Convolution power

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If \ x is a function and \ n is a natural number, then one can define the convolution power as follows:

 x^{*n} \triangleq \underbrace{x * x * x * \cdots * x * x}_n

where * denotes the convolution operation.

More generally, we can define convolution power for any complex number \ c:

\ x^{*c} \triangleq \mathcal{F}^{-1}\Big\{\mathcal{F}\{x\}^c\Big\}

where \mathcal{F} denotes the Fourier transform, and \mathcal{F}^{-1} the inverse Fourier transform. The convolution theorem shows that the previous definition for natural numbers holds.

Contents

[edit] Convolution root

We define the convolution root as

\ \sqrt[*c]{x} \triangleq x^{*(1/c)}

This obeys the following property:

\ \left(\sqrt[*c]{x}\right)^{*c} = x

[edit] Convolution exponential and logarithm

We define the convolution exponential and convolution logarithm as follows:


\begin{align}
  \exp^*(x) &\triangleq \mathcal{F}^{-1}\Big\{\exp\big(\mathcal{F}\{x\}\big)\Big\} \\
  \ln^*(x) &\triangleq \mathcal{F}^{-1}\Big\{\ln\big(\mathcal{F}\{x\}\big)\Big\}
\end{align}

Using the convolution theorem and the Taylor series expansions for \ \exp(x) and \ \ln(x), we find that the convolution exponential and logarithm may also be expressed as


\begin{align}
  \exp^*(x) &= \sum_{k=0}^\infin \frac{x^{*k}}{k!} \\
  \ln^*(x) &= \sum^{\infin}_{n=0} \frac{(-1)^n}{n+1} (x - \delta)^{*(n+1)}
\end{align}

where \ \delta is the Dirac delta.

The convolution exponential and logarithm obey many of the same properties of the standard exponential and logarithm, but with multiplication replaced by convolution:


\begin{align}
  \exp^*(x * y) &= \exp^*(x) * \exp^*(y)  \\
  \ln^*(x * y) &= \ln^*(x) + \ln^*(y) \\
  \exp^*(\ln^*(x)) &= x
\end{align}

[edit] Other definitions

We may also define the following, where \ x and \ y are both functions:


\begin{align}
  \ x^{*y} &\triangleq e^{\log^*(x) * y}  \\
  \sqrt[*x]{y} &\triangleq e^{\log^*(y) * x^{*-1}} \\
  \log^*_x(y) &\triangleq \frac{\ln^*(y)}{\ln^*(x)}
\end{align}

[edit] Convolution inverse

Again using the convolution theorem and the Taylor series expansion for \ (1+x)^{-1}, we find the following relationship:

\ x^{*-1} = \sum_{k=0}^\infin (x-\delta)^{*k} \,

This is the convolution inverse, such that \ x * \sum_{k=0}^\infin (x-\delta)^{*k} = \delta.

Proof:


\begin{align}
  x * \sum_{k=0}^\infin (x-\delta)^{*k}
       &= \mathcal{F}^{-1} \left\{ \mathcal{F} \{x\} \, \mathcal{F} \left\{ \sum_{k=0}^\infin (x-\delta)^{*k} \right\} \right\} \\
       &= \mathcal{F}^{-1} \left\{\mathcal{F}\{x\} \, \sum_{k=0}^\infin \big(\mathcal{F}\{x\}-\mathcal{F}\{\delta\}\big)^{k} \right\} \\
       &= \mathcal{F}^{-1} \left\{\mathcal{F}\{x\} \, \sum_{k=0}^\infin \big(\mathcal{F}\{x\}-1\big)^{k} \right\} \\
       &= \mathcal{F}^{-1} \left\{\frac{\mathcal{F}\{x\} }{ \mathcal{F}\{x\}} \right\} \\
       &= \delta
\end{align}

[edit] Derivatives

From the properties of convolution, the following holds:

\ \mathcal{D}\big\{x^{*n}\big\} = x  * \mathcal{D}\big\{x^{*(n-1)}\big\}

where \mathcal{D} denotes the derivative operator.

[edit] See also

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