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
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.
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.
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
Michael, at 12:50, did you mean to say “contrapositive” instead of “converse”?
Life would be so easy if the converse were true....
Yes, I should have said "contrapositive". Thanks for catching this!
Binod
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 UA-cam. Great work
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
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.
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.
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
You are absolutely right :)
Referencing Abbott I believe the finite sum should be a_m+1 +...+a_n rather than a_m+1 +...+ a_m+n
@@davidmoss9926 I believe so as well.
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.
I am making these videos to support a full semester course in Real Analysis that I am teaching this Fall.
Binod
Can you also make more videos on advanced topics in Real Analysis such as the Contraction Principle, and so forth?
Best explanation!
Why if we set n=m+1, the sum collapses shouldn’t it be |(a_m+1 +a_m+2+.....a_2m+1)|?
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|.
Incredible stuff! Thanks.
12:20, appreciate the effort 😂
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
You say its koshi but can you parve it?
XD I'm wheezing.
14:37
Binod
14:36