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Discuss the problem of determining a polynomial of degree at most $2$ for which $p(0)$, $p(1)$, and $p'(\zeta)$ are prescribed, $\zeta$ being any preassigned point.

So, I started by using the interpolation polynomials that I know:

  • Newton's Interpolation Polynomial: $$p(x)=c_0+c_1(x-0)+c_2(x-0)(x-1),$$ where $c_0=y_0=p(0)$, $c_1=\dfrac{y_1-y_0}{x_1-x_0}=p(1)-p(0)$, and $c_2=\dfrac{x_2-P_1(x_2)}{(x-x_0)(x-x_1)}$ with $P_1(x_2)=p(0)+p(1)x_2-p(0)x_2$.

But the problem here is that I do not know $x_2$.

  • Lagrange Polynomial: $$p(x)=y_0\ell_0(x)+y_1\ell_1(x), \text{ where } \ell_i(x)=\prod_{j=0 \& j\neq i}^2\dfrac{x-x_j}{x_i-x_j}.$$ So, $$p(x)=p(0)\left(\dfrac{x}{1-0}\right)+p(1)\left(\dfrac{x-1}{0-1}\right)$$

But the problem here is that $p$ is a polynomial of degree 1 and not 2.

Now the one piece of information I haven't used is that fact I have $p'(\zeta)$. But I don't know how to use it.

Username Unknown
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1 Answers1

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I would draw a secent from $x=0$ to $x=2\zeta$. With a quadratic function this secant will be parallel to the tangent of the curve at the midpoint coordinate $x=\zeta$, thus:

$p(2\zeta)-p(0)=2\zeta p'(\zeta)$

Then you have a third point for a Newton interpolation.

Note this implies $\zeta\ne(1/2)$. Can you see why?

Oscar Lanzi
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