Math
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Math
There is a sentence in math that says that if lim(as n->infinity)of An=L than the lim(as n->infinity) of the arithmetical average of the numbers in An is also L.
The opposite is not true.
If i am asked to show the opposite is not true all i need to do is give an example when it isn't true?or do I need to give some sort of general proof?
If I need a general proof can anyone roughly explain the way to prove it?
The opposite is not true.
If i am asked to show the opposite is not true all i need to do is give an example when it isn't true?or do I need to give some sort of general proof?
If I need a general proof can anyone roughly explain the way to prove it?
Re: Math
You are given a proof "if A, then B" (A ==> B). You want to show that the opposite "if B, then A" (B ==> A) is not true. You can do that by finding a B such that A is false; in other words, a counter-example. If (and only if) you cannot think of such an example you must prove "if B then not A" in general.
- CarRepairer
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Re: Math
Is capital A a function? I don't understand the notation An if so. I am used to f(n).
Last edited by CarRepairer on 01 Dec 2009, 20:06, edited 1 time in total.
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Re: Math
Guess what he means is A_n (in tex-notation)CarRepairer wrote:Is capital A a function? I don't understand the notation An if so. I am used to f(n).
- CarRepairer
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Re: Math
Kids these days and their tex. In my day we used mechanical pencils.Auswaschbar wrote:Guess what he means is A_n (in tex-notation)CarRepairer wrote:Is capital A a function? I don't understand the notation An if so. I am used to f(n).
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Re: Math
I guess you were waiting for an answer... I'm interpreting from your wording that you're given the fact that the if first formula converges to a limit then the second converges to the same limit. Your job is just to prove that if the second formula converges to a limit then the first formula won't necessarily converge to the same limit (it might converge to a different limit or diverge). You gave the example of a[n] = -1^n which converges to a limit in the second but diverges in the first (from a glance anyway). That counterexample is all you need to prove what was asked.
Re: Math
But what if i wanted to prove it in a general form.how would i go about doing that?CarRepairer wrote:I guess you were waiting for an answer... I'm interpreting from your wording that you're given the fact that the if first formula converges to a limit then the second converges to the same limit. Your job is just to prove that if the second formula converges to a limit then the first formula won't necessarily converge to the same limit (it might converge to a different limit or diverge). You gave the example of a[n] = -1^n which converges to a limit in the second but diverges in the first (from a glance anyway). That counterexample is all you need to prove what was asked.
Re: Math
You can't really prove it in 'general', just because there ARE cases when it's true (so any 'general' proof would be false just because of them). Your example of -1^n is proof enough, you don't need more.
Idea is this: if you want to prove some assumption is false, you only need 1 example of it being false. If you want to prove an assumption is true however, THEN you need a general proof.
(at least those are based on my memories of math courses, that was some years ago so I may be missing something)
Idea is this: if you want to prove some assumption is false, you only need 1 example of it being false. If you want to prove an assumption is true however, THEN you need a general proof.
(at least those are based on my memories of math courses, that was some years ago so I may be missing something)
Re: Math
Well the definitions of limits use stuff like epsilon so maybe i need to choose a certain epsilon and show there is a contradiction in some way...im just not sure how.This really calls for someone that remembers differential and integral math.
Re: Math
Its all in hebrew...
Ill try to be precise.
I need to show that if the limit of the arithmetical average((a1,a2,a3....an)/n) of the numbers in a general sequence({an} is L(when n runs to infinity) it does not mean that the limit of an(when n runs to infinity) is also L.
Ill try to be precise.
I need to show that if the limit of the arithmetical average((a1,a2,a3....an)/n) of the numbers in a general sequence({an} is L(when n runs to infinity) it does not mean that the limit of an(when n runs to infinity) is also L.