Just wanted to say these videos are incredibly helpful. If I need a refresher on something I can pop in for 20 minutes and get a cogent, informative lecture. Really appreciate your work!
I am self-studying real analysis using a book, Understanding Analysis, and there are some points here and there that sort of lack clear explanations in the book which really confuses me sometimes. But your are making things super clear and explain things in the terms that are easily comprehensible for someone like me who is not used to the language of mathematics in general, so I cannot appreciate enough Michael! Thank you so much!!
I definietly agree with Gabriella. Good spot! The logic in the 3rd example is definitely flawed in the way Mat presents it. Some assumptions are made in the sketchwork which Mat has presented as logic because they happen to pan out in the proof. But it's really not precise enough to say that epsilon and n are different mathematical objects (ie. integer and real number) and so the rules relating them via the operators do not apply. For the mathematician to claim this, there must be a supporting Theorem for which ΕχιΜιμζ has presented none. So in my opinion the proof works, it just was arrived at via a bit more luck than presented in the video. It would be great to hear from Mat about this. Maybe me and Gabriella are missing something, but we won't know for sure without a but more of an explanation from Mat.
He's doing it backwards. He's showing that (2-n)/(n^2) < epsilon. This would mean that (2-n)/(n^2 +n) < epsilon. Since ,the first expression is greater than the second. He's proving 7 < 8 which implies that 5 < 8 since 5 < 7.
18:10 you don't even have to worry about the maximum, just pretend your sequence doesn't have the first term, I.e. it starts at n=2, because for any convergent sequence, the new sequence achieved by eliminating the first k elements will have the same limit since the rest of the infinitely many terms behave the same way anyway. More precisely lim of a(n) = lim of a(n+k) for any natural number k, which can easily be proven
Remember although epsilon is supposed to be very small it still needs to be positive and "large" enough to contain the absolute term. Put it in another way if you instead makes the absolute term smaller then you choose the epsilon to be bigger than your final smaller absolute term, then the epsilon you choose can't contain the original absolute term because it's too small.
Thank you so much for your explanation it was amazing. I would improve my skill in epsilon N definition could you please give me some advice or where I could find some exercise with solution if it is possible
Thanks for your videos. Speaking of limits of sequences, would it be possible that, in a future video, you address the limit of S(x)=\sum_{n=0}^\inf (-1)^n x^{2^n}, when x tends to 1 from below?
I wondered that as well but it turns out the ceiling function does not suffice: The ceiling function does not change integer values (e.g. ⌈9⌉ = 9) but we want a strict inequality (e.g. n > 1/Ɛ²). To elaborate on that: At around 5:00 we see that we need n > 1/Ɛ² to be held by all n >= N. Then in the proof we could have said N = ⌈1/Ɛ²⌉ (without the +1). If now we choose e.g. Ɛ = 1/3 our N would be exactly N = ⌈1/(1/3)²⌉ = ⌈9⌉ = 9. Therefore our smallest n is also n = 9 but since 1/Ɛ² = 9 the inequality n > 1/Ɛ² does not hold.
So am I wrong or does this definition say "if after some point (N) in a sequence (a_n within real numbers) the value of the sequence is always contained within some 'arbitrarily small' region (epsilon) that continuously approaches a value (L), then the limit as a point in the sequence approaches infinity is equal to the value being approached by the epsilon region" and the whole n >= N and N = ceil(something containing epsilon) + 1 business is to 'seed out' this point N? Also, why does it matter that N is a natural number? I hope I am at least on the right track in terms of a starting point
I think of it as insurance that you are far enough out into the sequence. He mentions that its not super necessary if you use a strict inequality, but considering some books do and some books don't adding 1 makes it certain you will be sufficiently "deep"
In one of the proofs you choose N = ceil(15/(4.epsilon) - 5/2) + 1. But that doesn't guarantee that ceil(15/(4.epsilon) - 5/2) is positive. I think you should add +3 instead of +1?
I’m a bit confused about how choosing L (the limit value) cannot be just some arbitrary number. Would the proof not work if the incorrect L was chosen?
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Just wanted to say these videos are incredibly helpful. If I need a refresher on something I can pop in for 20 minutes and get a cogent, informative lecture. Really appreciate your work!
I can second this. Michael's channel has been a great supplement to reading 'Understanding Analysis' by Abbott.
I am self-studying real analysis using a book, Understanding Analysis, and there are some points here and there that sort of lack clear explanations in the book which really confuses me sometimes. But your are making things super clear and explain things in the terms that are easily comprehensible for someone like me who is not used to the language of mathematics in general, so I cannot appreciate enough Michael! Thank you so much!!
Great professor.Your explanation is so clear!
I am Sam ,always watching you from Ghana .. I am huge fan of your real analysis lectures ... it is really preparing me for my exams 😊
19:17
i don't get it, 2-n/(n²+n)
Epsilon is an infinitesimal value and (2-n)/(n^2) is a rational number, read from bottom to up or observe the following proof stated by Michael.
im still confused can anyone be more elaborate ?
I definietly agree with Gabriella. Good spot! The logic in the 3rd example is definitely flawed in the way Mat presents it. Some assumptions are made in the sketchwork which Mat has presented as logic because they happen to pan out in the proof. But it's really not precise enough to say that epsilon and n are different mathematical objects (ie. integer and real number) and so the rules relating them via the operators do not apply. For the mathematician to claim this, there must be a supporting Theorem for which ΕχιΜιμζ has presented none. So in my opinion the proof works, it just was arrived at via a bit more luck than presented in the video. It would be great to hear from Mat about this. Maybe me and Gabriella are missing something, but we won't know for sure without a but more of an explanation from Mat.
He's doing it backwards. He's showing that (2-n)/(n^2) < epsilon. This would mean that (2-n)/(n^2 +n) < epsilon. Since ,the first expression is greater than the second. He's proving 7 < 8 which implies that 5 < 8 since 5 < 7.
18:10 you don't even have to worry about the maximum, just pretend your sequence doesn't have the first term, I.e. it starts at n=2, because for any convergent sequence, the new sequence achieved by eliminating the first k elements will have the same limit since the rest of the infinitely many terms behave the same way anyway. More precisely lim of a(n) = lim of a(n+k) for any natural number k, which can easily be proven
Also, since the ceiling of 1/epsilon is greater than or equal to 1, N is greater than or equal to 2. So we don’t need to worry about that regardless.
YEE!! The best explanation I have seen till now.
In the second example I would just choose N = ⌈15/4𝛆⌉. You only need to find one such N and not necessarily the best one.
I appriciate your effort and work really much. You do a difference because you are SO great at explaining things. :-)
If you took a strict inequality in the definition of the limit, "for all n>N", you wouldn't need to constantly deal with this N=ceiling+1 thing, no?
great teacher
15:20 can someone help me understand how making the abs(An - L) term larger will preserve the fact that it is less than epsilon?
Because if |1/n| > |(2- n) / (n^2+n)|
And n > 1/epsilon
Then | 1/ 1/e | = e > | an -1 |
Remember although epsilon is supposed to be very small it still needs to be positive and "large" enough to contain the absolute term.
Put it in another way if you instead makes the absolute term smaller then you choose the epsilon to be bigger than your final smaller absolute term, then the epsilon you choose can't contain the original absolute term because it's too small.
6:15 i think the floor( x) N should be enough to say that n> 1/epsilon^2
I think you got floor and ceiling mixed up.
15:20 can't understand why this inequality holds
Thank you so much for your explanation it was amazing.
I would improve my skill in epsilon N definition could you please give me some advice or where I could find some exercise with solution if it is possible
thank you math Neil Patrick Harris
Great work, Mic!
Which book you recommend to use to study this topic?
Thanks for your videos. Speaking of limits of sequences, would it be possible that, in a future video, you address the limit of S(x)=\sum_{n=0}^\inf (-1)^n x^{2^n}, when x tends to 1 from below?
Why add 1 to equality? I didn't get that.
Limit of several examples --> 19:22 = 3.
I'm a little confused why you need to add + 1 in order to guarantee N sufficiently big. Shouldn't the ceiling function suffice?
I wondered that as well but it turns out the ceiling function does not suffice:
The ceiling function does not change integer values (e.g. ⌈9⌉ = 9) but we want a strict inequality (e.g. n > 1/Ɛ²).
To elaborate on that:
At around 5:00 we see that we need n > 1/Ɛ² to be held by all n >= N.
Then in the proof we could have said N = ⌈1/Ɛ²⌉ (without the +1).
If now we choose e.g. Ɛ = 1/3 our N would be exactly N = ⌈1/(1/3)²⌉ = ⌈9⌉ = 9.
Therefore our smallest n is also n = 9 but since 1/Ɛ² = 9 the inequality n > 1/Ɛ² does not hold.
@@aboyhya612 Could you point out what part of the proof you mean. I don't know what you're refering to.
@@DarkCloud7 Thank you, actually i got what i want.
So am I wrong or does this definition say "if after some point (N) in a sequence (a_n within real numbers) the value of the sequence is always contained within some 'arbitrarily small' region (epsilon) that continuously approaches a value (L), then the limit as a point in the sequence approaches infinity is equal to the value being approached by the epsilon region" and the whole n >= N and N = ceil(something containing epsilon) + 1 business is to 'seed out' this point N? Also, why does it matter that N is a natural number? I hope I am at least on the right track in terms of a starting point
Why do we need to add a 1 next to the (1 over epsilon)? Why do we need to make sure it is big enough? Thanks in advance!
I think of it as insurance that you are far enough out into the sequence. He mentions that its not super necessary if you use a strict inequality, but considering some books do and some books don't adding 1 makes it certain you will be sufficiently "deep"
In one of the proofs you choose N = ceil(15/(4.epsilon) - 5/2) + 1. But that doesn't guarantee that ceil(15/(4.epsilon) - 5/2) is positive. I think you should add +3 instead of +1?
Yes. It would be simpler to get rid of 5/2 from the start. Or just say: take N strictly greater than the quantity you need, then for all n>=N ...
I don't get why he adds the +1. Doesn't the proof work with 1/eps^2 ?
Please take an Indian jee question.
Didn't we need to take floor rather than the ceiling in the first question ❓
Floor+1 works but ceiling+1 also works. I guess he chose ceiling to avoid showing floor(x)+1>x, as ceiling(x)+1>x is obvious.
How's (2 -n) equal to (n - 2)
Its absolut value is.
🔥🔥🔥
ok great
👍😀
Find a,b,c if a+b+c=13 and a×b×c=56
7, 4, 2
Right 👍
What is real analysis, please someone tell me ?
basically the proof of calculus, carefully checking why things are true
Illuminati intensifies
I’m a bit confused about how choosing L (the limit value) cannot be just some arbitrary number. Would the proof not work if the incorrect L was chosen?
It wouldn't work