Converge pointwise with full Fourier series

[A.Magnus]

I am working on a simple PDE problem on full Fourier series like this:

Given this piecewise function,

$f(x) = \begin{cases} e^x, &-1 \leq x \leq 0 \\ mx + b, &0 \leq x \leq 1.\\ \end{cases}$​

Without computing any Fourier coefficients, find any values of $m$ and $b$, if there is any, that will make $f(x)$ converge pointwise on $-1 < x < 1$ with its full Fourier series.

I know for sure that if $f(x)$ is to converge pointwise with its full Fourier series, then $f(x)$ has to be piecewise smooth, meaning that each piece of $f(x)$ has to be differentiable.

(a) Is this the right way to go?
(b) If it is, how do you prove $e^x$ and $mx + b$ differentiable? By proving $f'(x) = \lim_{x \to c}\frac{f(x) - f(c)}{x - c}$ exists?

Thank you for your time.

 

[Svein]

1. f(x) has to be continuous at x = 0.
2. f'(x) has to be continuous at x = 0.

 

[A.Magnus]

1. f(x) has to be continuous at x = 0.
2. f'(x) has to be continuous at x = 0.

I think I am confused with the word "piecewise smooth." I had always thought it means "smooth piece by piece," meaning that $f(x) = e^x$ is smooth individually and then the next $f(x) = mx +b$ is smooth individually also. But your response implies that both parts of $f(x)$ have to be smooth as one big piece. So I am wrong on this? Let me know and thank you!

 

[Dick]

I think I am confused with the word "piecewise smooth." I had always thought it means "smooth piece by piece," meaning that $f(x) = e^x$ is smooth individually and then the next $f(x) = mx +b$ is smooth individually also. But your response implies that both parts of $f(x)$ have to be smooth as one big piece. So I am wrong on this? Let me know and thank you!

No, you are right. That means there are no conditions on m and b. Notice there is a difference between saying "the series converges pointwise" and "the series converges pointwise to f(x)". If it's the latter you have a condition.

 

[A.Magnus]

No, you are right. That means there are no conditions on m and b. Notice there is a difference between saying "the series converges pointwise" and "the series converges pointwise to f(x)". If it's the latter you have a condition.

What do you mean by "there are no conditions on $m$ and $b$"? Thanks. [Nice to see you again! See, I had to tend one course after another! :-) ]

 

[Dick]

What do you mean by "there are no conditions on $m$ and $b$"? Thanks. [Nice to see you again! See, I had to tend one course after another! :-) ]

I mean that it's piecewise smooth no matter what m and b are. Nice to see you!

 

[A.Magnus]

I mean that it's piecewise smooth no matter what m and b are. Nice to see you!

Thanks! I think it means $m, b$ are good for any real numbers. You are always omniscience from A to Z, omnipresent, and omni-helpful, if that is the right word.

 

[Ray Vickson]

I am working on a simple PDE problem on full Fourier series like this:

Given this piecewise function,

$f(x) = \begin{cases} e^x, &-1 \leq x \leq 0 \\ mx + b, &0 \leq x \leq 1.\\ \end{cases}$​

Without computing any Fourier coefficients, find any values of $m$ and $b$, if there is any, that will make $f(x)$ converge pointwise on $-1 < x < 1$ with its full Fourier series.

I know for sure that if $f(x)$ is to converge pointwise with its full Fourier series, then $f(x)$ has to be piecewise smooth, meaning that each piece of $f(x)$ has to be differentiable.

(a) Is this the right way to go?
(b) If it is, how do you prove $e^x$ and $mx + b$ differentiable? By proving $f'(x) = \lim_{x \to c}\frac{f(x) - f(c)}{x - c}$ exists?

Thank you for your time.

Your statement " ... $f(x)$ has to be piecewise smooth..." is false: it does not have to be piecewise smooth. It just has to obey the Dirichlet conditions; see, eg.,
http://en.wikipedia.org/wiki/Dirichlet_conditions . These do not involve smoothness or differentiablility.

So, with no restrictions on $m,b$ your function's Fourier series will converge pointwise on $-1 \leq x \leq 1$, and will converge to $f(x)$ for $-1 < x < 1, x \neq 0$. For some $m,b$ it will also converge to $f(0)$ when $x = 0$, but for some other choices of $m,b$ it will converge to something else at $x = 0$ (but still converge).

 

[LCKurtz]

@A.Magnus: I would almost bet that the original problem wants you to find m and b such that the FS converges pointwise to f(x). Otherwise there isn't much point to the problem. That would require specific values of m and b.

 

[A.Magnus]

@A.Magnus: I would almost bet that the original problem wants you to find m and b such that the FS converges pointwise to f(x). Otherwise there isn't much point to the problem. That would require specific values of m and b.

I have uploaded the page that has the original problem 9, see the attached file. The text is "Introduction to Applied PDE" by John Davis, let me know if I got it very wrong in the first place, I will happily stand to be corrected. Also do let me know how should I go ahead if I was wrong. Thank you!

PS:The text is extremely cut and dry, on top of that this is an online class, we get only reading assignments and homework, no lectures. Never complaining, so I take this site as crowd-teaching forum!

 

[Ray Vickson]

I have uploaded the page that has the original problem 9, see the attached file. The text is "Introduction to Applied PDE" by John Davis, let me know if I got it very wrong in the first place, I will happily stand to be corrected. Also do let me know how should I go ahead if I was wrong. Thank you!

PS:The text is extremely cut and dry, on top of that this is an online class, we get only reading assignments and homework, no lectures. Never complaining, so I take this site as crowd-teaching forum!

The pdf displays upside-down on my screen, and I cannot rotate it (and so cannot read it). Anyway, have you read post #8?

 

[A.Magnus]

The pdf displays upside-down on my screen, and I cannot rotate it (and so cannot read it). Anyway, have you read post #8?

Yes, I did see #8, I am about to response. For the file, I will attached another one, give me just a second. Thanks, Ray!

 

[A.Magnus]

The pdf displays upside-down on my screen, and I cannot rotate it (and so cannot read it). Anyway, have you read post #8?

Ray, here is the corrected file. Feel free to

crowd-teach me. Thanks.

 

[A.Magnus]

Your statement " ... $f(x)$ has to be piecewise smooth..." is false: it does not have to be piecewise smooth. It just has to obey the Dirichlet conditions; see, eg.,
http://en.wikipedia.org/wiki/Dirichlet_conditions . These do not involve smoothness or differentiablility.

So, with no restrictions on $m,b$ your function's Fourier series will converge pointwise on $-1 \leq x \leq 1$, and will converge to $f(x)$ for $-1 < x < 1, x \neq 0$. For some $m,b$ it will also converge to $f(0)$ when $x = 0$, but for some other choices of $m,b$ it will converge to something else at $x = 0$ (but still converge).

Ray, here is what I copy down verbatim from the John Davis' text, page 88:

Theorem 3.2 (Pointwise Convergence of Fourier Series).
If $f$ is piecewise smooth on $(-l, l)$, then the Fourier series of $f$ given by the above (3.8) converges pointwise on $(-l, l)$ and

$\frac{1}{2} a_0 + \sum_{n=1}^{\infty} [a_n \cos (n \pi x/l) + b_n \sin(n \pi x/l] = \frac{f(x^+) + f(x^-)}{2}, \quad x \in (-l, l).$

Here, $f(x^+) := \lim_{w \to x^+} f(w)$ and $f(x^-) := \lim_{w \to x^-} f(w)$, and (3.8) is referring to this: $f(x) = \frac{1}{2} a_0 + \sum_{n=1}^{\infty} [a_n \cos (n \pi x/l) + b_n \sin(n \pi x/l], \quad -l < x < l.$

Let me know what I got wrong. Thanks again and again.