Образователни технологии
SOME NUMERICAL SQUARE ROOTS (PART TWO)
Резюме. The present article is a continuation of the homonymous publication in the journal “Mathematics and Informatics”, issue 5, 2017. The theoretical considerations in it cover all cases when calculating the value of a finite and infinite number of nested square radicals.
Ключови думи: infinite sequence; nested radical; convergence; mathematical induction
In the first part of this research we examined the convergence of the numerical series \(c_{n}=\sqrt{a+b c_{n-1}}, c_{1}=\sqrt{a}\), where \(a \gt 0\) and \(b \gt 0\). The aim of the second part is to determine conditions for the limit existence including the case 0.
For a starting base and background we will use again some basic skills and knowledge in the field of:
Infinite numerical series;
Convergence of infinite numerical series;
Limits of infinite numerical series.
On this theoretical basis we will apply some specific and non-standard skills for proving scientifically determined dependencies.
Using theorem 1 from Part one (Nikolaev, Milkova \& Petkov, 2017) for example we can easily find the value of the expression \(L=\sqrt{3+2 \sqrt{3+2 \sqrt{3+\ldots}}}\).
Concerning this, a question arises - can we solve the next problem in a similar way?Determine the value of the expression \(L=\sqrt{3-2 \sqrt{3-2 \sqrt{3-\ldots}}}\).
The answer seems to be positive and the solution is based on the statements mentioned in the present paper.
We know from theorem 1 that if \(t \gt b \gt 0\), then \(\sqrt{t(t-b)+b \sqrt{t(t-b)+b \sqrt{\ldots}}}=t\).
We will formulate and prove the following theorem:
Theorem 2. If \(b \lt 0\) and \(L \gt \tfrac{b(1-\sqrt{5})}{2}\), then \(\sqrt{L(L-b)+b \sqrt{L(L-b)+b \sqrt{\ldots}}}=L\).
It is obvious that \(L \gt 0\).
We examine the numerical series \(c_{n}=\sqrt{L(L-b)+b c_{n-1}}\), with its first element \(c_{1}=\sqrt{L(L-b)}\).
This numerical series can be divided into two subseries – the first one consisting of the odd elements \(\left\{c_{2 k-1}\right\}_{k=1}^{\infty}\) and the other one ofh the even elements, i.e. \(\left\{c_{2 k}\right\}_{k=1}^{\infty}\).
We will formulate and prove two more theorems which will help in proving Theorem 2.
Theorem 3. The subseries \(\left\{c_{2 k-1}\right\}_{k=1}^{\infty}\) converges and when \(\mathrm{b} \lt 0\) its limit is equal to \(L \gt \tfrac{b(1-\sqrt{5})}{2}\).
Theorem 4. The subseries \(\left\{c_{2 k}\right\}_{k=1}^{\infty}\) converges and when \(b \lt 0\) its limit is equal to \(L \gt \tfrac{b(1-\sqrt{5})}{2}\).
To prove theorem 3 in an easier way we will formulate three lemmas.
Lemma 1.The subseries \(\left\{c_{2 k-1}\right\}_{k=1}^{\infty}\) is limited from below.
Proof: We use the principle of mathematical induction.
1. \(k=1, c_{1}=\sqrt{L(L-b)}\).
We will prove that \(\sqrt{L(L-b)} \gt L\).
\(\sqrt{L(L-b)} \gt \left.L\right|^{2}\);
\(L(L-b) \gt L^{2}\);
\(-L b \gt 0\), since \(L \gt 0, b \lt 0\) (by condition)
\(\Rightarrow \mathrm{c}_{1} \gt L-\) c1 > L - correct (1)
\(c_{3}=\sqrt{L(L-b)+b c_{2}}\)
We will prove that \(c_{3}=\sqrt{L(L-b)+b c_{2}} \gt L\).
\(\sqrt{L(L-b)+b c_{2}} \gt \left.L\right|^{2}\);
\(L(L-b)+b c_{2} \gt L^{2} ;\)
\(b c_{2} \gt L b \mid: b \lt 0\);
\(c_{2} \lt L ;\)
\(c_{2}=\sqrt{L(L-b)+b c_{1}}\).
Let us check if \(c_{2} \lt L\) :
\[ c_{2}=\sqrt{L(L-b)+b c_{1}} \lt L \Leftrightarrow L(L-b)+b c_{1} \lt L^{2} \Leftrightarrow b c_{1} \lt L b . \]
Since \(b \lt 0\), then \(c_{1} \gt L\)- correct, concluded from (1).
2. Let us assume that for each \(k=1,2 \ldots n\) it is true that \(c_{2 k-1} \gt L\).
3. We will prove that \(k=\mathbf{u} n+\quad c_{2(n-1) 1}=c_{2 n 1} \gt L\)
If \(c_{2 n+1}=\sqrt{L(L-b)+b c_{2 n}} \gt L\), \(c_{2 n+1}=\sqrt{L(L-b)+b c_{2 n}} \gt \left.L\right|^{2} \Leftrightarrow L^{2}-L b+b c_{2 n} \gt L^{2} \Leftrightarrow b c_{2 n} \gt L b \mid: b \lt 0 \Leftrightarrow c_{2 n} \lt L ;\) \(c_{2 n}=\sqrt{L(L-b)+b c_{2 n-1}} \lt \left.L\right|^{2} \Leftrightarrow L^{2}-L b+b c_{2 n-1} \lt L^{2} \Leftrightarrow b c_{2 n-1} \lt L b \mid: b \lt 0 \Leftrightarrow c_{2 n-1} \gt L\), which is correct according to the assumption in point 2.
\(c_{2 k-1} \gt L\). In such a way we proved that the subseries \(\left\{c_{2 k-1}\right\}_{k=1}^{\infty}\) is limited from below, i.е.
Lemma 2. The subseries \(\left\{c_{2 k-1}\right\}_{k=1}^{\infty}\) is decreasing monotonously.
Proof: Once again we will use the principle of mathematical induction.
1. \(k=1\).
We will check whether \(c_{1} \gt c_{3}\).
\[ \sqrt{L(L-b)} \gt \sqrt{L(L-b)+b c_{2}} \Leftrightarrow L(L-b) \gt L(L-b)+b c_{2} \Leftrightarrow 0 \gt b c_{2} . \] Since \(b \lt 0\), then \(c_{2}\) must be positive.
\(c_{2}=\sqrt{L(L-b)+b c_{1}}\), and for the value under the radical to be defined we need \(L(l-b)+b \sqrt{L(L-b)} \gt 0\).
\[ \begin{aligned} & L(l-b) \gt -\left.b \sqrt{L(L-b)}\right|^{2} \quad-b \gt 0, L(L-b) \gt 0 ; \\ & {[L(L-b)]^{2} \gt b^{2} L(L-b) \mid: L(L-b) \gt 0 ;} \\ & L(L-b) \gt b^{2} ; \\ & L^{2}-L b-b^{2} \gt 0 \Leftrightarrow L \in\left(-\infty ; L_{1}\right) \cup\left(L_{2} ;+\infty\right), \end{aligned} \] where \(L_{1}\) and \(L_{2}\) are the roots of the equation \(L^{2}-L b-b^{2}=0\) :
\[ \begin{aligned} & L_{1,2}=\tfrac{b \pm \sqrt{b^{2}+4 b^{2}}}{2}=\tfrac{b(1 \pm \sqrt{5})}{2} \\ & L_{1}=\tfrac{b(1+\sqrt{5})}{2} \lt 0 ; \quad L_{2}=\tfrac{b(1-\sqrt{5})}{2} \gt 0 \\ & \Rightarrow L \in\left(-\infty ; \tfrac{b(1+\sqrt{5})}{2}\right) \cup\left(\tfrac{b(1-\sqrt{5})}{2} ;+\infty\right) . \end{aligned} \]
In theorem 2 we stated that \(L \gt \tfrac{b(1-\sqrt{5})}{2}\), and in such a way we can make the conclusion that \(c_{1} \gt c_{3}\).
2. We assume that \(c_{2 k-1} \gt c_{2 k+1}\), for each \(k=1,2 \ldots n\).
3. We will prove that this is satisfied also for \(k=2 n+1: \quad c_{2 n+1} \gt c_{2 n+3}\).
\(c_{2 n+3}=\sqrt{L(L-b)+b c_{2 n+2}} \lt c_{2 n+1} \Leftrightarrow L(L-b)+b c_{2 n+2} \lt c_{2 n+1}^{2} \Leftrightarrow\)
\(\Leftrightarrow \cancel{L(L \subset b)}+b \sqrt{L(L-b)+b c_{2 n+1}} \lt \cancel{L(L -b)}+b c_{n} \mid: b \lt 0\)
\(\Leftrightarrow \sqrt{L(L-b)+b c_{2 n+1}} \gt c_{n}=\sqrt{L(L-b)+b c_{2 n-1}}\)
\(\Leftrightarrow b c_{2 n+1} \gt b c_{2 n-1} \mid: b \lt 0\)
\(\Leftrightarrow c_{2 n+1} \lt c_{2 n-1}\),
which is correct according to the assumption of the induction.
This way we proved that the subseries \(\left\{c_{2 k-1}\right\}_{k=1}^{\infty}\) is decreasing monotonously.
Lemma 3. The limit of the series \(\left\{c_{2 k-1}\right\}_{k=1}^{\infty}\) is equal to \(L\).
Proof: From Lemma 1 and Lemma 2 we can conclude that \(\left\{c_{2 k-1}\right\}_{k=1}^{\infty}\) has a limit denoting it by \(l\). Then:
\(\lim _{k \rightarrow \infty} c_{2 k-1}=l\) and \(\lim _{k \rightarrow \infty} c_{2 k+1}=l\).
From the recurrence
\(c_{2 k+1}=\sqrt{L(L-b)+b c_{2 k}}=\sqrt{L(L-b)+b \sqrt{L(L-b)+b c_{2 k-1}}}\), after a limit transition, we receive an irrational equation for \(l\) :
\(l=\sqrt{L(L-b)+b \sqrt{L(L-b)+b l}} \Leftrightarrow l^{2}-L(L-b)=b \sqrt{L(L-b)+b l} \Leftrightarrow \\ \left.\left.\begin{array}{l} \left\lvert\, \begin{array}{l} l^{2}-L(L-b) \leq 0 \\ L(L-b)+b l \geq 0 \\ {\left[l^{2}-L(L-b)\right]^{2}=b^{2}(L(L-b)+b l)} \end{array}\right. \end{array} \Leftrightarrow \right.\, \left|\begin{array}{l} l \in[-\sqrt{L(L-b)} ; \sqrt{L(L-b)}] \\ l \in\left(-\infty ; \cfrac{L(L-b)}{-b}\right]\\ \left[ l^2-L(L-b)\right]^2=b^2(L(L-b)+bl) \end{array}\right.\right.\)
If we consider from Theorem 2 that \(L \gt \tfrac{b(1-\sqrt{5})}{2} \gt 0\) together with Lemma 1 then the limit \(l\) will be positive. After some transformation, the last system turns to be equivalent to:
\(\left| \begin{aligned} &l\in \left( 0;\sqrt{L(L-b)} \right] \\ &l^4-2(L^2-Lb)l^2-b^3l+(L^2-Lb)(L^2-Lb-b^2)=0 \end{aligned} \right.\)
Then we obtain immediately that \(l_{1}=L\) and \(l_{2}=b-L\) are roots of the above mentioned equation. After factorising its left hand side we get the equivalent form:
\[ (l-L)(l-b+L)\left(l^{2}+b l+L b-L^{2}+b^{2}\right)=0 \] and consequently \(l_{3}=\tfrac{-b-\sqrt{4 L^{2}-3 b^{2}-4 L b}}{2}\) and \(l_{4}=\tfrac{-b+\sqrt{4 L^{2}-3 b^{2}-4 L b}}{2}\). But from \(b \lt 0\) and \(L \gt \tfrac{b(1-\sqrt{5})}{2} \gt 0\), we have
\[ l_{1}=L \in(0 ; \sqrt{L(L-b)}], l_{2}=b-L \lt 0, \quad l_{3}=\tfrac{-b-\sqrt{4 L^{2}-3 b^{2}-4 L b}}{2} \lt 0 \] and \(l_{4}=\tfrac{-b+\sqrt{4 L^{2}-3 b^{2}-4 L b}}{2} \gt \sqrt{L(L-b)}\). Hence \(l=L\).
So, from the proved Lemma 1, Lemma 2 and Lemma 3 in advance we conclude that thr assertion in Theorem 3 is true.
For proving Theorem 4 we will formulate three more lemmas.
Lemma 4. The subseries \(\left\{c_{2 k}\right\}_{k=1}^{\infty}\) is limited from above.
Proof: We will use again the principle of mathematical induction.
1. \(k=1, c_{2}=\sqrt{L(L-b)+b c_{1}}\).
We will prove that \(c_{2}=\sqrt{L(L-b)+b c_{1}} \lt L\).
\[ \begin{aligned} & \sqrt{L(L-b)+b c_{1}} \lt L .\left.\right|^{2} \\ & L^{2}-L b+b c_{1} \lt L^{2} \\ & L b \gt b c_{1}, \text { since } L \gt 0, b \lt 0 \quad \text { (by condition) } \\ & \Rightarrow \mathrm{c}_{1} \gt L \text { - true to Lemma } 1 . \end{aligned} \]
If \(k=2, c_{4}=\sqrt{L(L-b)+b c_{3}}\).
We will prove that \(c_{4}=\sqrt{L(L-b)+b c_{3}} \lt L\)
\[ \begin{aligned} & \sqrt{L(L-b)+b c_{3}} \lt \left.L\right|^{2} \\ & L^{2}-L b+b c_{3} \lt L^{2} \\ & b c_{3} \lt L b \mid: b \lt 0 \\ & c_{3} \gt L-\text { true to Lemma } 1 \end{aligned} \]
2. We assume that for each \(k=1,2 \ldots n\) it is true that \(c_{2 k} \lt L\).
3. We will prove that if \(k=n+1 \quad c_{2(n+1)}=c_{2 n+2} \lt L\).
And if \(c_{2 n+2}=\sqrt{L(L-b)+b c_{2 n+1}} \lt L\),
\(c_{2 n+2}=\sqrt{L(L-b)+b c_{2 n+1}} \lt \left.L\right|^{2} \Leftrightarrow L^{2}-L b+b c_{2 n+1} \lt L^{2} \Leftrightarrow b c_{2 n+1} \lt L b \mid: b \lt 0 \Leftrightarrow c_{2 n+1} \gt L\), which can be concluded by Lemma 1.
So we proved that the subseries \(\left\{c_{2 k}\right\}_{k=1}^{\infty}\) is limited from above by the constant \(L\), i.e. \(c_{2 k} \lt L\).
Lemma 5. The subseries \(\left\{c_{2 k}\right\}_{k=1}^{\infty}\) is increasing monotonously.
Proof: We will use once again the principle of mathematical induction.
1. \(k=1\).
We will check whether \(c_{4} \gt c_{2}\).
\(\sqrt{L(L-b)+b c_{3}} \gt \sqrt{L(L-b)+b c_{1}} \Leftrightarrow L(L-b)+b c_{3} \gt L(L-b)+b c_{1} \Leftrightarrow b c_{3} \gt b c_{1}\).
Because of \(b \lt 0\), we have \(c_{3} \lt c_{1}\), which can be concluded by Lemma 2.
2. We assume that \(c_{\mathrm{u} k} \gt 2_{k-}\), for each \(k=1,2 \ldots n\).
3. We will prove that this is also correct for \(k=n+1: c_{2 n+2} \gt c_{2 n}\).
\(c_{2 n+2}=\sqrt{L(L-b)+b c_{2 n+1}} \gt c_{2 n} \Leftrightarrow L(L-b)+b c_{2 n+1} \gt c_{2 n}^{2}=L(L-b)+b c_{2 n-1}\) \(\Leftrightarrow b c_{2 n+1} \gt b c_{2 n-1} \mid: b \lt 0\)
\(\Leftrightarrow c_{2 n+1} \lt c_{2 n-1}\), which is a consequence of Lemma 2.
Hence, Lemma 5 is proved.
Lemma 6. The limit of the series \(\left\{c_{2 k}\right\}_{k=1}^{\infty}\) is equal to \(L\).
Proof: From Lemma 4 and Lemma 5 we can make the conclusion that the subseries \(\left\{c_{2 k}\right\}_{k=1}^{\infty}\) has a limit and denote it by \(l\). Then:
\(\lim _{k \rightarrow \infty} c_{2 k}=l\).
From the recurrence
\(c_{2 k}=\sqrt{L(L-b)+b c_{2 k-1}}\), after a limit transition and applying Theorem 3, we can conclude that
\(l=\lim _{k \rightarrow \infty} c_{2 k}=\lim _{k \rightarrow \infty} \sqrt{L(L-b)+b c_{2 k-1}}=\lim _{k \rightarrow \infty} \sqrt{L(L-b)+b L}=L\).
Thus, \(l=L\).
Theorem 4 follows by Lemma 4, Lemma 5 and Lemma 6.
Coombining Theorem 3 and Theorem 4 we receive the statement of Theorem 2.
Following the definition of the square root, the following conclusions could be made from Theorem 2:
Conclusion 1: If \(L(L-b) \lt 0\), then the expression \(\sqrt{L(L-b)+b \sqrt{L(L-b)+b \sqrt{\ldots}}}\) is not defined in the set of the real numbers.
Proof: \(c_{1}=\sqrt{L(L-b)}\) does not exist because \(L(L-b) \lt 0\).
The following conclusion is also obvious.
Conclusion 2: If \(\mathrm{b} \lt 0\) and \(L(L-b) \lt 0\), then the expression
\(\sqrt{L(L-b)+b \sqrt{L(L-b)+b \sqrt{\ldots}}}\)
is not defined in the set of the real numbers.
Conclusion 3: If \(b \lt 0\) and \(L \lt \tfrac{b(1-\sqrt{5})}{2}\), then the expression
\[ \sqrt{L(L-b)+b \sqrt{L(L-b)+b \sqrt{\ldots}}} \] is not defined in the set of the real numbers.
Proof: The elements of the subseries \(\left\{c_{2 k}\right\}_{k=1}^{\infty}\) have negative values under the square root.
Theorem 5. Let \(b \lt 0\) and \(L=\tfrac{b(1-\sqrt{5})}{2}\). Then
\[ \sqrt{L(L-b)+b \sqrt{L(L-b)+b \sqrt{\cdots}}}=\left\{\begin{array}{l} -b, \text { if the number of roots is a finite odd number, } \\ 0, \text { if the number of roots is a finite even number, } \\ \text { indefinite, if the number of roots is infinite. } \end{array}\right. \] Proof: We will use again the principle of mathematical induction.
1. \(n=1\).
\[ \begin{aligned} & c_{1}=\sqrt{L(L-b)}=\sqrt{\tfrac{b(1-\sqrt{5})}{2}\left(\tfrac{b(1-\sqrt{5})}{2}-b\right)}=\sqrt{\tfrac{b(1-\sqrt{5})}{2}\left(\tfrac{b(-1-\sqrt{5})}{2}\right)}=\sqrt{b^{2}}=|b| \\ & b \lt 0 \Rightarrow c_{1}=-b . \\ & c_{2}=\sqrt{L(L-b)+b c_{1}}=\sqrt{L(L-b)+b(-b)}=\sqrt{L(L-b)-b^{2}}=\sqrt{b^{2}-b^{2}}=0 . \end{aligned} \]
2. Let us assume that \(c_{2 n-1}=-b\) and \(c_{2 n}=0\) for each \(n=1,2 \ldots k\).
3. We will prove that for \(n=k+1\) it is true that \(c_{2 k+1}=-b\) and \(c_{2 k+2}=0\).
\[ \begin{aligned} & c_{2 k+1}=\sqrt{L(L-b)+b c_{2 k}}=\sqrt{b^{2}+b \cdot 0}=\sqrt{b^{2}}=|b|=-b \\ & c_{2 k+2}=\sqrt{L(L-b)+b c_{2 k+1}}=\sqrt{b^{2}-b^{2}}=0 \end{aligned} \]
If the number of the roots is infinite, then the value of the expression \(\sqrt{L(L-b)+b \sqrt{L(L-b)+b \sqrt{\ldots}}}\) cannot be determined.
The proved statements complete the possibilities to calculate expressions of the type \(\sqrt{a+b \sqrt{a+b \sqrt{\ldots}}}\).
Applying the proved statements, we will examine three problems.
Problem 1. Determine the value of the expression \(L=\sqrt{3-2 \sqrt{3-2 \sqrt{3-\ldots}}}\).
Solution:
\[ \begin{aligned} & b=-2, a=L(L-b)=L(L+2)=3 ; \\ & L^{2}+2 L-3=0 ; \\ & L_{1}=-3, L_{2}=1 ; \\ & L=\tfrac{b(1-\sqrt{5})}{2}=\tfrac{-\not 2(1-\sqrt{5})}{\not 2}=\sqrt{5}-1 . \end{aligned} \]
We compare \(L_{1}=-3\) and \(L_{2}=1\) with \(\sqrt{5}-1\).
Since \(b \lt 0, L \lt \tfrac{b(1-\sqrt{5})}{2}\) and the number of roots is infinite, by Conclusion 3 we get that the value of the expression does not exist.
Problem 2. Determine the value of the expression \(\sqrt{4-2 \sqrt{4-2 \sqrt{4-2 \sqrt{\ldots}}}}\), if:
а) the number of the roots is 2017;
b) the number of the roots is 2018.
Solution:
\[ \begin{aligned} & b=-2, a=4=L(L-b) \\ & L^{2}+2 L-4=0 \\ & L_{1}=-1-\sqrt{5}, L_{2}=-1+\sqrt{5} \end{aligned} \]
By Theorem 5 we deduce that \(c_{2 k-1}=-b=2\) and \(c_{2 k}=0\).
a) \(c_{2017}=c_{2 k-1}=2\);
b) \(c_{2018}=c_{2 k}=0\).
Problem 3. Determine the value of the expression \[ \sqrt{2017-17 \sqrt{2017-17 \sqrt{2017-17 \sqrt{\ldots}}}} \]
Solution:
\(b=-17, L(L-b)=L(L+17)=2017\);
\(L^{2}+17 L-2017=0 ;\)
\(L_{1,2}=\tfrac{-17 \pm \sqrt{8357}}{2}\).
From \(L \gt 0 \Rightarrow L=\tfrac{\sqrt{8357}-17}{2}\).
We compare \(\tfrac{\sqrt{8357}-17}{2}\) with \(\tfrac{-17(1-\sqrt{5})}{2}=\tfrac{17 \sqrt{5}-17}{2}=\tfrac{\sqrt{1445}-17}{2}\)
\(\Rightarrow \tfrac{\sqrt{8357}-17}{2} \gt \tfrac{\sqrt{1445}-17}{2}\).
From Theorem \(2 \Rightarrow L=\tfrac{\sqrt{8357}-17}{2}\), i.e.
\[ \sqrt{2017-17 \sqrt{2017-17 \sqrt{2017-17 \sqrt{\ldots}}}}=\tfrac{\sqrt{8357}-17}{2} . \]
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