0
$\begingroup$

I am asked to construct an interpolating polynomial of at most degree one and two. I am attempting to use Lagrange polynomials, but I have already tried using general polynomials that lead to a system of equations and Taylor series. My function is $$f(x)= \sin(\pi x)$$ and my x values are $\{1,1.25,1.6\}$ and I am approximating $$f(1.4)$$ My question is how can I use Lagrange polynomials to calculate a polynomial of at most one degree? For example, I know that $$ l_0(x) = \frac{(x-x_1)(x-x_2)}{(x_0 - x_1)(x_0-x_2)}$$ and $$ l_1(x) = \frac{(x-x_0)(x-x_2)}{(x_1 - x_0)(x_1-x_2)}$$ and so forth, but clearly, neither of these equations will produce a polynomial of at most degree 1. So which point to I leave out and why? It seems a good choice would be to leave out the $(x-x_1)$ term since it is between $1$ and $1.6$. So, is this the correct choice and is the reasoning for choosing it correct ? $$ l_0(x) = \frac{(x-x_2)}{(x_0 - x_1)}$$

$\endgroup$
1
$\begingroup$

I did the average between $1.25$ and $1.6$ getting $1.425$ so my data are

$\{(1, 0), (1.425, -0.97237)\}$

and the line has equation

$q(x)=2.28793\, -2.28793 x$

Computing at $x=1.4$ I got

$q(1.4)\approx -0.915172$

The actual value of $\sin \pi x$ is

$f(1.4)\approx -0.951057$

and the value with the second degree polynomial

$P(x)=3.55238 x^2-10.8213 x+7.2689$

$P(1.4)\approx -0.918228$

Not much better than the first degree

Doing as you wanted to do leads to the polynomial

$r(x)=1.58509-1.58509x$

and

$r(1.4)\approx -0.634038$

The value is very poor. If you look at the graph you will understand the reason why

Hope this helps

$$...$$

enter image description here

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.