Serouj Ghazarian 0^0 is an indeterminate form, which is why limits with this form can equal many different values and exist. This is different from the expression not being defined. If a limit ever is evaluated to equal an undefined arithmetic or algebraic expression, then the limit does not exist. There is a much more rigorous way to state this, but I’m not going to bother with on a short TH-cam comment. Needless to say, there is much evidence that suggests that 0^0 is a necessary convention in mathematics, and in most fields of study, 0^0 = 1 is assumed even if it is unprovable.
@@angelmendez-rivera351 exactly! LIMITS! you know why 0^0 is inderterminate? Because it takes you straight to 0×infinity (lim x->0+ e^(xlnx)) and I'm pretty sure that ln(0) is undefined. Although, if we think like that, 0 itself is undefined
@@angelmendez-rivera351 also, while you're at it, why don't you take sin(0)/0=1, (cos(0)-1)/0=0, log(b=1)(1)=1 (actually, the last one can be any real number)
RUSapache Read my comment above, where I derived a symmetry formula for all integers n, extending H for negative inputs, and I also demonstrated that every half-integers is a root of H(x) given this definition of H.
well, if we go from n to 1 so it looks like this: 1 П k^k k=n Then it would work, actually. H(-1) in this case is (-1)^(-1)*0^0*1^1 We got a 0^0 problem here. But if to fix it knowing that lim x->0 (x^x) =1, then it's (-1)^(-1)*1*1 = -1*1*1 = -1. Seems legit? I think it does.
Craftist Except you are not allowed to do that, and that is not the definition of the hyper-factorial. In particular, if you have a product over an ordered sequence, then the indexing is not allowed to happen over a variable boundary if there is no function over the boundary and if the upper boundary is a fixed constant which is lower. By your argument, H(n) = 0 everywhere except for H(1) = 1, which is absurd. No, we simply define H via the recursion formula, and if the value H(x) exists for some x, then it must satisfy the recursion. Then we say that H(0) = 1 by the recursion, and if we use a limit or adopt the convention 0^0 = 1 - and Let me tell you that there is far more reason to adopt than to not adopt it - then we can prove easily that H(-1) by the recursion. However, H(n) for n < 0 could not be calculated using products as you tried it because products are undefined for negative integers. Product and summation notation are a pain in the ass and they only add obstacles unnecessarily to the problem, even if they are more intuitive.
Easy. H(1)=1=H(0)×1, so H(0)=0^0= conventionally 1. Then H(0)=0^0 × H(-1)=H(-1), and from there H(-n)=H(n)×H(-1)×-1^(f), and the H(-1) goes away, being equal to 1. f must be worked out to agree with the parity of H(-n), I'm sure there's a closed form. The reason H(n) appears is because on the left side it gets multiplied by values ^ negatives, and so you can see that (-2)^(-2)=1/2^2, with the only difference being a factor of -1 for odd bases, and 1/2^2 on the left becomes 2^2 on the right
H(0) = 0^0 * H(-1) 1 = 0^0 * H(-1) 0^0 is undefined, so H(-1) is undefined. The question, then, is what about noninteger values of n? What about complex values of n?
When computing H(5), you can group the factors as follows: H(5) = 5⁵·4⁴·3³·2²·1¹ [2² = 4, so put it in with the 4's] = 5⁵·4⁵·3³ = 20⁵·27 = 3,200,000·27 = 86,400,000 Fred
@@nikodempatrycjuszswiercz4064 This is exactly what I thought he was going to do when he says he's going to rearrange. I did, and you can compact the formula to: *H(n) = (n!)^n / sf(n-1).* sf(n) is the superfactorial, I just looked it up, and it's sf(n) = n! · (n-1)! · (n-2)! · ... · 2! · 1!
New topic 😊 😊for me.but after i watched this video I did some of the problems on this topic and they seems to be too interesting .thanks,I easily catched the concept right here😊😊😊😊😊😊😊
Hyperfactorial of a negative number is equal to zero, as the exponent gets closer to minus infinity, it gets divided by a larger number, which means each index gets smaller and smaller, and tends to zero. And if you multiply a number by zero you get zero.
Sorry, I didn't explain the infinity thing. Take H(-3) for example. H(-3)=-3^-3*(-3-1)^(-3-1)*(-3-2)^(-3-2)•••=-3^-3*-4^-4*-5^-5••• the exponent doesnt get closer to zero but closer to minus infinity, as it gets subtracted one each time. A negative exponent is the same as 1/the number to that exponent. If the exponent is minus infinity it is the same as 1/infinity which tends to zero. If you multiply by zero, you get zero.
I'm curious as to what the use of this might be, as basically any value n greater than 10 is already incomprehensibly large. As in near universe-breaking
Well if you believe in your formula you don't have to write H(n) in terms of H(n-1) to get H(0). By convention, an empty product (product of zero terms) is equal to the neutral element for the multiplication law, so it is 1. In the same way, an empty sum is by convention equal to 0, which is the neutral element for addition. These conventions ensure that any reccurence formula you will get for your sum or product remains true at 0
Even here mah boi the gamma function is present. Because H(z) can be defined as follows: H(z)=K(z+1) where K is the, oh wonder, K-Function. And K(z) is defined as K(z)=(2pi)^((1-z)/2)*exp[(z choose 2) + integral from 0 to (z-1) of ln(Γ(t+1))dt]. Boi Edit: So you just need to plug in z=z+1 to get a definition for H(z) in terms of the Gamma-Function. I'm using z here because you can plug everything, even complex numbers, into the Gamma-Function.
I’m trying to think of where hyperfactorial would actually come up in a real life combinatorics problem but can’t think of one and didn’t find anything off-hand on Google. (Comparatively factorial n! is extremely useful and subfactorial !n is the number of derangements of n objects). I’m wondering if this came up in the course of a real world calculation or combinatorial proof or if it was just a function someone thought up that just seemed like it might have some interesting numerical properties.
Well for negatives actually, due to the recursive, H(n)=H(n+1)/(n+1)^(n+1), and for negatives the n+1 is negative or zero, and you can flip the denominator and get a multiplying chain. H(-5)=H(5)×H(-1) (from odd H to odd H the ± is the same), and now we just need to define H(-1), and since H(0)=H(1)/1, we can understand 0^0 to be 1. This leaves 0^0 H(-1) = H(0), and therefore H(-1) is also 1, since 0^0 is 1
Acman There are two ways I can answer your question. If you use the Barnes G-function (look it up) there is a way to relate the gamma function and the Hyperfactorial with the statement: H(z-1)*G(z) = e^((z-1)*(log (Γ(z)))). Another answer is that the Hyperfactorial is given by the integral: H(z) = (2π)^(-z/2) * e^(((z+1)C(2) + (integral from t=0 to t=z of ln(Γ(t+1)) dt))) where (x)C(y) represents the choose function.
Judging by the title I was thinking it would be something like: f(n) = n ^ f(n - 1) So if n = 4 then f(4) = 4 ^ 3 ^ 2 ^ 1 = 4 ^ 3 ^ 2 = 4 ^ 9 = 262,144 Following the same pattern, f(5) = 5 ^ 262,144, or roughly 6 * 10 ^ 183,230. And f(6) would be far too stupidly big to comprehend.
How about integrating (x^3)/cos(2x) by expressing it with De Jonquiere's Polylogarithm (the Li_s(z) function)? I wanna see you labor and sweat with tears, see you have to use the full rainbow of pens, and/or see to it that they run out of ink in the middle of a derivation, and see you to have to recruit extra blackboards! I want drama! :g:
I have a different formula for the hyperfactorial. assume n = 5 for this example Step 1: H(5) = 5^5*4^4*3^3*2^2*1^1 Step 2 (still equal to H(5), but I won't put it there because of the width of a space): 5*5*5*5*5* 4*4*4*4* 3*3*3* 2*2* 1 Step 3: first column is 5!/0!, second one is 5!/1!, third is 5!/2! and so on... Step 4: H(5) = (5!/0!)*(5!/1!)*(5!/2!)*(5!/3!)*(5!/4!) Step 5 (again, not writing H(5)=): 5 Π 5!/(k-1)! k=1 Step 6: replace 5 with n n H(n) = Π n!/(k-1)! k=1 EDIT: in step 6 I absolutely had to write H(n)=
Hey blackpenredpen. Recently I have been watching dr peyam’s show, and mainly his videos on the half derivative of x and imaginary derivative of x. I would really like to see you try things like half intergrals and imaginary intergrals. Please like if you agree.
0^0=1 by convention in hyperfactorial as well. : )
Just like in power series, see this video th-cam.com/video/rJil85GHEyc/w-d-xo.html
I thought 0^0 was undefined
Serouj Ghazarian 0^0 is an indeterminate form, which is why limits with this form can equal many different values and exist. This is different from the expression not being defined. If a limit ever is evaluated to equal an undefined arithmetic or algebraic expression, then the limit does not exist. There is a much more rigorous way to state this, but I’m not going to bother with on a short TH-cam comment. Needless to say, there is much evidence that suggests that 0^0 is a necessary convention in mathematics, and in most fields of study, 0^0 = 1 is assumed even if it is unprovable.
@@angelmendez-rivera351 exactly! LIMITS! you know why 0^0 is inderterminate? Because it takes you straight to 0×infinity (lim x->0+ e^(xlnx)) and I'm pretty sure that ln(0) is undefined. Although, if we think like that, 0 itself is undefined
@@angelmendez-rivera351 also, while you're at it, why don't you take sin(0)/0=1, (cos(0)-1)/0=0, log(b=1)(1)=1 (actually, the last one can be any real number)
I found this formula for negative hyperfactorial
H(-n) = (-1)^(n+1)*(n-1)^(n-1)*H(-n+1)
Cool jacket! Did you start a motorcycle gang for mathematicians? You could call yourselves The Unbounded Functions.
lol, that will be cool!
Speed Without Limit!
Painted with black pen
5!= number of seconds in 2 minutes.
H(5)=number of milliseconds in a day.
Wow that escalated quickly.
I think 10! is the number of seconds in six weeks
@@Gold161803 yup! Numberphile did a video on it.
And what is H(10)?
@@stewartzayat7526 Number of Planck times in 11.63 seconds.
Once upon a time there was a factorial, he drinks a lot of coffee and eats a lot of sugar, he is... HYPER. Haha
Pervy Sage lolll
XD
I think with enough brute force it is possible to do H(-1)
RUSapache Read my comment above, where I derived a symmetry formula for all integers n, extending H for negative inputs, and I also demonstrated that every half-integers is a root of H(x) given this definition of H.
well, if we go from n to 1 so it looks like this:
1
П k^k
k=n
Then it would work, actually.
H(-1) in this case is (-1)^(-1)*0^0*1^1
We got a 0^0 problem here. But if to fix it knowing that lim x->0 (x^x) =1, then it's (-1)^(-1)*1*1 = -1*1*1 = -1. Seems legit? I think it does.
Craftist Except you are not allowed to do that, and that is not the definition of the hyper-factorial. In particular, if you have a product over an ordered sequence, then the indexing is not allowed to happen over a variable boundary if there is no function over the boundary and if the upper boundary is a fixed constant which is lower. By your argument, H(n) = 0 everywhere except for H(1) = 1, which is absurd. No, we simply define H via the recursion formula, and if the value H(x) exists for some x, then it must satisfy the recursion. Then we say that H(0) = 1 by the recursion, and if we use a limit or adopt the convention 0^0 = 1 - and Let me tell you that there is far more reason to adopt than to not adopt it - then we can prove easily that H(-1) by the recursion. However, H(n) for n < 0 could not be calculated using products as you tried it because products are undefined for negative integers. Product and summation notation are a pain in the ass and they only add obstacles unnecessarily to the problem, even if they are more intuitive.
Easy. H(1)=1=H(0)×1, so H(0)=0^0= conventionally 1. Then H(0)=0^0 × H(-1)=H(-1), and from there H(-n)=H(n)×H(-1)×-1^(f), and the H(-1) goes away, being equal to 1. f must be worked out to agree with the parity of H(-n), I'm sure there's a closed form. The reason H(n) appears is because on the left side it gets multiplied by values ^ negatives, and so you can see that (-2)^(-2)=1/2^2, with the only difference being a factor of -1 for odd bases, and 1/2^2 on the left becomes 2^2 on the right
Is there any real continuation of the hyperfactorial? Similar to how we have the Gamma function for the regular factorial.
If we reverse the input to n and n+1, etc, and input negative numbers, it equals (H(n))^-1.
Plz integrate
ln(ln(ln(lnx)))
Gourav Madhwal Cannot be done in terms of elementary functions, but...
extreme integration math 1052 xD
@@DiegoMathemagician just put the horseshoe dude
Isn't that horseshoe math?
@@ZelForShort yes it is
so the hyperfactorial of 0 represents the number of milliseconds in one millisecond.
also, better notation would be "n?"
That's so much better!
Isn't it?
AndDiracisHisProphet it is!
Hmmm, what should I draw next? I did fish, shark, what's next??
@@blackpenredpen a dolphin
Octopus?
Ever since I started watching you, you've reignited my passion for mathematics. Thank you :)
xusui _ : )
Hyperfactorial is cool...BUT...it is not as cool as your jacket, isn't it?
Arthur Clay lol thank you!!
0^0=1 :thinking:
Really digging the new fish/shark themed thumbnails.
these thumbnails mannnn
: ))))))
Superb right ?
@@yrcmurthy8323 thanks!
*@blackpenredpen No need to say thanks sir,. You deserve an 👏 applause*
Nice video. What about "tetrafactorial" : T(n) = n^(n-1)^(n-2)^...^1 ?
^...^
That would be 2^10^44 if T(5) was the input
Which is around 10^10^43.6
hyperfactoreos
Vincent William Rodriguez sounds tasty
That's a slick jacket! Love it!
Lightning Fast thank you!!!
I just started the video. I can't wait to learn about hyperfactoreos!
H(0) = 0^0 * H(-1)
1 = 0^0 * H(-1)
0^0 is undefined, so H(-1) is undefined.
The question, then, is what about noninteger values of n? What about complex values of n?
hyperfactoreos :D
When computing H(5), you can group the factors as follows:
H(5) = 5⁵·4⁴·3³·2²·1¹ [2² = 4, so put it in with the 4's]
= 5⁵·4⁵·3³ = 20⁵·27 = 3,200,000·27 = 86,400,000
Fred
There's another formula for it I've found (using math not google):
H(n) = n!/0! * n!/1! * n!/2! * ... * n!/(n-2)! * n!/(n-1)!
Cool!
This one actually uses regular factorials showing the „hyperness” of the hyperfactorial
@@nikodempatrycjuszswiercz4064 This is exactly what I thought he was going to do when he says he's going to rearrange. I did, and you can compact the formula to: *H(n) = (n!)^n / sf(n-1).*
sf(n) is the superfactorial, I just looked it up, and it's sf(n) = n! · (n-1)! · (n-2)! · ... · 2! · 1!
You should put the exclamation mark in the exponent so you’d have n^! Now THAT is what I call hyper.
Is there an integral-based formula for a hyperfactorial like there is for a normal factorial (i.e. the pi function for a normal factorial)?
Is there an Ultrafactorial? like using tetration: U(5)=high5^low5 * 4^4 * 3^3 * ...
tetration**
@@caloz.3656 thx, didn't notice it
@@nietschecrossout550 you're welcome!
Aden Tate he said tetation.
Next is hypersupremeover9000megatron factorial 😂. Love your videos btw!!!
Really enjoy these videos man!
"He eats a lot of coffee and drinks a lot of sugar." Had me laughing for a while.
Why don't you try the integral of the tetration of sin(x)
Shivam Malluri ok
@@blackpenredpen great👍
Think a bit about the Q function; if it does not blow your mind then think harder ;-)
I have more to say about it later
New topic 😊 😊for me.but after i watched this video I did some of the problems on this topic and they seems to be too interesting .thanks,I easily catched the concept right here😊😊😊😊😊😊😊
Hyperfactorial of a negative number is equal to zero, as the exponent gets closer to minus infinity, it gets divided by a larger number, which means each index gets smaller and smaller, and tends to zero. And if you multiply a number by zero you get zero.
Sorry, I didn't explain the infinity thing. Take H(-3) for example. H(-3)=-3^-3*(-3-1)^(-3-1)*(-3-2)^(-3-2)•••=-3^-3*-4^-4*-5^-5••• the exponent doesnt get closer to zero but closer to minus infinity, as it gets subtracted one each time. A negative exponent is the same as 1/the number to that exponent.
If the exponent is minus infinity it is the same as 1/infinity which tends to zero. If you multiply by zero, you get zero.
Introducing the K-function/Barne’s G-function
I was literally thinking about it yesterday, and you make a video today, wow
Where did it came across
@@karthikrambhatla7465 Just popped up when doing maths
Wow! What class?
@@blackpenredpen 12th but it was not in the book😂😀
1:54 what happened?
Crazy.. heard this for the first time. Thank you❤️. But BlackpenRedPen where this hyperfactorial is used
"Eats a lot of coffee and DRINKS a lot of sugar" That's how hyper it is
Hyper factorial has a lot in common with hyper dashing, there is tremendous horizontal ground covered.
Loving that jacket!
Dan Dart thank you!!!
lim n -> -1 H(n)=1 where we approach from right. Is it right...?
Dwaraganathan Rengasamy Yes
The natural extension would be for 0^0 = 1, same as 0! = 1. That way you just multiply all the results by one. Therefore, H(0) = 1.
Can you tell what is method of Multipliers in linear PDE?
Check out Foias Ewing constant.
YES DO INFINITE NEGATIVE CONVERGING SUMS - NEGATIVE HYPERFACTORIALS!
Is there such a thing as a hyper-gamma function? By that I mean an analytic extension of the hyperfactorial to some well behaved subset of R or C?
Learn how a factorial can be such a hypebeast
I'm guessing the negatives would work but it depends if they are even or odd.
Make more videos about factorials!!
Can you make a video with hyperfactorial arrow notation please
Anyway great video! Now do superfactorials!
Pl. Integrate
If a>1 then
dx /( x^(2) + 2ax + 1 ) is equal to
Isn’t this just every factorial from n! to 1! multiplied together? Pretty cool!
Analytic continuation of hyperfactorial?
Woah, I'm drooling
I'm curious as to what the use of this might be, as basically any value n greater than 10 is already incomprehensibly large. As in near universe-breaking
Well if you believe in your formula you don't have to write H(n) in terms of H(n-1) to get H(0). By convention, an empty product (product of zero terms) is equal to the neutral element for the multiplication law, so it is 1. In the same way, an empty sum is by convention equal to 0, which is the neutral element for addition.
These conventions ensure that any reccurence formula you will get for your sum or product remains true at 0
Wait, isn't H(n)=[(n!)^n]/[(n-k)!] ? (k
Even here mah boi the gamma function is present. Because H(z) can be defined as follows:
H(z)=K(z+1) where K is the, oh wonder, K-Function. And K(z) is defined as K(z)=(2pi)^((1-z)/2)*exp[(z choose 2) + integral from 0 to (z-1) of ln(Γ(t+1))dt].
Boi
Edit: So you just need to plug in z=z+1 to get a definition for H(z) in terms of the Gamma-Function. I'm using z here because you can plug everything, even complex numbers, into
the Gamma-Function.
It is said that bprp is still hearting people and will continue to do so forever...
Harsh Srivastava
Isn't it?
If factorial is for arrangement then what is superfactorial and hyperfactorial for??
Wondering the same🤔
For *fun*
you can rewrite it as the product k=1 till n of n!/k!. so you get the product of permutations*. maybe hyperarrangement isn´t so stupid.
Lok 739 you beat me to it
(k-1)!**
What about H(1/2)?
I’m trying to think of where hyperfactorial would actually come up in a real life combinatorics problem but can’t think of one and didn’t find anything off-hand on Google. (Comparatively factorial n! is extremely useful and subfactorial !n is the number of derangements of n objects). I’m wondering if this came up in the course of a real world calculation or combinatorial proof or if it was just a function someone thought up that just seemed like it might have some interesting numerical properties.
So... What exactly would you use this for? Apart from H(5) of course
Hi. Can you write H (n) such as integral
Is there any identety that conects the regular factorial to the hyper one? #yay!
BenHaim Yes. Product from m = 0 to m = n of n!/(n - m)!
Well for negatives actually, due to the recursive, H(n)=H(n+1)/(n+1)^(n+1), and for negatives the n+1 is negative or zero, and you can flip the denominator and get a multiplying chain. H(-5)=H(5)×H(-1) (from odd H to odd H the ± is the same), and now we just need to define H(-1), and since H(0)=H(1)/1, we can understand 0^0 to be 1. This leaves 0^0 H(-1) = H(0), and therefore H(-1) is also 1, since 0^0 is 1
The whole "from odd H" is wrong, there's a rule but it doesn't resolve itself to me at the moment
Is there a definition of hyperfactorial which is an improper integral, like with factorial (gamma function) and derangements/subfactorial?
Acman There are two ways I can answer your question. If you use the Barnes G-function (look it up) there is a way to relate the gamma function and the Hyperfactorial with the statement: H(z-1)*G(z) = e^((z-1)*(log (Γ(z)))). Another answer is that the Hyperfactorial is given by the integral: H(z) = (2π)^(-z/2) * e^(((z+1)C(2) + (integral from t=0 to t=z of ln(Γ(t+1)) dt))) where (x)C(y) represents the choose function.
I know that the second one includes an integral inside an integral but please bear with that.
my favorite note is H flat
More like hypersharktorial.
The thumbnail says it
Judging by the title I was thinking it would be something like:
f(n) = n ^ f(n - 1)
So if n = 4 then
f(4) = 4 ^ 3 ^ 2 ^ 1 = 4 ^ 3 ^ 2 = 4 ^ 9 = 262,144
Following the same pattern, f(5) = 5 ^ 262,144, or roughly 6 * 10 ^ 183,230. And f(6) would be far too stupidly big to comprehend.
Are you able to make an app over it ??
How about integrating (x^3)/cos(2x) by expressing it with De Jonquiere's Polylogarithm (the Li_s(z) function)? I wanna see you labor and sweat with tears, see you have to use the full rainbow of pens, and/or see to it that they run out of ink in the middle of a derivation, and see you to have to recruit extra blackboards! I want drama! :g:
Should try ¡n! Instead of H(n)
Rafael Dubois
Good idea
Isn't the conclusion in the end mean that 0^0 equals 1?
Where r u frm
Im lovin the smilie's 😂😂😂
I have a different formula for the hyperfactorial. assume n = 5 for this example
Step 1:
H(5) = 5^5*4^4*3^3*2^2*1^1
Step 2 (still equal to H(5), but I won't put it there because of the width of a space):
5*5*5*5*5*
4*4*4*4*
3*3*3*
2*2*
1
Step 3:
first column is 5!/0!, second one is 5!/1!, third is 5!/2! and so on...
Step 4:
H(5) = (5!/0!)*(5!/1!)*(5!/2!)*(5!/3!)*(5!/4!)
Step 5 (again, not writing H(5)=):
5
Π 5!/(k-1)!
k=1
Step 6:
replace 5 with n
n
H(n) = Π n!/(k-1)!
k=1
EDIT: in step 6 I absolutely had to write H(n)=
H(5) = 86,400,000. Also, how about using ¡ as a factorial symbol?
Wait, so ñ' isn't a proper notation?
You can't have H(-1). Following that argument, H(0)=0^0*H(-1), and since 0^0 is undetermined, H(-1) is undetermined too.
How about H(1/2)? 😁😁
Hey blackpenredpen. Recently I have been watching dr peyam’s show, and mainly his videos on the half derivative of x and imaginary derivative of x. I would really like to see you try things like half intergrals and imaginary intergrals. Please like if you agree.
We need a hypergamma function!
4:27
Now I wonder if there is a factorial for each version of street fighter.
hyperfactorial
superfactorial
turbofactorial
alphafactorial
arcadeeditionfactorial
...
Write it as !!n instead of H(n)
Explostion in my mind : 5:26
Not the next hyper-operation up? Lets try a few ranks.
F(0, 4) = 5
F(1, 4) = 10
F(2, 4) = 24
F(3, 3) = 9
or maybe
= ^{3, 2, 1} = elements of {9, 8, 1}
Sir i think we can do its write .....
isnt H(n) equal to n! x (n-1)! x (n-2)! x ... x 2! x 1! ?
We have factorial, hyperfactorial, and…
Hold up…
Superfactorial??? 🤔😮
Me: Calling Superman for explaining this type of factorial. 😁
There's also subfactorial.
does this mean that there is a hyper gamma function for with this? (:V)
So we have factorial of 1/2, subfactorials of 1/2 but what about the hyperfactorial and superfactorial of 1/2 ?!
So 0 to the 0th power is 1... Intresting as bronzes
HYPER GAMMA FUNCTION
hyperfactorial is about hype
H(-1) approaches infinity
I think that we can do hyperfactorial of a negative number.
How? Idk, I just think that it is posibble
Is there a Hypersubfactorial, or Subhyperfactorial?
Took any question from iit subjective pllzz
please solve ln (e + ln ( e + ln e + ... )) = ?
in one minute!!!
Titan Unspeakable
Do you have the answer in exact form?
@@blackpenredpen i only it is ln (e+x) = x will be one of the step
e^x = e+x
Yes, I noticed that.
But do you have the answer in exact form?
There are two solutions to the equation but I only found one.
x = -e - W(-(e^(-e)))