anonymous
  • anonymous
What is the difference between divergence and convergence?
Mathematics
jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
I can give you an example to explain if necessary
amistre64
  • amistre64
|dw:1327426598793:dw|
amistre64
  • amistre64
divergence goes off into infinity; and convergence settles down to something that we can define

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amistre64
  • amistre64
determining "d" or "c" can be a pain tho
anonymous
  • anonymous
wait just gotta show u something
anonymous
  • anonymous
\[\int\limits_{1}^{\infty}1/x ^{2}\]
amistre64
  • amistre64
convergent I believe; since it gets real small real quick
anonymous
  • anonymous
and lets say \[\int\limits_{1}^{\infty}1/\sqrt{x}\]
amistre64
  • amistre64
x^(-1/2) gets big
amistre64
  • amistre64
i think i got that right lol
anonymous
  • anonymous
ya u did lol
anonymous
  • anonymous
but they are both getting smaller
amistre64
  • amistre64
1/x is a dividing line; 1/x diverges, but anything just shy of it bigger, like x^(-1.00000000001) converges
amistre64
  • amistre64
the sqrt doesnt do it fast enough
amistre64
  • amistre64
1/x goes to gets smaller buth the sums never settle down
anonymous
  • anonymous
oh ok
amistre64
  • amistre64
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
anonymous
  • anonymous
oh ok get it now thanks :D
amistre64
  • amistre64
youre welcome :)
anonymous
  • anonymous
\[\int_a^{\infty} \frac{1}{x^r}dx=\int_a^{\infty}x^{-r}dx\] converges if \[r>1\] diverges if \[r\leq 1\]

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