I think there is some thing wrong

[MIB]

I think there is some thing wrong in this exercise which I met by chance in a book of Calculus and analysis while looking for rigorous definition for angle , it says

let f : (a,b) → ℝ be a differentiable function , suppose that f' is bounded , and that f has a root r in (a,b) . suppose that for x ≠ r , J_x denote the open interval between x and r , where if f(x) > 0 then f is convex on J_x , and if f(x) < 0 then f is concave on J_x.Then prove that for any x₀ in (a,b) the Newton sequence converges to a root where x₀ is it initial point .

The problem here is that we can take the initial point x₀ where f'(x₀)=0 .for the simplicity consider f(x) := x^3 for all x in (-4,4)

 

[MIB]

I think we must replace "f' is bounded on (a,b)" by "f' is non-zero throughout (a,b)"

 

[Mute]

I think there is some thing wrong in this exercise which I met by chance in a book of Calculus and analysis while looking for rigorous definition for angle , it says

let f : (a,b) → ℝ be a differentiable function , suppose that f' is bounded , and that f has a root r in (a,b) . suppose that for x ≠ r , J_x denote the open interval between x and r , where if f(x) > 0 then f is convex on J_x , and if f(x) < 0 then f is concave on J_x.Then prove that for any x₀ in (a,b) the Newton sequence converges to a root where x₀ is it initial point .

The problem here is that we can take the initial point x₀ where f'(x₀)=0 .for the simplicity consider f(x) := x^3 for all x in (-4,4)

There is no problem in this particular example because the initial point you chose is the root of your function, so you've already converged to the root and aren't going anywhere.

Try coming up with an example where the derivative vanishes at a point that is not the root of the function. Be sure to double-check that the concavity conditions of the theorem are still met in this case. If you can show that you can't pick a function with f'(x₀) = 0 for which the concavity requirements are satisfied, then there is no problem with the theorem as stated.

 

[MIB]

There is no problem in this particular example because the initial point you chose is the root of your function, so you've already converged to the root and aren't going anywhere.

Try coming up with an example where the derivative vanishes at a point that is not the root of the function. Be sure to double-check that the concavity conditions of the theorem are still met in this case. If you can show that you can't pick a function with f'(x₀) = 0 for which the concavity requirements are satisfied, then there is no problem with the theorem as stated.

I know it can be very hard to think an example where the derivative vanishes at a point which is not a root , and I think it is impossible the problem is that it says " for any x0 in (a,b) the Newton sequence converges to a root where x0 is it initial point " , and then this wrong ,because the root is in (a,b) .

 

[MIB]

OK I restated it as following

let f : (a,b) → ℝ be a differentiable function , suppose that f' is bounded , and that f has a root r in (a,b) . suppose that for x ≠ r , Jx denote the open interval between x and r , where if f(x) > 0 then f is convex on Jx , and if f(x) < 0 then f is concave on Jx.Then prove that for any x0 in (a,b) the Newton sequence converges to a root where x0 is it initial point where the derivative doesn't vanish at x0.

And I proved it easily then .