I know what is newton-Raphson's method, i.e., to approximate the value of roots using numerical analysis. But, when there is no function and only data is given, then how to find actual function. Question is: how to find second degree polynomial if data is:
$$
\begin{array}{c|cccc}
x & 0.0 & 0.2 & 0.4 & 0.6 \\
\hline
f(x) & 15.0 & 21.0 & 30.0 & 51.0
\end{array}
$$
And also to estimate the value of $f(0.1)$ and $f(0.5)$. I searched rigorously, but couldn't understand more advanced topics of calculus etc. I think two posts are quite helpful, but too difficult for me to properly grasp the ideas. {https://math.stackexchange.com/questions/1616995/finding-all-roots-of-multivariate-polynomial-using-newtons-method}, {https://math.stackexchange.com/questions/742871/newtons-method-for-roots-of-polynomials}. If any one guide me properly I will be highly thankful.
Finding second degree polynomial using newton’s method
calculusnumerical methodspolynomials
Related Solutions
We are given the Rosenbrock Function, sometimes called the banana function with $a = 1, b = 100$ and asked to find the minimum using Newton's method with a starting point $X_0 = (x, y) = (1.2, 1.2)$ as:
$$f(x,y)=(1-x)^2+100(y-x^2)^2$$
Newton's method can be extended for the minimization of multivariable functions.
The iteration is given by:
$$X_{n+1} = X_n - [J_n]^{-1} \nabla f_n$$
where: $$ \nabla f = \begin{bmatrix} \dfrac{\partial f}{\partial x} \\ \dfrac{\partial f}{\partial y} \end{bmatrix} = \begin{bmatrix} 2 (1 - x) - 400 x (-x^2 + y) \\ 200 (-x^2 + y) \end{bmatrix}$$
$$J = \begin{bmatrix} \dfrac{\partial^2 f}{\partial x^2} & \dfrac{\partial^2 f}{\partial x \partial y} \\ \dfrac{\partial^2 f}{\partial y \partial x} & \dfrac{\partial^2 f}{\partial y^2} \end{bmatrix} = \begin{bmatrix} 2 + 800 x^2 + 400 (x^2 - y) & 400 x \\ -400 x & 200 \end{bmatrix}$$
The iterates are:
- $X_0 = \begin{bmatrix} 1.2 \\ 1.2 \end{bmatrix}$
- $X_1 = \begin{bmatrix} 1.19592 \\ 1.4302 \end{bmatrix}$
- $X_2 = \begin{bmatrix} 1.00065 \\ 0.963172 \end{bmatrix}$
- $X_3 = \begin{bmatrix} 1.00058 \\ 1.00115 \end{bmatrix}$
- $X_4 = \begin{bmatrix} 1. \\ 1. \end{bmatrix}$
- $X_5 = \begin{bmatrix} 1. \\ 1. \end{bmatrix}$
Here is a graphical representation of the Newton iterates:
Your whole equation is $f$. More precisely, $y_{n+1}$ is the solution of $$ 0=f(y)=y-y_n-\frac h2(y^2+y_n^2-ϵ(y^3+y_n^3)) $$ closest to $y_n$.
The derivative is then $$ f'(y)=1-\frac h2(2y-3ϵy^2) $$ so that the Newton iteration is $$ y_+=y-\frac{f(y)}{f'(y)}=\frac{y_n-\frac h2(y^2-y_n^2−ϵ(2y^3-y_n^3))}{1-\frac h2(2y-3ϵy^2)} $$
Best Answer
As suggested by @AlexBurdin it has 4 points so it is of third degree polynomial i.e., $4_{points}-1 = 3_{degree}$. It can be solved using the Lagrange polynomial method. $$ L(x) = \sum_{j=0}^{k=data\: points}y_jl_j(x) \\ l_j(x) = {\prod_{0\le m \le k}} \frac{x-x_m}{x_j-x_m} $$ For my case it is four points, so j = 0-3. Where, $m \ne k$ and $f(x) = y$. $$ f(x) = y_0 \cdot \frac{x-x_1}{x_0-x_1} \cdot \frac{x-x_2}{x_0-x_2} \cdot\frac{x-x_3}{x_0-x_3} + y_1 \cdot \frac{x-x_0}{x_1-x_0} \cdot \frac{x-x_2}{x_1-x_2} \cdot\frac{x-x_3}{x_1-x_3} \\+ y_2 \cdot \frac{x-x_0}{x_2-x_0} \cdot \frac{x-x_1}{x_2-x_1} \cdot\frac{x-x_3}{x_2-x_3}+ y_3 \cdot \frac{x-x_0}{x_3-x_0} \cdot \frac{x-x_1}{x_3-x_1} \cdot\frac{x-x_2}{x_3-x_2} $$ Plugging the values: $$ f(x) = 15 \cdot \frac{x-0.2}{0-0.2} \cdot \frac{x-0.4}{0.0-0.4} \cdot\frac{x-0.6}{0.0-0.6}+ 21 \cdot \frac{x-0}{0.2-0} \cdot \frac{x-0.4}{0.2-0.4} \cdot\frac{x-0.6}{0.2-0.6}\\ + 30 \cdot \frac{x-0}{0.4-0} \cdot \frac{x-0.2}{0.4-0.2} \cdot\frac{x-0.6}{0.4-0.6}+ 51 \cdot \frac{x-0}{0.6-0} \cdot \frac{x-0.2}{0.6-0.2} \cdot\frac{x-0.4}{0.6-0.4} \\ $$ $$ f(x)=\frac{-312}{2}(x-0.2)(x-0.4)(x-0.6) + \frac{2625}{2}(x-0.0)(x-0.4)(x-0.6)\\ + \frac{-1875}{1}(x-0.0)(x-0.2)(x-0.6)+ \frac{2125}{2}(x-0.0)(x-0.2)(x-0.4) $$ Solving and we get: $$ f(x) = \frac{375}{2}x^3-\frac{150}{2}x^2+\frac{75}{2}x + 15 $$ By putting 0.1 and 0.5 values into the function we get: $$ f(0.1) = \frac{375}{2}0.1^3-\frac{150}{2}0.1^2+\frac{75}{2}0.1 + 15 = 18.1875\\ f(0.5) = \frac{375}{2}0.5^3-\frac{150}{2}0.5^2+\frac{75}{2}0.5 + 15 = 38.1844 $$ Proof: If we put the given x data values into the function then we get the function values given in the table: $$ f(0.0) = \frac{375}{2}0.0-\frac{150}{2}0.0^2+\frac{75}{2}0.0 + 15 = 15\\ f(0.2) = \frac{375}{2}0.2^3-\frac{150}{2}0.2^2+\frac{75}{2}0.2 + 15 = 21 $$ and so on. Hope this helps.