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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 82,724

No, doing those dishes. All the dishes and glasses and everything else was in the sink. I got sick of looking at that.

**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**

Oh, that. Yes, it takes me an hour to do the dishes sometimes... only because I am thorough.

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 82,724

Takes me long because I let them pile up.

**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 think that is human nature...

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 82,724

I hate housework...

Does it have a residue command?

**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 haven't been able to find one. So I tried looking for a Laurent series command or something (then I could just look for the 1/(z - c) power), but I could not find that either...

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 82,724

Okay, I was just wondering. Perhaps taylor can do it?

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 tried that too but I cannot get it to give me negative powers (well it gives me something else that isn't making sense to me)...

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 82,724

What one did you try?

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**

taylor(exp(x), x, -5, 5);

that ends up giving me a Taylor series with powers of e in the denominators and powers of (x+5) in the numerators...

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 82,724

That is correct according to Mathematica.

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**

What is correct?

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 82,724

That expansion is correct.

Try this one and you will see that taylor can get Laurents too.

taylor(1/(1-x^2),x,-1,5);

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**

That is interesting, so sometimes it works, sometimes it doesn't.

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 82,724

When did it not?

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**

Hmm, never mind, I was confused why it was not working for e^x -- it seems e^x does not have a Laurent series?

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 82,724

Hi;

Correct , it does not.

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 feel like I have a powerful tool (Maxima) in an idiot's hands (me).

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 82,724

I have been working with many CAS since 1998. I have only scratched the surface. Welcome to the New Math!

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 thought you said you knew 16 programming languages or something...

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 82,724

At one time I knew 10 or 12. The last being C++ which was my favorite. Then we discovered Mupad lite. I could not believe my eyes. Everything we were trying to write our program to do they had already done and then some.

I threw away those languages and my TI-92 and became what is known as a package jockey.

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**

Hmm. So, is there any use in knowing lots of programming languages these days? Or can a package likely do all of that for you?

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 82,724

A package can do all of that and more. Combining mathematics and programming produces a powerful weapon.

The newest idea is the revolutionary program called geogebra. It combines geometry with the rest of mathematics and programming.

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've heard of that one. Never downloaded it though, I hate geometry...

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 82,724

I did too, until geogebra made it fun. What you hate is how they teach it.

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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