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Fourier transform of a gaussian is another gaussian. Fourier/Laplace transforms of $\frac{1}{\sqrt t}$ is something like $\frac{1}{\sqrt \omega}$.

I realize that we can't call these eigenfunctions since the variable is also changed, but is there a special name, are there other interesting stuff to know about these?

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    $\begingroup$ The change of the variable's "name" should be disregarded in this context. While in some physical situations a "transform" may have qualitatively incomparable source and target, the mathematically abstracted question about eigenfunctions effectively ignores those interpretations. $\endgroup$ Jun 17, 2015 at 14:11

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I'll just add some detail to to Cameron William's answer. Cameron's answer is correct.

The term eigenvector is usually reserved for a linear a operator $A : X\rightarrow X$ on some particular linear space $X$; $x$ is an eigenvector of $A$ if $Ax=\lambda x$ for some scalar $\lambda$. If the linear space consists of functions, then eigenfunction is used interchangeably with eigenvector.

The Fourier transform $\mathscr{F}$ maps $X=L^{2}(\mathbb{R})$ to itself. So it makes sense to talk about eigenfucntions of $\mathscr{F}$ on $L^{2}(\mathbb{R})$. $\mathscr{F}$ is unitary, meaning that (a) $\|\mathscr{F}f\|=\|f\|$ for all $f\in L^{2}$, and (b) $\mathscr{F}$ is surjective. It turns out that $\mathscr{F}$ has an orthonormal basis of eigenfunctions, and the eigenvalues are $1,i,-1,-i$. The normalized eigenfunction are the Hermite functions $$ h_{n}(x) = \frac{(-1)^{n}e^{x^{2}/2}}{(2\pi)^{1/4}2^{n/2-1/4}\sqrt{n!}}\frac{d^{n}}{dx^{n}}e^{-x^{2}},\;\;\; n=0,1,2,3,\cdots . $$ The Hermite functions satisfy $$ \mathscr{F} h_{n} = i^{n}h_{n},\;\;\|h_{n}\|_{L^{2}}=1,\;\; n=0,1,2,3,\cdots . $$ These functions also satisfy the Hermite differential equation $$ -h''+x^{2}h = 0. $$ This equation is invariant under the Fourier transform.

The Laplace transform of $1/\sqrt{t}$ is \begin{align} \int_{0}^{\infty}\frac{1}{\sqrt{t}}e^{-st}dt & = \frac{1}{\sqrt{s}} \int_{0}^{\infty}\frac{1}{\sqrt{st}}e^{-st}d(ts) \\ & = \frac{1}{\sqrt{s}}\int_{0}^{\infty}\frac{1}{\sqrt{u}}e^{-u}du \\ & = \Gamma(1/2)\frac{1}{\sqrt{s}} = \sqrt{\pi}\frac{1}{\sqrt{s}} \end{align} In this context, it is unusual to think of $1/\sqrt{x}$ as an eigenfunction because it is difficult to formulate the Laplace transform $\mathscr{L}$ in such a way that it maps some space to itself. It is more natural to think of $\mathscr{L} : L^{2}[0,\infty) \rightarrow H^{2}(\Pi_{+})$ where $H^{2}(\Pi_{+})$ is the Hardy space of holomorphic functions on the right half-plane $\Pi_{+}$ with $L^{2}$ boundary values; this correspondence is also unitary from one Hilbert space to a different Hilbert space, but, in this context, eigenfunction doesn't make sense.

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Eigenfunction is exactly the word you're looking for. It does not matter what the "variable" is. The functional form is the important part. Eigenfunctions play an extremely similar role to eigenvectors for matrices. There are marked differences and the biggest issues are the domain/range issues associated to operators on function spaces. The Fourier transform - in its most natural setting - is densely defined on $L^2(\Bbb R)$.

The domain and range of the Laplace transform are not quite as nice as this so the notion of "eigenfunction" is not quite as clear. The functional form of $\frac{1}{\sqrt{t}}$ is preserved under the Laplace transform, but its interpretation in the domain and range are quite different so some might take issue with calling it an eigenfunction. In cases like this, it might be better to say that $\frac{1}{\sqrt{t}}$ is (form) invariant under the Laplace transform.

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Eigenvector/function as vocabulary is somewhat reserved (although not entirely) for linear functions. In the sense that that is the place you'll heat of them most commonly. It certainly wouldn't be wrong here.

The terminology I would use is is that the set of Gaussian functions is invariant under the Fourier transform. That is, call the set of Gausians $U$ and the fourier operator $F$. Then $F(U)\subset U$.

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    $\begingroup$ hear heat, potato, potahto $\endgroup$ Jun 14, 2015 at 18:32
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That transforms of gaussians are gaussians ( but with other parameters ) is not a pure eigen-relation, but in a more general sense it is. You can block-diagonalize such that the set of gaussian functions are mapped exclusively within the same block. In some sense this is a generalization of the eigenvector concept. If you study more algebra you will be able to specify and quantify such relations more in detail.

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The derivative of sin is cos and cos is -sin (for any particular frequency) so if we make a suitable base change to pairs of such sin and cos will be able to describe this differentiation in block matrix form with for each 2x2 block: $$\left(\begin{array}{rr}0&1\\-1&0\end{array}\right)$$ This is not a pure "eigen" decomposition, since is not a diagonal matrix, but it is block-diagonal, since there will be 2x2 blocks looking like the above all along the watchtower and nowhere else. For gaussians in fourier transforms there will instead be blocks looking like $$\left(\begin{array}{rr}0&1\\1&0\end{array}\right)$$

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