Recent content by AutumnWater

Thanks, that clears it all up for me now.

Thanks, I realize this is the case for double integral volumes and non volume triple integrals. But I was specifically thinking when the integrands are 1 for double integral areas and triple integral volumes with boundaries that are in the negative regions. Does this situation allow the (1...

So when finding the Area from a double integral; or Volume from a triple integral: If the curve/surface has a negative region: (for areas, under the x axis), (for volumes, below z = 0 where z is negative) What circumstances allow the negative regions to be taken into account as positive when...

Thanks, I'll have a look at those too. I'm currently reading Victor Bryant's "yet another introduction to analysis" in preparation for Rudin's text. When reading those kinds of texts, I sometimes feel guilty about looking at the solution too soon for their proof exercises, but other times feel...

Actually I didn't see the phrase "Note 8: Students in a double degree course can replace one level 1 science listed unit with a level 2 or level 3 science listed unit.". So that settles that, I just withdrew from MAT1830 and am enrolling in either ETC2410 or ETC2450, which should I do first? is...

Wow what do you know, it's a small world isn't it ;) really appreciate such elaborate advice. So MTH3051 before MTH3060, but should MTH3060 precede PDEs? And Complex analysis before Time series gotchya. So should the three 3rd year probability courses be done before or after Applied...

oops, the p^2/2 < epsilon earlier didn't include the $$1/(2^{3/2})$$ so if it was $$(1/(2^{3/2})) * (p^2/2) < epsilon ==> p^2/2 < 2sqrt(2)epsilon ==> p < 2 (2^{1/4})\sqrt(epsilon)$$ and $$\sqrt(5)p/2 < delta ==> p < 2(delta)/\sqrt(5)$$ make ps equal, and I got...

would it? whilst there will be a coefficient of $$1/(2^{3/2})$$ on the numerator of the f(x) only when doing the z substitution, the equation for delta wouldn't be p=sqrt(x^2 + y^2) anymore, instead it would turn out to be sqrt(x^2 + (z^2)/2) wouldn't it? so $$|(x^{2} + (z^{2})/2))^{1/2}| <=...

Ok thanks. Here I got 2 different deltas, would it be safe to say both are correct, only one is even more closer than the other one? If looking at the initial equation directly: We have: $$|(xy^3)/(x^2+2y^2)| = [(p^2*(sin(theta))^2*sin(2theta))/2(1+(sin(theta))^2)] <= (p^2)/2 < epsilon$$ so we...

yes I meant 1/(x^2+2y^2) <= 1/(2y^2) implies (xy^3)/(x^2+2y^2) <= (xy^3)/(2y^2)

so we have: f(x,y) = (xy^3)/(x^2+2y^2) since 1/(x^2+2y^2) <= 1/2y^2 f(x,y) <= (xy^3)/2y^2 where (xy^3)/2y^2 = xy/2 = p^2 sin(2theta) / 4 (let that = g(p,theta)) since sin(2theta) is between [-1, 1], -p^2/4 <= g(p,theta) <= p^2/4 and f(x,y) is <= g(p,theta) does that mean I should set delta =...

thanks, I'll go review my polar coordinates chapters first :P

Homework Statement Use the epsilon delta definition to show that lim(x,y) -> (0,0) (x*y^3)/(x^2 + 2y^2) = 0 Homework Equations sqrt(x^2) = |x| <= sqrt(x^2+y^2) ==> |x|/sqrt(x^2+y^2) <= 1 ==> |x|/(x^2+2y^2)? The Attempt at a Solution This limit is true IFF for all values of epsilon > 0, there...

sin(1/0) is undefined, but limit of sin(1/x) as x approach 0, should be 0, since f(a) does not equal lim x->a f(x), it's not continuous? to prove the limit: we need to find a delta such that: |x| < delta implies |sin(1/x)| < epsilon; since sin(1/x) is always between -1, to 1, |sin(1/x)| <= 1...

Alright gotcha thanks!