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...
So I currently have the following: (All B.S.)
A degree in Math specializing in Graduate studies
A degree in Physics specializing in medical physics
and currently pursuing a degree in chemistry.
I am pursuing the third because I have a pattern in my classes.
Intro level with little...
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.
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...