Question about taking the difference between two square root expression

  • Thread starter zaybu
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  • #1
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In Ramakrishna's paper, http://arxiv.org/ftp/arxiv/papers/1111/1111.1922.pdf , he derived equation (9), page 5.

It is a difference of two square-roots using an approximation method. Can anyone help in how this is done?

Thanks
 

Answers and Replies

  • #2
Simon Bridge
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Lets see - the equation would have been:

[tex]
\frac{1}{c}\sqrt{
H^2 + \frac{1}{4}\left (
\frac{D}{2}-v\Delta t
\right )^2
} - \frac{1}{c}\sqrt{
H^2+\frac{D^2}{4}
} \approx -\frac{1}{2}\frac{v \delta t}{c}\frac{D}{\sqrt{H^2 + \frac{D^2}{4}}}
[/tex]

The text says this is "to first order in [itex]v\Delta t[/itex]" ... which means he took the Taylor Series to first order, putting [itex]x=v\Delta t[/itex] about [itex]x=0[/itex].

Taking the LHS for f(x), the 1st order Taylor approx is:

[tex]f(x) \approx f(0) + x{f^\prime}(0) [/tex]

He just rewrote the delta-tee to emphasize it's small size.

I havn't actually crunched the numbers but it looks about the right shape - the deriverative gets you the -1/2 and pulls the vδt outside the root - putting the root itself in the denominator.
 
  • #3
53
2
Lets see - the equation would have been:

[tex]
\frac{1}{c}\sqrt{
H^2 + \frac{1}{4}\left (
\frac{D}{2}-v\Delta t
\right )^2
} - \frac{1}{c}\sqrt{
H^2+\frac{D^2}{4}
} \approx -\frac{1}{2}\frac{v \delta t}{c}\frac{D}{\sqrt{H^2 + \frac{D^2}{4}}}
[/tex]

The text says this is "to first order in [itex]v\Delta t[/itex]" ... which means he took the Taylor Series to first order, putting [itex]x=v\Delta t[/itex] about [itex]x=0[/itex].

Taking the LHS for f(x), the 1st order Taylor approx is:

[tex]f(x) \approx f(0) + x{f^\prime}(0) [/tex]

He just rewrote the delta-tee to emphasize it's small size.

I havn't actually crunched the numbers but it looks about the right shape - the deriverative gets you the -1/2 and pulls the vδt outside the root - putting the root itself in the denominator.
Thanks, I appreciate.
 
  • #4
Simon Bridge
Science Advisor
Homework Helper
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I suspect that 1/4 in the first square root is not supposed to be there.
If you leave it off, then the zeroth order term vanishes at x=0 and

[tex]f^\prime (v\Delta t) = \frac{1}{c}\frac{1}{2}\left [ H^2 + \left ( \frac{D}{2} -v\Delta t \right )^2 \right ]^{-1/2} 2 \left ( \frac{D}{2} -v\Delta t \right ) (-1)[/tex]

(Leaving it messy to show the working.) So the first order term, [itex](v\delta t) f^\prime(0)[/itex], comes out as shown.

Otherwise you end up with:

[tex]
f(v\delta t) \approx

\frac{1}{c}\sqrt{H^2 + \frac{1}{4}\frac{D^2}{4}} -
\frac{1}{c}\sqrt{H^2 + \frac{D^2}{4}} -

\frac{1}{8}\frac{v\delta t}{c}\frac{D}{\sqrt{H^2 + \frac{1}{4}\frac{D^2}{4}}}

[/tex]

I have not found any reason for the division by 4 in the equation for PDsat-B(B:2) since it should just be the distance from B to the satellite (in position 2) by Pythagoras.
 
  • #5
53
2
Indeed, the 1/4 in the PDsat-b(B:2) term is a typo error.

Thanks again
 

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