Calculating a Square Root Without Calculator

  • Thread starter bjnartowt
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  • #1
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Hi all, I am trying to find a nice back-of-the-envelope way to numerically-calculate nasty things like square roots and stuff. For instance:
[itex]\sqrt {4.51} = ?[/itex]

I tried to use a linearization:
[itex]\sqrt x = \sqrt {n + \varepsilon } = \sqrt {\left[ {{\rm{integer}}} \right] + \left[ {{\rm{decimal part}}} \right]} = \sqrt n + L(\sqrt x ) \cdot \varepsilon [/itex]
…where you “linearly” count your way from [itex]\sqrt {n + \varepsilon } [/itex] to [itex]\sqrt x [/itex] by way of the linearization, [itex]L(\sqrt x )[/itex], namely, the derivative:
[itex]L(\sqrt x ) = {\left. {\frac{d}{{dx}}\sqrt x } \right|_{x = n}} = \frac{n}{{2\sqrt n }}[/itex]

But that’s another square root to calculate, which would further-roughen my approximation. Are there any other numerical-approximations you’d suggest?

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The Attempt at a Solution

 

Answers and Replies

  • #2
berkeman
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Hi all, I am trying to find a nice back-of-the-envelope way to numerically-calculate nasty things like square roots and stuff. For instance:
[itex]\sqrt {4.51} = ?[/itex]

I tried to use a linearization:
[itex]\sqrt x = \sqrt {n + \varepsilon } = \sqrt {\left[ {{\rm{integer}}} \right] + \left[ {{\rm{decimal part}}} \right]} = \sqrt n + L(\sqrt x ) \cdot \varepsilon [/itex]
…where you “linearly” count your way from [itex]\sqrt {n + \varepsilon } [/itex] to [itex]\sqrt x [/itex] by way of the linearization, [itex]L(\sqrt x )[/itex], namely, the derivative:
[itex]L(\sqrt x ) = {\left. {\frac{d}{{dx}}\sqrt x } \right|_{x = n}} = \frac{n}{{2\sqrt n }}[/itex]

But that’s another square root to calculate, which would further-roughen my approximation. Are there any other numerical-approximations you’d suggest?

Homework Statement





Homework Equations





The Attempt at a Solution

I tried googling the title of your thread, and got what look to be lots of good hits:

http://www.google.com/search?source...=Calculating+a+Square+Root+Without+Calculator

I didn't click into any of them -- are they helpful?
 
  • #3
marcusl
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How accurate do you need? There are a lot of strategies if high accuracy is not the goal (and in any case, if I'm trying to estimate in my head I shoot for 5-10% accuracy). In the present case, I know sqrt(4) = 2 and sqrt(5) = 2.23 (which we had to memorize in junior high school). Taking the mean gives the estimate sqrt(4.51) = 2.12 (approx.).

OR, sqrt(4.5) = sqrt(9) * sqrt(0.5) = 3 * .707 = 2.12.

Actual answer is 2.124.

I'll often use logs, too. I wouldn't use them for this problem, but I'll do it to show how it works.
I've memorized four values, namely,
log10 of {2, 3, 5, 7} = {0.3, 0.48, 0.7, 0.85}.
I can figure out other simple ones from these: obviously log10(4)=0.6 and log10(9)=0.96, for instance.

Applied to present problem, x = sqrt(4.5)=10^(1/2*log10(4.5)).

First, 1/2 * log10(4.5) = 1/2 * log10(9*5/10) = 1/2 * (0.96 + 0.7 - 1) = 0.36

Second, find x=10^0.36. Looking at the values I know above, I find that 0.36 = 0.96 - 0.6
Using my known values in reverse (antilogs), x = 10^0.96 / 10^0.6 = 9 / 4 = 2.25. That's within 6%.

Logs are harder for this problem, but are easier for others (like finding cube roots, e.g.).
 
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  • #4
vela
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You could use the binomial expansion of (1+x)1/2≈1+x/2 to get an estimate. Using your example, you'd do something like

[tex]\sqrt{4.51} = \sqrt{4+0.51} = \sqrt{4}\sqrt{1+0.1275}[/tex]

to get it into the right form, and then expand the latter square root. In this case, you'd get an answer of 2.1275, which is pretty close to the actual answer, to four decimal places, of 2.1237.
 

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