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agentx5

  • 3 years ago

Topic: \(\textbf{Calculus 2}\) Q: Radius of convergence and interval of convergence for: \(\large \sum_{n=1}^{\infty} n!(2x-1)^n\) Ratio Test for Divergence \(\lim_{n \rightarrow \infty} \left| \frac{a_{n+1}}{a_n} \right|\) = L if ( L < 1 ) then {convergent} if( L > 1 ) then {"divergent"} else {"inconclusive"}

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  1. agentx5
    • 3 years ago
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    \[\lim_{n \rightarrow \infty} \left| \frac{a_{n+1}}{a_n} \right|\] \[\lim_{n \rightarrow \infty} \left| \frac{(n+1)!(2x-1)^{n+1}}{n!(2x+1)^n} \right|\] \[\lim_{n \rightarrow \infty} \left| \frac{\cancel{n!}(n+1)(2x-1)^{\cancel{n}+1}}{\cancel{n!}\cancel{(2x+1)^n}} \right| = \infty\]

  2. agentx5
    • 3 years ago
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    (trying to copy over from the related discussion about the trick shown in the last step to get rid of the factorial)

  3. Spacelimbus
    • 3 years ago
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    @agentx5, it looks right to me, if you are only asking for a double check.

  4. agentx5
    • 3 years ago
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    Interval of convergence I have correct, -0.5 < x < 0.5 by solving the conjunction" |2x-1|<0

  5. agentx5
    • 3 years ago
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    But the radius? Why is it zero? (correct answer) I would have said the radius is either the same as the interval or all real numbers...

  6. agentx5
    • 3 years ago
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    See what I mean @Spacelimbus ? This is the second time I've seen this type of problem, so there's some concept here about radius of convergence I'm not understanding I guess...

  7. agentx5
    • 3 years ago
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    I made an educated guess on zero after 1/9 and \(\infty\) failed to be correct

  8. agentx5
    • 3 years ago
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    Graphing this on a TI-83 shows it variates, it's not constant at zero.

  9. Spacelimbus
    • 3 years ago
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    well what makes sense to me in the example you just posted @agentx5 is that the term (9x-1)^n approaches it's limit faster then n! , but then I would have said that this is only the case if it it smaller than 1.

  10. experimentX
    • 3 years ago
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    man this is bugging me!!

  11. experimentX
    • 3 years ago
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    if x=1/n, the the sum converges!! but i haven't seen radius of interval in terms of n

  12. Spacelimbus
    • 3 years ago
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    yes @experimentX, usually they neatly cancel out.

  13. experimentX
    • 3 years ago
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    yeah they do ... still, for any numerical value of x, the series clearly diverges no matther how small ... but for variable value 1/n ... this series converges. I haven't seen radius of convergence expressed interms of n either ... and can't show from from ratio test either

  14. experimentX
    • 3 years ago
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    1/9

  15. experimentX
    • 3 years ago
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    that's the only value for which the series converges and it converges to 0 .. there's no interval ... just a point

  16. Spacelimbus
    • 3 years ago
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    I believe what @experimentX said would work for both answers @agentx5

  17. agentx5
    • 3 years ago
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    Oh bigger picture question to this then: What IS the radius of convergence, really?

  18. Spacelimbus
    • 3 years ago
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    In your previous example. x=1/2 is the only point where the series converges. And that is one individual point, so no radius.

  19. agentx5
    • 3 years ago
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    Like in a practical or visual sense so I can understand it, what is it? Interval is the range of inputs in which it would converge

  20. experimentX
    • 3 years ago
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    it's a point

  21. Spacelimbus
    • 3 years ago
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    A radius of convergence would mean, any value which is inside that radius (think of a circle) would make the sum converge, but in the examples you have posted, there is only one point that makes that happen, a point has no radius.

  22. agentx5
    • 3 years ago
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    |dw:1343149575540:dw|

  23. experimentX
    • 3 years ago
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    radius of convergence is an interval for which the the series converges, for eg sum, 2^n x^n if |x| < 1/2 .. the series converges -1/2 < x < 1/2 in out case radius of convergence is 1/9 <= x <= 1/9

  24. agentx5
    • 3 years ago
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    Ah and by the squeeze theorem it's just 1/9. :-)

  25. agentx5
    • 3 years ago
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    Thank you for your assistance gentlemen :-)

  26. experimentX
    • 3 years ago
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    well ... the best would be to use Cauchy root test than D'Alembert's ratio test here!! yw .. if it helped!!

  27. agentx5
    • 3 years ago
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    Whoa! Cauchy what?! D'Alembert what?! Never heard of those lol >_<

  28. agentx5
    • 3 years ago
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    I'm impressed!

  29. experimentX
    • 3 years ago
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    lots of and lots of convergence test http://en.wikipedia.org/wiki/Convergence_tests

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