# 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.

• Presumably, for the interpolation of degree one, you only need $x_1$ and $x_2$. – Raskolnikov Sep 12 '12 at 21:27
• So I would interpolate 2 polynomials in this case, one for x_1 and x_2 and the other for x_0 and x_1 ? – math101 Sep 12 '12 at 21:29
• So you chose those 2 points since we need to approximate f(1.4)? – math101 Sep 12 '12 at 21:32
• To your last question: yes, indeed! – Raskolnikov Sep 12 '12 at 21:39
• Yup that was correct, I got the right answer. Thanks :)))) – math101 Sep 12 '12 at 21:55

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 ;).
• First construct three polynomials $l_i(x)$, $i=1,2,3$ such that $l_i(x_j)=0$ if $i\ne j$ and $l_i(x_i)=1$. Then set $$P(x) = l_0(x) f(x_0) + l_1(x) f(x_1) + l_2(x) f(x_2)$$ Now to construct the $l_i$'s just do as such (example with $i=0$) : $$l_0(x) = \frac{(x-x_1)(x-x_2)}{(x_0-x_1)(x_0-x_2)}$$ – vanna Sep 12 '12 at 22:57
• What am I doing incorrectly? $L_0(x)= \frac{3} {20}(x^2-2.85x+2)\cdot\sin( \pi \cdot1) = 0, L_1(x) = \frac{-7} {20}(x^2-2.6x+1.6)\cdot\sin( \pi *1.25), L_2 = \frac{21} {100}(x^2-2.25x+1.25)\cdot\sin( \pi \cdot1.6)$ – math101 Sep 12 '12 at 23:11
• $L_0$ is fine but it is $\frac{20}{3}$ not its inverse. I didn't check the other ones. – vanna Sep 12 '12 at 23:16
• @math101: $$f(1)\frac{20}{3}(x^2-2.85x+2) -f(1.25)\frac{80}{7}(x^2-2.6x+1.6) +f(1.6)\frac{100}{21}(x^2-2.25x+1.25)$$ – robjohn Sep 12 '12 at 23:58