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
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
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...