# Finding the Sine Representation of an Odd Function Using Fourier Series

• Morbidly_Green
In summary, you are trying to find the sine representation of a cosine function, but are having problems because the function is odd. To solve the problem, you use a specific formula to calculate the series expansion of the function, and then plot the series. However, you need to start from the first term of the series and ignore the terms after that. If you include the terms before that, you will get an incorrect result.
Morbidly_Green
Homework 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.

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

BvU said:
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

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 ?

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

BvU said:
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 ?

Morbidly_Green said:
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##

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:

LCKurtz said:
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 !

In particular ##b_1## ! Why did you start at ##n=3## ?

BvU said:
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

OK, thanks. All in all: well done !

Morbidly_Green said:
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.

## 1. What is a Fourier sine series?

A Fourier sine series is a mathematical representation of a periodic function using a sum of sine functions with different frequencies and amplitudes.

## 2. How is a Fourier sine series different from a Fourier cosine series?

A Fourier sine series only includes sine functions, while a Fourier cosine series only includes cosine functions. They are both used to represent periodic functions, but the choice between sine and cosine depends on the symmetry of the function being represented.

## 3. What is the purpose of using a Fourier sine series to represent cos 2x?

The purpose of using a Fourier sine series to represent cos 2x is to break down the function into simpler components, making it easier to analyze and manipulate. It also allows for the function to be represented as a sum of sine functions, which can be useful in certain applications.

## 4. How is the Fourier sine series of cos 2x calculated?

The Fourier sine series of cos 2x is calculated by using the formula: f(x) = a0/2 + Σ(ansin(nx) + bncos(nx)), where a0 is the average value of the function, an and bn are coefficients that can be calculated using integrals, and n is the frequency of the sine function.

## 5. What are the applications of Fourier sine series?

Fourier sine series have many applications in mathematics, physics, engineering, and other fields. They are used to represent periodic functions in signal processing, image compression, and solving differential equations. They are also used in analyzing vibrations, sound waves, and other physical phenomena.

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