0

I was given the following problem to solve and I am not confident with my solution.

I have tried finding a variable alpha that would help me create $p(x)$ as a Hermite polynomial, which would be of $2nd$ degree at most and unique by definition.

The values that I've found for alpha are 0 and 1, but I am not sure that this answer is correct based on feedback I've received from peers in class (The answer is supposedly for every alpha not equal 0.5).

Could somebody please explain how to satisfy this condition in the right way + answer the second part which talks about the case that the polynomial is not unique?

Thanks !!! enter image description here enter image description here

RandyMarsh
  • 85

1 Answers1

1

You can rescale and shift the domain so that $x_0=0$, $x_1=1$ and $x_α=α$.

You can remove a linear function from $f$ and the polynomial so that $f(0)=f(1)=0$.

The quadratic polynomials now satisfying the conditions at the boundaries are $p(x)=cx(1-x)$ with $p'(x)=c(1-2x)$. Now decide when $c$ can be uniquely determined for any value of $f'(α)$.

Lutz Lehmann
  • 126,666
  • The only thing that's a little bit fuzzy with this way of explaining it is that the requirement for when a non-unique solution exists (in the "singular" case) should really be imposed on the unshifted data. – Ian May 05 '21 at 18:19
  • I think this solution might be too specific because they never said I could pick values for these points, I think they want me to find alpha for general points x0 and x1. – RandyMarsh May 06 '21 at 08:26
  • @RandyMarsh : But you can transform every setup to this situation and reconstruct the general solution from the one of the reduced problem. – Lutz Lehmann May 06 '21 at 08:41