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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,181

This will be a thread for discussing GFs,their applications and uses,cool identities that can be proven using them and pretty much anything concerning them.Any questions about them are welcome and desirable.

Here is a starter topic:"Using GFs to represent binary trees.How can this be done?"

*Last edited by anonimnystefy (2012-04-28 11:26:59)*

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
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This is a small tree but it is perfect for an example.

The resulting nodes of the tree are compactly represented by the generating function.

**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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**anonimnystefy****Real Member**- From: The Foundation
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I assume the T and H are tales and heads.

I know this will sound too much,but,what would the graph and the GF look like for a coin that has the probability of heads change according to some known formula?

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
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Sometimes that is possible with a GF but only by a trick or two. It might be a little more like a Markov chain but that can be tricky too.

**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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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
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It rhymes.

Ok. Could we get an example for that graph you posted using a Markov Chain?

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**bobbym****Administrator**- From: Bumpkinland
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Hi;

Sorry about that, I do not remember how. Checked my notes and found I did not write it down.

**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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**anonimnystefy****Real Member**- From: The Foundation
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I think this is the one:

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**bobbym****Administrator**- From: Bumpkinland
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When you square that you get the same matrix.

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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**anonimnystefy****Real Member**- From: The Foundation
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Then it is correct.

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**bobbym****Administrator**- From: Bumpkinland
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Apparently, I was trying to get it to do the calculation on the right.

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

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**anonimnystefy****Real Member**- From: The Foundation
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No,squaring the matrix gives the probability of getting from the given state to another in two steps.

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**bobbym****Administrator**- From: Bumpkinland
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That is what I am saying. The answer on the right needs something more than a matrix multiplication. But it is the tree answer. So my original statement that the tree can be represented entirely by a markov chain and matrix operations is not accurate.

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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**anonimnystefy****Real Member**- From: The Foundation
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Yes,I agree.

What if we multiply by a vector?

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**bobbym****Administrator**- From: Bumpkinland
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Sometimes there is an initial state vector. What vector do you want to use?

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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**anonimnystefy****Real Member**- From: The Foundation
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Yes,I know. But it doesn't help. What matrix would you like as your output?

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**bobbym****Administrator**- From: Bumpkinland
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You could do a scalar multiplication by A^(1) x A^(2)

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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**anonimnystefy****Real Member**- From: The Foundation
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What is A?

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**bobbym****Administrator**- From: Bumpkinland
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A is your matrix. A^(1) being the first level of the tree, A^(2) being the second level.

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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**anonimnystefy****Real Member**- From: The Foundation
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I don't understand what you want to do with that.

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**bobbym****Administrator**- From: Bumpkinland
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Trees are computed by mutiplying nodes down the branches.

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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**anonimnystefy****Real Member**- From: The Foundation
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Ok,but what does that have to do with matrices?

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**bobbym****Administrator**- From: Bumpkinland
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Instead of the usual multiplication of 2 matrices, we could use an element by element multiplication of 2 matrices.

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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**anonimnystefy****Real Member**- From: The Foundation
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I don't think it would work for three steps.

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**bobbym****Administrator**- From: Bumpkinland
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Sure it would. Element by element would give 1 / 8 for each branch.

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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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
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But there would be 8 branches! We only have a 2x2 matrix!

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