SOLUTION: Let P(x)=2009x^9+a₁x^8+...+a₉ such that {{{P(1/n)=1/(n^3)}}}, n = 1,2,...,9. Find {{{P(1/10)}}}.

Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: Let P(x)=2009x^9+a₁x^8+...+a₉ such that {{{P(1/n)=1/(n^3)}}}, n = 1,2,...,9. Find {{{P(1/10)}}}.      Log On


   

Question 1195271: Let P(x)=2009x^9+a₁x^8+...+a₉ such that
P%281%2Fn%29=1%2F%28n%5E3%29, n = 1,2,...,9.
Find P%281%2F10%29.
Found 2 solutions by MathLover1, ikleyn:
Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!

P%281%2Fn%29=1%2F%28n%5E3%29, n+= 1,2,...,9.

Find P%281%2F10%29=> so, 1%2Fn=1%2F10=>n=10
P%281%2F10%29=1%2F%2810%5E3%29
P%281%2F10%29=1%2F1000
P%281%2F10%29=+0.001

Answer by ikleyn(52792) About Me  (Show Source):
You can put this solution on YOUR website!
.
Let P(x) = 2009x^9 + a₁x^8 + . . . + a₉ such that
P%281%2Fn%29=1%2F%28n%5E3%29, n = 1,2,...,9.
Find P%281%2F10%29.
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            The solution by @MathLover1 is  INCORRECT  and  IRRELEVANT.

            Do not spend your time looking in it;  I also will not spend my and your time,
            explaining why it is wrong and irrelevant.

            Simply madam does not understand the problem,  does not know the subject
            and is not in the theme.

            Read my solution below. 


You are given a polynomial P(x) of degree 9 with the leading coefficient 2009;
all 9 coefficients of lower degrees are unknown.


You also are given the values of the polynomial in 9 points

    P(1) = 1,  P%281%2F2%29 = 1%2F2%5E3,  P%281%2F3%29 = 1%2F3%5E3, . . . , P%281%2F9%29 = 1%2F9%5E3.


Consider another polynomial

    Q(x) = 2009%2A%28x-1%29%2A%28x-1%2F2%29%2A%28x-1%2F3%29%2A+ellipsis+%2A%28x-1%2F9%29 + x%5E3.


This polynomial has the degree of 9 and has the same values  1,  1%2F2%5E3,  1%2F3%5E3, . . . , 1%2F9%5E3
in all 9 given points  1,  1%2F2,  1%2F3, . . . , 1%2F9,  as polynomial P(x) has.  

You can easily see it from the formula.



    In addition, polynomial Q(X) has the same value 
    of the leading coefficient 2009 as polynomial P(X) has.



It  highlight%28highlight%28IMPLIES%29%29  that polynomials P(x) and Q(X) are identical as polynomials of x.


THEREFORE,  P%281%2F10%29 = Q%281%2F10%29 =  + 1%2F10%5E3.



      This expression is the  highlight%28highlight%28expected%29%29  highlight%28highlight%28ANSWER%29%29  to the problem.


Solved,  answered,  explained and completed.


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This problem is of a  Math  Olympiad level or an advanced  Math  Circle level,

and my solution is written adequately to this level.