Honestly, all the confusions and stress about these theorems proofs,that was in my mind got washed out by your teaching 😊😊😊....keep it high my honoured......,,👍👍👍
So glad they've been helpful for you! I cannot recommend Jay Cummings' long form math textbooks enough. I have always tried to explain proofs lucidly, and his analysis text has definitely been a helpful guide for these videos.
I hear real analysis is known for being a very difficult course. I'm glad to have the opportunity to begin learning it with these detailed explanations before ever doing an analysis class. I'm gonna be so far ahead of my classmates (assuming I get accepted).
Thanks for watching and I hope you will get accepted! It is certainly a challenging course. Personally, I found both Calc II and III to be harder, but I'm more of a proof-minded guy so Real Analysis is definitely more my style anyways. I will be uploading many more videos to my Real Analysis playlist this year, I hope you will continue to find the explanations helpful. Whatever text your future course ends up using, I would strongly recommend Real Analysis by Jay Cummings as a fantastic text both in its quality of explanations and in its price. It's also very funny, and is in large part the book whose structure I am basing my analysis playlist on! Good luck!
Make fun of them when they struggle with concepts like limits, and solve crosswords during lectures, and act smart when your teacher gets upset at your inattentiveness.
@@WrathofMath I still don't fully understand why it's possible to use N = max(N1,N2) for the definition of the limit of the sum. Should N not be an even larger value to accomodate the fact that epsilon now bounds |a_n-a+b_n-b|0. In the case where I have the limit of the sum, I will have to look for an N(epsilon) which makes it so for all n>N then |a_n-a+b_n-b|
Also, is it possible to add the two definitions together, in the same way that one would add an inequality together. So if I have two definitions: 1: that for all epsilon greater than 0, there exists an N=max(N_1,N_2) such that for all n>=N, then |a_n-a|=N, then |b_n-b|=2N, then |a_n-a|+|b_n-b|
I cant tell you how insightful your videos have been🔥before i was trying so hard to have the defintions down but now I can literally write a proof without thinking twice
Honestly Ty so much, you are saving me from my uni lectures, i literally only know whats going on because of ur videos. Thanks so much. Wish u were my lecturer
Today i was feeling good for almost completing 30 videos . I have been studying this playlist for 2 months now. But you had to drop this 50 minute video haha ! This will be tough!
@@WrathofMath oh! How many more videos are left? ( approximately) ? I think there are around 70 videos right now. I thought I have to complete these 70 videos and I will be done with real analysis
If only! I'd recommend using a textbook, and this playlist for clarification where necessary. When it is done, it will certainly be more thorough than a typical course just because courses have time restrictions which I do not have in this playlist. The major topics I still have to cover are: Continuity, Differentiation, Integration, and Sequences and Series of Functions. I'd guess 200 videos or so when the essentials are all done. Hopefully will make a lot of progress this summer!
Thank you so much @Wrath of math for all your real analysis videos they are saving my life even if there are proofs in the study guides I always come here to understand them.
Hey! Thanks for the video. It was insightful. ❤ I have a question regarding it. At 30:13 , we added 1 to the 2c to make sure if c was 0, we didn't face the division by zero problem. However, can we say c could be -1/2, thus 2c = -1 and 2c + 1 = 0? And, we face the division by zero problem again... Is this right? And, how can we make sure it doesn't happen? Thank you again. ❤
Hey, thanks for the videos. They're much appreciated. When you say "we can make the absolute value of |a_n - a| as small as we want", what exactly do you mean by that?
What I mean, exactly, is that - given any epsilon > 0, there is an N sufficiently large such that |a_n - a| < epsilon for all n > N. Meaning if we go far enough in the sequence, we can make the terms as close to the limit as we desire.
at 27:28 cant we do just like we did in 18:48 (for both bn and a). That move also doesn't require" bn is converges to C" information ı guess? so the proof will be more short. if ım wrong please tell me.
Thanks for watching and I think I see what you mean, good idea! So you could replace them both with L, and then you'd replace |a_n - a| with ε/2L and |b_n - b| with ε/2L. That works just fine, and some might find it more slick since we then have two terms with the same denominator, and taking an L > 0 eliminates the need to have +1 in the denominator. Cool!
At 27:09 is it possible to just say since |a| is a constant, it converges by def as the lim(n->inf) a = a so then just claim |a| < 1/e and also have |bn-b|
Thank you very much for the videos you create. I am wondering if you can help me with this question. Determine if the sequence is convergent of divergent. ( 2+ (-1)^n)/n
The sum and subtraction of bounded sequences are both bounded, I'll leave you to try to prove it, it's pretty straightforward if you just give some names to the bounds! Products will not hold. For example, imagine one bounded sequence converging to 0 and another bounded sequence converging to 1. Quotients have similar problems.
Working on it! I'll be using the theorem connecting functional limits to sequential limits to prove these laws, and a proof of that theorem is covered in my next video releasing tonight.
Honestly, all the confusions and stress about these theorems proofs,that was in my mind got washed out by your teaching 😊😊😊....keep it high my honoured......,,👍👍👍
I can't thank you enough for these videos. You take the mystery out of proof writing in a way I have not seen anywhere else.
So glad they've been helpful for you! I cannot recommend Jay Cummings' long form math textbooks enough. I have always tried to explain proofs lucidly, and his analysis text has definitely been a helpful guide for these videos.
@@WrathofMath Thanks for the recommendation. I've just ordered a couple. They are affordable too!
I hear real analysis is known for being a very difficult course. I'm glad to have the opportunity to begin learning it with these detailed explanations before ever doing an analysis class. I'm gonna be so far ahead of my classmates (assuming I get accepted).
Thanks for watching and I hope you will get accepted! It is certainly a challenging course. Personally, I found both Calc II and III to be harder, but I'm more of a proof-minded guy so Real Analysis is definitely more my style anyways. I will be uploading many more videos to my Real Analysis playlist this year, I hope you will continue to find the explanations helpful. Whatever text your future course ends up using, I would strongly recommend Real Analysis by Jay Cummings as a fantastic text both in its quality of explanations and in its price. It's also very funny, and is in large part the book whose structure I am basing my analysis playlist on! Good luck!
Make fun of them when they struggle with concepts like limits, and solve crosswords during lectures, and act smart when your teacher gets upset at your inattentiveness.
@@WrathofMath I still don't fully understand why it's possible to use N = max(N1,N2) for the definition of the limit of the sum. Should N not be an even larger value to accomodate the fact that epsilon now bounds |a_n-a+b_n-b|0.
In the case where I have the limit of the sum, I will have to look for an N(epsilon) which makes it so for all n>N then |a_n-a+b_n-b|
Also, is it possible to add the two definitions together, in the same way that one would add an inequality together.
So if I have two definitions: 1: that for all epsilon greater than 0, there exists an N=max(N_1,N_2) such that for all n>=N, then |a_n-a|=N, then |b_n-b|=2N, then |a_n-a|+|b_n-b|
I cant tell you how insightful your videos have been🔥before i was trying so hard to have the defintions down but now I can literally write a proof without thinking twice
That's awesome to hear! Thanks for watching!
@@WrathofMath it has been more than a pleasure 🔥Please keep impacting others by just doing what you do, all the best in your endeavours
Honestly Ty so much, you are saving me from my uni lectures, i literally only know whats going on because of ur videos. Thanks so much. Wish u were my lecturer
So glad to help - thanks for watching and good luck!
Today i was feeling good for almost completing 30 videos . I have been studying this playlist for 2 months now. But you had to drop this 50 minute video haha ! This will be tough!
That's awesome - congrats on all your hard work! And I apologize the playlist is still far from done - but I am always working. Good luck!
@@WrathofMath oh! How many more videos are left? ( approximately) ? I think there are around 70 videos right now. I thought I have to complete these 70 videos and I will be done with real analysis
If only! I'd recommend using a textbook, and this playlist for clarification where necessary. When it is done, it will certainly be more thorough than a typical course just because courses have time restrictions which I do not have in this playlist. The major topics I still have to cover are: Continuity, Differentiation, Integration, and Sequences and Series of Functions. I'd guess 200 videos or so when the essentials are all done. Hopefully will make a lot of progress this summer!
This video is pure golden gold!
Thank you so much @Wrath of math for all your real analysis videos they are saving my life even if there are proofs in the study guides I always come here to understand them.
brilliant explanation. saving my January exams
Glad it helped!
soooo helpful i’d much rather watch these vids than read my lecturers notes lol u explain it more clearly tysm!!!
Happy to help!
Hey! Thanks for the video. It was insightful. ❤ I have a question regarding it. At 30:13 , we added 1 to the 2c to make sure if c was 0, we didn't face the division by zero problem. However, can we say c could be -1/2, thus 2c = -1 and 2c + 1 = 0? And, we face the division by zero problem again... Is this right? And, how can we make sure it doesn't happen? Thank you again. ❤
Quick question how would you prove the limit law for {an^p} -> a^p?
i'm gonna be in early middle school next year. time to confuse my teacher in my tests' explanation by taking out all the proofs I learnt :))
Amazing, so good.
Thank you very much for the detailed explanations of the proofs. It helped a lot.
Awesome to hear, thanks for watching!
Thank you!
You're welcome! thanks for watching!
Hey, thanks for the videos. They're much appreciated. When you say "we can make the absolute value of |a_n - a| as small as we want", what exactly do you mean by that?
What I mean, exactly, is that - given any epsilon > 0, there is an N sufficiently large such that |a_n - a| < epsilon for all n > N. Meaning if we go far enough in the sequence, we can make the terms as close to the limit as we desire.
@@WrathofMath thank you!
Do we have to show the scratch work in an exam so that the lecturer knows why N=max{N1,N2}?
it was so heplful thanks from korea
That was really helpful, thanks
Love from India ❤️
So glad it was helpful! Thanks a lot for watching and much love back from the USA!
at 27:28 cant we do just like we did in 18:48 (for both bn and a). That move also doesn't require" bn is converges to C" information ı guess? so the proof will be more short. if ım wrong please tell me.
26:16 Here I choose an L > 0 such that both |a| < L and |bₙ| < L.
Thanks for watching and I think I see what you mean, good idea! So you could replace them both with L, and then you'd replace |a_n - a| with ε/2L and |b_n - b| with ε/2L. That works just fine, and some might find it more slick since we then have two terms with the same denominator, and taking an L > 0 eliminates the need to have +1 in the denominator. Cool!
This helped a lot ,thanks.
Glad to hear it!
At 27:09 is it possible to just say since |a| is a constant, it converges by def as the lim(n->inf) a = a so then just claim |a| < 1/e and also have |bn-b|
Thank you very much for the videos you create. I am wondering if you can help me with this question. Determine if the sequence is convergent of divergent. ( 2+ (-1)^n)/n
u saved my ass fr
come teach at my college plz
Glad to help, I have to stay put though - gotta keep making math videos!
You are helping me so much with university work - I really hope these videos get more views! : )
Thank you, I'm glad to help!
Could you show the same things for a bounded sequence an and bn
Thanks for the question Leyla, could you specify exactly what you mean? Do you mean "If a_n and b_n are bounded then a_n + b_n is bounded?" And so on?
@@WrathofMath yes, if an and bn are bounded then are the sum, product subtraction, of an and bn are also bounded
The sum and subtraction of bounded sequences are both bounded, I'll leave you to try to prove it, it's pretty straightforward if you just give some names to the bounds! Products will not hold. For example, imagine one bounded sequence converging to 0 and another bounded sequence converging to 1. Quotients have similar problems.
@@WrathofMath thank you very much💕
18:48 can't we just add 1 to the denominator and fix this problem?
Sure, that’d work too!
Do you have a video like this but for functions?
Working on it! I'll be using the theorem connecting functional limits to sequential limits to prove these laws, and a proof of that theorem is covered in my next video releasing tonight.