# Fourier sine series of cos 2x

#### Morbidly_Green

Problem Statement
Find the Fourier sine series of cos 2x on [0, π]
Relevant Equations
$$b_n = \dfrac{2}{l} \int^\pi _0 f(x) sin(\dfrac{n\pi x}{l})dx$$
I am attempting to find the sine representation of cos 2x by letting

$$f(x) = \cos2x, x>0$$ and $$-\cos2x, x<0$$

Which is an odd function. Hence using $$b_n = \dfrac{2}{l} \int^\pi _0 f(x) \sin(\dfrac{n\pi x}{l})dx$$ I obtain $$b_n = \dfrac{2n}{\pi} \left( \dfrac{(-1)^n - 1}{4-n^2} \right)$$

which when I plot the sine series

$$\sum^\infty _{n=3} \dfrac{2n}{\pi} \left( \dfrac{(-1)^n - 1}{4-n^2} \right) \sin(n x)$$

I don't get the fn of cos 2x on [0, π]. I don't understand where I went wrong any help would be great.

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#### BvU

Homework Helper
Don't follow: you make an odd continuation of $f(x)$ but don't actually use it: your bounds are 0 and $\pi$

#### Morbidly_Green

Don't follow: you make an odd continuation of $f(x)$ but don't actually use it: your bounds are 0 and $\pi$
Yes, this is true, but I used the formula for $$b_n$$ that assumes I do indeed have an odd function

#### BvU

Homework Helper
So basically you have only forced $\ f(0) = f(\pi)=0\$ as is necessary with a sine series.

(and integrating from $\pi$ to $2\pi$ gives the same as from 0 to $\pi$ anyway)

I tried a cheat and got (for n up to 9):

so it looks your result is fine. Why do you think otherwise ?

#### Morbidly_Green

When I plotted the Fourier series on top of the function cos2x they did not match

So basically you have only forced $\ f(0) = f(\pi)=0\$ as is necessary with a sine series.

(and integrating from $\pi$ to $2\pi$ gives the same as from 0 to $\pi$ anyway)

I tried a cheat and got (for n up to 9):

View attachment 241922
so it looks your result is fine. Why do you think otherwise ?

#### BvU

Homework Helper
When I plotted the Fourier series on top of the function cos2x they did not match
Like 'not at all', 'not very well', 'not perfectly' ?

Did you check the link ? it has a plot too ...

To get a good match you need an awful lot of terms: the function has to jump from 0 to 1 in a step of 'size zero' at $x=0$

#### LCKurtz

Homework Helper
Gold Member
Edit[added]: I didn't notice your sum started at n=3. Anyway here's a plot to compare with.

@Morbidly_Green: Your problem is your formula for $b_n$ when $n=2$. You need to calculate $b_2$ separately, and when you do, you will find than $b_2=0$. Once you fix that, your plot should look better. Here's what I get for the first 10 terms:

#### Morbidly_Green

Edit[added]: I didn't notice your sum started at n=3. Anyway here's a plot to compare with.

@Morbidly_Green: Your problem is your formula for $b_n$ when $n=2$. You need to calculate $b_2$ separately, and when you do, you will find than $b_2=0$. Once you fix that, your plot should look better. Here's what I get for the first 10 terms:
View attachment 241924
Ah yes! Thank you, I completely forgot that you can't just ignore the first terms !

#### BvU

Homework Helper
In particular $b_1$ ! Why did you start at $n=3$ ?

#### Morbidly_Green

In particular $b_1$ ! Why did you start at $n=3$ ?
Since n=2 would yield a division by zero I started from the first calculable term, n=3, and as I said I forgot how important the terms before are

#### BvU

Homework Helper
OK, thanks. All in all: well done !

#### Ray Vickson

Homework Helper
I am attempting to find the sine representation of cos 2x by letting

$$f(x) = \cos2x, x>0$$ and $$-\cos2x, x<0$$

Which is an odd function. Hence using $$b_n = \dfrac{2}{l} \int^\pi _0 f(x) \sin(\dfrac{n\pi x}{l})dx$$ I obtain $$b_n = \dfrac{2n}{\pi} \left( \dfrac{(-1)^n - 1}{4-n^2} \right)$$

which when I plot the sine series

$$\sum^\infty _{n=3} \dfrac{2n}{\pi} \left( \dfrac{(-1)^n - 1}{4-n^2} \right) \sin(n x)$$

I don't get the fn of cos 2x on [0, π]. I don't understand where I went wrong any help would be great.
You need to keep $a_1, a_3, a_4, \ldots$. You could also include $a_2$, but you cannot use the general form for $a_n$: you need to put $a_2=0$ manually.

"Fourier sine series of cos 2x"

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