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
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.🎉😊❤
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)
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.
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.
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 --> ∞.
I did problem 1 a bit differently: For problem 1, if you take the cube out and write the limit as 1 over its reciprocal, you get the cube of the limit of 1/(1 + x/arctan(x)). We know tanx/x goes to 1 as x->0, i.e. tan(x) is approximately x. Then arctan(x) is also approximately x in the limit, so the original limit becomes (1/(1+1))^3 = 1/8.
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
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.
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)
Literally did in mind (JEE aspirant) first I simplified it as n^2/(n-1)! Then wrote it as n^2-1+1/(n-1)! Then it is n+1/(n-2)! +1/(n-1)! Which is 1/(n-3)! +3/(n-2)! +1/(n-1)! And its sum is 5e
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)
I've solved your equation like 1^x = 2 without seeing your video (by seeing just question) . But I got only principal value, i.e, without getting any "n" on my answer. Could you please tell me, is it necessary to get only general solutions?
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
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...
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.
3. sum(x^n/n!,n=0..infinity) = exp(x) sum(nx^n/n!,n=0..infinity) = x*exp(x) (by differentiation once and shifting) sum(n^2x^n/n!,n=0..infinity) = x*d/dx(x*exp(x)) = x*(x+1)exp(x) sum(n^3x^n/n!,n=0..infinity) = x*d/dx((x^2+x)*exp(x)) sum(n^3x^n/n!,n=0..infinity) = x*((2x+1)*exp(x)+(x^2+x)exp(x)) sum(n^3x^n/n!,n=0..infinity) = x*(x^2+3x+1)exp(x) and then if we plug in x=1 we will get 5e
![ออร่า (AURA) - JUEPAK feat. SARAN [OFFICIAL MV] | เอแคลร์ จือปาก](http://i.ytimg.com/vi/RY9j7Lzp-FI/mqdefault.jpg)
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
Yeah, also solved in this way. Much simplier )
I did it the same way !
i was looking for this comment, this approach looked obvious to me
3:45 no way, I'm in the video. It makes me so happy😊. Very clever solution btw
Thanks!
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.
@@IamACrafter fr, just rationalise it
Please don't retire so soon, you're a rare and excellent internet teacher.
Managed to solve all of them with relative ease - super fun exercise
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.
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)
Clear, simple, best.
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
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
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.
@@programmingpi314 can I see it?
Nice job!
Thank you!
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 --> ∞.
You can also solve the last sum using power series and solving the deferential equations you get from it
I did problem 1 a bit differently:
For problem 1, if you take the cube out and write the limit as 1 over its reciprocal, you get the cube of the limit of 1/(1 + x/arctan(x)). We know tanx/x goes to 1 as x->0, i.e. tan(x) is approximately x. Then arctan(x) is also approximately x in the limit, so the original limit becomes (1/(1+1))^3 = 1/8.
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
3. use the expansion of e^x to do(d/dx→*x)3 times, and plug-in x=1 will finish it.
thanks
awesome video
great video thanks ! the solutions seem so obvious once you've shown them
note that sum(n(n-1)...(n-k)/n!) is always e, then write n^3 as n(n-1)(n-2) + 3n(n-1)+n, you immediately get the result 5e.
The last one is 🤯🤯🤯🤯
Number 3 can be solve using touchard polynomial
I know your channel is huge, but I thought it was so cool that you can be identified by Akinator(the game).
Thanks for doing what you do like always!❤
i saw wwangs solution few days ago, but this is brilliant
In the first one i just divided up and down by arctan and use that x/arctan x tends to 1
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
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.
For (1): recall arctan(x)~x, when X->0: you immediately get the result. 😊
man if only india had these, it would be so epic!
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)
Please clarify. dy/dx Can be written as Dy. That means D=d/dx. My questions: does D in general means d/dx or it means d/dx because y=f(x)
You are correct.
Banger questions
b/c when x goes to 0, x/arctanx=1, expression=[1/(1+1)]^3=1/8
Sir can you do integral tan^-1(sqrt(x+1)) using D I method
Very interesting infinite sum. This is related to OEIS A000110.
3:03 Please teach me how you solve that
8:15 so now we can take the derivative.. so now we can take the derivative
7:58 Siri trying to beatbox:
Literally did in mind (JEE aspirant) first I simplified it as n^2/(n-1)!
Then wrote it as n^2-1+1/(n-1)! Then it is n+1/(n-2)! +1/(n-1)! Which is 1/(n-3)! +3/(n-2)! +1/(n-1)! And its sum is 5e
3:42 I just watched it literally 5 minutes ago 😂
The +C
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!
Sum(n^3)*Sum(1/n!) = Sum 1/(factor)(n-3)!
I didn't get why d(e^nx/n!)/dx = ne^nx/n! and the 1/n! factor wasn't part of the diferentiation
I've solved your equation like 1^x = 2 without seeing your video (by seeing just question) . But I got only principal value, i.e, without getting any "n" on my answer. Could you please tell me, is it necessary to get only general solutions?
which grade are these questions for
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...
Would you like to spend 5 hours solving hard integrals?
😁😁 interesting problems
3)...derivo 2 volte e^x=Σx^n/n...S=2e+6...no,ho rifatto i calcoli S=5e
The last one is just EEEEEEEE
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 😂
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.
hello !!!!
3.
sum(x^n/n!,n=0..infinity) = exp(x)
sum(nx^n/n!,n=0..infinity) = x*exp(x) (by differentiation once and shifting)
sum(n^2x^n/n!,n=0..infinity) = x*d/dx(x*exp(x)) = x*(x+1)exp(x)
sum(n^3x^n/n!,n=0..infinity) = x*d/dx((x^2+x)*exp(x))
sum(n^3x^n/n!,n=0..infinity) = x*((2x+1)*exp(x)+(x^2+x)exp(x))
sum(n^3x^n/n!,n=0..infinity) = x*(x^2+3x+1)exp(x)
and then if we plug in x=1 we will get 5e
INDIANS ATTENDANCE HERE 😂😂
👇
رائع
👍
Why did he solve such an easy problem in a ridiculous overcomplicated way?
1.3 million subs but 23k views... sad
1st view 😊