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@nullspace_xxii. Жыл бұрын
your intuitive explanations are a force to be reckoned with!
@WrathofMath Жыл бұрын
I do my best - thanks for watching!
@XahhaTheCrimson2 жыл бұрын
These days I review my undergraduate mathematics and this video is just perfect for that.... Thank you for clear explanation!
@WrathofMath2 жыл бұрын
Glad to help, thanks for watching!
@Joshua748012 жыл бұрын
using this for my grad class. thanks
@_kainya3 жыл бұрын
Thank you so much for this video, it sure helped me a lot!!! Your channel is highly underrated and your content is great! Your videos are really good!! I'm glad I found your channel!
@WrathofMath2 ай бұрын
Thank you!
@vernitasutton42152 жыл бұрын
Your videos help me understand the writings of my real analysis book. You should totally write a text!
@WrathofMath2 жыл бұрын
Thanks so much, I'm glad to hear that! One day I'd love to write a text, though I wouldn't start with Real Analysis. Currently, my analysis playlist is mostly based on Real Analysis by Jay Cummings and Understanding Analysis by Abbott. I recommend both strongly!
@johnvonneumannsdaddy820711 ай бұрын
tysm, my book proved the theorem in a really interesting way, but this clarifies it so much!
@WrathofMath11 ай бұрын
You're very welcome!
@A7medK-wb5lk Жыл бұрын
A great video! What drawing app did you use in this video?
@oyeajugando72626 ай бұрын
Alabado seas!!! Te amoooo, me re ayudaste, Gracias!!!
@WrathofMath6 ай бұрын
¡De nada! ¡Muchas gracias por mirar!
@lunkadapoorv9853 жыл бұрын
Why the assumption of monotonicity in monotone convergence theorem is necessary? Can you give me a detailed answer of this?
@WrathofMath3 жыл бұрын
Thanks for watching and good question! Monotonicity is necessary because if a sequence is only bounded, it may not converge. Bounded sequences can still behave strangely - they can oscillate. Consider (-1)^n. This sequence is bounded below by -1 and above by 1, but it does not converge since it isn't monotone, it oscillates between 1 and -1. If a sequence is bounded, then it being monotone forces it to get closer and closer to its bound, and so we get convergence!
@lunkadapoorv9853 жыл бұрын
@@WrathofMath ohh that was a wonderful explanation sir. Thank you so much, you made my day.
@FullerHob3 жыл бұрын
We use the completeness property of the reals to conclude that sup(a n) exists when (a n) is increasing, but what do we use to conclude inf(a n) exists when the sequence is decreasing?
@WrathofMath3 жыл бұрын
Thanks for watching and great question! We can use the completeness axiom to prove that a nonempty subset of the reals that is bounded below has an infimum. So I was taking that result for granted. It will be the next lesson I record, so give the proof a try yourself - it's fairly straightforward with some contradiction - and I'll reply with a link to the lesson when it is out!
@FullerHob3 жыл бұрын
@@WrathofMath Thanks a lot for responding. Great video btw, you deserve a lot more views :-) I ended up working around it by taking a bounded increasing sequence that we just proved converges and timsing it by -1 so that it decreases, then concluding it converges to a limit which is the same as the increasing sequence’s limit timesed by -1 using the algebraic limit theorem
@vedantsinha7374 Жыл бұрын
you're a life saver. Thank You
@WrathofMath Жыл бұрын
Glad to help, thanks for watching!
@rayxxkaiser3586 Жыл бұрын
i don't understand for the summation of 1/n, where n is an integer from 1 to infinite. While ther series of 1/n to infinite decreases and tends to zero, the Integral test tell us the summation of 1/n is smaller than ln(n) - ln(1) = infinite. However, it fulfills the monotone convergence theoerm conditions, bounded and decreasing. Have I already confused these Mathematic ideas? I'm learning Integral and Series, so please help......many thanks
@thomaspickin9376 Жыл бұрын
If you're looking at the infinite series 1/n (the summation), the sequence you're looking at is not (1, 1/2, 1/3, 1/4, 1/5, ... ), this just tells you what happens as n->inf in 1/n, not in a summation. This is monotone decreasing and bounded and so converges. The sequence you should be looking at is (1, 1+1/2, 1+1/2+1/3, 1+1/2+1/3+1/4, 1+1/2+1/4+1/5, ...), also known as the sequence of partial sums, this tells you what is happening in the sum. This is increasing and unbounded so diverges.
@JTinfected Жыл бұрын
Thank you soooo much sir🙏
@antonincasadebaig29223 жыл бұрын
Very helpful! thanks
@WrathofMath3 жыл бұрын
Glad to hear it, thanks for watching! If you're looking for more real analysis, check out my analysis playlist and let me know if you have any video requests! kzbin.info/aero/PLztBpqftvzxWo4HxUYV58ENhxHV32Wxli
@a.nelprober49712 жыл бұрын
Bounded does not imply convergent surely? Consider (-1)^n
@WrathofMath2 жыл бұрын
Correct, that is why the monotone convergence theorem also requires a function be monotone.
@a.nelprober49712 жыл бұрын
@@WrathofMath i would be so lost without your videos
@WrathofMath2 жыл бұрын
Glad to help! Let me know if you have any video requests!
@music_lyrics-ni7ks7 ай бұрын
MONOTONE bounded. The sequence you mentioned, oscillates between 1 and -1, and hence cannot be monotone.
@krasimirronkov173 жыл бұрын
Didnt really understand why for increasing sequence the limit should be the supremum
@bobek61553 жыл бұрын
If the increasing sequence is unbounded -> diverges then the limit is infinity and so is the supremum. Bounded increasing sequence converges to something, that something is the limit therefore the limit is "the boundary" = supremum. Look up the definition of a supremum and it should become pretty clear.