Exactly, there may be negative powers depending on your choice of m, but there certainly shouldn't be infinitely many negative terms as was suggested in the OP.
When m<0 there is clearly a pole at z=0, not an essential singularity.
Thanks!
So I suppose that with your notation:
V = span\{e_1,e_2,...\}\supseteq span\{e_2,e_3,...\}\supseteq... will give me the infinite decreasing chain as desired.
Thanks again for your help!
EDIT: On second thought, can I be assured that such a chain will not stabilize? It is...
Building of your idea of using integration by parts to obtain:
(e^x)*arctan(x) - ∫(e^x)/(x^2+1)dx
If you consider the function in the integral as a function of a complex variable, you may note that it is holomorphic on open sets U that stay away from i and -i. Using Cauchy's Integral...
I have a question concerning subspaces of infinite dimensional vector spaces. Specifically given any infinite dimensional vector space V, how might one construct an infinite decreasing chain of subspaces?
That is:
V=V0\supseteqV1\supseteq... , where each Vi is properly contained in Vi-1...
I would prefer:
ax=a
ax-a=0
a(x-1)=0
which implies that a=0 or x-1=0 (since R contains no zero divisors). Since a is given as non-zero we conclude that x-1=0 which gives x=1.
I should note that this proof avoids the use inverses and thus proves this result not only for any field but for...
If you have already been doing group theory then you should be aware that
C forms an abelian group under addition and the nonzero elements of C forms an abelian group under multiplication.
So, obviously C is a commutative ring with 1. Now every nonzero element is in C's multiplicative...
I just checked the above expression and you did multiply correctly. It is much easier to just use the properties of the modulus instead though.
\left|(1-3i)^5(\sqrt{2}+i\sqrt{3})^7|= |1-3i|^5|\sqrt{2}+\sqrt{3}i\right|^7
\left=(\sqrt{1^2+(-3)^2} )^5(\sqrt{\sqrt{2}^2+\sqrt{3}^2 )^7\right...
Instead of using induction try a simple congruence argument.
24 divides 2(7)^n + 3(5)^n - 5 iff 2(7)^n + 3(5)^n = 5 mod 24
Now if n is even this is true since 7^2 = 5^2 = 1 mod 24
If is n is odd the result is also immediate.
You can't assume that there exists points in A such that d(x,A) = d(x,a) and similarly for y. The easiest example is if A is open. Then the inf is most likely not in A.