generating function: prove ( 1 - x - x^2 - x^3 - x^4 - x^5 - x^6)^-1 is the number of way which 'n' achieved by rolling the dice any time.

[sna_gz]

hi there I have tried to solve this problem but I think I am a bit wrong so I cant get the answer.

 

[topsquark]

How did you get
[imath]1 - n - n^2 - n^3 - n^4 - n^5 - n^6 = \dfrac{1 - n^6}{1 + n}[/imath]

That isn't true for all n!

-Dan

Addendum: It is true that the sum you listed is [imath]1 - \dfrac{n^6 - 1}{n - 1}[/imath].

 

[sna_gz]

How did you get
[imath]1 - n - n^2 - n^3 - n^4 - n^5 - n^6 = \dfrac{1 - n^6}{1 + n}[/imath]

That isn't true for all n!

-Dan

Addendum: It is true that the sum you listed is [imath]1 - \dfrac{n^6 - 1}{n - 1}[/imath].

i was trying to put -1 as 'a' but I forgot that 'a' also have power

 

[sna_gz]

How did you get
[imath]1 - n - n^2 - n^3 - n^4 - n^5 - n^6 = \dfrac{1 - n^6}{1 + n}[/imath]

That isn't true for all n!

-Dan

Addendum: It is true that the sum you listed is [imath]1 - \dfrac{n^6 - 1}{n - 1}[/imath].

how do you think is this gonna prove the question.
[imath](1 - \dfrac{x^6 - 1}{x - 1})^n[/imath]
I am still cannot find good relation between the problem and its generating function.

 

[JeffM]

I cannot even figure out what you are asking: “number of way in which n achieved by rolling the dice any time” makes no sense in English.

 

[topsquark]

how do you think is this gonna prove the question.
[imath](1 - \dfrac{x^6 - 1}{x - 1})^n[/imath]
I am still cannot find good relation between the problem and its generating function.

Neither can I. I was simply pointing out a mistake.

-Dan

 

[pka]

Wherever you got this problem or whoever wrote the problem is mistaken.
If a die (standard cubic die, numbered 1 to six) is tossed \[imath]n[/imath] times then
the generating function [imath](x+x^2+x^3+x^4+x^5+x^6)^n[/imath] tells us the number of ways a sum of the faces can be made. SEE THIS
For example a gamming die is tossed eight them [imath]{\left( {\sum\limits_{k = 1}^6 {{x^k}} } \right)^8}[/imath] contains the term
[imath]3368x^{41}[/imath]. Which tell s that the sum [imath]41[/imath] can be gotten in [imath]3368[/imath] ways so the probability is [imath]\dfrac{3368}{6^8}[/imath]. We also use that sum if we roll [imath]n[/imath] dice.

 

[sna_gz]

I cannot even figure out what you are asking: “number of way in which n achieved by rolling the dice any time” makes no sense in English.

me neither, even the source question was hard to understand for me.

 

[sna_gz]

Wherever you got this problem or whoever wrote the problem is mistaken.
If a die (standard cubic die, numbered 1 to six) is tossed \[imath]n[/imath] times then
the generating function [imath](x+x^2+x^3+x^4+x^5+x^6)^n[/imath] tells us the number of ways a sum of the faces can be made. SEE THIS
For example a gamming die is tossed eight them ( {\sum\limits_{k = 1}^6 {{x^k}} } \right)^8}[/imath] [imath]{\leftcontains the term [imath]3368x^{41}[/imath]. Which tell s that the sum [imath]41[/imath] can be gotten in [imath]3368[/imath] ways so the probability is [imath]\dfrac{3368}{6^8}[/imath]. We also use that sum if we roll [imath]n[/imath] dice.

thanks, the dice problem is a known subject and if we wanna look at the other ways of showing generating function :

[imath]\prod_{i=a}^{n} \frac{1-x^6}{1-x}[/imath]
which I am not sure but can be a bit calculatable by some choosing drama.
I was trying to be sure about this so I can contact the author of the book who seems to be an expert.
I will notify you if I got useful information.
finally, I would be thankful if anybody has a nice and rich source of generating function.