Tuesday, 2 August 2016

real analysis - convergence of a recursive sequence with parameter a



How can you determine if the following recursive sequence converges:



xn+1=12(a+x2n)



where 0a1 and x1=0




I know that the limit x (if it exists) satisfies the following equation:



x=12(a+x2)



since limxxn=limxxn+1. Therefore x=1±1a



Thank you in advance :)


Answer



That is a really useful obervation. Notice that by induction you can easily prove that xn1 which means that the option 1+1a is definitely not the limit for a<1, since then it's strictly greater than 1. If a=1 then 1+1a=11a. That means that if the limit exists, it's equal to 11a.




Now, the case for a=0 is obvious since then xn=0 for all n so we will assume a>1. Notice that if we prove that the sequence is bounded from above and increasing, we will get convergence and from the above argument the limit. We can show inductively that the sequence is bounded from above by 11a. The base case x0<11a is obviously fulfilled. Now if xk<11a, then



xk+1=12(a+x2k)<12(a+(11a)2)=12(a+121a+1a)=11a.



That complete the induction. Now to prove the sequence is increasing, you can notice that xn<12(a+x2n) for all 0xn<11a (it's a quadratic in x) which means xn+1=12(a+x2n)>xn. That completes the proof that the seqence (xn) is convergent. As we stated earlier, convergence implies that the limit is 1+1a, which means we're done.


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