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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
2 years, 11 months ago

## Comments

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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.) · 2 years, 11 months ago

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Comment deleted Apr 20, 2014

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I agree since it is defined it doyle be kequals n+1/n+2 · 2 years, 11 months ago

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No, f(k) is only true for k=1,2,...,n. No one said it was true for k=n+1 · 2 years, 11 months ago

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? · 2 years, 11 months ago

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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 · 2 years, 11 months ago

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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... · 2 years, 11 months ago

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They actually do. Go read up more. This is a past year Singapore competition problem... · 2 years, 11 months ago

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The degree n part is important. · 2 years, 11 months ago

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I see! I have finally understood. I will post a solution. · 2 years, 11 months ago

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That's great! · 2 years, 11 months ago

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I just started writing it and hit a dead end. I think I understand this problem! And it sucks! :O · 2 years, 11 months ago

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