Find all natural numbers satissfying the equation

[@Akenfelds1]

You are the best mathematics educator I've ever seen on KZbin. You neither overexplain nor underexplain. Every step is 100% clear.

[@gusmath1001]

Nice presentation! Note, however, that it’s not necessary to use math induction. Claim: If n>3, then n! >(n(n+1))/2. Multiplying both sides by 2 and dividing by n, we get the equivalent inequality 2(n-1)!>n+1. As 2(n-1)!>2(n-1), to prove the claim it will suffice to show that 2(n-1)>n+1, for n>3. But this is immediate: 2(n-1)>n+1 iff 2n-2>n+1 iff 2n-n>2+1 iff n>3.

[@PrimeNewtons]

I realized that while watching the video. Thanks.

[@puneetkumarsingh1484]

Great application of the Principal of Mathematical Induction. We almost forget how powerful it can be at times 😅

[@jay_13875]

Because n! ≥ n*(n-1)*(n-2) and n*(n+1)/2 < n*(n+1) for all n≥4, all we need to show is that (n-1)*(n-2) > n+1 n² - 3n + 2 > n + 1 n² + 1 > 4*n n + 1/n > 4, which is true for n ≥ 4 since 1/n > 0

[@jpl569]

Excellent lecture ! Induction works smartly, with ± heavy writings… Let’s try directly : in order to prove that n ! > n (n + 1) / 2 for any n ≥ 4, equivalent to 2 (n-1) ! > n + 1, we notice that (for n ≥ 4 ) : 2 (n-1) ! > 2^(n-1) because 2 (n-1) ! = 2x2x3x…x(n-1), and 2^(n-1) > n + 1 by studying f(x) = 2^x - x - 2 for x ≥ 3 (easy stuff…). Then we’ve got it… Thank you for your interesting videos ! 🙂

[@jpl569]

An other way is : let U_n = 2 (n-1) ! and V_n = n +1. Then U_n+1 / U_n = 2 n and V_n+1 / V_n = (n+2) / (n+1). As U_3 = V_3 = 4, and for n ≥ 1, U_n+1 / U_n > V_n+1 / V_n, Then V_n < U_n for n ≥ 4.

[@kingsgamer2019]

Never stop learning, nice lecture, you are genious.

[@mahmoudalbahar1641]

I am thankful for your efforts, your videos are always nice and filled with benefits, it's our pleasure to watch your videos.

[@PrimeNewtons]

I am grateful for you

[@bashkimelbasani2382]

Very methodical explanation profesor! I'm very content to You! Bravo!

[@adamcionoob3912]

Great video. While comparing n^2 + n with n + 2, you could also see that since n >= 4, n^2 >= 16 so n^2 + n >= n + 16 > n + 2

[@s.n.mishra501]

Nice point

[@Your_choise]

This can be done without using the formula for Σi=(n)(n+1)/2 since If n=1, then 1=1, If n=2, then LHS=3, RHS=2 if n=3, RHS=LHS=6 And if n ≥4, 1+2+3+…+n n-1+1=n, n! ≥n(n-1)*2>n*n=n^2> 1+2+…+n.

[@michaelz2270]

You have n(n+1)/2 = n! iff (n + 1)/2 = (n-1)!. But one has (n - 1)! >= n - 1 > (n +1)/2 for all n > 3. So you don't have to check beyond n = 3.

[@glorrin]

Something bugged me, in the induction, since we only needed n>=2 why couldn't we start at 2 ? well, 15 years after formerly learning about induction, I finaly understand how important the initial step is (base case). if we start at 2, initial step would be 2! = 2 2+1 = 3 2 is not greater than 3 So we can't start with 2, And we can't start with 3 either since we have shown it is equal. This is a marvelous Induction.

[@dan-florinchereches4892]

Hello sir, Very interesting approach to the problem. But if we are looking for the condition of the equality happening is it not easier? What needs to happen so n*(n+1)/2=n! ? Since n!=0 we can divide by n so (n+1)/2=(n-1)! n+1=2(n-1)! n=2(n-1)!-1 If we replace n by k+1 to have a nicer number inside the factorial k=2k!-2 so k=2(k!-1) which means that k>k!-1 or k+1>k! and we can easily verify that this proposition is only true for very small values

[@maxhagenauer24]

9:42 where did that new n+1 on the RHS come from?

[@maxhagenauer24]

@Salko_ So to get (n+l)! > (n+1)(n+2)/2, he took n! > n(n+1)/2 and replaced n with n+ 1? And then to get (n+1)n! > (n+1)n(n+1)/2, he took n! > n(n+1)/2 and multiplied both sides by n+1?

[@Myhair0_0]

​@Salko_ but are we not proving that n! > n(n+1)/2 so using that fact in our proof is circular logic?

[@KavyaVINOCHA]

we multiply both sides of eq at 7:11 by n+1

[@jaime9927]

@Myhair0_0 That's why we have to check whether the inequality holds for the base case (n=4 in this example) Then, the proof from the video, which @Salko_ summarized, shows that if the inequality holds for any integer >=4, then it holds for the next integer. Thus, after manually verifying that the inequality holds for n=4 and completing the short proof, we know that the inequality holds for n=4, n=4+1=5, n=5+1=6, and so on. Hope this helps

[@maxhagenauer24]

@@Salko_ I don't know why the freak my response got deleted last night, I was asking this: Are you saying (n+1)! > (n+1)(n+2)/2 come from replacing n with n+1 in the original n! > n(n+1)/2? And does (n+1)n! > (n+1)n(n+1)/2 cone from the original but after multiplying both sides by n+1?

[@dengankunghacharles1115]

Excellent job

[@benjaminvatovez8823]

Thank you for your video. It is possible not to use induction: as (n-1)! = (n+1)/2 is in Z, n must be odd and as (n-1)! = (n-1).(n-2)..2.1 > (n-1)(n-2).4 for any n >=7, we get (n-1)(8n-17) < 2 which is impossible as n>6.

[@prajjawaltiwari9566]

10:25 only if n>1, which is understood here...

[@AinomugishaAllan]

Thank you my best professor but I didn't understand the end part of the solution especially on how you got that of n+n to be equal to (n+1)!

[@pizza8725]

n²+n isn't always bigger than n+n as it is smaller at n=(0,1)

[@AnesMechekak]

but n is greater then 3

[@pizza8725]

I know but i said that it isn't always, but it is in that case

[@VisionXu-y6k]

can you use gama function to figure out the equation ∑n=n!？

[@Rahul.G.Paikaray27]

It's really interesting sir make more videos like this sir 💯💯💯❣️💫✨🌟

[@FortuneMachaka]

You are the best sir

[@yurenchu]

From the thumbnail: the solutions are n=1 and n=3 . For any n>3, the cumulative product is greater than and also will be increasing faster than the cumulative sum. I'm watching the video to see if you consider n=0 a solution or not (and why).

[@robertveith6383]

@ Prime Newtons n = 1 *OR* 3, not 1 "and" 3.

[@itsphoenixingtime]

Very rough, not rigorous idea but... I figured that because the factorial grows much faster than the quadratic, after some point there won't be any more solutions, so there isn't any need to check every case. I remembered the 1 + 2 + 3 = 1 x 2 x 3 meme, so n = 3 n also = 1. That was basically my reasoning, because after n = 3 the factorial grows much faster than the quadratic so they will never ever intersect again, the only need is to check for answers within that range. I think the proof of the factorial outlasting the triangular numbers for n >= 4 was rigorous to help cement that idea that they can never be equal and hence no solutions for that region.

[@ThomasMeeson]

That’s obvious and as you said not rigorous

[@stevenwilson5556]

I immediately know of 1, and 3. I think those are the only 2 but proving that is a whole different issue. I might be able to do that but not sure exactly what I'd do maybe induction or contradiction but would take awhile to prove it.

[@NobleTheThinkingOne678]

All you have to do is prove that n! grows faster than n(n+1)/2 for all n>=4. You can prove that and thus since 24>10. That is 4!>10 and that n! grows faster than n(n+1)/2 then I can show that n(n+1)/2 can never catch up to n! thus only 1 and 3 are true

[@ishanpurkait9124]

sir , can you record me some books to learn advanced mathematics

[@Tommy_007]

Begin with "Algebraic Geometry" by Hartshorne.

[@ishanpurkait9124]

@Tommy_007 thank you

[@Tommy_007]

@@ishanpurkait9124 It's a VERY difficult book. I'll recommend that you start with basic books about calculus, elementary number theory, classical plane geometry, linear algebra, and abstract algebra.

[@Tommy_007]

An older series of books that are recommended for interested high school students: New Mathematical Library.

[@ishanpurkait9124]

thank you ,can you help me choose between calculus by stewart ( republished by clegg and watson ) and thomas , both the early transcendental version

[@juergenilse3259]

I thin, the o natural number satisfying this equation are n=1 and n=3. The left side can be substituted b n*(n+1)/2 (according to gauss forula for sumof the first n natural numbers) while the rigtside is equall to n! (accordingto the definition of factorial).So we seach values for n with n*(n+1)/2=n!. Since n! is per definition n*(n-1)!, we can transform this equation to (n+1)/2=(n-1)!. n=1 and n=3 are possiblesolutionsforthhis equation, because (1+1)/2=(1-1)! and (3+1)/2=(3-1)!. Equal numbers can not fullfill the equation, because the right side is alwas a natural number, while the left side is for even values of n neer an integer. For all odd numbers n greater than 3, (n-1)! is greater than (n+1)/2, so n=1 and n=3 are the only solutions. For me, it was obvious,that (n-1)! is greater than (n+1)/2 for an n>3, but nice, that you gave a proof ...

[@abulfazmehdizada]

There are exactly 77000 ordered quadruples (a,b, c,d) such that gcd (a,b, c,d) =77 and lcm (a,b, c,d) =n, What is the smallest possible value of n? Hello teacher. Could we look at this question? There were many solutions that i didn't understand well. I would like to see your approach

[@yurenchu]

What restrictions are placed on a, b, c, d ? For example, can a, b, c, or d be a negative integer?

[@abulfazmehdizada]

@yurenchu there's no restrictions I think

[@yurenchu]

@@abulfazmehdizada In that case, there is no possible solution for n ; because the number of quadruples has to be a multiple of 16 ; and 77000 is not a multiple of 16 . A quadruple (a,b,c,d) cannot contain any 0, because in that case, n=0 and in that case there are an infinite number of quadruples, for example of the form (0, 77, 77, 77k) where k is any non-zero integer. Therefore, |a| , |b| , |c| and |d| must be positive (i.e. nonzero). Now, suppose (a, b, c, d) = (t, u, v, w) satisfies gcd(a,b,c,d) = 77 and lcm(a,b,c,d) = n , and t, u, v, w are positive integers. Then (t, u, v, w) represents a set of 16 _distinct_ quadruples that each satisfy the gcd and lcm conditions, namely (a, b, c, d) = (t, u, v, w), (t, u, v, -w), (t, u, -v, w), (t, u, -v, -w), (t, -u, v, w), (t, -u, v, -w), (t, -u, -v, w), (t, -u, -v, -w), (-t, u, v, w), (-t, u, v, -w), (-t, u, -v, w), (-t, u, -v, -w), (-t, -u, v, w), (-t, -u, v, -w), (-t, -u, -v, w), or (-t, -u, -v, -w). This is true for any quadruple (t, u, v, w) of positive integers that satisfies the conditions, hence the total number of quadruples must equal {16 times the number of distinct quadruples of only positive integers}.

[@yurenchu]

@@abulfazmehdizada In that case, there is no possible solution for n ; because the number of quadruples has to be a multiple of 16 ; and 77000 is not a multiple of 16 . A quadruple (a,b,c,d) cannot contain any 0, because in that case, n=0 and in that case there are an infinite number of distinct quadruples, for example of the form (0, 77, 77, 77k) where k is any integer. Therefore, |a| , |b| , |c| and |d| must be positive (i.e. nonzero). Now, suppose (|a|, |b|, |c|, |d|) = (t, u, v, w) satisfies gcd(a,b,c,d) = 77 and lcm(a,b,c,d) = n , and t, u, v, w are positive integers. Then (t, u, v, w) represents a set of 16 _distinct_ quadruples that each satisfy the gcd and lcm conditions, namely (a, b, c, d) = (t, u, v, w), (t, u, v, -w), (t, u, -v, w), (t, u, -v, -w), (t, -u, v, w), (t, -u, v, -w), (t, -u, -v, w), (t, -u, -v, -w), (-t, u, v, w), (-t, u, v, -w), (-t, u, -v, w), (-t, u, -v, -w), (-t, -u, v, w), (-t, -u, v, -w), (-t, -u, -v, w), or (-t, -u, -v, -w). This is true for any quadruple (t, u, v, w) of positive integers that satisfies the conditions, hence the total number of quadruples must equal {16 times the number of distinct quadruples of only positive integers}.

[@abulfazmehdizada]

@@yurenchu answer is 27720

[@QuickStories_123]

Can I send you a question

[@Khaled-kardashev]

Thanks!:)

[@guyhoghton399]

Hence _tan⁻¹(1) + tan⁻¹(2) + tan⁻¹(3) = 180°_

[@Fereydoon.Shekofte]

Best wishes for you and your family Professor 🎉🎉😊😊❤❤ In year 2025

[@stottpie]

Let's get into the video

[@MatondoMaduhu-s9d]

We need to carve a statue for this guy because he is really skilled in mathematics. 😂🎉😂🎉🎉😂

[@maths01n]

Ready for it

[@AnesMechekak]

thank you teach us some strong induction

[@CaioFINNIN]

This is amazing!!! 🔥🫶😜

[@OffiicalComedyClips]

"Satisfying" spelt wrong in the title.

[@guyhoghton399]

*Suppose **_∃n ≥ 4 : _Σ⁽ⁿ⁾ᵢ₌₁{i} = Π⁽ⁿ⁾ᵢ₌₁{i}_* ⇒ _½n(n + 1) = n!_ ⇒ _½(n + 1)!/(n - 1)! = n!_ ⇒ _2n!(n - 1)! = (n + 1)!_ ⇒ _2n[(n - 1)!]² = (n + 1)! = (n + 1)n(n - 1)!_ ⇒ _2(n - 1)! = n + 1 = (n - 1) + 2_ ⇒ _2(n - 2)! = 1 + 2/(n - 1) < 2_ since _n ≥ 4_ ⇒ *_(n - 2)! < 1_** which is impossible for any factorial.* ∴ *_Σ⁽ⁿ⁾ᵢ₌₁{i} ≠ Π⁽ⁿ⁾ᵢ₌₁{i} ∀n ≥ 4_* By inspection equality holds when *_n = 1 or 3 but not 2._*

[@koutarousatomi1552]

Wrong notation. Why i? That is n.

[@elliott2501]

It’s can be any variable as long as you define it that way.

[@topquark22]

By inspection, there is only one answer, n=3

[@BTRequiemOfficial]

n=1

[@maxhagenauer24]

What about n = 1?

[@thomazsoares1316]

n = (1;3)

[@anestismoutafidis4575]

If i=1, then n(Σ) and n(Π)= {1- ♾️ \♾️ }ℕ

[@aaravgamingboy225]

Sir pls make video on fermat's last theorem ❤ lost of love from India ❤❤

[@nanamacapagal8342]

ATTEMPT: By inspection, N = 1, 3 1 = 1 1 + 2 + 3 = 1 * 2 * 3 = 6 N = 2 doesn't work. 1 + 2 = 3, 1 * 2 = 2 For N >= 4: N! > N(N-1) = N^2 - N = (N^2)/2 + N/2 + (N^2)/2 - 3N/2 >= (N^2)/2 + N/2 + 8 - 6 > (N^2)/2 + N/2 = N(N+1)/2 Which is the sum of all natural numbers up to N. Therefore for all natural N >= 4, 1 + 2 + 3 + ... + N < 1 * 2 * 3 * ... * N, and so the two sides cannot be equal. The only solutions are N = 1 and N = 3.

[@robertlunderwood]

The one slight issue is the substitution of n = 4 in the (n²-3n)/2. We would just need to show that (n²-3n)/2 is bigger than 0 for n ≥ 4. But that's easy.

[@robertveith6383]

n = 1 *or* 3.

[@DarkBoo007]

Me: "Obviously its n = 1 or n = 3" *Trying to prove that these are the ONLY values* Me: You got me there LMAO I thought about using induction to prove it since I saw that n = 4 didn't work and I knew for sure n > 4 didn't work either but I was a bit apprehensive knowing that it would've required some work.

[@robertveith6383]

No, it's n = 1 *OR* 3.

[@dieuwer5370]

By observation: n can be 1 and 3. But not 2, 4....

[@robertveith6383]

No, n can be 1 *or* 3.

[@holyshit922]

n=1 and n=3 and in my opinion that's all possibilities

[@robertveith6383]

n = 1 *or* 3

[@holyshit922]

@@robertveith6383 you are right, yes n=1 xor n=3 As a fact I can write you that in my language there is a word "albo" which suits the best here and it is equivalent to exclusive or which you don't use

[@ErickOliveira-i3w]

log( 1 + 2 + 3 ) = log(1) + log(2) + log(3)

[@leonz-g8l]

that's actually true

[@PrimeNewtons]

Those are not natural numbers

[@ErickOliveira-i3w]

log( j1 x j2 x j3 x ... jn) = log(j1) + log(j2) = log(j3) + ...+ log(jn), from j1 + j2 + j3 + ... + jn = j1 x j2 x j3 x ... jn, log( j1 + j2 + j3 + ... + jn) = log(j1) + log(j2) = log(j3) + ...+ log(jn)

[@maxvangulik1988]

n!=n(n+1)/2 n!=(n+1)!/2(n-1)! (n-1)!=(n+1)/2 Ř(n)=(n+1)/2 d/dn((n+1)/2)=1/2 Ř'(1)=-ř≈-.57 Ř'(2)=1-ř≈.43 Ř'(3)=3-2ř≈1.86>1/2 Ř(3)=2 4/2=2 the only solutions are n=1 and n=3

[@maxvangulik1988]

n! | T(n) 1!=1 | T(1)=1 2!=2 | T(2)=3 3!=6 | T(3)=6 4!=24 | T(4)=10 5!=120 | T(5)=15 6!=720 | T(6)=21 7!=5040 | T(7)=28 8!=40320 | T(8)=36 n! is already 3 orders of magnitude larger than T(n)

[@Kakarot-kr]

For a, b, c, d, e ∈N If a + b + c + d + e = abcde Find the maximum possible value of max {a, b, c, d, e } Sir pls explain it You've explained it before but there are different solutions at different places and the answer got by u also don't satisfy it

[@PrimeNewtons]

I will redo it

[@Kakarot-kr]

@@PrimeNewtons thnx a lot ❤️

[@robertveith6383]

This is not partial texting. Spell out "You've " and "you."

[@Kakarot-kr]

@@robertveith6383 ok but it's all about conveying information 😅💗💞

[@karamsedighi]

my friend, please answer! ! Are you from South Africa ? I LOVE MANDELA AND BLACK PEOPLE !!!!!

[@PrimeNewtons]

Nigeria 🇳🇬

[@robertveith6383]

Original poster, stop yelling in all caps. A color of a person is not to be loved. That is not logical.

[@AmilQarayev41]

THE INTEGRAL. 1/(1+x⁴). THE THIRD WAY. WHERE IS ITTT??

[@robertveith6383]

Stop yelling your post in all caps.

[@AmilQarayev41]

@robertveith6383 I just wonder what the 3rd way is😞😞