I love how you don’t cut out any mistakes you make. It makes me feel like I’m in college again listening to a real lecture. Just like every professor ends up doing. Good memories.
He did make mistakes. He assumed x was not 0, and then showed that x could not be 0. Then he did this again with x not being 1. If you have x!, you can only write x! = x * (x-1)! if x ≠ 0. He did the 0 check after he already made the assumption, which is backward.
He is like the Bob Ross of painting, I swear that voice of his is so calming, plus he's soooo good at teaching maths, I wish I wasn't prepping for exams that just focus on you to learn as much as possible but rather be taught on what you wish to learn from him
I saw another video that did this same problem a little while ago and the solution was clunky and inelegant (and wildly and unnecessarily more complicated than this) and didn't explain much about the properties of the functions in question or justify the manipulations being made so it really was better just to do it by inspection as you note at the beginning. This was much nicer, thank you!
@@user-om1te4qo7k wdym? If you mean looking at a figure, that's not at all the same as testing values for a possible soulution, then showing given value solves the equation. His proof is not illegitimate.
If you go 1 step further, you can factor out the n: n(n-1)! = 1+(3/n) -> since we know that a factorial always gives an integer answer, 3/n must be an integer value as well. Therefore n can only be 3 or 1 as 3 is prime.
@@aryandas5643there's infinite ways to do it, but one of them is known as Euler's Gamma Function, can define the factorial for any real x>0. Being the function: integral from 0 to inf(t^x*e^-t*dt)
I went even further when you had n!=n+3, you can say that n!-n=3 => n((n-1)!-1)=3 => either n=3 and (n-1)!-1=1 or n=1 and (n-1)!-1=3 or (because I also assumed that factorial was defined for negative numbers as well) n=-3 and (n-1)!-1=-1 or n=-1 and (n-1)!-1=-3
@@yyy76yyvhxxffb32 the way to get x from n is the same as in the video, which is why I didn't explain it here, just as I didn't explain how I defined n
Another great video, thanks for posting. Alternatively, noting that 1 and 2 are not solutions of x! = x^3 - x = (x + 1)x(x - 1), this can be rewritten as (x - 2)! = x + 1, and then as (x - 2) [(x - 3)! - 1] = 3. If we are assuming that x is a non-negative integer greater than 2, then since 3 is prime, the two factors on the left must be 1 and 3 (not necessarily in that order), so either (x - 2) = 1 and [(x - 3)! - 1] = 3, which is inconsistent, or (x - 2) = 3 and [(x - 3)! - 1] = 1, which is consistent and gives x = 5.
6:34 You don’t need to use trial-and-error/inspection. n ≠ 0 because 0! ≠ 0 + 3. This means that n | n + 3 (n divides/is a factor of n + 3) because n | n!. As a result, n | 3. The only positive factors/divisors of 3 are 1 & 3. However n ≠ 1 because 1! ≠ 1 + 3. Checking for n = 3, we get 3! = 3 + 3. Hence n = 3. Therefore x = 5. No trial-and-error used, just simple number theory and divisibility.
I love the use of number theory, but I don’t think we can say n | 3 simply because n | (n+3). As a counterexample, we know 6 = 2 + 4 6|6 obviously but 6 doesn’t divide 2 or 4
@@elijahcriswell1658 you aren’t following the n | n + 3 rule. Since n | n + 3, this means that (n + 3)/n = m, where m is an integer. Simplifying (n + 3)/n gives us m = 1 + 3/n, so 3/n = m - 1. Because m - 1 is an integer, 3/n must be an integer, so n | 3. As simple as that. These are basic divisibility rules. Your counterexample is wrong! 😑
The point of this video isn't really the answer. it's the clever method used and to teach people that factorial equations can be simplified by dividing by 'x'.
The proof is wrong. The answer is right, but the proof does things in the wrong order. As a combinatorial design theorist with a PhD in math, I would give this solution probably a 7/10 because of the mistakes.
@@dangboor4277 The very FIRST thing you need to do is show that x ≠ 0. As soon as you write x! = x * (x-1), you've already worked under the assumption that x ≠ 0, which he hasn't demonstrated yet: if x = 0. then (x-1)! = (-1)! is undefined and the equation is incorrect. So, before doing anything else, you first check if x = 0 is a valid solution. LHS: x! = 0! = 1 RHS: x^3 - x = 0^3 - 0 = 0 Thus, LHS is not equal to the RHS, so x = 0 is not a solution. Thus, x ≥ 1. Now you can proceed with his proof without violating the factorial function. This may not seem important to you, but it is absolutely important that every step in your proof follows in a valid way from the former steps, which in this proof, it does not quite do.
@@dangboor4277 Oh, reading the proof again (I just watched it again), he makes the same mistake again. Even after you prove that x ≠ 0, in order to expand (x-1)! to (x-1)(x-2)!, you. need to check that x is not 1. He does this backwards again. The idea of the proof is right, but the steps are out of order.
6:00 You don't need to guess. Since LHS is divisible by _n_ , RHS has to be as well. _n | n + 3_ Since _n | n_ , it must be _n | 3_ . And as _3_ is a *prime numbers* _n_ can only be either _3_ or _1_ Since we have already proved _n ≠ 1_ , _n = 3_ _n = x - 2_ *_x = 5_* I know guessing can be much easier here, but guessing does not always come in handy. I personally am not a fan of guessing. The main problem with it is that it does not prove if that is the only possible solution. We do not always become lucky like with this one.
When you have n!=n+3 you don't have to guess. Just solve it: n!-n=3 n[(n-1)!-1]=3. Since n is natural, from factorization you have two options: 1. n=1 and (n-1)!-1=3 or 2. n=3 and (n-1)!-1=1. From 1. you have no solutions and from 2. you have n=3 and hence, x=5.
n! = n + 3. Divide by n: (n-1)! = 1 + 3/n. We only want integer solutions and we know (n-1)! is integer. Since 3 only has two divisors (1 and 3), we only need to checkout n=1 and n=3. n = 3 holds at equality and n=1 doesn't. So we know n=3 => x = 5 is the only solution.
It is also quite easy to find the solution graphically, by simply plotting a rough graph of the two functions z = (x-2)! and g = x + 1. The function g is linear, y = (x-2)! we consider to be defined only in the domain of positive integers. So by simply plotting it we can quickly find the single intersection point (x,y) = (5, 6). However, if we dig a little deeper, we notice that (x-1)! is the gamma function, which has a rather complicated graph. At this point we have (x-1)! = x^2-1 - if you try to plot these functions, you will see that there are additional intersections of the parabola x^2-1 and the gamma function G(x) = (x-1)!, which is defined in the negative range. These solutions can be found numerically, which is obviously much broader than the scope of this video. But just so you know... :)
Not a method you could use every time, but you can easily guess X=5. This is because 5 is the only number that has similar values for factorial and cube. (Excluding 1 and 0, obviously) 2! = 2, 2^3 = 8 6, 27 24, 64 120, 125 720, 216 X! For X>6 is far larger than x^3. But the method in this video with solve for any case where is an integer. Great video 3
Is there any way to find n that is not by inspection? It just feels like this was such a cool puzzle, but then we just have to brute-force it at the end which feels like a bit of a let-down.
Cribbing from another comment. We have n! = n + 3. We can divide both sides by n to give (n-1)! = (n+3)/n. Since we know that a factorial must be an integer, we can deduce that n is divisible by 3, which narrows down the search space.
I did the work on my own and I managed to get to the answer on my own, but it ended up with some supplemental guess and check after i got to (x-2)! = x+1. I see why the n’s were used, but it was easier without the extra step for me
We want a more general solution, because even if the value of the solution is a little bit larger, the answer can't be obtained by substituting natural numbers one by one. And most of those not-general solutions are not useful.
I feel like from a practical perspective, factorial grows sufficiently faster than a cubic function and the domain of factorial is only countably infinite so if you just find an upper bound for x, you can just test a few values and boom
¡Hey, I enjoyed this video a lot! (true) I've learned a couple of criteria to deal with the factorial function, but also I enjoyed your fresh and natural attitude in developing the solution. Even with the little mistakes.Those little pauses you took to think showed me that you were not just repeating a mechanical procedure. ¡Really fine Math class! Best regards
Just a minor comment.....Not only the Gamma function, but also the factorial function can be evaluated in the complex plane ( as well as for not integer values) using the integral representations of these functions.
oh, this took some thinking; i was confused by the x+1 factor on the right until i realized it had to be a product of two smaller factors on the left, at which point 3*2=6 seemed like the most reasonable option it's nice to see a well organized version of solving it, instead of my "stare at the wall for a while" approach
From n! = n + 3 we have n( (n-1)! -1) = 3 So n is a factor of 3, since 3 is a prime it only has 2 factors: 1 or 3. n=1 does not solve the eqn so n must be 3 which works.
I'm watching and learning precalculus to help prepare me for my calculus class in the fall this was a very informative video, will be subscribing and adding you to my daily math practice, currently going through a precalc course with python which is a lot of gun for me as I'm a programming major
I equated x and (x - 2)! - 1, but the result is the same. The other solution in reals at ~1.37439 has to be found numerically, of course. You'll need a numerics library that can handle gamma functions or s symbolic program like Mathematica. The same goes for the infinitely many complex solutions (an example being ~9.349 + 11.327 i), which are rather beautifully ramified in the right hand side of the complex plane.
7:35 This makes each side monotonic, and it's clear that there's only one crossing point, and so you can place bounds and know for sure that the answer is within those, and that your answer is unique
The equation X! = (x^3) - x is a factorial equation, where X! denotes the factorial of X. However, the right-hand side of the equation, (x^3) - x, is a polynomial expression. For small values of X, we can try to find a solution: - X = 0: 0! = 1, but (0^3) - 0 = 0, so X = 0 is not a solution. - X = 1: 1! = 1, and (1^3) - 1 = 0, so X = 1 is not a solution. - X = 2: 2! = 2, but (2^3) - 2 = 6, so X = 2 is not a solution. - X = 3: 3! = 6, and (3^3) - 3 = 24, so X = 3 is not a solution. - X = 4: 4! = 24, and (4^3) - 4 = 60, so X = 4 is not a solution. - X = 5: 5! = 120, and (5^3) - 5 = 120, so X = 5 is a solution! Indeed, 5! = 120, and (5^3) - 5 = 125 - 5 = 120. Therefore, the solution to the equation X! = (x^3) - x is X = 5.
I have never really loved math, only kinda liked it. Until I saw this guy. He brings a sense of joy and wonder when he talks about math. I love that. I love this guy. Keep up the good work. ❤
from n!=n+3 instead of guessing all the numbers i went down the idea... well n! is divisible by n... then n+3 must be divisible by n as well... n is ofc divisible by n... then 3 must be divisible by n as well, otherwise their sum won't be divisible by n... what are the divisors of 3? 1 and 3... then we try 1... 1! =1 which does not equal 1+3=4, then it must be 3... 3! = 6 = 3+3 , yes i just rulled out 2 , which is not that much, but imagine if the original equation was x!=x^3 - 5 instead, then following the same logic we would only have to try out 1 and 5
There will actually be two intersections in quadrant one since desmos uses the gamma function when you graph the factorial and there is a 2nd positive solution if you allow for non integer answers
Mathematician here. While you have proven 3 is a solution, it doesn’t prove it’s the only one (in the natural numbers). One way to prove this (that I have not done but my intuition says) is by induction proving that from 4 onwards, n!>n+3 (this is obvious but still needs to be proven in order to give a complete solution). While extensions of a factorial function may be given, I really don’t think problems like this need to go there.
[6:35] I would change By Inspection for other way: by passing n to left, n! - n = 3 >> n(n-1)! - n = 3 >> n[(n-1)! - 1] = 3 So, let n an positive integer ('cause x is). The only way to n times [(n-1)! - 1] equals 3 is 1 and 3 n = 1 >> [(n-1)! - 1] = 3 >> (n-1)! = 4 >> n not integer 🚫 n = 3 >>[(n-1)! - 1] = 1 >>(n-1)! = 2 >> n - 1 = 2 >> n = 3 ✅
My first reaction is that x! goes skywards at a very rapid rate, so trial and error works. You can reduce the problem space a bit by dividing by "x", to get (x-1)!=x^2-1. It's obvious that (1-1)! Clearly x=1 obviously doesn't work, nor does 2, 3 or 4 but 5 does. Factorial 4 is 24 and 5 squared is 25. Subtract the 1 from 25 and you have the 24. It's simple enough to do in your head in a few seconds.
A caution on dealing with factorials: Expanding x! = x (x - 1)! is only possible for x ≥ 1. While this is implied here _afterwards_ by checking for x ≠ 0, and later again with x - 1, it's always worth mentioning at the expansion step itself, in the same manner as checking for not dividing by 0, which you presented very well. Consider that in other situations one might not immediately attempt to cancel x and thus be prompted to check for x≠ 0.
I find your lecture excellent, although I find it essential you check the final result and demonstrate, in this case publicly ,whether it fits or not. .
Good question, but this is actually much harder than one thinks, 5 is not the single answer. If you use Gamma function, there are one more positive and infinite many more negative
At the moment when (x-2)!=x+1 we can rewrite the equation as: (x-2)!=x-2+3 (x-3)!=1+(3/( x-2)) Now our solutions are limited to numbers that, when subtracting "-2", are multiples of three, but not more than the three itself This is 3, 5 Also, given that 0! = 1, we can conclude that 1+n cannot equal 0! Hence the answer is 5 This may not be such a pretentious solution, but it avoids iterating over numbers - the only answer (well, or one of two)
I really surprised with your explanation teacher, coz I've never thought about equation of factorial and polynomial. I want to ask you one thing. Is it meaningful to expand and process previous equations to get final equation? It seems like you put most likely value that meets n! = n + 3. Why don't we just do it from the start then?
There's a more rigorous way than proof by inspection. Notice that n!-n = 3, and n divides the LHS, so n divides RHS = 3. Then n = 1 or 3. 3 works, 1 doesn't.
I feel like I went through the same steps but thought about it differently at the last step: (x-2)! = x+1 tells us that x+1 is the product of at least 2 consecutive numbers, so going through those, I tried 2, then 6.
You could subtract x-2 from both sides (x-3)!-1=1 => (x-3)! = 2 => x-3 = 2 (since 2=2!) => x=5. But I'm sure you knew that and had you reason for just inspecting.
Bro, you have created a phenomenon whereby many crazed UA-cam viewers have shouted "it's x+1", "it's x+1"! At their phones and screens😂❤🎉 Apologies in advance to unsuspecting companions that we may startle with our exclamations. You needn't worry, we're just watching math again😂
Before watching the solutions here is my approach: x!=x³-x (x)(x-1)(x-2)(x-3)....3×2×1=x(x²-1) =>(x)(x-1)(x-2)(x-3)....3×2×1=x(x-1)(x+1) Now cancelation of two terms i.e x and (x-1) on both sides We got (x-2)!=(x+1) By hit and trial method Substitution of x=5 we can get LHS and RHS (Answer applicable only in case of whole numbers)
First time ever that I have come across a channel and after one video, have I decided to immediately subscribe. His happiness is everything and his enthusiasm makes me keep going. Thank you!
Why "any natural number in the power of 3 minus this number" is always a multiple of 6? 3^3-3=6 can be divided by 6, it will be 1 4^3-4=60 can be divided by 6, it will be 10 5^3-5=120 can be divided by 6, it will be 20 6^3-3=210 can be divided by 6, it will be 35 And so on And even 31 for example: 31^3-31=29760 can be divided by 6, it will be 4960
x^3 - x = x(x-1)(x+1) So for an integer x greater than 1 we have the product of 3 consecutive positive integers. we know that in a set of 3 consecutive integers at least one will be even so their product is divisible by 2. Additionally every third integer is divisible by 3 so one of the three numbers will be divisible by 3 meaning the product is also divisible by 3. Since the product is divisible by both 2 and 3 it must be divisible by 6
Watch a redo here:
ua-cam.com/video/m_Q_2IUv_P0/v-deo.html
I love how you don’t cut out any mistakes you make. It makes me feel like I’m in college again listening to a real lecture. Just like every professor ends up doing. Good memories.
I wish my professor actually gives lecture like this. All he does is pull up his pre made slideshow from 4-5 years ago and just shows us and says gl.
He did make mistakes. He assumed x was not 0, and then showed that x could not be 0. Then he did this again with x not being 1.
If you have x!, you can only write x! = x * (x-1)! if x ≠ 0. He did the 0 check after he already made the assumption, which is backward.
Public chalkboard math is 1000x harder than private paper math. I can't prove it but it is true
@@joevero4568 In maths, if you cant prove it, then its probably not true.
And this is what I really agreee!
He is like the Bob Ross of painting, I swear that voice of his is so calming, plus he's soooo good at teaching maths, I wish I wasn't prepping for exams that just focus on you to learn as much as possible but rather be taught on what you wish to learn from him
Factorial is a happy little function . . .
Proof by inspection 🗿
No srsly inspection doesnt count tho
Its not a proof. Its a solve. Inspection is allowed.
He should have started by this
I saw another video that did this same problem a little while ago and the solution was clunky and inelegant (and wildly and unnecessarily more complicated than this) and didn't explain much about the properties of the functions in question or justify the manipulations being made so it really was better just to do it by inspection as you note at the beginning. This was much nicer, thank you!
I think I did too. This way is really nice
I watched that video too.
Smooth❤❤
Love your energy - one can tell that you really enjoy teaching the subject and that goes a long way towards engaging your students!
6:27 .. n! = n + 3 =>
(n-1)! = (n+3)/n : is an integer => n is a multiple of 3.
=> start guessing n = 3 and go up in steps of 3.
(n+3)/n=1+3/n so n=1 or 3 are only possibilities.
@@jacobgoldman5780 yup
n! = n + 3 implies that 3 must be a multiple of n which means that n has to be a factor of 3, 1 is wrong so the answer is 3
This is even better. It shows as well that no other solutions except 5 is possible. Great job!
This feels wrong
Bro made "by inspection" sound like a mathematical proof lmao
It is, indeed.
@@PrimeNewtons then I can prove given triangle is equilateral by *inspection*
@@user-om1te4qo7kyes well if you measure all the sides with a tool , this is used a lot in questions
@@user-om1te4qo7k wdym? If you mean looking at a figure, that's not at all the same as testing values for a possible soulution, then showing given value solves the equation. His proof is not illegitimate.
@@user-om1te4qo7kif you inspect it and two of the angles are 60° You're done.
His smile is contagious 😆
If you go 1 step further, you can factor out the n: n(n-1)! = 1+(3/n) -> since we know that a factorial always gives an integer answer, 3/n must be an integer value as well. Therefore n can only be 3 or 1 as 3 is prime.
In natural nambers,
x!=x³-x⇔x=5
However,in real numbers
x>0,x=1.37....,5
x
how do you even define factorials for real numbers that are not positive integers?
@@aryandas5643 Gamma function. It's really interesting so look at it
@@aryandas5643Г-function
@@aryandas5643 Gamma function!😊
@@aryandas5643there's infinite ways to do it, but one of them is known as Euler's Gamma Function, can define the factorial for any real x>0. Being the function: integral from 0 to inf(t^x*e^-t*dt)
I went even further when you had n!=n+3, you can say that n!-n=3 => n((n-1)!-1)=3 => either n=3 and (n-1)!-1=1 or n=1 and (n-1)!-1=3 or (because I also assumed that factorial was defined for negative numbers as well) n=-3 and (n-1)!-1=-1 or n=-1 and (n-1)!-1=-3
Brilliant!
That's what I should have done. Thanks
Brilliant the fact that you forgot about the x
You know you could have done the equation with (x-2) but its easier with n
@@yyy76yyvhxxffb32 the way to get x from n is the same as in the video, which is why I didn't explain it here, just as I didn't explain how I defined n
Wow you're smart
Great didactics and an elegant presentation
Another great video, thanks for posting. Alternatively, noting that 1 and 2 are not solutions of x! = x^3 - x = (x + 1)x(x - 1), this can be rewritten as (x - 2)! = x + 1, and then as (x - 2) [(x - 3)! - 1] = 3. If we are assuming that x is a non-negative integer greater than 2, then since 3 is prime, the two factors on the left must be 1 and 3 (not necessarily in that order), so either (x - 2) = 1 and [(x - 3)! - 1] = 3, which is inconsistent, or (x - 2) = 3 and [(x - 3)! - 1] = 1, which is consistent and gives x = 5.
6:34 You don’t need to use trial-and-error/inspection. n ≠ 0 because 0! ≠ 0 + 3. This means that n | n + 3 (n divides/is a factor of n + 3) because n | n!. As a result, n | 3. The only positive factors/divisors of 3 are 1 & 3. However n ≠ 1 because 1! ≠ 1 + 3. Checking for n = 3, we get 3! = 3 + 3. Hence n = 3. Therefore x = 5. No trial-and-error used, just simple number theory and divisibility.
I love the use of number theory, but I don’t think we can say n | 3 simply because n | (n+3).
As a counterexample, we know
6 = 2 + 4
6|6 obviously but 6 doesn’t divide 2 or 4
@@elijahcriswell1658 you aren’t following the n | n + 3 rule. Since n | n + 3, this means that (n + 3)/n = m, where m is an integer. Simplifying (n + 3)/n gives us m = 1 + 3/n, so 3/n = m - 1. Because m - 1 is an integer, 3/n must be an integer, so n | 3. As simple as that. These are basic divisibility rules. Your counterexample is wrong! 😑
Thanks for this! I intuitively knew that the nature of prime factors in n! would make other solutions unlikely. Your work confirms it.
Factorial dominates x^3 - x polynomial for x>5. Then you just have to try the cases:
x=0,1,2,3,4,5
x=5 works.
The real question is, is it faster to start at 5 or to start at 0. Or 3. Hmmm…
he did acknowedge that he just wanted to show how to do it algebraically
The point of this video isn't really the answer. it's the clever method used and to teach people that factorial equations can be simplified by dividing by 'x'.
How did you know that the factorial dominates for x > 5? I knew it would dominate at some point, how did you get to the 5?
@@3snoW_ i guess because 4x3x2 is less than 5^2, but 5x4x3x2 is way bigger than 6^2
man your vibes are so positive, if i’d have had a teacher like you in grade school it wouldn’t have taken me until college to fall in love with math
I especially like the proof format! I can see how this method greatly reinforces the concept that math needs to be rigorous and logical.
The proof is wrong. The answer is right, but the proof does things in the wrong order. As a combinatorial design theorist with a PhD in math, I would give this solution probably a 7/10 because of the mistakes.
@@vorpal22Why
@@dangboor4277 The very FIRST thing you need to do is show that x ≠ 0. As soon as you write x! = x * (x-1), you've already worked under the assumption that x ≠ 0, which he hasn't demonstrated yet: if x = 0. then (x-1)! = (-1)! is undefined and the equation is incorrect.
So, before doing anything else, you first check if x = 0 is a valid solution.
LHS: x! = 0! = 1
RHS: x^3 - x = 0^3 - 0 = 0
Thus, LHS is not equal to the RHS, so x = 0 is not a solution.
Thus, x ≥ 1.
Now you can proceed with his proof without violating the factorial function. This may not seem important to you, but it is absolutely important that every step in your proof follows in a valid way from the former steps, which in this proof, it does not quite do.
@@dangboor4277 Oh, reading the proof again (I just watched it again), he makes the same mistake again. Even after you prove that x ≠ 0, in order to expand (x-1)! to (x-1)(x-2)!, you. need to check that x is not 1. He does this backwards again. The idea of the proof is right, but the steps are out of order.
6:00 You don't need to guess.
Since LHS is divisible by _n_ , RHS has to be as well.
_n | n + 3_
Since _n | n_ , it must be _n | 3_ .
And as _3_ is a *prime numbers* _n_ can only be either _3_ or _1_
Since we have already proved _n ≠ 1_ ,
_n = 3_
_n = x - 2_
*_x = 5_*
I know guessing can be much easier here, but guessing does not always come in handy. I personally am not a fan of guessing.
The main problem with it is that it does not prove if that is the only possible solution. We do not always become lucky like with this one.
Bro got them math rizz, your lecture are so enjoyable lol
Handwriting so good for a math teacher. 👏
What a marvelous joy and energy you have... I'm gonna take you as an example while helping my sons homework ❤
When you have
n!=n+3
you don't have to guess. Just solve it:
n!-n=3
n[(n-1)!-1]=3.
Since n is natural, from factorization you have two options:
1. n=1 and (n-1)!-1=3 or
2. n=3 and (n-1)!-1=1.
From 1. you have no solutions and from 2. you have n=3 and hence, x=5.
I think I would enjoy you teaching any subject because of your talent for presenting and curiosity. Bravo!
n! = n + 3. Divide by n: (n-1)! = 1 + 3/n. We only want integer solutions and we know (n-1)! is integer. Since 3 only has two divisors (1 and 3), we only need to checkout n=1 and n=3. n = 3 holds at equality and n=1 doesn't. So we know n=3 => x = 5 is the only solution.
It is also quite easy to find the solution graphically, by simply plotting a rough graph of the two functions z = (x-2)! and g = x + 1. The function g is linear, y = (x-2)! we consider to be defined only in the domain of positive integers. So by simply plotting it we can quickly find the single intersection point (x,y) = (5, 6). However, if we dig a little deeper, we notice that (x-1)! is the gamma function, which has a rather complicated graph. At this point we have (x-1)! = x^2-1 - if you try to plot these functions, you will see that there are additional intersections of the parabola x^2-1 and the gamma function G(x) = (x-1)!, which is defined in the negative range. These solutions can be found numerically, which is obviously much broader than the scope of this video. But just so you know... :)
5:24
We can use induction to prove that n!>n+3 for all n>3
For natural numbers.
@@TheLukeLsd obviously
the factorial function is limited to natural numbers as a domain, so induction is the best idea
Brooooo, you are GENIUS 🤌👏
Jackpot. A Math- UA-camr who uses a blackboard. Thank you.
My favorite phrase: "by inspection..."
Not a method you could use every time, but you can easily guess X=5. This is because 5 is the only number that has similar values for factorial and cube. (Excluding 1 and 0, obviously)
2! = 2, 2^3 = 8
6, 27
24, 64
120, 125
720, 216
X! For X>6 is far larger than x^3.
But the method in this video with solve for any case where is an integer. Great video
3
Is there any way to find n that is not by inspection? It just feels like this was such a cool puzzle, but then we just have to brute-force it at the end which feels like a bit of a let-down.
Cribbing from another comment.
We have n! = n + 3. We can divide both sides by n to give (n-1)! = (n+3)/n. Since we know that a factorial must be an integer, we can deduce that n is divisible by 3, which narrows down the search space.
@@qwertyTRiG Going further: distribute (n+3)/n to get 1+3/n. 3 is only divisible by 1 or 3, so n is 1 or 3.
I did the work on my own and I managed to get to the answer on my own, but it ended up with some supplemental guess and check after i got to (x-2)! = x+1. I see why the n’s were used, but it was easier without the extra step for me
We want a more general solution, because even if the value of the solution is a little bit larger, the answer can't be obtained by substituting natural numbers one by one. And most of those not-general solutions are not useful.
I feel like from a practical perspective, factorial grows sufficiently faster than a cubic function and the domain of factorial is only countably infinite so if you just find an upper bound for x, you can just test a few values and boom
You have excellent presentation my friend. I'm a math graduate and loved professors like you!
Simple and beautiful, basic algebra working to solve different equations! This kind of problem is perfect to "polish" the definitions and equalities
¡Hey, I enjoyed this video a lot! (true) I've learned a couple of criteria to deal with the factorial function, but also I enjoyed your fresh and natural attitude in developing the solution. Even with the little mistakes.Those little pauses you took to think showed me that you were not just repeating a mechanical procedure. ¡Really fine Math class! Best regards
This can easily be done in your head. 😊
You're a legend , I like the way you transfer us your enthusiasm about math
Just a minor comment.....Not only the Gamma function, but also the factorial function can be evaluated in the complex plane ( as well as for not integer values) using the integral representations of these functions.
Gonna be honest, the moment I saw the thumbnail I guessed 5.
oh, this took some thinking; i was confused by the x+1 factor on the right until i realized it had to be a product of two smaller factors on the left, at which point 3*2=6 seemed like the most reasonable option
it's nice to see a well organized version of solving it, instead of my "stare at the wall for a while" approach
From n! = n + 3 we have
n( (n-1)! -1) = 3
So n is a factor of 3, since 3 is a prime it only has 2 factors: 1 or 3. n=1 does not solve the eqn so n must be 3 which works.
I'm watching and learning precalculus to help prepare me for my calculus class in the fall this was a very informative video, will be subscribing and adding you to my daily math practice, currently going through a precalc course with python which is a lot of gun for me as I'm a programming major
Honestly, the best part is how clear the speech and pronounciation was
I equated x and (x - 2)! - 1, but the result is the same. The other solution in reals at ~1.37439 has to be found numerically, of course. You'll need a numerics library that can handle gamma functions or s symbolic program like Mathematica. The same goes for the infinitely many complex solutions (an example being ~9.349 + 11.327 i), which are rather beautifully ramified in the right hand side of the complex plane.
Cool! But it troubles that all these or geometrical ones start with dividing by some rsndom shit or drawing some random line or circle or smth
7:35 This makes each side monotonic, and it's clear that there's only one crossing point, and so you can place bounds and know for sure that the answer is within those, and that your answer is unique
The equation X! = (x^3) - x is a factorial equation, where X! denotes the factorial of X.
However, the right-hand side of the equation, (x^3) - x, is a polynomial expression.
For small values of X, we can try to find a solution:
- X = 0: 0! = 1, but (0^3) - 0 = 0, so X = 0 is not a solution.
- X = 1: 1! = 1, and (1^3) - 1 = 0, so X = 1 is not a solution.
- X = 2: 2! = 2, but (2^3) - 2 = 6, so X = 2 is not a solution.
- X = 3: 3! = 6, and (3^3) - 3 = 24, so X = 3 is not a solution.
- X = 4: 4! = 24, and (4^3) - 4 = 60, so X = 4 is not a solution.
- X = 5: 5! = 120, and (5^3) - 5 = 120, so X = 5 is a solution!
Indeed, 5! = 120, and (5^3) - 5 = 125 - 5 = 120.
Therefore, the solution to the equation X! = (x^3) - x is X = 5.
me at 4:14 : "oh no no no no no please"
me at 5:08 : "oh, thank god"
watching this video makes me feel smart in mathematic but when i tried to solve an easy looking algebra, i don't even know what to do first 😂😂
I have never really loved math, only kinda liked it. Until I saw this guy. He brings a sense of joy and wonder when he talks about math. I love that. I love this guy. Keep up the good work. ❤
Wow, thanks!
the fact that, you can prove x! > x^3-x for all interger x>5, by induction. Therefore, you only need to check x=1,2,3,4,5 and x=5 is the solution
en los naturales simplemente es mas facil acotar...
(x-1)!>x²-1, cuando x≥6.(esto es cierto por inducción)
Luego 0
those who stop learning stop living 🙂
Thank you so much for the lecture!
Sending love from Brazil 🫶
Love your teaching style. Wish kids had more teachers like you.
from n!=n+3 instead of guessing all the numbers i went down the idea... well n! is divisible by n... then n+3 must be divisible by n as well... n is ofc divisible by n... then 3 must be divisible by n as well, otherwise their sum won't be divisible by n... what are the divisors of 3? 1 and 3... then we try 1... 1! =1 which does not equal 1+3=4, then it must be 3... 3! = 6 = 3+3 , yes i just rulled out 2 , which is not that much, but imagine if the original equation was x!=x^3 - 5 instead, then following the same logic we would only have to try out 1 and 5
We can use desmos to graph the functions y = x! & y = x + 3, and the solution will be the intersection of the functions in quadrant 1
There will actually be two intersections in quadrant one since desmos uses the gamma function when you graph the factorial and there is a 2nd positive solution if you allow for non integer answers
Mathematician here. While you have proven 3 is a solution, it doesn’t prove it’s the only one (in the natural numbers). One way to prove this (that I have not done but my intuition says) is by induction proving that from 4 onwards, n!>n+3 (this is obvious but still needs to be proven in order to give a complete solution).
While extensions of a factorial function may be given, I really don’t think problems like this need to go there.
شكرا استاد. برايم انا عبدالله. من سوريا. لقد اعجبني اعطائك وشرحك ❤. هذه التمارين موجودة عندنا في سوريا وبشكل اصعب
I just discovered this channel :) I like the way of explaining and his voice and im going to check some other videos of this cannel, great video!!
[6:35] I would change By Inspection for other way: by passing n to left, n! - n = 3 >> n(n-1)! - n = 3 >> n[(n-1)! - 1] = 3
So, let n an positive integer ('cause x is). The only way to n times [(n-1)! - 1] equals 3 is 1 and 3
n = 1 >> [(n-1)! - 1] = 3 >> (n-1)! = 4 >> n not integer 🚫
n = 3 >>[(n-1)! - 1] = 1 >>(n-1)! = 2 >> n - 1 = 2 >> n = 3 ✅
My first reaction is that x! goes skywards at a very rapid rate, so trial and error works. You can reduce the problem space a bit by dividing by "x", to get (x-1)!=x^2-1. It's obvious that (1-1)! Clearly x=1 obviously doesn't work, nor does 2, 3 or 4 but 5 does. Factorial 4 is 24 and 5 squared is 25. Subtract the 1 from 25 and you have the 24. It's simple enough to do in your head in a few seconds.
Absolutely love your explanation, very simple yet so easy to understand, thank you for this!
I was so happy that I got it by myself, thank you for the class.
A caution on dealing with factorials: Expanding x! = x (x - 1)! is only possible for x ≥ 1. While this is implied here _afterwards_ by checking for x ≠ 0, and later again with x - 1, it's always worth mentioning at the expansion step itself, in the same manner as checking for not dividing by 0, which you presented very well.
Consider that in other situations one might not immediately attempt to cancel x and thus be prompted to check for x≠ 0.
Thanks for explaining it in a well organized way! appreciate it man
X!=x³-x =x(x²-1)=(x+1)(x)(x-1)
For factorial to be defined x-1 could be 1,2
x-1=1,2
X=2,3
This is one of my new favorite channels. The way you explain and talk is awesome.
Man you so good broooo
I find your lecture excellent, although I find it essential you check the final result and demonstrate, in this case publicly ,whether it fits or not. .
Good question, but this is actually much harder than one thinks, 5 is not the single answer. If you use Gamma function, there are one more positive and infinite many more negative
I think you also have to prove by induction that there are no other solutions for n! = n + 3, for the answer to be complete
solve it in my mind but it definitelly was an interesting one
x! = x^3 - x
x(x-1)! = x(x^2 - 1)
A factorial is one less than a square
25 less 1 is 4! and so the answer is 5
At the moment when (x-2)!=x+1 we can rewrite the equation as:
(x-2)!=x-2+3
(x-3)!=1+(3/( x-2))
Now our solutions are limited to numbers that, when subtracting "-2", are multiples of three, but not more than the three itself
This is 3, 5
Also, given that 0! = 1, we can conclude that 1+n cannot equal 0!
Hence the answer is 5
This may not be such a pretentious solution, but it avoids iterating over numbers - the only answer (well, or one of two)
I really surprised with your explanation teacher, coz I've never thought about equation of factorial and polynomial.
I want to ask you one thing. Is it meaningful to expand and process previous equations to get final equation? It seems like you put most likely value that meets n! = n + 3. Why don't we just do it from the start then?
Nice video and thanks for the problem. I would take the last equation mod n and you get the solution immediately
Since the right hand side is a cubic equation then the given situation should have 3 solutions (some in complex world can be there also) .
There's a more rigorous way than proof by inspection. Notice that n!-n = 3, and n divides the LHS, so n divides RHS = 3. Then n = 1 or 3. 3 works, 1 doesn't.
i am brazilian, i found your vídeo in explorar, i like lots of your vídeo, its a nice lesson teacher, you have a new subscriber
Or you can just do
n!=n+3
n!-n=3
n[(n-1)!-1]=3 , since n ≠0
(n-1)!-1=3/n
So n divide 3 so n =1 or n=3
And because n≠1 so n=3
I feel like I went through the same steps but thought about it differently at the last step: (x-2)! = x+1 tells us that x+1 is the product of at least 2 consecutive numbers, so going through those, I tried 2, then 6.
You could subtract x-2 from both sides (x-3)!-1=1 => (x-3)! = 2 => x-3 = 2 (since 2=2!) => x=5. But I'm sure you knew that and had you reason for just inspecting.
Bro, you have created a phenomenon whereby many crazed UA-cam viewers have shouted "it's x+1", "it's x+1"! At their phones and screens😂❤🎉 Apologies in advance to unsuspecting companions that we may startle with our exclamations. You needn't worry, we're just watching math again😂
Haha so true
Very nice video ! Thx.
I'm subscribing !
Send this to the kidnapper so he can finally find the factorial of the x
Hoping you guys not taking it in the wrong way. Happily surprised to see a non chinese/caucasian person doing this type of content!
Before watching the solutions here is my approach:
x!=x³-x
(x)(x-1)(x-2)(x-3)....3×2×1=x(x²-1)
=>(x)(x-1)(x-2)(x-3)....3×2×1=x(x-1)(x+1)
Now cancelation of two terms i.e x and (x-1) on both sides
We got (x-2)!=(x+1)
By hit and trial method
Substitution of x=5 we can get LHS and RHS (Answer applicable only in case of whole numbers)
First time ever that I have come across a channel and after one video, have I decided to immediately subscribe. His happiness is everything and his enthusiasm makes me keep going. Thank you!
Welcome aboard!
"Those stop learning, stop living" 👏👏👏 By the way, your channel is super.
you have a way of speaking and conveying a message that is very satisfying to watch
Sir thank you for your fantastic demonstration and working through the problem without edits. I look forward to future problems explored by you.
Also you can say that n should be lesser than 3, so that the factorial is positive.
you should show that for n=2 n!n+3 and factorial is growing faster than n+3...
That intro looking like a teaser ☠️
Why "any natural number in the power of 3 minus this number" is always a multiple of 6?
3^3-3=6 can be divided by 6, it will be 1
4^3-4=60 can be divided by 6, it will be 10
5^3-5=120 can be divided by 6, it will be 20
6^3-3=210 can be divided by 6, it will be 35
And so on
And even 31 for example:
31^3-31=29760 can be divided by 6, it will be 4960
x^3 - x = x(x-1)(x+1)
So for an integer x greater than 1 we have the product of 3 consecutive positive integers. we know that in a set of 3 consecutive integers at least one will be even so their product is divisible by 2. Additionally every third integer is divisible by 3 so one of the three numbers will be divisible by 3 meaning the product is also divisible by 3. Since the product is divisible by both 2 and 3 it must be divisible by 6
@@deananderson7714 thanks. I didn't know that equasion x^3-x=x(x-1)(x+1).
So since X+1 Is now n+ 1 and n=3 now x= 3+2 =5?
Well someone finally made this understandable ! Thank you