Raising Cauchy's integral formula to some power $n-1$
For a course on Complex Analysis I was checking the solution to some problems the professor gave us. One such problem was calculating $int_{|z|=1}e^z z^{-n}$ and the solution goes on to state the following:
Taking $f(z)=e^z$ then
$$
f^{(n-1)}(z) = frac{(n-1)!}{2pi i}int_{|z|=1}frac{f(zeta)}{(zeta-z)^n}dzeta
$$
I can see why, through Cauchy's integral formula we could have something like this but I don't yet understand where the factorial comes from or why we would have the denominator inside the integral raised to $n$. It seems strange to me.
integration complex-analysis
asked Dec 2 '18 at 23:08
D. Brito
358111
add a comment |
For a course on Complex Analysis I was checking the solution to some problems the professor gave us. One such problem was calculating $int_{|z|=1}e^z z^{-n}$ and the solution goes on to state the following:
Taking $f(z)=e^z$ then
$$
f^{(n-1)}(z) = frac{(n-1)!}{2pi i}int_{|z|=1}frac{f(zeta)}{(zeta-z)^n}dzeta
$$
I can see why, through Cauchy's integral formula we could have something like this but I don't yet understand where the factorial comes from or why we would have the denominator inside the integral raised to $n$. It seems strange to me.
integration complex-analysis
asked Dec 2 '18 at 23:08
D. Brito
358111
Differentiate the Cauchy integral formula with respect to $z$ enough times, noting that you can commute the derivative and integral.
– T. Bongers
Dec 2 '18 at 23:09
Ohh man, I thought it was raising to a power instead of derivation. Thanks for pointing that out.
– D. Brito
Dec 2 '18 at 23:11
Oh, ok. You could write a variant of the Cauchy integral formula for the holomorphic function $f(z)^{n - 1}$, but the power would be on $f$ instead of the denominator.
– T. Bongers
Dec 2 '18 at 23:12
add a comment |
For a course on Complex Analysis I was checking the solution to some problems the professor gave us. One such problem was calculating $int_{|z|=1}e^z z^{-n}$ and the solution goes on to state the following:
Taking $f(z)=e^z$ then
$$
f^{(n-1)}(z) = frac{(n-1)!}{2pi i}int_{|z|=1}frac{f(zeta)}{(zeta-z)^n}dzeta
$$
I can see why, through Cauchy's integral formula we could have something like this but I don't yet understand where the factorial comes from or why we would have the denominator inside the integral raised to $n$. It seems strange to me.
integration complex-analysis
asked Dec 2 '18 at 23:08
D. Brito
358111
For a course on Complex Analysis I was checking the solution to some problems the professor gave us. One such problem was calculating $int_{|z|=1}e^z z^{-n}$ and the solution goes on to state the following:
Taking $f(z)=e^z$ then
$$
f^{(n-1)}(z) = frac{(n-1)!}{2pi i}int_{|z|=1}frac{f(zeta)}{(zeta-z)^n}dzeta
$$
I can see why, through Cauchy's integral formula we could have something like this but I don't yet understand where the factorial comes from or why we would have the denominator inside the integral raised to $n$. It seems strange to me.
integration complex-analysis
integration complex-analysis
asked Dec 2 '18 at 23:08
D. Brito
358111
asked Dec 2 '18 at 23:08
D. Brito
358111
edited Dec 2 '18 at 23:20
asked Dec 2 '18 at 23:08
D. Brito
358111
asked Dec 2 '18 at 23:08
D. Brito
358111
asked Dec 2 '18 at 23:08
D. Brito
358111
358111
Differentiate the Cauchy integral formula with respect to $z$ enough times, noting that you can commute the derivative and integral.
– T. Bongers
Dec 2 '18 at 23:09
Ohh man, I thought it was raising to a power instead of derivation. Thanks for pointing that out.
– D. Brito
Dec 2 '18 at 23:11
Oh, ok. You could write a variant of the Cauchy integral formula for the holomorphic function $f(z)^{n - 1}$, but the power would be on $f$ instead of the denominator.
– T. Bongers
Dec 2 '18 at 23:12
add a comment |
Differentiate the Cauchy integral formula with respect to $z$ enough times, noting that you can commute the derivative and integral.
– T. Bongers
Dec 2 '18 at 23:09
Ohh man, I thought it was raising to a power instead of derivation. Thanks for pointing that out.
– D. Brito
Dec 2 '18 at 23:11
Oh, ok. You could write a variant of the Cauchy integral formula for the holomorphic function $f(z)^{n - 1}$, but the power would be on $f$ instead of the denominator.
– T. Bongers
Dec 2 '18 at 23:12
Differentiate the Cauchy integral formula with respect to $z$ enough times, noting that you can commute the derivative and integral.
– T. Bongers
Dec 2 '18 at 23:09
Differentiate the Cauchy integral formula with respect to $z$ enough times, noting that you can commute the derivative and integral.
– T. Bongers
Dec 2 '18 at 23:09
Ohh man, I thought it was raising to a power instead of derivation. Thanks for pointing that out.
– D. Brito
Dec 2 '18 at 23:11
Ohh man, I thought it was raising to a power instead of derivation. Thanks for pointing that out.
– D. Brito
Dec 2 '18 at 23:11
Oh, ok. You could write a variant of the Cauchy integral formula for the holomorphic function $f(z)^{n - 1}$, but the power would be on $f$ instead of the denominator.
– T. Bongers
Dec 2 '18 at 23:12
Oh, ok. You could write a variant of the Cauchy integral formula for the holomorphic function $f(z)^{n - 1}$, but the power would be on $f$ instead of the denominator.
– T. Bongers
Dec 2 '18 at 23:12
add a comment |
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Differentiate the Cauchy integral formula with respect to $z$ enough times, noting that you can commute the derivative and integral.
– T. Bongers
Dec 2 '18 at 23:09
Ohh man, I thought it was raising to a power instead of derivation. Thanks for pointing that out.
– D. Brito
Dec 2 '18 at 23:11
Oh, ok. You could write a variant of the Cauchy integral formula for the holomorphic function $f(z)^{n - 1}$, but the power would be on $f$ instead of the denominator.
– T. Bongers
Dec 2 '18 at 23:12