Let $f(x)$ be a polynomial of degree $11$ such that $f(x)=\frac{1}{x+1}$,for $x=0,1,2,3.......,11$.
Then what is the value of $f(12)?$

My attempt at this is:
Let $f(x)=a_0+a_1x+a_2x^2+a_3x^3+......+a_{11}x^{11}$


$f(1)=\frac{1}{1+1}=\frac{1}{2}=a_0+a_1+a_2+a_3+......+a_{11} $

$f(2)=\frac{1}{2+1}=\frac{1}{3}=a_0+2a_1+4a_2+8a_3+......+2^{11}a_{11} $
$f(11)=\frac{1}{11+1}=\frac{1}{12}=a_0+11a_1+11^2a_2+11^3a_3+......+11^{11}a_{11} $

for calculating $f(12)$, I need to calculate $a_0,a_1,a_2,....,a_11$ but I could solve further.Is my approach right,how can I solve further or there is another right way to solve it.



$(C)0 $


which one is correct answer?



Let $(x+1)f(x)=1+A\prod_{r=0}^{11}(x-r)$ where $A$ is an arbitrary constant

  • 6
    $\begingroup$ A is not an arbitrary constant. It is a fixed constant whose value we happen not to know. $\endgroup$ – Mariano Suárez-Álvarez Jul 26 '15 at 3:09
  • $\begingroup$ @MarianoSuárez-Alvarez, See any arbitrary finite constant will satisfy the given condition, right? $\endgroup$ – lab bhattacharjee Jul 26 '15 at 3:10
  • 6
    $\begingroup$ But f is a fixed, given polynomial. Your equation can only hold for exactly one scalar A. $\endgroup$ – Mariano Suárez-Álvarez Jul 26 '15 at 3:11
  • $\begingroup$ Solving for $f(x)$ from your equation will in general yield a rational function - there is a unique choice of $A$ for which it simplifies to a polynomial. Can you find that value for $A$? $\endgroup$ – anon Jul 26 '15 at 3:13
  • $\begingroup$ @MarianoSuárez-Alvarez, I was pointing to the given condition. But, from that $A$ can assume any arbitrary finite scalar value, right? $\endgroup$ – lab bhattacharjee Jul 26 '15 at 3:13

Somewhat less vague...

$(x+1)f(x)-1$ is a polynomial of degree 12 with roots at every integer in $[0,11]$, so could be $$(x+1)f(x)-1 = A \prod_{c=0}^{11} x-c$$ for some/any (nonzero) constant $A$.

When $x=-1$, we have $(0)f(-1)-1 = A (-1)^{12} 12!$, or $-1 = A \, 12!$ and discover only $A = \frac{-1}{12!}$ is consistent with the givens. (Why $-1$? Because it is the only choice we haven't already used (we have used the integers 0, ... 11) that makes some expression containing $x$s zero.)

Hence, $13 f(12) - 1 = \frac{-1}{12!} 12!$ and $f(12) = \frac{-1+1}{13} = 0$.


For convenience, let us shift the variable: $g(y)=\dfrac1y$ for $y=1,2,\cdots12$.

Then $$\frac{A(y-1)(y-2)\cdots(y-12)+1}y$$ is a polynomial of degree $11$ and coincides with $\dfrac1y$ at the given points, provided that the numerator has no independent term, i.e. $12!A+1=0$.

From this,



Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.