How I should deal with it? I’ve tried so far to find an obvious pattern finding some terms of the sequence.
$$x_{n+1}=x_n - x_{n}^2$$ $$x_0=\frac 12$$
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1What's the initial condition? That matters – Dec 27 '19 at 15:50
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3Also, what did you try? – Martund Dec 27 '19 at 15:51
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2$x_0\in\Bbb R$ doesn't tell us anything. – Martund Dec 27 '19 at 15:54
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1What is the source of the problem ? You seem to be missing an important detail. – The Demonix _ Hermit Dec 27 '19 at 15:55
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1How do you know there is a closed formula for it? – nonuser Dec 27 '19 at 15:57
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2$x_n=\frac{1}{2^{n+1}}$ if $x_0=\frac{1}{2}$. – Tuvasbien Dec 27 '19 at 16:04
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@Tuvasbien how do you came up with that result? – benj2k1 Dec 27 '19 at 16:06
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@Tuvasbien I get $x_2 = \frac{3}{16}$ not $\frac1{8}$ – The Demonix _ Hermit Dec 27 '19 at 16:07
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@TheDemonix_Hermit yah I got the same thing, but once again it doesn’t solve my problem. – benj2k1 Dec 27 '19 at 16:11
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By calculating the first terms, you can easily see the pattern, but I think $0,\frac{1}{2}$ and $1$ are the only values of $x_0$ for which there exists an easy formula for $x_n$. – Tuvasbien Dec 27 '19 at 16:11
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@Tuvasbien $x_n=1/2^{n+1}$ doesn't seem correct – bjorn93 Dec 27 '19 at 16:17
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You're right, calculus errors. – Tuvasbien Dec 27 '19 at 16:20
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Despite getting an explicit formula, one can show that $x_n\sim\frac{1}{n}$ for all $x_0\in]0,1[$. – Tuvasbien Dec 27 '19 at 16:26
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The earliest appearance of this sequence in MSE seems to be MSE question 506723 "Find asymptotic of recurrence sequence". Later appearances include questions 517099, 1558592, 2471982, 2861768, 3088856, 3338930. – Somos Dec 27 '19 at 18:17
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1Does this answer your question? Find asymptotic of recurrence sequence – Somos Jan 03 '20 at 00:27
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There is no formula for the general term of the recurrence $\,x_{n+1}=x_n-x_n^2\,$ however, there is some specific results about the sequence with $\,x_0=1/2.\,$ The denominator of $\,x_n\,$ is $\,2^{2^n}.\,$ OEIS sequence A076628 is the sequence of numerators of $\,x_n\,$ where $\,x_0=\frac12,\, x_1=\frac14,\, x_2=\frac3{16},\,x_3=\frac{39}{256},\, x_4=\frac{8463}{65536},\,\dots.$
In 1999 I found the result (where the constant $c$ depends on $x_0$) $$x_n =1/( n + c + \log(n + c-1/2 + \log( \\ n + c-17/24 + \log(n + c-919/1152 + \dots )))). \tag{1}$$ With $y_n:=1/x_n$ and $L_n:=\log(n)$, then $y_{n+1}=y_n^2/(y_n-1).\;$ The general result is $$ y_n = n + (c_0+c_1L_n) + (c_2+c_3L_n)/n + \\(c_4+c_5L_n+c_6L_n^2)/n^2 + \dots \tag{2}$$ for some constants $c_0,c_1,c_2,\dots$ which depend on $x_0$. Now first replace $n$ with $n+1$ in $y_n$ to get $$ y_{n+1}\!=\!n\!+\!(1\!+\!c_0\!+\!c_1L_n)\!+\!(c_1\!+\!c_2\!+\!c_3L_n)/n \!+\!(c_4\!+\!c_3\!-\!c_2\!-\!c_1/2\!+\!(c_5\!-\!c_3)L_n\!+\!c_6L_n^2)/n^2 \!+\!\dots \tag{3}$$ while using the recursion gives $$y_{n+1}\!=\!n\!+\!(1\!+\!c_0\!+\!c_1L_n)\!+\!(1\!+\!c_2\!+\!c_3L_n)/n \!+\!(c_4\!-\!c_0\!+\!1\!+\!(c_5\!-\!c_1)L_n\!+\!c_6L_n^2)/n^2 \!+\!\dots \tag{4}$$ and equating the two expressions for $y_{n+1}$ we must have $$c_1=c_3=1,c_2=c_0-1/2,c_5=3/2-c_0,c_6=-1/2,c_4=(-5+9c_0-3c_0^2)/6. \tag{5}$$ Making the substitutions for the constants we get $$y_n\!=\!n\!+\!(c_0\!+\!L_n)\!+\!(c_0\!-\!1/2\!+\!L_n)/n\! +\!((-5\!+\!9c_0\!-\!3c_0^2)/6+(3/2-c_0)L_n\!-\!1/2L_n^2)/n^2+\dots \tag{6}$$ and $$x_n=1/n+(-c_0-\log n)/n^2+(c_0^2-c_0+1/2+ \\ (2c_0-1)\log n+(\log n)^2)/n^3+\dots \tag{7}$$ which still depends on $c_0$ which comes from $x_0$.
Somos
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I decided to copy the contents of this answer to answer essentially the same question at MSE question 506723. – Somos Mar 28 '23 at 21:55