Why is the right part of the Fourier series periodic with period L?

[mathmari]

Hey!

The Fourier series of $f$ is
$$f(x) \sim \frac{a_0}{2}+ \sum_{n=1}^{ \infty} {(a_n \cos{(\frac{2 n \pi x}{L})}+b_n \sin{(\frac{2 n \pi x}{L})})}$$

How do we know that the series of the right part of the above relation is periodic with period $L$?

 

[Evgeny.Makarov]

How do we know that the series of the right part of the above relation is periodic with period $L$?

Because each trigonometric function on the right has a period $L$.

 

[mathmari]

Because each trigonometric function on the right has a period $L$.

I got stuck... (Worried) How do we know that the period is $L$?

 

[I like Serena]

I got stuck... (Worried) How do we know that the period is $L$?

Hey! (Mmm)

Suppose we fill in $n=1$.
What does the right side of the expression look like then?
And what is its period?

 

[Evgeny.Makarov]

How do we know that the period is $L$?

Functions $\cos(2\pi nx/L)$ and $\sin(2\pi nx/L)$ have many periods, so $L$ is only a period. Checking that it is a period is done by definition and is straightforward. You know the definition, don't you?

 

[mathmari]

Suppose we fill in $n=1$.
What does the right side of the expression look like then?
And what is its period?

For $n=1$:
$\frac{a_0}{2}+ (a_1 \cos{(\frac{2 \pi x}{L})}+b_1 \sin{(\frac{2 \pi x}{L})})$

Is the period $L$, because $\cos{(\frac{2 \pi (x+L)}{L})}=\cos{(\frac{2 \pi x}{L}+2 \pi)}=\cos{(\frac{2 \pi x}{L})}$?

Functions $\cos(2\pi nx/L)$ and $\sin(2\pi nx/L)$ have many periods, so $L$ is only a period. Checking that it is a period is done by definition and is straightforward. You know the definition, don't you?

The definition is: $L$ is a period when $g(x+L)=g(x)$, isn't it?

 

[Evgeny.Makarov]

Yes to both questions.

 

[mathmari]

Nice! Thank you! (Smile)