Prove Little o
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- Опубліковано 5 вер 2024
- Prove f(n) is little o of g(n)
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Hey Rand, I think you're doing a great job with your videos and I feel like you deserve a lot more recognition.
A suggestion to improve your content slightly may be creating separate thumbnails that show exactly whats going on in the video or that simply show the title of the video (maybe not necessarily in handwriting). I think this would make the videos more attractive for people who have not seen any of your content yet. (this video actually is an example of how you could do it, but I'd suggest using a screenshot of the computer-written text as I mentioned before. You could also add your logo or a picture that stands out)
PS: Have a nice day :)
Lost me with the limits
Hey Obose,
Limits are usually taught in calculus classes. A limit is the value that a function (or sequence) "approaches" as the input (or index) "approaches" some value -Wikipedia (en.wikipedia.org/wiki/Limit_(mathematics))
If you understand limits, it makes it easier to use them to prove Big O, Little O, etc. than some of the more conventional proofs like in my other videos (ua-cam.com/video/DjfYhHSkWqo/v-deo.html).
Limits aren't too bad though, the limit as n approaches 2 for the function "1/n" = 1/2,
and the limit as n approaches 3 for the function "1/n" = 1/3,
and the limit as n approaches 100 for the function "1/n" = 1/100.
So if n keeps getting bigger and bigger, the whole function "1/n" gets increasingly smaller and smaller and closer to the value 0.
So the limit as n approaches infinity for the function "1/n" = 0.
Once you know a functions limit or "L" for short, you can use the cases in this video (ua-cam.com/video/CjAiht0yt1s/v-deo.html) to match the limit and determine if it's Big O, little o, Big Omega etc. of another function.
In this video I was able to simplify the function to into the limit as n approaches infinity of (3/n + 1/n^2). The limit as n approaches infinity of 3/n = limit as n approaches infinity of (3 * (1/n) ) and we know the limit as n approaches infinity of "1/n" is 0 so we get limit as n approaches infinity of 3 * 0 = 0, and the limit as n approaches infinity of 1/n^2 is also 0 because as n increases the function 1/n^2 decreases and approaches the value 0.
Maybe this site can help you understand my example of the limit as n approaches infinity for "1/n" (mathcentral.uregina.ca/QQ/database/QQ.02.06/evan1.html).
Thanks for watching even more of my videos Obose!
@@randerson112358 hi can u please explain to prove 3^(2^n)=o (2^(3^n)) little o notation use pls
@@vaishnavipuli7152 lol
I crumbled my paper up so quick lool