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I know the moments of a real-valued function $f$: $$\int_{-\infty}^{\infty}x^{n}f(x)dx=\begin{cases} 1 & ,\ n=0\\ 0 & ,\ n=2,...q-1\\ c & ,\ n=q \end{cases} $$ What can I say about the wrapped function $g$: $$g(x)=\sum_{j=-\infty}^{\infty}f(x+j) $$

$g(x)$-is a periodic function with Fourier coeffitients: $$a(s)=\int_{0}^{1}g(x)\exp(-2\pi ixs)dx=\int_{-\infty}^{\infty}f(x)\exp(-2\pi ixs)dx$$ Can I use inverse Fourier transform on this step? If so, then

$$f(x)=\int_{-\infty}^{\infty}\overline{a(s)}\exp(-2\pi isx)ds $$ $$\int_{-\infty}^{\infty}x{}^{n}f(x)dx=\frac{1}{(2\pi i)^{n}}\frac{\partial^{n}}{\partial s^{n}}\left[\overline{a(s)}\right]_{s=0} $$ (I used Integral with Fourier transform)

From the definition of Fourier coefficients: $$\frac{\partial^{n}}{\partial s^{n}}\left[\overline{a(s)}\right]=\int_{0}^{1}(2\pi ix)^{n}g(x)\exp(2\pi isx)dx $$ Hence:$$\int_{0}^{1}x^{n}g(x)dx=\int_{-\infty}^{\infty}x{}^{n}f(x)dx $$

But it is not true!(I've plotted it). Where is my mistake?

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2 Answers 2

up vote 3 down vote accepted

Since $g(x)$ is periodic, the integral formula for its Fourier coefficients is only correct for integer index $s$. For example, we can write down the same integral, but exponentials with random coefficient in the exponent aren't orthogonal on $[0,1]$, and so on. Thus, differentiation in that parameter doesn't really have the sense we'd might have thought. The coefficient(s) $a(s)$ for $g$ are truly functions of $s\in\mathbb Z$, so differentiation has no natural sense. It is true that the $n$th Fourier coefficient of the $k$th derivative of $g$ is $(2\pi in)^k$ times the $n$th Fourier coefficient of $g$, but that's a different question.

Edit: in response to further comment... Yes, the Fourier coefficient $a(s)$ of $f$ is defined on the whole real line, and the identity about differentiating it with respect to $s$ is correct. However, to "wind up" (or whatever verb one wants) that identity does not give the integral of $g$ against $x^n$, but, instead $\int_0^1 \sum_{\ell\in\mathbb Z} (x+\ell)^n\,f(x+\ell)\,dx$. That is, the thing inside the integral is not $x^n \sum_\ell f(x+\ell)$, because the $x^n$ has to get summed over translates, too.

Edit 2: In response to second comment... it is not the case that the $n$th derivative $a^{(n)}(s)$ of $a(s)$ is $\int_0^1 x^n\,g(x)\,dx$ (with or without constants). Yes, the corresponding identity does hold with $f$ instead of $g$, but when you try to deduce the $g$-assertion from the $f$-assertion, the winding-up fails, because $x^n$ is not periodic, unlike the exponential.

Just for context, I remember when first learning calculus, thinking that I'd found an easier way to solve high-degree polynomial equations: for example, to solve $x^2=10$, _take_derivatives_ (haha!) of both sides, supposedly getting $2x=0$, so $x=0$??? No. Sure, one can take the derivative of an expression, but that does not guarantee harmony with the context, even if it sorta makes sense. The problem with the pollynomials was that $x$ is just a name for an unknown constant, so it makes no sense to take a derivative with respect to it. There's a similar problem, in part, in this Fourier business.

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In my particular case I can extend $a(s)$ on Real line. Somehow the very last step seems to be wrong :( – Katja Jul 21 '11 at 16:27
I am still confused.I do understand that the last formula is wrong. But it follows from the second and the third formulas from below. Hence some of the previous formulas should be wrong. $$\int_{-\infty}^{\infty}x{}^{n}f(x)dx=\frac{1}{(2\pi i)^{n}}\frac{\partial^{n}}{\partial s^{n}}\left[\overline{a(s)}\right]_{s=0} $$ and $$\frac{1}{(2\pi i)^{n}}\frac{\partial^{n}}{\partial s^{n}}\left[a(s)\right]_{s=0}=\int_{0}^{1}x^{n}g(x)dx $$ – Katja Jul 21 '11 at 22:51
in your example $x^2=10$ we have different functions from both sides and we are looking for the interception. Does it mean that $\int_{0}^{1}g(x)\exp(-2\pi ixs)dx$ and $\int_{-\infty}^{\infty}f(x)\exp(-2\pi ixs)dx$ are not precisely the same? Or the problem is in the differentiation of $a(s)=\int_{0}^{1}g(x)\exp(-2\pi ixs)dx$?(for example I can't change the order: first integration, then differentiation because of some continuously issues)? – Katja Jul 22 '11 at 8:03

If you want to compute $\int_{0}^{1}x^{n} g(x)dx$, i.e., consider the wrapped function over one period, you should better consider $h(x) = w(x) g(x) $ where $w(x)$ is a rectangular window ($w(x)=1$ in $[0,1]$, zero elsewhere).

$$ h(x) = w(x) g(x) = w(x) [ f(x) \star \sum_k \delta(x-k) ]$$

where $\star$ is a convolution. Then, if we call $F(\omega)$ the FT of $f(x)$, we have

$$ H(\omega) = W(\omega) \star \sum_n \sqrt{2 \pi} \; F(\omega_n) \delta(\omega-\omega_n) = \sqrt{2 \pi} \; \sum_n F(\omega_n) W(\omega - \omega_n) $$

where $\omega_n = 2 \pi n$ and $W(\omega)$ has a $sinc()$ shape.

We are interested in the $n$ derivative of $H(\omega)$ at $\omega=0$ But

$$ H'(0)= \sqrt{2 \pi} \; \sum_n F(\omega_n) W'(-\omega_n)$$

Hence it's seen that the first derivative (and the higher ones) cannot be expressed in terms of the derivatives of $F(\omega)$ at $\omega=0$.

I think, BTW, that this clarifies the confusion from other comments: how it's that we can derivate with respect to the $frequency$, when at the same time the fourier transform of the (unwindowed) wrapped signal is a fourier series, defined only in discrete values.

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