- #1

MOHD ZAKI

- 1

- 0

d is a variable. X known constant.

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- #1

MOHD ZAKI

- 1

- 0

d is a variable. X known constant.

- #2

- 22,178

- 3,305

[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

- 36,106

- 13,051

If d

- #4

Svein

Science Advisor

- 2,245

- 767

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

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