1.10.1 Comparison of Functions #1
ฝัง
- เผยแพร่เมื่อ 8 ก.ย. 2024
- Comparing Complex Functions
Two functions can be compared by
1.Substitution
2. Applying Log
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wish i had watched this series in lockdown...
we are currently in lockdown only dude
@@devildemon7511 😂😁🤘
So true bro
2024 almost feels the same as lockdown :(
Literally AMAZING. Calm, comfortable, tidy and step-by-step explanation and reaching the point in a short time! this is what a perfect teacher means 👍🏼 MANY THANKS!
The last two questions are of GATE and I was so frustrated as I could not solve it. I just searched how to compare functions asymptotically and sir your video came up. Such a perfect explanation. Sir thanks a lot.
lol ...bhai ..
I like how you left this blooper in 1:34 Lol
Thank you for these amazing videos. I am learning a lot :)
bug or feature?
@@oliverchen4079 depends on the user :)
this was necessary to give you in-person class experience! hahaa
On behalf of the CBSE students, THANK YOU Abdul Sir. Very fortunate to get to know about this channel of yours. Thank you once again.
thanks a lot sir, you did give me a lot of knowledge than the professor at my university did ! you deserved a big thumb from the bottom of my heart!!
rupees and paise great way of explaining
but large number of paisa can become rupee
@@qvikk333 Yes even for n = 8 and base 2 wont hold
@@qvikk333 but that large number would be on both sides
No? That power 10 was only on one side. @@036karan
I have been going through your videos and am pretty much impressed with the way you ease up the subject. Hats off. That is truly a talent. Will keep watching your videos.
It shows that sub-continent has some of the best teachers on planet. I'm from Pakistan, and I remember I had teacher with a same name "Abdul Bari" and a master of his subject (Mathematics). Learn a lot from your videos sir, please keep helping us absorb that complex knowledge making it simple enough. Huge respect from Pakistan!!
☺☺
I wish I had found these tutorials a few months ago. I could have done far better in my first test. Simply you are a genius. You know how to explain these things well.
Sir in the previous problem u have comapred 2logn and logn and said that 2logn is greater(they have just different coefficient) but not in the last question
Truely amazing!! All your tutorials are a tremendous help in understanding correlations that seemed so hard to decode - but with your explanation all becomes logical and cristal clear! Thank you so much!!
The best instructor I have ever seen in my life
Your teaching style is really very simple and easy to understand. Thanks a lot for the best lectures.
Tomorrow is my exam. If your videos would not been on TH-cam I would definitely fail. Thank you so much Sir. Rise and shine...May allah bless you abundantly. Kindly increase your playlists 🙏
dividing f(n) and g(n) and apply limit n -> ∞ will be more easy to estimate.
6:55 rupees and paisa perfect explanation
Excellent progression and crystal clear content. Thanks!
Best content on youtube 🙌🏼👍🏼
You are the most amazing tutor!
Thank you, sir, for this amazing video series............
Do we not ignore the coefficient for logarithmic terms as well?
Ex. 3logn > 2logn but 3n = 2n, why is it so?
You're a life saver sir!
Sir thanks for these amazing videos
So when comparing functions, are we using asymptotic behaviour or actual values? I don't understand the logN > loglogN so you can ignore the contribution of the 10 loglogN and focus only on 2logN.
My example is n^2+n vs 7n. According to the video's argument, no comparison n^2 wins. But if we compare actual values (like the end of video says), 7n is bigger for n = 1,2 and only loses when n > 2, so which values are we comparing?
awesome Sir Make a series of programming tutorial for non-programmer I am data analyst currently using R programming
Professor, I see why 2logn > logn but why is 10log(logn) less than log(logn)?
Shouldn’t it be also larger just like 2logn > logn?
I also have the same doubt. I can only add that perhaps logn is a increasing f. while loglogn is a decreasing f.( by sampling), not sure about the overall nature of loglogn .
What do you think?
Yeah man ....its embarrassing 2nd function is greater
He did not say that 10log(logn) is less than log(logn). But we are comparing the bigger n of the two functions. logn is more than log(logn), so we can ignore the log(logn) and instead compare the logn with the highest n of the other function, which is 2logn.
Forget about the first example, lets talk about the second. In the first example f(n) is greater than g(n), but in second case sir said, we are ignoring 2 and f(n) = g(n). I have a doubt on that. If anyone reading this, can explain.
no one replied so I am commenting.
log(logn) is much smaller than logn
why?
consider n = 16
then log(log16) = log(log(2^4)) = log4 = log(2^2) = 2
and logn = log16 = 4
hence we can say logn is larger than log(logn)
now suppose i travel a distance say 3km and 10m and you travel say 2km and 50m
we compare the bigger unit right? so I travelled more 3km>2km we ignore the 10 and 50m because they are smaller units.
The same applies here
i spent hours look for an explanation of this... thank you so much !!!
a gold mine in youtube
Your videos are amazing, you actually make Algorithms fun 😄
Your videos are too awesome. In this video around 7m time, you are comparing two functions, how f(n) can be bigger compared to g(n) as , let say n=4, f(n)=32, but g(n) will be 4*2^10; if we are considering log base 2
Спасибо Мистер
comparison at 1:40 wont hold when n=1 or 0
I respect this man
Nicely explained with examples. Thank you sir.
6:45 kya bat hai kya cheez hai paisa paisa lol
7:00 is the comparison correct? loglogn^10 isn't equal to log10logn?
What sir said is wrong. What you have written in the comment is correct
thank u sir, u r godsend!!!
sir in the example n^2(logn) and n(logn)^10 the first function is greater for some values and after that the second function is greater .I checked it by putting values in both the functions.
Yeah, you can check in desmos, second function becomes greater after x = 28 or something like that.
@@animerank4907 No worries, sir. The professor's got it right. Just remember, "n" is a natural number, so it's gotta be "n >= 1." And for any value of "n," "f(n)" is always gonna be more than "g(n)." Plot it on a graph and take a look from "n >= 1."
for the second example. shouldn't the left hand side equation greater than the right one starting from n= 28,1?
Superb sir u keep continue thnx sir
Very good videos, they are helping me a bunch!
OMG I never knew how to compare functions in 4 years of industry experience.
Very clear explanation... Thanks
great comeback at 1:38 sir sir sir
Thank you for your efforts
an excellent lesson
sir, if the value of log is greater than 1 it will be always greater than g(n) please reply
example 2 {time: 4:43}
Great..
f(n) is big omega of g(n)... @2
Great examples
Great Effort Sir!
TYSM
Date - 17/6/22
Imp note - a raise to log c base b == b raise to log c base a.
Perfect👌
I think that g(n) is greater than f(n) in the first example 4:42, I think Prof. Bari mistakenly says f(n) is bigger.
No worries, sir. The professor's got it right. Just remember, "n" is a natural number, so it's gotta be "n >= 1." And for any value of "n," "f(n)" is always gonna be more than "g(n)." Plot it on a graph and take a look from "n >= 1."
0:00 first method
1:39 sec method
Sir, I have a question. If your last question (the one with the root) has both functions asymptotically same, then why not the second one? Both functions have log n + log log n... Just different coefficients. So can they be said asymptotically same??
Nice explanation sir
Thankyou so much sir,
Very helpful
Thanks🌹
make a tutorial of computer fundamental and basic programming concept e.g data structure & algorithm
@7:00 sir how would we know that log(logn) should be ignored because it has lower value on both sides. if we dont know its value.
thank you so much sir today is my exam
Sir if you take n=100 then g(n) will become greater
Plesse include all subjects of CS🙏
2:58 log formulae
thanks
Thank you
I watched it in 2017 and also in 2022.
الهندي تاعنا
why is the base by default 2? shouldn't it be 10?
@@abdul_bari oh I see. thanks for answering
7:04 answer is maybe wrong
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At 7:05 where you say f(n) > g(n) I don’t believe this is the case. If you graph both of those functions we can see g(n) approaches infinity faster than f(n) so isn’t f(n) < g(n)?
You probably have taken log with a base 10 while graphing. Here the base is 2.
@@kera2403 It's still true for base 2.
i was making the same mistake,zoom out the graph and you will see f(n) >g(n)
4- law is -> a^log(c) is c^log (a) not as you wrote it also first example is not correct. from section 3.2 in introudction to algorithms book
sir dark mode use kijiye na
1:34 😄
Sir aap ne c++ ka course q nhi banaya
Bana dejiye premium hoga to chalega
At 6:37....why 2logn is greater than logn....they are assymptotically same naa!
@@abdul_bari thanks a lot sir!
because 2logn signifies n^2 and logn signifies n.
can 'n' be ever negative?
How do I derive rule 4?
where it helps in DAA
Do a iready lesson
actually I am wrong for n > 10000000000000000 f(n) is bigger!
Sir, Comparison question at 7:05 is not correct, you can check yourself using calculator, beyond 2.7 for any value the g(n) will be larger than f(n) , rupees and paisa explanation fails at least here , just a suggestion
Okkk sorry sir i got it, u were not talking about the value but the order was being compared.. Right na?
What is the base of the logarithm that you are taking Sir ?
it doesn't really matter on which base you take the log. it should just be any valid base.
it doesn't really matter on which base you take the log. it should just be any valid base.
@@subhrapaladhi5888 but why
as we are only going to compare. just make sure you take both of them on the same base
black bord , black t shirt
haters guna hate.
Your way of explaining is very much similar to Khaleel sir ❤
He was student of khaleel sir😊
Who is Khaleel?
math sucks
Which book you follow sir?
Please tell me the book name you follow?
@abdul bari