Calculating a Square Root Without Calculator

[bjnartowt]

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?

 

[berkeman]

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?

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?

 

[marcusl]

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.).

 

[vela]

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.

 

[rwisz]

Hello! Calculating square roots without a calculator can be a challenging but interesting task. Your approach using linearization is a good start, but as you mentioned, it can lead to further approximation errors.

One approach you could try is the Babylonian method, also known as Heron's method, which is based on the idea of repeatedly averaging the current estimate with the number divided by the current estimate. This can be written as:

x_{n+1} = \frac{1}{2}\left(x_n + \frac{a}{x_n}\right)

Where x_n is the current estimate and a is the number whose square root we are trying to find. This method converges quickly and can give a more accurate estimate than linearization.

Another approach is the bisection method, which involves repeatedly dividing the interval containing the square root into two halves and discarding the half that does not contain the root. This method is slower than the Babylonian method but can be useful for finding approximations with a desired level of precision.

Overall, there are various numerical approximation methods that can be used to calculate square roots without a calculator. It's important to keep in mind that these methods will always involve some level of approximation and may not give an exact answer. It's always a good idea to compare your result with a calculator to check for accuracy. I hope this helps!