# Fourier Series Coefficient Symmetries

• Marcus95
In summary: It is also convenient to write ##\sum_{k=0}^{\infty}a_{2k+1}\cos(2k+1)x## as ##\sum_{k=0}^{\infty}a_{2k+1}\cos(2k+1)x=\sum_{k=0}^{\infty}(a_{2k+1}\cos(2k+1)x+b_{k+1}\sin(2k+1)x)-b_1/2##. It is then convenient to write ##\sum_{k=0}^{\infty}b_{2k+2}\sin(2k+2)x## as ##\sum
Marcus95

## Homework Statement

Let ## f(x) = \frac{a_0}{2} + \sum_{n=1}^{\infty} (a_n \cos nx + b_n \sin nx) ##
What can be said about the coefficients ##a_n## and ##b_n## in the following cases?
a) f(x) = f(-x)
b) f(x) = - f(-x)
c) f(x) = f(π/2+x)
d) f(x) = f(π/2-x)
e) f(x) = f(2x)
f) f(x) = f(-x) = f(π/2-x)

## Homework Equations

Sine is odd and cosine is even. Even functions can only contain cosine terms and odd functions only sine terms in their Fourier Series.

## The Attempt at a Solution

Here is how far I have gotten:
a) All b=0 because even function
b) All a=0 because odd function
c) Funtion has period π/2 (?)
d) Funtion has period π/2-2x (?)
e) Function has period x (not neccessarily the shortest period)
f) All b=0 because even function.

But here I am stuck, because I have no idea how to relate the coefficients to the period in c)-f), and in c)-e) I don't know if the funtion is odd or even.
Many thanks for any help! :)

Marcus95 said:

## Homework Statement

Let ## f(x) = \frac{a_0}{2} + \sum_{n=1}^{\infty} (a_n \cos nx + b_n \sin nx) ##
What can be said about the coefficients ##a_n## and ##b_n## in the following cases?
a) f(x) = f(-x)
b) f(x) = - f(-x)
c) f(x) = f(π/2+x)
d) f(x) = f(π/2-x)
e) f(x) = f(2x)
f) f(x) = f(-x) = f(π/2-x)

## Homework Equations

Sine is odd and cosine is even. Even functions can only contain cosine terms and odd functions only sine terms in their Fourier Series.

## The Attempt at a Solution

Here is how far I have gotten:
a) All b=0 because even function
b) All a=0 because odd function
c) Funtion has period π/2 (?)
d) Funtion has period π/2-2x (?)
e) Function has period x (not neccessarily the shortest period)
f) All b=0 because even function.

But here I am stuck, because I have no idea how to relate the coefficients to the period in c)-f), and in c)-e) I don't know if the funtion is odd or even.
Many thanks for any help! :)

Use the identities for sin and cos:
## sin(x+y)=sin(x)cos(y)+sin(y)cos(x) ##
## cos(x+y)=cos(x)cos(y)-sin(x)sin(y) ##

for the different for terms in the sum. What is happening with ##a_n## and ##b_n## for odd and even values of ##n##?
By putting different values for ##y## you can study the different cases in c), d), e) and f).

Marcus95
eys_physics said:
Use the identities for sin and cos:
## sin(x+y)=sin(x)cos(y)+sin(y)cos(x) ##
## cos(x+y)=cos(x)cos(y)-sin(x)sin(y) ##

for the different for terms in the sum. What is happening with ##a_n## and ##b_n## for odd and even values of ##n##?
By putting different values for ##y## you can study the different cases in c), d), e) and f).

Alright, I tried using the formulas in case c):
##\sin(\pi/2 +x) = \cos x##
##\cos(pi/2 +x) = -\sin x##
which is as expected. However, I am confused about how this should be applied to the problem. We have a general function ##f(x)## and I can't really see how the above properties reveal anything about the coefficients.

Marcus95 said:
Alright, I tried using the formulas in case c):
##\sin(\pi/2 +x) = \cos x##
##\cos(pi/2 +x) = -\sin x##
which is as expected. However, I am confused about how this should be applied to the problem. We have a general function ##f(x)## and I can't really see how the above properties reveal anything about the coefficients.

Yes, but this is only valid for ##n=1##. What do you get for ##cos n(\pi/2+x)## and ##sin n(\pi/2+x)##, for ##n## odd and even? You can then write the Fourier serie for ##f(\pi/2+x)## and simplify the sum. An advice is to split into odd and even ##n##. I hope this helps.

Marcus95
eys_physics said:
Yes, but this is only valid for ##n=1##. What do you get for ##cos n(\pi/2+x)## and ##sin n(\pi/2+x)##, for ##n## odd and even? You can then write the Fourier serie for ##f(\pi/2+x)## and simplify the sum. An advice is to split into odd and even ##n##. I hope this helps.
Alrigth, I feel like the solution is close but I am still not really there. I have found the following:
##\sin n(\pi/2 +x) = \pm \cos nx## for odd n
##\sin n(pi/2 +x) = \pm \sin nx## for even n
##\cos n(\pi/2 +x) = \pm \cos nx## for even n
##\cos n(pi/2 +x) = \pm \sin nx## for odd n

But to me these are still "just" some trigonometric identities which I don't know how to apply to the function.
## f(x) = \frac{a_0}{2} + \sum_{n=1}^{\infty} (a_n \cos nx + b_n \sin nx) ##

Marcus95 said:
Alrigth, I feel like the solution is close but I am still not really there. I have found the following:
##\sin n(\pi/2 +x) = \pm \cos nx## for odd n
##\sin n(pi/2 +x) = \pm \sin nx## for even n
##\cos n(\pi/2 +x) = \pm \cos nx## for even n
##\cos n(pi/2 +x) = \pm \sin nx## for odd n

But to me these are still "just" some trigonometric identities which I don't know how to apply to the function.
## f(x) = \frac{a_0}{2} + \sum_{n=1}^{\infty} (a_n \cos nx + b_n \sin nx) ##
'
You can write the above relations as
##\sin n(\pi/2+x)=(-1)^{(n-1)/2}\cos nx ##
##\sin n(pi/2 +x) = (-1)^{n/2} \sin nx## for even n
##\cos n(\pi/2 +x) = (-1)^{(n-1)/2} \cos nx## for even n
##\cos n(pi/2 +x) =-(-1)^{n/2} \sin nx## for odd n

Furthermore, we can write
##f(\pi/2+x)=\frac{a_0}{2} + \sum_{n=1}^{\infty} (a_n \cos n(\pi/2+x) + b_n \sin n(\pi/2+x)) ##,

Can you now see how to use the above trignometric identies?

It is also convenient to split the sum into odd and even values of ##n##, i.e. something like ##\sum_{n=1}^{\infty}=\sum_{odd\: n}+\sum_{even\: n}##. It is then convenient to introduce a variable ##k## through ##n=2k+1## (for the sum over odd ##n##) and similarly ##n=2k## (for even n).

Marcus95
eys_physics said:
'
Can you now see how to use the above trignometric identies?

It is also convenient to split the sum into odd and even values of ##n##, i.e. something like ##\sum_{n=1}^{\infty}=\sum_{odd\: n}+\sum_{even\: n}##. It is then convenient to introduce a variable ##k## through ##n=2k+1## (for the sum over odd ##n##) and similarly ##n=2k## (for even n).

Hmm, I think there migth be an algebraic miss in the exponents in your previous post. shouldn't it be:
##\sin n(\pi/2+x)=(-1)^{n/2}\cos nx ## for odd n
##\sin n(\pi/2 +x) = (-1)^{(n-1)/2} \sin nx## for even n
##\cos n(\pi/2 +x) = (-1)^{n/2} \cos nx## for even n
##\cos n(\pi/2 +x) =-(-1)^{(n-1)/2} \sin nx## for odd n?

Nevertheless,
##f(x) = f(\pi/2+x)=\frac{a_0}{2} + \sum_{n=1}^{\infty} (a_n' \cos n(\pi/2+x) + b_n' \sin n(\pi/2+x)) ##,
Supposed we have odd n, that would imply: ##a_n' \cos n(\pi/2 +x) = a_n'( -(-1)^{(n-1)/2} \sin nx)## and ##b_n' \sin n(\pi/2+x)= b_n'(-1)^{n/2}\cos nx ##. But exactly what is this telling us? Just that:
##a_n = b_n'(-1)^{n/2}## and ##b_n = -a_n'(-1)^{(n-1)/2)}##? (I use prime to note that it is the coefficients for ##f(\pi/2+x)##, and unprimed for ##f(x)##)

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Marcus95 said:
Hmm, I think there migth be an algebraic miss in the exponents in your previous post. shouldn't it be:
##\sin n(\pi/2+x)=(-1)^{n/2}\cos nx ## for odd n
##\sin n(\pi/2 +x) = (-1)^{(n-1)/2} \sin nx## for even n
##\cos n(\pi/2 +x) = (-1)^{n/2} \cos nx## for even n
##\cos n(\pi/2 +x) =-(-1)^{(n-1)/2} \sin nx## for odd n?

Nevertheless,
##f(x) = f(\pi/2+x)=\frac{a_0}{2} + \sum_{n=1}^{\infty} (a_n' \cos n(\pi/2+x) + b_n' \sin n(\pi/2+x)) ##,
Supposed we have odd n, that would imply: ##a_n' \cos n(\pi/2 +x) = a_n'( -(-1)^{(n-1)/2} \sin nx)## and ##b_n' \sin n(\pi/2+x)= b_n'(-1)^{n/2}\cos nx ##. But exactly what is this telling us? Just that:
##a_n = b_n'(-1)^{n/2}## and ##b_n = -a_n'(-1)^{(n-1)/2)}##? (I use prime to note that it is the coefficients for ##f(\pi/2+x)##, and unprimed for ##f(x)##)

Yes, I did a typo while writing my previous post. Thanks for pointing this out. First, first I don't think you should not introduce other coefficients for ##f(\pi/2+x)## since the function ##f## is the same on both sides. By comparing term by term the series for ##f(\pi/2+x)## and ##f(x)## you get as said
##-(-1)^{(n-1)/2}a_n=b_n## and ##(-1)^{(n-1)/2}b_n=a_n## if ##n## is odd. But this can only be satisfied if ##a_n=b_n=0##. Now, for even ##n## one has ##(-1)^{n/2}a_n=a_n## and ##(-1)^{n/2}b_n=b_n##. So, we must have ##(-1)^{n/2}=1## or ##a_n=b_n=0##. Therefore, we have only non-zero ##a_n## and ##b_n## for ##n## being a multiple of 4. We have thus
##f(x)=a'_0/2+\sum_{n=1}^{\infty}(a'_n\cos(4nx)+b'_n\sin(4nx))##, where
##a'_n=a_{4n}## and ##b'_n=b_{4n}##. As is discussed on the page http://mathworld.wolfram.com/FourierSeries.html this correspond to a function with period ##\pi/2## which you already concluded from the beginning. This was maybe a long way to see this. But, I hope you are now able to proceed with the other parts of the exercise.

eys_physics said:
This was maybe a long way to see this. But, I hope you are now able to proceed with the other parts of the exercise.
Thank you very much for the help! I managed to solve all the problems except for the case when ## f(x) = f(2x)##, because here the trigonometric relations are much more complicated with squares and mixed terms. Do you have any last advice for this case?

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## 1. What is a Fourier Series Coefficient Symmetry?

Fourier Series Coefficient Symmetry refers to the property that the coefficients in the Fourier series expansion of a function exhibit certain symmetries. These symmetries can be in the form of even or odd functions, half-wave symmetries, or quarter-wave symmetries.

## 2. How are Fourier Series Coefficient Symmetries useful in scientific research?

Fourier Series Coefficient Symmetries can provide valuable insights into the properties of a function, which can be useful in fields such as signal processing, image processing, and quantum mechanics. They can also help simplify complex calculations and provide a better understanding of the underlying mathematical structure of a function.

## 3. Can Fourier Series Coefficient Symmetries be used to approximate any function?

No, Fourier Series Coefficient Symmetries can only be used to approximate functions that exhibit certain symmetries. For example, functions that are discontinuous or have sharp corners may not have any symmetries and therefore cannot be approximated using Fourier series coefficients.

## 4. How do Fourier Series Coefficient Symmetries relate to the Fourier Transform?

The Fourier Transform is a generalization of the Fourier series, which allows for the analysis of non-periodic functions. Fourier Series Coefficient Symmetries are a special case of the Fourier Transform, where the function being analyzed is periodic. However, the concept of symmetries still applies in both cases.

## 5. Can different types of symmetries coexist in a function's Fourier series expansion?

Yes, a function's Fourier series expansion can exhibit multiple symmetries simultaneously. For example, a function can have both even and half-wave symmetries, which would result in both cosine and sine terms in its Fourier series coefficients.

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