Newton Raphson Method

[reardear]

Suppose \(\displaystyle k\) is a constant. Show that if we apply Newton's method to equation \(\displaystyle \dfrac {1} {x}-k=0\), we get the iterative formula \(\displaystyle x_{n+1}=2x_{n}-kx_{n}^{2}\) for finding reciprocals without dividing.

So with that, I got these (correct me if wrong, maybe it's one of the reasons I'm not getting it).
\(\displaystyle f\left( x\right) = \dfrac {1} {x}-k\)
\(\displaystyle f^\prime\left( x\right) = -\dfrac {1} {x^{2}}-1\)

And the formula:
\(\displaystyle x_{n+1}=x_{n}-\dfrac {\dfrac {1} {x_{n}}-k} {-\dfrac {1} {x_n^{2}}-1}\)

So I assumed that I just try to simplify it to get it to arrive at the formula \(\displaystyle x_{n+1}=2x_{n}-kx_{n}^{2}\), but no matter what I always ended up with a fraction somewhere.

 

[pappus]

So with that, I got these (correct me if wrong, maybe it's one of the reasons I'm not getting it).
\(\displaystyle f\left( x\right) = \dfrac {1} {x}-k\)
\(\displaystyle f^\prime\left( x\right) = -\dfrac {1} {x^{2}}-1\) <--- if you differentiate a constant you'll get zero

And the formula:
\(\displaystyle x_{n+1}=x_{n}-\dfrac {\dfrac {1} {x_{n}}-k} {-\dfrac {1} {x_n^{2}}-1}\)

So I assumed that I just try to simplify it to get it to arrive at the formula \(\displaystyle x_{n+1}=2x_{n}-kx_{n}^{2}\), but no matter what I always ended up with a fraction somewhere.

Since k is a constant the 1st derivation should be:

\(\displaystyle f' \left( x\right) = -\dfrac {1} {x^2}\)

 

[reardear]

Since k is a constant the 1st derivation should be:

\(\displaystyle f' \left( x\right) = -\dfrac {1} {x^2}\)

Got it! Thanks!

 

[Bob Brown MSEE]

Suppose [FONT=MathJax_Math]k[/FONT]
is a constant. Show that if we apply Newton's method to equation[FONT=MathJax_Main]1[/FONT][FONT=MathJax_Math]x[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Math]k[/FONT][FONT=MathJax_Main]=[/FONT][FONT=MathJax_Main]0[/FONT], we get the iterative formula [FONT=MathJax_Math]x[/FONT][FONT=MathJax_Math]n[/FONT][FONT=MathJax_Main]+[/FONT][FONT=MathJax_Main]1[/FONT][FONT=MathJax_Main]=[/FONT][FONT=MathJax_Main]2[/FONT][FONT=MathJax_Math]x[/FONT][FONT=MathJax_Math]n[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Math]k[/FONT][FONT=MathJax_Math]x[/FONT][FONT=MathJax_Main]2[/FONT][FONT=MathJax_Math]n[/FONT] for finding reciprocals without dividing.

The above demonstrations are by application of Newton.
We can check results this way...

Suppose that the iterative formula does converge, Then any future iterations will not change the value. So X_n+1 = X_n.
formula becomes: x=2x-kx²

solving 1/x = 2/x -k when x not 0
0 = 1/x -k Checks!