- #1
Alephu5
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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.
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