jagr2713
  • jagr2713
Help.
Mathematics
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katieb
  • katieb
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jagr2713
  • 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?
jagr2713
  • jagr2713
@ganeshie8 @nincompoop @geerky42
anonymous
  • anonymous
oh what fun. -_-

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anonymous
  • anonymous
Note that\[f_{n-1} (x)=\frac{n}{x-n+1} f_{n} (x) \] \[f_n(x+2)=\frac{(x+1)(x+2)}{(x-n+1)(x-n+2)} f_n (x)\]
anonymous
  • anonymous
sum of all the roots of the functional equation?
anonymous
  • 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)
anonymous
  • 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 :)))
jagr2713
  • jagr2713
Hm i see :D So what is your answer
jagr2713
  • jagr2713
Nope
jagr2713
  • jagr2713
hm
jagr2713
  • jagr2713
Igtg but i will see this later :D
anonymous
  • anonymous
oops, let me retry
anonymous
  • anonymous
$$\frac{(x)_n}{n!}+\frac{(x)_{n-1}}{(n-1)!}=\frac{(x+2)_n}{n!}\\(x)_n+n(x)_{n-1}=(x+2)_n\\(x)_{n-1}\left(x-n+1+n\right)=(x+2)(x+1)(x)_{n-2}\\(x)_{n-2}(x-n+2)(x+1)-(x)_{n-2}(x+1)(x+2)=0\\(x)_{n-2}(x+1)\left[(x-n+2)-(x+2)\right]=0\\n(x)_{n-2}(x+1)=0$$so 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:$$\frac12(n-2)(n-3)-1=8514\\(n-2)(n-3)=2\cdot8515 $$
anonymous
  • anonymous
now notice \(\sqrt{2\cdot8515}\approx130.499\) so \(n-2=131\implies n=133\)
anonymous
  • anonymous
yeah, the above should be correct; are these Brilliant problems or something
jagr2713
  • jagr2713
Correct @oldrin.bataku :D

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