Relationships between Fourier coefficients

In summary, the conversation discusses the relation between two functions, f(t) and g(t), given by infinite sums with different periods. The goal is to show that the coefficients of g(t) are equal to the coefficients of f(t) when the index is multiplied by 2. The conversation then goes on to discuss possible approaches to solving this problem, including using the orthogonality of the functions and integrating over different periods.
  • #1
binbagsss
1,325
12

Homework Statement



I have ## f(t) = \sum\limits^{\infty}_0 a_{n} e^{2 \pi i n t} ## [1]
and ## g(t) = \sum\limits^{\infty}_0 b_{n} e^{ \pi i n t} ## [2]

I want to show that ##b_n = a _{2n} ##

Homework Equations



see above.

The Attempt at a Solution


[/B]
So obviously you want to use the orthogonality to obtain the Fourier coeffients, integrating the LHS multiplied by the negative exponential over the period. Period in [1] is ##2\pi## and the period in [2] is ##\pi ##.

So I have:

from [1]: ##\int^{2 \pi } _{0} f(t) e^{- 2 \pi i n t} dt= a_{n} ##

and

from [2]: ##\int^{\pi } _{0} g(t) e^{- \pi i n t} dt = b_{n} ##

I'm unsure what to do next,
Many thanks in advance.
 
Last edited:
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  • #2
binbagsss said:
Period in [1] is 2π and the period in [2] is π

Hi, how is possible that the same function ##f## in one case has period ##2\pi## and in the other case ##\pi##?
 
  • #3
Ssnow said:
Hi, how is possible that the same function ##f## in one case has period ##2\pi## and in the other case ##\pi##?
edited. ta. typo.
 
  • #4
binbagsss said:

Homework Statement



I have ## f(t) = \sum\limits^{\infty}_0 a_{n} e^{2 \pi i n t} ## [1]
and ## f(t) = \sum\limits^{\infty}_0 b_{n} e^{ \pi i n t} ## [2]

I want to show that ##b_n = a _{2n} ##

Homework Equations



see above.

The Attempt at a Solution


[/B]
So obviously you want to use the orthogonality to obtain the Fourier coeffients, integrating the LHS multiplied by the negative exponential over the period. Period in [1] is ##2\pi## and the period in [2] is ##\pi ##.

So I have:

from [1]: ##\int^{2 \pi } _{0} f(t) e^{- 2 \pi i n t} dt= a_{n} ##

and

from [2]: ##\int^{\pi } _{0} f(t) e^{- \pi i n t} dt = b_{n} ##

I'm unsure what to do next,
Many thanks in advance.

If you have ##\pi##s in the exponentials you cannot have them in the integration limits. Anyway, in the integrations the exponents should go from 0 to ##2 \pi n##, so:
$$ (1) \; a_n = c_a \int_0^1 f(t) e^{-2 \pi i n t} \, dt\\
(2) \; b_n = c_b \int_0^2 f(t) e^{-i \pi n t} \, dt
$$
Here, ##c_a## and ##c_b## are some appropriate constants.

(1) describes a function having period 1, while (2) describes a function having period 2; but of course, 2 periods of length 1 make up 1 period of length 2.
 
Last edited:
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Likes Ssnow
  • #5
Hi, as @Ray Vickson noted sometimes the definition of the Fourier transform is with the factor ##2\pi## and other times is with ##\pi##, changing variable it is possible to pass from one notation to another ...

Ssnow
 
  • #6
Ray Vickson said:
If you have ##\pi##s in the exponentials you cannot have them in the integration limits. Anyway, in the integrations the exponents should go from 0 to ##2 \pi n##, so:
$$ (1) \; a_n = c_a \int_0^1 f(t) e^{-2 \pi i n t} \, dt\\
(2) \; b_n = c_b \int_0^2 f(t) e^{-i \pi n t} \, dt
$$
Here, ##c_a## and ##c_b## are some appropriate constants.

(1) describes a function having period 1, while (2) describes a function having period 2; but of course, 2 periods of length 1 make up 1 period of length 2.

okay thanks, that's clarified up some basics
so ##c_a = 1 ##, c_{b} = 1/2##
but i sitll don't know what to do, i want to show it explicitly.
 
  • #7
Ray Vickson said:
If you have ##\pi##s in the exponentials you cannot have them in the integration limits. Anyway, in the integrations the exponents should go from 0 to ##2 \pi n##, so:
$$ (1) \; a_n = c_a \int_0^1 f(t) e^{-2 \pi i n t} \, dt\\
(2) \; b_n = c_b \int_0^2 f(t) e^{-i \pi n t} \, dt
$$
Here, ##c_a## and ##c_b## are some appropriate constants.

(1) describes a function having period 1, while (2) describes a function having period 2; but of course, 2 periods of length 1 make up 1 period of length 2.

e.g so I send ##n## to ##2n## in [2] and then I need to look at how my integration limits are affected by this.
How would I go about this?
Can I do this - does that make sense to do?
 
  • #8
You must start write what is ##b_{2n}## that is ##=c_{b}\int_{0}^{2}f(t)e^{-i\pi 2nt}dt## and by substitutions try to write the integral in the form of ##a_{n}## ...
 
  • #9
binbagsss said:

Homework Statement



I have ## f(t) = \sum\limits^{\infty}_0 a_{n} e^{2 \pi i n t} ## [1]
and ## g(t) = \sum\limits^{\infty}_0 b_{n} e^{ \pi i n t} ## [2]

I want to show that ##b_n = a _{2n} ##

Homework Equations



see above.

The Attempt at a Solution


[/B]

Many thanks in advance.

Note that in exponential form the Fourier series runs from ##n = -\infty## to ##n = +\infty##, not just from ##0## to ##\infty##. So anyway, we have
$$0 = b_0 - a_0 + b_1 e^{i \pi t} + (b_2-a_1) e^{i \pi 2 t} + \cdots + \text{similar terms in } n < 0.$$
In other words, ##\sum_n c_n e^{i \pi n t} \equiv 0##, where ##c_n = b_n## for odd ##n## and ##c_{2m} = b_{2m} - a_m## for ##n = 2m## even.
 
  • #10
Ssnow said:
You must start write what is ##b_{2n}## that is ##=c_{b}\int_{0}^{2}f(t)e^{-i\pi 2nt}dt## and by substitutions try to write the integral in the form of ##a_{n}## ...

So a substitution like ## t* = t/2 ## would fix the limits to ##1## and ##0## as in ##a_{n}##, however I want to keep the exponent with a ##2 \pi ## factor don't I? (rather than ##\pi## ? )
 
  • #11
binbagsss said:

Homework Statement



I have ## f(t) = \sum\limits^{\infty}_0 a_{n} e^{2 \pi i n t} ## [1]
and ## g(t) = \sum\limits^{\infty}_0 b_{n} e^{ \pi i n t} ## [2]

I want to show that ##b_n = a _{2n} ##
Are ƒ(t) and g(t) related to each other in some fashion?
 
  • #12
Ray Vickson said:
If you have ##\pi##s in the exponentials you cannot have them in the integration limits. Anyway, in the integrations the exponents should go from 0 to ##2 \pi n##, so:
$$ (1) \; a_n = c_a \int_0^1 f(t) e^{-2 \pi i n t} \, dt\\
(2) \; b_n = c_b \int_0^2 f(t) e^{-i \pi n t} \, dt
$$
Here, ##c_a## and ##c_b## are some appropriate constants.

(1) describes a function having period 1, while (2) describes a function having period 2; but of course, 2 periods of length 1 make up 1 period of length 2.
should the second one be ##c_b \int_0^2 g(t) e^{-i \pi n t} \, dt ## ?? where does ## f(t) ## come from here?
 
  • #13
Ray Vickson said:
Note that in exponential form the Fourier series runs from ##n = -\infty## to ##n = +\infty##, not just from ##0## to ##\infty##. So anyway, we have
$$0 = b_0 - a_0 + b_1 e^{i \pi t} + (b_2-a_1) e^{i \pi 2 t} + \cdots + \text{similar terms in } n < 0.$$
In other words, ##\sum_n c_n e^{i \pi n t} \equiv 0##, where ##c_n = b_n## for odd ##n## and ##c_{2m} = b_{2m} - a_m## for ##n = 2m## even.

okay thanks that's making some sense, so you have done that ##g(t)-f(t)=0##, and equated coefficients, however, I do not understand why this is ##0##...

many thanks
 
  • #14
binbagsss said:
okay thanks that's making some sense, so you have done that ##g(t)-f(t)=0##, and equated coefficients, however, I do not understand why this is ##0##...

many thanks

Look at my post #4, where you see your original question echoed in a panel along with my response. You will see that originally you had
$$ (1) \; f(t) = \sum_{0}^{\infty} a_{n} e^{2 \pi i n t} \\
(2) \; f(t) = \sum_{0}^{\infty} b_{n} e^{ \pi i n t}
$$
When did the second ##f(t)## become a ##g(t)##?

It is obvious that if ##f## and ##g## are just two unrelated functions there cannot be any predictable relationship between the ##a_{2n}## and ##b_n##.
 

FAQ: Relationships between Fourier coefficients

What are Fourier coefficients?

Fourier coefficients are mathematical representations of the amplitudes and phases of the different frequency components that make up a periodic function.

How are Fourier coefficients related to relationships between two functions?

The Fourier coefficients of two functions can be compared to determine the similarity or difference between them. If the coefficients are similar, it indicates a strong relationship between the two functions.

What is the significance of the magnitude and phase of Fourier coefficients?

The magnitude of the Fourier coefficients represents the amplitude of a specific frequency component in the function, while the phase represents the position of that component within the function. Together, they provide a complete picture of the frequency composition of a function.

Can Fourier coefficients be used to analyze non-periodic functions?

Yes, Fourier coefficients can also be used to analyze non-periodic functions by extending the concept to the Fourier transform, which allows for the analysis of non-periodic signals as well as signals with changing frequencies.

Are there any practical applications of relationships between Fourier coefficients?

Yes, the relationships between Fourier coefficients have many practical applications in fields such as signal processing, image and audio compression, and data analysis. They are also used in various engineering and scientific fields to analyze and understand complex systems and phenomena.

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