The Fall 2023 Berkeley Math Tournament will be held on the UC Berkeley campus on Saturday, Nov. 4th and I will be there! For more info and registration, check out berkeley.mt/events/bmt-2023/.
This is why I love this channel. Takes an integral that, quite frankly, almost makes me wanna cry, and turns it into something where I just need to brush up on infinite series to understand.
@@blackpenredpen No problem! Also I wanted to point out that I’ve been watching you for a while now and never expected you to see any of my comments, let alone reply to one! Thanks so much for replying, it made my day ❤️
That was fun. Thanks. I thought it was interesting that the integral itself wasn't that hard, but rather the simplification of the series. Have you ever thought about getting refillable pens? They work great.
I love your videos. Over the years you’ve always been so helpful, so direct, and so charismatic. Even if I don’t understand what’s happening it’s still entertaining. Thank you
My solution went basically the same as yours though figuring out everything took like 45mins , I'm just glad I was actually able to do one of these more advanced integrals.
EXCELLENT VIDEO, thank you for this. I struggled so much with infinite series, it's nice to see a fairly practical application of it to solve an incredibly difficult problem
Really Integration Bee very interesting. Thanks for video. For MIT integration bee recently it was published a book with solutions to problems of Qualifying Tests from 2010 to 2023
this is a beautiful one , like after seeing how its done it dosen't seem hard but it would probably take me days before actually seeing the Maclaurin Series for e^x with the ln x being there, props to the guys that found this beauty
Amazing mathmatical thought！using known theory to present unknown situation,for example we have learned the sum of x divided by n! is e to the power of x,then lnx is e to the power of lnx which means x,that simplified this question a lot! I 've learned a lot via watching your video!
Let y = f(x), where f(x) is the integrand. Then after some tedious manipulation one has ln y =-1 + e^(ln x) = -1+ x So y = e^(x-1) The integration can be simplified as Integrate[e^t, {t, -1, 1}] = e^t, {t, -1, 1} = e - e^(-1) Hence, Integrate[f(x), {x, 0, 2}] = e - 1/e.
Hello Professor. I applied to UC Berkeley this year so I might see you on campus next year if I get in. If I do I’ll definitely join the Berkeley math tournament.
I used to be able to solve questions of this difficulty level 10 years ago. I forgot most of the things and only remember some basic integral and derivatives formulas and concepts now :P Anyway this was cool to watch :)
if you convert each x (in the bases) to e^lnx, the series you get requires less finagling, as the power of lnx matches the index of the respective factorial
Oh great question, not that difficult btw. I thought it would require some sort of wizardry, but when I wrote the question and simplified to the power of x, I quickly saw e^ln x, just needed to multiply and divide by ln x and create the e^x series, then it was pretty simple :) Once again, a great and simple question which emphasizes on finding patterns
You could also use expansion of taylor series of e^x (its n-th derivative is always e^x) so that sum of those powers of x equals (x-1)/lnx and then substitute it back into the integral. Thats how i got it. Sorry for my english.
Here's how I did it: 1. Make the substitution u=log_2(x). This gives x=2^u => dx=2^u*ln(u)du. The bounds become -inf to 0. Our integral then becomes Integral(-inf to 0)(2^[u]*2^u*ln(2))du=ln(2)*Integral(-inf to 0)(2^[u]*2^u)du 2. Rewrite the integral as an infinite sum of smaller integrals. We are going to do so the following way: ln(2)*(Integral(-1 to 0)(2^[u]*2^u)du+Integral(-2 to -1)(2^[u]*2^u)du+Integral(-3 to -2)(2^[u]*2^u)du+…) =ln(2)*Sum(k=-inf to -1)(Integral(k to k+1)(2^[u]*2^u)du). The key insight here lies in the definition of the greatest integer function: for some integer k, [x]=k iff k
Integrals like this are rarely hard. They are just time consuming because they require a bit of algebra. Once you simplify the integrand by knowing a power series, there are no integration techniques required. This problem is more like a race than a puzzle. But that's what most bee integrals are. You can't put many genuinely difficult integrals in an integral bee because of the format.
This was an easy problem, (no offense), but the first thought in my mind was trying through u-substitution but I guess it would be easier to use Taylor series for defining e I mean it was pretty tough for some speed solving like an integration bee, but problems like these make you know about the essence of series...
Very interesting and not so difficult to solve if you know series. Honestly I haven’t expected to solve that without substitution or partial integration
The Fall 2023 Berkeley Math Tournament will be held on the UC Berkeley campus on Saturday, Nov. 4th and I will be there! For more info and registration, check out berkeley.mt/events/bmt-2023/.
This is why I love this channel. Takes an integral that, quite frankly, almost makes me wanna cry, and turns it into something where I just need to brush up on infinite series to understand.
I’m laughing when he said “I know this has the N but that’s ln” his facial expression and the way he delivered is killing me😂.
Same
Lol😂
HAHAA
SAMEE
😂
Ya, and what makes it extra funny to me is that these types of jokes he makes might have come from IRL mistakes he saw at some point :D
Looks so complex yet it’s so simple! Thanks for the tutorial, you never disappoint. 😊
Thanks!
@@blackpenredpen No problem! Also I wanted to point out that I’ve been watching you for a while now and never expected you to see any of my comments, let alone reply to one! Thanks so much for replying, it made my day ❤️
One of most unique integrals I’ve seen in a while
That was fun. Thanks. I thought it was interesting that the integral itself wasn't that hard, but rather the simplification of the series. Have you ever thought about getting refillable pens? They work great.
I wouldn't have been able to do this on my own, but man is this one satisfying
I did this integral after 100 integrals lol
full video th-cam.com/video/jQz1gQ24OHc/w-d-xo.html
1,000,000 SUBS
Hi.
I love your videos. Over the years you’ve always been so helpful, so direct, and so charismatic. Even if I don’t understand what’s happening it’s still entertaining. Thank you
Thank you!
🤣🤣7:49.... He has done his work & there's no reward except this mad behavior.
Feels great to be able to solve it while being in high school.
Thanks bprp.
You started my year old obsession with integrals🙇♂️.
That was fun! I have not done calculus classes since 1980 as a mathematics major but I could follow this! Great work and explanations!
It's so simplistic and beautiful, what an awesome integral
On a side note, we would be screwed if we didn't have Euler's symbol
I passed the hardest topics in Calculus thanks to you. Love how detailed your videos are.
My solution went basically the same as yours though figuring out everything took like 45mins , I'm just glad I was actually able to do one of these more advanced integrals.
Not doing maths anymore but it is a sheer joy to watch you doing all these interesting problems
Same here. His joy is relatable, for 20-year old me.
I am currently learning series and sequences in high school and this was so cool to see!
you have series and sequences in Highschool curriculum? Isn’t it college level?
@@kpax9284 its taught in AP bc calc, taught in high school
@@kpax9284 Pre-calc junior year most likely.
It is probabbly the most beautiful integral I've ever seen.
EXCELLENT VIDEO, thank you for this. I struggled so much with infinite series, it's nice to see a fairly practical application of it to solve an incredibly difficult problem
Very fun integral, adding it to the list!
This is incredible solution 👍
Bro didn't know how to solve the integration
He is the integration
7:50 "And that's a good place to stop."
One of the best integrals I've ever seen. :)
It means you have not even seen 1% of the integral problems...
THAT WAS AWESOME! thanks bprp :)
This is such an easy solution when you see someone else do it, but I would have never solved it by myself!
The way he speaks softly and calmly, and is reassuring, reminds me of Steve from Blues Clues, except with integrals. :B
Really Integration Bee very interesting.
Thanks for video.
For MIT integration bee recently it was published a book with solutions to problems of Qualifying Tests from 2010 to 2023
What a beatiful integral and mathematics!
That was a beautiful solution,clever
this is a beautiful one , like after seeing how its done it dosen't seem hard but it would probably take me days before actually seeing the Maclaurin Series for e^x with the ln x being there, props to the guys that found this beauty
And even better, e - 1/e is W^-1(1) + W^-1(-1)
W(x) is the lambert W function
2sinh(1) is even better imo
@@aadhavan7127 😂😂😂
FISH
This is glorious!
Amazing mathmatical thought！using known theory to present unknown situation,for example we have learned the sum of x divided by n! is e to the power of x,then lnx is e to the power of lnx which means x,that simplified this question a lot! I 've learned a lot via watching your video!
Gotta love when things cancel out and you get a neat and manageable solution.
Crazy how I know all the tools he used and still I coudn't solved it myself.
Let y = f(x), where f(x) is the integrand.
Then after some tedious manipulation one has
ln y =-1 + e^(ln x) = -1+ x
So
y = e^(x-1)
The integration can be simplified as
Integrate[e^t, {t, -1, 1}] = e^t, {t, -1, 1} = e - e^(-1)
Hence,
Integrate[f(x), {x, 0, 2}] = e - 1/e.
Amazing !
This is just beautiful!!! Love your work!!
I'm a math tutor at my college, I gotta practice that two pen technique! I love using different colors when I'm helping students.
Amazing video! So much fun to watch!
This was a fun one!
Hello Professor. I applied to UC Berkeley this year so I might see you on campus next year if I get in. If I do I’ll definitely join the Berkeley math tournament.
I don’t teach at UC Berkeley. I live in the SoCal area but I go back to UCB campus whenever I can 😃
@@blackpenredpen Oh. Where do you teach then?
@@rhversity5965 he teaches in kenya
I used to be able to solve questions of this difficulty level 10 years ago. I forgot most of the things and only remember some basic integral and derivatives formulas and concepts now :P Anyway this was cool to watch :)
Integrating like a boss.
My first instinct was "this probably simplifies to some kind of summation involving e," but I have no clue how to get there, lol.
Thats such a cool question
if you convert each x (in the bases) to e^lnx, the series you get requires less finagling, as the power of lnx matches the index of the respective factorial
Fun. watching these videos make you learn methods to solve problems.
Love even the Klein bottle behind you!
Calculus is so fascinating
Beautiful! So easy to follow along with your argument. But, to come up with it on my own? Sheesh, it would take me months, maybe infinite months!
I was missing your videos. =)
I solved it without watching the video! I feel so proud!
To the folks who want a more compact form: 2sinh(1). The hyperbolic sin function.
Sir I m ur die hard fan Really amazing content delivered
I love your content man. Now that I’ve solved it it seems almost trivial - of course it’s just an infinite series, :p
Oh great question, not that difficult btw. I thought it would require some sort of wizardry, but when I wrote the question and simplified to the power of x, I quickly saw e^ln x, just needed to multiply and divide by ln x and create the e^x series, then it was pretty simple :)
Once again, a great and simple question which emphasizes on finding patterns
You could also use expansion of taylor series of e^x (its n-th derivative is always e^x) so that sum of those powers of x equals (x-1)/lnx and then substitute it back into the integral. Thats how i got it. Sorry for my english.
You deserve 500m subscribers
You are the best! 🔥🔥🔥
Good job sir black 👍
Your are the best bro 👍
You too
Integral from 0 to 1
f(x)=2^[(logx)]
Where [ . ] denotes the greatest integer function.
Note: it is logx with base 2.
Here's how I did it:
1. Make the substitution u=log_2(x). This gives x=2^u => dx=2^u*ln(u)du. The bounds become -inf to 0. Our integral then becomes Integral(-inf to 0)(2^[u]*2^u*ln(2))du=ln(2)*Integral(-inf to 0)(2^[u]*2^u)du
2. Rewrite the integral as an infinite sum of smaller integrals. We are going to do so the following way:
ln(2)*(Integral(-1 to 0)(2^[u]*2^u)du+Integral(-2 to -1)(2^[u]*2^u)du+Integral(-3 to -2)(2^[u]*2^u)du+…)
=ln(2)*Sum(k=-inf to -1)(Integral(k to k+1)(2^[u]*2^u)du).
The key insight here lies in the definition of the greatest integer function: for some integer k, [x]=k iff k
I burned 1 billion brain cells trying to understand this monster integral, and failed.
In √-1 put e^iπ and take out the root of -1
pretty easy ngl did it as soon as i thought of writing the radicals in terms of powers and the x^a must been some series expansion :)
Perfect as always 👏👏
Goddamn was that a beauty of an equation
You could have started the power series from 0 as it already had the ln^0 x term, I feel it would have reduced some work. Amazing video as always...
He is a jewel
The braingasm at the end
I Didn't expect to solve this myself in 5min lol
HES POPPUNG OFF
Sir math is love for you and me❤️
Mom! A new bprp is here!
Integrals like this are rarely hard. They are just time consuming because they require a bit of algebra. Once you simplify the integrand by knowing a power series, there are no integration techniques required. This problem is more like a race than a puzzle. But that's what most bee integrals are. You can't put many genuinely difficult integrals in an integral bee because of the format.
I didn’t understand any of that but it was entertaining
So good! 😁
Making of the thumbnail........this guy....give me that fucking medal 🏅
This was an easy problem, (no offense), but the first thought in my mind was trying through u-substitution but I guess it would be easier to use Taylor series for defining e
I mean it was pretty tough for some speed solving like an integration bee, but problems like these make you know about the essence of series...
did this one in my head
Before the Solution :
Integral is easier but that series to be formed is main thing which is like (2^n)/(n^2)
the laugh at the end...
Amazing, that was amazing
Despite having more than a 1M subscribers, why does he still continue to hold the mic 🤔
Your are great sir ❤❤😊
Mathematics is the beginning of opening the mind and knowledge of the world and the universe ❤️❤️❤️
7:50 when u forget u r mathematician not martial artist.
I needed a u substitution, e^u=x. To see the pattern, but when I got to e^(e^u+u-1)du, I converted back to the x world.
Me watching this as an 8th grader and not knowing calculas; hmm interesting
First time I’ve seen x to the power of a series! Lol
this dude can quick-select between a red and blue whiteboard marker
How to find the sum of these series (Nc0)³+(Nc1)³+(Nc2)³+....+(NcN)³.is it possible?
I did this on my own just messed up a lil in the last . yay
amazingly good
Very interesting and not so difficult to solve if you know series. Honestly I haven’t expected to solve that without substitution or partial integration
Take my 👍. Great job
Please do the derivation of ln(-1)
this was so easy. wow.