Solve for 0<b<a: Proving Inequality & Approximation Error

[Alephu5]

Homework Statement

Firstly, I'd just like to point out that this is not actually a course related question. I have been trying to teach myself mathematics, and have been grappling with this for a couple of days. The book has no answer at the back for this particular question.

Variables:
$0<x<1$
$0<b<a$

Show that $(1-\frac{1}{2}x^{2})^{2} < 1-x < (1-\frac{1}{2}x)^{2}$. Hence show that if $0<b<a$, the error in taking $a-\frac{b^{2}}{2a}$ as an approximation to $\sqrt{a^{2}-b^{2}}$ is positive and less than $\frac{b^{4}}{2a^{3}}$.

Homework Equations

N/A

The Attempt at a Solution

The first part is relatively easy:

Expansion of the inequality involving x gives:

$1-x-\frac{3}{4}x^{2}+\frac{1}{2}x^{3}+\frac{1}{4}x^{4}<1-x<1-x+\frac{1}{4}x^{2}$​

Due to the fact fact that

$0<x<1$​

The following is true:

$x^{n}>x^{n+1}$​

This concept can be used to prove that

$\frac{3}{4}x^{2}>\frac{1}{2}x^{3}+\frac{1}{4}x^{4}$​

The last part is more straightforward, it is simply due to the fact that:

$\frac{1}{4}x^{2}>0$​

I have no idea how to connect the statement involving $a$ and $b$ to this set of inequalities, however from what I understand the initial statement is:

$0<a-\frac{b^{2}}{2a}-\sqrt{a^{2}-b^{2}}<\frac{b^4}{2a^3}$​

I have attempted a bit of algebra jiggling, which gives:

$2a^{2}+b^{2}<3a^{4}$​

Evidently, this is only true when $a>1$

Any help would be much appreciated! I would really love to put this to rest, so that I can move beyond page 34... there are about 450 more to go.

 

[awkward]

Hint: Let $$x = \left( \frac{b}{a} \right) ^2$$