Matematické Fórum

vanok · 10. 10. 2011 02:54

Let $(x_n)_ {n\in\ N}$ dedined by $x_0= 5$ and $x_{n+1}= x_n + \frac 1{x_n}$
Find an interval I of length 1 such as $x_{1000} \in I$

Good luck
Vanok

Marian · 10. 10. 2011 08:52 — Editoval Marian (10. 10. 2011 12:43)

↑ vanok:

From the recurrence equation we get

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vanok · 10. 10. 2011 10:03

Hi ↑ Marian:: nice
Challenge:Find an interval $I_1$ of length 0,1 such as $x_{1000} \in I_1$

Good luck
Vanok

Orel · 11. 10. 2011 09:58 — Editoval Orel (11. 10. 2011 20:32)

$\[\begin{array}{l} x_{n + 1}^2 = x_n^2 + 2 + \frac{1}{{x_n^2}},\\ x_{n + 1}^2 - x_n^2 = 2 + \frac{1}{{x_n^2}},\\ \sum\limits_{n = 0}^{99} {(x_{n + 1}^2 - x_n^2)} = 2{\mkern 1mu} 00 + \sum\limits_{n = 0}^{99} {\frac{1}{{x_n^2}}} \\ x_{100}^2 - x_0^2 = 200 + \sum\limits_{n = 0}^{99} {\frac{1}{{x_n^2}}} \\ x_{100}^2 = 225 + \sum\limits_{n = 0}^{99} {\frac{1}{{x_n^2}}} \\ \sqrt {225} \le {x_{100}} \le \sqrt {225 + 100\cdot\frac{1}{{25}}} \\ 15 \le {x_{100}} \le 15.14\\ \sum\limits_{n = 0}^{999} {(x_{n + 1}^2 - x_n^2)} = 2{\mkern 1mu} 000 + \sum\limits_{n = 0}^{99} {\frac{1}{{x_n^2}}} + \sum\limits_{n = 100}^{999} {\frac{1}{{x_n^2}}} \\ x_{1000}^2 - x_0^2 = 2000 + \sum\limits_{n = 0}^{99} {\frac{1}{{x_n^2}}} + \sum\limits_{n = 100}^{999} {\frac{1}{{x_n^2}}} \\ x_{1000}^2 \le 2025 + \sum\limits_{n = 0}^{99} {\frac{1}{{{5^2}}}} + \sum\limits_{n = 100}^{999} {\frac{1}{{{{15}^2}}}} = \\ = 2025 + \frac{{100}}{{25}} + \frac{{900}}{{225}} = 2025 + 4 + 4 = 2033\\ 45 = \sqrt {2025} \le {x_{1000}} \le \sqrt {2033} \le 45.09\\ 45 \le {x_{1000}} \le 45.09 \end{array}\]$
$\[{{x}_{1000}}\in \left\langle 45\ ,\ 45.09 \right\rangle \]$
Mucha suerte
Orel

vanok · 11. 10. 2011 15:20

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vanok · 16. 10. 2011 13:51

In a week I would put here my solution of challenge

Vanok

vanok · 23. 10. 2011 00:38 — Editoval vanok (23. 10. 2011 01:51)

In my demonstration the first part: $45 < x_{1000}$ is identical to those of Marian and Orel.

We put for all x in N $u_n=x_n^2$ ,
immediately for all x in N $u_{n+1}= u_n + \frac 1{u_n} +2$
and by recurrence we have for all x \in N $u_n= u_0 + \sum_ {k=0} ^{n-1} \frac 1{u_k}+ 2n$
As $u_n$ are positive, or all x \in N $u_n\ge u_0 + 2n$

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Let $k\geq 0$ , we have
$\frac 1 {2(k+1)+u_0} \leq {\int_k^{k+1}{\frac1{2t+u_0}}\mathrm{d}t} \leq \frac 1 {2k+u_0}$
By adding of $k=0$ to $k=n-1$ , we obtain
$\sum_ {k=1} ^n \frac 1 {2k+u_0} \leq {\int_0^n{\frac1{2t+u_0}}\mathrm{d}t}=\frac12\ln(1+\frac{2n}{u_0}) \leq \sum_ {k=0} ^{n-1} \frac 1 {2k+u_0}$
Thus
$\sum_ {k=0} ^{n-1} \frac 1{u_k}\leq \sum_ {k=0} ^{n-1} \frac 1 {2k+u_0} \leq\frac12\ln(1+\frac{2n}{u_0}) +\frac1{u_0}-\frac1{u_0 +2n}$
Finally,
for all n in N $u_0+2n \leq u_n \leq u_0 + \frac1{u_0} +2n +\frac12\ln(1+\frac{2n}{u_0})$
So for $n=1000$ and $x_0=5$ we have
$45^2= 2025 \leq u_1000 \leq 2025 +0,04 +\frac12\ln 81$
and finally $x_1000 \in I_1=[45; 45,1]$

Remark 1:ln is natural logarithm
Remark 2:sequence $x_n$ is divergent
Remarque 3: $x_n \thicksim \sqrt {2n}$ in infinity

Matematické Fórum

#1 10. 10. 2011 02:54

real sequence

#2 10. 10. 2011 08:52 — Editoval Marian (10. 10. 2011 12:43)

Re: real sequence

#3 10. 10. 2011 10:03

Re: real sequence

#4 11. 10. 2011 09:58 — Editoval Orel (11. 10. 2011 20:32)

Re: real sequence

#5 11. 10. 2011 15:20

Re: real sequence

#6 16. 10. 2011 13:51

Re: real sequence

#7 23. 10. 2011 00:38 — Editoval vanok (23. 10. 2011 01:51)

Re: real sequence

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