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Thread: Infinitely many differenciable+compact support. Electromagnetism exercise

  1. #1
    MHF Contributor arbolis's Avatar
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    Infinitely many differenciable+compact support. Electromagnetism exercise

    Verify that the function $\displaystyle f_a(x)=0$ if $\displaystyle |x|\geq |a|$ and $\displaystyle f_a (x)= exp \{ \frac{-1}{x+a}+ \frac{-1}{x-a} \}$ if $\displaystyle |x|< |a|$, $\displaystyle x \in \mathbb{R}$ is infinitely many times differentiable and has a compact support.

    I have this exercise in an electromagnetism course and I never heard about compact support before. I've been told that if I can bound the function inside a sphere of a finite radius, then the function is of compact support. But reading the wikipedia article on it, I'm more lost than before.
    Furthermore if I take the definition of a limit of a function, $\displaystyle f'_a(x)$ doesn't seem definied in $\displaystyle a$.
    It might be some kind of distribution function and I never took any course on the subject so I'm really lost.
    Any help is appreciated.
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    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by arbolis View Post
    Verify that the function $\displaystyle f_a(x)=0$ if $\displaystyle |x|\geq |a|$ and $\displaystyle f_a (x)= exp \{ \frac{-1}{x+a}+ \frac{-1}{x-a} \}$ if $\displaystyle |x|< |a|$, $\displaystyle x \in \mathbb{R}$ is infinitely many times differentiable and has a compact support.

    I have this exercise in an electromagnetism course and I never heard about compact support before. I've been told that if I can bound the function inside a sphere of a finite radius, then the function is of compact support. But reading the wikipedia article on it, I'm more lost than before.
    Furthermore if I take the definition of a limit of a function, $\displaystyle f'_a(x)$ doesn't seem definied in $\displaystyle a$.
    It might be some kind of distribution function and I never took any course on the subject so I'm really lost.
    Any help is appreciated.
    It's clearly differentiable except possibly at $\displaystyle x=\pm a$. For example it's clear that $\displaystyle \displaystyle \lim_{x\to a^-}\frac{f_a(x)f_a(a)}{x-a}=\lim_{x\to a^-}(x-a)^{-1}\exp\left(\frac{-1}{x+a}+\frac{-1}{x-a}\right)=0$ (since as $\displaystyle x\to a$ one has that $\displaystyle \frac{-1}{x-a}\to -\infty$ so that the exponential goes to zero faster than the polynomial) which agrees with $\displaystyle \lim_{x\to a^+}\frac{f_a(x)-f_a(a)}{x-a}=\lim_{x\to a^+}\frac{0}{x-a}=0$ and so $\displaystyle f_a(x)$ is differentiable at $\displaystyle a$. The exact same methodology shows that it is differentiablea at $\displaystyle -a$. Now, for $\displaystyle f_a$ to have compact support means that $\displaystyle \overline{f_a^{-1}\left(\mathbb{R}-\{0\}\right)}$ is compact. But, sine $\displaystyle \overline{f_a^{-1}\left(\mathbb{R}-\{0\}\right)}=[-a,a]$ the result is clear here.
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  3. #3
    MHF Contributor arbolis's Avatar
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    Thanks a lot mathstud28. I did the calculation of the limit on my draft too fast... I looked only at the denominator and since it tends to 0 I thought the limit would be undefinied but as you say, the exponential tends to 0 faster than any polynomial and so the limit is well defined. Ok for the rest, I get it.
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