Fourier series question baffles me

[mainguy]

Homework Statement

This question tests your ability to find, and evaluate, a sine Fourier series of a function.
f(x) = 3
Find the Fourier series for this function in the form
Ʃbⁿsin(nx∏/3) from n = 1 to infinity

Where b_n = 2/3∫f(x)sin(nx∏/3).dx where the integral is from 0 to 3

What is the value of the series at x = -6 - (1/2)

Homework Equations

Where b_n = 2/3∫f(x)sin(nx∏/3).dx where the integral is from 0 to 3

The Attempt at a Solution

I correctly determined b_n as 6/(n∏) - 6cos(n∏)/(n∏)
Then tried to sub n = 1 and x = -6.5 (no idea why the question gives it in that weird form) into the equation they gave for the Fourier series, but the answer was incorrect.Any help would be much appreciated! It will really help me understand the Fourier series, I feel this is a small hurdle to overcome and probably something obvious.

Thanks for taking your time to read this

 

[kai_sikorski]

Why are you substituting n=1? The variable n is a summation index.

Actually you can get the answer to the last question without doing any calculations. First think about what the value of the series is for 0 < x < 3; the entire point of the exercise is to approximate a certain function over this interval so the answer should be clear.

Now what is the value of the series for -3 < x <0. Use what you know about the sin function. Now you should be able to figure out the value at -6 < x < -3 and finally at -9 < x < -6.

 

[vela]

How can you substitute n=1? The variable n is the summation index.

 

[kai_sikorski]

Haha... Great minds think alike?

 

[vela]

Pretty funny the first paragraphs were almost identical.

 

[mainguy]

Why are you substituting n=1? The variable n is a summation index.

Actually you can get the answer to the last question without doing any calculations. First think about what the value of the series is for 0 < x < 3; the entire point of the exercise is to approximate a certain function over this interval so the answer should be clear.

Now what is the value of the series for -3 < x <0. Use what you know about the sin function. Now you should be able to figure out the value at -6 < x < -3 and finally at -9 < x < -6.

Okay, you seem to have shed some light on it. So the sign of the function if obviously changing, I assume b_n is the same for -6.5 as 6.5.

b_nsin(nx∏/3) taking x as 6.5 sure this means the value of the latter half of the equation, sin(nx∏/3), = -1/2?

I'm still a bit confused here, as I didnt think you could approximate a function at a point with fourier...This is my first time learning it though :P so its still a little weird

 

[kai_sikorski]

You are correct in that there is a distinction between approximating a function by a Taylor series which can often only be valid locally around a certain point; and approximating it via a Fourier series which should be valid across an entire interval. But you can still evaluate your Fourier series at a specific point. Why don't you actually write out the entire expression you got for the Fourier series approximation of f(x) = 3. It should be a function of x. While the expression has ns in it, it is not a function of n, because that is just a dummy variable. I think you're getting confused on this point.