Question

let a subscript 1 be a positive real number. Define a sequence a subscript n recursively by a subscript n +1 = a^2 subscript n -1. show that a subscript n does not converge to zero. Is there a subscript 1 such that the sequence a subscript n converges to some non-zero value?

Answer 1

wa ha, you have a big help. !!!

Answer 2

$$a_n>0\\a_{n+1}=a_{n-1}^2$$is that what you're telling us?

Answer 3

yes

Answer 4

Well, presume our sequence converges to some non-zero value \(L\), i.e. \(\lim\limits_{n\to\infty}a_n=L\). Observe, then, that taking our recurrence relation in the limit gives:$$\lim_{n\to\infty}a_{n+1}=\lim_{n\to\infty}a_{n-1}^2\\L=L^2$$therefore either \(L=0\) or \(L=1\)

Answer 5

err converges to some value \(L\) **

Answer 6

We know that if \(a_1>1\) such a limit cannot exist, since then \(a_5>a_3>a_1\) etc. If \(a_1=1\) we observe that \(a_{2k+1}\) for integer \(k\) must be \(1\) and hence \(L=1\); for such a limit to exist we also require \(a_{2k}\to1\).

Answer 7

If we let \(0<a_1<1\) we observe \(a_5<a_3<a_1\) etc. hence \(a_{2k+1}\to0\); since this limit corresponds to \(L=0\) we then infer that \(a_{2k}\to0\) as well.

Answer 8

Intuitively then if \(a_1=1\) our sequence should converge to \(1\), which is nonzero.

Answer 9

what was this part ("err converges to some value L **)?

Answer 10

Note we took advantage of subsequence convergence...

Answer 11

@mathtutoring22 well if we presume \(a_n\to L\) for some real value \(L\) we find our recurrence relation tells us \(L=L^2\) which has only two solutions, \(L=0\) and \(L=1\)

Answer 12

I was correcting what I said earlier about \(L\) being nonzero

Answer 13

ok, thanks.

Answer 14

How do I prove that the limit ( as n goes to infinity) of 1/(nsin(n)) = 0? using l'hopital's or another rule

Answer 15

$$\lim_{n\to\infty}\frac1{n\sin n}$$observe that since \(-1\le\sin n\le1\) we have \(-n\le n\sin n\le n\) for \(n\ge 0\). It follows that \(-\dfrac1n\ge\dfrac1{n\sin n}\ge\dfrac1n\). Now since in the limit as \(n\to\infty\) we have \(-\dfrac1n,\dfrac1n\to0\) so it follows that \(\dfrac1{n\sin n}\to 0\) -- this uses the 'squeeze' theorem

Answer 16

wow. thanks. i tried using l'hospital's rule but i couldn't prove it.

Answer 17

@mathtutoring22 are you sure it's \(a_{n+1}=(a_{n-1})^2\)... ? not, say, \(a_n=(a_{n-1})^2\) or \(a_{n+1}=a_n^2\)

Answer 18

\[_{a _{n=1}} = ^{a^2_{n}} -1\]

Answer 19

(a^2 subscript n) -1

Answer 20

@oldrin.bataku

Answer 21

OOOOOOOH

Answer 22

Alright, so we have \(a_{n+1}=a_n^2-1\).
Observe that if \(a_n\to L\) for some finite \(L\) in the limit as \(n\to\infty\) our recurrence relation tells us:$$\lim_{n\to\infty}a_{n+1}=\lim_{n\to\infty}(a_n^2-1)\\L=L^2-1\\L^2-L-1=0$$Ah, this problem is going to be fun ;-) can you recognize that polynomial?

Answer 23

(x - (1+√5)/2) (x- (1-√5)/2)?

Answer 24

well, one of its roots is \(\varphi\), the golden ratio, and its other is \(-\Phi\):$$L=\frac{1\pm\sqrt{1-4(-1)}}2=\frac{1\pm\sqrt5}2\\\varphi=\frac{1+\sqrt5}2,-\Phi=\frac{1-\sqrt5}2$$

Answer 25

So clearly in the limit our value is either of \(\frac12(1\pm\sqrt5)\) if our sequence converges, and neither of those are \(0\), so we conclude our limit cannot conclude to \(0\)

Answer 26

oops -- actually, if \(a_1<1\) the above occurs; if \(a_1=1\) our series becomes alternating and fails to converge: \(1,0,-1,0,-1,0,-1,0,-1,\dots\)

Answer 27

and thats shows it?

Answer 28

well i showed the first part now relax

Answer 29

For the second part all we need to observe is that \(\varphi\) satisfies \(x=x^2-1\) (as we did above for \(L\)) so if we consider a constant sequence i.e. \(a_{n+1}=a_n\) we see that \(a_n=\varphi\) will give us a solution (so \(a_1=\varphi\)). If we could pick a negative \(a_1\), we could just use \(a_n=-\Phi\).

http://www.wolframalpha.com/input/?i=a_1%3Dgolden+ratio%2C+a_%7Bn%2B1%7D%3Da_n%5E2+-+1

http://www.wolframalpha.com/input/?i=a_1%3Dconjugate+golden+ratio%2C+a_%7Bn%2B1%7D%3Da_n%5E2+-+1

Answer 30

@oldrin.bataku you make us like high school students. hehehe. .. which school did you graduate?

Answer 31

@Loser66 I am a high school student :-p

Answer 32

if so, tell me what the school is, I turn down my degree, back there apply to be a high school student like you.

Answer 33

I have two final question.

Answer 34

the sequence \[_{a _{n}}\]

Answer 35

is define in the following way: for every positive integer n we set a subscript n to be the sum of the positive divisors of n, for example a subscript 10 = 1+2+5+10 = 18. show that lim(as n goes to infinity) a subscript n / n is not = 2

Answer 36

@oldrin.bataku

Answer 37

this is my second and last question:

Answer 38

@Loser66 haha I haven't learned this in school... but I have begun taking college classes as a high school student

Answer 39

$$a_n=\sum d_k$$where \(d_k\) is the \(k\)-th positive divisor of \(n\).

Answer 40

Well, first note we have \(a_p=p+1\) for all primes \(p\). Consider the infinite subsequence \(\{a_p\}\) of primes \(p\) and we see that since we have a infinitude of primes, \(p\) can grow arbitrarily large and thus \((p+1)/p=1+1/p\) will grow arbitrarily close to \(1\). Since thsi is not \(2\) we know \(\{a_n\}\) cannot converge to \(2\).

Answer 41

Well consider that for \(n=1\) we have \((1+x)(1+x^2)=1+x+x^2+x^3\).

Answer 42

Now, let's say it holds for some integer \(n\), i.e.$$(1+x)(1+x^2)\cdots(1+x^{2^n})=\sum_{k=0}^nx^k$$Consider that we can multiply by \((1+x^{2^{n+1}})\):$$(1+x)(1+x^2)\cdots(1+x^{2^{n+1}})=(1+x^{2^{n+1}})\sum_{k=0}^nx^k=\sum_{k=0}^nx^k+\sum_{k=0}^nx^{k+2^{n+1}}$$

Answer 43

oops i messed up

Answer 44

what the hell is \(m\)? from what I can tell we can let \(m=2^{n+1}-1\)

Answer 45

Okay, let's rewrite:$$(1+x)(1+x^2)\cdots(1+x^{2^n})=\sum_{k=0}^mx^k\\(1+x)(1+x^2)\cdots(1+x^{2^{n+1}})=(1+x^{2^{n+1}})\sum_{k=0}^mx^k=\sum_{k=0}^m x^k+\sum_{k=0}^mx^{k+2^{n+1}}$$note we can rewrite our latter sum:$$\sum_{k=0}^mx^{k+2^{n+1}}=\sum_{k=2^{n+1}}^{m+2^{n+1}}x^k$$

Answer 46

Since \(m=2^{n+1}-1\) we see our latter summation is really:$$\sum_{k=m+1}^{2m+1}x^k=\sum_{k=m+1}^{2^{n+2}-2+1}x^k=\sum_{k=m+1}^{2^{n+2}-1}x^k$$Combine our two sums:$$(1+x)(1+X^2)\cdots(1+x^{2^{n+1}})=\sum_{k=0}^{2^{n+2}-1}x^k$$and we see \(m_{n+1}=2^{n+1}-1\) giving us the form seen above; by the principle of induction we're done

Answer 47

thanks again @oldrin.bataku @loser66

Answer 48

I did nothing, me helpless, friend.

Answer 49

Where does the 2^n+1 comes from again, @oldrin.bataku @loser66

Answer 50

@mathtutoring22 it's just the "m" that satisfies the prompt

Answer 51

so replace 2^n+1 by m?

Answer 52

oops i wrote that backwards

Answer 53

What about this:
prove that for every n>=, the number 11^n+1 + 12^2n-1 is divisible by 133. ( using math induction) I tried doing it but i made a mistake in the last part