# Thread: How do I go from A to B in this expression?

1. ## How do I go from A to B in this expression?

This is from my physics homework. It has to do with the relativistic doppler effect. I've simplified it.

They claim that
A: (1/sqrt(1-(x^2)))*sqrt((1+x)/(1-x))

is equal to

B: 1/(1-x)

How do they go from A to B?

2. ## Re: How do I go from A to B in this expression?

$\dfrac{1}{\sqrt{1-x^2}} \cdot \sqrt{\dfrac{1+x}{1-x}}$

$\dfrac{1}{\sqrt{(1-x)(1+x)}}\cdot \sqrt{\dfrac{1+x}{1-x}}$

$\dfrac{1}{\sqrt{(1-x)^2}}$

$\dfrac{1}{1-x}$

3. ## Re: How do I go from A to B in this expression?

$\displaystyle \frac{1}{\sqrt{1-x^2}}\cdot\sqrt{\frac{1+x}{1-x}}=\frac{1}{\sqrt{(1+x)(1-x)}}\cdot\sqrt{\frac{1+x}{1-x}}=\frac{1}{\sqrt{1-x}}\cdot\sqrt{\frac{1}{1-x}}=\frac{1}{\sqrt{(1-x)^2}}=\frac{1}{|1-x|}$

Since we must have $\displaystyle x<1$ for the original expression to be defined, this simplifies to:

$\displaystyle \frac{1}{1-x}$

4. ## Re: How do I go from A to B in this expression?

Originally Posted by romsek
$\dfrac{1}{\sqrt{1-x^2}} \cdot \sqrt{\dfrac{1+x}{1-x}}$

$\dfrac{1}{\sqrt{(1-x)(1+x)}}\cdot \sqrt{\dfrac{1+x}{1-x}}$

$\dfrac{1}{\sqrt{(1-x)^2}}$

$\dfrac{1}{1-x}$
Originally Posted by MarkFL
$\displaystyle \frac{1}{\sqrt{1-x^2}}\cdot\sqrt{\frac{1+x}{1-x}}=\frac{1}{\sqrt{(1+x)(1-x)}}\cdot\sqrt{\frac{1+x}{1-x}}=\frac{1}{\sqrt{1-x}}\cdot\sqrt{\frac{1}{1-x}}=\frac{1}{\sqrt{(1-x)^2}}=\frac{1}{|1-x|}$

Since we must have $\displaystyle x<1$ for the original expression to be defined, this simplifies to:

$\displaystyle \frac{1}{1-x}$
Thank you both! This really helped me!

5. ## Re: How do I go from A to B in this expression?

Originally Posted by SciSil
This is from my physics homework. It has to do with the relativistic doppler effect. I've simplified it.
They claim that
A: (1/sqrt(1-(x^2)))*sqrt((1+x)/(1-x)) is equal to B: 1/(1-x)
How do they go from A to B?
$\dfrac{1}{\sqrt{1-x^2}} \cdot \sqrt{\dfrac{1+x}{1-x}}$
Here is a correction. In order for the above to be defined it is necessary that, $x^2<1$ or $-1<x<1$