Hey /sci/, please help a brainlet out and explain me what the

what's the issue, it's a_n= 2n choose n

>>9132848
I know, my issue is how to find the left hand side from the a_n= 2n choose n

>>9132853
The 1/sqrt(1-4x) i mean

>>9132857
that's its taylor series on the right

>>9132853
So you want to know how you can find 1/sqrt(1-4x^2) if you have 1+2x^2+6x^4+...? Normally that's not possible.

>>9132940
Well that's what my teacher is asking me to do, i need to find 1/sqrt(1-4x) from 2n choose n

>>9132954
(1+2x+6x^2+20x^3+...)^2
=1+4x+16x^2+64x^3...
=1/(1-4x) by geometric series formula

so 1+2x+6x^2+20x^3+...= 1/sqrt(1-4x)

>>9132962
how the fuck does 6 squared equal 16 and 20 squared = 64

>>9133003
>how the fuck does 6 squared equal 16 and 20 squared = 64
it doesn't. do you not know how to multiply?

>>9132843
Try to write a_(n+1) in terms of a_n.

(1) I get (n+1)*a_(n+1)=(4n+2)*a_n.

Let F(x)=a_0*x^0+a_1*x^1...

Multiply eq (1) by x^(n+1) then sum n from 0 to infinity.

LHS gets xF'(x)
(notice the sum leaves out a_0, but the derivative chops it off anyway)

RHS gets 4x(xF'(x))+2xF(x)

You get the differential equation xF'(x)=4xxF'(x)+2xF(x) with F(0)=1.
Re-arrange to get F'(x)/F(x)=2/(1-4x).
log(F(x))=C-log(1-4x)/2.
F(x)=C/sqrt(1-4x).
From F(0)=1 you get C=1.