Differentiating Fourier Series

psholtz · 2007 March 5th, 02:10

Suppose we have the (exponential) function f(x) = e^x

Suppose this function is 2pi-periodic, or equivalently consider the function only on the interval -pi to +pi. This function can be expanded in a Fourier series as follows:

f(x) = Sum(-inf,+inf) c(n)*e^(inx)

("sum" is my designation for the sigma symbol).

c(n) is the Fourier coefficient, and e^(inx) is the nth Fourier basis function for the expansion.

Differentiating this expression term-by-term, we obtain:

f(x) =Sum(-inf,+inf) (in)*c(n)*e^(inx)

In other words, the LHS of the equation stays f(x) since the derivative of the exponential is the exponential, while in the RHS we bring down the factor of (in) for each term in the Fourier expansion.

However, comparing this w/ the first equation, we are left w/:

c(n) = i*n*c(n)

Obviously this is not true.

What is the problem?

Clem · 2007 March 5th, 15:00

FS can be tricky. Because the FS gives the periodic extension of e^x, the derivative of the FS is a different function than the original FS.
The FS has a discontinuities at + and - pi. Therefor its derivative has delta functions at + and - npi. The FS of a delta function is a constant, which is what your inc_n is approaaching for large n.

psholtz · 2007 March 5th, 18:14

Quote:

Originally Posted by Clem

FS can be tricky. Because the FS gives the periodic extension of e^x, the derivative of the FS is a different function than the original FS.
The FS has a discontinuities at + and - pi. Therefor its derivative has delta functions at + and - npi. The FS of a delta function is a constant, which is what your inc_n is approaaching for large n.

Ah yes, excellent point!

Yes, I think that's the answer... the LHS derivative is valid; i.e., the Fourier coeffient for the derivative really is i*n*c(n).. where the differentiation above breaks down is that the "derivative" of "e^x" is not really e^x because the function we're expanding is not really e^x.. it's a periodic analog of the exponential function defined on -pi to pi..

OK, that much makes sense.

The one question I have, regarding what you posted above, is you say that i*n*c(n) approaches a constnat for large n. Is this true? Doesn't c(n) decay to zero qute rapidly as n -- >inf? would i*n*c(n) really remain a constant? This would mean that c(n) tends to zero no faster than 1/n, and in general I'm not sure this is true?

One more question: we can take a trignometric function, or a polynomial, and obtain its Fourier series on [-pi,pi]. We can then differentiate the function and (termwise) differentiate its Fourier Series, and the two answes still correspond (even if there are discontinuities at -pi and +pi). So why does it break down in the case of the exponential function?

Clem · 2007 March 8th, 08:34

c_n is easy to calculate. I get c_n~[(-1)^n]/(n+i). The alternating sign leads to convergence.

The derivative of the FS will look like e^x away from +/- pi, but the convergence will be much slower. The same thing will happen for any function whose periodic extension is discontinuous. The Fourier coefficients of the derivative of the FS will not be the same as the Fourier coefficients of the derivative. This is because the convergence is not uniform, if you know what that means (if not, forget about it).

psholtz · 2007 March 13th, 19:41

Quote:

Originally Posted by Clem

c_n is easy to calculate. I get c_n~[(-1)^n]/(n+i). The alternating sign leads to convergence.

The derivative of the FS will look like e^x away from +/- pi, but the convergence will be much slower. The same thing will happen for any function whose periodic extension is discontinuous. The Fourier coefficients of the derivative of the FS will not be the same as the Fourier coefficients of the derivative. This is because the convergence is not uniform, if you know what that means (if not, forget about it).

Two questions:

1) if the convergence of the FS to the function *is* uniform, will the Fourier coefficients of the derivative of the FS be the same as the Fourier coefficient of the derivative of the function?
2) f(x) = e^x is infinitely differentiable.. presumably, the Fourier series of e^(x) is *not* infinitely differentiable, correct?

Clem · 2007 March 13th, 21:26

Quote:

Originally Posted by psholtz

Two questions:

1) if the convergence of the FS to the function *is* uniform, will the Fourier coefficients of the derivative of the FS be the same as the Fourier coefficient of the derivative of the function?
2) f(x) = e^x is infinitely differentiable.. presumably, the Fourier series of e^(x) is *not* infinitely differentiable, correct?

1) Yes, but the convergence is only uniform when the original periodic function is continuous. Actually, the possible non-uniform convergence of Fourier series is an asset, because it allows them to expand a discontinuous function.

2) Even the first derivataive of the FS will be discontinuous.

psholtz · 2007 March 16th, 02:03

Quote:

Originally Posted by Clem

1) Yes, but the convergence is only uniform when the original periodic function is continuous. Actually, the possible non-uniform convergence of Fourier series is an asset, because it allows them to expand a discontinuous function.

2) Even the first derivataive of the FS will be discontinuous.

Yes indeed.. in fact I can see that my question (2) was a bit trite... in fact, not only is the first derivative not continuous, but the original function isn't even continuous(!).. LOL

Indeed, the derivative of a FS can be quite sensitive .. like you're saying the function whose derivative is being taken has to be continuous, and moreover the coefficients of the derived series tend to increase by a factor of (n) each time you take a derivative, which raises the risk of whether or not the derived series is going to converge of not.

Specifically, for the function f(x)=e^x defined on [0,2*pi] (2-pi periodic), the Fourier coefficients I arrived at are:

a(0) = (e^(2pi)-1)/(2pi)
a(n) = (e^(2pi)-1)/(pi*(1+n^2))
b(n) = (-n)*(e^(2pi)-1)/(pi*(1+n^2))

where a(n) is the coefficient for cos(n*x) and b(n) is the coefficient for sin(n*x). Plotting a couple sums for this series out, the series seems to converge to e^x (if slowly .. probably b/c of the b(n) coefficients which only tend to zero as 1/n). All of which is to say, differentiating this series term-by-term, we see that the "derivated" b(n) (which is actually going to be the coefficent for cos(n*x) in the derived series) has a n^2 in both the numerator and denominator.. In other words, its not clear that by taking the term-by-term derivative of the function will produce a series that even converges.

Conversely, though, integration is not nearly as sensitive an operation to perform on FS as differentiation is it? The function whose FS is being integrated does not need to be continuous, and the coefficients of the integrated series will be ~ (1/n) the size of the coefficients of the original series, so the terms will tend to zero (i.e., the series will converge) even faster than the series for the original function.

This raises a couple question for me:

1) Suppose we generate a "FS" for periodic function, but FS doesn't even converge. Can we "integrate" this function (and integrate the FS term-by-term), and arrive at a FS expression for the integrated function that *does* converge (perhaps uniformly, if the integrated function is continuous, even though the original function might not have been)?

2) I tried to apply this to the expression I derived above for e^x.. which is to say, integrate term-by-term the expression I obtained for the FS expansion of e^x on [0,2pi]. The FS converges, but it doesn't really seem to converge to anything that resembles e^x on [0,2pi]. Is the problem the same thing that was encountered going in the other direction? Namely that the integral of e^x on [0,2pi] is *not* e^x? Or is there something else preventing the term-by-term integration of this function? Or did I just do the calculation incorrectly?

Clem · 2007 March 17th, 14:31

Quote:

Originally Posted by psholtz

1) Suppose we generate a "FS" for periodic function, but FS doesn't even converge. Can we "integrate" this function (and integrate the FS term-by-term), and arrive at a FS expression for the integrated function that *does* converge (perhaps uniformly, if the integrated function is continuous, even though the original function might not have been)?

2) I tried to apply this to the expression I derived above for e^x.. which is to say, integrate term-by-term the expression I obtained for the FS expansion of e^x on [0,2pi]. The FS converges, but it doesn't really seem to converge to anything that resembles e^x on [0,2pi]. Is the problem the same thing that was encountered going in the other direction? Namely that the integral of e^x on [0,2pi] is *not* e^x? Or is there something else preventing the term-by-term integration of this function? Or did I just do the calculation incorrectly?

1) The FS will converge for any finite function with a finite number of finite discontinuities in one period. The convergence will not be uniform if the function is discontinuous. The convergence can be very slow.
2) Integration usually improves convergence.
I think it should work in your case.