Getting the coefficient.

bobbym · 2013-12-14 02:24:17

Hi;

At the request of anonimnystefy:

Part 1

Sure generating functions are great because they turn a combinatorics problem into a computational one. But how do you get those coefficients? I am not talking about the baby problems they post in books that they can do by hand I am talking about industrial strength ones. Let's see how a CAS would do it.

Now there are several means, spot the pattern, recurrence relations, complex analysis and some others. We will try them all.

Let's look at a baby problem first and see how it is done.

A die is thrown 10 times, what is the coefficient of x^25?
This can restated as a die is thrown 10 times what is the probability that the sum of the die equals 25?

The gf is :

Without going into residues one way to do this is

Im[NIntegrate[Evaluate[
    				(( (x + x^2 + x^3 + x^4 + x^5 + x^6)/6)^10/x^26 /. 
       x -> E^(I \[CurlyPhi])) I E^(I \[CurlyPhi])],
   		{\[CurlyPhi], 0, 2 \[Pi]}, Method -> Trapezoidal, 
   WorkingPrecision -> 35, PrecisionGoal -> 16, MaxRecursion -> 12, 
   AccuracyGoal -> 16]]/(2 \[Pi])

Running that we get the immediate answer of 0.0137465944596860234720317024843773815. The exact answer by expanding the expression is

We did well as you can see.

Did you follow up to here?

anonimnystefy · 2013-12-14 02:48:26

Yes, that is the method from the other thread, right?

bobbym · 2013-12-14 02:55:36

Yep, but never shown. Let's see a larger one.

Let's say we flip the die 100 times. We want the coefficient of x^500.

Im[NIntegrate[Evaluate[
    				(( (x + x^2 + x^3 + x^4 + x^5 + x^6)/6)^100/x^501 /. 
       x -> E^(I \[CurlyPhi])) I E^(I \[CurlyPhi])],
   		{\[CurlyPhi], 0, 2 \[Pi]}, Method -> Trapezoidal, 
   WorkingPrecision -> 50, PrecisionGoal -> 25, MaxRecursion -> 13, 
   AccuracyGoal -> 25]]/(2 \[Pi])

We get:

Is that right?

anonimnystefy · 2013-12-14 02:59:44

Hm, by expanding, I get

Last edited by anonimnystefy (2013-12-14 03:01:08)

bobbym · 2013-12-14 03:04:45

This is what you should have:

anonimnystefy · 2013-12-14 03:50:34

Yeah.

bobbym · 2013-12-14 03:57:30

You see it now?

anonimnystefy · 2013-12-14 04:17:19

Wait, that's the residue method?

bobbym · 2013-12-14 04:25:15

I think that is what it is called but no matter it works.

Do you notice how much faster it is than expanding?

anonimnystefy · 2013-12-14 04:26:13

Hm, I've already seen this method.

bobbym · 2013-12-14 04:36:07

I do not think you have, I have kept this a secret because of its speed over the original routine.

Let's do a real sized problem now and see how it does.

How about 10000 throws and what is the coefficient of x^50000?

Im[NIntegrate[
   Evaluate[(((x + x^2 + x^3 + x^4 + x^5 + x^6)/6)^10000/x^50001 /. 
       x -> E^(I \[CurlyPhi])) I E^(I \[CurlyPhi])], {\[CurlyPhi], 0, 
    2 \[Pi]}, Method -> Trapezoidal, WorkingPrecision -> 150, 
   PrecisionGoal -> 75, MaxRecursion -> 14, 
   AccuracyGoal -> 75]]/(2 \[Pi])

I get:

This is the end of Part 1.

anonimnystefy · 2013-12-15 08:23:25

Hi bobbym

I cannot expand it, so I cannot check. I am sure that that value is close.

bobbym · 2013-12-15 13:33:57

How do we usually verify a result in numerical work like this?

anonimnystefy · 2013-12-15 18:25:49

With another method?

bobbym · 2013-12-15 22:36:06

That is correct! That brings us to Part 2.

anonimnystefy · 2013-12-16 01:23:55

Shoot!

bobbym · 2013-12-16 03:23:07

Part 2

Another way is to form a recurrence relation for the coefficients.

We will start with the toy problem given in post #1.

We wish to get the coefficient of x^25.

We form the recurrence for the numerator:

We will need to prime it with the first 5 values.

Series[(x + x^2 + x^3 + x^4 + x^5 + x^6)^k, {x, 0, 5}] // Normal

This is the ouput:

x^k (1 + k x + 1/2 (k + k^2) x^2 + 1/6 (2 k + 3 k^2 + k^3) x^3 + 
   1/24 (6 k + 11 k^2 + 6 k^3 + k^4) x^4 + 
   1/120 (24 k + 50 k^2 + 35 k^3 + 10 k^4 + k^5) x^5)

We eliminate the variable k.

x^k (1 + k x + 1/2 (k + k^2) x^2 + 1/6 (2 k + 3 k^2 + k^3) x^3 + 
     1/24 (6 k + 11 k^2 + 6 k^3 + k^4) x^4 + 
     1/120 (24 k + 50 k^2 + 35 k^3 + 10 k^4 + k^5) x^5) /. 
  k -> 10 // Expand

This yields:

Compute the recurrence:

RecurrenceTable[{a[n] == 
    1/(-10 + 
      n) (65 a[-5 + n] - n a[-5 + n] + 54 a[-4 + n] - n a[-4 + n] + 
       43 a[-3 + n] - n a[-3 + n] + 32 a[-2 + n] - n a[-2 + n] + 
       21 a[-1 + n] - n a[-1 + n]), a[10] == 1, a[11] == 10, 
   a[12] == 55, a[13] == 220, a[14] == 715}, a, {n, 10, 25}] // Last

this yields 831204. Now we divide by 6 ^ 10.

which we know is correct. Can you follow?

anonimnystefy · 2013-12-16 17:49:11

I understand everything except hiw you formed the recurrence.

bobbym · 2013-12-17 00:47:02

You have not followed our comments earlier on how to form a recurrence from a gf?

anonimnystefy · 2013-12-17 01:26:56

Hm, found it. It is very old.

bobbym · 2013-12-17 01:33:58

I have a little routine that does these automatically.

Agnishom · 2013-12-17 01:46:41

But you have to write the routine manually, unless you use butterflies....

bobbym · 2013-12-17 01:49:30

I have written it long ago. The great DZ says you do not truly understand a piece of math until you program it.

Agnishom · 2013-12-17 02:00:05

Does it mean that Euclid, Euler, Pythagoras, Fibonacci did not know any math?

bobbym · 2013-12-17 02:12:35

Agnishom:That is exactly what he means. had they been around today they would program their ideas and even be smarter.

Anonimnystefy: Can you do the steps to get the recurrence because where I am going with this is to test the answer to the 10000 die problem which I believe is incorrect. Thus proving that method is kaboobly doo as stated in the other thread.

Math Is Fun Forum

#1 2013-12-14 02:24:17

Getting the coefficient.

#2 2013-12-14 02:48:26

Re: Getting the coefficient.

#3 2013-12-14 02:55:36

Re: Getting the coefficient.

#4 2013-12-14 02:59:44

Re: Getting the coefficient.

#5 2013-12-14 03:04:45

Re: Getting the coefficient.

#6 2013-12-14 03:50:34

Re: Getting the coefficient.

#7 2013-12-14 03:57:30

Re: Getting the coefficient.

#8 2013-12-14 04:17:19

Re: Getting the coefficient.

#9 2013-12-14 04:25:15

Re: Getting the coefficient.

#10 2013-12-14 04:26:13

Re: Getting the coefficient.

#11 2013-12-14 04:36:07

Re: Getting the coefficient.

#12 2013-12-15 08:23:25

Re: Getting the coefficient.

#13 2013-12-15 13:33:57

Re: Getting the coefficient.

#14 2013-12-15 18:25:49

Re: Getting the coefficient.

#15 2013-12-15 22:36:06

Re: Getting the coefficient.

#16 2013-12-16 01:23:55

Re: Getting the coefficient.

#17 2013-12-16 03:23:07

Re: Getting the coefficient.

#18 2013-12-16 17:49:11

Re: Getting the coefficient.

#19 2013-12-17 00:47:02

Re: Getting the coefficient.

#20 2013-12-17 01:26:56

Re: Getting the coefficient.

#21 2013-12-17 01:33:58

Re: Getting the coefficient.

#22 2013-12-17 01:46:41

Re: Getting the coefficient.

#23 2013-12-17 01:49:30

Re: Getting the coefficient.

#24 2013-12-17 02:00:05

Re: Getting the coefficient.

#25 2013-12-17 02:12:35

Re: Getting the coefficient.

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