Can you solve these 3 calculus tiebreakers the fastest? (2023 Berkeley Math Tournament)
Вставка
- Опубліковано 9 жов 2024
- Let's do the calculus tiebreaker from the 2023 Berkeley Math Tournament (BMT). We must evaluate a limit, an integral, and an infinite series. The first two questions are quite straight forward but the last one is a bit tricky! Try the other parts of BMT here berkeley.mt/ar...
If you are participating in the BMT and you want to honor your coach, then apply for the BMT x BPRP Student Mentor Appreciation Scholarship here: berkeley.mt/ne...
Big thanks to all my subscribers and Patreon patrons for making this donation to BMT possible!
Check out @tongabonga video • An Infinite Sum from t...
----------------------------------------
Big thanks to my Patrons for the full-marathon support!
Ben D, Grant S, Mark M, Phillippe S. Michael Z, Jan P. Devun C. Stefan C. Ethan BW Didion S. NN Minkyu Y Brandon F Levon M shortsleeve Jack P Gabriel G David H Mateo F Emma M
💪 Support this channel and get my math notes by becoming a patron: / blackpenredpen
🛍 Shop my math t-shirt & hoodies: amzn.to/3qBeuw6
----------------------------------------
#blackpenredpen #math #calculus #apcalculus
If you are participating in the BMT and you want to honor your coach, then apply for the BMT x BPRP Student Mentor Appreciation Scholarship here: berkeley.mt/news/black-pen-red-pen-award/
The last sum can be cheesed neatly. For n = 0 we have n³/n! = 0, so we can start with n = 1 and then reduce: n³/n! = n²/(n-1)! therefore reducing the power by 1. We then go with n-1 = k which is legit since now n starts at 1. We then get ∑(k+1)²/k! starting with k = 0. This is same as ∑k²/k! + 2∑k/k! + ∑1/k!. The last one is just e, the one in the middle is also e: discard k = 0 term (it's 0) and reindex. As for the first: we just do the same trick with power reduction again, finding it's 2e. So ∑n³/n! = 2e + 2e + e = 5e
or in a different way: by differentiating x e^x three times we get Sum(n^3-n)x^n/n!
this question was in my notes copy, im literally in 11th grade
3:45 no way, I'm in the video. It makes me so happy😊. Very clever solution btw
Managed to solve all of them with relative ease - super fun exercise
Hi. I solved the integral problem myself. I am in University, I saw your notification, slide it down, and started solving on my Tablet. In a nutshell I found the answer(1-π/4), when I Rechecked, it matches your answer. Very Happy.🎉😊❤
my unborn child could do this question without a tablet. in its sleep.
congrats bro! its good to see another person who got excited after solving an integral like me (for the original comment)
nobody asked (for the reply above me)
@@mohammadfahrurrozy8082 yup.
Please don't retire so soon, you're a rare and excellent internet teacher.
YEEESSSSS, conjugating and using the area of a partial circle is the best and fastest way to solve it!!!!
hello Mr sexy speed integrator
@@fxrce6929 oh shizz wasabi!! xD
limits talk about the limiting behaviour, for the 1st problem its important to note that arctan x ~x as x approaches 0.( similar to sinx) so for extremely small values of x the expression reduces to 1/2, cubing it we get 1/8. The knowledge of limiting behavior saves us from the hazarad of blindy applying the differentiation rule all the times.
Thank you!
Nice job!
The last one can also (somewhat tediously) be solved by changing the numerator such that it as much as possible with the denominator (n³=n(n-1)(n-2)+3n(n-1)+n lets you split the sum into multiple ones, all related to sum of 1/n! from 0 to infty; one should also take the first two terms out before doing that to not cause negative number shenanigans)
For the last one, you can just aim to cancel the largest term of the factorial and then add and subtract the same value to cancel the next largest term, etc. until you only have constants in the numerator.
I'll skip the bookkeeping of the index of the sum, but it works out with the first term always being 0 - allowing you to shift things up as needed and sticking to the factorial of natural numbers. Here is the math for the summand:
n^3/n! = n^2/(n-1)! = (n^2-1+1)/(n-1)! = (n+1)(n-1)/(n-1)! + 1/(n-1)! = (n+1)/(n-2)! + 1/(n-1)! = (n-2+3)/(n-2)! + 1/(n-1)! = 1/(n-3)! + 3/(n-2)! + 1/(n-1)! = 1*e +3*e + 1*e = 5e.
Noiiiiiiiceee
there are too many e's
😂
5 of them
😂
My solution is several times faster and easier, and it only even mentions e in the final answer.
Question nr 2 you can also use X=sin(t) which will lead to ∫cos(t)- (cos(t))^2 use the double formula for the second part and fill in π/2 and 0 which will give 1-π/4
Probably a useless observation, but #3 can be expressed as something like 1/k! times the convolution of the sum of (k-n)^3 and the sum of the gamma function, taking its limit as k --> ∞.
Another easy way to solve the last sum is to play with the developpement of exp(x) to show that sum(x^n*n^3/n!)=d/dx (x*d/dx (x*exp(x)))=exp(x)*(1+3x+x^2)
For (1): recall arctan(x)~x, when X->0: you immediately get the result. 😊
I find it hard to believe these are Berkeley math tournament questions. I don’t even do math competitions (too stressful) and yet I still managed all three in a combined time of 4 minutes.
The limit was easy, the integral was something an algebra student could do (if they understood what the question was asking, which they probably wouldn’t. My point being it’s mostly area of a circle.), and the final one looks so close to the Taylor series definition of e^x it’s bound to set off alarms for anyone who’s taken calculus II.
The last one is 🤯🤯🤯🤯
Number 3 can be solve using touchard polynomial
3. use the expansion of e^x to do(d/dx→*x)3 times, and plug-in x=1 will finish it.
The third one is the most intuitive tbh, you only need to know the fact that the sum from 0 to infinity of 1/n! =e
man if only india had these, it would be so epic!
You can also solve the last sum using power series and solving the deferential equations you get from it
Very interesting infinite sum. This is related to OEIS A000110.
b/c when x goes to 0, x/arctanx=1, expression=[1/(1+1)]^3=1/8
i saw wwangs solution few days ago, but this is brilliant
Banger questions
In the first one i just divided up and down by arctan and use that x/arctan x tends to 1
I was very happy that I made a Law and Is correct.
I just call it my law because I don't see it anywhere.
" Law of Usage "
Usage is the amount of percentage you must consume in right amounts
FORMULA:
% consume = amount / grams * 100
Inequality:
grams, amount ≠ 0
grams > amount
I just use chatgpt to confirm because most of the person said it's very accurate(but sometimes make mistake). So the formula and law is reasonable for me because it seems like it.
What do you suggest to Improve the " Law " ?
I'm just a Grade 9 student so don't say I'm bad because I don't even know most of the things in School.
grams > amount => amount / grams < 1
=> % consume < 100%
What your law states is that what you consume is less than a 100%
A nice observation, but what your law essentially states is that optimal consumption is less than 100%. Since we cannot consume more than 100% of what we are given, it is mathematically obvious.
@@n.rv..n Thanks
I went to Cal. The math professors there are murder. They do this kind of thing for fun and most problems were much worse than these, and took quite some time for the teaching assistants to explain. A lot more time than this video. We poor science and engineering students never had it so rough.
Bring some more fun and hard problems....
HERE IS A QUESTION FOR YOU!
Find the sum of all the subsets from the set {1,2,3,4........2020} for which sum of their elements is divisible by 5.
IT'S SUPER INTERESTING PROBLEM!!!
I know that you got this question from 3b1b the only difference is yours is upto 2022 and he's was 2000 and i believe he's answer was
1/5(2^2000+4×2^400) so yours will be 1/5(2^2020+4×2^404)
@@zimamalmuntazir6258 yeahh...
RIGHTT!!
IT WAS A WONDERFUL SOLUTION!
For the first question I just said "sin=x, tanx=x and tan inverse(x) =x. Hence (x)/(x+x) so 1/2 and 1/2^3 = 1/8 😂
3)...derivo 2 volte e^x=Σx^n/n...S=2e+6...no,ho rifatto i calcoli S=5e
3:03 Please teach me how you solve that
The +C
Why did he solve such an easy problem in a ridiculous overcomplicated way?
8:15 so now we can take the derivative.. so now we can take the derivative
hello !!!!
Would you like to spend 5 hours solving hard integrals?
4:32 how is that true for x=0?
You would get 1=1, with the convention that 0^0=1 in power series and 0!=1
👍
i wonder will you be able to do it, but, let's say we have f(x)=2^x, and g(g(x))=f(x), what's g(x)?
Let f^1/2(x) = g(x)
To find g(x)
So f(x) = g^2(x)
Put in equation you provided:
g(g(x)) = g^2(x)
Put t = g(x)
g(t) = t²
Reuse variable x (cuzwe need g(x))
Put t = x
g(x) = x²
I am assuming that g(x) is bijective here
@@Anmol_Sinha f(x) is equal to 2^x, not x^2, and if you'll put x² into f^1/2(f^1/2(x)), you'll get f^1/2(x²) and then x⁴, not 2^x
first of all what is f^1/2(x)? is it the square root of f(x)? or just a different function altogether?
@@chinmay1958 yeah, you can consider it as different function, as i said, f^1/2(f^1/2(x))=f(x), how √f(√f(x)) can be always equal to f(x)? lol
i've atempted a similar question for a while: finding a continous version of f^n(x) where f(x)=e^x. (continous in n).
Then i found that some guy named Kneser did this decades ago, but the solution is behind paywalls...
😁😁 interesting problems
1st view 😊