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

In summary: But if the function is odd, then the "half-range" is limited to the interval (-1,1), which is not smooth at all.
  • #1
Niles
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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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  • #2
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:
 
  • #3
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?
 
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  • #4
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:
 
  • #5
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
That's right. In this case, there are two pieces, and in each piece, f and f' are both continuous.
 
  • #7
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
Everything that's smooth is a fortiori (all the more so) piecewise smooth! :biggrin:
 
  • #9
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.
 

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

What is a half-range sine expansion?

A half-range sine expansion is a mathematical technique used to approximate a periodic function using a series of sine functions. It is often used in signal processing and other areas of science and engineering.

Why is a half-range sine expansion useful?

A half-range sine expansion allows for a more efficient representation of a periodic function, as it only requires half the number of sine terms compared to a full-range expansion. This can save computational resources and make calculations easier.

How is a half-range sine expansion calculated?

To calculate a half-range sine expansion, the function is first expressed as a Fourier series with only odd terms. Then, the coefficients of the sine terms are modified to account for the half-range. The final expression is a sum of sine functions with different frequencies and amplitudes.

What are the limitations of a half-range sine expansion?

A half-range sine expansion can only approximate periodic functions that are symmetric about the origin. It also assumes that the function has a finite range, and may not accurately represent functions with sharp changes or discontinuities.

How can a half-range sine expansion be applied in real-world situations?

A half-range sine expansion can be used in various applications, such as in the analysis of electrical signals, sound waves, and vibrations. It can also be used in image processing and data compression to efficiently represent periodic patterns or signals.

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