petegutz
Whats the difference between diverge and converge???
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hba
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go to the physics portion
TuringTest
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what? no it's a math Q
petegutz
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but its in algebra 2
hba
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k sorry
pratu043
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Diverge means separate, converge means meet.
TuringTest
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converge is to "settle down" on a finite falue
TuringTest
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value*
TuringTest
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oh jeez, an algebra2 definition of "diverge" ?
I kind of would like to know the context
hba
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@pratu043 it is maths not physics
petegutz
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heres what the question is 1/5+1/25+1/125+1/625... Does the infinite geometric series converge or diverge?explain
Monkeyball
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Does it go to infinity?
Monkeyball
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Or does it go to a specific value?
TuringTest
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or does it settle on no value at all and oscillate forever?
nbouscal
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Algebra 2 is an interesting time to learn about this. You should at least see the formal definition, so here's that (in my words): A sequence \(a_n\) converges to a limit \(L\) iff for any given \(\epsilon\), there exists an \(N\) such that \(n>N\implies |a_n-L|<\epsilon\). That's the formal one. In english, that means, as you go on to infinity, the sequence gets as close as you like to a given value. Basically, like others have said, the sequence "settles down" to a value. Diverges simply means "does not converge."
petegutz
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it goes infinitly
Monkeyball
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no it does not.....the reason being that the denominator is getting bigger which means that the overall number is getting smaller.
TuringTest
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@nbouscal I wish they taught me that in algebra2, but I don't think so...
Monkeyball
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If it went to infinity then the sequence will increase exponentially
nbouscal
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Oh no, I'm sure they don't. They would never do something crazy like teach real mathematics to secondary school students :P
nbouscal
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Not saying I know pedagogy better than they do, but couldn't they at least flash it on the board? One slide of a powerpoint? No? *sigh* oh well.
TuringTest
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exactly^
@petegutz do you have a specific formula to use? there are a few...
petegutz
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no all that it gave me was what i put up
TuringTest
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\[\sum_{n=1}^\infty ar^{n-1}=\sum_{n=0}^\infty ar^n=\frac a{1-r}\]is maybe a formula you can use?
TuringTest
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the above is only true for \(|r|<1\), otherwise the series diverges
so you must identify }r| in your series
TuringTest
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identify |r| *
petegutz
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is it 1/5?
nbouscal
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Yes, r=1/5
petegutz
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so it converges and it has a sum right?
nbouscal
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Yes
petegutz
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Thanks!