0
$\begingroup$

In the following lecture:

http://www.ece.mcmaster.ca/~xwu/interp_1.pdf

the model (formula) for solving the linear interpolation problem (1D) given at p.5 is:

$f(x)= a_1x_1 + a_0x_0$

solve for $a1,a0$

the system:

$f(0)=a_i*0 + a_0*1$

and

$f(1)=a_i*1 + a_0*1$

I don't understand where this formula comes from. Even though deriving in all possible ways I could think of, I cannot get the equation of a line $y=mx+b$ ( also written as: $f(x)= mx+b$, where m is the slope and b is the y-intercept) to look something like $f(x)= a_1x_1 + a_0x_0$. Any idea/proposals/discussions etc.?

$\endgroup$
3
$\begingroup$

To my eye, the equations in the slides look like $f(x) = a_1 x^1 + a_0 x^0 = a_1 x + a_0$. So it's an exponent on the $x$, not an index. As long as you are interpolating linear, the only exponents that you encouter are 1 (linear) and 0 (constant). When you go to quadratic, you'll see $x^2$ terms appearing. Now, $a_1$ is the coefficient of the linear term (your slope $m$) and $a_0$ is the coefficient of the constant term (the y-intercept $b$).

This should explain where the system of equations for $a_1$ and $a_0$ comes from.

| improve this answer | |
$\endgroup$
  • $\begingroup$ yes thanks, I should have thought about that... its clearly and exponent as you say,. so the x^0 = 1. must have been tired... I was doing nonsense $\endgroup$ – Machupicchu Aug 19 '19 at 12:32
  • $\begingroup$ They want to find the coefficients slope a1 and intercept a0 that work for both points x=0 and x=1 $\endgroup$ – Machupicchu Aug 19 '19 at 12:41

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.