# Compute Sin Coeffs for f(x)=e^(-x^2) on [0,2π]

• glebovg
In summary, the conversation revolves around computing the sine coefficients for the function f(x)=e^{-x^{2}} on the interval [0,2\pi], and whether this means that f(x+2\pi k)=f(x) for k\in\mathbb{Z}. It is emphasized that before calculating the coefficients, one must know the periodic extension being used and that the function is defined on [0,2\pi]. The instructor's suggestion to only consider the positive x-axis is not feasible as a Fourier series requires a periodic function. The half-range sine series will only converge to e^{-x^2} on (0,\pi), with the odd 2\pi periodic extension outside this interval.
glebovg
Compute the sine coefficients for $f(x)=e^{-x^{2}}$ on the interval $[0,2\pi]$. Does this mean $f(x+2\pi k)=f(x)$, $k\in\mathbb{Z}$? Can $x\in[0,\infty)$?

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glebovg said:
Compute the sine coefficients for $f(x)=e^{-x^{2}}$ on the interval $[0,2\pi]$. Does this mean $f(x+2\pi k)=f(x)$, $k\in\mathbb{Z}$? Can $x\in[0,\infty)$?

You have to know what periodic function you are expanding in a FS before you can calculate the Fourier coefficients. Do you mean to calculate the coefficients for the odd periodic extension of ##f(x)##? Do you know what the graph of the odd periodic extension would look like? Does that answer your last question?

It says on the interval $[0,2\pi]$. Does this mean $f(x+2\pi k)=f(x)$, $k\in\mathbb{Z}$?

A Fourier series is an expansion of a periodic function in terms of an infinite sum of sines and cosines. A real-valued function $f(x)$ of a real variable is called periodic of period $T>0$ if $f(x+T) = f(x)$ for all $x\in\mathbb{R}$.

So, can $x\in[0,\infty)$?

glebovg said:
It says on the interval $[0,2\pi]$. Does this mean $f(x+2\pi k)=f(x)$, $k\in\mathbb{Z}$?

A Fourier series is an expansion of a periodic function in terms of an infinite sum of sines and cosines. A real-valued function $f(x)$ of a real variable is called periodic of period $T>0$ if $f(x+T) = f(x)$ for all $x\in\mathbb{R}$.

So, can $x\in[0,\infty)$?

I think you have answered your own question. Of course if a function is defined on ##[0,2\pi]## and periodic with period ##2\pi## it is defined for all x. But if you want to calculate the coefficients for your f(x) you need to know what periodic extension you are using. That's why I asked you if you mean to use the odd periodic extension of ##e^{-x^2}## to calculate the coefficients.

I need to use the the half-range sine expansion. Correct?

LCKurtz said:
I think you have answered your own question. Of course if a function is defined on ##[0,2\pi]## and periodic with period ##2\pi## it is defined for all x. But if you want to calculate the coefficients for your f(x) you need to know what periodic extension you are using. That's why I asked you if you mean to use the odd periodic extension of ##e^{-x^2}## to calculate the coefficients.

I need to use the the half-range sine expansion. Correct? However, the problem does not state that the function is periodic, nor that it is defined on $[0,2\pi]$. After all, the Gaussian function is not periodic. My instructor said that I should only consider $[0,\infty)$, but then this would not satisfy the definition of a periodic function.

glebovg said:
I need to use the the half-range sine expansion. Correct? However, the problem does not state that the function is periodic, nor that it is defined on $[0,2\pi]$. After all, the Gaussian function is not periodic. My instructor said that I should only consider $[0,\infty)$, but then this would not satisfy the definition of a periodic function.

That's right. But if you are going to find a FS that represents ##e^{-x^2}## on some interval, you must decide what interval. You must know that to use the appropriate half range formulas. I would suggest you ask your instructor what interval he wants. It can't be ##(0,\infty)## unless you are talking about a Fourier transform, not a FS.

In this case $T=\pi$, but he said that I should work on the positive x-axis only. That's what bothers me. You can't do that, right? $f(x)$ would not be periodic and it would not work because a Fourier series is an expansion of a periodic function in terms of an infinite sum of sines and cosines.

glebovg said:
In this case $T=\pi$, but he said that I should work on the positive x-axis only. That's what bothers me. You can't do that, right? $f(x)$ would not be periodic and it would not work because a Fourier series is an expansion of a periodic function in terms of an infinite sum of sines and cosines.

That's right. It isn't going to work on the positive x axis. The half range sine series will only converge to ##e^{-x^2}## on ##(0,\pi)##. What the FS will converge to outside of that interval is the odd ##2\pi## periodic extension of ##e^{-x^2}##, except for multiples of ##\pi##. Your teacher may be referring to the fact that the formula for the coefficients only uses positive values of ##x##.

## 1. What is the purpose of computing Sin Coeffs for f(x)=e^(-x^2) on [0,2π]?

The purpose of computing Sin Coeffs for f(x)=e^(-x^2) on [0,2π] is to find the coefficients of the Fourier series representation of the given function. These coefficients can then be used to approximate the function and analyze its behavior.

## 2. How are Sin Coeffs for f(x)=e^(-x^2) on [0,2π] calculated?

The Sin Coeffs for f(x)=e^(-x^2) on [0,2π] are calculated using the Fourier series formula, which involves integrating the function over the given interval and multiplying it by the appropriate sine function.

## 3. What does the function e^(-x^2) represent?

The function e^(-x^2) represents a Gaussian or normal distribution curve. It is commonly used in statistics and probability to model random variables.

## 4. Why is the interval [0,2π] chosen for computing Sin Coeffs?

The interval [0,2π] is chosen because it represents one full period of the function e^(-x^2). Using a full period ensures that the Fourier series representation accurately captures the behavior of the function.

## 5. Can Sin Coeffs be calculated for any function?

Yes, Sin Coeffs can be calculated for any function that is periodic and satisfies certain mathematical criteria. However, the convergence of the Fourier series representation may vary for different functions.

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