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How To Write A Case Study Analysis

Sherri Said:

Who is Clever to to help me with my home work thanks very much brothers and Sisters?

We Answered:

1. The company is, therefore, must function. Give us your money.
2. User's requirements in plain English: Give us your money. We set a price, later tack on hidden costs, then after your check clears, demand more money or we scrap the project.
3.An old laptop, cheap video camera, a desktop running Win98 and a pencil and pad.
4. Wordpad, checkers, hearts and spades. All come free with Win98.
5. If you have to ask, you can't afford it.
6. The final evaluation is if the customer doesn't like it, he can lump it.

That was fun.

Michelle Said:

Help me with home work ? when i say help me .not mean you do home work for me means give short idea as ur kno

We Answered:

What!!?? I'm sorry pal Your on ya own....! even if i wanted to try id be here all friggin night... :) good luck anyway...!

Marilyn Said:

how to concentrate better on an airplane. please help?

We Answered:

I agree that noise canceling headphones are a good idea. Bose makes a pretty good one.

Cory Said:

What does all this mean? Write the principal part of the function at its isolated singular point and ...?

We Answered:

For each function, look at the Laurent expansion around the singularity. The principal part of such an expansion is the series in terms of 1 / (z - w)^k for positive k, where w is our singularity.

1. f(z) = z exp(1/z) = z Sum (1/z)^k / k! = z + 1 + Sum (1/z)^k / (k + 1)!,

where the last series is summed from k = 1 to infinity. This last series is, in fact the prinicipal part of the expansion. Since the principal part is an infinite sum, the pole at z= 0 is an essential singularity (singular point).

2. g(z) = z^2 / (1+z) = z - 1 + [1/(z + 1)],

(this is simply a matter of arithmetic). Thus the principal part is

1 / (z + 1),

and so g has a pole of order 1 at z = -1.

3. h(z) = sinz/z

Write out the power series for sin z (this is a standard result) and, away from 0, divide through by z. You'll find that the resulting series has 0 singular part, and so the function has a removable singularity (ie. it's a power series which tends to a limit as z tends to 0).

Deborah Said:

This is my coursework and I have no clue where to start or where to finish.?

We Answered:

Get cracking. Good Luck to you.

ss

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