If $P(x)$ is a unique cubic polynomial for which $P(x_0)=f(x_0),P(x_2)=f(x_2),P^{'}(x_1)=f^{'}(x_1),P^{''}(x_1)=f^{''}(x_1)$,$f(x)$ is a given function differentiable $4$ times.

Show that $f(x)-P(x)=\dfrac{x^4-1}{4!}f^{4}(c) $ where $x_0=-1,x_1=0,x_2=1,c\in (-1,1) $


Let us assume that $P(x)=a_0+a_1x+a_2x^2+a_3x^3.$
Then $$\begin{aligned}P(-1)=a_0-a_1+a_2-a_3&=f(-1)\quad &(1)\\ P(1)=a_0+a_1+a_2+a_3&=f(1) \quad &(2)\\ P^{'}(0)=a_1&=f^{'}(0)\quad &(3)\\ P^{''}(0)=2a_2&=f^{''}(0)\quad &(4) \end{aligned}$$ How can I prove the given fact using the $4$ relations I have found here?

Please help.

  • $\begingroup$ You presumably have some typo in the last line of the box, I guess $x_0=-1,x_1=0,x_2=1$? Also presumably $P''(x)=f''(x)$ should also only be at one point, but you seem to have sorted that out in the main body of the post. $\endgroup$ – Ian Dec 16 '18 at 15:13
  • $\begingroup$ @Ian,Thanks for the info.I edited it $\endgroup$ – user596656 Dec 17 '18 at 7:21
  • $\begingroup$ you've got typos.. $x_0,x_1,x_2$ do not correspond to the ones in the problem and equations 1 thru 4. $\endgroup$ – Ahmad Bazzi Dec 17 '18 at 7:24
  • $\begingroup$ @AhmadBazzi;I think it is fixed now $\endgroup$ – user596656 Dec 17 '18 at 8:26

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