Interpolating polynomials for $f(x)=\sin(\pi x)$ I am having difficulties with interpolating polynomials to approximate. I am familiar when the problem involves 2 points but this particular problem involves 3 points. Can you explain the process to me?
For the given function $f(x)=\sin(\pi x)$ let $x_0=1,x_1=1.25,x_2=1.6$. Construct interpolation polynomials of degree at most 1 and at most 2 to approximate $f(1.4)$ and find the absolute error.
 A: Interpolation of degree at most $1$ is linear interpolation. You only need the value of $f$ evaluated in $x_1$ and $x_2$ since $x_1<1.4<x_2$. If $P$ denotes the polynomial then
$$ P(1.4) = \frac{f(x_2)-f(x_1)}{x_2-x_1}(1.4-x_1) + f(x_1) $$
Which yields
$$ \varepsilon = |P(1.4)-f(1.4)| = 0.139399849 $$
Then for interpolation of degree at most $2$ you search $P(x) = a x^2 + b x +c $ such that $P(x_i)=f(x_i)$ for $i=0,1,2$. This gives you a system of three linear equations to solve. More generally this polynomial is known as Lagrange polynomial. Solving the system yields
$$ P(x) = 3.5523798019047517 x^2 + -10.821281678285672 x + 7.2689018763809194 $$
and thus
$$ \varepsilon = |P(1.4) - f(1.4) | = 0.0328284548$$
So second order interpolation is slightly better than naive linear interpolation ;).
You can also interpolate using a piecewise polynomial as well but in this case you get fixed degree of smoothness. For example cubic splines, which are piecewise cubic polynomials define a twice differentiable function, and not more ! The nice thing about these interpolation polynomials is that between two knots they have degree at most 3, preventing Runge's phenomenon that Lagrange interpolation polynomial present when increasing the number of knots. 
