# Math Help - Sequence relation problem

1. ## Sequence relation problem

Let a be a real number...?

Let 'a' be a real number such that a 1 < a < 2. a(sub n) is the sequence defined by:
a1 = a,
an+1 = |an| -1
n= 1,2,3......

And put Sn= a1+a2....+an

Find. a4, a5 a6 a7
Find S2, S4, S6

I'm not following the problem at all O.o

the answers are a4 = 1-a
a5 = a -2
a6 = 1 -a
a7 = a-2

Maybe if i just get an example to solve a4 I might be able to figure out the rest...

2. Originally Posted by gundanium
Let a be a real number...?
Let 'a' be a real number such that a 1 < a < 2. a(sub n) is the sequence defined by:
a1 = a,
an+1 = |an| -1
n= 1,2,3......
And put Sn= a1+a2....+an
Find. a4, a5 a6 a7
Find S2, S4, S6
the answers are a4 = 1-a
a5 = a -2
a6 = 1 -a
a7 = a-2
Maybe if i just get an example to solve a4 I might be able to figure out the rest...
From $1 it follows that $a-1>0,~1-a<0,~a-2<0$
$a_1=a,~a_2=|a_1|-1=a-1>0$
$a_3=|a_2|-1=a-1-1=a-2<0$

3. So how do i deal with the ones that are <0? multiply the quantity by -1?

4. Originally Posted by gundanium
Let a be a real number...?

Let 'a' be a real number such that a 1 < a < 2. a(sub n) is the sequence defined by:
a1 = a,
an+1 = |an| -1
n= 1,2,3......

And put Sn= a1+a2....+an

Find. a4, a5 a6 a7
Find S2, S4, S6

I'm not following the problem at all O.o

the answers are a4 = 1-a
a5 = a -2
a6 = 1 -a
a7 = a-2

Maybe if i just get an example to solve a4 I might be able to figure out the rest...
The recursive relation defining the sequence can be written as...

(1)

The function f(x) is illustrated here...

There is only one 'attractive fixed point' in and, because f(x) crosses the x axis with slope -2 any initial value a will produce a sequence asyntotically oscillating around - 1/2. Regarding the 'sum' is...

(2)

Kind regards

$\chi$ $\sigma$

5. Originally Posted by gundanium
So how do i deal with the ones that are <0? multiply the quantity by -1?
$a_3=a-2<0$ therefore, $|a_3|=2-a$.
So $a_4=|a_3|-1=2-a-1=1-a<0$.

6. Thats right! Thanks man :d