Fourier series question
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Can you find(calculate):
\(\displaystyle \frac{d}{dx}\left[ \frac{cos(n*x)}{n}\right] = ?\)
Assuming you can, how is that related to the integral you are "troubled" with?
Thank you for your reply!
so the integral of sin(n*x)=(-1/n)cos(n*x)
When applied to my problem I get the first answer shown within the bounds of pi and 0. After substituting in the bounds, I am struggling to simplify to get [1/(n-1) -1/(n+1)] outside? And how the power n-1 has been eliminated.
Dr.Peterson
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So, you're saying your difficulty now is not in the integral, but in evaluation? It would be really helpful to see your work, so we could tell what is going wrong.
What do you get for [MATH]\frac{\cos((n-1)x)}{n-1}[/MATH] evaluated at [MATH]\pi[/MATH], and at 0?
Do you see that [MATH]\cos((n-1)\pi)[/MATH] is +1 if n is odd, and -1 if n is even?
Where do you see an exponent of n-1?
Again, showing your work step by step and marking the spots you are wondering about will make it a lot easier for us.
I managed to evaluate the integral within pi and 0 as shown in my workings below. I then tried the cases where n is odd and when n is even. Found that for odd n, an=0. Finally, I got a final answer for the cosine series. It is different from the solutions I was given so I am unsure as to where I went wrong?
Dr.Peterson
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What was the answer you were given? You only showed the solution up to a certain point that you have now gone past in your work, so I want to see what you are comparing your answer to. (I do see something wrong at the end of your work, but I think it would be obvious, so maybe I'm missing something, as I haven't tried doing all the work on my own.)
Dr.Peterson
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Okay, you've just missed something small. Here is your work:
Here is theirs:
You pointed out that the series uses only n = 2, 4, 6, ... (but I don't know why you used 4x, 8x, 12x in the sum, which is the one difference between the answers at that point); but in your sigma notation, you show n = 1, 2, 3, ... . You tried to adjust for this by replacing n with 2n in one place, but left it as n in the denominator.
You need to make a substitution, letting [MATH]n = 2m[/MATH], so that m = 1, 2, 3, ... . Then the term [MATH]\frac{\cos nx}{n^2-1}[/MATH] becomes [MATH]\frac{\cos (2m)x}{(2m)^2-1} = \frac{\cos 2mx}{4m^2-1}[/MATH]. And that's just what they have.
Here is theirs:
You pointed out that the series uses only n = 2, 4, 6, ... (but I don't know why you used 4x, 8x, 12x in the sum, which is the one difference between the answers at that point); but in your sigma notation, you show n = 1, 2, 3, ... . You tried to adjust for this by replacing n with 2n in one place, but left it as n in the denominator.
You need to make a substitution, letting [MATH]n = 2m[/MATH], so that m = 1, 2, 3, ... . Then the term [MATH]\frac{\cos nx}{n^2-1}[/MATH] becomes [MATH]\frac{\cos (2m)x}{(2m)^2-1} = \frac{\cos 2mx}{4m^2-1}[/MATH]. And that's just what they have.
ok, I understand now! I didn't realize I needed to let n=2m but that makes sense now, thank you!