My First Berkeley Math Tournament Problem

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.

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[JayOnDaCob]

After watching your calc videos for abt a year, I finally am in calc 1

[flashaymen]

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

[jschnei]

If Berkeley felt like being evil they could've made the official solution just a verification of the answer without a derivation.

[SOTminecraft]

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.

[xinpingdonohoe]

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).

[blackpenredpen]

I might show up to BMT on 11/5/2022 just to say hi.

[danitigre]

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 👊

[Mutual_Information]

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

[usualhacker]

At 5:30 if we use l'hopital for the second limite then we will get f'(x)=0

[HCGamingTV]

I'm really enjoying these. Takes me back to uni days.

[SlipperyTeeth]

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.

[pitapockets]

11:07 OMG THAT CURVE HE MADE ON THE BOARD IS SOO MAJESTIC

[xiaanubhabgoswami]

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

[alejrandom]

I love how in the comments we all got different approaches

[j.u..n]

Ahh! Whenever i go to new places, I always first go to a hotel

[maalikserebryakov]

Determining a function based on an equation containing the function and ifs inverse and some other relation, and a point on its derivative.

[blackpenredpen]

The first-place team wins a $1000 bprp scholarship and I will be there!

[guilhermerocha]

I started watching your videos when you did them in thazt building. Such great memories. <3

[chazzbunn]

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.

[giovannicaiolo]

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