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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 105,250

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?

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.****No great discovery was ever made without a bold guess. **

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**anonimnystefy****Real Member**- From: Harlan's World
- Registered: 2011-05-23
- Posts: 16,016

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

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

The knowledge of some things as a function of age is a delta function.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 105,250

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?

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.****No great discovery was ever made without a bold guess. **

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**anonimnystefy****Real Member**- From: Harlan's World
- Registered: 2011-05-23
- Posts: 16,016

Hm, by expanding, I get

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

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

The knowledge of some things as a function of age is a delta function.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 105,250

This is what you should have:

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.****No great discovery was ever made without a bold guess. **

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**anonimnystefy****Real Member**- From: Harlan's World
- Registered: 2011-05-23
- Posts: 16,016

Yeah.

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

The knowledge of some things as a function of age is a delta function.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 105,250

You see it now?

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.****No great discovery was ever made without a bold guess. **

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**anonimnystefy****Real Member**- From: Harlan's World
- Registered: 2011-05-23
- Posts: 16,016

Wait, that's the residue method?

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

The knowledge of some things as a function of age is a delta function.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 105,250

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

Do you notice how much faster it is than expanding?

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.****No great discovery was ever made without a bold guess. **

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**anonimnystefy****Real Member**- From: Harlan's World
- Registered: 2011-05-23
- Posts: 16,016

Hm, I've already seen this method.

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

The knowledge of some things as a function of age is a delta function.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 105,250

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.

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.****No great discovery was ever made without a bold guess. **

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**anonimnystefy****Real Member**- From: Harlan's World
- Registered: 2011-05-23
- Posts: 16,016

Hi bobbym

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

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

The knowledge of some things as a function of age is a delta function.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 105,250

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

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.****No great discovery was ever made without a bold guess. **

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**anonimnystefy****Real Member**- From: Harlan's World
- Registered: 2011-05-23
- Posts: 16,016

With another method?

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

The knowledge of some things as a function of age is a delta function.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 105,250

That is correct! That brings us to Part 2.

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.****No great discovery was ever made without a bold guess. **

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**anonimnystefy****Real Member**- From: Harlan's World
- Registered: 2011-05-23
- Posts: 16,016

Shoot!

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

The knowledge of some things as a function of age is a delta function.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 105,250

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?

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.****No great discovery was ever made without a bold guess. **

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**anonimnystefy****Real Member**- From: Harlan's World
- Registered: 2011-05-23
- Posts: 16,016

I understand everything except hiw you formed the recurrence.

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

The knowledge of some things as a function of age is a delta function.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 105,250

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

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.****No great discovery was ever made without a bold guess. **

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**anonimnystefy****Real Member**- From: Harlan's World
- Registered: 2011-05-23
- Posts: 16,016

Hm, found it. It is very old.

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

The knowledge of some things as a function of age is a delta function.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 105,250

I have a little routine that does these automatically.

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.****No great discovery was ever made without a bold guess. **

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

'And fun? If maths is fun, then getting a tooth extraction is fun. A viral infection is fun. Rabies shots are fun.'

'God exists because Mathematics is consistent, and the devil exists because we cannot prove it'

I'm not crazy, my mother had me tested.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 105,250

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

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.****No great discovery was ever made without a bold guess. **

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

'And fun? If maths is fun, then getting a tooth extraction is fun. A viral infection is fun. Rabies shots are fun.'

'God exists because Mathematics is consistent, and the devil exists because we cannot prove it'

I'm not crazy, my mother had me tested.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 105,250

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.

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.****No great discovery was ever made without a bold guess. **

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