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Very difficult limit

  • Thread starter dirk_mec1
  • Start date
  • #1
761
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Homework Statement



http://www.sosmath.com/CBB/latexrender/pictures/f45d5e5b84cb5aecef3fd2f568739e32.gif [Broken]


Homework Equations



hint:

http://www.sosmath.com/CBB/latexrender/pictures/fee1863dab606636f986e63c261b565f.gif [Broken]
http://www.sosmath.com/CBB/latexrender/pictures/61d6e1cc712fcb6ac6f19b40851e85be.gif [Broken]
http://www.sosmath.com/CBB/latexrender/pictures/ae169a7ff6b1d2c09a977e5aad223650.gif [Broken]
http://www.sosmath.com/CBB/latexrender/pictures/cd14347d968bdb5be67f72e832558702.gif [Broken]

The Attempt at a Solution


I don't understand how the hint helps in this limit. Could someone point me in the right direction?
 
Last edited by a moderator:

Answers and Replies

  • #2
392
0
I didn't finish it, but here's my possible hint. If L=your limit, consider log L.

Edited to add: f(x) will be log(1+x). I "feel" like this is right.
 
  • #3
761
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Yes, that's helps a little bit but what do you mean by f(x) will be 1+x? How is that helpful?
 
  • #4
392
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what do you mean by f(x) will be 1+x?
It will turn out that you will want to use f(x)=log(1+x).

Anyway, I did finish it now, and it works out. Use the log of the limit idea as I first suggested and apply properties of logs. At some step you will get terms with log(n+k). Rewrite as log[n(1+k/n)]=log n + log(1+k/n). Eventually you'll have to decide what to use for x, and it's not too hard to guess. Write out what you can, and I bet you will get it. If not, post your work and I'll give you more hints. I predict from time to time you'll say things to yourself like "wow, I can't believe those terms canceled out exactly, this must be right."
 
  • #5
761
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[tex] \ln(L) = \sum_{k=0}^n \frac{1}{2^n} \left( \stackrel{n}{k} \right) \left( \ln(n) + \ln \left(1+\frac{k}{n} \right) \right) - \ln(n) [/tex]



Now if [tex]f(x) = \ln(1+x)[/tex] then B_n(1) is what I have, I only need to add an extra (1/2^n)* ln(2) and (subtract it later) to use the hint but for B_n(1)


[tex] \ \sum_{k=0}^n \left( \stackrel{n}{k} \right) \ln \left(1+\frac{k}{n} \right) - \ln(2) \leq \frac{1}{2 \sqrt{n}} \leq \frac{1}{2}[/tex]



[tex] \left( \frac{1}{2} \right) ^n \sum_{k=0}^n \left( \stackrel{n}{k} \right) \ln(n) = \ln(n) [/tex]

so this cancels the last term leaving only that annoying thing from above plus that term which I needed to substract:



[tex]\left( \frac{1}{2} \right) ^n \left[ \frac{1}{2} + \ln(2) \right] [/tex]


So the limit is 1? Am I going in the right direction?
 
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  • #6
392
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Looking good, but B_n(1) would be 0. I used a different value for x.

(Also, leave [tex]\frac{1}{2\sqrt{n}}[/tex] as is. Don't replace it by 1/2.)
 
  • #7
761
13
x=0? then

[tex]
\ \sum_{k=0}^n \left( \stackrel{n}{k} \right) \ln \left(1+\frac{k}{n} \right) - \ln(1) \leq \frac{1}{2 \sqrt{n}}
[/tex]

This results in:

[tex]
\left( \frac{1}{2} \right) ^n \left[ \frac{1}{2\sqrt{n}}\right] = \frac{ \frac{1}{2}^{n+1} }{\sqrt{n}} \rightarrow 0
[/tex]

as [itex] n \rightarrow \infty [/itex]

Conclusion: the limit is 1.
 
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  • #8
392
0
x=0?


[tex]B_n(x)=\sum_{k=0}^n \binom{n}{k} x^k (1-x)^{n-k} f(\frac{k}{n})[/tex]

so [tex]B_n(0)=\sum_{k=0}^n \binom{n}{k} 0^k 1^{n-k} f(\frac{k}{n})[/tex]

so [tex]B_n(0)=\binom{n}{0} 0^0 f(0)[/tex], which can't be helpful.

Also, [tex]B_n(1)=\sum_{k=0}^n \binom{n}{k} 1^k 0^{n-k} f(\frac{k}{n})[/tex]

so [tex]B_n(1)=\sum_{k=0}^n \binom{n}{k} 1^k 0^{n-k} f(\frac{k}{n})[/tex]

so [tex]B_n(1)=\binom{n}{n} 0^0 f(1)[/tex] which may not be 0 as I claimed earlier, but it still can't be helpful.

Oh well, average the two.....
 
  • #9
761
13
Oops :biggrin:

I see, if you take x =1/2 it matches my term exactly:



[tex]
B_n \left( \frac{1}{2} \right) =\sum_{k=0}^n \binom{n}{k} \left( \frac{1}{2} \right) ^n f \left(\frac{k}{n} \right)
[/tex]


Now

[tex] | B_n \left( \frac{1}{2} \right) - f \left(\frac{1}{2} \right)| \leq \frac{1}{2 \sqrt{n} } [/tex]


[tex]
\frac{1}{2 \sqrt{n}} + \ln \left( \frac{3}{2} \right) \longrightarrow \ln \left( \frac{3}{2}\ \right) \mbox{as}\ n \rightarrow \infty
[/tex]


So the limit is 3/2!
 
Last edited:
  • #10
392
0
So the limit is 3/2!
Hooray! Isn't that nice!

I wrote [tex]\ln \left( \frac{3}{2} \right)-\frac{1}{2 \sqrt{n} }\le B_n \left( \frac{1}{2} \right) \le \ln \left( \frac{3}{2} \right)+ \frac{1}{2 \sqrt{n} } [/tex] to make the final steps crystal clear.

Don't forget to verify [tex]|f(x)-f(y)|\le|x-y|[/tex] so that the hint applies.
 
  • #11
761
13
Hooray! Isn't that nice!
Yes it is :)

I wrote [tex]\ln \left( \frac{3}{2} \right)-\frac{1}{2 \sqrt{n} }\le B_n \left( \frac{1}{2} \right) \le \ln \left( \frac{3}{2} \right)+ \frac{1}{2 \sqrt{n} } [/tex] to make the final steps crystal clear.
Right, so the squeeze theorem applies.

Don't forget to verify [tex]|f(x)-f(y)|\le|x-y|[/tex] so that the hint applies.
I'm currently thinking on the proof for the hint but CS (after triangle inequality) gives me some nasty squares...
 

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