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Next: Extending the result Up: Deducing Previous: Differential equation for

Solving the differential equation

Since we know that $ F_n(x)$ is a polynomial of degree $ n$, we can put

$\displaystyle F_n(x) := \sum_{r=0}^n a_r x^r = a_n x^n + \sum_{r=0}^{n-1} a_r x^r$ (4)

and it follows that

$\displaystyle F_n'(x) = \sum_{r=0}^n r a_r x^{r-1} = \sum_{r=1}^n r a_r x^{r-1} =
\sum_{r=0}^{n-1} (r+1) a_{r+1} x^r
$

Substituting these into (3) gives
$\displaystyle a_n x^n + \sum_{r=0}^{n-1} a_r x^r + \sum_{r=0}^{n-1} (r+1) a_{r+1}
x^r$ $\displaystyle \equiv$ $\displaystyle x^n$  
$\displaystyle a_n x^n + \sum_{r=0}^{n-1} (a_r + (r+1) a_{r+1}) x^r$ $\displaystyle \equiv$ $\displaystyle x^n$  

Comparing coefficients of $ x^m$ on both sides, we find $ a_n = 1$ and

$\displaystyle a_r + (r+1) a_{r+1} = 0$ (5)

(for $ r = 0, \ldots, n-1$).

Using (5), we obtain

$\displaystyle a_n$ $\displaystyle =$ $\displaystyle 1$  
$\displaystyle a_{n-1}$ $\displaystyle =$ $\displaystyle -n a_n = -n$  
$\displaystyle a_{n-2}$ $\displaystyle =$ $\displaystyle -(n-1) a_{n-1} = (n-1) n$  
$\displaystyle a_{n-3}$ $\displaystyle =$ $\displaystyle -(n-2) a_{n-2} = -(n-2) (n-1) n$  

There is a clear pattern emerging and we can make the conjecture

$\displaystyle a_{n-k} = (-1)^k \frac{n!}{(n-k)!}$ (6)

To prove (6), we note that
$\displaystyle a_{n-k-1}$ $\displaystyle =$ $\displaystyle -(n - k) a_{n-k}$  
  $\displaystyle =$ $\displaystyle -(n - k) (-1)^k \frac{n!}{(n-k)!}$  
  $\displaystyle =$ $\displaystyle (-1)^{k+1} \frac{n!}{(n-k-1)!}$  
  $\displaystyle =$ $\displaystyle a_{n-k-1}$  

It follows that (6) is correct, by induction.

To find $ a_r$ we simply substitute $ r = n - k$ into (6) and immediately obtain

$\displaystyle a_r = (-1)^{n-r} \frac{n!}{r!}
$

Substituting into (4), we obtain the1 polynomial solution to (3),

$\displaystyle F_n(x) = \sum_{r=0}^n (-1)^{n-r} \frac{n!}{r!} x^r
$


next up previous
Next: Extending the result Up: Deducing Previous: Differential equation for
Alexander Frolkin 2001-06-02