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LogicalApple

  • 3 years ago

Riemann Sum . . . Attached is the problem of evaluating a series as n approaches infinity. I am stuck on how to continue. I know if I can sqrt(n / (n + 1)) in a different form where there is only one n, I can do the rest.

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  1. LogicalApple
    • 3 years ago
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  2. LogicalApple
    • 3 years ago
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    That is to say, if I can get sqrt(n / (n + k)) to a form with only one n in it.

  3. duplicitycheese
    • 3 years ago
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    @logicalApple, I'm not sure either with the rest of the expression but if you have 1/n at the front and n is approaching infinite, can the whole thing just be approaching zero?

  4. LogicalApple
    • 3 years ago
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    1/n would approach 0 but that wouldn't cause the sum to approach 0. What if what is inside the parentheses approaches infinity? Then we have an indeterminate form 0 * infinity. If we can frame this into a Riemann Sum then it becomes a (hopefully) simple integration problem.

  5. duplicitycheese
    • 3 years ago
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    hm, I didn't think of that. I have no idea then. Sorry!

  6. joemath314159
    • 3 years ago
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    Maybe change the expression to:\[\sqrt{\frac{n}{n+k}}=\sqrt{\frac{(n+k)-k}{n+k}}=\sqrt{\frac{n+k}{n+k}-\frac{k}{n+k}}\]\[=\sqrt{1-\frac{k}{n+k}}\]This expresion has only one n in it.

  7. LogicalApple
    • 3 years ago
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    Hm... now there are two k's. I wonder if it's possible to write it in a form where there is only one of each? Otherwise I am not sure how to continue. I will work on this later. Thanks for the suggestions!

  8. LogicalApple
    • 3 years ago
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    \[\frac{ n }{ n+k } = \frac{ 1 }{ a + b } --> \frac{ n+k }{ n } = a + b --> 1 + \frac{ k }{ n } = a + b\] So . . . \[\frac{ n }{ n+k } = \frac{ 1 }{ 1 + \frac{ k }{ n } }\] Now the series becomes more manageable as: \[\lim_{n \rightarrow \infty} \frac{ 1 }{ n } \left( \frac{ 1 }{ \sqrt{1 + \frac{ 1 }{ n }} } + \frac{ 1 }{ \sqrt{1 + \frac{ 2 }{ n }} } + \frac{ 1 }{ \sqrt{1 + \frac{ 3 }{ n }} } + ... + \frac{ 1 }{ \sqrt{1 + \frac{ n }{ n }} } \right)\] Now I can imagine a function, f(x) = 1/sqrt(1 + x) divided into n partitions.

  9. LogicalApple
    • 3 years ago
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    \[\lim_{n \rightarrow \infty} \frac{ 1 }{ n }\sum_{1}^{n}\left( f(\frac{ i }{ n }) \right), f(x) = \frac{ 1 }{ \sqrt{1+ x} }\]

  10. LogicalApple
    • 3 years ago
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    \[\int\limits_{0}^{1}\frac{ 1 }{ \sqrt{1+x} }dx = 2(\sqrt{2}-1)\]

  11. LogicalApple
    • 3 years ago
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    I'm pretty sure of it. I wonder if someone can verify. Wolframalpha does not evaluate infinite series as well as humans.

  12. nitz
    • 3 years ago
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    correct.......

  13. LogicalApple
    • 3 years ago
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    Whew, thank you!

  14. nitz
    • 3 years ago
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    till generalising it was fine...........but actually didnt get your last step.....

  15. nitz
    • 3 years ago
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    how can u integrate.......?

  16. LogicalApple
    • 3 years ago
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    \[\lim_{n \rightarrow \infty} \frac{ 1 }{ n }\sum_{i =1}^{n}\left( f(\frac{ i }{ n }) \right), f(x) = \frac{ 1 }{ \sqrt{1+ x} }\] \[f(x _{i})\] returns the ith element of the sum \[\frac{ 1 }{ n } = \Delta x\] This is our partition with b - a = 1, where a is the lower limit and b is the upper limit. Let a = 0 and b = 1, then 0 < 1/n < 2/n < ... < i/n < ... < (n-1)/n < b \[a = 0 < x _{1} = \frac{ 1 }{ n } < x _{2} = \frac{ 2 }{ n } < ... < x _{i} = \frac{ i }{ n }, ... < x _{n-1} = \frac{ n-1 }{ n } < 1 = b\] So all these points of the function are partitioned by 1/n. I think this represents a Riemann sum as \[\sum_{i = 1}^{n} f(x _{i}) \Delta x\] Which, as n approaches infinity, becomes the Reimann integral: \[\int\limits_{0}^{1} f(x) dx\] where \[f(x) = \frac{ 1 }{ \sqrt{1 + x} }\] Sorry if my explanation is bad. I was never good with summations.

  17. nitz
    • 3 years ago
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    ok.............right ........thanks

  18. nitz
    • 3 years ago
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    actually we had some other way of doing these questions

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