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anonymous

  • 4 years ago

What is the difference between divergence and convergence?

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  1. anonymous
    • 4 years ago
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    I can give you an example to explain if necessary

  2. amistre64
    • 4 years ago
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    |dw:1327426598793:dw|

  3. amistre64
    • 4 years ago
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    divergence goes off into infinity; and convergence settles down to something that we can define

  4. amistre64
    • 4 years ago
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    determining "d" or "c" can be a pain tho

  5. anonymous
    • 4 years ago
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    wait just gotta show u something

  6. anonymous
    • 4 years ago
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    \[\int\limits_{1}^{\infty}1/x ^{2}\]

  7. amistre64
    • 4 years ago
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    convergent I believe; since it gets real small real quick

  8. anonymous
    • 4 years ago
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    and lets say \[\int\limits_{1}^{\infty}1/\sqrt{x}\]

  9. amistre64
    • 4 years ago
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    x^(-1/2) gets big

  10. amistre64
    • 4 years ago
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    i think i got that right lol

  11. anonymous
    • 4 years ago
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    ya u did lol

  12. anonymous
    • 4 years ago
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    but they are both getting smaller

  13. amistre64
    • 4 years ago
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    1/x is a dividing line; 1/x diverges, but anything just shy of it bigger, like x^(-1.00000000001) converges

  14. amistre64
    • 4 years ago
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    the sqrt doesnt do it fast enough

  15. amistre64
    • 4 years ago
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    1/x goes to gets smaller buth the sums never settle down

  16. anonymous
    • 4 years ago
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    oh ok

  17. amistre64
    • 4 years ago
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    the convergence is called a least upper boundary, not that that makes much difference. It just says that we can add up the partial sums and they tend towards a nice number or they dont

  18. anonymous
    • 4 years ago
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    oh ok get it now thanks :D

  19. amistre64
    • 4 years ago
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    youre welcome :)

  20. anonymous
    • 4 years ago
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    \[\int_a^{\infty} \frac{1}{x^r}dx=\int_a^{\infty}x^{-r}dx\] converges if \[r>1\] diverges if \[r\leq 1\]

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