Fourier series functions homework

[ramses07]

Homework Statement

Which of the following functions, when extended as 2pi periodic function, are equal to their Fourier series?
a. F(x) = 2x, pi<x<-pi
b. f(x)= 3 abs value (x)

Homework Equations

none

The Attempt at a Solution

After i graphed the functions into periodic functions of 2 pi, my professor told me that the first one doesn't represent its Fourier series since it doesn't converge at its end points, and the second one does. The coefficients do not need to be calculated.
Can anybody explain what my professor meant by it doesn't converge at its end points and why the b. does represent it series.

 

[HallsofIvy]

the first one, extended periodically, is not continuous while the second one is. That is, just to the left of x= pi, say x= pi- .001, the function value is 2x= 2(pi- .001) while just to the write, at x= pi+ .001, it is 2(pi+ .001- 2pi)= 2(.001- pi) because you have moved to the next period. The graph is a series of separate, parallel, line segments.

Similarly, for the secoind function, if x= pi- .001, the function value is 3|pi- .001|= 3(pi- .001) since pi> .001. If x= pi+ .001, the function value is 3|pi+ .001- 2pi|= 3|-pi+ .001|= 3|pi- .001| the same as before because of the absolute value. The graph is a series of connected line segments or "sawteeth".

At points of "jump discontinuity", the Fourier series converges to a point halfway between the two left and right limits.