Why is a half-range sine expansion asked for a non-piecewise smooth function?

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Homework Help Overview

The discussion revolves around the half-range sine expansion of the function f(x) = 1 for the interval 0 < x < 2. Participants are questioning the classification of this function as piecewise smooth, given its characteristics.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants explore the definitions of piecewise smoothness and continuity, questioning whether the absence of discontinuities in both the function and its derivative qualifies it as piecewise smooth. They discuss the implications of smoothness in relation to the half-range sine expansion.

Discussion Status

The discussion is active, with participants providing insights into the definitions of smoothness and piecewise smoothness. Some suggest that the function can be considered piecewise smooth despite its lack of discontinuities, while others are exploring the implications of the half-range sine expansion on the function's classification.

Contextual Notes

There are references to specific definitions of piecewise smoothness and the requirements for sine series expansions, indicating that the participants are navigating through the constraints of the problem as presented in the textbook.

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Homework Statement


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?
 
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Niles said:
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?

Hi Niles! :smile:

It looks piecewise smooth to me …

what definition of "piecewise smooth" are you using? :smile:
 
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:
Niles said:
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?

f' = 0 on the interval 0 < x < 2 … that's continuous, isn't it? :wink:
 
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?
 
That's right. In this case, there are two pieces, and in each piece, f and f' are both continuous.
 
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?
 
Everything that's smooth is a fortiori (all the more so) piecewise smooth! :biggrin:
 
So "smooth > piecewise smooth"?

I just thought that there had to be some discontinuities.
 
  • #10
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.
 

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