Consider a function F(x,n) defined by:
F(x,n) = (d/dx)(4n+3)(x2+1)-1
Determine the value of F(1,n).
Since we are asked to evaluate derivatives at x=1, I decided to see what the series expansion at x=1 is like. Wolfram gave me this (slightly modified for emphasis):
1/(x^2+1)
= (1/2)*(x-1)^0 - (1/2)*(x-1)^1 + (1/4)*(x-1)^2
- (1/8)*(x-1)^4 + (1/8)*(x-1)^5 - (1/16)*(x-1)^6
+ (1/32)*(x-1)^8 - (1/32)*(x-1)^9 + (1/64)*(x-1)^10 + ...
Split into groups of three terms it seems obvious what the pattern is, and the relevance to this problem is that there are no terms for (x-1)^(4n+3), or equivalently the coefficient of any (x-1)^(4n+3) is zero. This then means F(1,n)=0 for all nonnegative integer n.
This is certainly not rigorous. I hesitate to call this a proper solution since all I did was plug this into Wolfram and see what happened.
Edited on June 23, 2023, 11:47 am
Answer
