# Relationships between Fourier coefficients

1. Dec 23, 2016

### binbagsss

1. The problem statement, all variables and given/known data

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 $b_n = a _{2n}$

2. Relevant equations

see above.

3. The attempt at a solution

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,

Last edited: Dec 23, 2016
2. Dec 23, 2016

### Ssnow

Hi, how is possible that the same function $f$ in one case has period $2\pi$ and in the other case $\pi$?

3. Dec 23, 2016

### binbagsss

edited. ta. typo.

4. Dec 23, 2016

### Ray Vickson

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: Dec 23, 2016
5. Dec 24, 2016

### 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

6. Dec 24, 2016

### binbagsss

okay thanks, that's clarified up some basics