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## LetsSolveMathProblems

United States

Joined Aug 31, 2014

Hello! Welcome to LetsSolveMathProblems, where we will dive into and explore the intricate, breathtaking realm of mathematics.

LetsSolveMathProblems is a mathematics channel dedicated to providing free, quality videos that anybody can use anywhere to learn.

Your support is a heartfelt source of encouragement that propels the channel forward. Any likes, subscriptions, comments, constructive criticisms, etc., are wholeheartedly appreciated.

LetsSolveMathProblems is a mathematics channel dedicated to providing free, quality videos that anybody can use anywhere to learn.

Your support is a heartfelt source of encouragement that propels the channel forward. Any likes, subscriptions, comments, constructive criticisms, etc., are wholeheartedly appreciated.

# Video

**Solution 99: Creatively Juggling Limits, Integrals, and Derivatives**

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**Challenge 101: "Almost" Inverse functions (ft. floor and fractional part)**

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**100 Weeks of Challenges (Recognitions) + Challenge 100**

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**Solution 98: Expected Value in Gambler's Ruin (Steal the Chips)**

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**Challenge 99: Infinite Sum of the Difference of Roots**

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**Solution 97: The Art of Summation Manipulation**

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**Solution 96: Hidden Cyclic Quadrilaterals**

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**Challenge 98: Steal the Chips**

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**Challenge 97: Summation Manipulation**

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**Solution 95: Limit of the Sum of the Bottom Row of a Matrix Divided by a Perfect Square**

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**Challenge 96: Can You Find the Sum of Two Angles?**

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**Solution 94: Partitions into Groups and Subgroup**

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**Challenge 95: Limit of the Sum of the Bottom Row of a Matrix Divided by a Perfect Square**

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**Solution 93: Does a "p-rapid" Sequence Converge?**

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**Challenge 94: Partitions into Groups and Subgroups**

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**Where is LetsSolveMathProblems headed?**

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**Solution 92: A (not necessarily) Trisecting Median**

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**Challenge 93: Does a "p-rapid" Sequence Converge?**

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**Solution 91: The Last Three Digits of a Phi-Exponent Tower**

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**Challenge 92: A Trisecting Median**

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**Solution 90: Logic Problem on Types of Cars**

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**Challenge 91: The Last Three Digits of a Phi-Exponent Tower**

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**Solution 89: Closed Form of a Generating Function**

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**Challenge 90: Three Types of Cars Riddle**

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**Solution 88: 1/n and Square Roots**

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**Challenge 89: Can You Find the Generating Function of {a_n}?**

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**Solution 87: Chasing the Angle HGI**

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**Challenge 88: 1/n and Square Roots**

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**Solution 86: Double Factorial and Roots of Unity Filter (Proof)**

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**Challenge 87: Can You Chase the Angle HGI?**

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**Solution 85: Maximizing the Difference of "Alternating Sums"**

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**Challenge 86: Can You Sum the Reciprocals of (6n)!! ?**

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**Vieta's with Triple Roots (2019 AIME I Problem 10)**

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**Adding a Bunch of 9's (2019 AIME I Problem 1)**

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**Solution 84: Pairs of Lattice Paths with Restrictions**

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**Challenge 85: Can You Maximize D(n) - D(n+111)?**

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**Two Putnam Appetizers (2009 A1 & 2000 A2)**

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**Solution 83: Sequence over Catalan (feat. Induction and Generating Functions)**

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**Challenge 84: Pairs of Lattice Paths with Restrictions**

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**Solution 82: Simple Bashing & Cauchy-Schwarz "Equality"**

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**Challenge 83: Sequence over Catalan**

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**Solution 81: Fascinating Interplay between 100 and 001**

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**Challenge 82: Cosine to the 2019th Power**

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**Solution 80: Slaying a Monstrous Determinant using Second Differences**

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**Challenge 81: Can You Count the Number of 001 and 100?**

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**Finding the Shaded Area + Proof that Medians are Concurrent (Solution 79)**

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**Challenge 80: Can You Evaluate This Determinant?**

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**Log_Heart(Diamond): 2019 AMC 12 A Problem 23**

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**Solution 78: Infinitely Many Tangent Circles**

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**Challenge 79: Can You Find the Area of This Hexagon?**

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**A Delightful Solution (77) - Manipulating Polynomials and Roots of Unity**

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**Challenge 78: Infinitely Many Tangent Circles**

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**Solution 76: Polygonal Numbers and GCD's**

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**Challenge 77: Product-Pairing 2019 Omegas**

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**Solution 75: Integrating Limit, Arithmetic Mean, and Trig**

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**Challenge 76: Polygonal Numbers and GCD's**

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**Solution 74: Introducing Symmetry Using Trig Identities**

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**Challenge 75: Limit, Arithmetic Mean, and Trig**

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**Solution 73: Schröder Path Pattern Avoidance**

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**Challenge 74: Summing from the Year 2014 to N**

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**Solution 72: Hexagon Inside a Nine-Point Circle**

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**Challenge 73: Schröder Path Pattern Avoidance**

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**Solution 71: Forcing the u-Substitution to Work**

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**Challenge 72: Hexagon Over Triangle**

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**A Gorgeous Solution via Trig Substitution (WMC 70)**

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**Challenge 71: Integral Squared is Eight**

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**Solution 69: States, Expected Values, Ants, and an Octahedron**

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**Challenge 70: Can You Solve This System of Equations?**

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**Intro to Chinese Remainder Theorem and Euler's Totient Theorem via a Challenging Problem**

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**Challenge 69: Three Ants on an Octahedron**

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**Solution 67: Substitution and Symmetry vs. a Functional Equation**

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**Challenge 68: Can You Find the Last Two Digits?**

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**Solution 66: sec(sum(csc(tan(cot(sin(cos(...)))))))**

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**Challenge 67: An Almost Symmetric Functional Equation**

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**Solution 65: Tangent Circles and Angle Chasing**

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**Challenge 66: sec(sum(csc(tan(cot(sin(cos(...)))))))**

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**AIME: Polynomial Functional Equation (Reupload) 2016 A Problem 11**

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**Solution 64: Dyck Path Pattern Avoidance (UDUDUD) and Containment**

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**Challenge 65: Tangent Circles and Angle Chasing**

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**Solution 63: Minimization with Floor and Ceiling**

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**Challenge 64: Can You Avoid This Dyck Path?**

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**Challenge 63: Minimization with Floor and Ceiling**

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**Challenge 62: Positive Odd Integers**

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**Challenge 61: Comparing Polynomials**

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**Putnam for Amusement: 1998 B1 and 1991 A2**

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**Solution 59: Recursion and Hidden Fibonacci Sequence**

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**Challenge 60: Cosine and Cyclic Quadrilateral**

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**Exponential Generating Function for Bell Numbers (Proof)**

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**🔔 Bell Numbers and Its Recurrence Relation (Proof)**

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**Solution 58: The Art of Logarithmic Simplification**

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**Challenge 59: Inequality and 9-Digit Numbers**

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**Does Derivative Have to be Continuous? (feat. x^2sin(1/x))**

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**Challenge 58: A Chain of Logarithms**

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**Solution 57: Remainders and Summations**

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**A Lovely Solution to "System of Combinations"**

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**Challenge: Remainders and Summations**

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Michael Rhodan2 hours agoWell done. I could not solve it on my own. Thanks for sharing.

Mars Truth7 hours ago🎖

Austin Lian23 hours agoAmazing so helpful thank you a lot

Richard CookDay agoI 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 c2 days agodude talks like the associate prof at my discussion sessions

Supriya Manna3 days agoJust solve the arithmatico geometric progression

Les Wright3 days agoFor 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.

Leif4 days agoSomeone PLEASE RESPOND AND CORRECT THIS SOLUTION 2013 is the exponent it's NOT THE ANSWER

435iak4 days agothat was beautiful

Les Wright4 days agoThis 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.

Leif4 days agoDoes anyone have the solution video?

Karthasis5 days agoFind the number of pairs of parallel diagonals in a regular 10-gon.

Valid Invalid5 days agoDoes e^2 is irrational?

Jeevan Roy6 days agoI can do it in easier way

Karthasis6 days agoIt can be both 963 and 182 mod 1000.

Itachi uchiha6 days agoHoly shit im only in highschool

Dabznite8 days agoWhy do u say words so long like allllll

Hao Li9 days agoWould it work if we use Euler’s eqn?

Adil Bhatti9 days agoTHATS WAS CRAZY COOL!!

Leif9 days agoThe 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 ..

Leif9 days agoDoes 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?

Leif9 days agoBut who WHO would ever think of doing that step at 6:00..can you please respond and tell me?

Leif10 days agoYou created this problem? How may I ask?

pr0dz10 days agoI really like the use of tangent. It just makes you use all kinds of identities just to solve X and y.

neuralwarp10 days agoBut why do you need to define an x!! function ? And why choose a notation that ought to represent x!! = (x!)!

Aaron Justin11 days agoAwesome, beautifully explained.

Satya Srinivas11 days agoIt's difficult but use first method. Cauchy method not good

Karthasis11 days ago6-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 Girl11 days agoWhy 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.

Leif13 days agoCan 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?

pyLz14 days agoYou can also formulate this in terms of a random walk: after how many steps are you n steps away from the origin.

Ramasamy15 days agoIs it you blackpenredpen?

ya yeet15 days agoactually quite easy to do once you use variables!

Lance Zhang16 days agoGreat video.

Smokie Bear 🔴🔵17 days agoWheeler

eric18 days agofuck math

Anmol Singhaniya19 days agoVery interesting 😋😋

seghiri moha19 days agoIsnt clear your methode is confuse

Albert Mendoza19 days agoWhat book would you suggest to me if I were to start exploring these types of topics?

Albert Mendoza19 days agoExcellent explanation btw 👌

pratman20 days agoThanks for this video!

Красивый туалет20 days ago1976! 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 Sarfraz20 days agoThis is amazing!

Sitanshu Chaudhary20 days agoBro why you leave this channel please begin again please

Sitanshu Chaudhary21 day agoIs this come in exam then we use calculator for know now about your assumption

Leif21 day agoAt 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

Leif21 day agoYou 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 ashtaputre22 days agoIf u are thorough with formulae and problem solving techniques, Solution starts at 8:45

Manuel Odabashian24 days agoWhat does the capital lambda sign mean?

Hjtunfgb24 days agoBeggining: "So, we're doing this integral..." End: "...and by that, we have proven the Riemann Hypothesis"

Alain Sebaoun24 days agoJ'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 Nguyen25 days agoWhy is the 1st part of x is an interger?

chinmay prakash26 days ago5:06 What has he done there can someone please explain ?

Dhruv Kumar26 days agoOof bud thanks for this simple / efficient explanation , it will probably help me in exam tomorrow for sure :)

jose valenzuela26 days agoproof of D_x[x^n]=nx^(n-1) brougth me here

Владимир Королёв26 days agoThe problems of this channel are easy, but boring and ugly. Meh.

Владимир Королёв26 days agoThis channel is so easy.

Sagartirtha Sengupta27 days agoIts so beautiful that I'm outta words!

Sagartirtha Sengupta27 days agoI gave the 1000th like

Sagartirtha Sengupta27 days agoTake some JEE advanced problems too.... they're too good!

Daniel Chapman27 days agoBut what if your sequence starts a_1 = phi, a_2 = -1?

Sigma 126 days agoThen it approches (1 - sqrt (5))/2 But it's just one case when it happens.

Raj Thomas28 days agoWhat 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 Pulagam8 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 Thomas8 days ago@Sai Karthik R Pulagam thank-you, I will take some time into this.

Sai Karthik R Pulagam8 days agoYea 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 Lopez29 days agoEsto es un chiste??

Hans Wurst29 days agoNice tutorial 👍👍 What kind of Software do you use to make your tutorials?

Matt29 days agothank you so much!! officially my favorite explanation.

Anmol Singhaniya29 days agoBy Gamma function also we can solved

Johnes Nyamohanga29 days agoThanks

One in a billionMonth agoImagine if there was no hint pi^2/6

Anmol SinghaniyaMonth agoIt just take me 6 sec to solved this question

Anmol SinghaniyaMonth agoBest Solution 😊😊🇮🇳💓

Anmol SinghaniyaMonth agoI just solve in 9 sec 🇮🇳🇮🇳💓 Trust me

Anmol SinghaniyaMonth agoIf limit is from n to 2 Than we can use special form of Integration ?

Chirayu JainMonth agoI 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 prakashMonth agoWow amazing problem Please keep uploading such intresting problems they are pretty good

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

Osir isrex17 days agoSitanshu Chaudhary You misinterpreted his explanation: he’s saying that (x + 1/x ) > 2 for all positive x not equal to 1

Sitanshu ChaudharyMonth agoAre you accomplish this channel please begin again

pinki daroga vs mangilalMonth agoThat is the most easiest question of an jee aspirant

Angelo TreviñosMonth agoToo easy

Angelo TreviñosMonth agoToo easy :,l

Sitanshu ChaudharyMonth agoBro try some jee advanced 2016 maths question also 2019

Mhammed TomyMonth agoAmazing conclusion

Kevin SomenziMonth agoMIC 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 HEROMonth agothat was pretty cool

miguel cernaMonth agoAbsolutely astonishing.

eliya sneMonth agoI love the way you approached the the number theory problem in the middle of the video. Great video!

rhythm mandalMonth agoans: 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 irfanMonth agoIntegration of e^√tanx

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

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

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

Max HaibaraMonth agoCute problem~

Bhaskar DasMonth agoWhat is rational algebraic curve?

One in a billionMonth agoHi, 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 ZhangMonth agoMultiple uses of angle bisector theorem

Sitanshu ChaudharyMonth agoPlease make video on 1988 imo problem 6

udic01Month agoAt 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 LechnerMonth agoAussprache lernen If science serves only for its own ends, it is no longer science, but insanity.

One in a billionMonth agoI shall force myself to do normal trig substitution

MarkMonth agoEveryone loves the gamma function

LetsSolveMathProblemsMonth agoDear 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 Jones21 day agoKrish harvard

Krish28 days agoGreat man! What college are you attending?

Sitanshu ChaudharyMonth agoNice problem even i was very close to the answer but i don't occupy answer