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

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

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## Comments

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TopNewestLet \(g(x)\) be the polynomial \[g(x) = (x + 1) f(x) - x.\] What can you say about \(g(x)\)? (Think roots and degree.) – Jon Haussmann · 2 years, 9 months ago

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– Mardokay Mosazghi · 2 years, 9 months ago

I agree since it is defined it doyle be kequals n+1/n+2Log in to reply

– Victor Loh · 2 years, 9 months ago

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

– Finn Hulse · 2 years, 9 months ago

?Log in to reply

– Victor Loh · 2 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+2Log in to reply

– Finn Hulse · 2 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...Log in to reply

– Victor Loh · 2 years, 9 months ago

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

– Victor Loh · 2 years, 9 months ago

The degree n part is important.Log in to reply

– Finn Hulse · 2 years, 9 months ago

I see! I have finally understood. I will post a solution.Log in to reply

– Victor Loh · 2 years, 9 months ago

That's great!Log in to reply

– Finn Hulse · 2 years, 9 months ago

I just started writing it and hit a dead end. I think I understand this problem! And it sucks! :OLog in to reply