- #1

- 1

- 0

d is a variable. X known constant.

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- #1

- 1

- 0

d is a variable. X known constant.

- #2

- 22,089

- 3,296

[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

mfb

Mentor

- 35,399

- 11,766

If d

- #4

Svein

Science Advisor

- 2,156

- 701

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

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

Share: