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I am currently trying to grasp spherical harmonics and try to digest that we proved that the sine and cosine functions are a basis for the $L^2$ space of the squared-integrable functions.

So as far as I have understood it, the functions that can be integrated with $$\int_0^1 \mathrm dx \, f^2(x)$$ are forming a vector space. Then all the $e_n = \cos(n \pi x)$ (and sine) form a basis for that space. So any (even, since I like to drop the sine terms) function $f$ can be represented as a linear combination of the basis vectors like: $$f = \sum a_n e_n$$

To get the coefficients $a_n$, I need to project the vector (i. e. the function) onto the basis vector (i. e. the sine) using the inner (dot) product, like so:

$$ a_n = \left\langle f(x), e_n \right\rangle_F = \int_0^1 \mathrm dx \, f(x) \cos(n \pi x)$$

Now I was wondering whether the Taylor series is such a representation with orthogonal functions $e_n = x^n$ as well. Is the “Taylor inner product” something like this then?

$$ a_n = \left\langle f(x), x^n \right\rangle_T = \frac{1}{n!} \left. \frac{\mathrm d^n f(x)}{\mathrm d x^n} \right|_{x = 0}$$

In the end, I will have a series like so :

$$ f = \sum a_n e_n =\sum\limits_{n} \frac{1}{n!} \left. \frac{\mathrm d^n f(x)}{\mathrm d x^n} \right|_{x = 0} x^n$$

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up vote 4 down vote accepted

No. In short, the reason is because Taylor series are a local description of a function, whereas Fourier series and spherical harmonics incorporate global data. More precisely, the Taylor series of a smooth function at $0$ doesn't change if you modify the function arbitrarily outside of a neighborhood of $0$.

You are free to introduce an inner product on, say, polynomials given by

$$\langle x^n, x^m \rangle = \delta_{nm}$$

which formally reproduces the Taylor series of a polynomial, but the above argument strongly suggests that this inner product can't be described via integration against a function supported away from $0$, and in addition any integral of the form

$$\langle x^n, x^m \rangle = \int_{\mathbb{R}} x^n x^m g(x) \, dx$$

would have the property that $\langle x^n, x^m \rangle = \langle 1, x^{n+m} \rangle$.

However, in the special case of holomorphic functions, Taylor series can be related to Fourier series using the Cauchy integral formula. Intuitively this is because the local behavior of holomorphic functions determine their global behavior.

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Okay, I have not thought about the local/global issue. So the Taylor series cannot be described with a convolution inner product like Fourier series and spherical harmonics since the polynomials would not be orthogonal in any sense? –  queueoverflow Oct 28 '12 at 16:53
    
@queueoverflow: yes. The local / global distinction also tripped me up when I was first learning about this material, but it is crucial. One issue is that you're used to Taylor series which converge to the functions they describe; this is not the typical situation (see en.wikipedia.org/wiki/Bump_function). On the other hand, Fourier series always converge in the $L^2$ sense to functions they describe. –  Qiaochu Yuan Oct 28 '12 at 16:54
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