# Show that $f(x)-P(x)=\frac{x^4-1}{4!}f^{4}(c)$

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 $$x_0=-1,x_1=0,x_2=1,c\in (-1,1)$$

MY TRY:

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?

• 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. – Ian Dec 16 '18 at 15:13
• you've got typos.. $x_0,x_1,x_2$ do not correspond to the ones in the problem and equations 1 thru 4. – Ahmad Bazzi Dec 17 '18 at 7:24