You prolly dont care but does someone know of a trick to get back into an instagram account? I was dumb lost my login password. I would love any help you can give me
@@Geolas88 It is a bot. There is supposed to be another bot that answer that question with something like: ''Hey, I know an app can help, it's called InstagramStalker, try it out!'' or something like that.
Very lucid presentation. Thanks. Very satisfying to understand. I used the equation (the one just before you make n substitutions) to prove to myself that the d/dx lnX= 1/X. I plugged a few x values into the n-less equation using 0.000001 for delta X and sure enough the resulting slope = 1/x. To me that was proof enough but I understand the intellectual challenge to show proof without having to plug a single X into the equation. One thing about the rest of your proof presentation (after n substitutions) that I found daunting even discouraging as a beginner calculus student is the seemingly spontaneous and intuitive choice to introduce n= delta X/X. I thought to myself, "man...I'll never get to the level where I can be that intuitive about proofs." But then I remembered recently watching another of your videos where you show that the derivative of e^x is e^x. You start out by defining e as the limit as n>infinity and n>0 with simple equations. Then your subsequent substitution of n into your proof made sense as you mentioned that you wanted to guide your proof to include some semblance of those introductory equations. So I'm encouraged that if I watch enough videos I'll start to build up a repertoire of "mini" proofs to follow even greater proofs.
e=(1+1/r)^r as r goes to infinite. Would you please explain how is it that (1+n)^(1/n) is also e? In all honesty, it checks out - I calculated it in R and it does approx. e. But, mathematically, how are they the same?
Out of respect and ignorance, lets suppose we didn't know what e^1 was, and we couldn't recognize that the lim as n approaches 0 of (1+n)^1/n was e^1. Then how else could we have proved the derivative of ln x?
you can safely substitute n in parenthesis (1+n)^1/n to 0, as, if n approaches zero, in that case you can consider it as zero itself, so you get (1)^1/n, further you can consider 1/n as something infinitely big, as 1/0,000...1 approaches infinity, but n will never get zero itself, so 1 in any power will give you 1, then (1+n)^1/n magically becomes 1
@@connorfr no its wrong. [1+(1/n)]^n as n-> inf = e. It's sad because I've been trying to wrap my mind around the proof and I thought I'd finally found it but this is wrong.
@@connorfr RE: after further research, I was wrong. Khan as usual is right. The solution lies in Binomial Theorum. Similar to the proof for the power rule where terms in the BT collapse to 0 as we differentiate x->0, the same thing happens when you differentiate (1+n)^(1/n). Using binomial expansion, you obtain a set of terms in the form: e - e/2*n + 11e/24*2n ..... All terms with a multiple of n collapse as n->0 leaving only e. Finally the derivate of e is e. Therefore 1/x * ln(e) = 1/x * 1 = 1/x. Khan was right. Wish he had explained that here, but probably beyond the scope of the video.
@@pointlesslylukesplainingpo1200 --The problem is that "h" is a variable in algebra, while "delta-x" is a distance in algebra, and distance is the key to understanding differentiation. /Lonewolf Liberties
I'm always amazed by seeing proofs like this. I feel like I could never think of a proof like this no matter how much time I had.
You prolly dont care but does someone know of a trick to get back into an instagram account?
I was dumb lost my login password. I would love any help you can give me
@@otismohammed5555 why are you asking here lmao
@@Geolas88 It is a bot. There is supposed to be another bot that answer that question with something like: ''Hey, I know an app can help, it's called InstagramStalker, try it out!'' or something like that.
I was trying to remember the reason for this result and I finally found the demonstration. Very well explained. Thanks!
This video helped me solve an algorithmic problem! Thank you Sal and Khan Academy! These videos will never get old!
Very lucid presentation. Thanks. Very satisfying to understand. I used the equation (the one just before you make n substitutions) to prove to myself that the d/dx lnX= 1/X. I plugged a few x values into the n-less equation using 0.000001 for delta X and sure enough the resulting slope = 1/x. To me that was proof enough but I understand the intellectual challenge to show proof without having to plug a single X into the equation. One thing about the rest of your proof presentation (after n substitutions) that I found daunting even discouraging as a beginner calculus student is the seemingly spontaneous and intuitive choice to introduce n= delta X/X. I thought to myself, "man...I'll never get to the level where I can be that intuitive about proofs." But then I remembered recently watching another of your videos where you show that the derivative of e^x is e^x. You start out by defining e as the limit as n>infinity and n>0 with simple equations. Then your subsequent substitution of n into your proof made sense as you mentioned that you wanted to guide your proof to include some semblance of those introductory equations. So I'm encouraged that if I watch enough videos I'll start to build up a repertoire of "mini" proofs to follow even greater proofs.
why is this so cool to me
Same
sameee
because math is cool.
awesome video. I just would have loved to see how you could take "limit" from outside to inside of ln.
Thanks for this material... I'm really appreciate it
Thanks for helpful videos you helped me survive my school year last year !!!
But it's circular proof of 1^infinite
It's wrong
It's just like that
Are you proof why 2+1=3
And assume 1+1=2
Ive been confused for a long time and now i understand
e=(1+1/r)^r as r goes to infinite. Would you please explain how is it that (1+n)^(1/n) is also e? In all honesty, it checks out - I calculated it in R and it does approx. e. But, mathematically, how are they the same?
Heuristically:
Let n = 1/r. Then your limit, lim(r->inf)[1+1/r]^r is the same as lim(n->0)[1+n]^(1/n)
@@ActicAnDroid oh you are right... I see it now
@@ActicAnDroid i don’t understand I keep getting back the number 1 for these equations
Beautiful.
Plz prove also
lim(1+1/n)^n
thanks bro, really helped me.
4:29 what if x is very close to zero, such as 0.00001?
It really helped me 😊👌👌👍👍
Amazing video
Georgeous sir.
the only thing not 100% clearly explained is moving the limit operation inside of the ln( ) function. Is this always OK? why?
It doesn't matter, As N only exists inside the ln()
Out of respect and ignorance, lets suppose we didn't know what e^1 was, and we couldn't recognize that the lim as n approaches 0 of (1+n)^1/n was e^1. Then how else could we have proved the derivative of ln x?
you can safely substitute n in parenthesis (1+n)^1/n to 0, as, if n approaches zero, in that case you can consider it as zero itself, so you get (1)^1/n, further you can consider 1/n as something infinitely big, as 1/0,000...1 approaches infinity, but n will never get zero itself, so 1 in any power will give you 1, then (1+n)^1/n magically becomes 1
I know a better proof ask if u wanna know.
@@sakshamdobriyal9952 plz this one isn't super clear
@@arsenron
Wrong lol
@@ActicAnDroid shut your fart up, it is true. u dont have counterarguments, son of a clown.
God bless you
Sir how to prove
(1+n) ^(1/n) =e
did anyone figure this out?
@@connorfr no its wrong. [1+(1/n)]^n as n-> inf = e. It's sad because I've been trying to wrap my mind around the proof and I thought I'd finally found it but this is wrong.
@@connorfr RE: after further research, I was wrong. Khan as usual is right. The solution lies in Binomial Theorum. Similar to the proof for the power rule where terms in the BT collapse to 0 as we differentiate x->0, the same thing happens when you differentiate (1+n)^(1/n).
Using binomial expansion, you obtain a set of terms in the form: e - e/2*n + 11e/24*2n .....
All terms with a multiple of n collapse as n->0 leaving only e. Finally the derivate of e is e.
Therefore 1/x * ln(e) = 1/x * 1 = 1/x.
Khan was right. Wish he had explained that here, but probably beyond the scope of the video.
love this
I am smarter, I am better
thankk youu
Thank you for using delta x instead of that silly variable “h” that modern texts use.
/Regards
lol
H makes it easier and less confusing imo but ok
@@pointlesslylukesplainingpo1200 --The problem is that "h" is a variable in algebra, while "delta-x" is a distance in algebra, and distance is the key to understanding differentiation.
/Lonewolf Liberties
A ton of thanks!!!!!!!!!!!!!!!!!!!!!!!!!
The derivative of ln(x) is not defined when x is negative but 1/x is defined for when x is negative
if were being specific its 1/x with domain (0, infinity)
Why is it that e is (1+n)^(1/n) and not (1+1/n)^n ?
7:48
See, the only problem I have is that there is a power of 1/0. How can you get rid of that?
there isn't a power of 1/0 because you aren't reaching 0
you're taking the limit as n goes closer and closer to 0, but you never get to 0.
the definition of e is (1+1/inf)^inf. inf can be viewed as lim x->0 1/x
after a day of thinking I searched up the proof, apparently I wasn't far of. I had done everything up to the n part :(
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@@du42bz Oh no I'm old now. Good luck with classes!
can you prove why dy/dx of ln(a) = a'/a?
Oh, no need. It suddenly came to me.
oH Am ghEE!
Good video