does this series converge or diverge?

E

erblina

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Jan 5, 2015
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test the convergence of the following series :
Code: 
x[SUB]n+1[/SUB]= x[SUB]n[/SUB]-sin(x[SUB]n[/SUB])       ,   0<=x[SUB]1[/SUB]​<pi
,

can somebody help me with this i tried to prove monotonic or bound , using xn+1- xn>0 or xn+1- xn<0 but i find nothing at all :???: .Thank you
 
erblina said:
test the convergence of the following series :
Code: 
x[SUB]n+1[/SUB]= x[SUB]n[/SUB]-sin(x[SUB]n[/SUB])       ,   0<=x[SUB]1[/SUB]​<pi
,
can somebody help me with this i tried to prove monotonic or bound , using xn+1- xn>0 or xn+1- xn<0 but i find nothing at all
Click to expand...
If \(\displaystyle 0\le x_1<\pi\) then \(\displaystyle x_{n+1}=x_n-\sin(x_n)\) is not a series. It is a sequence.
 
Obviously the answer to the question is yes - The series does converge or diverge.

Looking at the series, we note that the series obviously converges for x1 = 0 [xn = 0 for all n] and for x1=\(\displaystyle \pi\) [xn = \(\displaystyle \pi\) for all n]. Thus we need only consider the interval (0,\(\displaystyle \pi\)).

Consider the sequence
xn+1= xn-sin(xn)
or
0 < sin(xn) = xn - xn+1
on (0,\(\displaystyle \pi\))
Thus xn is a decreasing sequence on (0,\(\displaystyle \pi\)). Let
f(x) = x - sin(x)
then
f'(x) = 1 - cos(x)
and
f''(x) = sin(x).
Since cos(x) \(\displaystyle \le\) 1 and sin(x) is non negative on [0,\(\displaystyle \pi\)], f(x) is non-decreasing on [0,\(\displaystyle \pi\)]. Thus the minimum of x - sin(x) occurs at the endpoint 0. That is the minimum of x-sin(x) on [0,\(\displaystyle \pi\)] is zero. Thus the sequence is bounded below.

Since a decreasing sequence bounded below converges, xn+1= xn-sin(xn) converges.

EDIT: Since, if xn converges, xn+j converges to the same thing for any finite fixed j, if xn converges to x we must have
x = x - sin(x)
or, with x confined to [0,pi], x=0 or x=pi.
 
Last edited:
Ishuda said:
Obviously the answer to the question is yes - The series does converge or diverge.

Looking at the series, we note that the series obviously converges for x1 = 0 [xn = 0 for all n] and for x1=\(\displaystyle \pi\) [xn = \(\displaystyle \pi\) for all n]. Thus we need only consider the interval (0,\(\displaystyle \pi\)).

Consider the sequence
xn+1= xn-sin(xn)
or
0 < sin(xn) = xn - xn+1
on (0,\(\displaystyle \pi\))
Thus xn is a decreasing sequence on (0,\(\displaystyle \pi\)). Let
f(x) = x - sin(x)
then
f'(x) = 1 - cos(x)
and
f''(x) = sin(x).
Since cos(x) \(\displaystyle \le\) 1 and sin(x) is non negative on [0,\(\displaystyle \pi\)], f(x) is non-decreasing on [0,\(\displaystyle \pi\)]. Thus the minimum of x - sin(x) occurs at the endpoint 0. That is the minimum of x-sin(x) on [0,\(\displaystyle \pi\)] is zero. Thus the sequence is bounded below.

Since a decreasing sequence bounded below converges, xn+1= xn-sin(xn) converges.

EDIT: Since, if xn converges, xn+j converges to the same thing for any finite fixed j, if xn converges to x we must have
x = x - sin(x)
or, with x confined to [0,pi], x=0 or x=pi.
Click to expand...


Thank you very much it was a great explanation . :D
 
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