how do we get from here.

$\displaystyle x^2_x > 2 - \frac{1}{10^n} (2x_n + \frac{1}{10^n})$

to here?

$\displaystyle > 2 - \frac{5}{10^n}$

It's done in one step in my textbook, so I'm assuming it's obvious, but I can't see it.

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- Mar 6th 2013, 06:12 AMAlyoshaKazAnalysis inequality.
how do we get from here.

$\displaystyle x^2_x > 2 - \frac{1}{10^n} (2x_n + \frac{1}{10^n})$

to here?

$\displaystyle > 2 - \frac{5}{10^n}$

It's done in one step in my textbook, so I'm assuming it's obvious, but I can't see it. - Mar 6th 2013, 06:32 AMPlatoRe: Analysis inequality.
You seem to expect us to be mind readers, in that you have put us in the middle of a proof with no context whatsoever.

It seems that $\displaystyle 2 - \frac{1}{10^n} (2x_n + \frac{1}{10^n})>2 - \frac{5}{10^n}.$

That can only happen if $\displaystyle 2x_n<5-\frac{1}{10^n}$.

You should post the complete problem and proof. - Mar 6th 2013, 06:53 AMAlyoshaKazRe: Analysis inequality.
Ok, sorry. It's a proof for the Theorem:

"For each positive real number a and each integer n > 1, there is a unique real number b such that$\displaystyle b^n = a$

In the special case $\displaystyle b^2 = 2 $

The proof begins with

$\displaystyle 1^2 < 2 < 2^2$

$\displaystyle (1.4)^2 < 2 < (1.5)^2$

$\displaystyle (1.41)^2 < 2 < (1.42)^2$

etc.

to give decimal b^2 = (1.414...)^2 = 2

Then the method uses a table

b b b^2 1 1 1 1.4 1.4 1.96 1.41 1.41 1.9881 1/414 1/414 1.999396 etc.

to prove so.

Then to prove the least upper bound of the set of numbesr in the third column of the table is 2, we check that M=2 is an upper bound of E, which follows from the inequalities at the top. Then to show if M' < 2 there is a number in E that is greater then M'.

We put$\displaystyle x_0 = 1, x_1 = 1.4, x_2 = 1.41, ...$

Then we have

$\displaystyle 2 < (x_n + \frac{1}{10^n})^2 = x^2_n + \frac{2x_n}{10^n} + (\frac{1}{10^n})^2$

so

$\displaystyle x^2_n > 2 - \frac{1}{10^n}(2x_n + \frac{1}{10^n})$

$\displaystyle > 2 - \frac{5}{10^n}$ - Mar 6th 2013, 07:20 AMAlyoshaKazRe: Analysis inequality.
Also, on a side not, I'm not sure how the initial numbers are chosen for the inequalities at the beginning.

- Mar 7th 2013, 12:07 AMAlyoshaKazRe: Analysis inequality.
Is bumping allowed?