Fourier Transform of Trigonometric Polynomials | Learn How It Works

Bob
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Trigonometric Polynomials...

It's too difficult to understand...

Please tell me how a complex trigonometric polynomial works. I think real trigonometric polynomial is good enough.
T_{N}=\sum^N_{n=0}a_n cos(nx) +i\sum^N_{n=0}a_n*sin(nx)
T_{N} is postion at time x of an object moving along a line. seems have nothing to do with complex numbers.
 
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I like this one.

T_{N}=\frac{a_0}{2} +\sum^N_{n=0}a_n cos(nx) +\sum^N_{n=0}b_n sin(nx)
 
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Bob said:
I like this one.

T_{N}=\frac{a_0}{2} +\sum^N_{n=0}a_n cos(nx) +\sum^N_{n=0}b_n sin(nx)
Do you just want to know how to calculate the coefficients? Look up Fourier series. Hmm.. Now that I think about it this probably didn't help you much, but I can't quite figure out what your question is.
 
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I want to know how to use it in practice.
We can find the coefficients and a function f(x) from x0, x1, ...xn and f(x0), f(x1)...f(xn). These values come from observations. I am sure they are not complex numbers. So, the real trigonometric polynomial is good enough. The question is why we use the complex polynomial? Is it only for making things more complicated? :o
 
Bob said:
Trigonometric Polynomials...

It's too difficult to understand...

Please tell me how a complex trigonometric polynomial works. I think real trigonometric polynomial is good enough.
T_{N}=\sum^N_{n=0}a_n cos(nx) +i\sum^N_{n=0}b_n*sin(nx)
T_{N} is postion at time x of an object moving along a line. seems have nothing to do with complex numbers.
(Note: I've changed the second "an" to "bn[/b]". Surely you don't want to use the same notation for two different values.

You don't have to use complex numbers. In fact, in real applications, I've never seen it done. If everything in your application is real, then the bn would have to be imaginary in order to cancel that "i".

The form you give might be used if they were trying to make the point that the Fourier series can be written in terms of exponentials:
T_{N}=\sum^N_{n=0}c_n e^{nix}
where cn is, itself, a complex number.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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