I am struggling to understand this:
The problem asks me to find the lagrange error of the polynomial approximation given the nodes
$x_0 = 1, x_1 = 1.25, x_2 = 1.6$ with $x = 1.4$
The function I am approximating is $f(x) = e^{2x} - x$
So first I get the derivatives:
$f'(x) = 2e^{2x} - 1; f''(x) = 4e^{2x}$
and for the linear polynomial error I get
$|\frac{4e^{2\xi}}{2!}(x-1)(x-1.25)|$
and since we have $x$, just sub it in
$|\frac{4e^{2\xi}}{2!}0.06|$
But at this point I get lost. I know that $4e^{2\xi}$ is strictly increasing on [1,1.6], and that $1 \le \xi \le 1.6$. So if I replace everything I get
|$|\frac{4e^{2(1.6)}}{2!}0.06| \le 2.94390$
Except that the book has the answer as 1.47195, which just so happens to be half of my answer. I'm not sure where I went wrong here, and I'd like to get this solved so I can continue my homework. Can anyone help?
Thank you!
We need to use the points $1.25$ and $1.6$ since the $x$ is in between those and we are using linear interpolation.
You should have:
$$\left|\frac{4e^{2\xi}}{2!}(x-1.25)(x-1.6)\right|$$
At $x = 1.4$, this gives:
$$\left|\frac{4e^{2\xi}}{2!}(1.4-1.25)(1.4-1.6)\right| = \left|\frac{4e^{2\xi}}{2!}(0.03)\right|$$
At the max point $\xi = 1.6$, this yields:
$$\left|\frac{4e^{2(1.6)}}{2!}0.03\right| \le 1.47195$$