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anonymous

  • 5 years ago

Comparison test to determine if improper integral converges or diverges, (3+cos(x))/(x^3) from 1 to infinity

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  1. anonymous
    • 5 years ago
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    3+cos(x) varies between 4 and 2 as x approaches infinity. Since 4 >= 3+cos(x) always, the series of (3+cos(x))/x^3 will converge if 4/x^3 converges.

  2. anonymous
    • 5 years ago
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    f(x) >= g(x) >= 0 f(x) = 1/x^2 g(x) = (3+cos(x))/(x^3) because of f(x) is larger for all values of g(x) and f(x) converges to \[\int\limits_{1}^{∞} 1/x^2 dx = \lim t \rightarrow∞ \int\limits_{1}^{t} 1/x^2 = {-1/x}\] from t to1 = 1

  3. anonymous
    • 5 years ago
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    1/x^2 is not larger than (3+cos(x))(x^3) For quite a few values greater than 1.

  4. anonymous
    • 5 years ago
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    Infinitely many in fact.

  5. anonymous
    • 5 years ago
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    I dont understand how he got the 1/x^2 :P , also whats the difference if you select 2 instead of 4? say 2<=3+cos(x)?

  6. anonymous
    • 5 years ago
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    You want to select something that is greater than the function you want to solve. If you can find something which converges but which is strictly larger than the function you're looking to solve, then you know that your function must also converge.

  7. anonymous
    • 5 years ago
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    Oh ok, so when I get 4/x^3 do I have to evaluate the definite integral of that and find out if it converges or diverges?

  8. anonymous
    • 5 years ago
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    Oops i meant (3+cos(x))/ x^2 which is greater then all values of (3+cos(x))/ x^2

  9. anonymous
    • 5 years ago
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    ***(3+cos(x))/ x^3 sorry

  10. anonymous
    • 5 years ago
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    Ok nevermind, i got it, thanks!

  11. anonymous
    • 5 years ago
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    Yep. the point is to find a function that is infinitely larger and if that function converges then the smaller one converges. However if the function ( f(x) for example) you test is divergent then f(x) >= g(x) >= 0 then g(x) can't be said is divergent. but if the function ( g(x) in this case) you test for is smaller and divergent then f(x) is divergent.

  12. anonymous
    • 5 years ago
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    so what if the interval was not to infinity?

  13. anonymous
    • 5 years ago
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    if the function is convergent on the interval and larger (meaning on discontinuities), the function is only to be said convergent on that interval.

  14. anonymous
    • 5 years ago
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    No*

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