For the answer of this question, I am told that it is $a > 0$, but if $a$ is $1$ then would the resulting fraction not be $\frac{0}{2}$. Is it still converging then?
Am I wrong in thinking the answer should be $a > 1$ instead?
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2$\begingroup$ Three things: 1) Do you mean $u_{n+1} = \frac{a-1}{a+1} u_{n}$? 2) $0/2 = 0$, which is not undefined. 3) You essentially want the absolute value of the coefficient of $u_{n}$ to be $< 1$. To see this, note that by induction we have $u_{n+1} = \left( \frac{a-1}{a+1} \right)^{n} u_{0}$ $\endgroup$– mattosNov 20 '16 at 11:35
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$\begingroup$ Sorry I corrected that, 0/2 is indeed not undefined. And yes I do (for number 1). I'm a bit new at this, sorry. $\endgroup$– KyzenNov 20 '16 at 11:39
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$\begingroup$ All good, just check my edit to make sure it is correct. Also, in response to your updated question, if $a = 1$ then like you say the coefficient of $u_{n}$ is $0$ which means your recurrence is given by $u_{n+1} = 0 \cdot u_{n} = 0 \quad \forall n$. So the sequence is clearly convergent. $\endgroup$– mattosNov 20 '16 at 11:43
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1$\begingroup$ Ah, I get kind of get it now. Thanks for the answer and help in editing. $\endgroup$– KyzenNov 20 '16 at 11:46
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$\begingroup$ No worries. If you need anymore help, just ask. $\endgroup$– mattosNov 20 '16 at 11:49
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There is nothing wrong with a feaction like $0/2$ where just the numerator is zero; such a fraction is just zero. The problem is if the _ denominator_ is zero.
answered Nov 20 '16 at 11:38
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As commented, you have
$$u_n=\left(\frac{a-1}{a+1}\right)^nu_0$$
and the above converges (to a finite value, that is) iff ( since you're given $\;a>0\;$)
$$\left|\frac{a-1}{a+1}\right|\le1\iff |a-1|\le|a+1|= a+1$$
Now, we can do:
$$\begin{align*}&0<a< 1: \implies -a+1\le a+1\implies 2a\ge0\;,\;\;\text{and this is true in any such case}\\{}\\ &a\ge1: \implies a-1\le a+1\iff -1\le 1\,,\,\,\text{and again this is true in any such case}\end{align*}$$
Thus, the sequence $\;\{u_n\}\;$ converges to a finite value, which is either $\;u_0\;$ o $\;0\;$ , for any value of $\;a>0\;$ .
answered Nov 20 '16 at 11:51
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Hint for a geometric solution:
As a geometric sequence, it converges if and only if $\;\dfrac{\lvert a-1\rvert}{\lvert a+1\rvert}\iff\lvert a-1\rvert<\lvert a+1\rvert$.
Now interpret in terms of distance: if $\;b>c$, $$\lvert a-b\rvert <\lvert a-c\rvert\iff a > \frac{b+c}2.$$
