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There are four $(x,y)$ coordinates $(-5,-1), (0,0), (5,1), (8,4)$ and using linear and cubic interpolation I intended to find the value of y at x=3.

My solution

For linear interpolation I picked $(0,0)$ and $(5,1)$ as they are close to each other.

Then with the help of this equation $y_2$ = $\frac{(x_2 -x_1) (y_3 - y_1)}{(x_3-x_1)}$, I found $y_2$ to be $\frac{3}{5}$.

For cubic interpolation I picked all the $(x,y)$ points and used the following equation:

$f_3$(x) = $a_0+a_1x+a_2x^2+a_3x^3$.

I plugged in each of the points into the equation and solved the equation for the unknowns. I ended up with

$f_3$(x) = 1/130x + 1/130$x^3$

(as $a_0$ and $a_2$ are zero and $a_1$ and $a_3$ are 1/130 based on my calculations). Lastly, I plugged in the given x-value (x=3) into the equation and found that y=3/13.

I am wondering if this is correct considering that the linear and cubic interpolation result in pretty different y-values (3/5; 3/13 respectively).

Any help is greatly appreciated!

etet112
  • 443

1 Answers1

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You results are correct, perhaps a plot of the function will help why the difference is so large

enter image description here

The black points are the interpolation nodes, and the colored ones are the result of applying linear ($\color{red}{\rm red}$) and cubic ($\color{blue}{\rm blue}$) interpolation to $x = 3$

caverac
  • 19,345