Newton's Method Reversed (using iteration formula to figure out f(x))

Question

The iteration formula $x_{n+1} = x_n − \cos(x_n)\sin(x_n) + R\cos^2x_n$ , where $R$ is a positive constant, was obtained by applying Newton's method to some function $f(x)$. What was $f(x)$? What can this formula be used for?

So $f(x)/f'(x) = −\cos(x_n)\sin(x_n) + R\cos^2x_n$...but I tried out different equations and have no idea what to do from this point.

Answer 1

Let's see... your first mistake is already evident: the equation of the tangent line through $(x_n,f(x_n))$ is $$\frac{y-f(x_n)}{x-x_n}=f^{\prime}(x_n)$$ Setting $y_{n+1}=y=0$ and solving for $x_{n+1}=x$, we get $$x_{n+1}=x_n-\frac{f(x_n)}{f^{\prime}(x_n)}$$ That means our starting point should be $$\frac{f(x)}{f^{\prime}(x)}=\cos x\sin x-R\cos^2x$$ So you have a wrong sign and inconsistent notation ($x$ vs. $x_n$) in your question.

To solve, take reciprocals to get $$\frac{f^{\prime}(x)}{f(x)}=\frac d{dx}\ln f(x)=\frac1{(\sin x-R\cos x)\cos x}$$ Integrating both sides via the substitution $z=\tan\frac x2$, so $\cos x=\frac{1-z^2}{1+z^2}$, $\sin x=\frac{2z}{1+z^2}$, and $dx=\frac{2dz}{1+z^2}$ and partial fractions decomposition, we get $$\begin{align}\ln\left|f(x)\right|&=\int\frac{dx}{(\sin x-R\cos x)\cos x}=\int\frac{2(1+z^2)dz}{(1-z^2)(Rz^2+2z-R)}\\ &=\int\left(\frac{2z}{1-z^2}+\frac{2Rz+2}{Rz^2+2z-R}\right)dz\\ &=-\ln\left|1-z^2\right|+\ln\left|Rz^2+2z-R\right|+C_1\\ &=\ln\left|\frac{Rz^2+2z-R}{1-z^2}\right|+C_1\\ &=\ln\left|\frac{2z}{1+z^2}\frac{1+z^2}{1-z^2}-R\right|+C_1\\ &=\ln\left|\tan x-R\right|+C_1\end{align}$$ So $$f(x)=\pm e^{C_1}\left(\tan x-R\right)=C\left(\tan x-R\right)$$ Which may be checked by differentiation. This formula may be used to find $x=\tan^{-1}R$

Answer 2

Newton's Method is $$x_{n+1}=x_n - \frac{f(x_n)}{f'(x_n)}$$ so if the former iteration formula is $$x_{n+1} = x_n − \cos(x_n)\sin(x_n) + R\cos^2x_n$$ then the corresponding $f(x)/f'(x) = \cos(x_n)\sin(x_n) - R\cos^2x_n$ and the next step is to solve this ordinary differential equation, but it is too hard for ordinary people like me (maybe include u:), i use one python package named sympy to help me solve. The answer is as follows: $$f(x)=\begin{cases} \frac{-C_1\cdot\tan(x)}2, &\text{for $R$=0}\\[10pt] \frac{C1 \cdot R \cdot \tan^2(\frac{x}2) - C_1 \cdot R + 2 \cdot C_1 \cdot \tan(\frac{x}2)}{R \cdot \tan^2(\frac{x}2)-R}, & \text{otherwise} \end{cases}$$ I hope I made it clear, Please feed me back if anything doesn't make sense.