ganeshie8
  • ganeshie8
@DLS
Mathematics
  • Stacey Warren - Expert brainly.com
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katieb
  • katieb
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ganeshie8
  • ganeshie8
suppose \(G(x) = \sum\limits_{x=0}^{\infty} a_nx^n\)
ganeshie8
  • ganeshie8
can you express below in terms of \(G(x)\) ? \[\sum\limits_{x=0}^{\infty} \color{red}{a_{n+1}}x^n = ?\]
DLS
  • DLS
multiply divide by x ?

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ganeshie8
  • ganeshie8
Exactly, could you do it all the way..
ganeshie8
  • ganeshie8
\[\sum\limits_{x=0}^{\infty} \color{red}{a_{n+1}}x^n =\dfrac{1}{x}\sum\limits_{x=0}^{\infty} \color{red}{a_{n+1}}x^{n+1} \]
ganeshie8
  • ganeshie8
next what
DLS
  • DLS
next what? :o
ganeshie8
  • ganeshie8
i want to express above series in terms of G(x)
DLS
  • DLS
so it becomes G(x)/x
DLS
  • DLS
wait wait
ganeshie8
  • ganeshie8
That is very wrong
DLS
  • DLS
G(x) + a0 / x ?
ganeshie8
  • ganeshie8
G(x) has the term \(a_0\) but below has no \(a_0\) \[\sum\limits_{x=0}^{\infty} \color{red}{a_{n+1}}x^n =\dfrac{1}{x}\sum\limits_{x=0}^{\infty} \color{red}{a_{n+1}}x^{n+1} \]
ganeshie8
  • ganeshie8
Could you show me how
DLS
  • DLS
isn't the summation variable n ? :O
ganeshie8
  • ganeshie8
my mistake, fixed below \[\sum\limits_{n=0}^{\infty} \color{red}{a_{n+1}}x^n =\dfrac{1}{x}\sum\limits_{n=0}^{\infty} \color{red}{a_{n+1}}x^{n+1} \]
DLS
  • DLS
\[\sum\limits_{n=0}^{\infty} \color{red}{a_{n+1}}x^n =\dfrac{1}{x}\sum\limits_{n=0}^{\infty} \color{red}{a_{n+1}}x^{n+1} = \frac{1}{x}(G(x)+a_0)\]
ganeshie8
  • ganeshie8
that is wrong
ganeshie8
  • ganeshie8
my mistake, fixed below \[\sum\limits_{n=0}^{\infty} \color{red}{a_{n+1}}x^n =\dfrac{1}{x}\sum\limits_{n=0}^{\infty} \color{red}{a_{n+1}}x^{n+1} =\dfrac{1}{x}(a_1x^1+a_2x^2+\cdots ) \]
ganeshie8
  • ganeshie8
the stuff inside that parenthesis should be \(G(x) \color{red}{-}a_0\) right
DLS
  • DLS
yes
DLS
  • DLS
so we need to add 2*a0 ?
ganeshie8
  • ganeshie8
No
ganeshie8
  • ganeshie8
\[\sum\limits_{n=0}^{\infty} \color{red}{a_{n+1}}x^n =\dfrac{1}{x}\sum\limits_{n=0}^{\infty} \color{red}{a_{n+1}}x^{n+1} =\dfrac{1}{x}(a_1x^1+a_2x^2+\cdots ) = \dfrac{1}{x}(G(x)-a_0)\]
DLS
  • DLS
oh yes, sorry my bad
ganeshie8
  • ganeshie8
that is easy ok
ganeshie8
  • ganeshie8
using that, try solving below system : \(a_n = 3a_{n-1} + 2b_{n-1}\) \(b_n = a_{n-1} + 2b_{n-1}\)
ganeshie8
  • ganeshie8
notice, that system is same as \(a_{n+1} = 3a_{n} + 2b_{n}\) \(b_{n+1} = a_{n} + 2b_{n}\)
DLS
  • DLS
yes it is the same :)
ganeshie8
  • ganeshie8
You may start by letting the generating functions for \(a_n\) and \(b_n\) be \(G(x)\) and \(H(x)\) respectively
ganeshie8
  • ganeshie8
multiply \(x^n\) both sides of the recurrence relations, and sum them from 0 to infinity
DLS
  • DLS
\[\sum_{n=0}^\infty a_{n+1}x^n = \sum_{n=0}^\infty3a_{n}x^n +\sum_{n=0}^\infty 2b_{n}x^n\]
ganeshie8
  • ganeshie8
we get : \(\color{red}{\sum \limits_{0}^{\infty}}a_{n+1} ~x^n= \color{red}{\sum \limits_{0}^{\infty}}3a_{n} ~x^n+ \color{red}{\sum \limits_{0}^{\infty}}2b_{n}~x^n\) \(\color{red}{\sum \limits_{0}^{\infty}}b_{n+1} ~x^n= \color{red}{\sum \limits_{0}^{\infty}}a_{n}~x^n + \color{red} {\sum \limits_{0}^{\infty}}2b_{n}~x^n\)
DLS
  • DLS
\[\Large \frac{G(x) - a_0}{x} = 3 G(x) + 2 H(x)\]
ganeshie8
  • ganeshie8
substitute the previous formula formula
ganeshie8
  • ganeshie8
Yes!
DLS
  • DLS
\[\color{red}{\sum \limits_{0}^{\infty}}b_{n+1} ~x^n= \color{red}{\sum \limits_{0}^{\infty}}a_{n}~x^n + \color{red} {\sum \limits_{0}^{\infty}}2b_{n}~x^n \] \[\Large \frac{H(x) - a_0}{x} = G(x) + 2H(x)\]
ganeshie8
  • ganeshie8
we get : \(\dfrac{G(x)-a_0}{x}= 3G(x) + 2H(x)\) \(\dfrac{H(x)-b_0}{x}= G(x) + 2H(x)\)
ganeshie8
  • ganeshie8
Collect the terms G and H in each equation
ganeshie8
  • ganeshie8
Remember, our goal is to solve \(G(x)\) and \(H(x)\)
ganeshie8
  • ganeshie8
and we have a linear system of two equations in terms of G(x) and H(x) we can solve it easily
DLS
  • DLS
yes..suppose we get both G(x) and H(x)
ganeshie8
  • ganeshie8
we're done
ganeshie8
  • ganeshie8
once you have the generating functions G(x) and H(x) independently, you know what to do next to solve the recurrence relation
DLS
  • DLS
oh yes :) this was also easy :D
ganeshie8
  • ganeshie8
look at 2nd reply http://math.stackexchange.com/questions/369377/solving-two-simultaneous-recurrence-relations
ganeshie8
  • ganeshie8
btw there was a typo in 5th line in that reply, he wrote 2A(z) instead of 3A(z)
DLS
  • DLS
yeah I just noticed :P nonetheless the procedure is exactly the same
ganeshie8
  • ganeshie8
Yes
DLS
  • DLS
and in that question yesterday, where we were supposed to count the solutions using G((x) , things can get pretty nasty when there are constraints on all 5 variables, G(x) may nor may not be easy to solve..is there something that can be done in the exam using calculator or something ? :| or should I leave it as "The answer is the coefficient of x^blah blah in this equation) ?
ganeshie8
  • ganeshie8
they cookup problems such that finding the coefficient would be easy
ganeshie8
  • ganeshie8
they test whether you know the method or not
ganeshie8
  • ganeshie8
testing ur algebra skills is pointless in an exam
DLS
  • DLS
yeah I guessed so :D it should be easy then. BTW I have been trying this for a while : Prove using binomial theorem > \[\Large \sum_0^n C(n,r) 2^r = 3^n\] I am unsure how to proceed with BT in this proof actually, any hints ?
ganeshie8
  • ganeshie8
that is very easy
ganeshie8
  • ganeshie8
expand below using binomial theorem : \[(1+2)^n\]
ganeshie8
  • ganeshie8
\[3^n = (1+2)^n = \sum\limits_{r=0}^n C(n,r)1^{n-r}2^r = \sum\limits_{r=0}^n C(n,r)2^r \]
ganeshie8
  • ganeshie8
that is an one line proof
DLS
  • DLS
lol didn't think of that :O amazing :P why is this of 6 marks nevermind :p
ganeshie8
  • ganeshie8
using same reasong, try proving below : \[\sum\limits_{r=0}^n C(n,r)=2^n\]
DLS
  • DLS
yeah that's the expansion of (1+x)^n with x = 1
DLS
  • DLS
I guess I should avoid thinking too complex :|
ganeshie8
  • ganeshie8
Yep!
ganeshie8
  • ganeshie8
trust that your professor loves you and gives only problems that you're familiar with :)
DLS
  • DLS
hahah..its not true in case of my college, if they aren't in a good mood they would either make the paper difficult or there would be mistakes in the question. :| there was a mistake in a 2 questions in my maths exam, and I spent a lot of time wasting on them :/ hopefully tomorrow will be good :)

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