## root in denominator - how to get this result?

Complex numbers, rational functions, logarithms, sequences and series, matrix operations, etc.
Liz
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### root in denominator - how to get this result?

I'm trying to figure out how to get from point A to point B here. What are the steps in between? Thanks in advance! Liz

Point A: $1/sqrt(a^2 + b^2) - 1/sqrt (a^2)$

Point B: $1 - 1/sqrt(1 + b^2/a^2)$

maggiemagnet
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### Re: root in denominator - how to get this result?

I'm trying to figure out how to get from point A to point B here. What are the steps in between? Thanks in advance! Liz

Point A: $\dfrac{1}{\sqrt{a^2\, +\, b^2}}\, -\, \dfrac{1}{\sqrt{a^2}}$

Point B: $1\, -\, \dfrac{1}{\sqrt{1\, +\, \dfrac{b^2}{a^2}}$
I edited inside the quote to show what I think you meant in the 1st post. Did I get it right? If I messed up, please write back with corrections. Either way please write back showing what you did so far. Thanks!

Liz
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### Re: root in denominator - how to get this result?

Yes, that is what I meant. Sorry I do not know LaTex very well.
I can't even figure out how to get started. Can you give me a hint what would be Step 1?
Thanks, Liz

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Joined: Sun Feb 22, 2009 11:12 pm

### Re: root in denominator - how to get this result?

Point A: $\dfrac{1}{\sqrt{a^2\, +\, b^2}}\, -\, \dfrac{1}{\sqrt{a^2}}$

Point B: $1\, -\, \dfrac{1}{\sqrt{1\, +\, \dfrac{b^2}{a^2}}$

Yes, that is what I meant....

I can't even figure out how to get started. Can you give me a hint what would be Step 1?
When you have fractions, it can help to see what happens if you try to go to common denominators:

$\dfrac{1}{\sqrt{a^2\, +\, b^2}}\, -\, \dfrac{1}{\sqrt{a^2}}$

$\dfrac{\sqrt{a^2}}{\sqrt{a^2}\,\sqrt{a^2\, +\, b^2}}\, -\, \dfrac{\sqrt{a^2\, +\, b^2}}{\sqrt{a^2}\,\sqrt{a^2\, +\, b^2}}$

$\dfrac{\sqrt{a^2}\, -\, \sqrt{a^2\, +\, b^2}}{\sqrt{a^2}\,\sqrt{a^2\, +\, b^2}}$

OK this doesn't look like it's really working, so try something else. Maybe let's do rationalizing the denominators:

$\dfrac{1}{\sqrt{a^2\, +\, b^2}}\, -\, \dfrac{1}{\sqrt{a^2}}$

$\dfrac{1}{\sqrt{a^2\, +\, b^2}}\,\cdot \, \dfrac{\sqrt{a^2\, +\, b^2}}{\sqrt{a^2\, +\, b^2}}\, -\, \dfrac{1}{\sqrt{a^2}}\, \cdot \, \dfrac{\sqrt{a^2}}{\sqrt{a^2}}$

$\dfrac{\sqrt{a^2\, +\, b^2}}{a^2\, +\, b^2}\, -\, \dfrac{\sqrt{a^2}}{a^2}$

This doesn't look right either. Maybe let's try going backwards from Point B:

$1\, -\, \dfrac{1}{\sqrt{1\, +\, \dfrac{b^2}{a^2}}$

$1\, -\, \dfrac{1}{\sqrt{\dfrac{a^2}{a^2}\, +\, \dfrac{b^2}{a^2}}}$

$1\, -\, \dfrac{1}{\sqrt{\dfrac{a^2\, +\, b^2}{a^2}}}$

$1\, -\, \dfrac{1}{\left(\dfrac{\sqrt{a^2\, +\, b^2}}{\sqrt{a^2}}\right)}$

$1\, -\, \dfrac{\sqrt{a^2}}{\sqrt{a^2\, +\, b^2}}$

$\dfrac{\sqrt{a^2}}{\sqrt{a^2}}\, -\, \dfrac{\sqrt{a^2}}{\sqrt{a^2\, +\, b^2}}$

If we factor now we get close to Point A:

$\left(\dfrac{-\sqrt{a^2}}{1}\right)\left(\dfrac{1}{\sqrt{a^2\, +\, b^2}}\, -\, \dfrac{1}{\sqrt{a^2}}\right)$

But I don't see how this can go all the way back to Point A because that would mean that a = 1 and we don't have anything that says that. So maybe this isn't right in the first place? Try checking:

$a\, =\, 3,\, b\, =\, 4$

$\mbox{Point A: }\, \dfrac{1}{\sqrt{a^2\, +\, b^2}}\, -\, \dfrac{1}{\sqrt{a^2}}\, =\, \dfrac{1}{\sqrt{9\, +\, 16}}\, -\, \dfrac{1}{\sqrt{9}}\, =\, \dfrac{1}{5}\, -\, \dfrac{1}{3}\, =\, \dfrac{2}{15}$

$\mbox{Point B: }\, 1\, -\, \dfrac{1}{\sqrt{1\, +\, \dfrac{b^2}{a^2}}}\, =\, 1\, -\, \dfrac{1}{\sqrt{1\, +\, \dfrac{16}{9}}}\, =\, 1\, -\, \dfrac{1}{\sqrt{\dfrac{25}{9}}}\, =\, 1\, -\, \dfrac{1}{\left(\dfrac{5}{3}\right)}\, =\, 1\, -\, \dfrac{3}{5}\, =\, \dfrac{2}{5}$

So maybe that's the problem: they're not actually equal.

Liz
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### Re: root in denominator - how to get this result?

Thank you very much. This helps tremendously.
The problem came up as part of the derivation of an equation in physics. I could not figure out how they got Point B from Point A. I see now later in the text that they say a must be either 1 or -1.
Thanks,
Liz