I have been working on this problem longer than I would like to admit. I am to find the error for $P_1(x)$ for $f(x) = cos(x)$ on given the points $ x=0, x=0.6$ and $f(0.45)$ I know the formula $$ \left| { \frac{f^{n+1}(\xi(x))}{(n+1)!} \cdot (x-x_0)(x-x_1)} \right| = \left| \frac{ – \cos(\xi(x))}{2} \cdot (0.45-0)(0.45-0.6)\right| $$ So the term $$ \left| \frac{f^{n+1}(\xi(x))}{(n+1)!} \right| = \text{max}_{0,0.6} \left| \frac{f^{n+1}(\xi(x))}{(n+1)!} \right|$$ So how do you find the max value for this expression. My first attempt was to try $- \cos(0)/2$ since this has to be the maximum value given $ |\cos(x)| \le 1$. But that is not it. I tried taking the derivative, looking for it on the Ti-84 and examining it on Wolfram Alpha. I have also read through three text books and several other related questions and it does not seem to appear, explicitly, anywhere.
[Math] Compute the Error Bound for the Lagrange Interpolation Polynomial
interpolationlagrange-interpolation
Best Answer
So I think I have a solution:
We have $$ \left| \frac{f^{n+1}(\xi(x))}{(n+1)!} \right| = \left| \frac{-cos(\xi(x))}{2!} \right| $$ where $\xi(x)$ is the max value on $[0,0.6]$. The max value is $x=0$. This means that term is $$ \left| \frac{f^{n+1}(\xi(x))}{(n+1)!} \right| = \left| \frac{-cos(0)}{2!} \right| = \frac{1}{2} $$ For the rest of the terms $$\left|(x-x_0)*(x-x_1)\right|$$ we have for x = 0.45 $$\left|(0.45-0)*(0.45-0.6)\right| =0.0675 $$ so we get that $$ \left| \frac{f^{n+1}(\xi(x))}{(n+1)!} *(x-x_0)*(x-x_1)\right| = \frac{1}{2}*|0.0675|= 0.03375 $$ In the solutions manual, the number given as a solution is $0.135$ which is actually $0.0675*2=0.135$. Which leads me to believe the solutions manual is incorrect (for all future students). Please contradict me if I am wrong!