Proving All the Sequence Limit Laws | Real Analysis

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  • เผยแพร่เมื่อ 18 ธ.ค. 2024

ความคิดเห็น • 67

  • @adoxographer
    @adoxographer 2 ปีที่แล้ว +9

    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.

    • @WrathofMath
      @WrathofMath  2 ปีที่แล้ว +2

      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.

    • @adoxographer
      @adoxographer 2 ปีที่แล้ว

      @@WrathofMath Thanks for the recommendation. I've just ordered a couple. They are affordable too!

  • @ndafapheinrich7405
    @ndafapheinrich7405 8 หลายเดือนก่อน +3

    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......,,👍👍👍

  • @vh5244
    @vh5244 29 วันที่ผ่านมา +2

    6:48 this guy is wayyy ahead of his time

  • @CantTouchMe7
    @CantTouchMe7 หลายเดือนก่อน

    It feels so good to have so many proofs in video format. Thank you man!

  • @dieselguitar1440
    @dieselguitar1440 3 ปีที่แล้ว +17

    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).

    • @WrathofMath
      @WrathofMath  3 ปีที่แล้ว +12

      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!

    • @Channel-zb1fi
      @Channel-zb1fi 11 หลายเดือนก่อน +1

      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.

    • @Channel-zb1fi
      @Channel-zb1fi 11 หลายเดือนก่อน

      @@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|

    • @Channel-zb1fi
      @Channel-zb1fi 11 หลายเดือนก่อน

      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|

  • @tshepangjohnmolatedi7425
    @tshepangjohnmolatedi7425 2 ปีที่แล้ว +9

    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

    • @WrathofMath
      @WrathofMath  2 ปีที่แล้ว +4

      That's awesome to hear! Thanks for watching!

    • @tshepangjohnmolatedi7425
      @tshepangjohnmolatedi7425 2 ปีที่แล้ว +3

      @@WrathofMath it has been more than a pleasure 🔥Please keep impacting others by just doing what you do, all the best in your endeavours

  • @malawigw
    @malawigw 2 ปีที่แล้ว +2

    This video is pure golden gold!

  • @SacredSphynx
    @SacredSphynx ปีที่แล้ว +1

    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

    • @WrathofMath
      @WrathofMath  ปีที่แล้ว +1

      So glad to help - thanks for watching and good luck!

  • @alishash5497
    @alishash5497 11 หลายเดือนก่อน +1

    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. ❤

    • @vixguy
      @vixguy 2 หลายเดือนก่อน +1

      We can replace c and |c|

    • @CautionRamen
      @CautionRamen 20 วันที่ผ่านมา

      I believe that C is always considered positive, but as the other comment says using absolute value works too

  • @SILE75
    @SILE75 8 หลายเดือนก่อน

    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.

  • @Dupamine
    @Dupamine ปีที่แล้ว +1

    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
      @WrathofMath  ปีที่แล้ว +1

      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!

    • @Dupamine
      @Dupamine ปีที่แล้ว

      @@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

    • @WrathofMath
      @WrathofMath  ปีที่แล้ว +2

      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!

  • @cahlilthomas8187
    @cahlilthomas8187 ปีที่แล้ว +2

    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?

    • @WrathofMath
      @WrathofMath  ปีที่แล้ว +3

      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.

    • @cahlilthomas8187
      @cahlilthomas8187 ปีที่แล้ว +1

      @@WrathofMath thank you!

  • @christinesustraido3272
    @christinesustraido3272 3 หลายเดือนก่อน

    The best explanation I ever heard. Thank you!!

    • @WrathofMath
      @WrathofMath  3 หลายเดือนก่อน

      Glad it was helpful - thanks for watching!

  • @akihayakawa788
    @akihayakawa788 11 หลายเดือนก่อน

    soooo helpful i’d much rather watch these vids than read my lecturers notes lol u explain it more clearly tysm!!!

    • @WrathofMath
      @WrathofMath  11 หลายเดือนก่อน

      Happy to help!

  • @LianZhen-f3k
    @LianZhen-f3k 8 หลายเดือนก่อน +1

    Amazing, so good.

  • @begum9591
    @begum9591 ปีที่แล้ว +1

    Thank you very much for the detailed explanations of the proofs. It helped a lot.

    • @WrathofMath
      @WrathofMath  ปีที่แล้ว

      Awesome to hear, thanks for watching!

  • @clydefrog6974
    @clydefrog6974 2 ปีที่แล้ว

    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|

  • @okikiolaotitoloju2208
    @okikiolaotitoloju2208 ปีที่แล้ว +2

    Quick question how would you prove the limit law for {an^p} -> a^p?

  • @SimchaWaldman
    @SimchaWaldman 3 ปีที่แล้ว

    26:16 Here I choose an L > 0 such that both |a| < L and |bₙ| < L.

    • @WrathofMath
      @WrathofMath  3 ปีที่แล้ว +2

      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!

  • @haroldkim3479
    @haroldkim3479 2 ปีที่แล้ว

    it was so heplful thanks from korea

  • @ralph3295
    @ralph3295 11 หลายเดือนก่อน

    brilliant explanation. saving my January exams

    • @WrathofMath
      @WrathofMath  11 หลายเดือนก่อน

      Glad it helped!

  • @Jewisaosj
    @Jewisaosj 2 หลายเดือนก่อน

    when dealing with functions, is delta also have to be the max of delta 1 and delta 2? or minimum of delta 1 and 2?

  • @percy748
    @percy748 3 ปีที่แล้ว

    That was really helpful, thanks
    Love from India ❤️

    • @WrathofMath
      @WrathofMath  3 ปีที่แล้ว +1

      So glad it was helpful! Thanks a lot for watching and much love back from the USA!

  • @imeldatorres9386
    @imeldatorres9386 ปีที่แล้ว

    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

  • @sportmaster2586
    @sportmaster2586 2 ปีที่แล้ว

    You are helping me so much with university work - I really hope these videos get more views! : )

    • @WrathofMath
      @WrathofMath  2 ปีที่แล้ว +1

      Thank you, I'm glad to help!

  • @kartiku_
    @kartiku_ 2 ปีที่แล้ว

    This helped a lot ,thanks.

  • @aliakutami
    @aliakutami หลายเดือนก่อน

    Thank you

  • @SILE75
    @SILE75 8 หลายเดือนก่อน

    Do we have to show the scratch work in an exam so that the lecturer knows why N=max{N1,N2}?

  • @grande_schuylor6095
    @grande_schuylor6095 5 หลายเดือนก่อน

    Thank you!

    • @WrathofMath
      @WrathofMath  5 หลายเดือนก่อน

      You're welcome! thanks for watching!

  • @Leyla-et9nh
    @Leyla-et9nh 3 ปีที่แล้ว +1

    Could you show the same things for a bounded sequence an and bn

    • @WrathofMath
      @WrathofMath  3 ปีที่แล้ว +1

      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?

    • @Leyla-et9nh
      @Leyla-et9nh 3 ปีที่แล้ว

      @@WrathofMath yes, if an and bn are bounded then are the sum, product subtraction, of an and bn are also bounded

    • @WrathofMath
      @WrathofMath  3 ปีที่แล้ว +1

      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.

    • @Leyla-et9nh
      @Leyla-et9nh 3 ปีที่แล้ว

      @@WrathofMath thank you very much💕

  • @murongning9663
    @murongning9663 7 หลายเดือนก่อน

    u saved my ass fr
    come teach at my college plz

    • @WrathofMath
      @WrathofMath  7 หลายเดือนก่อน +1

      Glad to help, I have to stay put though - gotta keep making math videos!

  • @turokg1578
    @turokg1578 ปีที่แล้ว

    18:48 can't we just add 1 to the denominator and fix this problem?

    • @WrathofMath
      @WrathofMath  ปีที่แล้ว +1

      Sure, that’d work too!

  • @like_that4966
    @like_that4966 8 หลายเดือนก่อน

    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 :))

  • @thomasjefferson6225
    @thomasjefferson6225 ปีที่แล้ว

    Do you have a video like this but for functions?

    • @WrathofMath
      @WrathofMath  ปีที่แล้ว

      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.