Secret pattern of irrational numbers

Question

Are their any pattern in irrational numbers?, I know that there's been no hint as to the appearance of their infinite digits, but I discovered a pattern within themselves

$\sqrt1 = 1$

$\sqrt2 = 1.414....$

$\sqrt3 = 1.732....$

$\sqrt(4) = 2$

$\sqrt(5) = 2.236$

$\sqrt(6) = 2.449$

$\sqrt(7) = 2.645$

$\sqrt(8) = 2.828$

$\sqrt(9) = 3$

$\sqrt{10} = 3.162$

$\sqrt{11} = 3.316$

$\sqrt{12} = 3.464$ etc...

It turns out that the numbers are increasing with the same rate

Notice that in the section of $1 → 4$, it seems that the numbers increased by $0.3 (1/3) $, while in the section of $4 → 9$ there is an increase rate of $0.2 (1/5)$, while in the section of $9 → 16$, an increase by $ \sqrt(10)-3 , (5/7)$ and also the section of $16 → 25$, an increase by $ \sqrt(17)-4, (2/9)$

So this process keeps on happening and continues forevermore

To get the exert value, notice it's an A.P. I'll skip some details now to write the terms of the progression

$1, (x+2)/3 ,4, (x+6)/5 ,9, (x+12)/7 , 16, (x+20)/9 ,25,............$

I'll try to find an approximate formula for the square root of a number into fraction, using the idea of its nearest perfect square

Let's say a number $x$ lies in the number plane $x£Z$ as $0,1,2,3,.......,n,.........,x,...........,m,.........,∞$ Every number $x$ is always bounded by two perfect square, the one before it say $n$ and the after it say $m$

Now let's write out the set of square roots $0,1,1.414..,.....,2,....,3,......,\sqrt(n),......,\sqrt(x),........,\sqrt(m),........,∞$ Since $\sqrt(x)$ let's between $\sqrt(n)$ and $\sqrt(m)$, there must be a rate to with $\sqrt(n)$ increases from, even to $\sqrt(x)$ and until $\sqrt(m)$

By checking the progression, let's skip the calculations to write the basic term, Therefore $:$

$\sqrt(x) ≈ (x + (n+m-1)/2)/(m-n)$

Since $n$ and $m$ are closest perfect squares, say $n=k^2$ and $m=(k+1)^2$

$\sqrt(x) ≈(x+(k^2+(k+1)^2-1)/2)((k+1)^2-k^2)$

Which then becomes

$\sqrt(x) ≈ (x+k^2+k)/(2*k+1)$ Where $k^2 ≤ x$

In fact the statement can be true $\sqrt(x) = \lim(x→∞,k^2→x) (x+k^2+k)/(2*k+1)$

We can use this to approximate the square root of many numbers Examples:

$\sqrt(128) = 11.313....$

$\sqrt(128) ≈ (128+11^2+11)/(2*11+1) ≈ 260/23 ≈ 11.304.....$ because $11^2<128$

$\sqrt(571) = 23.895....$

$\sqrt(571) ≈ (571+23^2+23)/(2*23+1) ≈ 1123/47 ≈ 23.893....$ because $23^2<571$

$\sqrt(75) = 8.660.....$

$\sqrt(75) ≈ (75+8^2+8)/(2*8+1) ≈ 147/17 ≈ 8.647....$ because $8^2<75$

$\sqrt(1004) = 31.685.......$

$\sqrt(1004) ≈ (1004+31^2+31)/(2*31+1) ≈ 1996/63 ≈ 31.682....$ because $31^2<1004$

Question: What are the meaning of this, If there is a pattern within themselves, why no pattern in their digits?

It's not hard to conclude that something similar to this would exist for $\sqrt[3]()$, $\sqrt[4]()$, etc, then I'll suggest something more complicated than this exists of all Algebraic-irrational number

Since this numbers here belong to the same group, roots , is it possible that patterns can be found within transcendental numbers of same group

Answer 1

Let $k^2$ be the largest square less than or equal to $x$. Since $\sqrt{x}$ flattens out for large $x$, drawing a secant line between the points at $k^2$ and $(k+1)^2$ should give a good approximation of $\sqrt{x}$, i.e.$$ \sqrt{x} \approx k + \frac{x-k^2}{(k+1)^2 - k^2}\left(k+1 - k\right) = k + \frac{x - k^2}{2k+1} =\frac{x+k^2+k}{2k+1} $$ But how good of an approximation is it? Your calculations suggest it's a good one, but let's use a little calculus and get more precise: Notice that the approximation is exact for $x=k^2$ or $x=(k+1)^2$. We will find the maximum of the difference on the interval $[k^2,(k+1)^2]$ by looking for where the derivative is $0$: \begin{eqnarray} \frac{d}{dx} \left(\sqrt{x} - \frac{x+k^2+k}{2k+1}\right) &=& 0\\ \frac1{2\sqrt x} - \frac{1}{2k+1} &=& 0\\ \sqrt{x} &=& k+\frac12\\ x &=& (k+\frac12)^2 = k^2 + k +\frac14 \end{eqnarray} hence the most this approximation can be off by is at $x=k^2 + k +\frac14$, at which point: \begin{eqnarray} \sqrt{x} - \frac{x+k^2+k}{2k+1} &=& k+\frac12 - \frac{k^2 +k +\frac14 + k^2 + k}{2k+1} = k+\frac12 - \frac{2k^2 + 2k +\frac14}{2k+1}\\ &=& \frac{2k^2+k+k+\frac12 -(2k^2 + 2k +\frac14)}{2k+1} = \frac1{4(2k+1)} \end{eqnarray} Hence even for $k=1$, the approximation is not off by more than $1/12$, which is pretty good, and it gets better for larger $k$.

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In general, you can get a similarly good approximation by secants to any function that flattens off for large $x$, so the same behavior will be observed for higher order roots as well. This is not a special property of the individual numbers $\sqrt{6}$, $\sqrt7$, etc., so much as it is a special property of the square root function.

Answer 2

First: "patterns" in the (decimal) digits representation of numbers have practically no relationship with your approximations, which only deal with the values.

Your approximation is actually a linear interpolation.

In general, if we know the values of a function in two points $a,b$ we can attempt to approximate an intermediate value by

$$f(x) \approx f(a) + (x-a) \frac{f(b)-f(a)}{b-a}$$

Applying this to $f(x)=\sqrt{x}$ with $a=k^2$ and $b=(k+1)^2$ we get

$$ \sqrt{x}\approx \frac{x+k^2+k}{2k+1}$$

With respect to the increase rate, it can be obtained easily with Calculus: Letting $y_n= \sqrt{n}$ then

$$ \frac{y_{n+1}}{y_n}=\frac{\sqrt{n+1}}{\sqrt n}=\sqrt{1+1/n}\approx 1+\frac{1}{2n}=1+\frac{1}{2 y_n^2}$$