Newton Raphson Example help

Question

For h:= $\mathbb{R} \rightarrow \mathbb{R}, x \rightarrow e^{x}-x^2+1$ I know the formula as $$X_{n+1}=X_{n}-\frac{f(X_{n})}{f'(X_{n})}$$ so this would give me: $$ X_{n+1}=X_{n}-\frac{e^{x}-x^{2}+1}{e^{x}-2x}$$ However my lecture has in his solution the iteration map is: $$\frac{(xe^{x}-x^{2}+e^{2}-1)}{e^{x}-2x}$$

My question is how does he get this which is different to mine and dosen't look to similar to the Newton Raphson formula?

Answer 1

You haven't completely substituted. Using the formula, we have: $$\begin{align*}x_{n+1} &= x_n - \frac{e^{x_n}-x_n^2 +1}{e^{x_n} - 2x_n}\\&=\frac{x_n(e^{x_n} - 2x_n) -e^{x_n} + x_n^2 - 1}{e^{x_n} - 2x_n}\\&= \frac{x_ne^{x_n}-x_n^2-e^{x_n} - 1}{e^{x_n} - 2x_n}\end{align*}$$

So that the iteration map is

$$x \mapsto \frac{xe^x - x^2 - e^x - 1}{e^x - 2x}$$

Answer 2

Write $$x_{n+1}=x_{n}-\frac{e^{x_n}-x_n^{2}+1}{e^{x_{n}}-2x_{n}}$$ Over a common denominator $$x_{n+1} = \frac{x_{n}e^{x_n}-x_{n}^{2}-e^{x_{n}}-1}{e^{x_{n}}-2x_{n}} $$