I know im asking randomly but does anyone know of a trick to log back into an instagram account..? I somehow lost the login password. I appreciate any tips you can give me.
|bn-B| < |B|/2 means that the distance between bn and B is less than |B|/2. We can rewrite this as B-|B|/2 < bn < B+|B|/2. Basically bn is inside the interval (B-|B|/2, B+|B|/2). If B is positive, then B-|B|/2=B-B/2=B/2. Since that is the lower bound of the open interval, bn>B/2=|B|/2 If B is negative, then B+|B|/2=B-B/2=B/2. Since that is the upper bound of the open interval, bn
I'm still having trouble with #4. I understand that the limit doesn't exist if B = 0, but if I define a sequence of b values where at least one of the values is zero, but B is nonzero (for example, b[n] = 10-n), then even in the case where a[n] = 1, you have an individual value that is undefined (in this case a[10]/b[10]). It seems the restriction on the b sequence is stronger than just its limit being nonzero -- it can't contain _any_ zero values.
Many people will also say @ゴゴ Joji Joestar ゴゴ is correct. Here is how the two views come together: If b[n] =0 for some n, the sequence a[n]/b[n] is not well defined, so some would say it doesn't make sense to even speak of a limit for this not even defined sequence. However, if the limit of b[n] as n approaches infinity exists and is nonzero, b[n] can only equal 0 finitely many times. That means that there is an N so that b[n] is not zero for all n>=N. We can then consider the sequence a[n+N]/b[n+N] which you can think of as the sequence a[n]/b[n] but we have removed the first N terms. We can then take the limit of this new sequence, and its limit will be equal to the limit of a[n] divided by the limit of b[n] (provided both of these limits exist and the limit of b[n] is nonzero).
@@angelmendez-rivera351 This would sound correct, but I'm agreeing with Tracy H. You can't call a[n]/b[n] a sequence if for some values of n the result is undefined or infinite. If you allow infinity into the mix, then you break the rule than a sequence which converges is also bounded, and this unravels other parts of the presented proof.
No, we cannot because b_n is not equal to B for all n. Your way of reasoning is correct, but you have to say "b_n converges to B, so b_n is *close* to B, so |b_n| is bounded from above" (which is the bounded property)
@@vikaskalsariya9425 yep, sometimes it is probably easier I’d guess to do sessions of recording in different topics where he would record like 4 or so videos.
Well explained. More details than Rudin textbook for this topic I guess
Isnt it Rudin? Maybe you made a typo
Nathan Scott haha yes Rudin! That was a typo thanks for pointing out~
MathFlix Not a problem bro that happens
Abott>Rudin
Thanks for this insightful video on basic proofs of limt properties
I know im asking randomly but does anyone know of a trick to log back into an instagram account..?
I somehow lost the login password. I appreciate any tips you can give me.
I love all these kind of proofs!! I would very much appreciate a complete lhopital rule proof, there are many non rigorous out there
At 17:35 how did |bn-B|＜|B|/2 imply that |bn| >|B|/2
|bn-B| < |B|/2 means that the distance between bn and B is less than |B|/2.
We can rewrite this as B-|B|/2 < bn < B+|B|/2. Basically bn is inside the interval (B-|B|/2, B+|B|/2).
If B is positive, then B-|B|/2=B-B/2=B/2. Since that is the lower bound of the open interval, bn>B/2=|B|/2
If B is negative, then B+|B|/2=B-B/2=B/2. Since that is the upper bound of the open interval, bn
@@beaumaths a
Because of the inverse triangle inequality: | |x| - |y| |
Epsilon delta stuff is so cool. Very nice video.
a cool use of epsilon delta stuff are Cesaro-style theorems / lemmas.
I'm still having trouble with #4. I understand that the limit doesn't exist if B = 0, but if I define a sequence of b values where at least one of the values is zero, but B is nonzero (for example, b[n] = 10-n), then even in the case where a[n] = 1, you have an individual value that is undefined (in this case a[10]/b[10]). It seems the restriction on the b sequence is stronger than just its limit being nonzero -- it can't contain _any_ zero values.
Correct. a_n/b_n is not even defined if b_n=0 for any n.
Many people will also say @ゴゴ Joji Joestar ゴゴ is correct. Here is how the two views come together: If b[n] =0 for some n, the sequence a[n]/b[n] is not well defined, so some would say it doesn't make sense to even speak of a limit for this not even defined sequence.
However, if the limit of b[n] as n approaches infinity exists and is nonzero, b[n] can only equal 0 finitely many times. That means that there is an N so that b[n] is not zero for all n>=N. We can then consider the sequence a[n+N]/b[n+N] which you can think of as the sequence a[n]/b[n] but we have removed the first N terms.
We can then take the limit of this new sequence, and its limit will be equal to the limit of a[n] divided by the limit of b[n] (provided both of these limits exist and the limit of b[n] is nonzero).
@@angelmendez-rivera351 This would sound correct, but I'm agreeing with Tracy H. You can't call a[n]/b[n] a sequence if for some values of n the result is undefined or infinite. If you allow infinity into the mix, then you break the rule than a sequence which converges is also bounded, and this unravels other parts of the presented proof.
At 10:42, can we just choose |bn| = B since it converges to B, instead of using the bounded property and choose |bn| < M?
No, we cannot because b_n is not equal to B for all n. Your way of reasoning is correct, but you have to say "b_n converges to B, so b_n is *close* to B, so |b_n| is bounded from above" (which is the bounded property)
@@mohamedaminehadji6415 Thank you! Actl after one year as a Uni student, I kinda got that down alr. But thanks anyways!
20:41
Why are some of the videos private in the Real Analysis playlist?
Loved the video btw!
he records the videos in bunch and releases 1 per day. Almost every math channel does this.
@@vikaskalsariya9425 yep, sometimes it is probably easier I’d guess to do sessions of recording in different topics where he would record like 4 or so videos.
thank you
In 11:04, i think you have to exclude the case when A=0
It’s okey you’ve mentioned it later 😅😅
5:40
I dont't get why |an - A| + |bn - B| < ε.
Edit: Ok, now I see he is working in reverse.
Thankhs Pro
For the last result why not just use the fact that bn is bounded
How would you use it? Share your proof and we can discuss.
CAN YOU EXPLAIN HOW YOU SEE VERY EASILY THAT |Bn-Ｂ｜＜Ｂ/2 IMPLIES THAT |Bn| >|B|/2
Draw a picture. If the |bn| is within |B|/2 of |B| you can observe that the |bn| must be greater than |B|/2
By the triangle difference inequality we have ||b_n|-|B||
Yo 🤪from India.... Thanx sir