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 $b_n$.