Num.Methods - in Newtons Meth. how to get "m" in the formula ?

[artiny]

Newton metods,how to get "m" in formula ?
|alfa - x2| <= |f (x2) |/m - something with derivations.


example in book:
<1,5 2>
f = x^3 - x - 1 =0
f'= 3*x^2 - 1 = 0

the m is: m = 2 ,how to calculate the m?

 

[HallsofIvy]

Newton metods,how to get "m" in formula ?
|alfa - x2| <= |f (x2) |/m - something with derivations.


example in book:
<1,5 2>
f = x^3 - x - 1 =0
f'= 3*x^2 - 1 = 0

the m is: m = 2 ,how to calculate the m?

First, that is NOT the way I would write "Newton's method"- I would write it as \(\displaystyle x_{n+1}= x_n- \frac{f(x_n)}{f'(x_n)}\). It is not clear to me what your "alfa and "x2" are. My "\(\displaystyle x_n\)" and "\(\displaystyle x_{n+1}\)" are two successive terms in a sequence that, hopefully, converges to a solution to the equation f(x)= 0. The idea is to chose some starting point, say \(\displaystyle x_0\), and construct the tangent line to y= f(x) at \(\displaystyle (x_0, f(x_0)\), \(\displaystyle y= f'(x_0)(x- x_0)+ f(x_0)\) and find the point where that tangent line crosses the x-axis; solve \(\displaystyle f'(x_0)(x- x_0)+ f(x_0)= 0\). Since the equation is linear, that's easy. Subtract \(\displaystyle f(x_0)\) from both sides to get \(\displaystyle f'(x_0)(x- x_0)= -f(x_0)\), divide both sides by \(\displaystyle f'(x_0)\) to get \(\displaystyle x- x_0= -\frac{f(x_0)}{f'(x_0)}\) and, finally, add \(\displaystyle x_0\) to both sides to get \(\displaystyle x= x_0- \frac{f(x_0)}{f'(x_0)}\). That should be, hopefully, closer to the solution to \(\displaystyle f(x)= 0\) than the initial \(\displaystyle x_0\) was so call it "\(\displaystyle x_1\)" and continue.

In other words, your "m" is the derivative of f evaluated at the given \(\displaystyle x_0\).

Unfortunately, I don't know what your notation "< 1, 5 2>" is intended to mean. Is that what I would call (1.5, 2), the point where x= 1.5 and y= 2? If so then your "m" should be \(\displaystyle f'(1.5)= 3(1.5)^2- 1= 5.75\), not 2 so apparently that is NOT what you mean!
It is true that f'(1)= 3(1^2)- 1= 3- 1= 2 but I don't know what the ", 5" is intended to indicate.

 

[artiny]

From the presentation I found this:

​< link to objectionable page removed >

 

[HallsofIvy]

Okay, I don't know why they are using "Matlab" to do simple arithmetic but the denominator, what you are calling "m", is the slope or derivative of the original function evaluated at the given point. The given equation to be solved is \(\displaystyle f(x)= 5.3x^3+ 2.2x^2- 12.5x- 6= 0\) so the derivative is \(\displaystyle 15.9x^2+ 4.4x- 12.5\). That, evaluated at x= 1.5 is \(\displaystyle 15.9(1.5)^2+ 4.4(1.5)- 12.5= 35.775+ 6.6- 12.5= 29.875\), which was rounded to 30 since the original coefficients have only two significant figures.