Maths problem

[borisp]

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the appropriate generating function

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ahhh, this is how to do it. The relation asks for a square root of the generating function 1+4x+16x^2+...=1/(1-4x). So use binomial coefficients to compute the nth coefficient of (1-4x)^(-1/2), and that should be the answer. (I haven't actually done this, but it should work out.)

[borisp]

Ok, so to hell with real work: the nth coeffcient is

4^n(1/2 choose n) = (4^n/2^n) ((1*3*5*...*2n-1)/n!)=

=2^n*((2n!)/((2n)*(2n-2)*...*(4)*(2)*n!)=(2^n/2^n)(2n!/(n!n!))=

= 2n choose n

[blah_blah]

the idea is similar to what you do with the catalan number generating function/recurrence.

[borisp]

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the idea is similar to what you do with the catalan number generating function/recurrence.

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The catalan generating function would be the integral of this function, divided by x. Is there a nice combinatorial interpretation of derivatives and integrals of generating functions?

Forgive me, I am an analyst [img]/images/graemlins/smile.gif[/img]...haven't done this since undergrad.

[blah_blah]

idk, i too am an analyst (well, more of a probabilist these days ... but close enough)

[thylacine]

[ QUOTE ]
Ok, so to hell with real work: the nth coeffcient is

4^n(1/2 choose n) = (4^n/2^n) ((1*3*5*...*2n-1)/n!)=

=2^n*((2n!)/((2n)*(2n-2)*...*(4)*(2)*n!)=(2^n/2^n)(2n!/(n!n!))=

= 2n choose n

[/ QUOTE ]

There are some sign errors here.
Rough idea is right.

[borisp]

I left out the signs because they become +1 in the end. The (-1)^n from the -4 and the (-1)^n from the -1/2 join forces to become (-1)^2n = +1.

So there really aren't any sign errors, in that the result is correct. I was just lazy with the typing. Technically the first term should be (-4)^n*(-1/2 choose n).

[thylacine]

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I left out the signs because they become +1 in the end. The (-1)^n from the -4 and the (-1)^n from the -1/2 join forces to become (-1)^2n = +1.

So there really aren't any sign errors, in that the result is correct. I was just lazy with the typing. Technically the first term should be (-4)^n*(-1/2 choose n).

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No. (-1)^n (r choose n) does not equal (-r choose n). You can figure out what it does equal.

You are right that the first term should be (-4)^n*(-1/2 choose n).

You essentially got the solution I was thinking of. I was interested if there were any other approaches.

[jay_shark]

Has anyone thought of a combinatorial solution to this ?
I have tried and failed .

We wish to prove :

2nCn + 2c1*(2n-2)C(n-1) + 4c2*(2n-4)C(n-2) + ...+ 2nCn = 4^n

[thylacine]

[ QUOTE ]
Has anyone thought of a combinatorial solution to this ?
I have tried and failed .

We wish to prove :

2nCn + 2c1*(2n-2)C(n-1) + 4c2*(2n-4)C(n-2) + ...+ 2nCn = 4^n

[/ QUOTE ]

I don't know of one. I was thinking of the generating function method. But it would be interesting to know.