Keun Woong

Hint: Substitute sinx = (e^-x - e^(-x)) / 2i

I made the substituion but have no idea where to go from there.

My guess is that the improper integral is 0 since when transformed to contour integral, the answer is 0 with C being half circle above the real axis. I can't seem to understand what to do with the hint.

Any help would be greatly appreciated. Thanks in advance.

Construct an example accompanied by a brief explanation that illustrate the following fact:

If v is a harmonic conjugate of u in some domain D, then u is, in general, not a harmonic conjugate of v.

Here's my problem. The way I understand harmonic conjugates is two harmonic functions u and v that satisfy Cauchy-Riemann equation. If this is so, I don't see how it's possible for u to not be harmonic conjugate of v, when v is harmonic conjugate of u.

Thanks in advance for any help.

Given u(x, y) = xy + x + 2y and f(z) = u(x, y) + i*v(x, y), can the real part of f(2i) be -1?

Original problem:

Verify that the given function u is harmonic in an appropriate domain D. Find its harmonic conjugate v and find analytic function f(z) = u + i*v satisfying the indicated conditions.

u(x, y) = xy + x + 2y;

f(2i) = -1 + 5i

I've verified that u(x, y) is harmonic.

I've found that v(x, y) = 1/2 y^2 + y - 1/2 x^2 - 2x + c

When attempting to find the value of the constant c, I've run into above problem where the real parts are different. With c being real, I don't understand how this is possible.

Thanks in advance for any help.

Hello, I need a check on the validity.

The statement I'm trying to prove is this:

If (a_n) is a bounded sequence with property that every convergent sub-sequence converges to L, then (a_n) converges to L as well.

Proof:

Suppose the sequence does not converge to L; then there exists some epsilon > 0 where a_n is not an element of epsilon neighborhood of L for all n. Because (a_n) is bounded, let (a_n_k) be the sub-sequence that converges to L. This implies that for the given epsilon, there exists a_N_k, an element of (a_n) that belongs to the epsilon neighborhood of L, contradicting our assumption that no element of (a_n) belongs to the neighborhood. Therefore, (a_n) converges to L.

I think my proof is valid, but I have a doubt as the solution offered by the book is slightly different.

Solution:

Suppose (a_n) does not converge to L; then there exists epsilon such that abs[a_n - L] >_ epsilon. The book goes onto construct sub-sequence satisfying the above inequality. Because original sequence is bounded, the sub-sequence is also bounded, and by Bolzano-Weierstrass theorem, there exists sub-sub-sequence (sub-sequence of original sequence) that converges to L. But this cannot be as the sub-sub-sequence is formed of elements satisfying the inequality. Hence, (a_n) converges to L.

I don't see why we it's necessary to go onto sub-sub-sequence, instead of drawing the desired conclusion from the sub-sequence alone.

Thanks in advance. Any help would be much appreciated.

Trying to see if the following statement is possible.

A continuous function f : [0,∞) → R and a Cauchy sequence (xn) such

that f(xn) is not a Cauchy sequence

I came up with this example.

f(x) = 1 / (1 + x); (Xn) = (-n / (n + 1))

f(x) is continuous on [0, inf) and (Xn) is Cauchy sequence as it's convergent. But

f(Xn) = n + 1, which is unbounded, and hence not Cauchy.

This contradicts the answer key which states that an example of the statement is impossible.

Something must be wrong with my example, but I don't know what. Any help will be much appreciated and thanks in advance.

Hello, I have a few questions on the above topic.

Consider then set Q\{0}. Let (X_n) be a sequence contained in the set defined as X_n = 1/n.

1. Would you say that the sequence is convergent? If so, how would you describe the limit? In another words, can a sequence be convergent without the limit?

2. If the sequence is not convergent, doesn't it contradict Cauchy criterion for convergent sequence as well as Monotone Convergence Theorem?

Thanks in advance.

Hi!

My aunt owns a commercial building in downtown and recently she was charged with invoice for removal of illegal signs. Apparently, one of the tenants posted three signs without her knowing and consent. The city took down the signs and now is charging her the labor fee as well as fine. I want to know if the building owner is responsible to pay this fine even though the signs were posted without her knowing.

Thanks in advance.

I get confused whenever infinity is is involved.

Prove:

If the series from n = 0 to infinity of (2^n)(b sub 2^n) diverges, then so does the series from n = 1 to infinity of b sub n.

It's also given that the sequence (b sub n) is decreasing and b sub n >_ 0 for all n element of N.

I tried to show that partial sum sequence for 2^n(bn) is smaller than the one for bn hence if the former diverges, then the latter does as well. But then I got stuck. Am I on the right path? If it's not too much trouble, could you show the correct proof?

Thanks in advance.

Hello,

My grandmother received non-traffic ticket for evading fair by MTA. She was told by one of her acquaintance that she does not need to pay the fine since she had purchased senior pass prior to getting the ticket. I want to know if this is true. Does she need to do anything for this to take affect? Thanks in advance.

PS: This was in Los Angeles, BTW.