Real Analysis | Cauchy Criterion for Series

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Michael Penn

Michael Penn

Күн бұрын

Пікірлер: 28
@nicholasroberts2933
@nicholasroberts2933 4 жыл бұрын
Michael, at 12:50, did you mean to say “contrapositive” instead of “converse”?
@minime1235able
@minime1235able 4 жыл бұрын
Life would be so easy if the converse were true....
@MichaelPennMath
@MichaelPennMath 4 жыл бұрын
Yes, I should have said "contrapositive". Thanks for catching this!
@BinodTharu-co4by
@BinodTharu-co4by 4 жыл бұрын
Binod
@MrCentrax
@MrCentrax 4 жыл бұрын
Found your chanell when I was studying some basic number theory and I have to say your's one of the best math channels on KZbin. Great work
@adrianamor8472
@adrianamor8472 4 жыл бұрын
The theorem statement is just wrong. If you are considering both m,n >= N then you should have |a_n+1 + ... + a_m| < eps. The alternative version is to consider n >= N and p >= 1, then you have |a_n+1 + ... + a_n+p| < eps. This statement is mixing the two above, which is just not Cauchy criterion (perhaps still holds but i'd consider it wrong). Also, Just take p = 1 for the corollary proof, taking n = m + 1 doesn't give you a_n, it gives you a_m+1 + ... + a_(2m+1). Taking p=1 you get |a_n+1| < eps which is not |a_n| but you could just pick K = N+1 and then for all n >= K => |a_n| < eps
@thesecondderivative8967
@thesecondderivative8967 Жыл бұрын
14:13 I think the sequence of terms in the absolute value should end in n and not m+n. That way, it would collapse.
@mohameddaoud4885
@mohameddaoud4885 4 жыл бұрын
14:15 why would it collapse to |a_n|, I believe it should be |a_n, a_n+1, a_n+2, ...... a_2n-1|? Hope to clear my confusion. Thank you for everything
@stefanschroder4694
@stefanschroder4694 4 жыл бұрын
You are absolutely right :)
@davidmoss9926
@davidmoss9926 4 жыл бұрын
Referencing Abbott I believe the finite sum should be a_m+1 +...+a_n rather than a_m+1 +...+ a_m+n
@thesecondderivative8967
@thesecondderivative8967 Жыл бұрын
​@@davidmoss9926 I believe so as well.
@EmilianoPM6754
@EmilianoPM6754 4 жыл бұрын
There is an error in the theorem, as n does not necessarily have to be greater than N given the form of the theorem that is written. It would be fine if the sum inside the absolute values ended in a_n and not a_(n+m). In the form given we should only ask for n>=1.
@thenewest1
@thenewest1 4 жыл бұрын
Hey Michael, big fan of the RA series, was wondering how far you were planning on going with this series e.g. just a few videos or maybe following a textbook. I'm Preparing for the math GRE and brushing up on RA with your videos is a godsend.
@MichaelPennMath
@MichaelPennMath 4 жыл бұрын
I am making these videos to support a full semester course in Real Analysis that I am teaching this Fall.
@BinodTharu-co4by
@BinodTharu-co4by 4 жыл бұрын
Binod
@jimallysonnevado3973
@jimallysonnevado3973 4 жыл бұрын
Why if we set n=m+1, the sum collapses shouldn’t it be |(a_m+1 +a_m+2+.....a_2m+1)|?
@jimallysonnevado3973
@jimallysonnevado3973 4 жыл бұрын
There is a mistake in the index of the stated theorem, it should be |a_m+1+.....+a_n| not |a_m+1....+a_m+n|.
@asht750
@asht750 3 жыл бұрын
Can you also make more videos on advanced topics in Real Analysis such as the Contraction Principle, and so forth?
@victorserras
@victorserras 4 жыл бұрын
Incredible stuff! Thanks.
@thenewdimension9832
@thenewdimension9832 2 жыл бұрын
Best explanation!
@youssefbouhtouch6050
@youssefbouhtouch6050 4 жыл бұрын
I have a question : if we have two polynomials p and q in Q[x] and there exists a complex number z such that : p(z)=q(z)=0 (z is a root fo each of p and q) can we say that gcd(p,q) in Q[x] is not a constant and please why
@zakthayer9315
@zakthayer9315 Жыл бұрын
12:20, appreciate the effort 😂
@JB-ym4up
@JB-ym4up 4 жыл бұрын
You say its koshi but can you parve it?
@euclidtheorem4817
@euclidtheorem4817 4 жыл бұрын
XD I'm wheezing.
@________6295
@________6295 4 жыл бұрын
14:37
@BinodTharu-co4by
@BinodTharu-co4by 4 жыл бұрын
Binod
@goodplacetostop2973
@goodplacetostop2973 4 жыл бұрын
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