The mualphatheta link has (almost) this question set as a problem - the answer link did not work for me. Note that \(xp(x)-1\) is a degree \(9\) polynomial with \(1,2,3,4,5,6,7,8,9\) as zeros, and hence
\[
xp(x) - 1 \; = \; A(x-1)(x-2)(x-3)(x-4)(x-5)(x-6)(x-7)(x-8)(x-9)
\]
Putting \(x=0\) we see that \(-1 \,=\, -A\times9!\) and so \(A = \tfrac{1}{9!}\). Then \(10p(10)-1 \,=\, A\times9! = 1\), and hence \(p(10) = \tfrac{1}{5}\).
–
Mark Hennings
·
3 years, 2 months ago

Log in to reply

@Mark Hennings
–
As a side note, you should verify that \[ \frac{ 1 + \frac{1}{9!} \prod_{i=1}^9 (x-i) } { x } \]

is indeed a polynomial. This follows because the constant term in the numerator cancels out, so it is a multiple of \(x\).

It is possible (especially in scenarios where the exact roots are uncertain), for the function that you define to end up being a rational function, instead of a polynomial.
–
Calvin Lin
Staff
·
3 years, 2 months ago

Log in to reply

@Calvin Lin
–
I already did this when I evaluated \(A\), because I chose the value of \(A\) to be such that the value of
\[ 1 + A\prod_{j=1}^9(x-j) \]
when \(x=0\) was \(0\), which means that this polynomial has no constant term and hence can be safely divided by \(x\).
–
Mark Hennings
·
3 years, 2 months ago

Log in to reply

Try remainder theorem, although it will take many steps.
–
Siddharth Kumar
·
3 years, 2 months ago

Log in to reply

This will do it : http://www.mualphatheta.org/problemcorner/MathematicalLog/Issues/0402/MathematicalLogProblemistSpring02.aspx
–
Vikram Waradpande
·
3 years, 2 months ago

Log in to reply

you can use newton-ramphson techhnique, newton forward or backward formula . it will be easy .
–
Ashok Kumar
·
3 years, 2 months ago

## Comments

Sort by:

TopNewestThe

mualphathetalink has (almost) this question set as a problem - the answer link did not work for me. Note that \(xp(x)-1\) is a degree \(9\) polynomial with \(1,2,3,4,5,6,7,8,9\) as zeros, and hence \[ xp(x) - 1 \; = \; A(x-1)(x-2)(x-3)(x-4)(x-5)(x-6)(x-7)(x-8)(x-9) \] Putting \(x=0\) we see that \(-1 \,=\, -A\times9!\) and so \(A = \tfrac{1}{9!}\). Then \(10p(10)-1 \,=\, A\times9! = 1\), and hence \(p(10) = \tfrac{1}{5}\). – Mark Hennings · 3 years, 2 months agoLog in to reply

is indeed a polynomial. This follows because the constant term in the numerator cancels out, so it is a multiple of \(x\).

It is possible (especially in scenarios where the exact roots are uncertain), for the function that you define to end up being a rational function, instead of a polynomial. – Calvin Lin Staff · 3 years, 2 months ago

Log in to reply

– Mark Hennings · 3 years, 2 months ago

I already did this when I evaluated \(A\), because I chose the value of \(A\) to be such that the value of \[ 1 + A\prod_{j=1}^9(x-j) \] when \(x=0\) was \(0\), which means that this polynomial has no constant term and hence can be safely divided by \(x\).Log in to reply

Try remainder theorem, although it will take many steps. – Siddharth Kumar · 3 years, 2 months ago

Log in to reply

This will do it : http://www.mualphatheta.org/problem

corner/MathematicalLog/Issues/0402/MathematicalLogProblemistSpring02.aspx – Vikram Waradpande · 3 years, 2 months agoLog in to reply

you can use newton-ramphson techhnique, newton forward or backward formula . it will be easy . – Ashok Kumar · 3 years, 2 months ago

Log in to reply