The Student Room Group

help !!! generating function

If N is number of heads in n fair coin tosses, then show that
the generating function G X(s) = (2^(-n) )(1+s)^n

Scroll to see replies

Reply 1
Avatar for Zacken
Zacken
22
Original post by vanessa_tram
If N is number of heads in n fair coin tosses, then show that
the generating function G X(s) = (2^(-n) )(1+s)^n


What have you tried?
for my other exercises i've been doing, I think it should be (see attached file)
but then i dont know what to do next. i dont know if i should have E(s^N/tails)P(tails) in there or not as we only want number of heads.
CodeCogsEqn (3).gif2.6KB
Original post by Zacken
What have you tried?


for my other exercises i've been doing, I think it should be (see attached file)but then i dont know what to do next. i dont know if i should have E(s^N/tails)P(tails) in there or not as we only want number of heads.
CodeCogsEqn (3).gif
Original post by vanessa_tram
for my other exercises i've been doing, I think it should be (see attached file)but then i dont know what to do next. i dont know if i should have E(s^N/tails)P(tails) in there or not as we only want number of heads.
CodeCogsEqn (3).gif


No need to condition on anything here. The probability distribution for the number of heads thrown is Binomial with parameters n and 1/2. You then just calculate the sum resulting from the definition of a generating function.
Original post by Gregorius
No need to condition on anything here. The probability distribution for the number of heads thrown is Binomial with parameters n and 1/2. You then just calculate the sum resulting from the definition of a generating function.


thank you, i got to here, do u know how to sum this up ???
CodeCogsEqn.gif
Reply 6
Avatar for Zacken
Zacken
22
Original post by vanessa_tram
thank you, i got to here, do u know how to sum this up ???
CodeCogsEqn.gif


Do you know how to sum
Unparseable latex formula:

\sum_{x=0}^{n} \binom{n \choose x}

?

Think about the binomial series (1+x)n(1+x)^n with a certain value of xx.
Original post by Zacken
Do you know how to sum
Unparseable latex formula:

\sum_{x=0}^{n} \binom{n \choose x}

?

Think about the binomial series (1+x)n(1+x)^n with a certain value of xx.


i used this theorem NumberedEquation1.gif so i got (1+(1/2)s)^n , still cant get the answer
Reply 8
Avatar for Zacken
Zacken
22
Original post by vanessa_tram
i used this theorem NumberedEquation1.gif so i got (1+(1/2)s)^n , still cant get the answer


Huh? ∑x=0n(nx)(s2)n\sum_{x=0}^{n} {n \choose{x}} \left( \frac{s}{2} \right)^n =(s2)n∑x=0n(nx)=(s2)n(1+x)n= \left( \frac{s}{2} \right)^n \sum_{x=0}^{n} {n \choose{x}}= \left( \frac{s}{2} \right)^n (1 + x)^n for some suitable xx.
(edited 8 years ago)
Original post by Zacken
Huh? ∑x=0n(nx)(s2)n\sum_{x=0}^{n} {n \choose{x}} \left( \frac{s}{2} \right)^n =(s2)n∑x=0n(nx)=(s2)n(1+x)n= \left( \frac{s}{2} \right)^n \sum_{x=0}^{n} {n \choose{x}}= \left( \frac{s}{2} \right)^n (1 + x)^n for some suitable xx.


i got it wrong from the last step cos it was power of n, not x, so i cant use that formula, i dont get ur hint, sorry, please help me again, please. can i write like this ?
CodeCogsEqn.gif
Reply 10
Avatar for Zacken
Zacken
22
Original post by vanessa_tram
i got it wrong from the last step cos it was power of n, not x, so i cant use that formula, i dont get ur hint, sorry, please help me again, please. can i write like this ?
CodeCogsEqn.gif


(s2)n\left(\frac{s}{2}\right)^n is just a constant, so you can pull it out of your sum.

Then you're left with (s2)n∑x=0n(nx)\left(\frac{s}{2}\right)^n \sum_{x=0}^{n} {n \choose x}

where the sum evaluates to (1+1)n(1+1)^n, then multiply it by the constant.
Original post by vanessa_tram
thank you, i got to here, do u know how to sum this up ???
CodeCogsEqn.gif


Your equation isn't quite right yet. Remember that the definition of the generating function is

∑r=0nP(X=r)sr\displaystyle \sum_{r=0}^{n} \mathbb{P}(X=r) s^r

So you want

∑r=0n(nr)(12)r(12)n−rsr\displaystyle \sum_{r=0}^{n} \binom{n}{r} \left(\frac{1}{2}\right)^r \left(\frac{1}{2}\right )^{n-r} s^r

Not an sns^n as you have in your formula.
Original post by Zacken
(s2)n\left(\frac{s}{2}\right)^n is just a constant, so you can pull it out of your sum.

Then you're left with (s2)n∑x=0n(nx)\left(\frac{s}{2}\right)^n \sum_{x=0}^{n} {n \choose x}

where the sum evaluates to (1+1)n(1+1)^n, then multiply it by the constant.


but as we have s^n outside, no matter how we expand inside the bracket, how can we get the result with (1+s)^n
Reply 13
Avatar for Zacken
Zacken
22
Original post by vanessa_tram
but as we have s^n outside, no matter how we expand inside the bracket, how can we get the result with (1+s)^n


I assumed your result was correct, it's not - see Greg's post for more information.
Original post by Zacken
I assumed your result was correct, it's not - see Greg's post for more information.


thank you, i got the answer, made a little mistake , he just pointed it out
Original post by Gregorius
Your equation isn't quite right yet. Remember that the definition of the generating function is

∑r=0nP(X=r)sr\displaystyle \sum_{r=0}^{n} \mathbb{P}(X=r) s^r

So you want

∑r=0n(nr)(12)r(12)n−rsr\displaystyle \sum_{r=0}^{n} \binom{n}{r} \left(\frac{1}{2}\right)^r \left(\frac{1}{2}\right )^{n-r} s^r

Not an sns^n as you have in your formula.

thank you very much, i got it now.
Reply 16
Avatar for Zacken
Zacken
22
Original post by vanessa_tram
thank you, i got the answer, made a little mistake , he just pointed it out


Don't thank me, I was essentially useless - Greg can do with some praise :yep:
Original post by Zacken
Don't thank me, I was essentially useless - Greg can do with some praise :yep:


no, thank you u was trying ur best to help me. :smile:
Original post by Zacken
Don't thank me, I was essentially useless - Greg can do with some praise :yep:


That's OK, just throw rose petals and money...:u:
Oh God I bet this is A-level Maths/FM isn't it... I'm going to cry...

Quick Reply

Related discussions

Latest

Trending

Trending

Articles for you