INFINITE ROOT

This problem is also from my old notebook from school days. I loved deriving formulas in physics and mathematics because this activity was both exciting and often practically useful. One of the tasks I encountered then was the so-called "infinite root". In general, "games with infinity" are very interesting and educational in themselves. After all, it is difficult for a person to perceive the infinite...

This task, strictly speaking, is not a task. There is simply an endless series:

        _________________
       /     ____________
      /     /     _______
m = \/x + \/x + \/x + ...    (1)

The question is, what is the value of m? How to find out? When you look at this series, you are not at all sure that it converges to a finite number. After all, series in mathematics are sometimes convergent and sometimes not, and also in different ways. This series seems to grow endlessly, doesn't it? But that's not true.

So how to calculate the value of m depending on x? That is, what is m(x) equal to in the general case? To do this, we need to present this series in a slightly different form.

If we assume that
          _
m = x + \/x, then what happens in the next step? Obviously, this m needs to be added to x and placed under the root. And it will also turn out m! After all, the series is endless?

As a result we get the expression:
      ___
m = \/x+m, which needs to be solved for m. Obviously, squaring both sides, we get the equation m²=x+m

The quadratic equation, as is known, has two roots. In our case, only one of them will be needed:

        ____
    1+\/1+4x
m = --------    (2)
       2

It is clear that a minus in front of the root will give an incorrect result (m cannot be negative!). Likewise, negative values ​​of x will also make it impossible to calculate the result on the set of real numbers. But here's what's interesting!

What if x is negative, but less than one quarter? If it is strictly equal to "-1/4", then m="1/2". But if it is equal to, for example, "-8/9", what then? Then, according to the resulting formula (2), the value is m="2/3". And if we remember about the second value of m from equation (2), which we discarded? If we take it (with a minus in front of the root), we get m="1/3".

But in formula (1) the values ​​of the roots can also be negative! What then? Then formula (2) turns into formula (3):

         ____
    -1+\/1+4x
m = ---------    (3)
        2

Taking into account all these options, the results obtained become quite large. I will give here several examples of values ​​of m(x) for formula (2) and for positive integers x. For example:
m(1)=F₁ (see page "Fibonacci numbers")
m(2)=2 (which is interesting in itself!)
m(6)=3
m(12)=4
m(20)=5

See a pattern? If not, look for it, it’s easily noticeable!

And now another interesting point. Have you noticed that for fractional negative x less than "1/4" the formula gives a normal result? But it’s impossible to substitute negative numbers into formula (1), because the roots are there! And this is where the infinity of the series comes into force. Yes, you cannot calculate the root of "-1/4", but you can calculate the root of the expression "-1/4+1/2"! But it only becomes this way if formula (1) is truly infinite! In other words, it is impossible to calculate m numerically using a program for negative numbers in the same way as for positive x (and I checked, calculated the above values ​​using formula (1): everything converges!).

What if we move into the sphere of complex numbers? What if...

Formula (2) gives a formal result. Formula (3) too. But calculations using formula (1) for a sufficiently large number of root embeddings give accurate results. But it is not always possible to calculate this!

This is the beauty of mathematics: sometimes the impossible "step by step" becomes possible "in the end". But there are also dangers when the final formula ceases to work in particular cases. Does it always operate here, on all sets and on all variants? If you are interested, explore this issue yourself! Mathematics is a magical science... :)