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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 $f(x)=a_0+a_1x+a_2x^2+a_3x^3+......+a_{11}x^{11}$

$f(0)=\frac{1}{0+1}=1=a_0$

$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} $
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$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.

$(A)\frac{1}{13}$

$(B)\frac{1}{12}$

$(C)0 $

$(D)\frac{1}{7}$

which one is correct answer?

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HINT:

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

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    $\begingroup$ 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
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    $\begingroup$ 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
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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 $f(12) = \frac{-1+1}{13} = 0$.

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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,

$$f(12)=g(13)=\frac{A(13-1)(13-2)\cdots(13-12)+1}{13}=\frac{12!A+1}{13}=0.$$

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