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Author Topic: TT's Maths Thread  (Read 135312 times)  Share 

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kamil9876

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Re: TT's Maths Thread
« Reply #375 on: December 11, 2009, 12:36:32 pm »
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actually, I just thought about it and I think the cominatorial argument for is wrong, but the algebraic is still true. Simply because does not exactly count how many with at least pairs, but it actually overcounts this, so we cannot be sure.
Voltaire: "There is an astonishing imagination even in the science of mathematics ... We repeat, there is far more imagination in the head of Archimedes than in that of Homer."

TrueTears

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Re: TT's Maths Thread
« Reply #376 on: December 11, 2009, 04:16:25 pm »
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actually, I just thought about it and I think the cominatorial argument for is wrong, but the algebraic is still true. Simply because does not exactly count how many with at least pairs, but it actually overcounts this, so we cannot be sure.
How do you do it algebraically?

I tried comparing factors but didn't really work =S
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Ahmad

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Re: TT's Maths Thread
« Reply #377 on: December 11, 2009, 04:30:22 pm »
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Mandark: Please, oh please, set me up on a date with that golden-haired angel who graces our undeserving school with her infinite beauty!

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TrueTears

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Re: TT's Maths Thread
« Reply #379 on: December 11, 2009, 06:33:54 pm »
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Use a combinatorial argument to prove that
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Ahmad

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Re: TT's Maths Thread
« Reply #380 on: December 11, 2009, 09:10:10 pm »
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The right hand side summand counts the number of permutations with exactly r fixed points. Make sure you make use of your expression for derangements which you derived previously to interpret parts of the summand. :)
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TrueTears

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Re: TT's Maths Thread
« Reply #381 on: December 11, 2009, 09:45:02 pm »
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Oh yeah, I got it, thanks Ahmad =)

This may not be combinatorics but I just saw it in AnC but they don't give a proof of where it came from. Can anyone enlighten me?

The Fibonacci recurrence formula for and

However a more "closed-form" formula for any number in the Fibonacci sequence is given by

Is there an elementary way to prove that formula or is it beyond my capabilities?
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kamil9876

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Re: TT's Maths Thread
« Reply #382 on: December 11, 2009, 09:47:23 pm »
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actually, I just thought about it and I think the cominatorial argument for is wrong, but the algebraic is still true. Simply because does not exactly count how many with at least pairs, but it actually overcounts this, so we cannot be sure.
How do you do it algebraically?

I tried comparing factors but didn't really work =S

comparing factors does work. Ie you have to compare factors of and :







ie the expresion in the first set of square brackets comes from the numerator of with a 2 multipled into each factor (these lots of 2 come from ). The expression in the second pair of square brackets comes from And the denomonator comes from the denominator of which clearly increases as i increases.

Now all you need to do is compare numerators, and you can clearly spot the difference between them, it is that the bottom one has a 2n-2i factor in place of a 2n-i factor.
« Last Edit: December 11, 2009, 10:29:37 pm by kamil9876 »
Voltaire: "There is an astonishing imagination even in the science of mathematics ... We repeat, there is far more imagination in the head of Archimedes than in that of Homer."

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Re: TT's Maths Thread
« Reply #383 on: December 11, 2009, 09:54:04 pm »
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It's called Binet's formula, and you can prove it by induction. But that's if you know the result already. If you don't you can discover it using generating functions, or linear algebra, or perhaps a bunch of other ways. :)
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TrueTears

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Re: TT's Maths Thread
« Reply #384 on: December 11, 2009, 09:54:55 pm »
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Ahmah can you run me through deriving it by using generating functions? :P (I haven't studied it yet so I don't really know how to approach it myself lolz but I'm super interested)
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kamil9876

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Re: TT's Maths Thread
« Reply #385 on: December 11, 2009, 10:30:41 pm »
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Here is a good one. Enjoy
Voltaire: "There is an astonishing imagination even in the science of mathematics ... We repeat, there is far more imagination in the head of Archimedes than in that of Homer."

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Re: TT's Maths Thread
« Reply #386 on: December 11, 2009, 10:35:46 pm »
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Sure thing, there's nothing to it, really.

Step 1. Define the generating function where is the nth Fibonacci number. F is our clothesline upon which we hang up our sequence of numbers for display!

Step 2. Write down the recurrence equation, and multiply it by , then sum both sides for n from 0 to infinity! This step is used to turn our recurrence into an equation describing F.



The last term is F(x). It's not so obvious what to do with our second last term, but we ultimately want to write it in terms of F(x). Here's what we do:





But the RHS of this equation is just the second last term, so the second last term is . In a similar fashion, you find that .

So we've turned out recurrence equation into . We know that and , so we can use this to solve for F(x), giving . I expect you to perform this calculation.

Step 3. The idea of this stage is to expand our function as a series, to do that we'll be making use of the fact that this is just the geometric series formula, so keep this in mind. We can use partial fractions to write where r and r' are the roots of the denominator i.e. roots of , and I'll suppose r is the positive root. I'll leave you to work out what r, r', A and B are. Once we've done that we can use our geometric series formula in reverse:





So that

Equating coefficients gives us, , which is Binet's formula.
Mandark: Please, oh please, set me up on a date with that golden-haired angel who graces our undeserving school with her infinite beauty!

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Ahmad

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Re: TT's Maths Thread
« Reply #387 on: December 11, 2009, 10:52:08 pm »
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I didn't get to mention how one actually does this in practice, which is to use Mathematica or Maple, or some other CAS:

Mathematica:
Input: SeriesCoefficient[x/(1 - x - x^2), {x, 0, n}]
Output: (-(1/2 (1 - Sqrt[5]))^n + (1/2 (1 + Sqrt[5]))^n)/Sqrt[5]

So the key step really is finding the generating function, the rest is routine computer work, which I've done by hand before, but it takes a few minutes for me vs a few microseconds for the computer.  8-)
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TrueTears

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Re: TT's Maths Thread
« Reply #388 on: December 11, 2009, 11:13:18 pm »
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Sure thing, there's nothing to it, really.

Step 1. Define the generating function where is the nth Fibonacci number. F is our clothesline upon which we hang up our sequence of numbers for display!

Step 2. Write down the recurrence equation, and multiply it by , then sum both sides for n from 0 to infinity! This step is used to turn our recurrence into an equation describing F.



The last term is F(x). It's not so obvious what to do with our second last term, but we ultimately want to write it in terms of F(x). Here's what we do:





But the RHS of this equation is just the second last term, so the second last term is . In a similar fashion, you find that .

So we've turned out recurrence equation into . We know that and , so we can use this to solve for F(x), giving . I expect you to perform this calculation.

Step 3. The idea of this stage is to expand our function as a series, to do that we'll be making use of the fact that this is just the geometric series formula, so keep this in mind. We can use partial fractions to write where r and r' are the roots of the denominator i.e. roots of , and I'll suppose r is the positive root. I'll leave you to work out what r, r', A and B are. Once we've done that we can use our geometric series formula in reverse:





So that

Equating coefficients gives us, , which is Binet's formula.

Thanks Ahmath, I can't wait till I start generating functions.
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Interested in asset pricing, econometrics, and social choice theory.

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Re: TT's Maths Thread
« Reply #389 on: December 11, 2009, 11:14:21 pm »
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I didn't get to mention how one actually does this in practice, which is to use Mathematica or Maple, or some other CAS:

Mathematica:
Input: SeriesCoefficient[x/(1 - x - x^2), {x, 0, n}]
Output: (-(1/2 (1 - Sqrt[5]))^n + (1/2 (1 + Sqrt[5]))^n)/Sqrt[5]

So the key step really is finding the generating function, the rest is routine computer work, which I've done by hand before, but it takes a few minutes for me vs a few microseconds for the computer.  8-)

I've heard some universities provide require a computer math program. Do you need to buy it yourself or do they provide?