Relationships between Fourier coefficients

[binbagsss]

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.

 

[Ssnow]

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##?

 

[binbagsss]

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.

 

[Ray Vickson]

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.

 

[Ssnow]

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

 

[binbagsss]

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.

 

[binbagsss]

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?

 

[Ssnow]

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}## ...

 

[Ray Vickson]

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.

 

[binbagsss]

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## ? )

 

[SammyS]

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?

 

[binbagsss]

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?

 

[binbagsss]

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

 

[Ray Vickson]

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##.