chwala's latest activity

Arg function depends on direction not magnitude of complex number as you got the same result with him. He made use of it for easy...

@Gavran ...you left out ##50## i.e in the denominator... can you see that? ought to be ...$$=\arg(0.5+0.5i)=\dfrac{π}{4}$$ Cheers.

The substitution ## y_1=10^y ## simplifies the equation $$ 10^ydy=\frac{dx}{x\ln10} $$ to the equation ## dy_1=dx/x ##. The general case...

This is getting to be a very boring thread. It's as if someone thinks it could be possible to prove this stuff actually wrong!

For ##f_n(\theta)=e^{-n\sin \theta} ## on ##[0,π]## three cases to consider. Case 1 consider interior values, ##\sin \theta >0##...

A generalization is to fix ##g: [0,1] \to \mathbb{R}## and let $$f_n : [0,1] \to \mathbb{R} : x \mapsto \begin{cases} ng(nx) & x < \frac...

Will definitely look at it...

I was going to append this to @chwala's thread here, but thought it deserved a new thread. For ##n \geq 1##, define $$f_n : [0,1] \to...

I meant for uniform convergence, this must hold ##\lim _{n→∞} sup |f_n (x) - f(x) | =0## and in our case, ##\lim _{n→∞} sup |f_n (x)...

Thanks @WWGD , getting analysis slowly, in any case analysis is not that difficult as i had earlier thought, ... For uniform convergence...

Notice the limit function is discontinuous, which shows convergence isn't uniform. The limit of a sequence ##f_n \rightarrow f##...

Consider also ##e^{-n\sin \theta}## for ##\theta \in [0, \pi]##.

You should try to prove the non-uniformness of the two examples @WWGD and I gave to see that you cannot choose N independently of the...

clearer now, For ##f(x)=x^n ## on ##[0,1]## three cases to consider. Case 1 consider ##0 ≤x <1## ##x^n → 0, ∀ 0 ≤x <1 ## further...

Note that ##\sin(\frac{n\pi}{2}) = 0## if ##n## is even and alternates between ##\pm1## if ##n## is odd. Using the alternating series...