OK, I think I was able to derive this correctly. I really hope this isn't homework, as I put a lot of time into this for you to just be cheating off me. I'll see if I can explain this concisely, but it'll be tough. First, we'll write out the recursion relation explicitly:
da_n=da_{n-1}-\frac{1}{p^2} a_{n-1}
a_n = a_{n-1} + da_{n}
Where a0=1 and da0=0. Next we define:
\alpha \equiv -\frac{1}{p^2}
And we find:
da_n=\alpha(a_0+a_1+ ... +a_{n-1})
And similarly:
\begin{align*} a_n &= da_0+... + da_{n-1}\\<br />
&=1+ \alpha \left( (a_0) + (a_0+a_1) + ... + (a_0+...+a_{n-1})\right) \\<br />
&=1+ \alpha \sum_{k=0}^{n-1} (n-k) a_k \end{align*}
Which eliminates the need for the dan. Now, you can see by computing a few terms that the an will have the form:
a_n=\sum_{j=0}^n c_{nj} \alpha^j
For some constants cnj. Plugging this into the above equation:
\begin{align*} \sum_{j=0}^n c_{nj} \alpha^j &=1+ \alpha \sum_{k=0}^{n-1} (n-k) \sum_{j=0}^k c_{kj} \alpha^j \\<br />
&=1+\sum_{j=0}^{n-1} \alpha^{j+1}\sum_{k=j}^{n-1} (n-k)c_{kj} \end{align*}
Equating like powers of \alpha gives a recursion relation for cnj:
c_{nj}=\sum_{k=j-1}^{n-1} (n-k)c_{k j-1}
With initial conditions cn0=1. Now, it turns out these coefficients are (2j)th order polynomials in n. We will be taking the limit as \alpha goes to zero later on, and n will go to p=1/\sqrt{\alpha}, so we're only interested in the highest power of n. Thus, we will use the approximation:
c_{nj} \approx b_j n^{2j}
Also, the following approximation holds to first order (as you'll see, we actually approximate a series that doesn't start at 1 by this formula, but the limit justitifies this):
\sum_{k=1}^{n-1} k^q \approx \frac{1}{q+1} n^{q+1}
So that:
\begin{align*}b_j n^{2j} &= \sum_{k=j-1}^{n-1} (n-k) b_{j-1} k^{2j-2} \\<br />
&=n\sum_{k=j-1}^{n-1}b_{j-1} k^{2j-2}-\sum_{k=j-1}^{n-1} b_{j-1} k^{2j-1} \\<br />
&\approx \frac{1}{2j-1} n^{2j} b_{j-1}- \frac{1}{2j} n^{2j}b_{j-1} \\<br />
&=\frac{1}{(2j)(2j-1)} n^{2j} b_{j-1} \end{align*}
And so:
b_{j} = \frac{1}{(2j)(2j-1)} b_{j-1}
And together with b0=1, we have our first non-recursive results, even though they're only approximate:
b_j= \frac{1}{(2j)!}
c_{nj} \approx \frac{n^{2j}}{(2j)!}
a_n \approx \sum_{j=0}^n \frac{n^{2j}}{(2j)!} \alpha^j
We are interested in apx (I'm using x here where the OP used n) This is given by (putting back in p):
\begin{align*}a_{px} &\approx \sum_{j=0}^{px} \frac{{(px)}^{2j}}{(2j)!} \frac{(-1)^j}{p^{2j}}\\<br />
&=\sum_{j=0}^{px}(-1)^j \frac{x^{2j}}{(2j)!}\end{align*}
Which, in the limit p>>1/x (x>0) gives:
a_{px} &\approx \sum_{j=0}^{\infty} (-1)^j \frac{x^{2j}}{(2j)!}=cos(x)
The taylor series for cos(x). Anyone know an easier way?
