$A(x) = a_0 + a_1x + a_2x^2 +\cdots + a_nx^n + \cdots$

$B(x) = b_0 + b_1x + b_2x^2 +\cdots + b_nx^n + \cdots$

#### Solution

We start with $A(x) = 1 + 2x + 6x^2 + 20x^3 + \cdots + \binom{2n}{n}x^n + \cdots$

Differentiate both sides,

$A'(x) = 2 + 12x + 60x^2 + \cdots + \binom{2n}{n}nx^{n-1} + \binom{2n+2}{n+1}(n+1)x^n + \cdots$

$\frac{A' - 2A}{4x} = 2 + 12x + 60x^2 + \cdots $

The coefficient of $x^{n-1}$ in $\frac{A'-2A}{4x}$ is the coefficient of $x^n$ in $\frac{A'-2A}{4}$

$= \frac{1}{4}(\binom{2n+2}{n+1}(n+1) - 2\binom{2n}{n})$

$= \frac{1}{4}\binom{2n}{n}(\frac{(2n+2)(2n+1)}{n+1} - 2) = n\binom{2n}{n}$

which is the coefficient of $x^{n-1}$ in $A'$.

Thus, $\frac{A'(x)-2A(x)}{4x} = A'(x)$.

$A'(x)(1-4x)= 2A(x) \iff \frac{dA}{dx} (1-4x) = 2A$

$\frac{dA}{2A} = \frac{dx}{1-4x}$

Integrate both sides, with the boundary condition that $A(0)=0$

$\frac{\ln |A|}{2} = \frac{\ln (1-4x)}{-4} \iff A = (1-4x)^{-\frac{1}{2}}$

So $A(x) = \frac{1}{\sqrt{1-4x}}$

To find a compact expression for $B(x)$, we temporarily use $x^2$ as the generating variable.

Let $P(x) = B(x^2) = b_0 + b_1x^2 + b_2x^4 + \cdots + b_nx^{2n} + \cdots$

By inspection, we have

$(4-27x^2)P'' - 81xP' -24P = 0$

And its correctness can be verified by comparing the coefficient of $x^n$ directly. All that's left is to simply solve the differential equation.

I wasn't able to solve the actual equation, but I did find that if we let $y = -27xP + (4-27x^2)P'$ then $y' = -3P$

If you follow a similar methodology for B:

ReplyDeleteSet B + (p+qx)B' + (r+sx+tx^2)B'' = 0

Compare coefficients for particular k values..eg 0,1,2,3,4 to generate 5 equations

Solve the linear system...

You will get:

B + (-1/3 + 9x)B' + (-2x/3 + (9x^2)/2)B'' = 0

Perhaps this can be solved with the right substitution...?