Proof: Supremum of {n/(n+1)} = 1 | Real Analysis
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- Опубликовано: 22 авг 2024
- Today we prove the supremum of {n/(n+1)} is 1, using the Archimedean principle and the epsilon definition of supremum of a set.
The Real Analysis Playlist: • Real Analysis
Proof of Archimedean Principle: • Proof: Archimedean Pri...
Definition of Supremum and Infimum of a Set: • Definition of Supremum...
Epsilon Definition of Supremum and Infimum: • Epsilon Definition of ...
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Thanks for watching! Learn more in my Real Analysis playlist: ruclips.net/p/PLztBpqftvzxWo4HxUYV58ENhxHV32Wxli
thanks a lot. I messed up a real analysis quiz cuz i found the topic quite boring but now I discovered your channel and have a newfound interest in it. Might ace it this semester
That's really well explained and carried me through the first week of college
wow this was really awesome, im just starting out with real analysis and i found your general method of proofs to be super helpful!
Thanks, Filip! If you're looking for more real analysis, be sure to check out my playlist! ruclips.net/p/PLztBpqftvzxWo4HxUYV58ENhxHV32Wxli
Good luck as you continue your study and let me know if you ever have any questions! I have plenty more lessons to come in the analysis playlist. It is based largely on "Real Analysis" by Jay Cummings and "Understanding Analysis" by Stephen Abbott. And, to a lesser degree "A Friendly Introduction to Analysis" by Kosmala.
That first part of the explanation is so intuitive. Hope you make more on these advanced topics.
Thank you, I will
I wish my professor went over the details of proofs like this but he wanted to just show the definitions and let us explore as if we were pursuing a Ph.D in math. I almost tricked him and got him to give a solution to an exam problem one time lol. Anyway, great explanation.
Lol
Wow. This is amazing. Thank you so much!!!
You're very welcome, glad it helped! Be sure to check out the real analysis playlist, you may find more helpful videos there, and adding to it is my main focus for this year: ruclips.net/p/PLztBpqftvzxWo4HxUYV58ENhxHV32Wxli
Hey, yes I agree- you may be the best teacher I've found yet and I'm just saying-that's saying a lot. Even though, I don't understand any of it, there is hope that one day I might. In the meantime, could you please show a proof for what the sup and inf for e^x are? It seems like a basic question/proof but, it would be an amazing video I'm sure. Unless you already have one, but if you do- I can't find it. I love your videos. Keep going! You are a blessing to those of us who need things broken down so much.
also it is a cool thing how we have different perception of the math topics. I mean i have different school but we never did such approach to that topic. Cool one
Wow, cool sir am happy knowing this channel. Am new in Analysis. I wish we will ride together😅.
Watching from Africa❤
Thanks for the video : )
a) Can you say that S = {0.5, 2/3, 0.75....}, so the elements of S form an increasing sequence where lim as n-->infinity(1 + 1/n) = 1. So x x 0, 1 - epsilon > 1- 0.5epsilon which is an element of S, so 1 is the smallest upper bound of S. I don't think this is right though because we are only considering natural numbers N rather than the entire set of real numbers
when proving that for all epsilon>0 (e>0), 1-e is not an upper bound. would this work: we could fix e>0 but if e>1 then we just pick 1/2 which is in S to show an example of an element in S greater than 1-e, so we just fix 0
Thank you for your explanation. Really helpful.
Thanks for it , I understood it completely
How would you go about this problem when you have to find the sup/inf, not prove that a given one is indeed true? I have to find inf/sup of 2-(1/(n+1)) and i know that inf= 3/2 but i dont think I'm allowed to state this at the beginning of my proof since its not given. A reply would be very nice :)
Many thanks for your good videos.
No problem - looking forward to many more in 2021!
as always great video! i was just wondering, can I start the proof for second condition by choosing epsilon > 1/n+1 ? I have other problems but it couldn’t be broken down into n < …
It just doesnt feel like it’ll suffice since the second condition start with given any epsilon > 0. would choosing epsilon > something works?
Amazing video!
Thank you, Alexia! If you're looking for more analysis, check out my playlist!
ruclips.net/p/PLztBpqftvzxWo4HxUYV58ENhxHV32Wxli
hi
I am unable to understand your conclusion for the second part of the proof. Can you help me out?
Can you do the infimum of n/(n+1)
Hello sir, I'm happy that I have watched your video, but i have a problem of how I can find supA and infA given that A ={1/m -1/n}
am I right we could start that prooj from Archimedean principle and that would be clear one ? I mean no need for the first part except just to understand where it come from
Hello thanks for the vid (:
Why dont we just do limit of n/n+1 and n goes to + infinity and we will find it 1 ? Its not enough to prove that 1 is the sup? Thank you again
Great strategy
Supper teacher
Nice video keep it up
Thank you, I will!
Thank you very much.
My pleasure, thanks for watching! If you're looking for more analysis, check out my playlist and let me know if you ever have any video requests! ruclips.net/p/PLztBpqftvzxWo4HxUYV58ENhxHV32Wxli
Please can you help proof that Supremum {n/(2n+1), n= 1,2,3....}= 1/2 and infimum of same set = 1/3?
Thanks
hey do you know how to prove this without any use of limits, as in just by showing that any real number x with x < 1 is not an
upper bound, by finding an element of S larger than x?
love the vid by the way! just wondering if there is another way of proving it without epsilon :) xxx
Thanks for watching Jasmine! I'm not sure I understand your question because we do not use limits in this proof, and showing that any real number x < 1 is not an upper bound of the set is precisely what we did.
Any number less than 1 is certainly of the form 1 - ε for some positive ε, and we showed any number of that form is not an upper bound of the set, since there must exist a number n greater than 1/ε - 1, and such a number n will correspond to a term of the sequence n/(n+1) which is greater than 1- ε. In short, we show that any number less than 1 (which is obtained by subtracting some positive ε from 1) is not an upper bound of the set. Thus, since 1 IS an upper bound, it is the least upper bound and so is the supremum. Does that help?
If you want to avoid using ε you could just as well call it something else, but 1) that would be breaking with convention and 2) I know that is not what you're looking for haha!
thats so abstract !
Sir I usually take limits to infinity. How is it???
thankyou so much❤👌
You're very welcome! Thanks for watching and if you're looking for more real analysis, check out my analysis playlist: ruclips.net/p/PLztBpqftvzxWo4HxUYV58ENhxHV32Wxli
thank you
Glad to help!
How can n/(n+1) be equal to 1 the inequality sign is greater or equal too but it’s impossible for it to be equal to 1
The supremum doesn't need to be inside the set, it's just the 'least upper bound' , consider the open inteval A := (0, 2), sup(A) = 2, but 2 is not in the set A.
Thanks
My pleasure! Thanks for watching and let me know if you ever have any questions!
Somebody please ans me - glb and lub of mn/m^2 + n^2 , m belongs to reals and n belongs to natural number.
What's the infimum of this set
Is n/n+1 bounded above
Thanks for watching! Answer this, is n < n+1?
yeah im gonna accept that im gonna fail
Do not resign yourself! More studying and hard work shall raise your capacity and your spirits, the world is your oyster!
@@WrathofMath man im just mad, why im studying proofs in calculus 1 like calc is juse derivatives limits..
im just lost
Please can you help proof that Supremum {n/(2n+1), n= 1,2,3....}= 1/2 and infimum of same set = 1/3?
Thanks
what if n element of real numbers then infimum is -1 right?