I.e. is the function $y=b^{kx}$ an eigenfunction for the derivative operator $\frac{dy}{dx}$, where k is a constant because the derivative of such a function is ${k\ln(b)}b^{kx}$, which is a constant ($k\ln(b)$) times the original function ($b^{kx}$.) Does it even make sense to say this? If it does, are there any other such eigenfunctions for the derivative operator?
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Yes, it does make sense to think of it this way. The differentiation operator is (in one general sense) a linear transformation from the space of all smooth functions from $\mathbb{R}$ to $\mathbb{R}$, to the space of all smooth functions $\mathbb{R}$ to $\mathbb{R}$. Because it's a linear transformation from a space to itself, talking about eigenvectors, eigenvalues, etc., make perfect sense for it. You can show that, in this case, $e^{\alpha x}$ are the unique eigenvectors, up to linear dependence, by very easy ODE.
Duncan Ramage
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To find the eigenvalues and eigenfunctions of the derivative operator we need to solve the differential equation
$$\frac{dy}{dx} = \lambda y$$
which has solutions
$$y = ce^{\lambda x}$$
For any constant $c$, the function above is an eigenfunction with eigenvalue $\lambda$.
Now notice that the function $y = b^{k x}$ is an eigenfunction with eigenvalue $k \ln b$ since
$$y = b^{k x} = (e^{\ln b})^{k x} = e^{k \ln b x}$$
Adam Francey
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The solution of $y'=ay$ is $e^{ax}$, so $e^{ax}$ are the eigenvectors of the derivative operator.
Tsemo Aristide
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