Cannot simplify


by MathewsMD
Tags: simplify
MathewsMD
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#1
Jan20-14, 10:34 PM
P: 285
Hi,

I've attached a screenshot in which a solution simplifies an expression given x>>a. I have been trying to simplify and express the equation as shown, but keep failing to recognize how they reached their solution. If anyone could provide a step-by-step breakdown that would be greatly appreciated!
Attached Thumbnails
Screen Shot 2014-01-20 at 11.32.45 PM.png  
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Mentallic
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#2
Jan20-14, 10:58 PM
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When [itex]x>>a[/itex] then you can assume that [tex]x\pm a\to x[/tex] In other words, you just cancel the a when it's in a similar form to the above.

[tex]3x^2-a^2\approx 3x^2[/tex] for example.
MathewsMD
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#3
Jan21-14, 08:54 PM
P: 285
Quote Quote by Mentallic View Post
When [itex]x>>a[/itex] then you can assume that [tex]x\pm a\to x[/tex] In other words, you just cancel the a when it's in a similar form to the above.

[tex]3x^2-a^2\approx 3x^2[/tex] for example.
Okay. That is what I did...I may just be simplifying incorrectly but the expression never seems to simplify to the answer stated...I've tried expanding the given equation multiple times now but to no avail and anyone's steps to simplify this would be greatly appreciated!

Dick
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#4
Jan21-14, 10:51 PM
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Cannot simplify


Quote Quote by MathewsMD View Post
Okay. That is what I did...I may just be simplifying incorrectly but the expression never seems to simplify to the answer stated...I've tried expanding the given equation multiple times now but to no avail and anyone's steps to simplify this would be greatly appreciated!
I think you need to show what you tried before anyone can make a comment on where you are going wrong. Put the terms over a common denominator first and then work with the numerator.
vela
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#5
Jan21-14, 11:39 PM
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Quote Quote by Mentallic View Post
When [itex]x>>a[/itex] then you can assume that [tex]x\pm a\to x[/tex] In other words, you just cancel the a when it's in a similar form to the above.

[tex]3x^2-a^2\approx 3x^2[/tex] for example.
You need to be a little more careful. For example, suppose you have something like ##\frac{1}{x-a}##. You can factor the denominator and expand as a series to get
$$\frac{1}{x(1-a/x)} = \frac{1}{x}\left[1+\frac{a}{x}+\left(\frac{a}{x}\right)^2 + \left(\frac{a}{x}\right)^3 + \cdots \right].$$ If you say
$$\frac{1}{x-a} \rightarrow \frac{1}{x},$$ you're only taking the zeroth-order approximation. The first-order approximation would be
$$\frac{1}{x-a} \rightarrow \frac{1}{x}+\frac{a}{x^2}.$$ You generally want to keep enough terms to find the first non-vanishing order in some calculation.


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