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Let $u_n$ be a sequence defined by recurrence relation as : $u_{0} =1$ and $u_n=\frac{1}{3}u_{n-1}-1$, I'm not familiar to look for the closed form of $u_n $ , I have tried to find $u_n $ with a function of $n$ but I don't succeed because it's not defined as an arithmetic progression or a geometric progression. Then, is there any simple way to get it closed form ?

Thank for any help

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    $\begingroup$ Hint: $$u_{n+1}+\tfrac32=\tfrac13\left(u_n+\tfrac32\right)$$ $\endgroup$ – Did Feb 4 '18 at 11:56
  • $\begingroup$ Now, to understand the hint, you might want to wonder how to choose $c$ such that $$u_{n+1}=au_n+b$$ with $a\ne1$ is equivalent to $$u_{n+1}-c=a(u_n-c)$$ $\endgroup$ – Did Feb 4 '18 at 11:57
  • $\begingroup$ could you show me the method , because i seek for the way, try to share me an answer with detail $\endgroup$ – user517526 Feb 4 '18 at 11:57
  • $\begingroup$ "could you show me the method" Sorry but I just did, no? $\endgroup$ – Did Feb 4 '18 at 11:58
  • $\begingroup$ but i want u_n with a function of n ( I want to express u_n with n) $\endgroup$ – user517526 Feb 4 '18 at 11:59
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I was about to suggest generating functions (for example), but characteristic polynomials will be easier and it's always good to have an arsenal of various tools. $$3u_n=u_{n-1}-3 \iff 3=u_{n-1}-3u_n$$ same applies to $$3=u_n-3u_{n+1}$$ altogether $$u_{n-1}-3u_n=u_n-3u_{n+1} \iff 3u_{n+1}-4u_n+u_{n-1}=0$$ with the following characteristic polynomial $$3x^2-4x+1=0$$ with solutions $x_1=1$ and $x_2=\frac{1}{3}$, thus $$u_n=Ax_1^n+Bx_2^n=A+B\frac{1}{3^n} \tag{1}$$ Since $u_0=1 \Rightarrow A+B=1$ and given $u_1=-\frac{2}{3} \Rightarrow -\frac{2}{3}=A+\frac{B}{3}$ we obtains $A=-\frac{3}{2}$ and $B=\frac{5}{2}$ and finally from $(1)$ $$u_n=-\frac{3}{2} + \frac{5}{2}\frac{1}{3^n}=\frac{5-3^{n+1}}{2\cdot 3^n}$$

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You have to bring you back to a geometric progression, let say $v_n:=u_n+a$, with $v_n=\frac13 v_{n-1}$, $a$ to be determined. $$v_n=\frac13 v_{n-1} \quad \Rightarrow \quad u_n+a=\frac13 (u_{n-1}+a) \tag{1}$$ so (since $u_n=\frac13 u_{n-1} -1$ (2)), it comes ( (1)- (2)) $$ a = \frac13 a+1$$ so $$ a=\frac32 $$ $u_0=1$ so $v_0=\frac52$, so (since $v_n$ is a geometric progression) $$ v_n = \frac1{3^{n-1}}\frac52$$ and you can have $u_n$ : $$ u_n= \frac1{3^{n-1}}\frac52 -\frac32$$

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  • $\begingroup$ Your solution almost certainly contains a typo since it doesn't allow for $u_0=1$. $\endgroup$ – rtybase Feb 5 '18 at 19:45
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You can express $u_n$ through $u_0$ sequentially through $u_{n-1}$, $u_{n-2}$ and so on $$u_n = \frac{1}{3} \cdot u_{n-1} - 1$$ Substitute $u_{n-1}$ $$u_n = \frac{1}{3} \left( \frac{1}{3} \cdot u_{n-2} - 1 \right) - 1 = \frac{1}{3^2} \cdot u_{n-2} - \frac{1}{3} - 1$$ Then $u_{n-2}$ and so on $$u_n = \frac{1}{3^2} \left( \frac{1}{3} \cdot u_{n-3} - 1 \right) - \frac{1}{3} - 1 = \frac{1}{3^3} \cdot u_{n-3} - \frac{1}{3^2} - \frac{1}{3} - 1 = \dotsc$$ Finally $$u_n = \frac{1}{3^n} - \sum \limits_{k = 1}^n \frac{1}{3^{n-k}}$$

To check this formula we can use mathematical induction. The base case: $$n = 1: u_1 = \frac{1}{3} - 1$$ The inductive step: $$u_{n+1} = \frac{1}{3} \cdot u_n - 1 = \frac{1}{3} \left( \frac{1}{3^n} - \sum \limits_{k=1}^n \frac{1}{3^{n-k}} \right) - 1 = \frac{1}{3^{n+1}} - \sum \limits_{k=1}^n \frac{1}{3^{n+1-k}} - 1$$ We can substitute $$1 = \frac{1}{3^{n+1 - k}}$$ where $k = n+1$, thus $$u_{n+1} = \frac{1}{3^{n+1}} - \sum \limits_{k=1}^{n+1} \frac{1}{3^{n+1-k}}$$ So the formula is right

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