I need to check if sequence of functions $f_n(x):=\sqrt{x^2+\frac{1}{n}}$, $n\in \mathbb{N}$ converges (pointwisely, uniformly) in intervals:$[-1;1]$ and $\mathbb{R}$. Is there any algorithm how to solve this problem? Firstly I find its derivative, later maxima/minima and then what? Thanks for help
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You do not need to take any derivatives or find extrema. Can you guess which function $f_n$ converges to? Do you know the definition of pointwise and uniform convergence? – snar Sep 11 '14 at 19:47
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$x^2 + \frac{1}{n} \rightarrow x^2$, and the square root function is continuous on its domain, so certainly $f_n$ converges pointwise. (To what?) Once you know the limit, call it $f$, you can check uniform convergence by computing $\sup_x |f_n(x) - f(x)|$ and seeing if this sequence converges to zero as $n \rightarrow \infty$. – Sep 11 '14 at 19:49
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See here. – Mhenni Benghorbal Sep 11 '14 at 20:21
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Usually the steps go as follows:
Check for Pointwise convergence, by essentially just taking the limit with respect to n, while keeping $x$ fixed. In your example $\lim \sqrt (x^2 + 1/n) = \sqrt (x^2)=x$ or $-x$ if $x<0$ This can also be seen because $f_n (x)$ is bounded from below by $x^2$ and is decreasing with respect to n.
Uniform convergence is a bit trickier, as it tests if the maximum difference (with respect to $x$) of $f_n$, and it's pointwise limit $f$ goes to $0$ as $n$ goes to infinity. In your example $\sup_{x\in[0,1]} ( \sqrt (x^2 + 1/n) - x)$ or $\sup_{x\in[-1,0]} ( \sqrt (x^2 + 1/n) + x)$ One way to do this is to find the derivative of this new function, as test the boundaries to find the maximum value. If you do this you'll get $x=0$ which gives $ \sqrt (1/n))$ which goes to $0$ as $n\rightarrow \infty$. Hence it converges uniformly to $x$ or $-x$.
Emil Kerimov
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