Mathematics, 02.11.2019 03:31 Aidanme25

Find the closed form solutions of the following recurrence relations with given initial conditions. use forward substitution or backward substitution as described in example 10 in the text. (a) an = -an-1, a0 = 5 (b) an = an-1 + 3, a0 = 1 (c) an = an-1 - n, a0 = 4 (d) an = 2nan-1, a0 = 3

Answers

Answer from: Carlosruelas5409

a) $a_n=(-1)^n \cdot 5$

b) a_n = 1 + 3n

c) $a_n=4-\frac{n(n+1)}{2}$

d) $a_n=3 \cdot 2^{\frac{n(n+1)}{2}}$

Step-by-step explanation:

We solve this using backward or forward substitution.

a) We have this:

a_1= - a_0= (-1)a_0

then:

a_2= -a_1= (-1)(-1)a_0=(-1)^2 a_0

for n=3 we have:

a_3= -a_2= (-1)(-1)^2 a_0=(-1)^3 a_0

from this, we can see that a_n = (-1)^n a_0 is a solution for this recurrence relation, where a_0=5 . This is:

$a_n=(-1)^n\cdot 5$

b) We have $a_n=a_{n-1}+3$ with a_0=1 . Then:

$a_1=a_0+3=1+3=4\\a_2=a_1+3=4+3=7$ but at the same time

$a_2 = a_1 + 3 =(a_0+3)+3 = a_0+ 2 \cdot 3 [/text]for [tex]a_3$ we have:

a_3=a_2+3=7+10=4 or $a_3=a_2+3=(a_0+2\cdot 3)+3=a_0+3\cdot 3$

by the next:

$a_4 = a_3 + 3 = (a_0 + 3\cdot 3)+3 = a_0 + 4\cdot 3$

We can see that the recurrence rule is:

$a_n=a_0+n\cdot 3$

this is $a_n=1+n\cdot 3$

c)Note that:

$a_1-a_0 = (a_0 - 1)-a_0=-1\\a_2-a_1 = (a_1 - 2) -a_1 = -2\\a_3-a_2 = (a_2 - 3) - a_2 = -3\\\ldots\\a_n-a_{n-1} = (a_{n-1}-n)-a_{n-1} =-n$

taking all this we have to:

$a_n-a_0=\sum\limits_{k=1}^n (a_k - a_{k-1}) =\sum\limits_{k=1}^n -k = -\sum\limits_{k=1}^n k = - \frac{n(n+1)}{2}$

then:

$a_n=a_0-\frac{n(n+1)}{2}$

this is:

$a_n=4-\frac{n(n+1)}{2}$

d)We take $a_n=2^na_{n-1}$ . Then:

$a_n=2^na_{n-1}=2^n(2^{n-1}a_{n-2}) = 2^n\cdot 2^{n-1}(2^{n-2}a_{n-3}) = \dots =2^n\cdot2^{n-1}\cdot2^{n-2}\cdots 2^1 a_0=2^{n+(n-1)+(n-2)+\dots + 1}a_0=2^{\frac{n(n+1)}{2}}a_0$

replacing a_0=3 we have:

$a_n=3 \cdot 2^{\frac{n(n+1)}{2}}$

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Find the closed form solutions of the following recurrence relations with given initial conditions....