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

  1. #1
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    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?
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  2. #2
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    Re: How do I go from A to B in this expression?

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

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

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

    $\dfrac{1}{1-x}$
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  3. #3
    MHF Contributor MarkFL's Avatar
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    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:

    $\displaystyle \frac{1}{1-x}$
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    Re: How do I go from A to B in this expression?

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

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

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

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

    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!
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    Re: How do I go from A to B in this expression?

    Quote Originally Posted by SciSil View Post
    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 $-1<x<1$
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