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Suppose that R>0, $x_0>0$, and $$x_{n+1}= \frac{1}{2} (\frac{R}{x_n}+x_n), n \geq0$$

Prove by induction for $n \geq 1$, $$x_n>x_{n+1} > \sqrt{R} $$ and $$ x_n- \sqrt{R} \leq \frac{1}{2^n} \cdotp \frac {(x_0- \sqrt{R})^2} {x_0} $$

I have written pages on this, getting more and more lost. I considered first 2 propositions:

$P_n: x_n>x_{n+1} > \sqrt{R}$

$G_n: x_n- \sqrt{R} \leq \frac{1}{2^n} \cdotp \frac {(x_0- \sqrt{R})^2} {x_0} $

And I looked at the first one $P_n$:

$P_1: x_1>x_2 > \sqrt{R} \rightarrow x_1 > \frac{1}{2} (\frac{R}{x_1}+x_1)= \frac{R+{x_1}^2}{2x_1}> \sqrt{R}$

$P_2: x_2>x_3 > \sqrt{R} \rightarrow$

$$\frac{R+{x_1}^2}{2x_1}> \frac{R+{x_2}^2}{2x_2}> \sqrt{R}$$

$$\frac{R+{x_1}^2}{2x_1}> \frac{R+{\frac{R+{x_1}^2}{2x_1}}^2}{2(\frac{R+{x_1}^2}{2x_1})}> \sqrt{R}$$

I attempted to simplify with no satisfying result.

Am i going in the right direction? is there a different inductive approach more efficient?

gegu
  • 1,694

3 Answers3

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Hints:

  • $P_n\,$: $\;\;x_{n+1}-x_n = \cfrac{1}{2} \left(\cfrac{R}{x_n}-x_n\right)$

  • $G_n\,$: $\;\;x_{n+1}= \cfrac{1}{2} \left(\cfrac{R}{x_n} \color{red}{-2\sqrt{R}} + x_n + \color{red}{2\sqrt{R}}\right) \;\;\iff\;\; x_{n+1}-\sqrt{R} = \cfrac{1}{2}\cdot\cfrac{\left(x_n-\sqrt{R}\right)^2}{x_n}$

dxiv
  • 76,497
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Assume $P_n$ and $G_n$. Then $$x_{n+2} =\frac{1}{2}\left(x_{n+1} + \frac{R}{x_{n+1}}\right) < \frac{1}{2}\left(x_{n+1} + x_{n+1}\right) < x_{n+1}$$ since we have $x_{n+1} > \sqrt{R} \implies x_{n+1}^2 > R \implies x_{n+1} > \frac{R}{x_{n+1}} $.

Also we have $x_{n+2} = \frac{1}{2}\left(\frac{x^2_{n+1} + R}{x_{n+1}}\right) > \frac{1}{2}\left(\frac{2R}{\sqrt{R}}\right) = \sqrt{R}$ since the numerator is $>2R$ and the denominator is $<\sqrt{R}$ from assuming $G_{n}$, we use $x_0$ positive here. Hence we have $P_{n+1}.$

Now $$x_{n+1} - \sqrt{R} \leq \frac{1}{2}x_n -\sqrt{R} \leq \frac{1}{2^{n+1}}\cdot \frac{(x_0 - \sqrt{R})^2}{x_0}$$ and we are done again.

Moral: When given two statements to prove using induction, you often need to do them simultaneously.

Zain Patel
  • 16,802
1

$x_{n+1} = \frac 12 (\frac {R}{x_n} +x_n)$
Proposition:

$x_1>x_c>\cdots x_n> \sqrt R$

Base case. If $x_0> \sqrt R$ we can take this as the base case

If $x_0< \sqrt R$

$x_1 = \frac 12 (\frac {R}{x_0} +x_0)$

$\frac 12 (R + x_0^2) > x_0 \sqrt R$ by the AM-GM ineqality

$x_1 > \sqrt R$

Inductive hypothesis.

Suppose $x_{n-1}>x_n > \sqrt R$

we must show that: $x_{n}>x_{n+1} > \sqrt R$

$x_{n+1} > \sqrt R$ from the same AM-GM inequality used above.

We still need to show that $x_{n} > x_n+1$

$x_{n+1} = \frac 12 (\frac {R^2 + x_n^2}{x_n})\\ x_{n+1} - x_n = \frac 12 (\frac {R^2 - x_n^2}{x_n})$

$x_n > \sqrt R$ (from the inductive hypothesis)

$x_{n+1} - x_n < 0$

QED

$x_n - \sqrt R = \frac 1{2^n} \frac {(x_0-\sqrt R)^2}{x_0}$

base case $n=1$

$x_1 - \sqrt R = \frac {R + x_0^2-2x_0\sqrt R}{2x_0} = \frac 1{2} \frac {(x_0-\sqrt R)^2}{x_0}$

Inductive hypothesis:

Suppose, $x_n - \sqrt R \le \frac {(x_0-\sqrt R)^2}{2^nx_0}$

We must show that $x_{n+1} - \sqrt R \le \frac 1{2^{n+1}} \frac {(x_0-\sqrt R)^2}{x_0}$

$x_{n+1} - \sqrt R = \frac {(x_n-\sqrt R)^2}{2x_n} = \frac{(x_n - \sqrt R)}{2}\frac{(x_n-\sqrt R)}{x_n}$

With one factor we will say $\frac{(x_n-\sqrt R)}{x_n}< 1$

and with the other $(x_n - \sqrt R)\le \frac {(x_0-\sqrt R)^2}{2^nx_0}$ by the inductive hypothesis

$x_{n+1} - \sqrt R \le \frac 1{2^{n+1}} \frac {(x_0-\sqrt R)^2}{x_0}$

QED

Doug M
  • 57,877
  • Thx a lot. What does AM-GM mean? what about the fact that $n \geq 1$ in the proposition, is having $x_0$ a problem? – gegu May 21 '17 at 05:26
  • AM-GM inequality. The Arithmetic mean is always greater that the geometric mean. $\sum_\limits{i=1}^n \frac {x_i}{n} \ge \prod_\limits{i=1}^n x_i^{\frac 1n}$. Or in a simple case $(x-y)^2 > 0 \implies x^2 + y^2 > 2xy$ – Doug M May 22 '17 at 15:47