I know how to get the closed form of linear homogeneous recurrence relations such as
a_0 = 1, a_1 = 1
a_n = a_{n-1} + a_{n-2}
But I have no idea how to do this:
a_0 = 1, a_1 = 10,
a_n = \frac{a_n - 1}{a^5_{n-2}}, n >= 2
Can I get a clue on how to solve this one? Maybe a link to a similar relation getting solved?
Name:
Anonymous2009-12-10 0:59
Ah crap, I fucked it up. How do I delete posts on a text board?
Name:
Anonymous2009-12-10 1:54
Trying again.
I know how to get the closed form of linear homogeneous recurrence relations such as a$_{0}=1,$ a$_{1}$ = 1,
a$_{n}$ = a$_{n-1}$ + a$_{n-2}$
But I have no idea how to do this: a$_{0}=1,$ a$_{1}$ = 10,
a$_{\text{n}}$= $\frac{a_{n-1}}{a_{n-2}^{5}}$, n$\leq2$
Can I get a clue on how to solve this one? Maybe a link to a similar relation getting solved?
Name:
Anonymous2009-12-10 1:55
Trying again.
I know how to get the closed form of linear homogeneous recurrence relations such as
[TeX]a$_{0}=1,$ a$_{1}$ = 1,
a$_{n}$ = a$_{n-1}$ + a$_{n-2}$[/TeX]
But I have no idea how to do this:
[TeX]a$_{0}=1,$ a$_{1}$ = 10,
a$_{\text{n}}$= $\frac{a_{n-1}}{a_{n-2}^{5}}$, n$\leq2$[/TeX]
Can I get a clue on how to solve this one? Maybe a link to a similar relation getting solved?
Name:
Anonymous2009-12-10 2:17
>>2
There are no second chances in the realm of text boards.
Name:
Anonymous2009-12-10 2:38
Fuck it, here's a link to the relation that I'm having trouble with: http://i48.tinypic.com/29x97d1.png
I just have no fucking clue where to start.
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Anonymous2009-12-10 3:04
First, note that a_{n} is an integer power of 10 for all n. This is easy to show by induction.
Then define b_{n} = b_{n-1} - 5*b_{n-2}, with b_{0} = 0, b_{1} = 1. The closed form for a_{n} is 10^f(n), where f(n) is the closed form for b_{n}.
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Anonymous2009-12-10 12:17
Alright, I've gotten to the general solution of b_{n} as shown here: http://i46.tinypic.com/4t5xe8.png
I'm not supposed to use i in my answer, but I can't figure out how to do that since there is no x that satisfies (1 + sqrt(19)) / 2 = sin(x) + cos(x).
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4tran2009-12-10 22:57
>>8
Note that bn = (# - complex conjugate)/i -> it is real
Aside from factors of 2, you should be able to get away with Im[#], which should get you an overall factor and sine of some stuff.
Name:
Anonymous2009-12-12 22:47
>>8
Write (1\pm\sqrt{-19})/2 in polar form, \sqrt5\mathop{\rm cis}(\pm\tan^{-1}\sqrt{19}), and then you can cancel the cosines and is in the closed-form expression for b_n.
By the way, you used the math tags correctly the first time. You'll get your second chance once the admin(s) notice the \rm L\kern-.36em\raise.5ex{\scriptsize A}\kern-.15em\TeX plugin is broken. Also, don't start math mode in the middle of an equation!