Series of function...

**Also sprach Zarathustra** · June 27th 2010, 01:38 PM

Let, $f_n(x)=xarctg(nx)$ when $0\leq x<\infty$ .

I need to find:
1. Limit function (Which I found: $\frac{\pi}{2}x$ )

2. Checking the unitary convergence. (So, here is my problem)

Thank you all!

**roninpro** · June 27th 2010, 03:28 PM

Be careful! The limit function is not exactly $\frac{\pi}{2}x$ . Make sure you consider two cases: $x\geq 0$ and $x< 0$ .

Also, what do you mean by "unitary convergence"? Do you mean "uniform convergence"?

**Also sprach Zarathustra** · June 27th 2010, 03:59 PM

Yes, "uniform convergence".
We don't need x<0. x is always grater than 0 (or x=0)

Thank you!

**roninpro** · June 27th 2010, 06:52 PM

Oops, sorry about that. I didn't see the domain restriction.

For uniform convergence, did you try looking at the definition? Given $\varepsilon>0$ , is there an $N$ large enough so that $n>N$ implies $\left|x\arctan (nx)-\frac{\pi}{2}x\right|<\varepsilon$ for all $x\geq 0$ ?

**mohammadfawaz** · June 27th 2010, 11:01 PM

Note that $\frac{d}{dx}(x.arctan(n x)) = \frac{nx}{n^2 x^2+1}+arctan(n x)$ .
Look for the value of x for which the derivative of $|x.arctan(nx)-\frac{\pi}{2}x|$ (for x positive) is zero and draw the table of variations. Now, find the maximum value of that quantity over th given domain. You have uniform convergence if and only if the limit of the value you get is zero.

hope this helps

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