My First Berkeley Math Tournament Problem
ฝัง
- เผยแพร่เมื่อ 22 ส.ค. 2022
- We will solve my first Berkeley Math Tournament bmt.berkeley.edu/ calculus problem on UC Berkeley Campus. This question is from the calculus part of the 2020 BMT. We are given f'(1)=1 and f(y)-f(x)=x^x/y^yf(y^y/x^x) and we have to find f(2020^2)/f(2020). Since we are given f'(1)=1 and the property shows the difference between the function of two different inputs, we should proceed by using the definition of the derivative to solve this functional equation. Watch the find out the rest!
BMT 2022 will be held on the UC Berkeley campus on Saturday, November 5, 2022!
There will also be an informal asynchronous BMT online for those of you who are unable to make it in person.
For more info: bmt.berkeley.edu/
Past tournament tests & solutions: www.ocf.berkeley.edu/~bmt/arc...
Contact them at questions@bmt.berkeley.edu.
🛍 Shop math t-shirts & hoodies: bit.ly/bprpmerch
10% off with the code "WELCOME10"
----------------------------------------
**Thanks to ALL my lovely patrons for supporting my channel and believing in what I do**
AP-IP Ben Delo Marcelo Silva Ehud Ezra 3blue1brown Joseph DeStefano
Mark Mann Philippe Zivan Sussholz AlkanKondo89 Adam Quentin Colley
Gary Tugan Stephen Stofka Alex Dodge Gary Huntress Alison Hansel
Delton Ding Klemens Christopher Ursich buda Vincent Poirier Toma Kolev
Tibees Bob Maxell A.B.C Cristian Navarro Jan Bormans Galios Theorist
Robert Sundling Stuart Wurtman Nick S William O'Corrigan Ron Jensen
Patapom Daniel Kahn Lea Denise James Steven Ridgway Jason Bucata
Mirko Schultz xeioex Jean-Manuel Izaret Jason Clement robert huff
Julian Moik Hiu Fung Lam Ronald Bryant Jan Řehák Robert Toltowicz
Angel Marchev, Jr. Antonio Luiz Brandao SquadriWilliam Laderer Natasha Caron Yevonnael Andrew Angel Marchev Sam Padilla ScienceBro Ryan Bingham
Papa Fassi Hoang Nguyen Arun Iyengar Michael Miller Sandun Panthangi
Skorj Olafsen Riley Faison Rolf Waefler Andrew Jack Ingham P Dwag Jason Kevin Davis Franco Tejero Klasseh Khornate Richard Payne Witek Mozga Brandon Smith Jan Lukas Kiermeyer Ralph Sato Kischel Nair Carsten Milkau Keith Kevelson Christoph Hipp Witness Forest Roberts Abd-alijaleel Laraki Anthony Bruent-Bessette Samuel Gronwold Tyler Bennett christopher careta Troy R Katy Lap C Niltiac, Stealer of Souls Jon Daivd R meh Tom Noa Overloop Jude Khine R3factor. Jasmine Soni L wan na Marcelo Silva
----------------------------------------
💪 If you would also like to support this channel and have your name in the video description, then you could become my patron here / blackpenredpen
![แม่กลองสองใจ - ซีแกรม โตเกียว มิวสิค [ OFFICIAL MUSIC VIDEO ]](http://i.ytimg.com/vi/GSQfVjcr_f4/mqdefault.jpg)
If Berkeley felt like being evil they could've made the official solution just a verification of the answer without a derivation.
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
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
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
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.
11:07 OMG THAT CURVE HE MADE ON THE BOARD IS SOO MAJESTIC
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 😢
Ahh! Whenever i go to new places,I always first go to a hotel
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
That was really good. It seems like a hard differential equation and it is but you solved it. It's hard to know some times what direction to go. Like when you saw 2 directions to go earlier, but rather than choosing once, you made a hybrid of both. That was smart. I really wish I knew more about this stuff. Well, anyways great video! I look fwd to future ones.
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!
I love calculus for fun problems like that
wow, beautiful equation i liked it a lot
I started watching your videos when you did them in thazt building. Such great memories.
😃
Determining a function based on an equation containing the function and ifs inverse and some other relation, and a point on its derivative.
Nicely explained ✨
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?
I love how in the comments we all got different approaches
Super cool, one love
Awesome sir
this was really fun
Beautiful
Very cool!
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!
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
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.
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)
Cool funtional equation!
😄
love that chalkboard
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!
At 5:30 if we use l'hopital for the second limite then we will get f'(x)=0
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)
12:00... how about using log diff as take ln in both side? such as (ln f)' = f'/f
Very fun indeed
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.
UC Berk incoming freshman here. If this is what all math at Cal is like I am shivering in my boots!
Wow!!
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.
Like this problem
We can just apply lhopital rule as soon as we see 0/0 form
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).
this was neat
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
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.
What about l'hospital ? Wouldn't that be a shortcut?
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)
But this is very hard to know, it's complicated, can anyone know where can I get detailed explanation?
I was following along, but by the end of the video I was like wait, what was the question again? Lol
You could have just use LHoptial’s rule to evaluate the limit
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 ?
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...
Neat.
Where are you?? I am on campus and would love to meet you!!
I was there last week. Now I am back to LA.
You could have easily used l hospital rule and reached to the function very fast
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
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?
red chalk white chalk yaaay
😆
where you got stuck i used L' hospital rule due to 0/0 form. then it got easier ahead..
Holy crap...
Cool
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.
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?
Booommmm..
🤯🤯🤯🤯
7:33 How did he get tht note!!
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 😃
y is not a quantity, it is an arbitrary real number. In particular real numbers of the form "x+h" will work.
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
cant believe blackredpen was IN MY DISCUSSION SECTION ROOM! sad that I didnt see him :(
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
Why we let x=y=1?
I’m going to every room in evans till I find you
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 😆
Hey just got an idea ! Could you do next time 0^z=1 ? Thank you for your videos !!
7:16 - why so?
asnwer= 1 isit
I didn't understand this 😟
Why is x^x/(x+h)^(x+h)=1
中间那个替换有点不严格，上下同乘除因子即可！NⅰCe
7:39 The part about the composition is not convincing, some steps are missing I guess
Try ln(1+ln(x))=1
You should use L opital
Highly probable question for jee advanced
This more fun: 7/93 + 1 + 5 + 7 + 8 = 7/93 x 1 x 5 x 7 x 8.
Shoot, are you at Cal?
Last week.
Blackchalkredchalk
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
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
Even I didn't get the part u didn't get
hi
sir iam.like maths but english very past 😭😭😭
?
@@a_minor my english very poor iam solw talking english only understand
I thought you live in China 🤷♂️
tricky
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
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
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.