fantastic video! I'm finding that in reading proofs online or trying to prove them myself i'm left feeling hopeless in where the aspects of the proof come from like why use this law and not another - why create this inequality and not another - etc. This was a great explanation of both the proof and motivations of the proof so it solved this issue for me - thank you!
Great Video! Well explained and really well made! Also a quick question, wouldn't this method of proof require that we also proof the Limit Product Laws for convergent sequences? I've seen other proofs without that, but this seems like a lot more cleaner and simple proof. As always, thanks for the great video!
Thank you! Yes, I mention in the video that we use the limit product law which we previously proved (as in we did it in a previous video) and that the link is in the description. Here is a link to the proof of the limit product law: ruclips.net/video/Zvy860EOlVk/видео.html You can also check out this video which contains all the proofs of the basic sequence limit laws: ruclips.net/video/dWLhfYUb3wc/видео.html I am putting these lessons in their proper order as I upload them in my Real Analysis playlist: ruclips.net/p/PLztBpqftvzxWo4HxUYV58ENhxHV32Wxli Thanks for watching!
Then we get, (|bn - b|)/(|b|*|bn|) < (ε*|b|*|k|)/(|b|*|bn|). The numerator in the RHS of the inequality is larger than the denominator. We then need the |bn| term to cancel out so we're left with a strict inequality with only ε.
Thanks for watching and let's see if we can clear that up. Once you understand it conceptually, you may be unsettled by how much algebra we had to do to demonstrate such an obvious truth. We know b is nonzero, suppose it is 1. Then, if bn converges to 1, it gets arbitrarily close to 1, and so it certainly must exceed half of 1, otherwise it couldn't get arbitrarily close. The sequence may look like 0.4, 0.49, 0.499, 0.4999, 0.49999, for a while, but at some point it needs to start looking more like 0.9, 0.98, 0.999, 0.99989, otherwise it couldn't possibly converge to 1. Similarly, if b is -4, bn has to eventually be less than half of -4, and stay less than half of -4, since it converges to -4. But being less than half -4 means |bn| is greater than |(1/2)* -4|. So we use the absolute value to cover the positive and negative case at once. Hopefully that helps. It's simply that the sequence bn converges to b, so it's terms have to exceed half of b in order to get arbitrarily close. And of course, they might not exceed 0.5b for a while, but at some point they must exceed it and stay that way. Edit: Note when I say "bn must exceed half of 1" that's shorthand for saying there exists some N, so that |bn| > (1/2)b for all n > N.
My pleasure, thanks for watching and check out my analysis playlist if you're looking for more! Let me know if you have any questions. ruclips.net/p/PLztBpqftvzxWo4HxUYV58ENhxHV32Wxli
Not necessarily! For example, an = (1/n^3) and bn = (1/n^2). Even if the denominator is going to zero, the whole sequence could behave in various ways depending on how an relates to bn.
this is more satisfactory than that of abbott's approach. Thanks in advance!
Glad to help - thanks for watching!
fantastic video! I'm finding that in reading proofs online or trying to prove them myself i'm left feeling hopeless in where the aspects of the proof come from like why use this law and not another - why create this inequality and not another - etc. This was a great explanation of both the proof and motivations of the proof so it solved this issue for me - thank you!
Glad to help - thanks for watching!
Great Video! Well explained and really well made!
Also a quick question, wouldn't this method of proof require that we also proof the Limit Product Laws for convergent sequences? I've seen other proofs without that, but this seems like a lot more cleaner and simple proof.
As always, thanks for the great video!
Thank you! Yes, I mention in the video that we use the limit product law which we previously proved (as in we did it in a previous video) and that the link is in the description. Here is a link to the proof of the limit product law: ruclips.net/video/Zvy860EOlVk/видео.html
You can also check out this video which contains all the proofs of the basic sequence limit laws: ruclips.net/video/dWLhfYUb3wc/видео.html
I am putting these lessons in their proper order as I upload them in my Real Analysis playlist: ruclips.net/p/PLztBpqftvzxWo4HxUYV58ENhxHV32Wxli
Thanks for watching!
Thanks! The limit law videos are great.
Glad they've been helpful, thanks a lot, Tyler!
Wow. This really cleared it up
best explanation👍🏻 jazakallah Sir
Thank you!
There are some ambiguous fact which I'm yet to comprehend
Where were you when I was taking undergrad analysis??
How did you find the epsilon absolute of b squared over w
5:26 Can't you make |bn - b| smaller than ε*|b|*|k| where k is the upper bound of bn and you know |bn| < |k|?
Then we get, (|bn - b|)/(|b|*|bn|) < (ε*|b|*|k|)/(|b|*|bn|). The numerator in the RHS of the inequality is larger than the denominator. We then need the |bn| term to cancel out so we're left with a strict inequality with only ε.
I understand algebraically why |b_n| > |b| / 2 but not conceptually
Thanks for watching and let's see if we can clear that up. Once you understand it conceptually, you may be unsettled by how much algebra we had to do to demonstrate such an obvious truth. We know b is nonzero, suppose it is 1. Then, if bn converges to 1, it gets arbitrarily close to 1, and so it certainly must exceed half of 1, otherwise it couldn't get arbitrarily close. The sequence may look like 0.4, 0.49, 0.499, 0.4999, 0.49999, for a while, but at some point it needs to start looking more like 0.9, 0.98, 0.999, 0.99989, otherwise it couldn't possibly converge to 1.
Similarly, if b is -4, bn has to eventually be less than half of -4, and stay less than half of -4, since it converges to -4. But being less than half -4 means |bn| is greater than |(1/2)* -4|. So we use the absolute value to cover the positive and negative case at once. Hopefully that helps. It's simply that the sequence bn converges to b, so it's terms have to exceed half of b in order to get arbitrarily close. And of course, they might not exceed 0.5b for a while, but at some point they must exceed it and stay that way.
Edit: Note when I say "bn must exceed half of 1" that's shorthand for saying there exists some N, so that |bn| > (1/2)b for all n > N.
@@WrathofMath thanks for the detailed response, it is much clearer now!
i still don't get it.
because "∀ε>0", you can choose any radius. I doubt |bₙ| > ½|b| is always the case in anywhere
Thnx bro
My pleasure, thanks for watching and check out my analysis playlist if you're looking for more! Let me know if you have any questions. ruclips.net/p/PLztBpqftvzxWo4HxUYV58ENhxHV32Wxli
keep doing what you are doing , really good work
Thank you, I will!
👍
If b is zero then sequence an/ bn diverge?
Not necessarily! For example, an = (1/n^3) and bn = (1/n^2). Even if the denominator is going to zero, the whole sequence could behave in various ways depending on how an relates to bn.