I did it without Limits: Lets call the equation given to us (A): differentiate (A) with respect to y and set y=1 ( we also know f'(1)=1), we get: (B) f'(1/x^x) - x^x.f(1/x^x) = 1 Then set y=1 in (A) and differentiate it we get: (C) f'(x)=(lnx+1).( f'(1/x^x) - x^x.f(1/x^x) ) from (B) and (C) we can deduce that f'(x)=lnx+1 by integration we get f(x)=xlnx , and finally find that f(2020^2)/f(2020)=4040
at 9:16 you can't just replace the value by saying "they are the same" as the result is 0/0 and the linearity does not work. What you can do is divide by 1= (f(ugly)-f(1))/(ugly - 1) which will have the same effect. Actually lim a/b = 1 is stronger that lim a = lim b and is what you need, especially here when we know lim a = lim b = 0.
@@SOTminecraft Could you plz tell how to get the fomula in the red circle at 7:40. What I can imagine is that x is replaced by ugly in the fomula： f'(1)=limit x->1:f(x)-1/x-1 But why here to replace x by ugly is legal? Just because they get the same limit when h->0? Like x = limit h->0 ugly. Are there any theorems about this substitution or it‘s deduced in some way that I don't know about and not mentioned in the vedio?
That was good. I saw one shortcut though from the definition of logarithmic differentiation to skip the integral. If f'(x)=(d/dx(x^x))/(x^x), we know that f(x)=ln(x^x)=xln(x).
I think foing it by partial differentiation will be easier. Differentiate wrt x treating y as constant and then placing x=1 we get a pretty simple linear differential equation
lim(a)/lim(b)=1 is much stronger than lim(a)=lim(b) - especially when you have that lim(a)=0 or ±∞ (as you do here). You can't substitute one limit for another, but you can multiply it into the equation (or in this case divide it in) to achieve the same result.
I went to UC Berkeley's financial engineering program.. love that place. Beautiful campus and, of course, top tier mathematics! This video demonstrates how to properly visit the campus.. crush brain twisting problems
Ahhhh no wonder you made video on Kelly’s. Brain twisting problems have to be done in Evans, with the orange chairs around you Side note, Been some time since you upload video man 😢
Nice! Also missed you using a blackboard and chalk, such a good video. Btw I’ve been watching you the past two or three years, you’re amazing dude! Keep up the good work 👊
My solution was much different. First, I differentiated both sides with respect to x remembering that f'(u(x)) = f'(u)*u'(x), giving me -f'(x) = (1+lnx)[ (x^x/y^y)f(y^y/x^x) - f'(y^y/x^x) ]. Then I let x = 1, giving me -1 = f(y^y) - (1/y^y)f'(y^y), which is a differential equation that I can use later. Then I let y = 1 to get f(1) = 0. Then I went back to my differential equation letting v = y^y/x^x getting f'(v) - (1/v)f(v) = 1.Solving that using an integrating factor we get f(v) = vln(v) + cv, using f(1) = 0 we get c = 0, hence f(x) = xlnx.
I think maybe a simpler way to explain the step at 7:50 in your video is to say that the limit of that complex element Ilets call it z) of the formula is 1, so lim(h->0) of z-1 is the same limit as lim(h->0) of h (what was previously on the bottom).
Another aproach: let g(x)=f(x)/x. g(1)=0, g'(1)=1. And we get next equation: y*g(y) - x*g(x)=g(y^y/x^x). Also we have x*g(x)-y*g(y)=g(x^x/y^y) => g(x^x/y^y) = - g(y^y/x^x), so g(t) = - g(1/t). Pluging in y=1 we have x*g(x)=g(x^x) => g(x^x) + g(1/y^y) = g(x^x) - g(y^y) = x*g(x) - y*g(y) = g(x^x/y^y). Recall x^x = a, 1/y^y = b and we have: g(a)+g(b)=g(ab), g(1)=0, g'(1)=1 and f(x)=x*g(x). Let h(x) = exp(g(x)). => h(1)=1, h'(1)=1 and h(x)*h(y) = h(xy) Differentiate equation h(xy) = h(x)*h(y) in respect to x: y*h'(xy)=h'(x)*h(y) pluging in x=1 => => y*h'(y)=h(y) => ln(h(y)) = ln(y)+c => h(x)=Cx, h(1)=1 => => h(x)=x and f(x) = x*g(x) = x*ln(h(x)) = x*ln(x).
At one point you had: f'(x) = 1/x^x * d{x^x}/dx Isn't it allowed (and simpler) to just integrate directly after this step? Substituting u = x^x and integral 1/u du = ln(u)
Sure that was nostalgic for you to go back to same classroom at your alma mater! Will try the other definition of derivative, lim y approaches x of (f(y)-f(x))/(y-x)
I imagined the solution f(x) = x ln(x), and then check, as we do with the integrations, imaging the solution first and then derivating. But this is not math really. The math things have to be demostrated, of course. I did have problems to understand your demonstration, but I got it. Thanks.
Once you've got the second constraint: F(1)=0 just graph it with F'(1)=1 and it's pretty obviously related to ln x. Test a couple of values of x and x ln x is easy to see.
Hi blackpenredpen, recently i faced with an interesting paradox Consider solid that constructed by rotating function 1/x around x axis , for x>=1 This solid's surface area is infinite,however volume is finite. If you think just a moment, we can color its infinite surface area ,with finite amount of colour by filling it with color... Will be glad to hear your opinion
this is called the gabriel's horn paradox, and isn't really a paradox in my opinion. if you think about it, if you make the thickness of the colour to be infinitisemally small, its volume can still be finite and also fill up infinite surface area.
@@viscid2906 what d u mean by infinitismally and infinite volume, i5s numerically finite and if u fil that with water, surface will be(should) wet , i didnt get u really
@@ozaman-buzaman9300 thing is, you can colour a really large surface area with a fixed amount of paint, its just the thickness of the paint when u colour the area that matters
@@viscid2906 oh man, im talkking about filling it, thicknees doesnt matter there,u donnt want to paint only surface, inside also,when u filling a bottle u dont care thicnnee of water in the surface, thats a nonsense
@@ozaman-buzaman9300 the thickness of the surface of the horn is assumed to be 0 (it's a 2d manifold), so the outside surface area is equal to the inside and yes i shouldnt really care about the thickness, but what im trying to say is that u cant really compare 'area' with 'volume', because first of all the units are different, second of all, a surface is a 2d object so it has virtually no volume so there is no paradox
My approach: Differentiate both sides of the functional equation with respect to y (holding x constant), which gives a functional equation for f'(y). Substitute in y = 1, using f(1) = 0. This gives an ODE, which you can solve using f'(1) = 1 to get f(t) = t ln t. Then you can just plug in 2020 to get the solution. Detailed steps: Substitute x=y to get f(1)=0 Differentiate both sides of the functional equation with respect to y (holding x constant) → f'(y) = (ln y + 1) * ( f'(y^y/x^x) - x^x/y^y * f(y^y/x^x) ) Substitute y=1 by using f'(1)=1 → 1 = f'(1/x^x) - x^x * f(1/x^x) Let t=1/x^x → 1 = f'(t) - 1/t * f(t) Solve the ODE in t → f(t) = t ln t + c t, for some constant c Solve for c using f(1) = 0 → c = 0 f(t) = t ln t
why dont we use method of partial differentiation . just differentiate wrt to x and y taking other as constant then we will get finally y'/(1+lny)=const
Dear blackpenredpen can you please explain what you did in 7:50 how did you come up with that weird equation and how is it equal to f'(1) also that it does not matches with any defination of derivitaves i wonder how and why you changed the limit from x-1from h-0 please read my comment and answer my question 🙏🙏🤔🤔
Or one could just fix x > 0 and implicitly differentiate both sides with respect to y, obtaining: f'(y^y/x^x) (1 + ln y) = (f(y)-f(x)) (1 + ln y) + f'(y). Now, w.l.o.g., one can impose y = x and use that f'(1) = 1 to get f'(y) = 1 + ln y, that is f(y) = y ln y + c. We can determine c by imposing f(1) = 0, so ultimately f(y) = y ln y
Hey BlackPenRedPen what was the reason for you to do assume (x=y). I failed the task by not using this assumption. Was because we had given the property of f'(1) = 1 ?
We don't "assume" x=y, we "let" x=y For functional equations it is really common to just plug in numbers. If your equation is true for any pair of points (x,y), you can "choose" (x,x) or (y,y) just to see what happens...
I "solved" this problem in like 5 minutes or something but through educated guesses. The properties of f(x) really resembled the logarithmic functions and the fact that f'(1)=1 further tempted me to think that f(x) must have something to do with ln(x). f(x)=ln(x) didn't work so I thought it could be a nested function like f(x)=ln(g(x)), when I worked this out, it was kinda obvious that g(x)=x^x and it worked out pretty well. Should I feel guilty for doing this because I really feel like I cheated a little bit.
put y = x, so f(1) = 0; then divide both sides by (y - x) and take the limit y -> x; in the left you have f'(x); in the right 0/0, so apply l'Hopital rule (for f(y^y/x^x)/(y - x) with respect to y) and get 1 + ln x; so f'(x) = 1 + ln x; etc; piece of cake; LOL
Because it is given that the domain of the function is positive reals. Given any positive real x and any h that approaches 0 (see that x cannot be 0 so even if h approaches 0 from below we should not have any problem) it is always a valid input for the function
Is that really that much to think like assume and stuff? Simple hai it's a formulae and we are putting the value in it. The formula is the definition of derivative
Since the equation is true for all x and y in the domain, it must be true when you choose specific values of x and y. Any result you get from a specific x and y is then an implication of the original equation, but you also have to check whether that result is enough to satisfy the original equation. For example, if you find that the functional equation implies f(x) = x(C+ln x), then you know that all solutions must have that form but you still need to check whether it’s true for all values of C or only specific ones.
Bruh i just read the spring paper, and i almost thought this was IIT maths tournament cuz of the amount of Indian names in it, nikhil, Ajith, sumith, sreyas, how many indian students are in UC Berkeley anyway?
I think i need to turn off notifications on your videos cause i just lost an hour on one of these problems, an hour i was supposed be doing other stuff 😂😂😂
yeah that's not how you solve a problem ... changing the limit of a variable ... with a function that goes with the same variable..... the only thing you could have done was to divide and multiply with that magic denominator that you added... and prove that the limit goes to 1+ln(x)
I did it without Limits: Lets call the equation given to us (A):
differentiate (A) with respect to y and set y=1 ( we also know f'(1)=1), we get: (B) f'(1/x^x) - x^x.f(1/x^x) = 1
Then set y=1 in (A) and differentiate it we get: (C) f'(x)=(lnx+1).( f'(1/x^x) - x^x.f(1/x^x) )
from (B) and (C) we can deduce that f'(x)=lnx+1
by integration we get f(x)=xlnx , and finally find that f(2020^2)/f(2020)=4040
Are you a mathematician? Or are you in college.
@@shashwatjha3537 i am currently in university
@Luffy major in automation
can you explain how you got rid off the f'(x) from (A) in your (B) equation?
@@Shiynu in my comment i meant to write differentiate (A) with respect to y and set y=1 , try it that way it should make sense
After watching your calc videos for abt a year, I finally am in calc 1
Same, I think. Not sure what it would be where you live since I live in England but oh well
If Berkeley felt like being evil they could've made the official solution just a verification of the answer without a derivation.
at 9:16 you can't just replace the value by saying "they are the same" as the result is 0/0 and the linearity does not work. What you can do is divide by 1= (f(ugly)-f(1))/(ugly - 1) which will have the same effect. Actually lim a/b = 1 is stronger that lim a = lim b and is what you need, especially here when we know lim a = lim b = 0.
I'm gonna use 'ugly' every time I get some annoying expression that I don't feel like writing lol
@@spaceybread this is so useful (ugly)-(ugly)=0
(ugly)/(ugly)=1
Good to know
@@JTtheking134 😆 I mean, you understood what I meant xD
@@SOTminecraft yes ugly is ugly. And with loss of generality that is ugly too
@@SOTminecraft Could you plz tell how to get the fomula in the red circle at 7:40. What I can imagine is that x is replaced by ugly in the fomula：
f'(1)=limit x->1:f(x)-1/x-1
But why here to replace x by ugly is legal? Just because they get the same limit when h->0? Like x = limit h->0 ugly.
Are there any theorems about this substitution or it‘s deduced in some way that I don't know about and not mentioned in the vedio?
That was good. I saw one shortcut though from the definition of logarithmic differentiation to skip the integral. If f'(x)=(d/dx(x^x))/(x^x), we know that f(x)=ln(x^x)=xln(x).
Yeah that would’ve worked as the problem specified that x>0
I love how in the comments we all got different approaches
11:07 OMG THAT CURVE HE MADE ON THE BOARD IS SOO MAJESTIC
I think foing it by partial differentiation will be easier. Differentiate wrt x treating y as constant and then placing x=1 we get a pretty simple linear differential equation
lim(a)/lim(b)=1 is much stronger than lim(a)=lim(b) - especially when you have that lim(a)=0 or ±∞ (as you do here). You can't substitute one limit for another, but you can multiply it into the equation (or in this case divide it in) to achieve the same result.
Nice catch!
I went to UC Berkeley's financial engineering program.. love that place. Beautiful campus and, of course, top tier mathematics! This video demonstrates how to properly visit the campus.. crush brain twisting problems
Ahhhh no wonder you made video on Kelly’s. Brain twisting problems have to be done in Evans, with the orange chairs around you
Side note, Been some time since you upload video man 😢
Nice! Also missed you using a blackboard and chalk, such a good video.
Btw I’ve been watching you the past two or three years, you’re amazing dude! Keep up the good work 👊
I'm really enjoying these. Takes me back to uni days.
For more info and past tournaments, please visit: bmt.berkeley.edu/
I might show up to BMT on 11/5/2022 just to say hi.
sam sir : hello an neong ha se yo ah emducation isit sam best 🤣sam a reaver convert isit 100 man conturion
brain left the chat
😮That's awesome! We would love to see you here!
Ahh! Whenever i go to new places,I always first go to a hotel
I love calculus for fun problems like that
At 5:30 if we use l'hopital for the second limite then we will get f'(x)=0
Determining a function based on an equation containing the function and ifs inverse and some other relation, and a point on its derivative.
At 11.11 when you got f'(x)=1/x^x times d(x^x)/dx you could just have integrated both parts and got f(x)=ln(x^x) +c
My solution was much different. First, I differentiated both sides with respect to x remembering that f'(u(x)) = f'(u)*u'(x), giving me -f'(x) = (1+lnx)[ (x^x/y^y)f(y^y/x^x) - f'(y^y/x^x) ]. Then I let x = 1, giving me -1 = f(y^y) - (1/y^y)f'(y^y), which is a differential equation that I can use later. Then I let y = 1 to get f(1) = 0. Then I went back to my differential equation letting v = y^y/x^x getting f'(v) - (1/v)f(v) = 1.Solving that using an integrating factor we get f(v) = vln(v) + cv, using f(1) = 0 we get c = 0, hence f(x) = xlnx.
just by blindly trying I figured out the function, first logical is ln then you go to xlnx and it works out.
can't believe i missed that. looking at it now it's so obvious.
I started watching your videos when you did them in thazt building. Such great memories.
😃
I think maybe a simpler way to explain the step at 7:50 in your video is to say that the limit of that complex element Ilets call it z) of the formula is 1, so lim(h->0) of z-1 is the same limit as lim(h->0) of h (what was previously on the bottom).
Please can you explain this in detail i didn't got that part that how he came up a weird kind of equation and how does he know that's equal to f'(1)
UC Berk incoming freshman here. If this is what all math at Cal is like I am shivering in my boots!
Another aproach: let g(x)=f(x)/x. g(1)=0, g'(1)=1. And we get next equation: y*g(y) - x*g(x)=g(y^y/x^x).
Also we have x*g(x)-y*g(y)=g(x^x/y^y) => g(x^x/y^y) = - g(y^y/x^x), so g(t) = - g(1/t).
Pluging in y=1 we have x*g(x)=g(x^x) => g(x^x) + g(1/y^y) = g(x^x) - g(y^y) = x*g(x) - y*g(y) = g(x^x/y^y).
Recall x^x = a, 1/y^y = b and we have:
g(a)+g(b)=g(ab), g(1)=0, g'(1)=1 and f(x)=x*g(x).
Let h(x) = exp(g(x)). => h(1)=1, h'(1)=1 and h(x)*h(y) = h(xy)
Differentiate equation h(xy) = h(x)*h(y) in respect to x:
y*h'(xy)=h'(x)*h(y) pluging in x=1 =>
=> y*h'(y)=h(y) => ln(h(y)) = ln(y)+c => h(x)=Cx, h(1)=1 =>
=> h(x)=x and f(x) = x*g(x) = x*ln(h(x)) = x*ln(x).
At one point you had:
f'(x) = 1/x^x * d{x^x}/dx
Isn't it allowed (and simpler) to just integrate directly after this step? Substituting u = x^x and integral 1/u du = ln(u)
You can now register for BMT 2022 here: www.ocf.berkeley.edu/~bmt/bmt-2022/
The first-place team wins a $1000 bprp scholarship and I will be there!
We can just apply lhopital rule as soon as we see 0/0 form
wow, beautiful equation i liked it a lot
Didn't know you were a student at Berkely, pretty based. I just started my journey at Berkely as a freshman this year.
Cool funtional equation!
😄
Sure that was nostalgic for you to go back to same classroom at your alma mater!
Will try the other definition of derivative, lim y approaches x of (f(y)-f(x))/(y-x)
You could have just use LHoptial’s rule to evaluate the limit
I imagined the solution f(x) = x ln(x), and then check, as we do with the integrations, imaging the solution first and then derivating. But this is not math really. The math things have to be demostrated, of course. I did have problems to understand your demonstration, but I got it. Thanks.
Once you've got the second constraint: F(1)=0 just graph it with F'(1)=1 and it's pretty obviously related to ln x. Test a couple of values of x and x ln x is easy to see.
So arrogant
@@alejrandom6592 Try it. It works!
@@rodbhar6522 ok in what base ?
Hi blackpenredpen, recently i faced with an interesting paradox
Consider solid that constructed by rotating function 1/x around x axis , for x>=1
This solid's surface area is infinite,however volume is finite.
If you think just a moment, we can color its infinite surface area ,with finite amount of colour by filling it with color...
Will be glad to hear your opinion
this is called the gabriel's horn paradox, and isn't really a paradox in my opinion. if you think about it, if you make the thickness of the colour to be infinitisemally small, its volume can still be finite and also fill up infinite surface area.
@@viscid2906 what d u mean by infinitismally and infinite volume, i5s numerically finite and if u fil that with water, surface will be(should) wet , i didnt get u really
@@ozaman-buzaman9300 thing is, you can colour a really large surface area with a fixed amount of paint, its just the thickness of the paint when u colour the area that matters
@@viscid2906 oh man, im talkking about filling it, thicknees doesnt matter there,u donnt want to paint only surface, inside also,when u filling a bottle u dont care thicnnee of water in the surface, thats a nonsense
@@ozaman-buzaman9300 the thickness of the surface of the horn is assumed to be 0 (it's a 2d manifold), so the outside surface area is equal to the inside
and yes i shouldnt really care about the thickness, but what im trying to say is that u cant really compare 'area' with 'volume', because first of all the units are different, second of all, a surface is a 2d object so it has virtually no volume so there is no paradox
12:00... how about using log diff as take ln in both side? such as (ln f)' = f'/f
asnwer= 1 isit
You could have easily used l hospital rule and reached to the function very fast
7:16 - why so?
What about l'hospital ? Wouldn't that be a shortcut?
where you got stuck i used L' hospital rule due to 0/0 form. then it got easier ahead..
My approach: Differentiate both sides of the functional equation with respect to y (holding x constant), which gives a functional equation for f'(y). Substitute in y = 1, using f(1) = 0. This gives an ODE, which you can solve using f'(1) = 1 to get f(t) = t ln t. Then you can just plug in 2020 to get the solution.
Detailed steps:
Substitute x=y to get f(1)=0
Differentiate both sides of the functional equation with respect to y (holding x constant) → f'(y) = (ln y + 1) * ( f'(y^y/x^x) - x^x/y^y * f(y^y/x^x) )
Substitute y=1 by using f'(1)=1 → 1 = f'(1/x^x) - x^x * f(1/x^x)
Let t=1/x^x → 1 = f'(t) - 1/t * f(t)
Solve the ODE in t → f(t) = t ln t + c t, for some constant c
Solve for c using f(1) = 0 → c = 0
f(t) = t ln t
I really liked this question loll...I didn't understand it at ALL loll, but your explanation made it make a bit more sense to me.
why dont we use method of partial differentiation . just differentiate wrt to x and y taking other as constant then we will get finally y'/(1+lny)=const
Dear blackpenredpen can you please explain what you did in 7:50 how did you come up with that weird equation and how is it equal to f'(1) also that it does not matches with any defination of derivitaves i wonder how and why you changed the limit from x-1from h-0 please read my comment and answer my question 🙏🙏🤔🤔
And after changing the limit it is still equal to f'(1) howwwwwwwwwww?
Nicely explained ✨
You made an error at 8:51 minute.
7:39 The part about the composition is not convincing, some steps are missing I guess
Where are you?? I am on campus and would love to meet you!!
I was there last week. Now I am back to LA.
Awesome sir
Try ln(1+ln(x))=1
red chalk white chalk yaaay
😆
this was really fun
I just saw that f(1) must be 0 and arrived at the equation f(x^x)=(x^x)f(x), then guessed that f(x)=ln(x^x)=xln(x) and it worked out from there
Nice. If you graph f(1)=0 and f'(1)=1 you can see the answer is going to be close to ln x.
中间那个替换有点不严格，上下同乘除因子即可！NⅰCe
You should use L opital
Super cool, one love
Very fun indeed
love that chalkboard
I was following along, but by the end of the video I was like wait, what was the question again? Lol
Beautiful
I didn't understand this 😟
This more fun: 7/93 + 1 + 5 + 7 + 8 = 7/93 x 1 x 5 x 7 x 8.
Why is x^x/(x+h)^(x+h)=1
Or one could just fix x > 0 and implicitly differentiate both sides with respect to y, obtaining:
f'(y^y/x^x) (1 + ln y) = (f(y)-f(x)) (1 + ln y) + f'(y).
Now, w.l.o.g., one can impose y = x and use that f'(1) = 1 to get f'(y) = 1 + ln y, that is f(y) = y ln y + c. We can determine c by imposing f(1) = 0, so ultimately f(y) = y ln y
Hey BlackPenRedPen what was the reason for you to do assume (x=y). I failed the task by not using this assumption. Was because we had given the property of f'(1) = 1 ?
We don't "assume" x=y, we "let" x=y
For functional equations it is really common to just plug in numbers. If your equation is true for any pair of points (x,y), you can "choose" (x,x) or (y,y) just to see what happens...
Like this problem
I "solved" this problem in like 5 minutes or something but through educated guesses. The properties of f(x) really resembled the logarithmic functions and the fact that f'(1)=1 further tempted me to think that f(x) must have something to do with ln(x). f(x)=ln(x) didn't work so I thought it could be a nested function like f(x)=ln(g(x)), when I worked this out, it was kinda obvious that g(x)=x^x and it worked out pretty well.
Should I feel guilty for doing this because I really feel like I cheated a little bit.
I did the exact same thing lmao
that's the smarter answer
Shoot, are you at Cal?
Last week.
put y = x, so f(1) = 0; then divide both sides by (y - x) and take the limit y -> x; in the left you have f'(x); in the right 0/0, so apply l'Hopital rule (for f(y^y/x^x)/(y - x) with respect to y) and get 1 + ln x; so f'(x) = 1 + ln x; etc; piece of cake; LOL
Bruv you are a focking genius
Exactly how I solved it lol
But this is very hard to know, it's complicated, can anyone know where can I get detailed explanation?
I’m going to every room in evans till I find you
How can we 'assume' y = x + h as shown in the video?
Because it is given that the domain of the function is positive reals. Given any positive real x and any h that approaches 0 (see that x cannot be 0 so even if h approaches 0 from below we should not have any problem) it is always a valid input for the function
Because x+h and y has same domain...so we can plug in a function anything that comes into it's domain 😃
Is that really that much to think like assume and stuff? Simple hai it's a formulae and we are putting the value in it. The formula is the definition of derivative
y and x are any positive real numbers so x + any other positive real number will work as y
Since the equation is true for all x and y in the domain, it must be true when you choose specific values of x and y. Any result you get from a specific x and y is then an implication of the original equation, but you also have to check whether that result is enough to satisfy the original equation.
For example, if you find that the functional equation implies f(x) = x(C+ln x), then you know that all solutions must have that form but you still need to check whether it’s true for all values of C or only specific ones.
Very cool!
Why we let x=y=1?
Bruh i just read the spring paper, and i almost thought this was IIT maths tournament cuz of the amount of Indian names in it, nikhil, Ajith, sumith, sreyas, how many indian students are in UC Berkeley anyway?
I think i need to turn off notifications on your videos cause i just lost an hour on one of these problems, an hour i was supposed be doing other stuff 😂😂😂
It happens 😆
Blackchalkredchalk
this was neat
Highly probable question for jee advanced
Wow!!
cant believe blackredpen was IN MY DISCUSSION SECTION ROOM! sad that I didnt see him :(
Hey just got an idea ! Could you do next time 0^z=1 ? Thank you for your videos !!
Holy crap...
Booommmm..
🤯🤯🤯🤯
Neat.
sir iam.like maths but english very past 😭😭😭
?
@@a_minor my english very poor iam solw talking english only understand
Cool
Even I didn't get the part u didn't get
tricky
hi
I thought you live in China 🤷♂️
yeah that's not how you solve a problem ... changing the limit of a variable ... with a function that goes with the same variable..... the only thing you could have done was to divide and multiply with that magic denominator that you added... and prove that the limit goes to 1+ln(x)
ρɾσɱσʂɱ
Actually you don't even need calculus to solve the problem:
f(y) - f(x) = (xˣ/yʸ) · f(yʸ/xˣ)
⇒ f(x) - f(x) = (xˣ/xˣ) · f(xˣ/xˣ)
⇒ f(1) = 0
∴ f(z) - f(1) = f(z) = f(zᶻ)/zᶻ
⇒ f(y) - f(x) = f(yʸ)/yʸ - f(xˣ)/xˣ = (xˣ/yʸ) · f(yʸ/xˣ)
⇒ f(y)/y - f(x)/x = f(y/x)/(y/x)
⇒ f(z)/z ∝ log(z)
⇒ f(z) ∝ z · log(z)
⇒ f(z²) ∝ z² · log(z²) = 2z² · log(z)
∴ f(z²)/f(z) = (2z²/z) · (log(z)/log(z)) = 2z
So f(2020²)/f(2020) = 4040.
Knowing that f'(1) = 1 is only necessary to work out the base of the logarithm (i.e., that f(z) = z · ln(z)), but this isn't required to solve the problem.