I love seeing these comments of elders still learning math! It's an oddly wholesome and strange thing to me since I live in a country where elders are very old-fashioned and aren't very educated.
x! = x(x-1)! x^3-x = x(x^2-1) So we can cancel the x and get the equality: (x-1)! = x^2-1 (x-1)! = (x-1)(x-2)! x^2-1 = (x-1)(x+1) So we can cancel the (x-1) and get the equality: (x-2)! = (x+1) Assuming no gamma function x >= 2. Since 0! = 1! = 1 < x+1 we know that x-2 >= 2 and x >= 4. Factorials larger than 2! is even, so x+1 has to be even, thus x has to be odd. Factorial is multiplicative and quickly increasing and x+1 is additive and slowly increasing, so the solution has to be small. Try the first odd integer larger than 4. (5-2)! = 5+1 3! = 6 So x=5 is our solution.
Here's what I came up with. x!=x³-x x(x-1)!=x(x²-1) (x-1)(x-2)!=(x-1)(x+1) (x-2)(x-3)!=x+1 (x-3)!=(x+1)/(x-2) (x-3)!=1+(3/(x-2)) Now (3/(x-2)) must be an integer. Since 3 is prime, (x-2) must either equal 1 or 3, but (x-2) cannot equal 1 becauss x=3 would not satisfy the given equation. So x-2=3, and so x=5.
Levels in math in my country (Finland) have been declining for some time. Had we this guy in all our schools (so we need to invent cloning) I think we wouldn't have any issues at all.
Or...n!=n+3. So, n*(n-1)!=n(1+(3/n)). And (n-1)!=1+3/n. The only way to get an integer is when n=3. Thus, (3-1)!=1+(3/3)=2. Indeed, 2!=2. Therefore x=5.
I guessed 5 but this is a very beautiful proof that 5 is only natural solution (although for most parts of high school we don’t learn factorial for non natural numbers)
This is so cool. I was like, does it work with only 3 ? no it would work with any prime number... Actualy it would work with any number, just the number of cases to check would go up. So now I have a way to solve x! = x^n + a for any n in N and a in Z
As an engineer student, i say u should use newton rhapson method or any numerical method so u get numerical answer, but if u insist to get the analytical solution probably change factorial into stirling approximation will help
it doesnt matter that much but when you wrote (x-1)! = (×-1)(x-2)!, you needed to have checked that x-2 > 0. If for instance X was 1, you dve have written 0 * (-1)! I get that you proved this wasnt true when cancelling but I think it would have been important to remind people when rewriting x! that everything in the bracket must be above 0 as -1! is undefined
Plotting two graphs, one of x! and one of x^3-x reveal that there are two intersection points. one intersection is at obviously x = 5 and the other one is at x=1.374. Can someone please explain me where do we lose the other root?
factorial is a function onl defined for non negative whole numbers. You can extent factorial to a function, that is also defined for fractions, but that is the gamma function, whis is not the same as factorial, because factorial is only defined for non negative whole numbers. Because of that, there is onl 1 solution.
There is only one solution with respect to the positive integers. If we want to find all real solution, we need to consider the continuous version of the factorial function, namely the Gamma Function. Under the reals, this equation has more than one solution
@@PrimeNewtons I also got confused. But don't worry ,we all make mistakes , that's how we learn , and our videos are awesome. I am grateful for your videos.
factoring on the right and dividing both parts by x(x-1), x=0 and x=1 are clearly to be discarded, we immediately obtain: (x-2)! = x+1 ,with x>2 by trial and error starting from x=3 the solution x=5 is found correct.
Love your videos, but this ciould have been done much easier using guess and check at the point of (x-2)!= x+1 with maximum of 3 consecutive trials. This is because we kniw that (x-2)! has to be: 1. Z+ 2. >= 2
Hello, This is the English or American way of approaching solving the problem; nevertheless, it is not mathematics but trial and error, guessing and assuming. Guessing and presuming are not mathematical formulas. Truly,
@@PrimeNewtons Hello, Mathematics mimics physics to achieve formulas, which most of the time are incorrect or lies; if God used Sin, Cos, Log, Ln, e, g, G, π (3.14159265358979323846264338327950288419...→ .∞, ETC., the universe would have destroyed itself before its creation. Most of the mathematics, such as the above signs, are lies to be valid upon Crooks's agreements. Mathematics is a tool (formula) to solve problems, not to keep assuming paths over and over again to find solutions in immature methods. For instance, by using the same strategies to solve 77e⁷⁷!=1.81880373878e¹⁴⁵ rebelliously, it shall take many million years to acquire the accurate number. As mathematics communication, nuclear physics, nuclear cosmology, mechanical, electrical, and electronics scientists reject the human coherence of knowledge because of illogicality; people think they are alive and perceive the past's future, not the present! Genuinely,
Well... hmmm... There is no potential of generalization, as i'm trying to reverse the process. It works out just for the one value of "x == 3". Since there is no generalization potential, the argumentation with "the right way to show" has a smell of... let's call it "artefact hunting". By accident, for this unique small factor, the decomposition "works", but neither is the decomposition necessary because of the pettiness of the factor, nor would the method be generalizable for arbitrary resulting values. I must admit that driven by pure curiosity, i too took that same way of decomposition (since in the math channels i had subscribed to, this decompositions into binoms seem to be the stick horse of all authors), but nevertheless i am not impressed by the example.
@@davidgagen9856 If you - for example - try to setup an instance of this problem with arbitrary parameters (from the solution back to the problem presentation), you end up with constructs that are not decomposible as in this very specific case.
I believe the explanation shown is as rigorous as it has to be. The procedure used does not have to be generalizable, it is enough to get to the equation where an integer expression is equal to a prime number. While it's true that the two cases work only when the number is prime, you can still do a similar procedure if the number is not prime, just use prime decomposition. There would just be more cases, it doesn't mean that it's wrong. My only two nitpick with the video are: 1) the initial question should have "x is an integer". 2) it must be shown that (x-2)! Is allowed. The rest follows pretty much flawlessly, great video.
Great!! I just wish I was born 70 years later, then I would be learning maths in this wonderful way (aged six )!!😊
6?
@@GeezSus Yup. Since I'm 76 😊
I love seeing these comments of elders still learning math! It's an oddly wholesome and strange thing to me since I live in a country where elders are very old-fashioned and aren't very educated.
@@deepakchandra9600 Oh my bad, I thought you mistyped 60
@@GeezSus No problem! !😊🤪
Yeah this way you can also be sure there are no other solutions
Nice job! I always enjoy your content but that was a much more satisying ending
x! = x(x-1)!
x^3-x = x(x^2-1)
So we can cancel the x and get the equality: (x-1)! = x^2-1
(x-1)! = (x-1)(x-2)!
x^2-1 = (x-1)(x+1)
So we can cancel the (x-1) and get the equality: (x-2)! = (x+1)
Assuming no gamma function x >= 2.
Since 0! = 1! = 1 < x+1 we know that x-2 >= 2 and x >= 4.
Factorials larger than 2! is even, so x+1 has to be even, thus x has to be odd.
Factorial is multiplicative and quickly increasing and x+1 is additive and slowly increasing, so the solution has to be small.
Try the first odd integer larger than 4.
(5-2)! = 5+1
3! = 6
So x=5 is our solution.
This method is much better than the previous one. I understand this method easily and clearly. THANK YOU.
Love your energy, love your titles, love the chalk. Keep doing what you're doing.
Thanks sir, this class is so helpful to me👍😎
Very clear explanation!
Dude your voice is just perfect
Here's what I came up with.
x!=x³-x
x(x-1)!=x(x²-1)
(x-1)(x-2)!=(x-1)(x+1)
(x-2)(x-3)!=x+1
(x-3)!=(x+1)/(x-2)
(x-3)!=1+(3/(x-2))
Now (3/(x-2)) must be an integer. Since 3 is prime, (x-2) must either equal 1 or 3, but (x-2) cannot equal 1 becauss x=3 would not satisfy the given equation.
So x-2=3, and so x=5.
Really fantastic equation and fantastic explain too..... Cool!
Levels in math in my country (Finland) have been declining for some time. Had we this guy in all our schools (so we need to invent cloning) I think we wouldn't have any issues at all.
Or...n!=n+3. So, n*(n-1)!=n(1+(3/n)). And (n-1)!=1+3/n. The only way to get an integer is when n=3. Thus, (3-1)!=1+(3/3)=2. Indeed, 2!=2. Therefore x=5.
Very nice, very clever.
Nice one
I guessed 5 but this is a very beautiful proof that 5 is only natural solution (although for most parts of high school we don’t learn factorial for non natural numbers)
I just instantly recognized the solution as 5 since 5!=120 and 5^3 = 125. But this proof is definitely very elegant and interesting.
That works, except you then need to show that there are no other solutions.
Very nice explanation Sir. Thanks 🙏
👍
This is so cool.
I was like, does it work with only 3 ? no it would work with any prime number...
Actualy it would work with any number, just the number of cases to check would go up.
So now I have a way to solve
x! = x^n + a
for any n in N and a in Z
As an engineer student, i say u should use newton rhapson method or any numerical method so u get numerical answer, but if u insist to get the analytical solution probably change factorial into stirling approximation will help
@@abulhasbullah8120 sorry mate I do math for fun, the process is more important than the result.
I JUST LOVE THIS, YOU MADE MY DAY SIR, AND I LOVE ALL UR VIDeos
Glad you enjoyed it!
Great video
it doesnt matter that much but when you wrote (x-1)! = (×-1)(x-2)!, you needed to have checked that x-2 > 0. If for instance X was 1, you dve have written 0 * (-1)! I get that you proved this wasnt true when cancelling but I think it would have been important to remind people when rewriting x! that everything in the bracket must be above 0 as -1! is undefined
He had to check, if x>1,because for x=2, (x-2)° i*is* defined and has the value 1.
You are a fantastic teacher !
How many caps do you have ?
So nice. Thank u again ❤
10:08 Eyes are talking.
Thank you
Great job.
We could use induction to show that x³ - x < x!, where x >= 6 and then just try the cases x = 1 through 5 to see if they are solutions.
You mean the opposite right? When x=6, we already have 6³-6 210
@@niloneto1608
I changed it
Great ! ❤
excelente.
Plotting two graphs, one of x! and one of x^3-x reveal that there are two intersection points.
one intersection is at obviously x = 5 and the other one is at x=1.374.
Can someone please explain me where do we lose the other root?
factorial is a function onl defined for non negative whole numbers. You can extent factorial to a function, that is also defined for fractions, but that is the gamma function, whis is not the same as factorial, because factorial is only defined for non negative whole numbers. Because of that, there is onl 1 solution.
@@juergenilse3259
Oh okay got it, thanks man
There is only one solution with respect to the positive integers. If we want to find all real solution, we need to consider the continuous version of the factorial function, namely the Gamma Function. Under the reals, this equation has more than one solution
Correct
Cool❤
We can solve by line numbers
Some day i want to see Newtons using a blackboard eraser to wipe the board clean ASMR style
Some day
that took 12 minutes?
(-1)! exists ? What is it ?
Ha! You caught it. That was a slip. My mind drifted to inputs in the gamma function
@@PrimeNewtons I also got confused. But don't worry ,we all make mistakes , that's how we learn , and our videos are awesome. I am grateful for your videos.
(-1)! == gamma(0) does not exist, in the sense that is a simple pole of the gamma function (as are all integers
I have
3=0 🤦🏼♂️😂
Hello
happy ending 🤨🤨
Who decide to use gamma function instead integral sol
I thought the other video was quite clever.
X³+x=33 nasıl oluyor bir aydinlatsaniz beni
factoring on the right and dividing both parts by x(x-1), x=0 and x=1 are clearly to be discarded, we immediately obtain:
(x-2)! = x+1 ,with x>2
by trial and error starting from x=3 the solution x=5 is found correct.
Love your videos, but this ciould have been done much easier using guess and check at the point of (x-2)!= x+1 with maximum of 3 consecutive trials. This is because we kniw that (x-2)! has to be:
1. Z+
2. >= 2
The point of the video was to show a way without guessing and checking. He did another video on the same problem before with the method you described.
asnwer=2 isit
If you just used a brute force method of solving, it is faster, here you need to test x=0,1 and 2 and do a lot of maths.
Hello,
This is the English or American way of approaching solving the problem; nevertheless, it is not mathematics but trial and error, guessing and assuming. Guessing and presuming are not mathematical formulas.
Truly,
I have seen your posts on other platforms. Please give me a reason to take this comment seriously.
@@PrimeNewtons
Hello,
Mathematics mimics physics to achieve formulas, which most of the time are incorrect or lies; if God used Sin, Cos, Log, Ln, e, g, G, π (3.14159265358979323846264338327950288419...→
.∞, ETC., the universe would have destroyed itself before its creation. Most of the mathematics, such as the above signs, are lies to be valid upon Crooks's agreements. Mathematics is a tool (formula) to solve problems, not to keep assuming paths over and over again to find solutions in immature methods.
For instance, by using the same strategies to solve 77e⁷⁷!=1.81880373878e¹⁴⁵ rebelliously, it shall take many million years to acquire the accurate number.
As mathematics communication, nuclear physics, nuclear cosmology, mechanical, electrical, and electronics scientists reject the human coherence of knowledge because of illogicality; people think they are alive and perceive the past's future, not the present!
Genuinely,
Well... hmmm... There is no potential of generalization, as i'm trying to reverse the process. It works out just for the one value of "x == 3". Since there is no generalization potential, the argumentation with "the right way to show" has a smell of... let's call it "artefact hunting". By accident, for this unique small factor, the decomposition "works", but neither is the decomposition necessary because of the pettiness of the factor, nor would the method be generalizable for arbitrary resulting values. I must admit that driven by pure curiosity, i too took that same way of decomposition (since in the math channels i had subscribed to, this decompositions into binoms seem to be the stick horse of all authors), but nevertheless i am not impressed by the example.
I'm not sure I understand what you mean?
@@davidgagen9856 If you - for example - try to setup an instance of this problem with arbitrary parameters (from the solution back to the problem presentation), you end up with constructs that are not decomposible as in this very specific case.
I believe the explanation shown is as rigorous as it has to be. The procedure used does not have to be generalizable, it is enough to get to the equation where an integer expression is equal to a prime number. While it's true that the two cases work only when the number is prime, you can still do a similar procedure if the number is not prime, just use prime decomposition. There would just be more cases, it doesn't mean that it's wrong.
My only two nitpick with the video are: 1) the initial question should have "x is an integer". 2) it must be shown that (x-2)! Is allowed.
The rest follows pretty much flawlessly, great video.