Newton's method to find next iterate

Question

Let $f(x), g(x) : \mathbb{R} \to \mathbb{R}$ be smooth functions and consider the problem of finding the solution to $$ f(x) + g(x) = b.$$

Assume one is given an approximate solution value $x^{n-1}$, derive the formula for the next approximation, $x^n,$ that is obtained by using one step of Newton's method with starting iterate $x^{n-1}$ applied to the problem $$f(x) + g(x^{n-1}) = b.$$

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I know Newton's method, $$x^{n} = x^{n-1} - \frac{f(x^{n-1})}{f'(x^{n-1})},$$

but the wording of this problem has me thrown. Specifically how to use the last statement. My guess was to write $f(x) = b - g(x^{n-1}),$ and then the next iterate would be given by $$ x^{n} = x^{n-1} - \frac{b - g(x^{n-1})}{g'(x^{n-1})}. $$

Does anyone know if this is what the question is asking, or have a better grip on what they're intending on asking? Thanks in advance.

Answer 1

I think the problem is asking you to write $F(x)=f(x)+g(x^{n-1})-b$, and find the zero of $F(x)$. So

$$x^n=x^{n-1}-\frac{f(x^{n-1})+g(x^{n-1})-b}{f'(x^{n-1})}.$$

Answer 2

I think the question would've been clearer if they'd said something about why you would do this. One scenario is the following: you want to solve $f(x)+g(x)=b$, but because $g$ and/or $g'$ is much more expensive to compute than $f$ and $f'$, you don't want to do it by just defining $h(x)=f(x)+g(x)$ and solving $h(x)=b$ by the ordinary Newton method. Instead, you want to do a bunch of Newton iterations with the argument to $g$ "frozen", changing only the argument to $f$. Periodically you'll then change the argument of $g$ to the current argument of $f$ and repeat the procedure.

Thus in this single step, you have $F(x)=f(x)+g(x^{n-1})$ and you want to apply a Newton step to the equation $F(x)=b$ with the initial guess $x=x^{n-1}$.