n=?

[Anonymous No.15811103]

a friend of mine gave me this equation. its a while back that i did math at the university. with my logic it can only be 2 or 0, i have no more information about n like n element N or R. is there any way to get a answer for n other than 0 or 2 even tho n^n grows way faster than n! +n

[Anonymous No.15811109]

It can't be 0 because 0^0 is undefined.

https://www.wolframalpha.com/input?i=x%5Ex%3Dx%21%2Bx

[Anonymous No.15811117]

>>15811103
>n^n grows way faster than n! +n
That should give you the answer that it's a no. Here's n^n / (n!+n) from 0 to 9. It's not coming back to 1

[Anonymous No.15811120]

>>15811117
okay ty

[Anonymous No.15811124]

>>15811103
The factorial is only defined for natural numbers, so n must be a natural. For any n>2, n^n > n! + n. You can check that it's true for n=0 and n=2 and false for n=1; therefore, n=0 and n=2 are the only two solutions.

>>15811109
No.

[Anonymous No.15811131]

>>15811109
You should use wolfreealpha
>>15811124
Wrong. gamma function of a number x is the factorial of x-1, therefore factorial of every number, including complex exists. There are poles of some numbers tho where the integral isn't defined.

[Anonymous No.15811134]

>>15811131
The gamma function is defined for any complex number, and for positive integer-valid inputs x equals the factorial of x - 1, but that doesn't mean the factorial function itself is defined over all complex numbers. It is not. It could be extended in such a way, yes; but it isn't.

[Anonymous No.15811146]

>>15811131
The gamma function works as an analytic continuation of the factorial, but that's not the same thing as being the factorial.
We can add any multiple of [math]\sin(\pi n)[/math] to [math]\Gamma(n)[/math] and end up with just as valid a continuation (it's analytic, and it shares a value with both gamma and the factorial everywhere where the factorial is defined!), but you'd be a lunatic to imply that these are somehow the same function

[Anonymous No.15811156]

>>15811134
>>15811146
It is retard. For eg if you want π!, just find gamma (π+1) ~ 7.1. stfu with your schizo nonsense

[Anonymous No.15811166]

>>15811156
What if I prefer gamma(π+1)+sin(π^2) ~ 6.8?

[Anonymous No.15811185]

>>15811166
why did you insert sin my nigga? Gamma x + sin x =\= (x-1)! Prove that they're equal and I shall concede

[Anonymous No.15811193]

>>15811131
It doesn't matter. For negatives x^x will return side numbers (called imaginary by faggots like descartes), and n! + n will return regular numbers (called real by the same faggots) and undefined on integer negative points.
https://www.desmos.com/calculator/azcn8s2hsd

[Anonymous No.15811199]

Use induction to show the other values don't exist
This is a good idea because based on intuition it's pretty obvious

[Anonymous No.15811211]

>>15811185
That's exactly my point you illiterate mong
sin(πn)=0 at all integer n, which is exactly where n! is defined
so they're equal precisely at all points upon which n! is defined and thus it's just as good an analytic continuation even though it's very clearly NOT equal.
You cannot give any justification as to why gamma(n) is a better extension of n! than gamma(n)+sin(πn).

[Anonymous No.15811215]

>>15811211
If he was too retarded to get your point before, what makes you think he possesses enough intelligence to admit that you're right? You know the type of 2023 /pol/ poster who embarrasses himself in nearly every comment and will never admit he's wrong.

[Anonymous No.15811217]

>>15811193
Um they are called orthogonal numbers or as I like to call them orthonumerals.

[Anonymous No.15811229]

>>15811185
[math]\Gamma(x) + x \mathrm{mod} 1 = (x-1)! [/math]

[Anonymous No.15811449]

[math] ( n)^n = n( n - 1 + 1)(n-2+2)...(n-(n-1) + (n-1)) = n[(n-1)! + (n-1)! + k] = n! + n! + kn > n! + n [/math] k is positive.

[Anonymous No.15811816]

>>15811193
>>15811217
Lateral numbers?

[Anonymous No.15811850]

>>15811109
>because 0^0 is undefined
https://en.wikipedia.org/wiki/Empty_product

[Anonymous No.15811934]

Indeed, you are right

[Anonymous No.15812111]

>>15811131
the gamma function is not the factorial.

[Anonymous No.15812232]

>>15812111
>he doesn't know
who's going to tell him?

[Anonymous No.15812303]

What do I need to learn to understand this? Functional analysis?

[Anonymous No.15812306]

>>15812303
Complex analysis

[Anonymous No.15812323]

>>15812306
дякyю

[Anonymous No.15812434]

>>15811103
You are too sophisticated, the answer is 2 because:
2^2=2*1+2

[Anonymous No.15812732]

>>15811103
Here you go anon, 0, 2, -2.488, -2.734, -4.039, -4.992, -6.001, etc. They intercept each other close to every negative integer < -3.

[Anonymous No.15812737]

>>15811193
>https://www.desmos.com/calculator/azcn8s2hsd
-5^-5 is imaginary? Huh, must have missed that part of algebra 2

[Anonymous No.15812773]

>>15812732
It appears I have come down with a case of the retardation. Here anon, look at this:
https://samuelj.li/complex-function-plotter/#z%5Ez-gamma(z%2B1)-z
The solutions for your problem are at the center of positive gradients (-.589+-.582i, 0, 2, +2.749+-3.240i,...). If you zoom in on each negative integer you will see a pole and and accompanying nearby zero (which I believe is actually two zero pairs, but the precision isn't high enough)
https://samuelj.li/complex-function-plotter/#z%5Ez-gamma(z%2B1)%2B10*sin(pi*(z%2B1))-z
If you look at other continuations of the factorial function >>15811146
You get a similar pattern of solutions: negative integer poles and nearby zero pairs, 0, 2, and positive integer asymptotes where periodic zeros lie, just with some extra zeros interspersed, determined by the coefficient of sin(pi*(z+1))

[Anonymous No.15812775]

>>15811103
n=2
Trial and error testing 2 steps

[Anonymous No.15812814]

>>15812773
https://samuelj.li/complex-function-plotter/#(z-1)%5E(z-1)-diff(ln(gamma((1%2F2)-(z%2F2))%2Fgamma(1-(z%2F2)))%2C%20z)%2Fgamma(1-z)-z%2B1
Here's one using a different continuation of the factorial function, Hadamard's gamma function
https://www.luschny.de/math/factorial/hadamard/HadamardsGammaFunctionMJ.html

[Anonymous No.15812819]

>>15812814
One more Luschny's factorial:
https://samuelj.li/complex-function-plotter/#(z-1)%5E(z-1)-(z!*(1-(((z*(diff(ln(gamma(z%2F2%2B1%2F2))%2C%20z)-diff(ln(gamma(z%2F2))%2C%20z)))%2F2-1%2F2)*(sin(pi*z)%2Fpi*z))))-z%2B1
I probably fucked this one up because the definition for L(x) is disgusting

[Anonymous No.15812838]

x^x has a greater slope than x! for x>1, therefore it can have at most 1 intersection for x>1

[Anonymous No.15812843]

>>15811109
>It can't be 0 because 0^0 is undefined.
It depends, but in Algebra it's considered a 1.

[Anonymous No.15812860]

>>15812838
>x^x = x! + 1
whoops. I think it still comes out

[Anonymous No.15812937]

>>15811109
Euler defined 0^0 as 1

[Anonymous No.15812976]

>>15812814
This one is wrong and I have no idea why, the graph is similar to other complex plots of the Hadamard Gamma Function
>>15812819
This one needs to be:
https://samuelj.li/complex-function-plotter/#(z)%5E(z)-(z!*(1-(((z*(diff(ln(gamma(z%2F2%2B1%2F2))%2C%20z)-diff(ln(gamma(z%2F2))%2C%20z)))%2F2-1%2F2)*(sin(pi*z)%2Fpi*z))))-z

[Anonymous No.15813057]

>>15812838
I'm retarded can you explain the fourth line

[Anonymous No.15813102]

>>15813057
factorial is defined as x! = x(x-1)!, apply that recursively to get x(x-1)...(2)(1). Then (x+1)! = (x+1)x! and i can stop it there for desired algebra manipulation. Is that the line you meant?