Berkeley Math Tournament calculus tiebreaker
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- Опубликовано: 3 июн 2024
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This calculus tutorial contains the solution to the 2022 Berkeley Math Tournament calculus tiebreaker! It contains a limit, derivative, and an integral at calculus 2 (i.e. AP calculus BC) level. The time limit is 15 minutes but you will also have to race with other times during the tournament. I love these math competition problems because they really force you to think and they really challenge your problem-solving skills. It's okay if you can't solve these competition problems. I, too, can't solve many of them either. It really takes lots of effort and practice in order to be good at these. Finally, I just want to tell you, to be sure to enjoy the learning process and have fun!
0:00 2022 Berkeley Math Tournament calculus tiebreaker
0:08 Q1, evaluate a limit with an integral
3:10 Q2, find the 2022nd derivative of f(x)=e^x*sin(x) at x=0
7:36 Q3, evaluate the integral of arctan(x)/x from 1/e to e
#calculus #tutorial #berkeley #blackpenredpen
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Is this a spectator event? Can anyone come by and watch?
Can you solve the equation i+1=e^x? Thx
Write 1+I in polar form as sqrt(2)*exp(i pi/4) to find x= 1/2 ln(2)+ i pi (2k +1/4) for arbitrary integer k. I don't much like the sound of the 3 Ks in your name though, und ich spreche auch ein bisschen deutsch.
For the second problem, if you differentiate e^x*sinx 4 times, you end up getting -4e^x*sinx, meaning that every 4 derivatives, you're just multiplying by -4. Notice that 2022/4 is 505 with a remainder of 2, meaning we're going to be multiplying by -4 505 times, then differentiating twice more. So that gives us (-4)^505*e^x*sinx. If we differentiate that twice more, we get (-4)^505*2*e^x*cosx, which is our 2022nd derivative. Now we can plug in 0 for x, which leaves (-4)^505*2. We can simplify this a bit to get (-2^2)^505*2=(-1)^505*2^1011. Recall that -1 to an odd power is simply -1, so the answer is just -2^1011
this is how I did it tooo
this is how i would have figured it out too
yea i think this is the most basic solution that normal calculus student can do
This is what I did
Half my life ago I had the kind of mind that would have spontaneously started solving these. Now I sit and enjoy someone else's solution. I feel old but I am glad you exist!
💀 damn man
I often overlook these videos, just assuming that these problems would be faar too difficult, but I'm pleasantly surprised that I can follow along easier and figure out the answers myself.
RIGHT! THIS WAS ONE OF THE FIRST VIDS WHERE I WAS ABLE TO SOLVE THE QUESITONS
fr man same here except for the last one(i overcomplicated it a bit too much)
In the third question, another method would be substituting
X= 1/t
dx=-1/t² and proceed you will get answer π/2 easily.
w
Yeah when I was watching...
I thought of the same
Yeah, an equivalent way which I wrote below is to split the integration between 1/e and 1, and 1 and e, then do the same substitution in just one of the ranges.
At 10:50, we may use kings property and formula for arc tan a + arc tan b that would give arc tan infinity put (pi/2) there and just solve. Quite a bit easier that way
Great video! Good explanations. I'm glad for you and everyone attending the BMT.
This is really cool man, I’m currently attending UC Berkeley right now and I think it’s awesome that you sponsored our Math Tournament as well as graduated from here!
The last one I never would be able to solve. Thanks for the video!
I had another solution for the 2022 problem I set S=e^x sinx and noted that d^4/dx^4S= (-4)S. Then you can do the 2020 differentiations and get a factor of (-4)^2020 and do the last two differentiations by hand. I like your way more. At times the complex way ist just the simpler way;-)
That was brilliant!!!
For the third question, we can also proceed with another method from third step... we can write it as (it is a property of DI) ∫ 0 -> 1 (f(x) + f(-x)) dx then you will get something like arctan(x)+ arccot(x) which is equal to ∏/2.. ( I couldn't find the right pi 😅.. and yeah x here means the e to the power u.... Btw, noice integrals.. Would love to see more of such covered in your future videos.. 😎
i was able to do 1st and 3rd but the using complex number for 2nd was beautiful.
For the third integral, you can make the substitution u = 1/x. You find that the integral is equal to that of arccot(x)/ x on the same interval. So the integral is equal to 1/2 of ∫(arctanx + arccotx) / x dx. Then you just need that arctanx + arccotx = π/2 to find the answer quite simply.
For the second problem, that one Michael Penn video about linear algebra with derivatives was still fresh in my mind, so I'll be using that.
Notice that when you differentiate a linear combination of e^x*sin(x) and e^x*cos(x) you get another linear combination of the two functions.
Let the first number in the vector space be the e^x*sin(x) terms and the second the e^x*cos(x) terms.
Differentiation can be represented by this matrix:
D = [[1, -1]; [1, 1]].
The starting vector is S = [1; 0].
That means we need to find D^2022 * S.
Usually when a big power of a matrix shows up it's a good idea to diagonalize:
Eigenvalues: (1 + i), (1 - i)
Eigenvectors: [1; i], [1; -i]
D = A'XA
= [[-i, -1]; [-i, 1]][[1 + i, 0]; [0, 1 - i]][[1, 1]; [i, -i]]
D^2022 = A' * X^2022 * A
= A' * [[(1 + i)^2022, 0]; [0, (1 - i)^2022]] * A
= A' * [[-i(2^1011), 0]; [0, i(2^1011)]] * A
I'm in a bit of a rush but the next step should be multiplying out A' * X * A * S to get the final result
Note: I write my matrices inline like this:
A = [[R1C1, R1C2, ...]; [R2C1, R2C2, ...]; ...]. Commas separate elements in rows and semicolons separate rows.
For the 3rd one, I think the shortest way is to split up the range of integration from 1/e to 1 and from 1 to e, then substitute x=1/y in the first interval to get the integral from 1 to e of ( tan(x)+tan(1/x)) /x, which is pi/2 times the integral of 1/x in this range, giving the required answer.
Thanks PROF 👍
Please Make a Playlist with Competition Math problems ❤
The third question is actually simpler: you divide your integral in due part ( I=1/2*(I+I) ) and in the second you substitute t=1/x. After some calculations, the 2nd integral is similar to the 1st one, except arctan(x)-->arctan(1/x); if you put them together you have I=1/2*∫(arctan(x)+arctan(1/x))/x*dx. Now we know that sum is identically pi/2 (if x>0), therefore I=pi/4*∫1/x*dx, and after elementary calculations you obtain I=pi/2
That's essentially the same as what he ended up with, but avoiding the exponential substitution which he used to get to the identity for arctan (x) + arctan (1/x).
I think in last question integration from -1 to 1 tan( inverse) 1/e^v dv is not equal to π/2 - tan(inverse)e^v because limit of v is varry from -1 to 1 hence first we have to break it from -1 to 0 and 0 to 1 then apply tan( inverse) 1/e^v equal to -π +cot( inverse) e^v for first fraction of integral and for second cot(inverse) e^v and then apply tan( inverse) e^v +cot(inverse) e^v =π/2
In any case u see something like the last one, if its a definite integral ofc u gotta try kings rule.
I used leibniz's rule on the second one, its not as elegant but it is way more practical
第二道题是某本中国考研复习全书上的模拟题。也可以用找规律来做。最严谨的方法还是用老师的欧拉公式方法来做
You can do the 1st problem without evaluating the integral by observing at your 2nd step that ln x is a non-negative and increasing function in the domain [1,inf) so the integral goes to +inf, thus both numerator and denominator go to +inf, then you can apply L’Hopital’s rule (which nicely cancels off the integral of the numerator) to get lim n->inf (ln n) / (n ln n)’ = lim n->inf (ln n) / (1 + ln n) = 1.
Yes, I think that's a neater solution, but for some reason he said at the start that he didn't want to fo that way because he doesn't like l'Hopital's rule!
me: struggling with second derivative
meanwhile this guy: doing 2022nd derivative in 2022 milliseconds
Very cool!
Thanks, Scott!!
Feynman's Technique works really well for the 3rd one. Basically solves the problem instantly!
I'd like to see how you did that.
Could you recommend the problem books
with collections of these integral - for fun and
learning. Please give me some titles?????
Hope you have a great day today at Berkeley
I did!! Thanks!!
sir for the first question can we use use l'hosptial, and libneitz rule to differentiate the integral and ger the answer
idk what all those mean but i find the solving so satisying-
No 2
It is easy to use Leibniz's product rule
d^n/dx^n (e^{x}) = e^{x}
d^n/dx^n (sin(x)) = sin(π/2n + x)
So we will get
d^n/dx^n (exp(x)sin(x)) = sum_{k=0}^{n} {n \choose k }exp(x)sin(\frac{π}{2}k + x)
d^4/dx^4 (exp(x)sin(x)) = -4(exp(x)sin(x))
No 3
Integration by parts with
D I
+ arctan(x) 1/x
- 1/(1+x^2) ln(x)
We will get
arctan(e) - (-arctan(1/e)) - Int(ln(x)/(1+x^2),x=1/e..e)
Int(ln(x)/(1+x^2),x=1/e..e) = 1/2Int(ln(x)/(1+x^2),x=1/e..e) + 1/2Int(ln(x)/(1+x^2),x=1/e..e)
Int(ln(x)/(1+x^2),x=1/e..e) = 1/2Int(ln(u)/(1+u^2),u=1/e..e) + 1/2Int(ln(1/u)/(1+1/u^2)*(-1/u^2),u=e..1/e)
Int(ln(x)/(1+x^2),x=1/e..e) = 1/2Int(ln(u)/(1+u^2),u=1/e..e) + 1/2Int(ln(1/u)/(u^2+1),u=1/e..e)
Int(ln(x)/(1+x^2),x=1/e..e) = 1/2Int(ln(u)/(1+u^2),u=1/e..e) - 1/2Int(ln(u)/(u^2+1),u=1/e..e)
Int(ln(x)/(1+x^2),x=1/e..e) = 0
So we will get \frac{π}{2} as an answer because e > 0
Surely there must be a better way to format a RUclips comment than TeX.
@@xinpingdonohoe3978 i mix the TeX format with mathematica commands
@@xinpingdonohoe3978if there is one, it would be good to know it!
Where did the -ve sign go in the third one?
I like the trick in third one
Problem 2 can be solved using Lebniz's formula for (n)th derivative of a product of two functions 😀
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I completely agree 👍👍💯
All limits should include +/- epsilon if its required 😊
The third one is what I call "mathmagic".
In the third integral, they should have put the limits as (1/e² to e²) instead of (1/e to e) so that the answer comes out as just π.
For the last one : King's Properties
Solving integrals, this is like our generation's Bob Ross
My guy really said I don’t need all 15 minutes. I can do it and add in a sponsor for a math comp
Great refresher for an old bum like me. I tried it and I get an answer of 1. I ended up with lim[1+1/ln(n)-1/(nln(n)] as n goes to infinity equals to 1.
Well my answer is luckily right despite me messing up some + and - signs....
3:21 really got me rofl
for the second question also there is easier method.
The pen switching is high level here
Fehman trick in last question?
Can you please make video on the solution of the problem which is in our profile picture
Is there a list of the non-tie braker problems? Were they harder?
Yes. See the link in description for the BMT website.
maths is beautiful 😍😍
i love you dad
I beated two of three but I could not with the three stars level
Please tell is d/dx(Im(fx))=im(f'x) ?
Is there a closed form solution for the indefinite integral sec{sqrt(x)} dx.
Probably not.
Can you solve tan(x) = x without any approximation methods?
No, it is a transcendental equation. You can use Newton-Raphson iteration, or write the Taylor series for tan on the left, then after cancelling out x from both sides (excluding the obvious solution x=0) you can truncate at different orders in x to get a sequence of algebraic approximations.
For the solution of the last problem, how is it legal to combine the values of pi/2 - arctan(e^v) and arctan(e^u) inside the integrals? Aren't they both integrated with respect to two different variables? (u and v, respectively)
They’re not different variables. In the integral with v, you can replace all the v with u. Or any other variable, as long as you’re consistent and don’t replace some vwith x and then some others with t, for instance. That’s what “dummy variables” mean
@hapedisedivide1980 Yes, I realize that. Sorry for not being clear. But if you were to replace the variables as is defined in the solution method, you see that the result he obtains is not correct with this method. Because if you were to replace back the dummy variables for what they represent in the u world, then you get something like ∫-π/2 + arctan(e^-u)du from u = 1 to u=-1. Not only that but the top and bottom bounds for the integral changes. This results in ∫π/2 - arctan(e^-u) du from u =-1 to 1. When you do the addition, as mentioned in the final (and key) step of the problem, 2I = I + I = ∫π/2 - arctan(e^-u) + arctan(e^u) du, the arctan's don't cancel out. Dummy variables are helpful in solving for the answers but what I think happened was 2I = I + I = ∫π/2 - arctan(e^-u) du + arctan(e^v) du. Here, the arctan's don't cancel out like mentioned in the video. Which is why cancelling them out didn't really make sense to me.
@@chessamania5594 But you don’t replace v with -u, you replace v with u, because the transformation is already taken care of when we wrote in original integral in the world. If you do that again you’ll change the integral.
Think of it this way. Both the black I and the blue I represents the area under some curve, and also they are the same, so the areas are the same. But since they are just areas, when you add them up, it doesn’t matter what the names of the reference coordinate plane is called. You could call it the (x,y) plane or the (u,v) plane or whatever variables you like, but the areas won’t change. Just think of a triangle in the plane for example. This means for instance, the blue I is like the area under the curve in the (v,y) plane, but if you called that whole plane the (u,y) plane, the area, or the value of that blue I, wouldn’t change.
@@chessamania5594 But you don’t replace v with -u, you replace v with u, because the transformation is already taken care of when we wrote in original integral in the world. Besides, if you want to replace v with -u, you should replace the dv part too, which is exactly just a u-sub. This results in exactly the black I.
Think of it this way. Both the black I and the blue I represents the area under some curve, and also they are the same, so the areas are the same. But since they are just areas, when you add them up, it doesn’t matter what the names of the reference coordinate plane is called. You could call it the (x,y) plane or the (u,v) plane or whatever variables you like, but the areas won’t change. Just think of a triangle in the plane for example. This means for instance, the blue I is like the area under the curve in the (v,y) plane, but if you called that whole plane the (u,y) plane, the area, or the value of that blue I, wouldn’t change.
Put x=1/t
我在很多數學 RUclipsr 老師的頻道上都問過這個問題
3:40 尤拉公式 到底為什麼 e^ix = cosx + isinx
目前沒有人回答
難道因為是尤拉說是這樣就是這樣嗎 ?
為什麼我們不說 5^ix = cosx + isinx ? 為什麼一定是 e ?
如果老師知道為什麼的話能否為我解答一下, 謝謝
Power series of e
Hi Im Daniel , I would Love to see an Answer from this "Ood" function
If
f(x)=x^x
If we plug in negative number whats happened ?
Like
(-½)^(-½) =
That's not an odd function, since it is completely different in the negative domain. It doesn't even have real numbers, except specifically at the integers.
Jeff Tupper of Tupper's Self-Referential Formula, created a 3-dimensional graph of it. You can see the super-exponential in the real numbers, that U-turns and approaches 1 at x=0. Then it spirals in and out of the real numbers throughout the negative real inputs. At each negative real input, it shows up on the real numbered 2-dimensional graph as an individual point, but is undefined for all fractional inputs, since you can't take negative numbers to fractional exponents and get real solutions, except in special cases.
Last problem: arctan(x)+arctan(1/x)= -pi/2 when x0, why you don't separate the integral in two parts (one from -1 to 0 and the other from 0 to +1), I tried to do that but at the end I get -pi/2 + pi/2 which is 0... I don't get where I'm wrong, if someone can help me please !
Are you forgetting to half your result, and are adding the -π/2 instead of subtracting it?
@@xinpingdonohoe3978 Ok I got it, what he didn't tell is that because e^x >0 you're in the case where arctan(e^x)+arctan(1/e^x) = pi/2 in the [-1,1]... and of course you don't get 0 at the end.
maybe leibniz for the second one
Would be cool if there was a version of this for people who arent aliens. Like the same problems, but the expectations are far lower and its more just for fun than actual competition. you have, say an hour to play around with the problem and then whoever is closest "wins" that round. Also could give each person a calculus textbook (the same book for each person). not a book that contains the solution, but so you have a resource to gain some insight on the problem. Could even do it in teams.
Man after typing this out, I think this would be so much fun. now im sad that this doesnt exist. Petition to create the "Normal Person Math Tournament"
My prediction for each question:
Q1: 0
3:10 Q(2)
Can someone please explain why I'm wrong?
So I differentiated e^x(sinx) 4 times to arrive at the 4th derivative=-2(e^xsinx)
Then since the initial function is e^x sinx
When we keep on differentiating for every cycle of 4 differentiations e^xsinx gets multiplied by a -2.
Therefore I divided 2022 into 4 to see how many cycles of 4 are ther and go 505 and two more differentiations more from there onwards since this repeats as cycles every 4 times at the 2020th derivative we get -2^505(e^xsinx)
And after differentiating 2 more times i got 2022nd derivative=2(-2^505)(e^xcox)
Now when you plug in 0 you get
The answer =2(-2^505)
Can someone please explain.
Hey! I a similar method, got there in the end,
Observing every other derivative like so,
f^{0}(x) = e^xsin(x)
f^{2}(x) = 2e^xcos(x)
f^{4}(x) = -4e^xsin(x)
f^{6}(x) = -8e^xcos(x)
f^{8}(x) = 16e^xsinx
...
We can discern immediately that f^{2022}(x) will contain 2^(1011).
Then it's just a matter of determining what trig function will be present, cos or sin, and if there is a negative sign or not.
2022 mod 8 = 6
Hence observing f^{6}(x) we know the function will contain a negative, and have a cos(x) part.
Therefore f^{2022}(x) = -2^(1011) e^x cos(x)
And at x = 0, we see the result -2^(1011)
Hope that helps!
The 4th derivative is -4(e^x*sinx), following by the same logic you'd get the final answer 2*(-4^505), which is the same as the one in the video. 😄
@@revxrsal
👍👍👍
Thank you all 😀
Man I just put -2 instead of -4 and messed up the whole answer😅😂
This is fine, but complex method is much easier. Also if you write 1+i in polar form as sqrt(2)*exp(i pi/4) then it is easier to raise to the power of 2022.
Does the goat answer?
😊
Which University level is this intended for?
These exercises look rather straightforward.
The exams are created by UC Berkeley students for high school students.
@@blackpenredpen Thank you for your response. Wow high school, that explains it.
Wow
At least the teacher has gone to college
There are rarely teachers who are under college age
Nggak usah Sok Pinter lah
Hallo sir my self Tanwar Singh Rathore I am from India
I was able to solve all the 3 problems correctly without any pen and paper. But I really like your solution for Q2. That was a new train of thought for me. Thanks bprp!!
Samak ekk
Its 2023 not 2022 😅
Nice Concept , But i used to do it in 1st grade
🥱
Whatbda faaaaqq
As a precalculus student, I can confidently say I am very scared of this stuff
This is from a Math competition, you typically don't see these types of problems in a Calculus Course.
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I HATE NOT LIVING IN THE US
Support Palestinian people 🇵🇸
sorry not going to support shariah supporters
Why?
No
Not the place for this buddy
Edit: take this ratio lol
Wrong place.