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# How to solve?

Let $$f(x)$$ be a polynomial of degree $$n$$ such that $$f(k)=\frac{k}{k+1}$$ for $$k=0,1,2,...,n$$. Find $$f(n+1)$$.

Help would be greatly appreciated.

Victor

Note by Victor Loh
3 years, 9 months ago

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Let $$g(x)$$ be the polynomial $g(x) = (x + 1) f(x) - x.$ What can you say about $$g(x)$$? (Think roots and degree.)

- 3 years, 9 months ago

Comment deleted Apr 20, 2014

I agree since it is defined it doyle be kequals n+1/n+2

- 3 years, 9 months ago

No, f(k) is only true for k=1,2,...,n. No one said it was true for k=n+1

- 3 years, 9 months ago

?

- 3 years, 9 months ago

f(k) = k/k+1 holds true when k=0,1,2,3,...,n. The question did not state that f(k)=k/k+1 holds true for k=n+1, so the answer need not be n+1/n+2

- 3 years, 9 months ago

I don't think that functions work like that. You're asking "This function only applies to these numbers. What happens if you plug in a different number?". That's a bit strange...

- 3 years, 9 months ago

They actually do. Go read up more. This is a past year Singapore competition problem...

- 3 years, 9 months ago

The degree n part is important.

- 3 years, 9 months ago

I see! I have finally understood. I will post a solution.

- 3 years, 9 months ago

That's great!

- 3 years, 9 months ago

I just started writing it and hit a dead end. I think I understand this problem! And it sucks! :O

- 3 years, 9 months ago