In Newton's method, if the residual delta x is iteratively updated ($X \leftarrow X + \Delta x$) at fixed slope, does it converge?
In newton's method, the estimation of X should be done iteratively. In each iteration the derivative (slope) of a function, the residual are calculated and used to update X. My question is if what will happen if we performed the derivative only one time.
More precisely, assume the following:
f(X)=A+B
my target is to estimate X
solution 1::
- outer loop calculate the slope of f(X)
- inner loop
- —-calculate A
- —-calculate B
- —-calculate delta x using f(x), A, B and the slope
- end inner loop
- X=X+ delta x
- end outer loop
solution 2::
- loop
- calculate the slope of f(X)
- calculate A
- calculate B
- calculate delta x using f(x), A, B and the slope
- X = X + delta x
- end loop
does the accuracy of X in both cases similar? and why?
Best Answer
This is an interesting question.
The "normal" iterative scheme is $$x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}$$ with a permanently updated slope.
You are asking about $$x_{n+1}=x_n-\frac{f(x_n)}{f'(x_0)}$$ This was used a lot at a time (I knew that time) where computers did not exist and everything had to be done by hand. It works quite well provided $\cdots$ a few things. For sure, the convergence is slower than the "normal" method.
For illustration purposes, I selected $f(x)=e^{-x}-x$ starting using $x_0=1$. Below are given the iterates for the two cases. $$\left( \begin{array}{ccc} n & \text{with update} & \text{without update}\\ 1 & 0.53788284273999024150 & 0.53788284273999024150 \\ 2 & 0.56698699140541323884 & 0.57158499872254489612 \\ 3 & 0.56714328598912294403 & 0.56650033488134524122 \\ 4 & 0.56714329040978386946 & 0.56723703773937297193 \\ 5 & 0.56714329040978387300 & 0.56712963572569512729 \\ 6 & 0.56714329040978387300 & 0.56714527957469274352 \\ 7 & 0.56714329040978387300 & 0.56714300064189749926 \\ 8 & 0.56714329040978387300 & 0.56714333262131691955 \\ 9 & 0.56714329040978387300 & 0.56714328426068052466 \\ 10 & 0.56714329040978387300 & 0.56714329130554557275 \end{array} \right)$$