I know how to get the equation of the form $x^2 = Ax + B$ and then from there get $a_k = C * x_1^k + D * x_2^k$ but doesn't the original $b_k$ equation have to be of the form $7b_{k-1} - 10b_{k-2}$ If it does, how to I get it to look like that?
Asked
Active
Viewed 188 times
1
-
Alternatively you can use generating functions as described in this answer, – user26486 Mar 30 '15 at 19:12
-
1It makes you think that there are typos in that book, but $7b_{k-1}-10b_{k-1}=-3b_{k-1}$. – Nathanson Mar 30 '15 at 19:12
-
@Nathanson I'm not sure what you mean. Could you explain a little more where the -3 came from? – bobert dillion Mar 30 '15 at 19:13
-
@bobertdillion Distributive property. $7b_{k-1}-10b_{k-1}=(7-10)b_{k-1}=-3b_{k-1}$. – user26486 Mar 30 '15 at 19:14
-
Okay, so then would I just have an equation of $x^2 = 3x$ ? – bobert dillion Mar 30 '15 at 19:16
-
It would simply be a geometric progression. If $b_k=-3b_{k-1},\forall k\in\mathbb Z^+$, then $b_k=(-3)^kb_0$. – user26486 Mar 30 '15 at 19:19
-
Okay, So I have a similar question then, if i have the equation $p_k = 1/2p_{k-1} + 1/2 p_{k+1}$ How do I get to $p_{k+1} = 2p_k - p_{k-1}$ – bobert dillion Mar 30 '15 at 19:24
1 Answers
1
Yes, almost surely this is a typo and the term being subtracted should be $10b_{k-2}$ (as you imagined when you planned your solution). Notice that as written, plugging in $k=1$ to the recursion violates the given initial conditions!
Greg Martin
- 92,241
-
Well, it says 'for all $k\ge 2$', so considering what happens if $k=1$ is irrelevant. – user26486 Mar 30 '15 at 19:51
-
Mathematically true, but fails the common sense test of why they'd write a problem that way. – Greg Martin Mar 31 '15 at 04:31