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.
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.
@@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
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 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.
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'.
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.
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
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
¡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
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
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
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... :)
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.
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.
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
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 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.
Bro, you have created a phenomenon whereby many crazed RUclips 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😂
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.
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.
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
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. ❤
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.
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
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!
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
U have to replace the ! Symbol for the factorial by the gamma function. So that it is well defined on the real line rather than the non-negative integers IN U {0}.
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
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 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.
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)
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.
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.
Math joke: Teach find X. Me: Y do you want me to keep finding your X? Did you loose her? That’s a missing persons situation. Go to the police! Get a good lawyer. If she left, then buddy just move on. Let her go! There are others out there. 😂😂😂😂
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.
Arrived at a similar step and due to university brainrot, started doing some asymptotic magic to try to cap an answer, until I stopped for a moment and realized, that n! grows at such a rate, that the answer has to be a small number. Great video :)
I saw the thumbnail and decide to pick a random number so i picked 5 and yea 5!=120 and 5³-5=125-5=120 and then i went👁👄👁 Edit:wait i was supposed to solve for x? Guess i got lucky with my guess
![How to Differentiate x^x ? [2 Different Methods]](http://i.ytimg.com/vi/FmkmaiesS9U/mqdefault.jpg)
Watch a redo here:
ruclips.net/video/m_Q_2IUv_P0/видео.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 . . .
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❤❤
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
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
His smile is contagious 😆
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.
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.
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
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.
Great didactics and an elegant presentation
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
What a marvelous joy and energy you have... I'm gonna take you as an example while helping my sons homework ❤
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.
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.
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
You're a legend , I like the way you transfer us your enthusiasm about math
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.
My favorite phrase: "by inspection..."
Gonna be honest, the moment I saw the thumbnail I guessed 5.
Jackpot. A Math- RUclipsr who uses a blackboard. Thank you.
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
This can easily be done in your head. 😊
Brooooo, you are GENIUS 🤌👏
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
¡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
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 😂😂
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
Simple and beautiful, basic algebra working to solve different equations! This kind of problem is perfect to "polish" the definitions and equalities
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
Honestly, the best part is how clear the speech and pronounciation was
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!
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... :)
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.
those who stop learning stop living 🙂
Man you so good broooo
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.
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
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
I was so happy that I got it by myself, thank you for the class.
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 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.
Bro, you have created a phenomenon whereby many crazed RUclips 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
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.
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.
Thank you so much for the lecture!
Sending love from Brazil 🫶
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!!
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
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
That intro looking like a teaser ☠️
Love your teaching style. Wish kids had more teachers like you.
solve it in my mind but it definitelly was an interesting one
We need to prove n!-n-3 > 0 for n>3 to prove there is no more solution for n>6.
Thanks for explaining it in a well organized way! appreciate it man
This is one of my new favorite channels. The way you explain and talk is awesome.
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.
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
Send this to the kidnapper so he can finally find the factorial of the x
Like the fact that wolframalpha proposes 2 decimal solutions, but not the 5.
Yes that puzzled me for a bit. Basically it didn't look hard enough. If you add the constraint x>3 it then finds 5.
en los naturales simplemente es mas facil acotar...
(x-1)!>x²-1, cuando x≥6.(esto es cierto por inducción)
Luego 0
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!
Absolutely love your explanation, very simple yet so easy to understand, thank you for this!
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
Looking at the thumbnail only, I though that the equation was a statement of equality for all N's
Hoping you guys not taking it in the wrong way. Happily surprised to see a non chinese/caucasian person doing this type of content!
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
U have to replace the ! Symbol for the factorial by the gamma function. So that it is well defined on the real line rather than the non-negative integers IN U {0}.
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
Very nice video ! Thx.
I'm subscribing !
Nice video and thanks for the problem. I would take the last equation mod n and you get the solution immediately
you have a way of speaking and conveying a message that is very satisfying to watch
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 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.
Sir thank you for your fantastic demonstration and working through the problem without edits. I look forward to future problems explored by you.
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)
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.
So since X+1 Is now n+ 1 and n=3 now x= 3+2 =5?
Well someone finally made this understandable ! Thank you
I thought bro was chewkz for a second
Cant wait “a math lecturer goes ghetto”
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
I looked at it and said, 4. No. 5. Yes, 5.
Took me 15 seconds.
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
Excellent explanation. Very nicely presented. Wish I knew you forty years ago in high school 😂
Math joke:
Teach find X.
Me: Y do you want me to keep finding your X? Did you loose her? That’s a missing persons situation. Go to the police!
Get a good lawyer.
If she left, then buddy just move on. Let her go! There are others out there. 😂😂😂😂
"Those stop learning, stop living" 👏👏👏 By the way, your channel is super.
I thought bro was chewks for a second
You are a genius! Im subscribing
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.
I am good at math. I don't speak English at all, but for some reason I could understand this video. I enjoyed it.
Arrived at a similar step and due to university brainrot, started doing some asymptotic magic to try to cap an answer, until I stopped for a moment and realized, that n! grows at such a rate, that the answer has to be a small number. Great video :)
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 saw the thumbnail and decide to pick a random number so i picked 5 and yea 5!=120 and 5³-5=125-5=120 and then i went👁👄👁
Edit:wait i was supposed to solve for x? Guess i got lucky with my guess
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) .