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:
[itex]0<x<1[/itex]
[itex]0<b<a[/itex]

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

Homework Equations

N/A

The Attempt at a Solution

The first part is relatively easy:

Expansion of the inequality involving x gives:

[itex]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}[/itex]​

Due to the fact fact that

[itex]0<x<1[/itex]​

The following is true:

[itex]x^{n}>x^{n+1}[/itex]​

This concept can be used to prove that

[itex]\frac{3}{4}x^{2}>\frac{1}{2}x^{3}+\frac{1}{4}x^{4}[/itex]​

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

[itex]\frac{1}{4}x^{2}>0[/itex]​

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

[itex]0<a-\frac{b^{2}}{2a}-\sqrt{a^{2}-b^{2}}<\frac{b^4}{2a^3}[/itex]​

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

[itex]2a^{2}+b^{2}<3a^{4}[/itex]​

Evidently, this is only true when [itex]a>1[/itex]

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 [tex]x = \left( \frac{b}{a} \right) ^2[/tex]