Discussion about math, puzzles, games and fun. Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ π -¹ ² ³ °

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**zetafunc.****Guest**

Maybe everyone is just bored of talking to me. And adriana is too lazy to go through the maths e-mails so she is putting off replying to them, probably.

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 82,563

Is she generally lazy?

**In mathematics, you don't understand things. You just get used to them.I have the result, but I do not yet know how to get it.All physicists, and a good many quite respectable mathematicians are contemptuous about proof.**

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**zetafunc.****Guest**

I'm not sure. She washed up all our dishes after we ate at her house, but another time I saw her just drop some food under the sofa (it was a bit of sweetcorn from her pizza) and she didn't bother to pick it up. That was pretty gross. A sign of laziness, maybe?

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 82,563

Girls are a lot sloppier in their habits then they let on.

**In mathematics, you don't understand things. You just get used to them.I have the result, but I do not yet know how to get it.All physicists, and a good many quite respectable mathematicians are contemptuous about proof.**

**Online**

**zetafunc.****Guest**

She was sloppy, in that case. But I can't think of other encounters where she was.

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 82,563

Sometimes they have closets full of junk. I mean from the floor to the ceiling just stuffed in there. Some of them just throw there clothes around.

**In mathematics, you don't understand things. You just get used to them.I have the result, but I do not yet know how to get it.All physicists, and a good many quite respectable mathematicians are contemptuous about proof.**

**Online**

**zetafunc.****Guest**

I don't know about that, but she does sometimes go to school in track bottoms or sometimes even pyjamas.

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 82,563

Pajamas? How can you tell?

I have the result, but I do not yet know how to get it.

All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

**Online**

**zetafunc.****Guest**

She told me. Although most of the time she sleeps in the track-bottoms then goes to school wearing the same ones. She said she doesn't care...

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 82,563

Seems like she is on the carefree side.

I have the result, but I do not yet know how to get it.

All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

**Online**

**zetafunc.****Guest**

I guess so...

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 82,563

Yes, the modern gal is getting stranger and stranger.

I have the result, but I do not yet know how to get it.

All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

**Online**

**zetafunc.****Guest**

She was nice to talk to though, even though she is strange. Made many of my otherwise boring days interesting.

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 82,563

She might come back, who knows. In the meantime forget her.

I have the result, but I do not yet know how to get it.

All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

**Online**

**zetafunc.****Guest**

Easier said than done, I even think about F sometimes.

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 82,563

I heard she was abducted by aliens and whisked away to Reticulum 4. You might as well forget about her.

I have the result, but I do not yet know how to get it.

All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

**Online**

**zetafunc.****Guest**

I just need to move on to another girl who interests me, although there aren't any at the moment.

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 82,563

Then turn off the thinking about them. Find something else to occupy your thoughts until one shows up.

I have the result, but I do not yet know how to get it.

All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

**Online**

**zetafunc.****Guest**

Maths reminds me of F sometimes. I can do that though.

I am trying to get to grips with contour integrals but I'm not too sure how they work, my knowledge is pretty limited. Do you know a lot about it?

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 82,563

Not really, I always have had trouble with them too. We could try them together maybe solve a few.

First I have to wash this stack of dishes so I will be back then.

I have the result, but I do not yet know how to get it.

All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

**Online**

**zetafunc.****Guest**

Is the residue of some function f(z) always the co-efficient of 1/(z - a) in its Laurent series? A lot of the time I just end up finding the residue to be zero...

Okay, see you later.

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 82,563

Hi;

They say that it is the coefficient of the

coefficient of the Laurent series, so I would say yes.

I have the result, but I do not yet know how to get it.

All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

**Online**

**zetafunc.****Guest**

And is it also true that if the residue is 0 then the contour evaluates to 0?

I thought that we had to evaluate 2*pi*i*(sum of the residues)... if the sum of the residues is 0, then...

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 82,563

Hi;

I do not know if that is true.

I have the result, but I do not yet know how to get it.

All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

**Online**

**zetafunc.****Guest**

Okay...

I guess I could try doing something simple like

We know this evaluates to π, so we could maybe try using a contour to try and get the answer too.

Considering instead

where C is a simple closed curve -- a semi-circle in the upper half of the complex plane, from negative infinity to positive infinity, so it encloses the entire real integral, including one of the poles (z = i).

However, I asked Wolfram for the series expansion of 1/(1 + z^2), and it's giving me a series expansion at z = i, and another at z = -i. Since my contour encloses the pole at z = i, I'm interested in the residue of that series.

But I don't know how they found that Laurent series... can you tell me how they found that series expansion that includes the negative power of (z - i)?