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A practice problem from my textbook (not homework/assignment). Show that there is a unique polynomial $P_{3}(x)$ where

$p_{3}(x_{0}) = f(x_{0}),\space p_{3}(x_{2}) = f(x_{2}), p'_{3}(x_{1}) = f'(x_{1}), p^{''}_{3}(x_{1}) = f^{''}(x_{1})$ where $f(x)$ is a given function.

I'm looking towards using Hermite Interpolation but I"m unsure of how to start the problem. If anyone could just provide a hint to push me in the right direction, it would be appreciated.

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    Don't you have derivative of $f$ available at all points specified $x_{0},x_{1},x_{2}$? – 2017-02-26
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    Also add some reference/source to your question so that other users can view and help better! – 2017-02-26
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    I thought for Hermite splines, when solving for a, b, c, and d in the form of a+b(x-xi) + ci(x-xi)^2 + di(x-xi)^3 we need the points in some form of [xi, xi+1], but i'm unsure of how the second derivative really helps. I know that to solve for a, b, c, and d there are general formulas, but for some of them we need to know what yi+1 and yi are, but in this case we don't know since we are only given the derivative at some points. – 2017-02-26
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    and furthermore, I only have the derivative of x1 and second derivative of x1. I thought for Hermite interpolation, i require the following : pi(xi) = yi pi(xi+1) = yi+1 p'i(xi) = si p'i(xi+1) = si+1 for each sub interval, which in our case would be from x0-x1, x1-x2? – 2017-02-26

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Let $p_3(x) = ax^3 + bx^2 + cx + d$ and without restricting generality suppose that $x_0 < x_1 < x_2$. Then in order for $p_3(x)$ to satisfy the given equations, its coefficients must satisfy the linear system $$ \begin{bmatrix} 1 & x_0 & x_0^2 & x_0^3\\ 1 & x_2 & x_2^2 & x_2^3\\ 0 & 1 & 2x_1 & 3x_1^2\\ 0 & 0 & 2 & 6x_1 \end{bmatrix} \begin{bmatrix} d\\ c\\ b\\ a \end{bmatrix} = \begin{bmatrix} f(x_0)\\ f(x_2)\\ f'(x_1)\\ f''(x_1) \end{bmatrix}, $$ where each row corresponds to one of the equations. Row reducing the coefficient matrix to a partial RREF we obtain $$ \begin{bmatrix} 1 & x_0 & x_0^2 & x_0^3\\ 0 & 1 & x_2+x_0 & x_2^2 + x_0x_2 + x_0^2\\ 0 & 0 & 2x_1-x_2-x_0 & 3x_1^2 - (x_2^2 + x_0x_2 + x_0^2)\\ 0 & 0 & 2 & 6x_1 \end{bmatrix} $$ From here it is easy to compute the determinant, which is $$ 6x_1^2 + 2x_2^2 + 2x_0^2 + 2x_0x_2 -6x_1x_2-6x_0x_1 = 3(x_1-x_0)^2 + 3(x_2 - x_1)^2 - (x_2-x_0)^2 $$ Now if you let $d_1 = x_1 - x_0 > 0$ and $d_2 = x_2 - x_1 > 0$, then $x_2 - x_0 = d_1 + d_2$. Thus the determinant may be written as $$ 3d_1^2 + 3d_2^2 - (d_1 + d_2)^2 = 2d_1^2 - 2d_1d_2 + 2d_2^2 = (d_1 - d_2)^2 + d_1^2 + d_2^2 > 0 $$ Since the determinant is nonzero, the linear system must have a unique solution. Hence, there exists a unique polynomial $p_3(x)$ that satisfies the given equations.

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    Just to clarify for myself, so for all cubic polynomials, we can always put them in some sort of system of equations? I understand that for the last two columns, namely (f'(x1), f''(x1)) we're saying that we should be using the derivative of p3 and the second derivative of p3 in order to to return the first and second derivative of x1 which makes sense. But why is the second matrix b and a in that case? as in, the first derivative still has the a constant but we're solving for the b constant in that case? – 2017-02-26
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    furthermore, why are we allowed to just treat the f(x)'s as constants? – 2017-02-26
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    It is not exactly clear what you are asking. Each row of the matrix corresponds to one of the equations. You can treat the $f(x)$ terms as constants because that is exactly what they are in this case. The equations constrain the values and the derivatives of the polynomial $p_3(x)$ to equal the corresponding values and derivatives of the function $f(x)$ at certain points in the function`s domain. – 2017-02-27
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    Ahh, i see, our review session went over the question today. Thanks for the clarification. – 2017-02-28