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[SOLVED] Fourier Transform

dwsmith

Well-known member
Feb 1, 2012
1,673
Let $g:[-\pi,\pi]\to\mathbb{R}$ be a continuous function. Define the fourier transform of $g$ as
$$
G(z)=\int_{-\pi}^{\pi}e^{zt}g(t)dt, \quad \text{for all} \ z\in\mathbb{C}.
$$
Prove that $G(z)$ is an entire function.

That means $G$ has to have no singularities, but other than that I am lost. We have to continuous functions multiplied together but then what?
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,680
Let $g:[-\pi,\pi]\to\mathbb{R}$ be a continuous function. Define the fourier transform of $g$ as
$$
G(z)=\int_{-\pi}^{\pi}e^{zt}g(t)dt, \quad \text{for all} \ z\in\mathbb{C}.
$$
Prove that $G(z)$ is an entire function.

That means $G$ has to have no singularities, but other than that I am lost. We have to continuous functions multiplied together but then what?
You want to show that $G(z)$ is differentiable at each point $z_0\in\mathbb{C}.$ If you differentiate $G(z)$ naively: $\displaystyle\frac d{dz}\int_{-\pi}^{\pi}e^{zt}g(t)\,dt$ (not worrying about how to justify slipping the differentiation operator past the integral sign), then you see that $G'(z_0)$ "ought" to be equal to $H(z_0)$, where $\displaystyle H(z) = \int_{-\pi}^{\pi}te^{zt}g(t)\,dt.$

To prove that that is indeed the case, go back to the definition of derivative and show that $$\left|\frac{G(z)-G(z_0)}{z-z_0} - H(z_0)\right| \to0\text{ as }z\to z_0.$$

That should be a reasonably straightforward exercise in epsilons and deltas, using the fact that $G$ is continuous, hence bounded, on $[-\pi,\pi].$
 

dwsmith

Well-known member
Feb 1, 2012
1,673
You want to show that $G(z)$ is differentiable at each point $z_0\in\mathbb{C}.$ If you differentiate $G(z)$ naively: $\displaystyle\frac d{dz}\int_{-\pi}^{\pi}e^{zt}g(t)\,dt$ (not worrying about how to justify slipping the differentiation operator past the integral sign), then you see that $G'(z_0)$ "ought" to be equal to $H(z_0)$, where $\displaystyle H(z) = \int_{-\pi}^{\pi}te^{zt}g(t)\,dt.$

To prove that that is indeed the case, go back to the definition of derivative and show that $$\left|\frac{G(z)-G(z_0)}{z-z_0} - H(z_0)\right| \to0\text{ as }z\to z_0.$$

That should be a reasonably straightforward exercise in epsilons and deltas, using the fact that $G$ is continuous, hence bounded, on $[-\pi,\pi].$
When you differentiated G, why didn't you differentiate via the product? It appears as if you only differentiated e^{zt} and treated g(t) as a constant.

So I need to do:
Given $\epsilon > 0$.

$ \left|\frac{G(z)-G(z_0)}{z-z_0} - H(z_0)\right| < \epsilon \ \ \text{whenever} \ \ 0<|z - z_0|<\delta $
 
Last edited by a moderator:

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,680
When you differentiated G, why didn't you differentiate via the product? It appears as if you only differentiated e^{zt} and treated g(t) as a constant.
When you are differentiating with respect to $\color{red}z$, $g(t)$ is indeed a constant.
 

dwsmith

Well-known member
Feb 1, 2012
1,673
When you are differentiating with respect to $\color{red}z$, $g(t)$ is indeed a constant.
I check that the C.R. equations are satisfied. Could that work as well?
 

dwsmith

Well-known member
Feb 1, 2012
1,673
If you differentiate $G(z)$ naively: $\displaystyle\frac d{dz}\int_{-\pi}^{\pi}e^{zt}g(t)\,dt$ (not worrying about how to justify slipping the differentiation operator past the integral sign), then you see that $G'(z_0)$ "ought" to be equal to $H(z_0)$, where $\displaystyle H(z) = \int_{-\pi}^{\pi}te^{zt}g(t)\,dt.$
Why can we slip the differentiation operator past the integral sign?
 

dwsmith

Well-known member
Feb 1, 2012
1,673
Let $g:[-\pi,\pi]\to\mathbb{R}$ be a continuous function. Define the fourier transform of $g$ as
$$
G(z)=\int_{-\pi}^{\pi}e^{zt}g(t)dt, \quad \text{for all} \ z\in\mathbb{C}.
$$
Prove that $G(z)$ is an entire function.

That means $G$ has to have no singularities, but other than that I am lost. We have to continuous functions multiplied together but then what?
Ok so $e^{zt} = \sum\limits_{n=0}^{\infty}\frac{(zt)^n}{n!}$.

Then
$$
G(z) = \int_{-\pi}^{\pi}\sum\limits_{n=0}^{\infty}\frac{(zt)^ng(t)}{n!}dt
$$

$e^{zt}$ converges uniformly on $[-\pi,\pi]$ so we can write
$$
G(z) = \sum\limits_{n=0}^{\infty}\int_{-\pi}^{\pi}\frac{(zt)^ng(t)}{n!}dt
$$

How can I perform the ratio test on this expression?
 
Last edited: