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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, 6 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.)

Jon Haussmann - 3 years, 6 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

Mardokay Mosazghi - 3 years, 6 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

Victor Loh - 3 years, 6 months ago

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@Victor Loh ?

Finn Hulse - 3 years, 6 months ago

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@Finn Hulse 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

Victor Loh - 3 years, 6 months ago

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@Victor Loh 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...

Finn Hulse - 3 years, 6 months ago

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

Victor Loh - 3 years, 6 months ago

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@Victor Loh The degree n part is important.

Victor Loh - 3 years, 6 months ago

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@Victor Loh I see! I have finally understood. I will post a solution.

Finn Hulse - 3 years, 6 months ago

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@Finn Hulse That's great!

Victor Loh - 3 years, 6 months ago

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

Finn Hulse - 3 years, 6 months ago

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