Help with Newton root approximation proof

[zzmanzz]

Homework Statement

Suppose we have:
$f(x) = x^2 - b$
$b > 0$
$x_0 = b$

And a sequence is defined by:
$x_{i+1} = x_i - \frac{f(x_i)}{f'(x_i) }$

prove
$\forall i \in N ( x_i > 0 )$

Homework Equations

The Attempt at a Solution

a)Here I tried solving for $x_1$ as:
$b > 0$ given in problem
$x_1 = b - \frac{(b^2 - b)}{2b} = b - \frac{1}{2} b - \frac{1}{2}$
$x_1 = \frac{1}{2} b -\frac{1}{2} = \frac{1}{2} (b - 1)$

I think I made a mistake here because it looks like x_1 can be negative which doesn't make sense in real root approximation. I understand that this sequence converges to $\sqrt(b)$ but I'm not sure how to prove that every value is positive. Thanks for the help

 

[Mark44]

Homework Statement

Suppose we have:
$f(x) = x^2 - b$
$b > 0$
$x_0 = b$

And a sequence is defined by:
$x_{i+1} = x_i - \frac{f(x_i)}{f'(x_i) }$

prove
$\forall i \in N ( x_i > 0 )$

Homework Equations

The Attempt at a Solution

a)Here I tried solving for $x_1$ as:
$b > 0$ given in problem
$x_1 = b - \frac{(b^2 - b)}{2b} = b - \frac{1}{2} b - \frac{1}{2}$

You have a sign error in the line above.

$x_1 = \frac{1}{2} b -\frac{1}{2} = \frac{1}{2} (b - 1)$

I think I made a mistake here because it looks like x_1 can be negative which doesn't make sense in real root approximation. I understand that this sequence converges to $\sqrt(b)$ but I'm not sure how to prove that every value is positive. Thanks for the help

 

[tnich]

Homework Statement

Suppose we have:
$f(x) = x^2 - b$
$b > 0$
$x_0 = b$

And a sequence is defined by:
$x_{i+1} = x_i - \frac{f(x_i)}{f'(x_i) }$

prove
$\forall i \in N ( x_i > 0 )$

Homework Equations

The Attempt at a Solution

a)Here I tried solving for $x_1$ as:
$b > 0$ given in problem
$x_1 = b - \frac{(b^2 - b)}{2b} = b - \frac{1}{2} b - \frac{1}{2}$
$x_1 = \frac{1}{2} b -\frac{1}{2} = \frac{1}{2} (b - 1)$

I think I made a mistake here because it looks like x_1 can be negative which doesn't make sense in real root approximation. I understand that this sequence converges to $\sqrt(b)$ but I'm not sure how to prove that every value is positive. Thanks for the help

Try a proof by induction.

 

[Ray Vickson]

Homework Statement

Suppose we have:
$f(x) = x^2 - b$
$b > 0$
$x_0 = b$

And a sequence is defined by:
$x_{i+1} = x_i - \frac{f(x_i)}{f'(x_i) }$

prove
$\forall i \in N ( x_i > 0 )$

Homework Equations

The Attempt at a Solution

a)Here I tried solving for $x_1$ as:
$b > 0$ given in problem
$x_1 = b - \frac{(b^2 - b)}{2b} = b - \frac{1}{2} b - \frac{1}{2}$
$x_1 = \frac{1}{2} b -\frac{1}{2} = \frac{1}{2} (b - 1)$

I think I made a mistake here because it looks like x_1 can be negative which doesn't make sense in real root approximation. I understand that this sequence converges to $\sqrt(b)$ but I'm not sure how to prove that every value is positive. Thanks for the help

First: carry out a complete simplification of the expression for $x_{i+1}$, by using the explicit forms for $f$ and $f'$.

BTW: Positivity of $x_i$ holds for any $x_0 > 0$, not just for $x_0 = b$.