Recent content by sr3056

Thanks that's really helpful. Is it because the ones that don't exist aren't going to zero fast enough? And what about non-negative powers - are there any special cases where these can be integrated to infinity that I should be aware of? e.g. ∫x2

But 1/x2 etc. is ok?

I want to integrate γβγx/(β + x)γ+1 from 0 to ∞ (given β and γ are both > 0) So for large x the integrand is approximately proportional to x-γ So for which values of γ is the integral defined? Surely for any γ > 0 the integrand tends to zero as x tends to infinity? Thanks

Thanks. Is there no way of proving convergence from the Leibniz formula though?

Thank you very much

Thank you for your responses. Inverting, I get e(x/x-1) < 1 - x < e-x Letting x = 1/60 I then get e-1/59 < 59/60 < e-1/60 This is starting to look a bit more like it, but I'm not sure how my f(60) and f(infinity) values are going to come in...

Given f(n) = (1 - (1/n))n I calculate that the limit as n -> infinity is 1/e. Also given that x/(1-x) > -log(1-x) > x with 0<x<1 (I proved this in an earlier part of the question) I want to show that: 1 > (f(60)/f(infinity)) > e-1/59 > 58/59 I have tried using my value for f...

Given f(x) = xe-x2 I can differentiate once and use Leibniz to show that for n greater than 1 f(n) = -2nf(n-2) - 2xf(n-1) I want to show that the Maclaurin series for f(x) converges for all x. At x = 0, the above Leibniz formula becomes f(n) = -2nf(n-2) I know that f(0) = zero so...

I think I've got it now.. Let ζ = u+iv so u²+v²=1 because |ζ| = 1 2/ζ + ζ = 2 / (u+iv) + (u+iv) = 2(u−iv) / (u²+v²) + (u+iv) = 3u−iv ∴ x+iy = 3u−iv and so u=x/3, v=−y From u²+v² = 1 this yields (x/3)²+y² = 1, an ellipse Thanks for your help

I want to show that if the complex variables ζ and z and related via the relation z = (2/ζ) + ζ then the unit circle mod(ζ) = 1 in the ζ plane maps to an ellipse in the z-plane. Then if I write z as x + iy, what is the equation for this ellipse in terms of x and y? Any help would be...