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# Video

Challenge 98: Steal the Chips
Views 1.8K3 months ago

• Michael Rhodan 2 hours ago

Well done. I could not solve it on my own. Thanks for sharing.

• Mars Truth 7 hours ago

🎖

• Austin Lian 23 hours ago

Amazing so helpful thank you a lot

• Richard Cook Day ago

I literally watch these videos in my spare time, I love these integration bee questions. If there are any newer ones you should start doing them as well.

• wf c 2 days ago

dude talks like the associate prof at my discussion sessions

• Supriya Manna 3 days ago

Just solve the arithmatico geometric progression

• Les Wright 3 days ago

For anyone interested in the similar approach to proving the transcendence of pi, this paper is good, and is very accessible once you work through this video for e, as many of the steps are similar: sixthform.info/maths/files/pitrans.pdf The only additional stuff you need to know beyond the elementary calculus needed for the e proof is a tiny bit of algebraic number theory (i.e., if we assume Pi is algebraic so too must be its complex conjugate i Pi) and some basic stuff about symmetric polynomials. Once you have a grasp of that, the proof proceeds along similar lines, and you end up with a situation where algebraic Pi leads you to an "equation" where a nonzero integer must be equal a number that approaches zero as you increase an arbitrary parameter, leading to the required contradiction.

• Leif 4 days ago

Someone PLEASE RESPOND AND CORRECT THIS SOLUTION 2013 is the exponent it's NOT THE ANSWER

• 435iak 4 days ago

that was beautiful

• Les Wright 4 days ago

This is excellent. I am personally fond of Hilbert's 1893 proof which is at base similar to the Hermite but somewhat shorter and much more elegant. Hilbert also treats Pi in the same paper. It is very similar to the e proof except one has to wrap ones brain around symmetric polynomials and the like. Both proofs rely on the fact that the integral for the gamma function equates to an integer (a factorial in fact) for integer values of a certain parameter. In the end, Hilbert achieves what Hermite did--the assumption that e (or Pi) is algebraic leads to an equation where one side remains a nonzero integer whilst the other approaches zero as some arbitrary prime parameter gets large. This is, as Hilbert writes, "unmoeglich" :). PS I can send a scan of the original paper to anyone who writes me. My reading German is so-so, but I learned of the e proof in English first in Leveque's Fundamentals of Number Theory (in Dover reprint). With my basic knowledge of algebraic number theory ans symmetric polynomials I could make out the gist of the Pi proof too.

• Leif 4 days ago

Does anyone have the solution video?

• Karthasis 5 days ago

Find the number of pairs of parallel diagonals in a regular 10-gon.

• Valid Invalid 5 days ago

Does e^2 is irrational?

• Jeevan Roy 6 days ago

I can do it in easier way

• Karthasis 6 days ago

It can be both 963 and 182 mod 1000.

• Itachi uchiha 6 days ago

Holy shit im only in highschool

• Dabznite 8 days ago

Why do u say words so long like allllll

• Hao Li 9 days ago

Would it work if we use Euler’s eqn?

• Adil Bhatti 9 days ago

THATS WAS CRAZY COOL!!

• Leif 9 days ago

The limit of a product equals,the product,of the limits so x times lnx limit is zero? Why not just do it that way..you get zero times one which is zero ..

• Leif 9 days ago

Does f subscriptt,n mean a function that takes vues n but does NOT mean a function of n, is that correct? Only f(n) means,something is a function of n,correct?

• Leif 9 days ago

But who WHO would ever think of doing that step at 6:00..can you please respond and tell me?

• Leif 10 days ago

You created this problem? How may I ask?

• pr0dz 10 days ago

I really like the use of tangent. It just makes you use all kinds of identities just to solve X and y.

• neuralwarp 10 days ago

But why do you need to define an x!! function ? And why choose a notation that ought to represent x!! = (x!)!

• Aaron Justin 11 days ago

Awesome, beautifully explained.

• Satya Srinivas 11 days ago

It's difficult but use first method. Cauchy method not good

• Karthasis 11 days ago

6-π Inradius=3+4-5/2=1 Area of incircle=πr^2=π Area of triangle=0.5*3*4=6 Area of shaded region=6-π

• Lovely Girl 11 days ago

Why is trig sub not going to work? When (sec x+tan x)^-n appears, a useful trick is to rewrite as (sec x-tan x)^n. In this case, it will give the expression (sec^2 x)(sec x-tan x)^2, which can be integrated nicely.

• Leif 13 days ago

Can someone PLEASE CLARIFY if 2003 is just the exponent, how does it tell you the number of,different solutions for the equation? If the question means for a particular ordered pair a and b how many tines can you raise to a power to get the complex conjugate, then it,makes sense..otherwise you could have an infinite number,of,solutions...Anybody,else see what I mean?

• pyLz 14 days ago

You can also formulate this in terms of a random walk: after how many steps are you n steps away from the origin.

• Ramasamy 15 days ago

Is it you blackpenredpen?

• ya yeet 15 days ago

actually quite easy to do once you use variables!

• Lance Zhang 16 days ago

Great video.

• Wheeler

• eric 18 days ago

fuck math

• Anmol Singhaniya 19 days ago

Very interesting 😋😋

• seghiri moha 19 days ago

Isnt clear your methode is confuse

• Albert Mendoza 19 days ago

What book would you suggest to me if I were to start exploring these types of topics?

• pratman 20 days ago

Thanks for this video!

• 1976! Let u(x) = floor(x)^4 + frac(x)^9 Then f(a) < g(b) only if u(a) < b. If n > 1. floor(n + 1/n^3) = n frac(n + 1/n^3) = 1/n^3 f(g(n + 1/n^3)) < g(f(n + 1/n^3)) f(sqrt(n) + 1/n) < g(n^2 + 1/n^9) u(sqrt(n) + 1/n) < n^2 + 1/n^9 If n is a square, then u(sqrt(n) + 1/n) = n^2 + 1/n^9 If n is not a square, then floor(sqrt(n))^2 + frac(sqrt(n) + 1/n)^9 < n^2 + 1/n^9 floor(sqrt(n))^2 < n^2 int a < int b => a - b >= 1 Since frac(a)^9 < 1 If n is not a square, equasion is right! There are 2019 - floor(sqrt(2019)) or 1975 numbers of n, where n is not a square of an integer. If n = , then f(g(2)) < g(f(2)) 1975 + 1 = 1976!!!

• Iffat Sarfraz 20 days ago

This is amazing!

• Sitanshu Chaudhary 20 days ago

Bro why you leave this channel please begin again please

• Sitanshu Chaudhary 21 day ago

Is this come in exam then we use calculator for know now about your assumption

• Leif 21 day ago

At 6:05 ..you say you should've memorized that formula but nit the other one..why sounds a but hypocritical..memorization is boring in any case..they should just TELL YOU THESE STUPID FORMULAS ON THESE TESTS

• Leif 21 day ago

You never explain how to think of this in the firdt place!! How? Its,not intuitive or logical in that sense at all! Please elaborate and correct this!

• devvrat ashtaputre 22 days ago

If u are thorough with formulae and problem solving techniques, Solution starts at 8:45

• Manuel Odabashian 24 days ago

What does the capital lambda sign mean?

• Hjtunfgb 24 days ago

Beggining: "So, we're doing this integral..." End: "...and by that, we have proven the Riemann Hypothesis"

• Alain Sebaoun 24 days ago

J'ai du mal à imaginer que ce calcul, tout juste du niveau d'un élève de mpsi en milieu d'année, soit une trouvaille du "MIT" !!!!! C'est rigolo de voir que des connaissances assez banales peuvent être présentées de façon aussi "théâtrale" par des étudiants en manque de reconnaissance.

• Lam Nguyen 25 days ago

Why is the 1st part of x is an interger?

• chinmay prakash 26 days ago

5:06 What has he done there can someone please explain ?

• Dhruv Kumar 26 days ago

Oof bud thanks for this simple / efficient explanation , it will probably help me in exam tomorrow for sure :)

• jose valenzuela 26 days ago

proof of D_x[x^n]=nx^(n-1) brougth me here

• The problems of this channel are easy, but boring and ugly. Meh.

• This channel is so easy.

• Its so beautiful that I'm outta words!

• I gave the 1000th like

• Take some JEE advanced problems too.... they're too good!

• Daniel Chapman 27 days ago

But what if your sequence starts a_1 = phi, a_2 = -1?

• Sigma 1 26 days ago

Then it approches (1 - sqrt (5))/2 But it's just one case when it happens.

• Raj Thomas 28 days ago

What is the math class requirement for MIT Integration Bee? I am in calc 2 I am able to understand this and kind of solve it out.

• Sai Karthik R Pulagam 8 days ago

@Raj Thomas Are you studying at MIT right now or are you in highschool, and planning to go to MIT. If you are applying for universities as a senior right now, there are many universities that also have integration bee, and MIT integration bee is arguably the hardest of all the universities that have integration bee.

• Raj Thomas 8 days ago

@Sai Karthik R Pulagam thank-you, I will take some time into this.

• Sai Karthik R Pulagam 8 days ago

Yea calculus 2 should be enough, but this is very simple problem, but I recommend the the book Inside Interesting Integrals book for practice. MIT integration bee is very hard, not because you cant solve it, but people do it really fast in the competition, and to get as fast as them, you need to memorize as many integrals as you can, also know u substitution, integrals of powers and all trig functions, integration by parts, partial fractions, trig substitutions, completing the square, elimination of radicals by substitution, substituting the tangent of half angle, integrals of hyperbolic functions. In addition, you need to know all the double half angle formulas, and be able to do partial fractions in your head. If you can do all this, and have very fast mental math skills, then you should be ready to go. Mainly, practice Inside Interesting Integrals book, and mit integration problems as well.

• Edwar Lopez 29 days ago

Esto es un chiste??

• Hans Wurst 29 days ago

Nice tutorial 👍👍 What kind of Software do you use to make your tutorials?

• Matt 29 days ago

thank you so much!! officially my favorite explanation.

• Anmol Singhaniya 29 days ago

By Gamma function also we can solved

• Johnes Nyamohanga 29 days ago

Thanks

• One in a billion Month ago

Imagine if there was no hint pi^2/6

• Anmol Singhaniya Month ago

It just take me 6 sec to solved this question

• Anmol Singhaniya Month ago

Best Solution 😊😊🇮🇳💓

• Anmol Singhaniya Month ago

I just solve in 9 sec 🇮🇳🇮🇳💓 Trust me

• Anmol Singhaniya Month ago

If limit is from n to 2 Than we can use special form of Integration ?

• Chirayu Jain Month ago

I would have use the property Integral of ((even(x)) /1+b^odd(x) from - a to a =integral of even(x) from - a to a Inspired by @blackpenredpen

• chinmay prakash Month ago

Wow amazing problem Please keep uploading such intresting problems they are pretty good

• Ok please answer my easy question at which value of x ( x+1/x) this gives us 2 0<x<1/2

• Osir isrex 17 days ago

Sitanshu Chaudhary You misinterpreted his explanation: he’s saying that (x + 1/x ) > 2 for all positive x not equal to 1

• Are you accomplish this channel please begin again

• That is the most easiest question of an jee aspirant

• Angelo Treviños Month ago

Too easy

• Angelo Treviños Month ago

Too easy :,l

• Bro try some jee advanced 2016 maths question also 2019

• Mhammed Tomy Month ago

Amazing conclusion

• Kevin Somenzi Month ago

MIC drop at the end there..! This was great, seeing how you can try and tackle problems like this is incredible, I don't think I would have thought of differentiating the series to go back and integrate it. That was incredile! Thanks.

• CHROMIUM HERO Month ago

that was pretty cool

• miguel cerna Month ago

Absolutely astonishing.

• eliya sne Month ago

I love the way you approached the the number theory problem in the middle of the video. Great video!

• rhythm mandal Month ago

ans: 1975, for all n ∈ Z+ gof will always yield the input. for all n = d^2 form where 1<n<=2019, d ∈ Z+ fog = gof , in this case every 2<=d<=44 satisfies this a total of 43 number for which fog = gof now for all n = d^2 - 1, √n = d - 1 + Q here Q ∈ R and Q < 1 for the case of fog if Q + 1/n > 1 then fog > gof we can derive that: for all d where 2<=d<=45, since 45^2-1 > 2019 > 44^2 iff the following is satisfied (d^2 - 1)^3 > (d^3 - d - 1)^2 fog > gof only d = 2 satisfies this so, the number of elements for which fog < gof = 2019 - 43 - 1 = 1975

• Md irfan Month ago

Integration of e^√tanx

• Murat Karadag Month ago

I klicked the like button and it showed 2800 likes!! I love how many ways you are trying to wvaluate that integral 😘

• Elie Makdessy Month ago

In 12:24 why you didn't use Euler's theorem for 2018^2019^2020 mod25?

• Piyush Kumbhare Month ago

I finally understand everything in this video after doing my Calculus summer homework :) I finally understand it's true beauty

• Max Haibara Month ago

Cute problem~

• Bhaskar Das Month ago

What is rational algebraic curve?

• One in a billion Month ago

Hi, I'm thinking if I should take degree in Math and applied math or electrical engineering. May I ask what you are doing right now?

• William Zhang Month ago

Multiple uses of angle bisector theorem

• Please make video on 1988 imo problem 6

• udic01 Month ago

At 2:52 you forgot to multiply the second element by 2. Luckily you got the right result. Still i am enjoying your videos for over a year now (i decided to watch all of them)

• Kurt Lechner Month ago

Aussprache lernen If science serves only for its own ends, it is no longer science, but insanity.

• One in a billion Month ago

I shall force myself to do normal trig substitution

• Mark Month ago

Everyone loves the gamma function

• Dear marvelous members of LetsSolveMathProblems community, As you may know, I am currently a freshman undergraduate student in the United States, nervously yet eagerly striving to adjust to the new lifestyle, both in its social and academic aspects. To facilitate the transition, I have decided to take an approximately two-month break from USclip, which would help me not only learn how to allocate my schedule and workload for specific classes but also help me devote more time and energy to develop important relationships and interests. I do understand this is a long break, and I do sincerely apologize. I have much planned for when I return and beyond, such as perhaps starting a Khan Academy-style course for real analysis or a series on calculus or introductory problem-solving. I am also considering an option to convert my "class notes" for certain classes as USclip videos, given that I am successful in the course and I can obtain necessary permissions. Of course, I may be overestimating my ability to adapt to a new environment and responsibilities, and it may be the case that I return much faster than two months. Nevertheless, I thank you very much for your patience and your continued support. I wish you the best of luck in your academic or personal endeavors, as well. Best regards, LetsSolveMathProblems

• Bob Jones 21 day ago

Krish harvard

• Krish 28 days ago

Great man! What college are you attending?

• Nice problem even i was very close to the answer but i don't occupy answer