Uniform convergence

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nayel
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Uniform convergence

Unread post by nayel » Fri Oct 21, 2011 4:45 am

Determine whether the sequence of functions $\displaystyle f_n(x)=e^{-x^2}\sin\frac xn$ converges uniformly on $\mathbb R$.
"Everything should be made as simple as possible, but not simpler." - Albert Einstein

tanvirab
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Re: Uniform convergence

Unread post by tanvirab » Fri Oct 21, 2011 10:05 am

How about this :

for any $\epsilon$ we can choose $x_0$ = $inf\{x : e^{-x^2} < \epsilon \}$
now we can choose $n$ large enough so that $sin(\frac{x}{n}) < \epsilon$ for all $|x| \leq x_0$

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nayel
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Re: Uniform convergence

Unread post by nayel » Fri Oct 21, 2011 5:56 pm

Does it not need to hold for all $x\in\mathbb R$?
"Everything should be made as simple as possible, but not simpler." - Albert Einstein

tanvirab
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Re: Uniform convergence

Unread post by tanvirab » Fri Oct 21, 2011 6:03 pm

It does (right?)

For $|x| \leq x_0$ the $sin$ part is smaller than $\epsilon$ and for $|x|>x_0$ the $exp$ part is smaller than $\epsilon$. So the product is always smaller than $\epsilon$. (?)

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nayel
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Re: Uniform convergence

Unread post by nayel » Fri Oct 21, 2011 8:33 pm

That looks so easy compared to my one page long solution! :cry:
"Everything should be made as simple as possible, but not simpler." - Albert Einstein

tanvirab
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Location: Pasadena, California, U.S.A.

Re: Uniform convergence

Unread post by tanvirab » Sun Oct 23, 2011 4:51 am

:)


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