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If $f(x)$ is the polynomial (coefficient of leading term is unity) in 'x' of least degree such that $f(1)=5 , f(2)=4, f(3)=3, f(4)=2, f(5)=1$

Then $f(0)= ?$

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1 Answer 1

The naive approach via Lagrange yields $$f(0)=5\frac{(0-2)(0-3)(0-4)}{(1-2)(1-3)(1-4)}+4\frac{(0-1)(0-3)(0-4)}{(2-1)(2-3)(2-4)}+3\frac{(0-1)(0-2)(0-4)}{(3-1)(3-2)(3-4)}+2\frac{(0-1)(0-2)(0-3)}{(4-1)(4-2)(4-3)},$$ but this polynomial is just $-x+6$ ... Notice that the four points lay on a straight line hence the polynomial must have degree $4$ or more. Agent Vandermonde then finds $f(x)=x^4-10x^3+35x^2-51x+30$.

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This would be more convincing if you could show us the polynomial itself (just to be sure that its leading coefficient is really +1). –  TonyK Nov 18 '13 at 12:04
    
You're right, my first guess was wrong ... –  Michael Hoppe Nov 18 '13 at 12:35
    
how did you get this expression of f(x) ? –  maths lover Nov 18 '13 at 17:12
    
@mathslover Via agent Vandermonde: en.wikipedia.org/wiki/Unisolvence_theorem#unisolvence_theorem by computing $\begin{pmatrix}1&1&1&1\\ 8&4&2&1\\ 27&9&3&1\\ 64&16&4&1\end{pmatrix}^{-1}\begin{pmatrix}4\\ -12\\ -78\\ -254\end{pmatrix}.$ –  Michael Hoppe Nov 18 '13 at 17:44
    
i thing so f(5) is not equal to 1 –  maths lover Nov 19 '13 at 14:43

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