Investigating the linearity between squares and their roots

Question

I recently noticed that $\sqrt{128} = 11.31$ and that $128$ is $\approx 30\%$ between $121 = 11^2$ and $144=12^2$, that is: $$ \frac{128-121}{144-121} = \frac{7}{23} \approx 30\%$$ and $\sqrt{128} = 11.31$ is $\approx 30\%$ away from $11$.

This made me wonder if there is a linear relationship between the distance of $\sqrt{x}$ between $a$ and $a+1$ and the distance of $x$ between $a^2$ and $(a+1)^2$.

What follows is my "proof" that the relationship tends to be linear for large numbers:

Let $D(a, b, x)$ denote the distance of $x$ between $a$ and $b$: $$D(a, b, x) = \frac{x - a}{b - a}$$ We are investigating the relationship between $$U = D(a^2, (a + 1)^2, x) = \frac{x-a^2}{2a+1}$$ and $$L = D(a, a + 1, \sqrt{x}) = \sqrt{x}-a$$

Defining $F = \frac{U}{L}$, we see that $$F = \frac{a + \sqrt{x}}{2a + 1}$$ Using the inequality $$a < \sqrt{x} < a + 1$$

adding $a$: $$2a <a + \sqrt{x} < 2a + 1$$

and dividing by $(2a + 1)$: $$\frac{2a}{2a+1} < F < 1$$

Therefore: $F$ has a lower bound of $\frac{2a}{2a+1}$ and as $a$ increases $F$ will get closer and closer to $1$, making the relationship between $U$ and $L$ more linear, and that's what I wanted to show.

My question, however, is: is this correct? Is there an easier / simpler way to show this?

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Why this is useful (to me): It provides an easy and accurate way to estimate square roots with simple arithmetic.

Answer 1

Your arithmetic certainly looks right. If I'm reading what you're doing correctly, you would want to say that 128 is 7/23 of the way from 121 to 144, so that $\sqrt{128}$ is approximately 7/23 of the way from 11 to 12. So, essentially what you are doing is drawing the straight line between the points (121,11) and (144,12) on the graph of $y = \sqrt{x}$, and plugging 128 into the formula for that line. You're using that straight line as an approximation for the graph of $y = \sqrt{x}$. Yes, that approximation will be pretty good for large numbers, but it's better near $a^2$ and $(a+1)^2$, and less good in the middle.

You could do a similar approximation (with arithmetic just as easy) by using the tangent line through $(a^2,a)$, which has slope $1/2a$. (Your line has slope $1/2a+1$.) The tangent line approximation will be better close to $a^2$, but gets worse as you get closer to $(a+1)^2$.

Answer 2

Such a relationship holds only for large numbers. In fact, $\lim_{a\to\infty}F=1$, by the squeeze theorem, so $U=L$ as $a\to\infty$. It is correct. Your argument isvery simple - I don't know how to simplify it more, except that you do not have to define $D,U$ or $L$. Just define $F$ directly, and then you're done.

Answer 3

Yes, the relationship does become "more linear" as the squares become larger. This is because $x^2$ is differentiable, which means that the changes in the values near a point $x_0$ is linear , which rate of linear change given by $2x_0$, the slope of the function. As $x_0 \rightarrow \infty $ , $(x_0+1 )^2 -x_0^2=2x_0+1 \rightarrow $ , the 1 becomes small-enough to say that the change "is" $2x_0$ , because for $x_0$ large-enough, $x_0 +1 $ is very close ( ratio-wise , i.e., $\frac{x_0 +1}{ x_0} \rightarrow 1$ ). This is the large-scale version of the fact that $$Lim_{h \rightarrow 0} \frac {f(x+h)-f(x)}{h} $$

is linear when $f$ is differentiable . This is equivalent to saying that $x_2$ is differentiable with slope $2x$. And this result holds for all differentiable functions when the change $\frac{x+h}{x}$ in the values of the argument tends to 1.

Answer 4

What you are doing is more or less one step of the secant method for finding $\sqrt{x}$ given the squares straddling $x$

Answer 5

The ancient Mesopotamians knew that square roots of numbers that are close to perfect squares can be approximated using the formula $$ \sqrt{a^2 + b} \approx a + \frac{b}{2a} $$ This formula can be justified by squaring both sides and disregarding the quantity $\frac{b^2}{4a^2}$. You can also interpret this formula visually, as follows:
One can also regard the RHS as the first two terms in the Taylor expansion of the LHS, but Calculus is not really needed here.

If you think of $a$ as fixed and $b$ as a variable, this explains the approximately linear relationship you have found. As you note, it becomes more accurate when is $a$ is large relative to $b$ (this corresponds in the diagram above to the "missing corner" of the square becoming smaller and smaller relative to the size of the whole figure).

In your example, we have $\sqrt{128} = \sqrt{11^2 + 7} \approx 11 + \frac{7}{22}$.

Note, by the way, that $b$ can be negative in the above formula. For example, one can approximate $\sqrt{220} = \sqrt{15^2 - 5} \approx 15 - \frac{5}{30} = 14 \frac{5}{6}$. In either case ($b$ positive or negative) the approximation yields an overestimate.

You may be interested to know that cube roots can be approximated using a similar formula, namely: $$ \sqrt[3]{a^3 + b} \approx a + \frac{b}{3a^2}$$ There is a great scene in the movie Infinity, about the life of physicist Richard Feynman, in wich Feynman (played by Matthew Broderick) challenges a Chinese shopkeeper to a calculation race: the shopkeeper uses an abacus, and Feynman uses pencil-and-paper algorithms. The abacus user wins handily at multiplication and division, but Feynman pulls out the victory at computing a cube root. As he leaves the shop, he explains to his fianceé that he used the approximation formula above. (Feynman also makes the mistake many in this thread have made of thinking that you need Calculus to derive this formula, when in fact it was well-known thousands of years ago.)

Edited to add: As others have noted, you seem to be using a slightly different approximation formula than the one I have given above -- namely, you are using $ \sqrt{a^2 + b} \approx a + \frac{b}{2a+1} $. The formula I gave corresponds to a line drawn tangent to the graph of $y = \sqrt{x}$; since that graph is concave down, the tangent line gives an overestimate. Your formula corresponds to a secant line drawn between two points on the graph; since the graph is concave down, your formula gives an underestimate. (I suppose if you want to be really fancy you could compute both $\frac{b}{2a}$ and $\frac{b}{2a+1}$ and compute their average... or use $\frac{b}{2a+0.5} = \frac{2b}{4a+1}$.)

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