Discussion about math, puzzles, games and fun. Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ π -¹ ² ³ °

You are not logged in.

- Topics: Active | Unanswered

**zetafunc.****Guest**

Yes, I understand. I thought both diagrams on the right were representing the same thing, which confused me.

I'm not too sure what you mean by the circles though. If it is 'nearly singular', how do we know the size of the circle? Is it sort of like taking a 'limit' as the lines get close to being on top of each other, so there are 'almost' an infinite number of intersections?

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 81,665

The circles are my drawings on top of the lines. We do not need any rigor here just an intuitive feel that as the lines get closer and closer the point of intersection becomes fuzzier. For instance if all three drawings were on graph paper it would be easy to read off the point of intersection for the first one with better accuracy than the third one.

**In mathematics, you don't understand things. You just get used to them.I have the result, but I do not yet know how to get it.All physicists, and a good many quite respectable mathematicians are contemptuous about proof.**

**Online**

**zetafunc.****Guest**

I see. So the circle gets bigger as the lines get closer.

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 81,665

Yes, it represents a certain amount of error in the calculation. Like your eyes, algorithms that try to find the points of intersection of the third example will experience instability and have more round off error and consequently give poorer results. So lines that cross at nearly right angles are more accurate in a sense than lines that cross at low angles. That is the point.

**In mathematics, you don't understand things. You just get used to them.I have the result, but I do not yet know how to get it.All physicists, and a good many quite respectable mathematicians are contemptuous about proof.**

**Online**

**zetafunc.****Guest**

Ah, I see.

But... what does that have to do with Fourier series?

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 81,665

When we curve fit using ordinary polynomials x, x^2, x^3, x^4, x^5,...as the basis we can see by graphing how much they are like example 3. Look at that mess around the origin. All of them on top of each other. That is why it is not recommended to curve fit a function using powers higher than say 10. The accumulated error makes them very difficult to get accurate results.

**In mathematics, you don't understand things. You just get used to them.I have the result, but I do not yet know how to get it.All physicists, and a good many quite respectable mathematicians are contemptuous about proof.**

**Online**

**zetafunc.****Guest**

I understand that orthogonality is preferred since it gives you the least possible error. But I can't see how this relates to our Fourier series for e^x. Where are the orthogonal lines?

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 81,665

Hi;

When we use an orthogonal basis we get this plot. Notice the high angles of intersection.

I have the result, but I do not yet know how to get it.

All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

**Online**

**zetafunc.****Guest**

I see them. So the Fourier series have an orthogonal basis? And I am guessing Taylor series do not?

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 81,665

Taylor series are not orthogonal but they are osculating so they have some benefits.

Now in the plot of e^x and the Fourier series you will not see the orthogonal basis in the plot itself. They are present in the formation of the series.

I have the result, but I do not yet know how to get it.

All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

**Online**

**zetafunc.****Guest**

Oh, I see. Are they orthogonal because sine is 90° out of phase with cosine, and the Fourier series is a sum of sines and cosines?

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 81,665

That I do not know for sure. The Fourier fit is also least squares or minimax, I am not sure.

I have the result, but I do not yet know how to get it.

All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

**Online**

**zetafunc.****Guest**

What do you mean by least squares? I have heard the term thrown around for regression lines in statistics.

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 81,665

Least squares minimize the square of the error between the fit equation and the data. First you start with an overdetermined system.

I have the result, but I do not yet know how to get it.

All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

**Online**

**zetafunc.****Guest**

Oh okay... but, how can we get this from the Fourier series?

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 81,665

It happens automatically. It is inherent in the fit.

I have the result, but I do not yet know how to get it.

All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

**Online**

**zetafunc.****Guest**

Okay...

By the way, when you derived that Fourier series for e^x, did you use Mathematica or did you derive it yourself? When I did it myself, I did not appear to get the right answer. Should I post my working?

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 81,665

Most of the time I use Mathematica. Today algebra is sort of like square roots of numbers. If you had to evaluate √ (234.176253) you would turn to your calculator. If you needed to multiply 102536 * 776241 you would turn to your calculator. At one time people did them both with pencil and paper. Same thing now with mechanical symbolic math. When you are learning do it by hand, when you know it use a CAS!

I have the result, but I do not yet know how to get it.

All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

**Online**

**zetafunc.****Guest**

I am still learning though. For instance, I did not want to just type in "FourierSeries[e^x]" because that won't show me how they dealt with the problem I had in post #1, which was having an undefined term (division by 0) at n = 1. The worry I have, for example, is having my knowledge of the concept deteriorate. Now that I use a calculator quite often I do not have to do something like 357*762 in my head, so over time, I have got slower at doing mental calculations. I am worried that the same thing would happen with this, for example. I may forget how to find Fourier series because I am used to getting something to do it for me.

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 81,665

That is always a danger and some of that will definitely happen. Everyone has slowed down with mental calculation.

What is the upside?

1) You are freed up from the drudgery that takes much time.

2) You can think about the problem without getting sidetracked with the details of solving it.

3) By getting passed things you could not get passed before you now can delve into deeper mathematics because you can use your package to help read journal articles. Incidentally, they do not admit it in the article but most journal articles are now written with the help of a CAS.

4) And this is a big one you never make a careless mistake that blows a solution away. For instance you might have wondered about that faulty solution for many months.

I have the result, but I do not yet know how to get it.

All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

**Online**