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- Thread starter Mr Davis 97
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- #3

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But, just say that I want to prove that the sequence is decreasing. How would I know beforehand that I must prove that ##s_n > \sqrt{5}## for all ##n##? It seems to me that it is not immediately obvious that I must prove that ##s_n > \sqrt{5}##. Where does the ##\sqrt{5}## come from?

- #4

fresh_42

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You won't and it is not necessary.But, just say that I want to prove that the sequence is decreasing. How would I know beforehand that I must prove that ##s_n > \sqrt{5}## for all ##n##?

Probably from ##s=\dfrac{s^2+5}{2s}\,.## If ##s_{n+1} < s_n## and ##s_0=5##, then the limit must be around ##s## and it can only approach from above.It seems to me that it is not immediately obvious that I must prove that ##s_n > \sqrt{5}##. Where does the ##\sqrt{5}## come from?

- #5

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So instead of proving that ##\sqrt{5} < s_{n+1} < s_n \le 5## is true for all ##n##, could I just as well prove that ##0 < s_{n+1} < s_n \le 5## to prove convergence?You won't and it is not necessary.Any finite lower boundas e.g. zerowill doto prove convergence. If you want to prove what the limit is, you need a proposal. I haven't checked, but maybe ##\sqrt{5}## comes into play doing that.

Probably from ##s=\dfrac{s^2+5}{2s}\,.## If ##s_{n+1} < s_n## and ##s_0=5##, then the limit must be around ##s## and it can only approach from above.

If this is the case, why is it that author just goes ahead and uses ##\sqrt{5}##?

- #6

fresh_42

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For convergence, yes.So instead of proving that ##\sqrt{5} < s_{n+1} < s_n \le 5## is true for all ##n##, could I just as well prove that ##0 < s_{n+1} < s_n \le 5## to prove convergence?

Because this is the limit.If this is the case, why is it that author just goes ahead and uses ##\sqrt{5}##?

It's like saying: "I'm less than 20 ft tall." You would probably use a number which is closer to the truth, even if you only want to say that you're not infinitely tall.

- #7

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Just to clarify though, since I don't technically know that ##\sqrt{5}## is the limit beforehand, in practice would I choose something more obvious as a lower bound, like ##0##?For convergence, yes.

Because this is the limit.

It's like saying: "I'm less than 20 ft tall." You would probably use a number which is closer to the truth, even if you only want to say that you're not infinitely tall.

- #8

fresh_42

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I would do so. The crucial part is ##s_{n+1} < s_n##. The rest of a convergence proof is more or less obvious, because of ##(s_n)\subseteq [0,5]##.Just to clarify though, since I don't technically know that ##\sqrt{5}## is the limit beforehand, in practice would I choose something more obvious as a lower bound, like ##0##?

- #9

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Is the fact that 5 is an upper bound at all necessary to show convergence? Don't we just have to know that the sequence is decreasing and that there is a lower bound?I would do so. The crucial part is ##s_{n+1} < s_n##. The rest of a convergence proof is more or less obvious, because of ##(s_n)\subseteq [0,5]##.

- #10

fresh_42

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With monotony the first sequence element is automatically an upper bound, soIs the fact that 5 is an upper bound at all necessary to show convergence? Don't we just have to know that the sequence is decreasing and that there is a lower bound?

If you apply pure logic, then things are more complicated anyway. E.g. you need an argument why ##s_n < 5 ## and ##s_{n+1}< s_n## implies ##s_{n+1} < 5## (transitivity of the ordering plus induction, resp. Peano on the index set).

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The upper bound of 5 is not used anywhere in the proof of convergence. It is not even used in proving that the actual limit is ##\sqrt 5##.Is the fact that 5 is an upper bound at all necessary to show convergence? Don't we just have to know that the sequence is decreasing and that there is a lower bound?

The lower bound of ##\sqrt 5## is a key part of proving the sequence is

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- #12

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With monotony the first sequence element is automatically an upper bound, sonecessaryis a bit difficult to say, as it is as obvious. Personally, I like ##(s_n) \subseteq [0,5]## because one sees at once, that it is a compact set and thus the limits are within. This way it isn't necessary to think about the right end. It's more a matter of convenience to have a compact set than a necessity.

If you apply pure logic, then things are more complicated anyway. E.g. you need an argument why ##s_n < 5 ## and ##s_{n+1}< s_n## implies ##s_{n+1} < 5## (transitivity of the ordering plus induction, resp. Peano on the index set).

Makes sense. Last thing. What would be the best way to prove that ##s_{n+1} < s_n## in this case? Should I start on the LHS and make an argument that looks something like ##\displaystyle s_{n+1} = \frac{s_{n}^2+5}{2 s_{n}} < \cdots < s_n##, OR should I make an argument that starts with ##s_{n+1} < s_n \implies \frac{s_{n}^2+5}{2 s_{n}} < s_n \implies s_n > \sqrt{5}## and prove that ##s_n > \sqrt{5}## for all ##n##?

- #13

fresh_42

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It's equivalent - because we have only positive numbers - to ##\sqrt{5} < s_n## and that also answers your question, why the square root of five as lower bound comes into play. I would do it by induction. Whether it is theMakes sense. Last thing. What would be the best way to prove that ##s_{n+1} < s_n## in this case? Should I start on the LHS and make an argument that looks something like ##\displaystyle s_{n+1} = \frac{s_{n}^2+5}{2 s_{n}} < \cdots < s_n##, OR should I make an argument that starts with ##s_{n+1} < s_n \implies \frac{s_{n}^2+5}{2 s_{n}} < s_n \implies s_n > \sqrt{5}## and prove that ##s_n > \sqrt{5}## for all ##n##?

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That would not lead anywhere. It is called 'affirming the consequent'. If ##A\to B## then proving ##B## tells us nothing about the truth or otherwise of ##A##. In any case, that ##s_n>\sqrt 5## has already been proved by the text.OR should I make an argument that starts with ##s_{n+1} < s_n \implies \frac{s_{n}^2+5}{2 s_{n}} < s_n \implies s_n > \sqrt{5}## and prove that ##s_n > \sqrt{5}## for all ##n##?

Look at the formula ##s_{n+1}=\frac{s_n{}^2+5}{2s_n}##. If it wasn't for that 5 in the numerator we could cancel out the ##s_n## in the denominator and get a simpler expression that might give us a useful inequality. Using one of the inequalities proved in the text, can we replace the 5 by something greater than or equal to 5 that allows us to cancel out the ##s_n## in the denominator?

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If we know that ##s_n^2 >5##, then ##2 s_n^2 > s^2_n + 5##, so ##s_{n+1} = \frac{s_n^2+5}{2s_n} < \frac{2s_n^2}{2s_n} = s_n##. I guess that makes sense... My question then is if I were to approach this problem on my own without the book, how would I know beforehand that ##s^2_n > 5##?That would not lead anywhere. It is called 'affirming the consequent'. If ##A\to B## then proving ##B## tells us nothing about the truth or otherwise of ##A##. In any case, that ##s_n>\sqrt 5## has already been proved by the text.

Look at the formula ##s_{n+1}=\frac{s_n{}^2+5}{2s_n}##. If it wasn't for that 5 in the numerator we could cancel out the ##s_n## in the denominator and get a simpler expression that might give us a useful inequality. Using one of the inequalities proved in the text, can we replace the 5 by something greater than or equal to 5 that allows us to cancel out the ##s_n## in the denominator?

- #16

fresh_42

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You would a) assume the sequence converges, b) therefore conclude that ##s_{n+1}-s_n## is narrowing down, so for big enough ##n## we c) get ##s=\dfrac{s^2+5}{2s}## with only a small error. But this means d) ##s=\sqrt{5}## which is e) our candidate for the limit. As we start with ##5## and ##\sqrt{5}<5## there must f) be a decrease somewhere. On the other hand, we g) don't have any hints that the sequence jumps back and forth, so h) ##s_{n+1} < s_n## is a natural assumption to be checked.My question then is if I were to approach this problem on my own without the book, how would I know beforehand that ##s^2_n > 5##?

- #17

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By the way, once we know the sequence is decreasing, we can directly prove the sequence has a limit of ##\sqrt 5## as follows. The reduction from one sequence element to the next is

\begin{align*}

s_n-s_{n+1}

&=

s_n-\frac{s_n{}^2 + 5}{2s_n}\\

&=

\frac{s_n{}^2 - 5}{2s_n}\\

&=

\frac{(s_n - \sqrt 5)(s_n + \sqrt 5)}{2s_n}\\

&=

(s_n - \sqrt 5)\left(\frac12 +\frac{\sqrt 5}{2s_n}\right)\\

&>

\frac12 (s_n - \sqrt 5)

\end{align*}

From which we observe that:

\begin{align*}

s_{n+1}-\sqrt 5

&= (s_n-\sqrt 5) - (a - b)\\

&<

(s_n-\sqrt 5) -\frac12 (s_n-\sqrt 5)\\

&=

\frac12 (s_n-\sqrt 5)

\end{align*}

So the distance of ##s_n## from ##\sqrt 5## is more than halving at every step, so it must go to that limit.

- #18

Svein

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As far as I can see, the formula is the recursive formula for solving the square root of 5...

- #19

fresh_42

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Yes, looks like Newton's method to take roots by hand.As far as I can see, the formula is the recursive formula for solving the square root of 5...

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