[Math] How to find a cube root of numbers

arithmeticradicals

While, I was solving a problem of Chemistry [Solid State] when I encountered an equation like : $$a^3 = 3.612 \times 10^{-23} $$

Where, a is just a quantity [Actually, it is the length of a cubic unit cell.]

I was required to find the cube-root of $a^3$ as I had to use $a$ to find another quantity which was actually my answer.

I know that the cube-root of $a^3$ here can be found by using logarithm.

$$\begin{align} & \log{a^3} = \log{(3.612 \times 10^{-23})} \\
& 3\log{a} = \log{(36.12 \times 10^{-24})} \\
& 3\log{a} = \log{(36.12)} + \log{(10^{-24})} \\
& 3\log{a} = \log{(36.12)} – 24 \\
& \log{a} = \cfrac{\log{(36.12)} – 24}{3} \end{align} $$

By using logarithm table, I found $\log{(36.12)} = 1.557$

Now, just putting this : $$\begin{align} & \log{a} = \cfrac{1.557 – 24}{3} \\ & \\ & \log{a} = \cfrac{-22.443}{3} \end{align} $$

Now, taking antilog both sides, this becomes : $$a = antilog(-7.481) $$

Now, by using anti-log table, I got this as $$\boxed{a = 3.303 \times 10^{-8} } $$

Now, as you all might have noticed that this is a lot longer method and this has larger probability of a student doing a mistake in the calculations etc.

Is there a shorter or any proper method to find cube-root of a number like stated above?

Best Answer

Given a log/antilog table and no calculator, your approach is probably optimal. Without such a table, I think you are better off proceeding as:

$$\left ( 3.612 \cdot 10^{-23} \right )^{1/3} \\ = \left ( 36.12 \cdot 10^{-24} \right )^{1/3} \\ = 36.12^{1/3} \cdot 10^{-8}$$

Then do bisection to find $36.12^{1/3}$. Note that $3^3=27<36.12$ and $4^3=64>36.12$. (If you can't find a small interval quickly, just use $[1,10]$.) So now you keep checking midpoints until you get your desired level of precision: $3.5^3 > 36.12$, $3.25^3<36.12$, $3.375^3>36.12$, etc. If at some point the arithmetic gets to be too messy, you can round, at the expense of some convergence speed. For example, you could check $3.4$ instead of $3.375$, then $3.4^3>36.12$, so you could check $3.3$. Then $3.3^3<36.12$, so you have $a=(3.35 \pm 0.05) \cdot 10^{-8}$.

You can also do Newton's method, which in this case is $x_{k+1} = \frac{2 x_k}{3} + \frac{c}{3 x_k^2}$ where you want $c^{1/3}$, but this is hard to do by hand, because the fractions quickly become very awkward, even with rounding.

Related Solutions

[Math] Calculate the root of a number without useing the root function or decimal numbers

Yes it can be done.
Suppose that you are given $n$, you want to calculate $\lceil \sqrt{n} \rceil$, so you can apply a binary search algorithm with $k^2 \geq n \land (k-1)^2 < n$ as the stopping condition. $k$ here is the variable over which you are doing the binary search. The algorithm would be like this:

low = 0
top = n
while(true)
  mid = (low+top)/2
  if(mid*mid >= n and (mid-1)*(mid-1) < n)
    return mid
  else
    if(mid*mid < n)
      low = mid + 1
    else
      top = mid-1
    end
  end
end

Arithmetic – How to Find the Square Root Using Pen and Paper

You can use the identity $(x+c)^2 = x^2 + 2xc + c^2$ to arrive at a "long-division" like method. Let me show you how it is done for 3838 before giving the algorithm.

Start with 3838
Write the digits in groups of 2, so write 3838 as 38,38.
For the "highest placed group of 2", find the biggest square number less than it. In this case, $6^2 = 36 < 38 < 49 = 7^2$. So you write down 6, and subtract 36 from the first group. You then get
2,38 = 38,38 - 36,00.
Take the 6 you wrote down before, multiply it by 20 (so you get 120). Now find the largest multiple of $6\times 20$ that is less than 238. You'll see that it is 1. So you write down 1 (so your number is now 61). Subtract from 238 120 to get 118.
Subtract from 118 the number $1 = 1^2$ to get 117. Now add two more digits to it (to form 117.00). Take 61, multiply it by 20, you get 1220. Find the largest multiple of 1220 less than 11700, which would be $1220 \times 9 = 10980$. So write down 9, and your number is 61.9, and subtract from 11700 the number 10980 to get 720. Subtract from 720 the number $9^2 = 81$ to get 639.

So you've arrived at, at this point, $3838 = (61.9)^2 + 6.39$. And you can continue the process indefinitely.

How does this work? Given a number $A$, you want to find its square root in base-10 representation. Suppose your square root looks like $$ a_{100}a_{10}a_1.a_{0.1}a_{0.0.1}\ldots $$ when expanded as a string of digits. Then you find the biggest $a_{100}$ such that $$ (a_{100}\times 100)^2 \leq A $$ (similar to how you do long division). Then to make solve for the next digit, you use that $$ (a_{100} \times 100 + a_{10}\times 10)^2 \leq A $$ (since you are truncating the decimal expansion, which makes the number smaller). So you solve for the best $a_{10}$ such that $$ (a_{100})^2\times 100^2 + \left[ 20 \times (a_{100}\times a_{10}) + (a_{10})^2\right] \times 100 + \leq A $$ The above expression shows why in the first step you want to group the digits in twos: in some sense we are thinking of $A$ as in base-100, the "square" of base-10.

At every step you use the identity $(x+c)^2 = x^2 + 2xc + c^2$ to compute the next digit correction to the square root.

Another method, as the Babylonians did it, was recently detailed by John Baez at his blog, which uses the "equality case" of the arithmetic-mean-geometric-mean inequality to power the iteration.

Best Answer

Related Solutions

[Math] Calculate the root of a number without useing the root function or decimal numbers

Arithmetic – How to Find the Square Root Using Pen and Paper

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