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

[MOHD ZAKI]

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.

 

[micromass]

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

$$\sqrt{1+ \frac{d^2}{x^2}}\approx 1 + \frac{1}{2} \frac{d^2}{x^2}$$

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.

 

[mfb]

@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.

 

[Svein]

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 $a=1+\frac{d^{2}}{x^{2}}$
2. Let $z_{0}=1+\frac{1}{2}\frac{d^{2}}{x^{2}}=\frac{1+a}{2}$
3. Then $z_{n+1}=\frac{1}{2}(z_{n}+\frac{a}{z_{n}})$ is a better approximation.

Stop when z_n+1 is sufficiently close to z_n.

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