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I have developed a calculation-by-hand method to do the square root of a given large number based on odd number sequence:

$$1 + 3 + 5 + 7 + \cdots + (2n-1) = n^{2}.$$

For example, $$16 - 1 - 3 - 5 - 7 = 0.$$ (remainder) We got $4$ steps of subtraction. Hence, the answer is $4$.

The method calculates the square root in much fewer steps of subtraction than one single odd number sequence.

For example, the square root of $2$ is $1.4142$ that the method needs only $12$ steps of subtraction.

Has it been published somewhere? If so, please give us a pointer.

Ѕᴀᴀᴅ
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GeoCoder
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    The fact that the sum of the first $n$ odd numbers is $n^2$ is well known - it's in many textbooks as an exercise on induction. I don't see how your method can be applied to something that's not already a perfect square. –  Feb 14 '18 at 00:02
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    Can you elaborate on how you would use your method to compute the square root of two in 12 steps of subtraction? – orlp Feb 14 '18 at 00:05
  • Possible duplicate of $\downarrow$ and $\longrightarrow$ https://math.stackexchange.com/questions/639068/sum-of-odd-numbers-always-gives-a-perfect-square ${}$ https://math.stackexchange.com/questions/606483/how-to-derive-the-formula-for-the-sum-of-the-first-n-odd-numbers-n2-sum-k?noredirect=1&lq=1 – Mr Pie Feb 14 '18 at 00:24
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    "For example, the square root of $2$ is $1.4142$"... except... it isn't. It couldn't be because it is well known and has been well known for thousands of years that the square root of two is an irrational number. $1.4142^2=1.99996164\neq 2$. – JMoravitz Feb 14 '18 at 00:28
  • My method can apply both perfect squares and imperfect squares. The fraction part can be any length as long as you want. I will be happy to elaborate on how it works once published if no such publishment is made. I still need to know if there is one. – GeoCoder Feb 14 '18 at 00:42
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    If your "method" that you wish to "publish" makes a claim such as "the square root of 2 is 1.4142" you will be laughed at and your submission will be immediately denied. Even skipping this and referring only to perfect squares, usually only new results or new approaches get published. The result you speak of is as mentioned by others certainly not new (probably also a few thousand years old) and you have not as of yet provided evidence that the method is sufficiently new or different than what is already known either. – JMoravitz Feb 14 '18 at 00:47
  • My so-called method might be already discovered. I just want to know where it is published. – GeoCoder Feb 14 '18 at 01:02
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    @GeoCoder let me stop you right there. We all already know that $\text{sqnumber}-1-3-5-7-\dots=0$ and the number of subtractions gives you the square root of a perfect square. We don't need another example, we should have all had to prove the generalization as an exercise in induction either in middleshcool, highschool, or undergraduate, wherever induction and proof writing was first introduced. You have not yet talked about your "method" and how it is somehow different than this observation, why it is useful, and how it might be different than what is already widely known. – JMoravitz Feb 14 '18 at 01:08
  • As for where this is currently published, it would be included as a homework exercise in many textbooks, not worthy enough of publication on its own. If you are looking for an earliest citation, I doubt that one exists. Even $1+2+3+\dots+n = \frac{n(n+1)}{2}$ was known well before Gauss (despite the anecdote of his days as a school boy, that was more in reference to his supposed ability to spot patterns quickly rather than having been the first to spot that pattern). – JMoravitz Feb 14 '18 at 01:11
  • The reason why I asked for its existence being "published" is that Wikipedia does not show a similar method as mine. If it is not worth to ask, it is fine with me. – GeoCoder Feb 14 '18 at 01:21
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    Computing $\sqrt{n}$ in about $\sqrt{n}$ steps is not a big conquest. The ancient babylonian method $x_{m+1} = \frac{x_m^2+n}{2x_m}$ (starting from a not terrible approximation $x_0$) is equivalent to Newton's method and performs much better. – Jack D'Aurizio Feb 14 '18 at 13:13

1 Answers1

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This discovery comes from the following formula:

$$\sum_{k=1}^n (2k-1) = n^2$$ which is read simply as, the sum of the first $n$ odd numbers from $1$ is equal to $n^2$.

We can prove this geometrically, where if we let $n = 1$, we have $1$ box. $$\blacksquare$$ Then when we let $n = 2$, we have to add the first two odd numbers $1$ and $3$ together, i.e. we add $3$ more boxes to our first box, and we will colour them in white to clearly demonstrate what I mean. $$\begin{align} \blacksquare&\,\,\,\Box \\ \Box&\,\,\,\Box\end{align}$$ Notice that now we have a square, $$\boxed{\begin{align}\blacksquare&\,\,\,\Box \\ \Box&\,\,\,\Box \end{align}}$$ so the number of boxes in total is a square number. Since this square is a $2\times 2$ square, then our new squared number is $2^2 = 4$.

Continue this pattern by adding $5$ more boxes in such a way where we make a $3\times 3$ square, and you get that the number of boxes is a squared number $3^2 = 9$. This is where the above formula comes from, and proves the truth of your method of square-rooting a squared number.

Go here for a further understanding if you are not familiar with the formula, and if you are familiar with the formula, go here instead.

Nice discovery though :)

Mr Pie
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