An algebra problem by Steven Lee

Algebra Level 3

Consider an n th n^\text{th} -degree polynomial p ( x ) p(x) that satisfies the following condition:

p ( k ) = k k + 1 , k = 0 , 1 , 2 , 3 , . . . , n . p(k) = \frac{k}{k+1}, ~~k = 0, 1, 2, 3, ..., n .

Find the value of p ( n + 1 ) p(n+1) when n = 101. n = 101.


The answer is 1.

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Steven Lee by Steven Lee , 22, Indonesia

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1 solution

Ravi Dwivedi
Aug 5, 2015

Let Q ( x ) = ( x + 1 ) P ( x ) − x Q(x)=(x+1)P(x)-x

( x + 1 ) P ( x ) − x = c x ( x − 1 ) ( x − 2 ) ⋅ ⋅ ⋅ ( x − n ) (x+1)P(x)-x=cx(x-1)(x-2)\cdot \cdot \cdot (x-n)

Put x = − 1 x=-1 and get 1 = c ( − 1 ) n + 1 ( n + 1 ) ! 1=c(-1)^{n+1}(n+1)!

P ( x ) = ( − 1 ) n + 1 x ( x − 1 ) ⋅ ⋅ ⋅ ( x − n ) / ( n + 1 ) ! + x x + 1 P(x)=\frac{(-1)^{n+1}x(x-1)\cdot \cdot \cdot (x-n)/(n+1)!+x}{x+1}

P ( n + 1 ) = { 1 for odd n , n n + 2 for even n P(n+1)=\begin{cases} 1 \quad \text{for odd n},\\ \frac{n}{n+2} \quad \text{for even n} \end{cases}

5 Helpful 0 Interesting 1 Brilliant 0 Confused

Moderator note:

This is the standard solution which uses the ideas of Remainder Factor Theorem to find the Polynomial Interpolation.

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