A few days ago, we had quiz and I solved 3/4 of questions, and I realized that the 4th question which was mentioned as wrong. Question was
FIND THE SMALLEST POSITIVE ROOT BY USING NEWTON-RAPHSON METHOD,
I asked what's wrong with my solution, so I had an argument with my calculus teacher, about Newton's method of finding roots, specifically about the initial guess
Example was this: $$\bf f(x)=\cos(3x) + x^3 -2$$ a root lies between 1 and 2
I said professor, that we can choose any points between them as initial guess $\bf x_0$ and had chosen 1.5, so solved.
But, professor said, that we can pick initial guess as following formula:
if f(a)f''(x) > 0 then x0 = aorif f(b)f''(x) > 0 then x0 = b,
I don't really know where she got these conditions, but I didn't reply he mail after this.
I searched google about 2 hours and I DIDN'T FIND any result like that. So, I wanted to ask you, who is right here? Me or her? Are we free to choose initial guess as we want, or we have to follow her instructions where it doesn't exist in real life? Or, I wrong, because even in wikipedia, there is no any info's about initial guess.
There is some very common exercise about the Newton method that guarantees convergence in the scalar case.
Consider as base case the one where the function has a sign change and is convex and increasing on the interval $[a,b]$.
Then the tangent $t_a(x)$ at $a$ (function value negative) could be almost horizontal, it could have a root almost anywhere, also outside of the interval. As an example consider $f(x)=x^2+0.1x-0.5$ on the interval $[a,b]=[0,1]$. The tangent in $a=0$ is $t_a(x)=0.1x-0.5$ and has its root $x=5$ well outside the interval. The tangent at $b=1$ is $t_b(x)=2.1(x-1)+0.6$ with its root at $x=\frac57$ inside the interval.
However, the tangent $t_b(x)$ at $b$ (function value positive) is more restricted in its slope and $x$-intercept. It is a supporting line of the convex curve, as is every tangent, thus always below it. Which means that at $a$ the tangent must have a negative value, $t_b(a)\le f(a)<0$, and thus its root $c$ is between $a$ and $b$, inside the interval.
The function value $0=t_b(c)\le f(c)$ at the root of the tangent has to be positive (or $c$ is already the root of $f$), so $b$ can be replaced with this root. The reduced interval $[a,b']=[a,c]$ now satisfies the same conditions. Repeating this observation gives a falling sequence of Newton iterates inside the interval converging to the root.
The other sign combinations of the first and second derivative (concave curve, falling curve) can now be obtained by reflections on the $x$ and $y$ axis, leading to the condition your teacher gave.