Define f0(x) = ex, fn+1(x) = x fn'(x). Show that ∑0∞ fn(1)/n! = ee.
Solution
The trick is to use the power series for ex. Then we have immediately that fn(x) = ∑ rn xr/r! Hence, ∑ fn(x)/n! = 1 + x/1! (1 + 1/1! + 12/2! + ... ) + x2/2! (1 + 2/1! + 22/2! + ... ) + ... = 1 + xe/1! + (xe)2/2! + ... = eex. [We assume the usual theorems about rearranging terms in absolutely convergent series.]
© John Scholes
jscholes@kalva.demon.co.uk
27 Jan 2001