Show that this is not a Sturm-Liouville problem

[mathmari]

To clarify something.. To show that the set of the eigenfunctions is not complete, by showing that not each function can be written as a linear combination of the eigenfunctions ( without using the sentence of square-integrable functions), do I have to show that a function cannot be written as:
$$f(x)= \sum_{n=1}^{\infty}{c_n \sin{(2 n \pi x)}}$$
or
$$f(x)= Cx+\sum_{n=1}^{\infty}{c_n \sin{(2 n \pi x)}}$$
?? (Wondering)

I mean that although the function $y(x)=x$ and $y_n(x)=\sin{(2 n \pi x)}$ are not orthogonal, do I have to use both of them to show that the set of the eigenfunctions is not complete? Or do I have to use just the orthogonal ones?

 

[I like Serena]

It means that the integral of the square of the absolute value of the function is finite, right?

Yes.
It's a little more specific though.
This will be required to properly define an inner product.

In your case, I guess the inner product is:
$$(v,u) = \int_0^1 v^*(x)u(x) dx$$
This inner product is only properly defined for functions that are square integrable on the domain.
That is why the functions have to be square integrable.
And then we can talk about a basis of functions and whether that basis is orthogonal and/or complete.

The execrise is:"...show that the eigenfunctions are orthogonal, but the eigenfunctions do not form a complete set". Since the eigenfunction $x$ is not orthogonal with the eigenfunctions $\sin{(2 n \pi x)}$, do I have to show maybe only that the set $\{ \sin{(2 n \pi x)} \}$ is not complete, as I did the post #22?

To clarify something.. To show that the set of the eigenfunctions is not complete, by showing that not each function can be written as a linear combination of the eigenfunctions ( without using the sentence of square-integrable functions), do I have to show that a function cannot be written as:
$$f(x)= \sum_{n=1}^{\infty}{c_n \sin{(2 n \pi x)}}$$
or
$$f(x)= Cx+\sum_{n=1}^{\infty}{c_n \sin{(2 n \pi x)}}$$
?? (Wondering)

I mean that although the function $y(x)=x$ and $y_n(x)=\sin{(2 n \pi x)}$ are not orthogonal, do I have to use both of them to show that the set of the eigenfunctions is not complete? Or do I have to use just the orthogonal ones?

As you have noticed, the phrasing of the exercise seems to be off.
We already found that the eigenfunctions are not orthogonal.
And if they were, it would actually be a (regular?) Sturm-Liouville problem.

So we're in the guessing arena here, which is very un-mathematical.

Can you perhaps ask your professor for clarification?

 

[mathmari]

Since $n$ is an integer they are indeed real.
That should be, according to your definition of the inner product (formally):
$$(y_m,y_n)=\int_0^1{y_m^* y_n}dx=\int_0^1{y_m y_n}dx$$

What is $\sin(2 \pi k )$ if $k$ is an integer?

Does it stand that
$$\int_0^1{y_m^* y_n}dx=\int_0^1{y_m y_n}dx$$
because the eigenfunctions $y_m$ are real?? (Wondering)

Let $f(x)=\cos{(2 \pi x)}-1$, $f(0)=0, f'(0)=f'(1)$.

$$\cos{(2 \pi x)}-1= \sum_{n=1}^{+\infty}{c_n \sin{(2 n \pi x)}}$$

$$c_n= \frac{2}{1} \int_0^{1}{[ \cos{(2 \pi x)}-1] \sin{({2 n \pi x})}}dx= \dots =0 $$

To calculate the coefficients $c_n$ with the Fourier series, do we have to expand the function $f$ in an odd way?? (Wondering)

 

[I like Serena]

Does it stand that
$$\int_0^1{y_m^* y_n}dx=\int_0^1{y_m y_n}dx$$
because the eigenfunctions $y_m$ are real?? (Wondering)

Yep! (Mmm)

To calculate the coefficients $c_n$ with the Fourier series, do we have to expand the function $f$ in an odd way?? (Wondering)

But... but... $f$ is not odd but even.
So there is no sine expansion, since a sine expansion is always odd. (Sweating)

 

[mathmari]

But... but... $f$ is not odd but even.
So there is no sine expansion, since a sine expansion is always odd. (Sweating)

How can I find then the coefficients using the Fourier series?? (Wondering)

 

[mathmari]

Yep! (Mmm)

How can we be sure that $\sin{(2 n \pi x)}$ is real?? Can it not be also complex?? (Wondering)

 

[I like Serena]

How can I find then the coefficients using the Fourier series?? (Wondering)

The Fourier expansion of some $f(x)$ is:
$$f(x)=A_0 + \sum A_k \cos(2\pi k x) + \sum B_k \sin(2\pi k x)$$
This expansion is unique.

The function:
$$\cos(2\pi x) - 1$$
is already in this form - this is the Fourier expansion!
Trying to find a coefficient for some $\sin(2\pi n x)$ can only result in zero. (Wasntme)

How can we be sure that $\sin{(2 n \pi x)}$ is real?? Can it not be also complex?? (Wondering)

This thread is a bit long... I've lost the reference to $x$. Can it be imaginary? (Thinking)

 

[mathmari]

The Fourier expansion of some $f(x)$ is:
$$f(x)=A_0 + \sum A_k \cos(2\pi k x) + \sum B_k \sin(2\pi k x)$$
This expansion is unique.

The function:
$$\cos(2\pi x) - 1$$
is already in this form - this is the Fourier expansion!
Trying to find a coefficient for some $\sin(2\pi n x)$ can only result in zero. (Wasntme)

We can write the function $\cos(2\pi x) - 1$ as a sum of the eigenfunctions,
$$\cos(2\pi x) - 1=\sum_{n=1}^{\infty}c_n \sin{(\frac{n \pi x}{L})}$$

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

Do we have now to compare the coefficients of these two series to calculate $c_n$?? (Wondering)

This thread is a bit long... I've lost the reference to $x$. Can it be imaginary? (Thinking)

At the probem $y''+\lambda y=0$, $$0 \leq x \leq L$$ How can I know if it can be also imaginary?? (Wondering)

 

[I like Serena]

We can write the function $\cos(2\pi x) - 1$ as a sum of the eigenfunctions,
$$\cos(2\pi x) - 1=\sum_{n=1}^{\infty}c_n \sin{(\frac{n \pi x}{L})}$$

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

Do we have now to compare the coefficients of these two series to calculate $c_n$?? (Wondering)

Yes. That would work. (Mmm)

At the probem $y''+\lambda y=0$, $$0 \leq x \leq L$$ How can I know if it can be also imaginary?? (Wondering)

The expression $0 \leq x \leq L$ implies that $x$ is not imaginary, because $\le$ is not defined for imaginary numbers.
Therefore $x$ is a real number. (Wasntme)

 

[mathmari]

Yes. That would work. (Mmm)

But how can I compare these two series since at the one there is $(\frac{2 n \pi x}{L})$ at the $\sin$ and at the other series there is $(\frac{ n \pi x}{L})$ ? (Wondering)

The expression $0 \leq x \leq L$ implies that $x$ is not imaginary, because $\le$ is not defined for imaginary numbers.
Therefore $x$ is a real number. (Wasntme)

Ahaa! Ok! I got it! (Mmm) (Yes)

 

[I like Serena]

But how can I compare these two series since at the one there is $(\frac{2 n \pi x}{L})$ at the $\sin$ and at the other series there is $(\frac{ n \pi x}{L})$ ? (Wondering)

Which series do you get if you write out, say, the first 3 terms of each summation? (Thinking)

Ahaa! Ok! I got it! (Mmm) (Yes)

Good! (Mmm)