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when i do that i get (1/sqrt(pi))*[-e^(2ixy-y^2) + integral((e^-(y^2))*2ix*e^(2ixy))dy]. Now what?

Well since the original integral is with respect to y and he wants me to differentiate the inside just with respect to x I would just differentiate e^2ixy which would give me 2iy*e^2ixy. Putting this all together we have... (1/sqrt(pi))*integral[2iy*(e^(2*ixy)*e^(-y^2)], y = -infinity ...

Thanks. But my mentor told me to do it a different way for some reason. He says to differentiate the inside of (1/sqrt(pi))*integral[(e^(2*ixy)*e^(-y^2)], y = -infinity .. infinity, with respect to x and then use integration by parts. I get stuck after I do that. Any ideas on what to do after that?

[b]1. Homework Statement [/b exp(-x^2) = (1/sqrt(pi))*integral[(e^(2*ixy)*e^(-y^2)], y = -infinity .. infinity how to justify this equality

Do you mean how can you express it? You can express an even permutation into a product of even number of 2-cycles. also the order of A(n) is n!/2

Homework Statement Show that every element in A(n)= set of even permutations, for n> or equal to 3 can be expressed as a 3-cycle or a product of three cycles. Homework Equations 3-cycle = (_ _ _). a permutation is a function from a set A to A that is bijective. The Attempt at a...

because in a group a(a^-1) must equal e. but if a^-1=a then a^2=e. ok so a should equal a^-1?

ab, a inverse and b inverse

that means a= a inverse which is not true

Well inorder to show something is a subgroup you have to show that it is closed under the operation and that if a belongs to the subgroup then a^-1 (inverse of a) belongs to the subgroup. but what i can't figure out is how to tie that into proving the sungroup has four elements in it. well wait...

Homework Statement Prove that an abelian group with two elements of order 2 must have a subgroup of order 4 Homework Equations The Attempt at a Solution Let G be an abelian group ==> for every a,b that belong to G ab=ba. Let a,b have order 2 ==> a^2 =e and b^2 = e. Since a...

Homework Statement T(2,1)---> (5,2) and T(1,2)--->(7,10) is a linear map on R^2. Determine the matrix T with respect to the basis B= {(3,3),(1,-1)} Homework Equations The Attempt at a Solution matrix = 5 7 2 10 ?

ok i see. i still am stuck though. i think that I am thinking about it too much. if i put c and d in the 2nd equation i have c(5n+8) +d(7n+11). is it enough to say that 5n+8 and 7n+11 are relatively prime because they have no other common factors other than 1.

ok using u and v i have (5n + 5+ 3)u + (7n +7 +4)v =1 we already know that (5n +3)u + (7n +4)v =1 by assumption. wats left over is 5u +7v. since 5 and 7 are relatively prime (5u + 7v) =1 also so the first equation equal two and it should equal one.

rite well like i said i tried to use induction to compute it for all n. the base case n=1 works and after you assume 5n+3 and 7n+4 are relatively prime, ie (5n+3)s + (7n+4)t =1 = gcd, then i tried to use that to prove for n+1, ie {5(n+1)+3}s + {7(n+1) +4}t =1. after expanding and using the...