Professor gave us this little bastard of a question and I'm at a complete loss about what to do. Some help or hints would be immensely appreciated, translated to the best of my abilities.
Let $x_0=0$, $x_1=1$, $y_0=0$ and $y_1=1$ and given the function $g(x)=\sin (x\pi/2) $
Let $p(x)$ be the polynomial for cubic Hermite interpolation of $g$ with the interval $[0, 1]$
a) Calculate $y'_0=g'(0)$, $y'_1=g'(1)$ and set up the interpolation problem with matrix form $Ax=b$ where $b^T = [y_0, y_1, y'_0, y'_1]$
b) Use the result from a) to determine the interpolation polynomial $p(x)$
Visual text that might be easier to read:
https://i.sstatic.net/XplpC.png
noticed a small mistake in the image. In a) It says $y'_1=y'(1)$ when it should say $y'_1=g'(1)$
Best Answer
You want to find a cubic interpolate, so right away you should be able to deduce that for some constant $a,b,c,d$ we have $$p(x)=a+bx+cx^2+dx^3.$$ We then get that $$p'(x)=b+2cx+3dx^2.$$ Since we want to interpolate $y$ on the nodes $0,1$ we get $$a=\sin(0\pi/2)=0,$$ $$a+b+c+d=\sin(\pi/2)=1.$$ We also want to interpolate the derivatives on the same nodes, so we get $$b=\frac{\pi}{2}\cos(0\pi/2)=\frac{\pi}{2},$$ $$b+2c+3d=\frac{\pi}{2}\cos(\pi/2)=0.$$ This gives us $$\begin{bmatrix}1&0&0&0\\1&1&1&1\\0&1&0&0\\0&1&2&3\end{bmatrix}\begin{bmatrix}a\\b\\c\\d\end{bmatrix}=\begin{bmatrix}0\\1\\\pi/2\\0\end{bmatrix}.$$
This gives you the answer for part $a$. You should be able to use the above expression to solve for your polynomial. Use Gaussian Eliminate, or find the inverse matrix and multiple. Any method to solve the system will work.
Personally, if you're going to work with Hermite Interpolations a lot, I recommend learning about Newton's Divided Difference Formula. It gives a very compact way to write the formula for a general Hermite Interpolating Polynomial.