Half-range sine expansion

1. Sep 20, 2008

Niles

1. The problem statement, all variables and given/known data
Hi all.

I have to find the half-range sine expansion of a function f(x) = 1 for 0<x<2. My question is: This function is not piecewise smooth, so why does the book ask me to do this?

2. Sep 20, 2008

tiny-tim

Hi Niles!

It looks piecewise smooth to me …

what definition of "piecewise smooth" are you using?

3. Sep 21, 2008

Niles

That f and f' must be piecewise continuous, i.e. that there are a finite number of discontinuities, where the limits exists.

f is "piecewise" continuous, so that is OK. But f' is defined on the interval 0 < x < 2 with no discontinuities?

Last edited: Sep 21, 2008
4. Sep 21, 2008

tiny-tim

f' = 0 on the interval 0 < x < 2 … that's continuous, isn't it?

5. Sep 21, 2008

Niles

Yes, but if there are no discontinuities for either f or f', then f is simply just smooth? Does this count as f being piecewise smooth?

6. Sep 21, 2008

bdforbes

That's right. In this case, there are two pieces, and in each piece, f and f' are both continuous.

7. Sep 21, 2008

Niles

According to my book, if a function f is defined on an interval 0 < x < p, and if f is piecewise smooth, then it has a sine series expansion.

In this case f is defined on 0 < x < 2. but it is only smooth - not piecewise smooth - on this interval?

8. Sep 21, 2008

tiny-tim

Everything that's smooth is a fortiori (all the more so) piecewise smooth!

9. Sep 21, 2008

Niles

So "smooth > piecewise smooth"?

I just thought that there had to be some discontinuities.

10. Sep 21, 2008

HallsofIvy

Staff Emeritus
Generally speaking "piecewise smooth" does include "smooth". But here, since you are talking about a "half-range sine expansion", you are assuming that the function is odd (as is sin(nx) for all x). That is, the full range is -2 to 2 and f(x)= -1 for -2< x< 0, 1 for 0< x< 2 and is periodic over all real numbers with period 4. That is surely "piecewise" smooth.