Fourier Series Review

[Johnny5]

I want to refresh my memory of fourier series. Can anyone explain what a Fourer series is for me?

 

Thank you

[Dave]

Basically, if you have some kind of periodic function then it can be represented as a sum of cosines and sines. There's certain restrictions on this, but basically if you have a [math](2\pi)[/math]-periodic function [math]f : [-\pi,\pi]\to \mathbb{R}[/math], then:

 

[math]f(x) = \frac{a_0}{2} + \sum_{n=1}^{\infty} a_n \cos(nx) + b_n \sin(nx)[/math]

[Johnny5]

[math](2\pi)[/math]-periodic function [math]f : [-\pi' date=\pi]\to \mathbb{R}[/math], then:

 

 

I want to make sure I'm reading this right...

 

f is a function which maps the inverval from negative pi to pi, onto the reals??

 

Also, does it matter whether or not n goes from 1 to infinity, shouldn't that be minus infinity to infinity.

 

When I took PDE's I derived that formula which you posted including the constant term a0/2, which is outside of the series.

 

I'd like to do it again.

 

There were some integral formulas, sine's weighted by sines and cosines, I don't remember them anymore.

 

As I recall, the derivation did make sense after I verified four true statements about integrals of sine times cosine, and so on...

 

I don't know how anyone figured it out, but I did verify the four integrals first, and then the rest just followed... as I recall.

[Dave]

Yes, it's mapping from the interval to the reals. I'm certain that it's from n=1 -> infinity, although there is a version that uses -infinity -> infinity (this uses complex exponentials, same difference).

 

I haven't derived this stuff, so I couldn't help you there

[Johnny5]

 

I haven't derived this stuff' date=' so I couldn't help you there [/quote']

 

LOL

 

Ok I will find it somewhere on the web.

 

Thanks for all the help Dave. Especially on symbolism. When I see your Latex coding, I learn the code for whatever.

 

Thanks very much

[ed84c]

Oo! this looks a good one,

 

http://mathworld.wolfram.com/FourierSeries.html

[Johnny5]

Oo! this looks a good one' date='

 

http://mathworld.wolfram.com/FourierSeries.html[/quote']

 

Yes with the n and m

 

Same letters that were in my PDE book. Yes those are the integral formulas there.

 

Thanks

[Dave]

Oh, you wanted the orthogonality relations? I could've told you those

[Johnny5]

Oh, you wanted the orthogonality relations? I could've told you those

 

Orthogonality relations, yes them.

 

I am going to work them out right now.

 

I know integration by parts, so...

 

As a quick example, I will integrate the natural log of x, using integration by parts.

 

[math] \int ln x \ dx = ? [/math]

 

Underneath ln x, write the derivative with respect to x of ln x, and underneath dx, write the integral of dx. In other words:

 

[math] \left \begin{array}{cc}

ln x \ \ dx \\

 

\frac{1}{x} \ \ x \\

 

\end{array} \right

[/math]

 

We can now write the following true statement:

 

[math] \int ln x \ dx = (ln x)(x) - \int (1/x)(x) dx

 

[/math]

 

Now look at the RHS.

 

There are four entries in the array

 

A11 A12

A21 A22

 

In this problem, A11 = ln x and A12 = dx

A21 = 1/x A22 = x

 

Using the integral by parts formula, we can show that the original integral is equivalent to:

 

(A11)(A22) - integral of (A21)(A22)

 

Finishing up we have:

 

[math] \int ln x \ dx = xln x - \int dx = xln x - x + C [/math]

 

Where C is an arbitrary constant of integration.

 

Ok so that is an example of integration by parts. We can also use it to evaluate the integrals at the wolfram site.

 

Let me have a look at the first integral we have to evaluate:

 

[math] \int_{x=- \pi}^{x=\pi} sin (mx) sin (nx) dx [/math]

 

In a problem like this, there is a quick way to do integration by parts.

Make two colums...

 

At the top of the first column, write sin (mx), and at the top of the second column write sin nx dx.

 

Underneath sin (mx) write the derivative of sin (mx) {which is m cos (mx)}, and underneath sin (nx)dx write the integral of sin (nx)dx {which is -cos(nx)/n}

 

Now do this one more time.

 

Underneath m cos (mx) write the derivative with respect to x of m cos (mx) which is -m² sin mx.

 

 

 

[math] \left \begin{array}{cc}

sin (mx) \ \ sin (nx) \\

m cos(mx) \ \ -\frac{cos(nx)}{n} \\

-m^2 sin (mx) \ \ -\frac{sin (nx)}{n^2} \\

 

\end{array} \right

[/math]

 

Notice that we have reached a row which is equivalent to the original row times a constant, so we can stop here.

 

Now change the sign of every other row:

 

[math] \left \begin{array}{cc}

sin (mx) \ \ sin (nx) \\

-m cos(mx) \ \ -\frac{cos(nx)}{n} \\

-m^2 sin (mx) \ \ -\frac{sin (nx)}{n^2} \\

 

\end{array} \right

[/math]

 

So here is the array:

 

A11 A12

A21 A22

A31 A32

 

Now, here is the pattern for integration by parts:

 

The integral of (A11)(A12) = (A11)(A22)+(A21)(A32) - integral of (A31)(A32)

 

So we can write out the answer as:

 

[math] \int sin (mx) \ \ sin (nx) \\ dx = [/math]

[Dave]

Whoa there tiger. A much easier way is to realise that:

 

[math]\sin(mx)\sin(nx) = \tfrac{1}{2}\left(\cos(m-n)x - \cos(m+n)x \right)[/math]

 

The proof will drop out fairly trivially then.

[Johnny5]

Whoa there tiger. A much easier way is to realise that:

 

[math]\sin(mx)\sin(nx) = \tfrac{1}{2}\left(\cos(m-n)x - \cos(m+n)x \right)[/math]

 

The proof will drop out fairly trivially then.

 

I'm sure there is, but I need to review integration by parts, it comes up like crazy during Fourier analysis.

 

Well let me have a look at what you wrote...

 

 

Oh I see, that's a trigonometric identity. Yeah that works good too. The only problem is I don't usually remember the product identities. But I have to know integration by parts, for other reasons.

 

But yes that works, it is equally good.

 

Thanks

[Dave]

Not a problem It's useful to know integration by parts, but things like this can save you a lot of time in the exam room. It's a pretty common question to be asked.

[Johnny5]

Not a problem It's useful to know integration by parts, but things like this can save you a lot of time in the exam room. It's a pretty common question to be asked.

 

 

yes you are right dave, I will derive that later. Let's see what it was again:

 

 

 

2 sin (mx) sin (nx) = cos (m-n)x - cos (m+n) x

 

That is what is to be derived.

 

Well...

 

2 sin (mx) sin (nx) = sin (mx) sin (nx) + sin (mx) sin (nx)

 

Therefore:

 

sin (mx) sin (nx) + sin (mx) sin (nx) = cos (m-n)x - cos (m+n) x

 

Let A = mx, let B = nx, therefore:

[Johnny5]

Here is are the trig identities I usually remember:

 

sin (A+B) = sinAcosB+sinBcosA

cos (A+B) = cosAcosB-sinAsinB

 

And then I can figure out these, because I know that sine is an odd function, and cosine is an even function, so I don't have to memorize them:

 

sin (A-B) = sinAcos(-B)+sin(-B)cosA = sinAcosB - sinBcosA

cos (A-B) = cosAcos(-B)-sinAsin(-B) = cosAcosB+sinAsinB

 

So

 

cos (A+B) - cos (A-B) = (cosAcosB-sinAsinB)-(cosAcosB+sinAsinB)

 

So..

 

cos (A+B) - cos (A-B) = -sinAsinB-sinAsinB = -2sinAsinB

 

So

 

2sinAsinB = -cos (A+B) + cos (A-B)

 

So

 

sinAsinB = 1/2[cos (A-B)-cos(A+B)]

 

There you go.

 

 

While we're on it, there is a good way to derive the formulas using the exponential relationship...

 

 

[math] e^{i \theta} = cos \theta + i sin \theta [/math]

 

You do something, then equate real part to real part, imaginary part to imaginary part, and you get some trig relations.

 

 

Let theta = A+B

 

[math] e^{i (A+B)} = cos (A+B) + i sin (A+B) [/math]

 

I'm trying to remember it.

 

 

[math] e^{iA} e^{iB} = e^{iA+iB} = e^{i(A+B)} [/math]

 

also

 

[math] e^{iA} e^{iB} = (cosA+isinA)(cosB+isinB) [/math]

 

Got it...

 

Therefore:

 

[math]

(cosA+isinA)(cosB+isinB) = cos (A+B) + i sin (A+B)

 

[/math]

 

Now the LHS of the equation above is equivalent to:

 

[math] cosAcosB+cosAsinB i + sinAcosB i - sinAsinB [/math]

 

Where I have made use of the algebraic fact that:

 

(A+B)(C+D) = AC+AD+BC+BD

 

Now, grouping the real parts together, and the imaginary parts together we have:

 

[math] (cosAcosB - sinAsinB) + (cosAsinB + sinAcosB) i [/math]

 

And the expression just obtained, is equivalent to the following expression:

 

[math] cos (A+B) + i sin (A+B) [/math]

 

At which point we can immediately see the trig identities for the sum of two angles:

 

[math] cos (A+B) = (cosAcosB - sinAsinB) [/math]

 

[math] sin (A+B) = (cosAsinB + sinAcosB) [/math]

 

To obtain the trig identities for the difference of two angles, just let B become -B in the formulas above, and use the fact that sine is an odd function and cosine is an even function, as I did earlier.

 

Thus, we can derive most of the trig relations, if we already know the following formula:

 

[math] e^{i \theta} = cos \theta + i sin \theta [/math]

 

This formula is easily provable using the series representation of e^x.

 

It is:

 

[math] e^x = \sum_{n=1}^{n= \infty} \frac{x^n}{n!} [/math]

 

Let x = i times theta.

 

[math] e^{i \theta} = \sum_{n=1}^{n= \infty} \frac{(i \theta)^n}{n!} [/math]

 

In order to finish the derivation, you have to know the series expressions for sine and cosine. They are:

 

[math] sin \theta = \sum_{n=1}^{n= \infty} (-1)^n \frac{x^{2n+1}}{(2n+1)!} [/math]

 

and

 

[math] cos \theta = \sum_{n=1}^{n= \infty} (-1)^n \frac{x^{2n}}{(2n)!} [/math]

 

And of course these formulas can be obtained using a Taylor series expansion.

 

Whew...