# Close approximation for absolute value function

I made a very acurate approximation function for $$\sqrt{n^{2}+1}$$

It is $$\sqrt{n^{2}+1}\approx\frac{2n(n^{2}+1)}{2n^{2}+1}+\frac{2n^{2}+1}{n(4(2n^{2}+1)^{2}+1)}$$

From this I can make a very close approximation of absolute value of n,

$$\mid n\mid\approx\frac{1}{\sqrt{n^{2}+1}}\left(\frac{2n^{2}(n^{2}+1)}{2n^{2}+1}+\frac{2n^{2}+1}{4(2n^{2}+1)^{2}+1}\right)$$

My question is if this is interesting in any way to anybody? It doesn't look like it would make any integral easier. Except maybe the square root approximation may make some integrals for arc length easier to solve with great accuracy, because you could just enter the derivative of your function as n in the equation. Seems though it would be more complicated most of the time.

Any thoughts?

Note: I have provided a somewhat detailed explanation of how this formula was derived in an answer below.

• The theory of Padé approximants is well-developed and well-known. Smooth uniform approximations for the absolute value function are exploited in some proofs (variation of Lebesgue's proof) of the Weierstrass approximation theorem. Is your approximation better to use than any other one for a particular reason? Commented May 14, 2016 at 15:36
• I'm not claiming any superiority in my approximation, I know they are a dime a dozen. Just wondering if it would help with an integral solution, particularly an arc length integral as a good approximation. Commented May 14, 2016 at 15:46
• @Jack D'Aurizio Now that you mention it, evaluate the square root formula a number, say 35. It's accurate to 19 decimal places! That's in one fraction, no iterations. And, the accuracy increases for higher numbers with a limit of exact at infinity. I think that's pretty good. I'm sure there are better approximations, I mean sure you can iterate all day and get whatever you want. But I think this one is pretty good by any standards. Commented May 16, 2016 at 7:21
• "I made a very accurate approximation function" - a note on how you came up with it would be nice. Commented May 16, 2016 at 19:02
• @J. M. That would be kind of long. I will have to answer that question in an answer. So, give me some short time and I will run it through in an answer to my own question. I will send you another comment when it is done. Commented May 16, 2016 at 19:11

By request, I am providing information on how I derived this formula here.

If you list the values of square roots of numbers from $1$ to whatever, the whole numbers place in this list can be seen to follow the sequence $2n+1$, where $n$ is the whole numbers place value of $\sqrt{x}.$ Meaning there are $3$ ones, $5$ two's, $7$ three's,..ect. So, if you wanted to know how many square roots of whole numbers begin with the number $3$, the equation would yield $2*3+1=7$, which is correct. Further, because each first new whole number is also the square root of the perfect square in the beginning of each interval of numbers, $n=\lfloor\sqrt{x}\rfloor$, which is the perfect square before any number $x.$ So, $n^2$ is this perfect square.

It is also intuitive to note that in each interval of values i.e.;(4,8), the values of the decimal places nearly have a uniform distribution with respect to the interval. Using this assumption, its possible to estimate the decimal place values with a fraction. We can represent the location of any number $x$ in an interval using the perfect square before it as, $x-n^{2}.$ Knowing the location of $x$, and the distance in the interval, this fraction can take the form of location over distance as follows,

$\frac{x-n^{2}}{2n+1}.$ This is a beginning approximation to the decimal expansion of $\sqrt{x}.$

Consequently, in knowing that the whole number value will be $n,$ a beginning approximation for $\sqrt{x}$ can be formulated as follows,

$\sqrt{x}\approx\frac{x-n^{2}}{2n+1}+n=\frac{x+n+n^{2}}{2n+1}=\frac{x+n(n+1)}{2n+1}.$

The graph of the error in this, which I will refer to as the "error function", is

$\sqrt{x}-\frac{x+n(n+1)}{2n+1}.$

You can find that the maximum error in the "error function" occurs at about $n(n+1).$ Simply make $x$ equal to $n(n+1).$ Now we have,

$\sqrt{n(n+1)}\approx\frac{2n(n+1)}{2n+1}.$

Now, if you divide the the distance $2n+1$ in each interval by the maximum error given by the error function, you will get a sequence of values that are approaching integers at a rapid rate. This sequence truncated is, ${37,101,197,325,...}$ This is a known sequence whose formula is $4(2n+1)^{2}+1.$ So, if the error divided by $2n+1$ gives almost perfectly $4(2n+1)^{2}+1,$ then of course the error is almost perfectly $\frac{2n+1}{4(2n+1)^{2}+1}.$

Thus, $\sqrt{n(n+1)}\approx\frac{2n(n+1)}{2n+1}+\frac{2n+1}{4(2n+1)^{2}+1}.$

The square root in my post and many other forms I've derived follow from the equation above using simple manipulation. For example, if you evaluate the above equation at $n=n^{2}$ and then divide the whole thing by $n,$ you get the formula in my original post.

That is a lot I know, but this was a long time to derive, research and make perfect. I'm trying to explain it all in one little post. In short, the equation starts from an approximation of the decimal expansion in a square root itself.

Note:In every form above that contained $x$ and $n$ in it together, $n=\lfloor\sqrt{x}\rfloor$ and for the ones containing only $n,$ its just n.

• @J. M. here is your answer. Commented May 16, 2016 at 21:00