|i Factorial| You Won't Believe The Outcome
Vložit
- čas přidán 4. 06. 2023
- 🎓Become a Math Master With My Intro To Proofs Course!
www.udemy.com/course/prove-it...
🛜 Connect with me on my Website
www.brithemathguy.com
🙏Support me by becoming a channel member!
/ @brithemathguy
#ComplexNumbers #GammaFunction #brithemathguy
This video was partially created using Manim. To learn more about animating with Manim, check out:manim.community
Disclaimer: This video is for entertainment purposes only and should not be considered academic. Though all information is provided in good faith, no warranty of any kind, expressed or implied, is made with regards to the accuracy, validity, reliability, consistency, adequacy, or completeness of this information. Viewers should always verify the information provided in this video by consulting other reliable sources.
🎓Become a Math Master With My Intro To Proofs Course!
www.udemy.com/course/prove-it-like-a-mathematician/?referralCode=D4A14680C629BCC9D84C
Is it true that you changed the title and thumbnail to show absolute value of i factorial rather that i factorial itself. Anyway, thanks for a brilliant video and being honest in the thumbnail!
This is pretty cool: (71^2) - 7! = 1.
71^2 = 5041
7! = 5040
Already on Brilliant.
@@iquitokay yeah, cool
Did you know i know what i+1 equals?
For those who wants the value of i! it's roughly 0.498 - 0.155i
Yeah, the missing bit of this video. Brilliant explanation, but it misses the step where it actually answers the question from the beginning.
That is the answer if you ask for the numerical value of i! in MATHEMATICA .
@@Misteribel I mean the thumbnail has the absolute value
but didn't we only calculate the magnitude of i! ? The magnitude is not the real value and shouldn't it be a real number? Or can there be a imagine number with a imagine magnitude? :D
The value should be a eal number not imaginary as in root( pi / sinh(pi)) = 0.521 564
That was absolute value. But what is the argument of `i!` ?
It’s around -0.30164
And what imaginary number correlates to the modulus of sqrt(π/sinh(π)) and the argument of -.30164…?
@@ValkyRiver thanks, I wonder if there is some expression like for absolute value.
@@deadheat1635 0.49802 - 0.15495i
Well we know the absolute value, meaning it has to lie somewhere on a circle.
Proof:
|a+bi| = n
√(a^2+b^2) = n
a^2 + b^2 = n^2
So this gives us the knowledge that i! must be somewhere on the circle we find.
i really like this style of teaching - showing us the tools we need and then applying them in the problem. This way it doesn't feel too overwhelming, but if we wanted to, we could still go study the proofs of those tools
Yes i agree,he is doing great job .
Fr, imagine learning how to cook ramen, then some dude starts babbling about how to grind wheat
Amazing video, the relationship between the Gamma function and pi is just incredible.
I really don't know how this guy doesn't have at least 1m sub for his good explanation. HE MADE LOVE MATH, THX!
I never thought of this question, thanks for you for giving me ideas to share to my viewers. Also, it is a real number with 'pi'es
The fact you don't write it's about 0.5215640468649398 at the end is scandalous.
And he misses the oportunity to say "finally, i! is 0.5215... /or just about a half/".
Has it a financial application/use ?
Actually, he gave the exact answer and any mathematician is happy with it rather than the approximation which is NOT the answer (except for a physicist or engineer…)
@@danielangulo2119 but that's the phase on the imaginary axis. There's also the bi-complex solution using Barnes G.
Beautiful! That was awesome!. Excellent presentation !.
make a follow up video in which you calculate arg(i!), having only the absolute value looks kinda incomplete
I feel like the arg might not have a closed expression. It doesn't have to have one, which is why the video went after finding the absolute value by itself. Still interesting tho
An easy way to represent or imagine what i is, is to consider it to be equivalent to either a 90 degree rotation or a rotation by PI/4 radians. Consider the following:
For some number x we can rotate it by i^n where i = sqrt(-1) and n is an integer value > 0. For each iteration of n, x will be rotated by 90 degrees in a counterclockwise rotation. Therefore we can see that the following expression holds true: f(x) = x.rotateBy(i^4) == x. We rotated the value of x by 90 degrees or PI/4 radians for each increment of n. Thus 90 + 90 + 90 + 90 = 360 degrees or PI/4 + PI/4 + PI/4 + PI/4 = PI.
The value of i isn't as imaginary or unreal as one would tend to think by its original definition. What's happening here is that i and -i are respectively orthogonal or perpendicular to their unit counterparts of 1 and -1. To further illustrate this we can consider the sine and cosine functions and compare their waveforms to see their similarities and their differences.
They both have the same shape, they both have the same domain and range, they both have the same periodicity of 2PI. These are their similarities. Where they differ is their corresponding inputs and outputs as they are 90 degrees, PI/4 radians, or i horizontal translations of each other. They are out of phase by 90 degrees from each other. Where does this phenomenon come from?
Let's look at their triangular definitions based on the properties of right triangles. We know that a right triangle as sides A, B, and C where A & B are the lengths or magnitudes of their two sides and C is the Hypotenuse or the side that has the longest magnitude or length which is opposite of the right angle between the other two side lengths. From this we are able to define the sine and cosine functions in this term based on the ratio or proportions of a given angle that is not the right angle with respect to one of those sides and the hypotenuse. Sine = opp/hyp and Cosine = adj/hyp. The common factor of the sine and cosine is the hypotenuse, their differences rely on the orthogonality of the two side lengths of A and B. We also know from Pythagorean's Theorem that A^2 + B^2 = C^2 has a direct relationship to that of the Trigonometric Functions. This is why we have a Pythagorean Identity amongst the trig functions.
When we extend our range and domain from the Reals or basic Euclidean Geometry into the Complex Plane or to Polar Coordinates we can easily see some wonderful properties emerge.
e^i*pi = -1, or e^i*pi +1 = 0
e^i*pi = i^2
i = +/- sqrt(e^i*pi)
e^i*x = cos x + i * sin x
This is all possible simply because 1+1 = 2. How and why? The simple expression of 1+1=2 is the unit circle with its center (h,k) located at the point (1,0) in the Cartesian plane. This is also why there is a direct relationship between the properties of vectors and the cosine function which we call the Dot Product. The orthogonality or pendicularness of numbers can be seen within the Cross Product between various vectors having equivalent unit basis components. This is also why other mathematical operations or functions are very efficient or optimized such as Quaternions, Octonions, Fast Fourier Transforms, and more.
Sure, I didn't get into the properties or concepts of Factorials, but that's what this video is for! I just wanted to show another way at looking at the value of i and what it is, what it represents. Yes we know it is defined as the sqrt(-1) by trying to solve for the roots of various polynomials, but this can be a non intuitive way of trying to understand it. If we look at i as being a 90 degree or PI/4 rotational transformation of some initial value where the result of applying this transformation has the effect of causing the output of that transformation on the original value to become orthogonal or perpendicular to its initial state is a better way of seeing the relationship that the complex numbers have in comparison to the real numbers.
A simple example considering we are working in the complex plane. If we take the value 1 and map it into the complex plane it will be the vector going from the origin (0,0) to the location (1,0). When we take the value 1 and apply the rotation of 90 degrees or PI/4 radians to it, then this point will translate to the point (0,i) in the complex plane. This is equivalent to saying that 1*i = i. When we apply a second translation of this point at (0,i) by another i we end up at the point (-1,0). This then shows that i^2 = 180 degrees or PI/2 radians or -1.
This is one of the many reasons why I love math! I just hope that this might bring some insight to others as another way at looking at something. Now, once one is able to understand the connections that I've made above, and understand what factorials are, then some of the things that were mentioned within this video might make more sense as to what is going on within various functions such as the Gamma function. Instead of trying to think of i as a linear value try to think of it as a curved value... Happy problem solving!
Such a beautiful result
Beautiful! That was awesome!
please go into crazy depth sometime, i'd love that!
I understand Euler's identity but this one's alluded me. I'll come back to it thanks.
Excellent presentation !
Well that just totally blew me away. Have really looked at this stuff from my college days and now I remember why. 😃
A bit above my level but got the gist of some of it. Thanks.
Very nice!
3:14 I guess the scale of the horizontal axis got stretched a bit 😅
|1-√3i| is pointing more to 0.5-√3i
3.14 :)
@@SPVLaboratories 👍
one of many flaws in this video....
animations are really cool!
Awesome explanation , the gamma function is great function
Hey i love your content as your content has Blackdrop background which use much less data so i can enjoy your content.. btw i am a maths lover thanks for this ❤
that doesn't make any sense, black backdrop only decrease the overall brightness of the video which would only help to save battery
Well, the fact it's pure black makes it very compressable, as any pure color background
whats subfactorial of i? and also can you do subfactorials of negative numbers or fractions?
This class is nothing short of breathtaking! Your ability to make complex concepts easily understandable is truly remarkable.
So, but what is the argument of i! ???
It’s around -0.30164
@@ValkyRiver aren't there analytic solutions?
@@tunafllsh That i dont think so. I dont think we can find the argument exactly, so we just have to settle for an approximate answer.
@@tunafllshthere is but it's dependebt on i!
And it is tan^-1(im(i!)÷re(i!))
@@monkey6114 the factorial function is not analytical
Nice
I always thought it the solution would just be passed as no solution, but to see that the answer is real and complex is quite intriguing 😮
But how can an answer be both real and complex? 🤔
@@CAustin582
I mean that the answer has a lot of somewhat difficult simplifying steps in order to get the real number. Also depending on what math solver you calculate it in, you might get a complex number.
@@test_dithered9860 The answer given is an absolute value. The actual solution is complex-valued. It's abit frustrating because the original question evaluating _i!_ is never answered but we do have *| **_i!_** |*
How could you not expect a real answer from a modulus lol
(IMO) your vids are just top quality the video is the most entertaining to watch for a idea, Remember: Its just My own Opinion on the suggestion, Advice; "Try getting used to making a opinion on a topic youre interested in works for *me works for anybody".
And I was gone for double integrals lol 💀💀. When I saw the title I tried my self and found
| i! | ^2 = integral {from 0 to infinity} of {cos(ln(s))/(1+s)^2 ds}.
The answer of your video killed me 😂
This could relate to Bose-Einstein condensates and to the midpoint in time between the solution to the Basel problem and its reciprocal.
CZcams is getting scary I thought about i factorial in my head this morning and then I get recommended this video
Nice video!
I just don't understand how you can explain what is the modulus of i after the three first high-level minutes of the video. You should stand in a certain level of complexity ^^
3:29 where did you get absolutes here?
The factorial function is simpler than the Gamma function. It's the same integral but with z instead of z - 1.
Awesome video. Though you do often confuse i! with |i!|. Saying that i! is something when really you mean |i!|
Quickly into the weeds. More power to you
you said click the video hard but the video isn't showing can you please add it too the description
|Gamma(n+1+ib)|²= (πb) Π k=1 to n (k²+b²)/sinh(πb) put b=1 and n=1 or you can find other values too.
I am too new to complex numbers to know how we got to hyperbolic space, as in your mention of hyperbolic sin. Help.
These are cleanest animations I've seen on any math channel! Thanks for the quality content
You’ve never watched 3Blue1Brown, I suppose
It's interesting to observe that (x i)! has a bell shape... The Gaussian distribution is an omnipresent invariant of mathematics!
but it isn't... it is of the form
sqrt(pi*x / sinh(pi*x))
so it looks similar to a gaussian, but it's not the same
So it's about 12/23. That's almost Christmas! :D
btw it looks fun to write
|i!|
I've been wondering about complex primes lately. Something like 5i can be dived by 5 and 1 and i and leave no remainder which suggests that a redefinition could be used for primes on the imaginary axis and the real axis. Can we define a complex prime that doesn't exist on either axis but still retains the basic property of a prime? Probably somebody has thought of this before.
Take a look at Gaussian Primes
Brilliant.
Reposting and slight editing of recent mathematical ideas into one post:
Split-complex numbers relate to the diagonality (like how it's expressed on Anakin's lightsaber) of ring/cylindrical singularities and to why the 6 corner/cusp singularities in dark matter must alternate.
The so-called triplex numbers deal with how energy is transferred between particles and bodies and how an increase in energy also increases the apparent mass.
Dual numbers relate to Euler's Identity, where the thin mass is cancelling most of the attractive and repulsive forces. The imaginary number is mass in stable particles of any conformation. In Big Bounce physics, dual numbers relate to how the attractive and repulsive forces work together to turn the matter that we normally think of into dark matter.
The natural logarithm of the imaginary number is pi divided by 2 radians times i. This means that, at whatever point of stable matter other than at a singularity, the attractive or repulsive force being emitted is perpendicular to the "plane" of mass.
In Big Bounce physics, this corresponds to how particles "crystalize" into stacks where a central particle is greatly pressured to break/degenerate by another particle that is in front, another behind, another to the left, another to the right, another on top, and another below. Dark matter is formed quickly afterwards.
Mediants are important to understanding the Big Crunch side of a Big Bounce event. Matter has locked up, with particles surrounding and pressuring each other. The matter gets broken up into fractions of what it was and then gets added together to form the dark matter known from our Inflationary Epoch. Sectrices are inversely related, as they deal with all stable conformations of matter being broken up, not added like the implosive "shrapnel" of mediants.
Ford circles relate to mediants. Tangential circles, tethered to a line.
Sectrices: the families of curves deal with black holes. (The Fibonacci spiral deals with how dark matter is degenerated/broken up and with supernovae. The Golden spiral deals with how the normal matter, that we usually think of, degenerates, forming black holes.) The Archimedean spiral deals with dark matter spinning too fast and breaking into primordial black holes, smaller dark matter, and regular matter. The Dinostratus quadratrix deals with the laminar flow of dark matter being broken up by lingering black holes.
Delanges sectrices (family of curves): dark matter has its "bubbles" force a rapid flaking off - the main driving force of the Big Bang.
Ceva sectrices (family of curves): spun up dark matter breaks into primordial black holes and smaller, galactic-sized dark matter and other, typically thought of matter.
Maclaurin sectrices (family of curves): older, lingering black holes, late to the party, impact and break up dark matter into galaxies.
Dark matter, on the stellar scale, are broken up by supernovae. Our solar system was seeded with the heavier elements from a supernova.
I'm happily surprised to figure out sectrices. Trisectrices are another thing. More complex and I don't know if I have all the curves available to use in analyzing them. But, I can see Fibonacci and Golden spirals relating to the trisectrices.
The Clausen function of order 2: dark matter flakes off, impacting the Big Bang mass directly and shocking the opposite side, somewhat like concussions happen. While a spin on that central mass is exerted, all the spins from all the flaking dark matter largely cancel out. I suspect that primordial black holes are formed by this, as well. Those black holes and older black holes, that came late to the Big Bounce, work together to break up dark matter.
Belows method (similar to Sylvester's Link Fan) relates to dark matter flaking off during a Big Bang event. Repetitious bisection relates to dark matter spinning so violently that it breaks, leaving smaller dark matter, primordial black holes, and other matter. Neusis construction relates to how dark matter is broken up near one of its singularities by an older black hole and to how black holes have their singularites sheared off during a Big Crunch.
General relativity: 8 shapes, as dictated by the equation? 4 general shapes, but with a variation of membranous or a filament? Dark matter mostly flat, with its 6 alternating corner/cusp edge singularities. Neutrons like if a balloon had two ends, for blowing it up. Protons with aligned singularities, and electrons with just a lone cylindrical singularity?
Prime numbers in polar coordinates: note the missing arms and the missing radials. Matter spiraling in, degenerating? Matter radiating out - the laminar flow of dark matter in an Inflationary Epoch? Connection to Big Bounce theory?
"Operation -- Annihilate!", from the first season of the original Star Trek: was that all about dark matter and the cosmic microwave background radiation? Anakin Skywalker connection?
I tried doing this for hours and got nowhere. That reflection formula seems to do all of the heavy lifting for this theorem, so now I just wanna go prove that haha.
The end solution |Γ(iκ)| = 0.521564 is the phase on the imaginary axis. The Barnes-G function gives a bi-complex permutation i!= 0.498015668 - 0.154949828 i. This uses G(i) Γ(i) = G(1+i) → Γ(i) = e^-ln G(i) + ln G(1+i).
These are misleading as i is a dimensionalizing operator, not an actual number. Actual number proportions do define it. It is specifically a unit logic for AND excluding the options of x and y with a rotation relative to y. Misleading doesn’t mean wrong. It means we’re standing on an interpretation landmine.
8:05 I don't understand this step, how did you magically make a 1 appear here: *i (IΓ(i)I)^2 = i (IΓ(1+i)I)^2* ?
It's not magic but mathematics.
As is stated before,
Γ(z + 1) = zΓ(z) (1)
where z is a complex number.
Letting z = i, we have that
Γ(i + 1) = Γ(1 + i) = iΓ(i) (2)
(complex numbers follow commutative property both for addition and for product).
Now take the absolute value of (2):
|Γ(1 + i)| = |iΓ(i)| (3)
Since
|ab| = |a|*|b|, a, b ∈ C (4)
where C is the set of complex numbers, we have that
|Γ(1 + i)| = |i|*|Γ(i)| (5)
The absolute value of a complex number z = x + yi is
|z| = +√(x² + y²) (6)
(I've put "+" before square root since square root can return both "+" values as "-" values. Since absolute value always returns positive real values, we need to put "+" before).
Since i = 0 + 1i, we have that
|i| = +√(0² + 1²) = +√(1) = +1 = 1 (7)
so we have that
|Γ(1 + i)| = 1*|Γ(i)| = |Γ(i)| (8)
Finally, squaring (8), we have that
|Γ(1 + i)|² = |Γ(i)|² (9)
Why is the absolute value |i!| the same as i! ? I don't get it... the question was for i! in the beginning, so I am missing one step in the end?
Bro assembled the infinity gauntlet of math concepts
You only showed the distance i! lies from the origin, that's not enough to know it's exact value
gosh now i need a 3b1b video explaining why pi is here and where the circle is
The circle comes from the fact that if we draw all the complex numbers with a given absolute value on an Argand diagram, it will form a circle
yeah that's just beautiful
In which world is absolute value good enough??????
You know what isn't brilliant? Going from a pleasing black background video to a white background sponsor segment.
Brilliant
You don't need the gamma function. There's a PI function for that.
It just doesn't have the -1 in the top.
Just write N! In integral form.
Gamma is just PI, but shifted to make identity relation easier.
The video starts with wondering what value i! has but ends up with calculating |i!| instead
the brilliant ad seared my eyeballs when the screen went white lol
Next the value of !|i| 😉
I understood some of those words.
Hyperbolic is where Goku trained to fight Cell.
That was fast... Like mind blazingly fast lol
Its true, I won't believe the outcome! Because the Gamma function is not unique among meromorphic functions in extending factorial from the natural numbers.
|amazing|
Alright, so we have an exact form for the magnitude of i!. Is there an exact form for the argument thereof?
Sadly no exact form exists...
You can use approximations for the gamma integral in order to get an approximate answer :(
But what does it mean? And how do these complex factorials behave? And what are they useful for?
I'm not sure who your intended audience is. You are trying to show people how you can derive this cool result, but you depend on all these properties that you tell people "just believe me on this because it's too complicated".
Is there an additiin version of the factorial?
I've had a use for it, from time to time, but no name for the process.
Isn't it y = 2x - 1 ?
Fibonacci
Do you mean triangle numbers? They are formed like this:
n + n-1 + n-2 + ... + 1.
Gauss famously derived a formula for these numbers. It's
n(n+1)/2.
I've not heard of anyone trying to extend these to non-integers. It might be interesting. I would suspect the formula would remain the same. So, like
triangle(i) = i(i+1)/2 = -1/2 + i/2
Besides real arguments, can the gamma function be calculated for any point in a non-numeric way?
The integral gets so much harder.
i ! is a complex number , you give the answer only for absolute value .However, I like your presentation .
For all those complaining about him not solving the argument, ill just say this.
If you tried to solve for i! exactly, we will just have to put it into the integral itself.
i! = integral t^i*e^-t, t=0 to inf
= integral (e^lnt)^i*e^-t
= integral [e^(i*lnt)]*e^-t
= integral (cos lnt+i*sin lnt)*e^-t
= integral e^(-t)*cos(lnt)
+ i*integral e^(-t)*sin(lnt).
The two integrals with give us the real and imaginary part. But try as you might, these two integrals are non-elementary, so we should not expect to be able to solve then in a nice exact closed form. You can even try on a online math solver like wolframalpha.
So the fact that he could find the exact value of the absolute value is already impressive, since we can’t even find exactly the real and imaginary parts individually.
Wolframalpha can also find the exact value for the absolute value but not the argument
We sometimes just have to accept that we can’t solve everything exactly
i like your funny words magic man
Why abs(i) is 1? Since by definition i^2 need to be -1.
What is -1/2!
(n+1)! = (n+1) x n! so
1/2! = 1/2 x (-1/2)!
Simple !
I legit spent a couple hours in office hours with my Professor talking about repeated exponentiation.
We called it the star operator! And x Star x increases WAY FASTER than x!
It’s so much fun to harass your professors =)
why do you call tetration "the star operator" instead of "tetration"
@@EnderLord99
We didn’t have a name for it (we didn’t know about it) so we just made up an operator, just for fun.
We didn’t know about the name (yes, including the Professor), he just only vaguely remembered it.
Oh believe me the rabbit hole goes way deeper than that. Knuth's up arrow notation and Graham's number come to mind. Numberphile have a few good videos on the ideas.
(Quizzical look) i is an integer.
You were right. I don't believe the outcome.
Bruh you calculated the magnitude of i!. But it's a complex number, what should it's argument be?
It’s around -0.30164
Sometimes we have to accept that we can’t solve everything exactly, and we have to just take an approximation.
@@Ninja20704 Well… we had no exact solution to x^2 = 2, before we invented square roots…
Mathematicians can always come up with ways to write exact solutions by means of definition.
Damn
3:23 Your wrong because if you do absolute value of one its positive one so you are still adding
I mean; technically, i is an integer, in the sense that it’s a whole number of units (1), it’s not a fraction, nor an irrational number. It’s just a complex integer. Yeah, you can’t have i things; but, by that definition, negative numbers shouldn’t (pardon the pun) count as integers, either. I mean, you can’t exactly have a negative number of things, either. You can’t have -3 apples, or anything.
“Not to get too technical” 😂
Me too 🙋🙋
So then i! by itself would be plus/minus the final result?
Nope because i! Is complex which means we need the argument (basically the angle between the direction of i! and the positive real axis)
The commercial gave me an opportunity to take a breath but I still hate it.
I used factorials today at my job in the military industrial complex. Weird.
BlackpenRedpen did this 4 years ago
I tried this on my own and took the following approach: |i!| = |i(-1 + i)(-2 + i)...| We know that (a e^ix)(b e^iy) = ab e^i(x + y), i.e., the magnitude of the product of complex numbers is just the product of the magnitudes. This gives us the product of k = 0 to infinity of sqrt(k^2 + 1), which is just infinity. Why doesn't this approach work? Maybe it is the infinity? I guess maybe this whole approach doesn't make sense, since you never reach 1 repeatedly subtracting 1 from 1 from i.
Where did you get the equation that you started with? |i!| = ...
@@KenJackson_US i - k = -k + i, so if you use the common meaning of factorial n(n - 1)(n - 2)... Plugging in i for n you get i(i - 1)(i - 2)... = i(-1 + i)(-2 + i)... This was just to put it in a + bi form which is more intuitive to reason about.
@@atrus3823 I think you're making an error with you _"common meaning of factorial"._ Normally, the products *stop* when they reach "1". But your complex products never, ever reach "1". If you just let it go to infinity, that's much different, which doesn't seem to fit the _"common meaning of factorial"._
@@KenJackson_US yeah, I said exactly that in my original comment
You have it at the end. It is definitely true that n! = n(n-1)! = n(n-1)(n-2)! and so on. But you needs some sort of starting point.
The same is why you can't start with a non-integer, and must use the gamma function (or similar) to first get a fractional factorial. You can't find (1/2)! with the usual formula, but you can then use (1/2)! to find (3/2)! or (-1/2)!
i!
Hey @brithemathguy, love your vids, but here it seems you forgot to answer the actual question: what is i!. Sure enough, it’s close to 0.498-0.1549i, but that isn’t obvious from your non-complex sinh result for the abs value.
Yeah approximate values. No closed form solution exists so this is more interesting :)
@@alexting827 ah, no closed form solution exists? I missed that in the video. Is that proven, or just the current state of affairs?
Oh, and the title has changed, it now includes abs
@@Misteribel It's not shown in the video. The way you would show that is by putting i into the gamma function, and then trying to evaluate it. You can show that it forms an integral with NO solution in terms of elementary functions. BPRP has a nice video on this: czcams.com/video/a9l1E-KhXC4/video.html
Hope that helps :)
(The actual "proof" that it cannot be evaluated is more difficult but it's not really necessary as there's just no integration technique that you can try that'll evaluate the integral)
Why does wolfram alpha show i! as a complex number?
Because in this vidéo, he only calculate the absolute value of i! which is always a positive real number
You're right. I dont believe it. Factorial of anything other than an integer is meaningless.
Now I understand why e^iπ + 1 =0
(IMO) this might be a amazing video to look for a idea, Disclaimer: Its
just My own Opinion on the suggestion, Advice;
"Be very proud of having a Opinion on something that counts for eachother including *me.”
You are factorial. Believe in yourself!
I am not convinced. How can you make the jump from n! = gamma(n+1) in the integers to complex numbers?
It works for all positive integers, so we just define it to be the solution of all complex numbers.
This is the same as how e^x = 1 + x + x^2/2 + … + x^n/n! for all real numbers, so we define it to be the answer for all complex numbers
You can take a look at wikipedia. The Gamma function is defined to be an analytic continuation of the factorial function to all complex numbers except the negative integers
Well, n! itself is also just in the natural numbers, so i! should be nonsense. However, if we allow it to continue to all complex numbers, to which the gamma function is a solution, then i! does make sense. It's the same as asking what happens when you do math such as 2,5*2,5. How can we multiply by a fraction, that doesn't make sense as multiplication is repeated addition? We just assume it works out and find a new definition that makes sense
@@potaatobaked7013 While that definition does work out to be very neat, the reservation by the OP is nonetheless a valid concern in my view. And while using a complex argument for x in the e-power series CAN be done, its convergence for all Complex numbers certainly merits some mathematical validation, though it is true.
Always fun to know that you don't even understand the question.
I thought it would be something like
(0+1i) * (0+2i) = -2
(-2+0i) * (0+3i) = (0-6i)
(0-6i) * (0+4i) = 24
(24+0i) * (0+5i) = (0+120i)
etc.
i!i!i!
(IMO) Imaginary Numbers is 1 of the topics to study for a idea, Note: Its just My own Opinion on the suggestion, Advice; "Feel free to exchange eachothers own Opinion even mine* to eachother".
👍👍👍👍👍👍