Convolution power

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.

Convolution root

We define the convolution root as

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

This obeys the following property:

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

Convolution exponential and logarithm

We define the convolution exponential and convolution logarithm as follows:

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

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

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

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:

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

Other definitions

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

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

Convolution inverse

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

$\ x^{*-1}=\sum _{k=0}^{\infty }(x-\delta )^{*k}\,$

This is the convolution inverse, such that $\ x*\sum _{k=0}^{\infty }(x-\delta )^{*k}=\delta$.

Proof:

${\begin{aligned}x*\sum _{k=0}^{\infty }(x-\delta )^{*k}&={\mathcal {F}}^{-1}\left\{{\mathcal {F}}\{x\}\,{\mathcal {F}}\left\{\sum _{k=0}^{\infty }(x-\delta )^{*k}\right\}\right\}\\&={\mathcal {F}}^{-1}\left\{{\mathcal {F}}\{x\}\,\sum _{k=0}^{\infty }{\big (}{\mathcal {F}}\{x\}-{\mathcal {F}}\{\delta \}{\big )}^{k}\right\}\\&={\mathcal {F}}^{-1}\left\{{\mathcal {F}}\{x\}\,\sum _{k=0}^{\infty }{\big (}{\mathcal {F}}\{x\}-1{\big )}^{k}\right\}\\&={\mathcal {F}}^{-1}\left\{{\frac {{\mathcal {F}}\{x\}}{{\mathcal {F}}\{x\}}}\right\}\\&=\delta \end{aligned}}$

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.

See also

• Convolution
• Convolution theorem
• Fourier transform
• Taylor series