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#26 2013-12-17 07:55:38

anonimnystefy
Real Member
From: The Foundation
Registered: 2011-05-23
Posts: 15,544

Re: Getting the coefficient.

Hi bobbym

I am getting


for the coefficient of
...

Last edited by anonimnystefy (2013-12-17 07:55:51)


“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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#27 2013-12-17 07:57:16

bobbym
Administrator
From: Bumpkinland
Registered: 2009-04-12
Posts: 87,238

Re: Getting the coefficient.

What are you getting for the coefficient of x^50000?


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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#28 2013-12-17 08:30:15

anonimnystefy
Real Member
From: The Foundation
Registered: 2011-05-23
Posts: 15,544

Re: Getting the coefficient.

Sorry, that is the coefficient of 50000.


“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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#29 2013-12-17 08:42:50

bobbym
Administrator
From: Bumpkinland
Registered: 2009-04-12
Posts: 87,238

Re: Getting the coefficient.

That is not what I am getting. See you in a bit.


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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#30 2013-12-17 08:56:52

anonimnystefy
Real Member
From: The Foundation
Registered: 2011-05-23
Posts: 15,544

Re: Getting the coefficient.

My recurrence is:

Last edited by anonimnystefy (2013-12-17 10:36:16)


“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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#31 2013-12-17 10:26:47

bobbym
Administrator
From: Bumpkinland
Registered: 2009-04-12
Posts: 87,238

Re: Getting the coefficient.

The recurrence should only be for the numerator.


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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#32 2013-12-17 10:30:16

anonimnystefy
Real Member
From: The Foundation
Registered: 2011-05-23
Posts: 15,544

Re: Getting the coefficient.

It does not matter. The answer will be the same.


“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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#33 2013-12-17 10:31:09

bobbym
Administrator
From: Bumpkinland
Registered: 2009-04-12
Posts: 87,238

Re: Getting the coefficient.

I suggest you try it first before saying that.


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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#34 2013-12-17 10:45:04

anonimnystefy
Real Member
From: The Foundation
Registered: 2011-05-23
Posts: 15,544

Re: Getting the coefficient.

The results are the same.


“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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#35 2013-12-17 10:49:41

bobbym
Administrator
From: Bumpkinland
Registered: 2009-04-12
Posts: 87,238

Re: Getting the coefficient.

Could be because your recurrence is wrong.

Use this one:

RecurrenceTable[{a[n] == 
     1/(-10000 + 
       n) (60005 a[-5 + n] - n a[-5 + n] + 50004 a[-4 + n] - 
        n a[-4 + n] + 40003 a[-3 + n] - n a[-3 + n] + 
        30002 a[-2 + n] - n a[-2 + n] + 20001 a[-1 + n] - 
        n a[-1 + n]), a[10000] == 1, a[10001] == 10000, 
    a[10002] == 50005000, a[10003] == 166716670000, 
    a[10004] == 416916712502500}, a, {n, 10000, 50000}] // Last;

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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#36 2013-12-17 10:53:33

anonimnystefy
Real Member
From: The Foundation
Registered: 2011-05-23
Posts: 15,544

Re: Getting the coefficient.

I know exactly what's wrong.


“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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#37 2013-12-17 10:55:01

bobbym
Administrator
From: Bumpkinland
Registered: 2009-04-12
Posts: 87,238

Re: Getting the coefficient.

What is wrong?


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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#38 2013-12-17 11:00:34

anonimnystefy
Real Member
From: The Foundation
Registered: 2011-05-23
Posts: 15,544

Re: Getting the coefficient.

I am using the wrong GF!


“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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#39 2013-12-17 11:04:22

bobbym
Administrator
From: Bumpkinland
Registered: 2009-04-12
Posts: 87,238

Re: Getting the coefficient.

Well, that will certainly cause a few problems. What will you do about that?

As soon as you get the correct answer you will see that there is an even bigger problem looming on the horizon. Also, then will come Part 3.


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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#40 2013-12-20 06:20:13

ElainaVW
Member
Registered: 2013-04-29
Posts: 303

Re: Getting the coefficient.

Hello smile

I see the problem! When will you post part 3?

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#41 2013-12-20 07:05:24

bobbym
Administrator
From: Bumpkinland
Registered: 2009-04-12
Posts: 87,238

Re: Getting the coefficient.

Part 3

Our problem is that the 2 methods provided agree for small problems but disagree on the large one. Why is this?

Before we can answer that we need to find out which answer is correct so we do the problem in a third way.

With 10000 throws of a die what is the coefficient of x^50000?

This is equivalent to the diophantine equation:

a1 + a2 + a3 + ... a10000 = 50000   

with 1 <= a1,a2,...a10000 <= 6

We make a change of variable.

a1 = a1+1, a2 = a2+1, a3 = a3+1, ...a10000=a10000+1

we get

a1 + a2 + a3 + ... + a10000 +10000 = 50000

this becomes

The equation becomes

a1 + a2 + a3 + ... + a10000 = 40000 

0 <= a1,a2,...,a10000 <= 5

with the answer being the coefficient of x^40000

The gf for that is:

You should recognize that

are the coefficients with no restrictions.

Now

Only the terms x^6  to  x^39996 could possibly contribute to the coefficient of  x^40000, making the final answer:

The whole process looks a lot like a PIE so it can be generalized to the following series. For,

with

In Mathematica speak:

r = 40000;
n = 10000;
m = 5;

ans = Binomial[r + n - 1, n - 1] + 
   Sum[(-1)^
     k Binomial[n, k] Binomial[(r + n - 1) - (m + 1) k, n - 1], {k, 1,
      Floor[r/(m + 1)]}];

To get the probability of the sum being 50000:

Which agrees with Part 2's answer. We can now safely say that the result in Part 1 is incorrect.

Okay, what went wrong?

The nice guy wrote:

Numerical methods are not infallible and some sort of round off error is destroying the precision of the complex analysis method. This happens and we should forgive those mathematicians who love it so much.

bobbym ( the mean guy) wrote:

That method is typical of guys who think their beloved complex analysis is truly amazing and naively think what they know from textbooks works in the real world. Kaboobly doo!


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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