Recent content by tt2348

Actually, i think this problem can be done without stirrings. For the top factorial, you have (2n-2) terms, giving a leading term of (2n)^(2n-2) for the expansion. Then the bottom will have (n^(n-2)+...)(n^n+...), giving a leading coefficient n^(2n-2)... This your limit becomes...

Homework Statement \lim _{ (x,y)\rightarrow (0^{ + },0^{ + }) }{ \frac { { e }^{ \sqrt { x+y } } }{ 4x+2y-5 } } Homework Equations eh. SO I did the problem. I usually sub 1/n for 0+ in most of these, but clearly the top goes to 1 from +inf, and the bottom goes to -5... Hence...

Just remember... \int _{ a }^{ b }{ f(x)dx } =F(b)-F(a) so let's call \int _{ 8 }^{ y }{ g(t)dt } =G(y)-G(8) where g(t)=\frac { 1 }{ 1+ t^{2}+sin^{2}t } then \int _{ 15 }^{ x }{ \int _{ 8 }^{ y }{ g(t)dt } } =\int _{ 15 }^{ x }{ G(y)-G(8){dy} } since the last term is a constant, we get...

I'm assuming Sp{x,y} means span of. Is invertibility allowed? Since given [x y] invertible, you can show that (0,1) and (1,0) are in the span of {x,y} and thus all combinations of (0,1) and (1,0) are in the span (R2)

I think assuming that the inf|P(x)|=/=inf(P(x)) in general is correct. unless there is a strict restrictions of P(x). you'll have two cases of x* that p(x*)=0 implies x* is in inf{|p(x)|} or the inf is the same.

I'd say separate your polynomial into it's even and odd powers. IE P2 is all even powered x (including a0) and P1 is your odd, with some type of uniform bound on the coefficients. like |ak|<M is this an analysis class? I assume so. we know P2>P1 at large x, and extremely small -x ( like -10^10...

I have a better idea.. Instead of going through and trying to formulate each of these piece by piece, try expressing x=x(u,v) and y=y(u,v) For instance y^{2}=4u(u-x)=4v(v+x)\Rightarrow 4(u^2-v^2)-4x(u+v)=0 \Rightarrow (u^2-v^2)=x(u+v) so for x>0, and v≠-u≠0, we get x=u-v Using this, we find...

consider this instead...I_{n} = \sum_{i=0}^{n} [\frac{(b-a)i}{n}+a](\frac{b-a}{n}) = \frac{a(b-a)}{n} \sum_{i=0}^{n} 1 + \frac{(b-a)^2}{n^2} \sum_{i=0}^{n} i now we know adding 1 n times... is n, how about adding numbers from 1 to n? know any relevant formulas for that? once you simplify...

alright, but what does |(x,y)-(a,b)|<δ look like? literally draw the neighborhood on paper |x-a|<r and |y-b|<r ... would look like a square right? think about how you would formulate your neighborhood

I posted this in the number theory forum to no success... so I figured maybe the homework help people would have some input Let x,y,z be integers with no common divisor satisfying a specific condition, which boils down to 5|(x+y-z) and 2*5^{4}k=(x+y)(z-y)(z-x)((x+y)^2+(z-y)^2+(z-x)^2) or...

yes, it is indeed an exclusive or. But you need to be careful with your formulation F=>F is true also F=>T is true Generally, logical propositions alone are not enough to be mathematically rigorous, and most professors want something using significantly less notation.

or it's converse, Q(x)&R(X)=> x isn't an integer

Using logical restrictions.. you can drop the P(x) requirement. since regardless, even or odd, there is no integer such that x=4k-1=4j+1 so it boils down to x is an integer=>~(Q(x)&R(x)) =>(~q)v(~r)

Let x,y,z be integers satisfying a specific condition, which boils down to 5|(x+y-z) and 2*5^{4}k=(x+y)(z-y)(z-x)((x+y)^2+(z-y)^2+(z-x)^2) or equivalently 5^{4}k=(x+y)(z-y)(z-x)((x+y-z)^2-xy+xz+yz) I want to show that GCD(x,y,z)≠1, starting with the assumption 5 dividing (x+y), (z-y), or...

did he not give you a specific treatment for continuity in higher dimension? and thinking back to the one dimensional case... your neighborhood around a point a (|x-a|<e) was an interval... how would you define points around a single point in 2 dimensions? maybe even consider all the points in...