- #1

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are we supposed to begin by guessing the limit and the bounds ??

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- Thread starter mansi
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- #1

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are we supposed to begin by guessing the limit and the bounds ??

- #2

shmoe

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Once you've got this 'guess' limit in hand, you should be able to prove convergence (hint:show it's an increasing & bounded sequence).

- #3

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so, it's an increasing sequence...

but i can't think of how we can show it's bounded...i mean,how do we use the recurrence relation?..and i guess, once we find the upper bound it would be easy to spot the limit of the sequence...

- #4

Galileo

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Maybe you could write out the first few terms, that might hint towards an upper bound.

- #5

shmoe

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mansi said:By mere observation, it's quite clear that a_1 < a_2 <.......<a_n.....

so, it's an increasing sequence...

You don't think you need any conditions on the terms to deduce this? What if, at some point, [tex]a_n=3[/tex]. Then what's [tex]a_{n+1}[/tex]? Can you guarantee this won't hapen?

mansi said:..and i guess, once we find the upper bound it would be easy to spot the limit of the sequence...

If you happened to find the

- #6

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In fact, I will tell you exactly what its value is. See my comments: https://www.physicsforums.com/showthread.php?t=73656

2cos(45) =[tex]\sqrt(2) [/tex]. 2cos(45/2)= [tex]\sqrt{2+\sqrt2}[/tex]

2cos(45/4) =[tex]\sqrt{2+\sqrt{2+\sqrt{2}}} [/tex]

2cos(45) =[tex]\sqrt(2) [/tex]. 2cos(45/2)= [tex]\sqrt{2+\sqrt2}[/tex]

2cos(45/4) =[tex]\sqrt{2+\sqrt{2+\sqrt{2}}} [/tex]

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- #7

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Robert,it can't be.The #-s on the right (the square roots) are increasing,while the sinus-es are decreasing (their arguments tends to 0)...

Daniel.

EDIT:You edited,and put "cos".Now it makes sense.Both are increasing.This "cosine" trick is really elegant...

Daniel.

EDIT:You edited,and put "cos".Now it makes sense.Both are increasing.This "cosine" trick is really elegant...

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- #8

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Wow! Thanks for the complement. I at first mistakenly put in the sin, which has this form:

2sin x = [tex]\sqrt{2-\sqrt{2+\sqrt{2+++}}}[/tex]

So that we get sin^2 + cos^2 = 1.

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- #9

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A1 = 2^(1/2) < 2

Assume An < 2

A(n+1) = (An + 2)^(1/2) < (2 + 2)^(1/2) = 2

By the principle of mathematical induction, An < 2 for all n.

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