Fourier Integral

[Kamykazee]

I'll start off with the actual text of the problem: Represent the following function using the Fourier Integral, \(\displaystyle f(t)=x*e^{-t^{2}}\)


To my knowledge, i would start it off like this \(\displaystyle \int_{-\infty}^{\infty}\frac {1}{2\pi}\int_{-\infty}^{\infty}f(t)*e^{i\theta(x-t)}\,dt\,d\theta\)

By taking it separately and integrating according to 't' first, i would get \(\displaystyle \int_{-\infty}^{\infty}t*e^{i\theta(x-t)-t^{2}}\,dt\). By making the substitution \(\displaystyle t^{2}=y\) that would result in \(\displaystyle \frac {1}{2}\int_{-\infty}^{\infty}e^{i\theta(x-\sqrt(y)-y}\,dy\). I do not know how to solve this from this point on. Also, am i on the correct path in order to solve what the initial problem was asking?



On a second note, somewhat related to this, a second problem is the following : Develop the following function into a Fourier Series, on the [0,2\pi] interval. The function is \(\displaystyle f(x)=\frac {\pi-x}{2}\). I have no idea where to begin with this. I have read the forum rules regarding not posting 'your homework' and just so others can solve it for you without any effort on your part, however, i think this is somewhat better than making another thread.

Thank you for your input on either of those.

 

[galactus]

On a second note, somewhat related to this, a second problem is the following : Develop the following function into a Fourier Series, on the [0,2\pi] interval. The function is \(\displaystyle f(x)=\frac {\pi-x}{2}\). I have no idea where to begin with this. I have read the forum rules regarding not posting 'your homework' and just so others can solve it for you without any effort on your part, however, i think this is somewhat better than making another thread.

Thank you for your input on either of those.

I think with this one we can use the leading coefficient as 0. Because:

\(\displaystyle a_{0}=\frac{1}{\pi}\int_{0}^{2\pi}\frac{\pi-x}{2}dx=0\)

\(\displaystyle a_{n}=\frac{1}{n}\int_{0}^{2\pi}\left(\frac{\pi-x}{2}\right)cos(nx)dx=0\)

\(\displaystyle b_{n}=\frac{1}{n}\int_{0}^{2\pi}\left(\frac{\pi-x}{2}\right)sin(nx)dx=\frac{cos(2n\pi)}{2n}-\frac{sin(2n\pi)}{2n^{2}{\pi}}+\frac{1}{2n}\)

The thing here is to notice that \(\displaystyle cos(2\pi n)=1\) for all n, \(\displaystyle sin(2n\pi)=0\) for all n.

Another to keep in mind is \(\displaystyle cos(n\pi)=(-1)^{n}\)

So, we are left with \(\displaystyle \frac{1}{2n}+\frac{1}{2n}=\frac{1}{n}\)

The series is \(\displaystyle \boxed{\sum_{n=1}^{\infty}\frac{sin(nx)}{n}}\)

If you plug in a value for x, it will converge to the same value as if you subbed into \(\displaystyle y=\frac{-x}{2}+\frac{\pi}{2}\).

The graph can be moved \(\displaystyle \pi\) units to the left. If \(\displaystyle y=\frac{-x}{2}+\frac{\pi}{2}\) is moved over

\(\displaystyle \pi\) units left so that it passes through the origin, we have \(\displaystyle y=\frac{-x}{2}\).

Then, we could use the interval \(\displaystyle [-\pi, \pi]\). The line, as is, is an odd function symmetric about

\(\displaystyle x=\pi\). If it were moved \(\displaystyle \pi\) units left, it is an odd function symmetric to the origin.

This appears to give the same result, though it is more work.

It is the same thing. Just shifted to the left.

For the first problem, I reckon this is a Fourier transform?.

\(\displaystyle e^{-x^{2}}\) is an even function. It looks like the normal distribution curve in stats.