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
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.
There are two problems in this solution; the first is the main one, the second is less important. (log is the natural logarithm in the following) First: the integration range is [0,2] but log(x) is defined for x> 0 only; so, strictly speaking, one needs to consider the convergence problem at x_0=0 from the right. So one has to take care of this problem. A possibility is to modify the problem, working , for example in [1/2,2]. Another one is to obtain the antiderivatives only for x>0 (all the computations here hold only for x>0) and then extend the definition of the integrand function to 0. Second: at some point, a factor 1/(log x) is used and this can be done for x different from 1 only (because log(1)=0 and we cannot divide by 0). This way of arguing is a problem for the modified integration range too. So, one should split the integration range in the sub intervals [1/2,1] and (1,2] and then “glue” together the two anti derivative functions that one gets. Since such two intervals are disjoint, this means that one should carefully choose the “arbitrary ” constants involved in the two antiderivatives obtained. The problem, though, can be easily fixed by avoiding to divide by log x.
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
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.
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!
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
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
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.
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
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.
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 :)
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
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...
so my friend gave me this equation and I found the real solution, which is x=2. I was wondering how to find the complex solutions to it if there are any. the equation is x^x^3=256
Let S^j be the sum from n = j to ∞ of (x^n)/n!, i.e., S^j = (x^j)/j! + [x^(j + 1)]/(j + 1)! + ... + (x^n)/n! + ... Note that j is not an exponent but a way to say that the sum starts at some value j of n. Note also that the sum after (x^j)/j! is just S^(j + 1), so we have that S^j = (x^j)/j! + S^(j + 1) The sum we want is the Maclaurin series of e^x (just the series for e^x that appears in the video), so j = 0 and then we have that S⁰ = (x^0)/0! + S¹ Now, x^0 = 1 and 0! = 1, so (x^0)/0! = 1/1 = 1 and then we have that S⁰ = 1 + S¹
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.
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 ❤️
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
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.
One of most unique integrals I’ve seen in a while
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.
🤣🤣7:49.... He has done his work & there's no reward except this mad behavior.
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!
Feels great to be able to solve it while being in high school.
Thanks bprp.
You started my year old obsession with integrals🙇♂️.
I passed the hardest topics in Calculus thanks to you. Love how detailed your videos are.
That was fun! I have not done calculus classes since 1980 as a mathematics major but I could follow this! Great work and explanations!
7:50 "And that's a good place to stop."
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
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.
Bro didn't know how to solve the integration
He is the integration
It is probabbly the most beautiful integral I've ever seen.
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.
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
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
My first instinct was "this probably simplifies to some kind of summation involving e," but I have no clue how to get there, lol.
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.
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.
I burned 1 billion brain cells trying to understand this monster integral, and failed.
Very fun integral, adding it to the list!
This is such an easy solution when you see someone else do it, but I would have never solved it by myself!
Gotta love when things cancel out and you get a neat and manageable solution.
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 :)
I solved it without watching the video! I feel so proud!
There are two problems in this solution; the first is the main one, the second is less important. (log is the natural logarithm in the following) First: the integration range is [0,2] but log(x) is defined for x> 0 only; so, strictly speaking, one needs to consider the convergence problem at x_0=0 from the right. So one has to take care of this problem. A possibility is to modify the problem, working , for example in [1/2,2]. Another one is to obtain the antiderivatives only for x>0 (all the computations here hold only for x>0) and then extend the definition of the integrand function to 0. Second: at some point, a factor 1/(log x) is used and this can be done for x different from 1 only (because log(1)=0 and we cannot divide by 0). This way of arguing is a problem for the modified integration range too. So, one should split the integration range in the sub intervals [1/2,1] and (1,2] and then “glue” together the two anti derivative functions that one gets. Since such two intervals are disjoint, this means that one should carefully choose the “arbitrary ” constants involved in the two antiderivatives obtained. The problem, though, can be easily fixed by avoiding to divide by log x.
Well , that's probably my new favorite integral in pure mathematics.
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
The way he speaks softly and calmly, and is reassuring, reminds me of Steve from Blues Clues, except with integrals. :B
Me watching this as an 8th grader and not knowing calculas; hmm interesting
To the folks who want a more compact form: 2sinh(1). The hyperbolic sin function.
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.
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!
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
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 incredible solution 👍
That was pure magic Steve.❤
This is just beautiful!!! Love your work!!
Integrating like a boss.
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
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'm a math tutor at my college, I gotta practice that two pen technique! I love using different colors when I'm helping students.
This is glorious!
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.
Amazing video! So much fun to watch!
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 :)
7:50 when u forget u r mathematician not martial artist.
This guy is a genius
Goddamn was that a beauty of an equation
I Didn't expect to solve this myself in 5min lol
That was a beautiful solution,clever
Despite having more than a 1M subscribers, why does he still continue to hold the mic 🤔
What a beatiful integral and mathematics!
This was a fun one!
The braingasm at the end
In √-1 put e^iπ and take out the root of -1
I didn’t understand any of that but it was entertaining
Fun. watching these videos make you learn methods to solve problems.
You deserve 500m subscribers
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!
Wtf
It seems so easy when I've seen the solution but I could never think of this during a competition
I love your content man. Now that I’ve solved it it seems almost trivial - of course it’s just an infinite series, :p
Amazing !
the laugh at the end...
Before the Solution :
Integral is easier but that series to be formed is main thing which is like (2^n)/(n^2)
this dude can quick-select between a red and blue whiteboard marker
Thats such a cool question
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 :)
You could have removed the X from the base and replaced it with e to the power of ln(x)
This
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
Sir I m ur die hard fan Really amazing content delivered
Love even the Klein bottle behind you!
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.
How to find the sum of these series (Nc0)³+(Nc1)³+(Nc2)³+....+(NcN)³.is it possible?
Calculus is so fascinating
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...
Making of the thumbnail........this guy....give me that fucking medal 🏅
HES POPPUNG OFF
Perfect as always 👏👏
I was missing your videos. =)
Mom! A new bprp is here!
When I saw tis thumbnail, I thought the answer was Nepal
so my friend gave me this equation and I found the real solution, which is x=2. I was wondering how to find the complex solutions to it if there are any.
the equation is x^x^3=256
I did this on my own just messed up a lil in the last . yay
Please do the derivation of ln(-1)
Can someone explain how modifying the series equation to fit ln(x) get a +1 for n = 0. 4:32
Let S^j be the sum from n = j to ∞ of (x^n)/n!, i.e.,
S^j = (x^j)/j! + [x^(j + 1)]/(j + 1)! + ... + (x^n)/n! + ...
Note that j is not an exponent but a way to say that the sum starts at some value j of n.
Note also that the sum after (x^j)/j! is just S^(j + 1), so we have that
S^j = (x^j)/j! + S^(j + 1)
The sum we want is the Maclaurin series of e^x (just the series for e^x that appears in the video), so j = 0 and then we have that
S⁰ = (x^0)/0! + S¹
Now, x^0 = 1 and 0! = 1, so (x^0)/0! = 1/1 = 1 and then we have that
S⁰ = 1 + S¹
Your are the best bro 👍
You too
and it is possible to find a primitive of this function?
He is a jewel
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...
I've never managed to find big enough square root in my garden. Thanks!
Good job sir black 👍
You are the best! 🔥🔥🔥