A polynomial problem

Algebra Level 5

Let $p(x)$ be a polynomial of degree $8$ , such that $P(k) = \frac1k$ , $k$ belongs to ${ 1,2,3,..,9 }$ . Then find the value of $P(10)$ .

The answer is 0.2.

1 solution

Priyanshu Mishra
Dec 30, 2014

Given that P(k) = 1/k, rewriting we get: k.p(k) = 1 Let replace k by x. Then, x. p(x) = C(x-1)(x-2)(x-3).....(x-9) + 1 where C is the leading coefficient. Put x =0 : 0. p(0) = C.-9! + 1 0 = C. -9! + 1 -1 = C.-9! C = 1/ 9! , now the polynomial p(x) becomes x.p(x) = (1/9!)(x-1)(x-2)(x-3)...(x-9) + 1 Now put x =10

p(10) = 1/9! . 9! + 1 10.p(10) = 1 + 1 10.p(10) = 2 p(10) = 2/10 = 1/5. Hence the result.

LONG SOLUTION BUT NICE ONE. OTHER SOLUTIONS ARE ALSO WELCOMED!

That's that standard approach to dealing with this question, by using the Remainder Factor Theorem to figure out what the polynomial is. The "trick" here is that we also know the value of $x p(x) - 1$ when $x = 0$ , so we actually know 10 values.

Calvin Lin Staff - 6 years, 5 months ago

There are more than $1$ solutions to this question, one of which is this:

$p(x)=\prod _{ k=1 }^{ 9 }{ \left( { x }^{ \frac { 8 }{ 9 } }-{ k }^{ \frac { 8 }{ 9 } } \right) } +\frac { 1 }{ x }$

Which makes $p(10)=17794.4150...$

You should specify more details on the polynomial.

Julian Poon - 6 years, 5 months ago

That is not a polynomial . A polynomial must have non-negative integer powers.

Your $p(x)$ has terms of the form $x ^{8/9}$ and $x^{-1}$ .

Calvin Lin Staff - 6 years, 5 months ago

@Calvin Lin – Oh... I thought a polynomial's definition was much more loose. Thanksfor clarifying anyway!

Julian Poon - 6 years, 5 months ago

A polynomial problem

The answer is 0.2.

1 solution

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