Approximation to sqrt(1+(d^2)/(x^2))

  • #1
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Main Question or Discussion Point

using binomial theorem can I write sqrt(1+(d^2)/(x^2)) = 1+ .5(d^2)/(x^2)?
d is a variable. X known constant.
 

Answers and Replies

  • #2
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If ##d^2 < x^2##, then you can indeed write an approximate equality

[tex]\sqrt{1+ \frac{d^2}{x^2}}\approx 1 + \frac{1}{2} \frac{d^2}{x^2}[/tex]

Whether this approximation is good enough depends entirely on the specifics.

If ##d^2 = x^2##, then it's a boundary case. You can still write the above approximaton, but this is very sensitive to errors. For example, if you measured ##d## or ##x## a bit incorrectly, then you could be in trouble easily.
 
  • #3
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@micromass: the sensitivity to d and x individually should not depend on the approximation made.
If d2=x2, then the left side is about 1.41 and the right side is 1.5, so we have some notable deviation.
 
  • #4
Svein
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using binomial theorem can I write sqrt(1+(d^2)/(x^2)) = 1+ .5(d^2)/(x^2)?
d is a variable. X known constant.
For a first approximation, yes. But you can do better:

  1. Calculate [itex]a=1+\frac{d^{2}}{x^{2}} [/itex]
  2. Let [itex]z_{0}=1+\frac{1}{2}\frac{d^{2}}{x^{2}}=\frac{1+a}{2} [/itex]
  3. Then [itex]z_{n+1}=\frac{1}{2}(z_{n}+\frac{a}{z_{n}}) [/itex] is a better approximation.
Stop when zn+1 is sufficiently close to zn.

In the example above (d=x); a = 2 and z0 = 1.5. Then z1 = 1.417 and z2 = 1.414
 

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