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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,522

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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Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and Punishment

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,376

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.Of course that result can be rigorously obtained, but who cares?Combinatorics is Algebra and Algebra is Combinatorics.**

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,522

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?

Here lies the reader who will never open this book. He is forever dead.

Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and Punishment

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,376

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.Of course that result can be rigorously obtained, but who cares?Combinatorics is Algebra and Algebra is Combinatorics.**

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,522

It rhymes.

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

Here lies the reader who will never open this book. He is forever dead.

Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and Punishment

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,376

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.Of course that result can be rigorously obtained, but who cares?Combinatorics is Algebra and Algebra is Combinatorics.**

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,522

I think this is the one:

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
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When you square that you get the same matrix.

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,522

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?

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
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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)

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,522

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.

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
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Ok,but what does that have to do with matrices?

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
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Instead of the usual multiplication of 2 matrices, we could use an element by element multiplication of 2 matrices.

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
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I don't think it would work for three steps.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
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Sure it would. Element by element would give 1 / 8 for each branch.

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,522

But there would be 8 branches! We only have a 2x2 matrix!

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