Recent content by chwala

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) - f(x) | =1 ≠ 0 ## therefore is not uniform convergent. The concept is dependent on the domain, like for this example if we change the limits slightly to ##x∈...

Thanks @WWGD , getting analysis slowly, in any case analysis is not that difficult as i had earlier thought, ... For uniform convergence to apply, The limit of a sequence (as n approaches infinity) for ##f_n \rightarrow f =0##. In our case, it does not. The limit as n approaches infinity, for...

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 from my reading as you analyse for ##x## values that are close to 1, the convergence rate slows down. Case 2 ##x=1,## ##1^n → 1## Case 3 ##x=0,## ##0^n → 0##...

Doing some reading ... I want to analyse my problem further by firstly looking at existence and uniqueness. Re-writing it as a first order; ##u_1=u## and ##u_2 =u^{\prime} ## ...i end up with ##u_2^{\prime\prime} = \dfrac{1}{x^2} u_1 - \dfrac{3}{x} u_2## The function is not continuous at...

ok man, I read that. The only thing remaining is for me to get to learn some analysis on this type of problem...

I’ve been exploring the limits of Cauchy–Euler type methods and constructed the following equation: ##4x^5 u'' + 12x^2 u' - 4u = 0## At first glance it resembles an Euler equation, but the usual substitution (u = x^m) does not reduce it to a constant characteristic equation. I’m interested in...

##10^y dy = \dfrac{dx}{x\ln 10}## ##\int 10^y dy = \int \dfrac{dx}{x\ln 10}## ... ##10^y = \ln x +C## ##10^y = \ln (ax)## ##y=\lg(\ln(ax))##

With complex numbers, it makes it even easier,...

done.

for part i. i guess the quadratic equation was not indicated. I will go ahead and finish on that; The quadratic equation will be ##x^2 -p(p^2+3c)x^2-c^3=0##

a different line for ##\tan 4S## ##\tan (4S)= \tan (2S +2S) = \dfrac{\tan 2S + \tan 2S}{1 - \tan^2 (2S)}## ##\tan (2S) = \dfrac{5}{12}## therefore, ##\tan 4S= \left[ \dfrac{\dfrac{5}{12} + \dfrac{5}{12}}{1-\left[\dfrac{5}{12}\right]^2}\right]## ##\tan 4S=...

I note that in my proof, i did not have the ##4S## and instead worked with ##S## ... that was wrong. Your working clearly shows that error on my part. Cheers.