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

1. Mar 25, 2015

### thorx440

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

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

3. 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

Last edited: Mar 25, 2015
2. Mar 25, 2015

### LCKurtz

Really??

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. Mar 25, 2015

### thorx440

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 dont understand what I am doing wrong.

edit: Im 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. Mar 25, 2015

### LCKurtz

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.

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. Mar 25, 2015

### LCKurtz

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. Mar 25, 2015

### thorx440

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. Mar 26, 2015

### LCKurtz

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.