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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.

d is a variable. X known constant.

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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.

d is a variable. X known constant.

- #2

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[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.

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mfb

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If d

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Svein

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For a first approximation, yes. But you can do better: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.

- Calculate [itex]a=1+\frac{d^{2}}{x^{2}} [/itex]
- Let [itex]z_{0}=1+\frac{1}{2}\frac{d^{2}}{x^{2}}=\frac{1+a}{2} [/itex]
- Then [itex]z_{n+1}=\frac{1}{2}(z_{n}+\frac{a}{z_{n}}) [/itex] is a better approximation.

In the example above (d=x); a = 2 and z

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