Why is there no unique solution for this interpolation task?

Question

Consider the space $span(g_0,g_1)$ with $g_0(x)=1,g_1(x)=x^2$. Look at an interpolation task for the following pair of values.

$(x_0, y_0) = (−1/2, 1); (x_1, y_1) := (1, 2)$

Why is there not always an unique solution if $x_0 \neq x_1$ is out of [-1,1] but there is if $x_0,x_1$ are out of [0,1]?

We have discussed the Neville algorithm in class. My best guess would be the fact that the interpolation solution is unique as long as all of the x-values are different from each other. But I'm not sure how I could integrate this into a proof?

Answer 1

It has nothing to do with $[-1,1]$ or $[0,1]$.

Note that $g_0$ and $g_1$, and thus their linear combinations, are even functions of $x$. So if $x_0 = -x_1$, there will either be infinitely many solutions or none. You always have a unique solution if $|x_0| \ne |x_1|$.

Answer 2

You're trying to find a linear combination of $g_0$ and $g_1$ (i.e., $f(x) = a + bx^2$) interpolating two points $(x_0, y_0)$ and ($x_1, y_1$). Plugging these in to $f(x) = y$ gives:

• $a + bx_0^2 = y_0$
• $a + bx_1^2 = y_1$

Some simple algebraic manipulation gives you $b = \frac{y_1 - y_0}{x_1^2 - x_0^2}$, which you can then plug into either of the original equations to get $a = y_0 - bx_0^2 = y_1 - bx_1^2$.

So you've got a unique closed-form solution — unless $x_1^2 - x_0^2 = 0$ (or equivalently, $x_1 = ±x_0$), in which case you get a divide-by-zero error.

Answer 3

You want a linear combination $\sum_{i=0}^1 \alpha_i g_i$ that takes prescribed values at $x_i$, $i=0,1$. The linear system obtained is $$\sum_{j=0}^1 \alpha_j g_j(x_i)= y_i$$ for $i=0,1$. The matrix of the system is $$(g_j(x_i))_{i=0,1; j=0,1}$$ ( this works in general for $n+1$ instead of $2$ functions $g_j$ ). Let's look at the matrx of the system $$(g_j(x_i))$$ The system has a unique solution for all $y = (y_i)$ if and only if the matrix of the system is invertible, that is $$\det (g_j(x_i))\ne 0$$ Note that $g_j(x) = x^{2j}$, so the matrix is in our case $$\left( \begin{matrix} 1& x_0^2 \\1& x_1^2 \end{matrix}\right)$$ with determinant $x_1^2 - x_0^2$. So, if $x_0^2 \ne x_1^2$ we have always a unique solution.

The conclusion is that $g_0(x) \equiv 1$, $g_1(x) = x^2$ form a Chebyshev system on every interval that does not contain $0$ in its interior.