Solve the recurrence $a_n=7a_{n-1}-10a_{n-2}$ where $a_{0}= 3$ and $a_{1}=3$
how can a $a_{0}$ and $a_{1}$ both equal $3$?
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1Why shouldn't $a_0$ and $a_1$ both equal $3$? – Gerry Myerson May 12 '15 at 13:17
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It means that $a_2 = 7\cdot 3 - 10\cdot 3 = -9$, and you just go from there. – Arthur May 12 '15 at 13:19
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Since $a_n$ depends from $a_{n-1}, a_{n-2}$ you need set two values, e.g $a_1, a_2$ – L F May 12 '15 at 13:20
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These are only initial conditions. This is how they are defined. – May 12 '15 at 13:20
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You only have to prove the case by induction with the initial conditions on $ a_0 $ and $ a_1 $. – May 12 '15 at 13:24
2 Answers
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Strategy for this type of relation:
Solve $x^2 = 7x -10$
You get $x=2$ and $x=5$.
Then your solution must be in the form $a_n = c(2)^n+d(5)^n$, and you get $c$ and $d$ by substituting the initial values. So $3 = c+d$ and $3 = 2c+5d$. Solve this linear system.
I would explain why this works, but that would involve linear algebra.
The initial values are defined independently of the recurrence relation, since you can't use the relation to guess what $a_1$ and $a_2$ are.
Ilham
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The recurrence states something that is implicit and may be the cause of your confusion:
The recurrence relation,
$$a_n = 7a_{n-1} - 10 a_{n-2}$$
Holds for all values of $n$ greater than $1$.
The equation says nothing about $a_0$ and $a_1$, so $a_0$ and $a_1$ can be anything. In your case, they equal $3$, and you can calculate $a_2$ by using the equation:
$$a_2 = 7a_1-10a_0=21-30=-9.$$
5xum
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Well, somebody was trigger happy on the downvotes. Does he care to explain how the answers thus provided are wrong? – 5xum May 12 '15 at 13:22