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**zetafunc.****Guest**

I've run into a problem with deriving the Fourier series for e^x.

I've got

putting it all together:

But, clearly this can't be valid, since plugging n = 1 into the RHS gives an undefined result for the first term. What can I do about this?

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 98,933

Hi;

Interesting to look at but you do know that Fourier series work best for periodic functions, which e^x is not.

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.** **Thinking is cheating.**

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**zetafunc.****Guest**

So what I have done is nonsense?

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 98,933

I did not say that. I will have to look at more. Just do not expect it to be a very good approximation.

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.** **Thinking is cheating.**

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**zetafunc.****Guest**

What if I defined it as having period 2π?

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 98,933

In some interval you can possibly fit a non periodic function with a Fourier series. Fourier series are notorious for having problems at the endpoints due to the Runge effect just like polynomial fits.

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.** **Thinking is cheating.**

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**anonimnystefy****Real Member**- From: Harlan's World
- Registered: 2011-05-23
- Posts: 15,954

It has a period of 2*i*pi not 2*pi.

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Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and Punishment

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 98,933

Yes, he may have used the wrong equations for the coefficients.

This yields:

Plotting that and e^x we get the typical Fourier fit of a function. Notice how the purple line ( fourier ) hugs e^x.

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.** **Thinking is cheating.**

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**zetafunc.****Guest**

So the period is 2iπ since you're thinking about it in the complex plane? (2iπ being one revolution)

Those are the equations I used for the co-efficients...

**zetafunc.****Guest**

Would you say that is a fairly good fit?

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 98,933

Hi;

No the period of my equations is either -2π to 2π or -π to π. I would never use anything for serious computation that contained i.

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.** **Thinking is cheating.**

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**zetafunc.****Guest**

Oh, okay. I thought it was 2π since a lot of Fourier series tend to be defined that way (f(x) = f(x + 2π)).

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 98,933

I would say that is a fairly useless fit. The good point about Fourier fits is that they are orthogonal and least squares at the same time.

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.** **Thinking is cheating.**

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**zetafunc.****Guest**

Where is the orthogonality in the diagram? Also, do you think there are any uses for this Fourier series? For example, finding the Fourier series for x^4 can get you the sum of 1/(n^4) (integer n).

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 98,933

Hi;

One question at a time because you have asked a big one. The orthogonality is in the basis.

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.** **Thinking is cheating.**

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**zetafunc.****Guest**

What do you mean by the basis? Are you referring to the peaks/troughs of the Fourier series, and the curve (e^x) it is approximating? And, why is this orthogonality useful?

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 98,933

To understand it as mathematicians do you have to understand basis vectors.

I understand it in a way more general, a geometric way.

We can discuss it all you want but we will have delve into some pretty bizarre concepts.

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.** **Thinking is cheating.**

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**zetafunc.****Guest**

Could you tell me the geometric way? Maybe that could then help me understand basis vectors.

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 98,933

You already know about something about basis vectors but I will draw some pictures that will make it clear to you.

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.** **Thinking is cheating.**

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**zetafunc.****Guest**

Okay, thank you.

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 98,933

First some background. When I frst met you in here you wanted to end up at Cern. You might find when you get there that they are doing a completely different type of mathematics than you were taught.

You are and were taught bookmath. It is important to understand that it is not real math. Meaning, it is not the real mathematics you might be called upon to do on a day to day basis. Most students do not even know this and even if you do not believe listen to it anyway.

1) In bookmath the number line is continuous, in practical math the number line has holes in it! These holes are numbers that are left out!

2) No one should ever use the quadratic formula to solve a quadratic equation.

3) We can not really subtract as accurately as we can add!

4) (a^2 - b^2) ≠ (a -b)(a+b)

5) a1 + a2 + a3 +...+an ≠ an +...+ a3 + a2 + a1

6) A matrix has three states, invertible, singular and nearly singular.

7) Newton's iteration is rarely the best way to go and should be rarely used.

.

.

.

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.** **Thinking is cheating.**

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**zetafunc.****Guest**

I don't understand #3-5...

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 98,933

It is okay for now, you do not need to for us to get back on track.

Take a look at this drawing.

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.** **Thinking is cheating.**

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**zetafunc.****Guest**

Hmm. So, AB, and CD, are orthogonal. But... I don't understand what's going on on the right-hand side. So you are saying that EF and GH are orthogonal (as you showed with AB and CD) but the bit about the line joining EG and FH has confused me...

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 98,933

Forget orthogonality for a second. It is easy for you to eyeball the intersection of AB and CD. It is well defined.It is a tiny point. GH is on top of EF and therefore it is impossible to pinpoint an intersection. There are zillions of them. The third one is the interesting case, it is what we call nearly singular. It is close to being on top but it is not. The circle shows how wide the possible point of intersection is. It is hard to pick it out by eye. Do you follow?

Going to get some food in me, see you in a bit.

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.** **Thinking is cheating.**

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