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I f be a function and n be a natural number , then one can define the convolution power as follows:
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f
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{\displaystyle f^{*(n)}=\underbrace {f*f*f*...*f*f} _{n}\,}
in which the * means the convolution operation Due to the convolution theorem , one can define the convolution power for any complex number c:
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{\displaystyle \ f^{*(c)}={\mathcal {F}}^{-1}{\big \{}{\mathcal {F}}\{f\}^{c}{\big \}}}
Convolution Root
Let c be a non-zero complex number , and let f and g denote functions , then one finds
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{\displaystyle g^{*(c)}=f\Leftrightarrow {\sqrt[{*(c)}]{f}}=g\,}
In which
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{\displaystyle {\sqrt[{*(c)}]{f}}=}
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{
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}
{\displaystyle \ {\mathcal {F}}^{-1}{\big \{}{\mathcal {F}}({\sqrt[{*(c)}]{f}}){\big \}}}
Convolution exponentiation and logarithm
The convolution exponentiation and convolution logarithm
Convolution exponentiation and logarithms can be defined as follows:
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k
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{\displaystyle e^{*(f)}=\sum _{k=0}^{\infty }{\frac {f^{*(k)}}{k!}}\,}
e
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∗
ln
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{\displaystyle e^{*(g)}=f\Leftrightarrow (*\ln ){f}=g\,}
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{\displaystyle f^{*(g)}=e^{(*\ln )(f)*g}\,}
f
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{\displaystyle f^{*(g)}=e^{(*\ln )(f)*g}\,}
g
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{\displaystyle {\sqrt[{*(f)}]{g}}=e^{(*\ln )(g)*f^{*(-1)}}\,}
Failed to parse (syntax error): {\displaystyle (*log)_{f}{g}=\frac{(*\ln)(g)}{(*\ln)(f)}} \,}
Convolution Inverse
f
∗
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n
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=
f
∗
f
∗
f
∗
.
.
.
∗
f
∗
f
⏟
n
{\displaystyle f^{*(n)}=\underbrace {f*f*f*...*f*f} _{n}\,}
g
∗
f
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=
δ
⇒
f
∗
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−
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{\displaystyle g*f^{*(n)}=\delta \Rightarrow f^{*(-n)}=g\,}
g
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n
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=
f
⇒
f
∗
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n
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=
g
{\displaystyle g^{*(n)}=f\Rightarrow {\sqrt[{*(n)}]{f}}=g\,}
e
∗
(
f
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=
∑
k
=
0
∞
f
∗
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k
)
k
!
{\displaystyle e^{*(f)}=\sum _{k=0}^{\infty }{\frac {f^{*(k)}}{k!}}\,}
e
∗
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g
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=
f
⇒
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∗
ln
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f
=
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{\displaystyle e^{*(g)}=f\Rightarrow (*\ln ){f}=g\,}
f
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e
(
∗
ln
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g
{\displaystyle f^{*(g)}=e^{(*\ln(f)*g}\,}
Via the convolution theorem, one can find:
f
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n
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=
F
−
1
{
F
{
f
}
n
}
{\displaystyle \ f^{*(n)}={\mathcal {F}}^{-1}{\big \{}{\mathcal {F}}\{f\}^{n}{\big \}}}
Identities concerning Taylor series still hold if one interchange both multiplication by convolution and exponentiation by convolution power (both roots and logarithms need to be replaced by their convolution equivalents too).
Examples:
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∞
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∞
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n
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f
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{\displaystyle (1+f)^{n}=\sum _{k=0}^{\infty }{n \choose k}f^{k}\Rightarrow (\delta +f)^{*(n)}=\sum _{k=0}^{\infty }{n \choose k}f^{*(k)}\,}
e
f
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∞
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k
!
⇒
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k
!
{\displaystyle e^{f}=\sum _{k=0}^{\infty }{\frac {f^{k}}{k!}}\Rightarrow e^{*(f)}=\sum _{k=0}^{\infty }{\frac {f^{*(k)}}{k!}}\,}
(
1
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−
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∞
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⇒
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∗
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∞
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{\displaystyle (1-f)^{-1}=\sum _{k=0}^{\infty }f^{k}\Rightarrow (\delta -f)^{*(-1)}=\sum _{k=0}^{\infty }f^{*(k)}\,}
ln
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{\displaystyle \ln(1+f)=\sum _{n=0}^{\infty }{\frac {(-1)^{n}}{n+1}}f^{n+1}\Rightarrow (*\ln )(1+f)=\sum _{n=0}^{\infty }{\frac {(-1)^{n}}{n+1}}f^{*(n+1)}}
![{\displaystyle g^{*(c)}=f\Leftrightarrow {\sqrt[{*(c)}]{f}}=g\,}](/media/api/rest_v1/media/math/render/svg/6cdd1c661cf24c1378cf38a6b8c211dfc1759cb7)
![{\displaystyle {\sqrt[{*(c)}]{f}}=}](/media/api/rest_v1/media/math/render/svg/5232512dfbbd21bd89784fa8ac51bf4112d89111)
![{\displaystyle \ {\mathcal {F}}^{-1}{\big \{}{\mathcal {F}}({\sqrt[{*(c)}]{f}}){\big \}}}](/media/api/rest_v1/media/math/render/svg/e419f3e85ce1fc3db765918847e3190099745905)
![{\displaystyle {\sqrt[{*(f)}]{g}}=e^{(*\ln )(g)*f^{*(-1)}}\,}](/media/api/rest_v1/media/math/render/svg/77a30cf102760b6d53754899a12684327dd92863)
![{\displaystyle g^{*(n)}=f\Rightarrow {\sqrt[{*(n)}]{f}}=g\,}](/media/api/rest_v1/media/math/render/svg/0b385de02bf559867dfe3354a421702b8510fa35)
