## jagr2713 one year ago Help.

1. jagr2713

For a positive integer n, define as n degree polynomials$f_{n}(x) \ as$ $f_{n}(x)=\frac{ x(x-1)(x-2)...(x-n+1) }{ n! }$ If the sum of all the roots of the functional equation $f_{n}(x)+f _{n-1}(x)=f _{n}(x+2)$ Is 8514, what is the value of n?

2. jagr2713

@ganeshie8 @nincompoop @geerky42

3. anonymous

oh what fun. -_-

4. anonymous

Note that$f_{n-1} (x)=\frac{n}{x-n+1} f_{n} (x)$ $f_n(x+2)=\frac{(x+1)(x+2)}{(x-n+1)(x-n+2)} f_n (x)$

5. anonymous

sum of all the roots of the functional equation?

6. anonymous

are you sure you don't mean sum of all the $$x$$ satisfying that equation in $$x$$ rather than talking about 'roots of a functional equation' (which would be presumably functions)

7. anonymous

Put these back in the functional equation to get$f_n(x) \left( 1+\frac{n}{x-n+1} -\frac{(x+1)(x+2)}{(x-n+1)(x-n+2)}\right)=0$Simplify and find the roots, That's all I'm sayin, I'll say no more :)))

8. jagr2713

9. jagr2713

Nope

10. jagr2713

hm

11. jagr2713

Igtg but i will see this later :D

12. anonymous

oops, let me retry

13. anonymous

$$\frac{(x)_n}{n!}+\frac{(x)_{n-1}}{(n-1)!}=\frac{(x+2)_n}{n!}\$$x)_n+n(x)_{n-1}=(x+2)_n\\(x)_{n-1}\left(x-n+1+n\right)=(x+2)(x+1)(x)_{n-2}\\(x)_{n-2}(x-n+2)(x+1)-(x)_{n-2}(x+1)(x+2)=0\\(x)_{n-2}(x+1)\left[(x-n+2)-(x+2)\right]=0\\n(x)_{n-2}(x+1)=0so let's find the sum of the roots of \((x)_{n-2}=x(x-1)(x-2)\cdots(x-(n-2)+1)=x(x-1)(x-2)\cdots(x-(n-3))$$:$$\sum_{k=0}^{n-3}k=\frac12(n-3)(n-2)$$and then our additional root of $$-1$$ from $$x+1=x-(-1)$$ gives us:$$\frac12(n-2)(n-3)-1=8514\$$n-2)(n-3)=2\cdot8515  14. anonymous now notice \(\sqrt{2\cdot8515}\approx130.499$$ so $$n-2=131\implies n=133$$

15. anonymous

yeah, the above should be correct; are these Brilliant problems or something

16. jagr2713

Correct @oldrin.bataku :D