{-\pi+a}^{-\pi}f(x)dx = \int_{\pi+a}^{\pi}f(x)dx = -\int_{\pi}^{\pi+a}f(x)dx

hence

\displaystyle\int_{-\pi+a}^{\pi+a}f(x)dx = \int_{-\pi}^{\pi}f(x)dx.

QED

Exercise 2: Show that if a series of complex numbers converges to s, then it is Cesàro summable to s.

Solution:

Let \sum c_n be a series converging to s. Denoting the partial sums of this series by s_n, we need to show that \sigma_N := \frac{s_0 + \cdots + s_{N-1}}{N} \rightarrow s as N \rightarrow \infty.

Since \sum c_n = s, for every \epsilon > 0 there exists m \in \textbf{N} such that |s_n - s| < \epsilon for all n \geq m. So for large enough N, we get

|\sigma_N - s| = \left|\displaystyle\frac{s_0 + \cdots + s_{N-1} - Ns}{N} \right|

= \left|\displaystyle\frac{1}{N} \sum_{n=0}^{N-1} (s_n-s) \right|

= \left|\displaystyle\frac{1}{N}\sum_{n=0}^{m-1}(s_n-s) + \frac{1}{N}\sum_{n=m}^{N-1}(s_n-s) \right|

\leq \displaystyle\frac{1}{N}\sum_{n=0}^{m-1}|s_n-s| + \frac{1}{N} \sum_{n=m}^{N-1}|s_n-s|

< \displaystyle\frac{mM}{N} + \frac{(N-1-m)\epsilon}{N}

where M is an upper bound for |s_n-s|. Since \frac{mM}{N} \rightarrow 0 and \frac{N-1-m}{N} \rightarrow 1 as N \rightarrow \infty, we have

\displaystyle\lim_{N \rightarrow \infty} |\sigma_N-s| < \epsilon

for any \epsilon > 0, hence \sigma_N \rightarrow s as wanted.

QED

Exercise 3: Show that Lagrange’s trigonometric identities hold. i.e.

\displaystyle\sum_{n=0}^{N} \cos(n\theta) = \frac{1}{2} + \frac{\sin((N+1/2)\theta)}{2\sin(\theta/2))}

and

\displaystyle\sum_{n=0}^N \sin(n\theta) = \frac{1}{2}\cot(\theta/2) - \frac{\cos((N+1/2)\theta)}{2\sin(\theta/2)}

Solution: I’ll actually only post the solution to the first identity. The idea is that 1 + \cos(\theta) + \cdots + \cos(n\theta) is the real part of 1+e^{i\theta} + \cdots e^{in\theta} which can be written as

\displaystyle\frac{1-e^{(n+1)\theta}}{1-e^{i\theta}} = \frac{1-\cos((n+1)\theta) + i\sin((n+1)\theta)}{1-\cos(\theta) + i\sin(\theta)}.

We then multiply both the numerator and denominator by the complex conjugate of the denominator. The denominator will then be real and given by:

(1-\cos(\theta))^2 + \sin^2(\theta) = 1-2\cos(\theta) + 1 = 2(1-\cos(\theta)) = 4\sin^2(\theta/2)

while the real part of the numerator will be given by

(1-\cos((n+1)\theta))(1-\cos(\theta)) + \sin((n+1)\theta)\sin(\theta)

=(1-\cos((n+1)\theta)(2\sin^2(\theta/2)) + \sin((n+1)\theta)2\sin(\theta/2)\cos(\theta/2)

=2\sin^2(\theta/2) - 2\sin(\theta/2)[\cos((n+1)\theta)\sin(\theta/2)-\sin((n+1)\theta)\cos(\theta/2)]

=2\sin^2(\theta/2) - 2\sin(\theta/2)[\sin(\theta/2-(n+1)\theta)].

Now, note that the expression in brackets is equal to

-\sin(\theta/2 + n\theta) = -\sin((N+1/2)\theta).

Putting the numerator over the denominator, we thus get

\displaystyle\sum_{n=0}^N \cos(n\theta) = \frac{2\sin^2(\theta/2) + 2\sin(\theta/2)\sin((N+1/2)\theta)}{4\sin^2(\theta/2)}

= \displaystyle\frac{1}{2} + \frac{\sin((N+1/2)\theta)}{2\sin(\theta/2)}.

QED

I actually use the other identity too in the next exercise but I am too lazy to write it down.

Exercise 4: Show that the Fejér kernel is given by

F_N(x) = \displaystyle\frac{1}{N}\frac{\sin^2(Nx/2)}{\sin^2(x/2)}.

Recall that the Fejér kernel is by definition F_N = \frac{1}{N}\sum_{n=0}^{N-1}D_n where D_n is the n-th Dirichlet kernel. Well, in Exercise 5, I’ll show that

D_N(x) = \displaystyle\frac{\sin((N+1/2)x}{\sin(x/2)}.

So we get

NF_N(x) = \displaystyle\sum_{n=0}^{N-1} \frac{\sin(Nx+x/2)}{\sin(x/2)}

= \displaystyle\frac{1}{\sin(x/2)}\sum_{n=0}^{N-1}(\cos(Nx)\sin(x/2) + \sin(Nx)\cos(x/2))

= 1- \displaystyle\frac{1}{2} + \frac{\sin((N-1+1/2)x)}{2\sin(x/2)} + \cot(\frac{x}{2})\left(\frac{1}{2}\cot(\frac{x}{2})-\frac{\cos((N+1/2)x)}{2\sin(x/2)}\right)

where the last equality is justified by the identities of Exercise 3. This is equal to

\displaystyle\frac{1}{2} + \frac{\sin((N-1/2)x}{2\sin(x/2)} + \frac{\cos^2(x/2)}{2\sin^2(x/2)} - \frac{\cos(x/2)}{\sin(x/2)}\frac{\cos((N-1/2)x)}{2\sin(x/2)}

=\displaystyle\frac{\sin^2(\frac{x}{2})+\sin(N-\frac{1}{2})x)\sin(\frac{x}{2})+\cos^2(\frac{x}{2})-\cos(\frac{x}{2})\cos((N-\frac{1}{2})x)}{2\sin^2(x/2)}

= \displaystyle \frac{1-\cos((N-1/2)x + x/2)}{2\sin^2(x/2)} = \frac{1-\cos(2(Nx/2))}{2\sin^2(x/2)}

= \displaystyle\frac{2\sin^2(Nx/2)}{2\sin^2(x/2)}

Which gives exactly what we wanted.

QED

Exercise 5: Show that the N-th Dirichlet kernel is given by

D_N(x) = \displaystyle\frac{\sin((N+1/2)x}{\sin(x/2)}.

Solution:

Recall that by definition D_N(x) = \sum_{n=-N}^N e^{inx}. Let \omega = e^{ix}. Then since

(1+\omega + \omega^2 + \cdots + \omega^N)(1-\omega) = 1-\omega^{N+1}

and

(\omega^{-N} + \omega^{-N+1} + \cdots + \omega^{-1})(1-\omega) = \omega^{-N} - 1,

we get

D_N(x) = \displaystyle\sum_{n=-N}^{N} \omega^n = \sum_{n=0}^{N}\omega^n + \sum_{n=-N}^{-1} \omega^n

= \displaystyle\frac{1-\omega^{N+1}}{1-\omega} + \frac{\omega^{-N}-1}{1-\omega}

=\displaystyle\frac{\omega^{-N} - \omega^{N+1}}{1-\omega}.

Multiplying by \frac{\omega^{-1/2}}{\omega^{-1/2}}, we thus get

D_N(x) = \displaystyle\frac{\omega^{-N-1/2} - \omega^{N+1/2}}{\omega^{-1/2} - \omega^{1/2}}

=\displaystyle\frac{e^{-i(N+1/2)x} - e^{i(N+1/2)x}}{e^{-ix/2} - e^{ix/2}}.

But since \sin(x) = \frac{e^{ix} - e^{-ix}}{2i} = \frac{e^{-ix} - e^{ix}}{-2i}, this last equality gives

D_N(x) = \displaystyle\frac{\sin((N+1/2)x}{\sin(x/2)}

as we wanted.

QED

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From → Analysis, Exercices

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