Find the actual sum of a fourier series at a given point

In summary: You are correct, the Fourier series does not represent the function on ##[0,1]##. This problem is asking for the Fourier series for the function on ##[0,1]##, not the ordinary Fourier series.
  • #1
thorx440
11
0

Homework Statement


You have series expansions of the function f(x) = 0 from 0 to .5, and 1 from .5 to 1 : the halfrange cosine series, the half-range sine series, and the Fourier series. For each of these series, find the actual sum of the series at x = 0, and x =1/2, and x =1

Homework Equations


1/2 [f(x+) + f(x-)] if discontinuous
f(x) if continuous

The Attempt at a Solution


I found the Fourier half range sine, and cosine expansions. But I was unsure of how to continue. Looking at the graph, I can see that the function is continuous from 0 to 1, then repeats, being discontinuous 0,1,2,3...etc.

For x = 0, it is discontinuous, so I used the first equation and found that f(x+) = 0 and f(x-) = 1, so the actual sum of the series at x = 0 is 1/2.

for x = 1/2, the point is continuous on the function, so the actual sum of the series is equal to f(1/2). Plugging f(1/2) into the Fourier series expansion that I found leaves me unable to solve it since I still have an infinite series.

The Fourier expansion is equal to

1/4 + summation from 1 to infinity of [(-1)^(n+1)/(n*pi) + cos(n*pi/2)/(n*pi)] cos(n*pi*x) + [- sin(n*pi/2)/(n*pi)]sin(n*pi*x)

(I apologize for the mess of equation, I have no idea how to format it so it looks nicer)

edit: looking at it again makes me start to think I did the first part wrong...unless the actual sum of the series at x=1 and x = 0 are the same and both equal 1/2
 
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  • #2
thorx440 said:

Homework Statement


You have series expansions of the function f(x) = 0 from 0 to .5, and 1 from .5 to 1 : the halfrange cosine series, the half-range sine series, and the Fourier series. For each of these series, find the actual sum of the series at x = 0, and x =1/2, and x =1

Homework Equations


1/2 [f(x+) + f(x-)] if discontinuous
f(x) if continuous

The Attempt at a Solution


I found the Fourier half range sine, and cosine expansions. But I was unsure of how to continue. Looking at the graph, I can see that the function is continuous from 0 to 1,

Really??

then repeats, being discontinuous 0,1,2,3...etc.

For x = 0, it is discontinuous, so I used the first equation and found that f(x+) = 0 and f(x-) = 1, so the actual sum of the series at x = 0 is 1/2.

for x = 1/2, the point is continuous on the function, so the actual sum of the series is equal to f(1/2). Plugging f(1/2) into the Fourier series expansion that I found leaves me unable to solve it since I still have an infinite series.
You have to realize that you are talking about two different functions in this problem. When you do a half range cosine expansion, that represents the even extension of your function to ##[-1,0]## and then extended periodically with period ##2##. Similarly, the half range sine expansion represents the odd periodic extension.

You should draw these two functions at least from ##-2## to ##2##. When you say above that "it is discontinuous", which function are you talking about? And show your calculations for ##f(0^+)## and ##f(0^-)##. Hint: What you have calculated isn't correct. Similarly, for each function do the calculations at ##1## and ##1/2##.

Also, there is no need to try to evaluate the sum of the series from the series itself. Your relevant equations give the answer for the sum.
 
  • #3
I drew out the function, and got basically the step function.and yes you are correct, the function is discontinuous at all intervals of 1/2. To find f(0+) and f(0-) I just looked at the graph f(0+) is 0 because the value coming in from the right hand side is 0, and f(0-) is 1 because the value coming in from the left side is 1. so I concluded that 1/2[f(0+) + f(0-)] = 1/2. I don't understand what I am doing wrong.

edit: I am looking at only the Fourier series first ^

I am still unsure of how to calculate the sum of the series. The only way I can think of is to just plug the values into the Fourier series and solve it.
 

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  • #4
thorx440 said:
I drew out the function, and got basically the step function.and yes you are correct, the function is discontinuous at all intervals of 1/2. To find f(0+) and f(0-) I just looked at the graph f(0+) is 0 because the value coming in from the right hand side is 0, and f(0-) is 1 because the value coming in from the left side is 1. so I concluded that 1/2[f(0+) + f(0-)] = 1/2. I don't understand what I am doing wrong.

edit: I am looking at only the Fourier series first ^

But the ordinary FS has nothing to do with this problem. You are given the function only on ##[0,1]##. So you can't even talk about ##f(0^-)## until you decide which extension you are using. And your first picture is neither the odd nor even extension. You need to pick each of those and answer the questions for it.

I am still unsure of how to calculate the sum of the series. The only way I can think of is to just plug the values into the Fourier series and solve it.

You use ##\frac{f(x^+) + f(x^-)}{2}## for any ##x##. You don't try to add up the series. You don't need to because that formula gives you the sum. That is the whole point of the theorem.
 
  • #5
Sorry, I just noticed that your problem did also ask about the full FS. So for your first picture, 1/2 is correct for the sum at x=0 for the full FS.
 
  • #6
The question does asks for the values at x = 0, x = 1/4, x = 1/2, and x = 1, for all three methods, the full FS, cosine and sine expansions.

For the full FS
I found that
f(0) = 1/2
f(1/4) = ??
f(1/2) = 1/2
f(1) = 1/2

The only one I am unsure of how to calculate now is f(1/4) because I cannot use the same formula since it is continuous at that point. Is my only option now to compute f(1/4) by plugging it into the expansion?
 

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  • #7
thorx440 said:
The question does asks for the values at x = 0, x = 1/4, x = 1/2, and x = 1, for all three methods, the full FS, cosine and sine expansions.

For the full FS
I found that
f(0) = 1/2
f(1/4) = ??
f(1/2) = 1/2
f(1) = 1/2

The only one I am unsure of how to calculate now is f(1/4) because I cannot use the same formula since it is continuous at that point. Is my only option now to compute f(1/4) by plugging it into the expansion?

At a point of continuity ##f(x^+)=f(x^-)=f(x)## so the formula just gives ##f(x)## so you just use the value of the function.
 

FAQ: Find the actual sum of a fourier series at a given point

1. What is a Fourier series?

A Fourier series is a mathematical tool used to represent a periodic function as a sum of sinusoidal functions with different frequencies and amplitudes.

2. How is a Fourier series calculated?

To calculate a Fourier series, the coefficients of the sinusoidal functions are determined by integrating the original function over one period and then plugging these coefficients into the series formula.

3. What is the purpose of finding the actual sum of a Fourier series at a given point?

The actual sum of a Fourier series at a given point is used to approximate the value of a periodic function at that point. This can be useful in analyzing and predicting the behavior of a function over time.

4. What factors can affect the accuracy of the actual sum of a Fourier series?

The accuracy of the actual sum of a Fourier series can be affected by the number of terms used in the series, the complexity of the original function, and the range of values over which the function is being evaluated.

5. How can the accuracy of the actual sum of a Fourier series be improved?

The accuracy of the actual sum of a Fourier series can be improved by using more terms in the series, using numerical methods for integration, and selecting a suitable range of values for evaluation.

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