EPTAnd to me, their intelligence level is not the same.
Calculus help - Page 2
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moriya
United States54 Posts
And to me, their intelligence level is not the same. | ||
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Malongo
Chile3479 Posts
On July 19 2009 11:37 moriya wrote: I need not u to show me that "both are the same". And to me, their intelligence level is not the same. Good luck in your life with epsilon/delta smart boy. | ||
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moriya
United States54 Posts
On July 19 2009 11:08 Plexa wrote: Math is teaching you to thinking logically  the any of the limit laws can easily be derived using anything from the sequential criterion for limits to straight up using epsilon-delta machinery. With that said, the arguments employed are exactly the same in both cases - except one has the complete solution and the other has the sketch solution. The latter, imo, is better for a thread like this. The reason being is that it gives the op the chance to complete the proof the e-d methods since from the outset it looked like he was confused about how to approach the problem. Nevertheless its a rather trivial problem in analysis any way you look at it | ||
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moriya
United States54 Posts
Have a brilliant future! Hope u can contribute sth new to the society! On July 19 2009 11:46 Malongo wrote: Good luck in your life with epsilon/delta smart boy. | ||
Plexa
Aotearoa39261 Posts
On July 19 2009 11:30 moriya wrote: Certainly i am aware of that "trivial" thing, using ur word, just lazy to input them. But u r so picky, so i add them, please prove that when lim(g)!=0. Uh, if you leave out that part of the hypothesis then your statement is not true... it is certainly not trivial + Show Spoiler + Let epsilon = e > 0 There exists an L and a delta = d1 such that if 0 < | x - c | < d1 then | f(x) - L | < e where L belongs to the reals. And there exists an M and a delta = d2 such that if 0 < | x - c | < d2 then | g(x) - M | < e where M belongs to the reals, M =/= 0. Consider f(x)/g(x). Choose d = min{d1, d2} (then clearly f converges to L and g converges to M for |x - c | < d). Assume 0 < | x - c | < d, define h(x) = 1/g(x) => lim(h(x)) = 1/M = m (since M =/= 0, the proof is easy, let me know if you want it). So, |f(x)*h(x) - Lm| = |(f(x)*h(x)-f(x)*m) + (f(x)*m - L*m)| =< |f(x)|*|(h(x) - m)|+|m|*|(f(x) - L)|. Since both h(x) and f(x) have a limit, they are bounded on the delta neighborhood of c. Hence, there exists and number Z such that |f(x)|*|(h(x) - m)|+|m|*|(f(x) - L)| <= Z*|h(x) - m| + Z*|f(x) - L|. Since e arbitrary, we can easily find conditions for which for 0 < | x - c | < d => |f(x) - L| < e/(2*Z) and |h(x) - m| < e/(2*Z). So we arrive at the result; Z*|h(x) - m| + Z*|f(x) - L| =< Z*e/(2*Z) + Z*e/(2*Z) = e. And we are done Messy, but it holds (I hate writing proofs out on the computer) | ||
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Muirhead
United States557 Posts
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moriya
United States54 Posts
But ur sketch solution implies u think the e-d part is trivial, while i think they are the core of the problem and ur statement is trivial. I just want to show that we have to touch the core of problem, sooner or later. If u cannot prove that for f/g, it means u need practice with e-d system more. If ur major is math, I would like to say sorry to you... | ||
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Plexa
Aotearoa39261 Posts
On July 19 2009 12:18 Muirhead wrote: Lolol Plexa can we put LaTeX support on teamliquid Would be very useful for these kinds of threads T_T On July 19 2009 12:20 moriya wrote: OK, i agree i shouldnot be so lazy when discussing a math problem. Maybe years of engineer student drag me too fay away lol. But ur sketch solution implies u think the e-d part is trivial, while i think they are the core of the problem and ur statement is trivial. I just want to show that we have to touch the core of problem, sooner or later. If u cannot prove that for f/g, it means u need practice with e-d system more. If ur major is math, I would like to say sorry to you... From the OP I was thinking about using a triangle inequality proof, but I have no clue where to start. Do you think its better to spell out the entire proof? Or to show him the path he should follow to arrive at the solution? | ||
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Faronel
United States658 Posts
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moriya
United States54 Posts
is easy. What i feel is lim1/g=1/limg is the core of this problem here. I use this problem to test so I know the method, lol funny that we always have different ideas about which is core and which is trivial... On July 19 2009 12:14 Plexa wrote: Uh, if you leave out that part of the hypothesis then your statement is not true... it is certainly not trivial + Show Spoiler + Let epsilon = e > 0 There exists an L and a delta = d1 such that if 0 < | x - c | < d1 then | f(x) - L | < e where L belongs to the reals. And there exists an M and a delta = d2 such that if 0 < | x - c | < d2 then | g(x) - M | < e where M belongs to the reals, M =/= 0. Consider f(x)/g(x). Choose d = min{d1, d2} (then clearly f converges to L and g converges to M for |x - c | < d). Assume 0 < | x - c | < d, define h(x) = 1/g(x) => lim(h(x)) = 1/M = m (since M =/= 0, the proof is easy, let me know if you want it). So, |f(x)*h(x) - Lm| = |(f(x)*h(x)-f(x)*m) + (f(x)*m - L*m)| =< |f(x)|*|(h(x) - m)|+|m|*|(f(x) - L)|. Since both h(x) and f(x) have a limit, they are bounded on the delta neighborhood of c. Hence, there exists and number Z such that |f(x)|*|(h(x) - m)|+|m|*|(f(x) - L)| <= Z*|h(x) - m| + Z*|f(x) - L|. Since e arbitrary, we can easily find conditions for which for 0 < | x - c | < d => |f(x) - L| < e/(2*Z) and |h(x) - m| < e/(2*Z). So we arrive at the result; Z*|h(x) - m| + Z*|f(x) - L| =< Z*e/(2*Z) + Z*e/(2*Z) = e. And we are done Messy, but it holds (I hate writing proofs out on the computer) | ||
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Plexa
Aotearoa39261 Posts
On July 19 2009 12:29 moriya wrote: not bad, although i see u spend a lot on proving lim(f*g)=limf*limg, while just say lim(1/g)=1/limg is easy. What i feel is lim1/g=1/limg is the core of this problem here. I use this problem to test so I know the method, lol funny that we always have different ideas about which is core and which is trivial... If I were proving the problem I would have the lim(1/g) = (1/M) part as a lemma before the problem began. You're suggesting proving two things in the same proof, I prefer to break things into bite sized chunks for the most part. I dont see why you are trying to "size up" or "attack" my ability with maths though I assume anyone writing here is competent. It would be wise if you mannered up your responses from being passive aggressive to something more constructive if you wish to remain a poster here. | ||
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Muirhead
United States557 Posts
I've seen some surprisingly advanced PHD-level posters here, but I didn't expect even the moderators to be mathy ^^ | ||
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moriya
United States54 Posts
BTW, It is a pity that we don't discuss that analogy. I want to know how u guys feel of that. | ||
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Plexa
Aotearoa39261 Posts
On July 19 2009 12:45 Muirhead wrote: I'm in my last year for my bsc in maths. Although next semester ill be taking honors papers to fill it up because i've done all the third year ones =/Hm Plexa just curious what is your maths background? I've seen some surprisingly advanced PHD-level posters here, but I didn't expect even the moderators to be mathy ^^ On July 19 2009 12:49 moriya wrote: Summary： I become so aggressive becoz i think some guy's solution is a typical BM in math proof. I make an analogy with and Pythagoras's Theorem and laws of cosine, to show why it is a BM solution. To be more "constructive", I propose a problem f/g and lol some one begin to work on it. BTW, It is a pity that we don't discuss that analogy. I want to know how u guys feel of that. mmm I disagree that the analogy is particularly relevant on the grounds that the argument in proving pythagoras is very different from setting theta = pi/2 in the cosine law, unlike the case we had here. | ||
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Muirhead
United States557 Posts
As for your analogy, I think it's not quite the same thing. Plexa just restated the problem as its contrapositive and claimed that contrapositive to be well-known (which it is). Using the Law of Cosines to prove the Pythagorean Theorem is bashing a problem with excessively powerful tools, but Plexa isn't doing that so much as simply saying that the problem is well-known. | ||
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moriya
United States54 Posts
What plexa used is f+g limit =lim f+ lim g, which is stronger. Just like using laws of cosine which is stronger to prove Pythagoras's Theorem. and i didnot see e-d at all so I feel its kinda BM. lol it is trivial and doesnot matter, sorry plexa. Maybe I get angry becoz i just get owned on USeast in some 3s4s game. ^_^ | ||
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Muirhead
United States557 Posts
if f and g limit both exist, then f+g limits exist without showing f+g limit =lim f+ lim g unlike in the Pythagorean Theorem case, which you can prove nicely without using the Law of Cosines. | ||
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moriya
United States54 Posts
What I mean is that law of cosines is based on pythagorean theorem, just like lim(f+g)=limf+limg is based on lim(f+g) exist. You cannot use the former to prove latter. But on the other hand, u guys are correct, the situation is delicate here because when proving lim(f+g) exist u automatically obtained lim(f+g)=limf+limg. lol i will go to see SPL playoff now. GL guys. | ||
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GreEny K
Germany7312 Posts
On July 19 2009 10:03 Muirhead wrote: Eh... I think it's cruel to equate asking for anlaysis help with asking for high school algebra help. The way I see it we're all here to help each other out here, and it's not easy to find people who know even moderately advanced math. Im just saying, plus it doesnt matter what level of math the question is it could be 2+2= ??? and it would still go against the rules. Just saying. | ||
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