Differentiation and Integration of Fourier Series Philippe B. Laval KSU Today Philippe B. Laval (KSU) Fourier Series Today 1 / 12
Introduction When doing manipulations with infinite sums, we must remember that many properties which hold for finite sums do not always hold for infinite sums. In this section, we will see that the derivative of a sum (infinite) is not always the sums of the derivatives. In other words, we cannot always differentiate a Fourier series term by term. In this section, we discuss continuity, differentiation and integration of Fourier series. Philippe B. Laval (KSU) Fourier Series Today 2 / 12
Linearity of Fourier Series Theorem (Linearity of Fourier Series) Suppose that f 1 and f 2 are piecewise smooth on [ L.L] and c 1 and c 2 are two constants. Then, the Fourier series of c 1 f 1 (x) + c 2 f 2 (x) is c 1 times the Fourier series of f 1 (x) plus c 2 times the Fourier series of f 2 (x). The theorem means that if for example we know the Fourier series of x and x 2 then the Fourier series of x + x 2 is the sum of the Fourier series of x and x 2. We don t have to go through the lengthy process of computing Euler s coeffi cients. Philippe B. Laval (KSU) Fourier Series Today 3 / 12
Continuity of Fourier Series From the convergence theorem on Fourier series, we know that where f is continuous, its Fourier series converges to f and is therefore continuous. The only points at which we need to worry about are the points where we have jump discontinuities. This can happen at points where the function itself has jump discontinuities. It can also happen at the endpoints of the interval under study. If f ( L) f (L) then the periodic extension of f will have a jump discontinuity there. We have the following theorem. Theorem (Continuity of Fourier Series) Suppose that f is piecewise smooth on the interval [ L, L]. The Fourier series of f is continuous and converges to f on [ L, L] if and only if f is continuous and f ( L) = f (L). We are now ready to discuss differentiation and integration of Fourier series. Philippe B. Laval (KSU) Fourier Series Today 4 / 12
Let us start with an example. Consider the Fourier series for f (x) = x on [ L, L]. First, find the series. Philippe B. Laval (KSU) Fourier Series Today 5 / 12
Let us start with an example. Consider the Fourier series for f (x) = x on [ L, L]. First, find the series. ( 1) n+1 2L We should have found x = sin nπx nπ L Philippe B. Laval (KSU) Fourier Series Today 5 / 12
Let us start with an example. Consider the Fourier series for f (x) = x on [ L, L]. First, find the series. ( 1) n+1 2L We should have found x = sin nπx nπ L What do we get if we differentiate each side of the equation, differentiating the right side term by term? Philippe B. Laval (KSU) Fourier Series Today 5 / 12
Let us start with an example. Consider the Fourier series for f (x) = x on [ L, L]. First, find the series. ( 1) n+1 2L We should have found x = sin nπx nπ L What do we get if we differentiate each side of the equation, differentiating the right side term by term? We should have found 1 = 2 ( 1) n+1 cos nπx L Philippe B. Laval (KSU) Fourier Series Today 5 / 12
Let us start with an example. Consider the Fourier series for f (x) = x on [ L, L]. First, find the series. ( 1) n+1 2L We should have found x = sin nπx nπ L What do we get if we differentiate each side of the equation, differentiating the right side term by term? We should have found 1 = 2 ( 1) n+1 cos nπx L Compute separately the Fourier series of 1, what do we get? Philippe B. Laval (KSU) Fourier Series Today 5 / 12
Let us start with an example. Consider the Fourier series for f (x) = x on [ L, L]. First, find the series. ( 1) n+1 2L We should have found x = sin nπx nπ L What do we get if we differentiate each side of the equation, differentiating the right side term by term? We should have found 1 = 2 ( 1) n+1 cos nπx L Compute separately the Fourier series of 1, what do we get? We should have gotten 1 = 1. Philippe B. Laval (KSU) Fourier Series Today 5 / 12
Let us start with an example. Consider the Fourier series for f (x) = x on [ L, L]. First, find the series. ( 1) n+1 2L We should have found x = sin nπx nπ L What do we get if we differentiate each side of the equation, differentiating the right side term by term? We should have found 1 = 2 ( 1) n+1 cos nπx L Compute separately the Fourier series of 1, what do we get? We should have gotten 1 = 1. What does this prove? Philippe B. Laval (KSU) Fourier Series Today 5 / 12
So, the above example shows us we cannot always differentiate term by term. We now give without proof the conditions under which we can differentiate term by term. Theorem If f is a piecewise smooth function and if f is also continuous, then the Fourier series of f can be differentiated term by term provided that f ( L) = f (L). Why did the above example fail? Philippe B. Laval (KSU) Fourier Series Today 6 / 12
Example Find a Fourier series for f (x) = 2x using the theorem above and remembering from the homework of the previous chapter that a Fourier series for 1 x 2 on [ 1, 1] is Two remarks: 1 x 2 = 2 3 + ( 1) n+1 4 (nπ) 2 cos nπx We can even find a Fourier series for f (x) = x if we divide what we 2 ( 1) n+1 just found by 2. We will obtain x = sin nπx which nπ agrees with what we found earlier in the case L = 1. One advantage of being able to differentiate term by term is to be able to derive new Fourier series from existing ones. This bypasses the lengthy process of computing Euler s coeffi cients. Philippe B. Laval (KSU) Fourier Series Today 7 / 12
y 2 1 1.0 0.8 0.6 0.4 0.2 0.2 0.4 0.6 0.8 1.0 x 1 2 Graphs of 2x (black), S 5 (x) (red), S 10 (x) (blue), and S 20 (x) (green) Philippe B. Laval (KSU) Fourier Series Today 8 / 12
y 2 1 1.0 0.8 0.6 0.4 0.2 0.2 0.4 0.6 0.8 1.0 x 1 2 Graphs of 2x (black), S 200 (x) (blue) Philippe B. Laval (KSU) Fourier Series Today 9 / 12
Integration of Fourier Series For integration, the situation is much simpler. Theorem (Integration of Fourier Series) A Fourier series of a piecewise smooth function f can always be integrated term by term and the result is a convergent infinite series that always converges to the integral of f for x [ L, L]. Remark The theorem says that a Fourier series can only be integrated term by term and that the result is a convergent infinite series which converges to the integral of f. Note that it does not say it will be a Fourier series. Indeed, it may not be the Fourier series of the function. Philippe B. Laval (KSU) Fourier Series Today 10 / 12
Integration of Fourier Series Example The Fourier series for 1 x 2 on [ 1, 1] is 1 x 2 = 2 3 + Integrate this series from 1 to x. ( 1) n+1 4 (nπ) 2 cos nπx Philippe B. Laval (KSU) Fourier Series Today 11 / 12
Exercises See the problems at the end of my notes on differentiation and integration of Fourier series. Philippe B. Laval (KSU) Fourier Series Today 12 / 12