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!
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
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
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.
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!
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…)
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
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)
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.
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 😂
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.
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.
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 ❤
@@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!_** |*
(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".
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.
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.
@@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"._
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)!
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.
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 ^^
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.
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.
(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".
(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.”
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.
@@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: ua-cam.com/video/a9l1E-KhXC4/v-deo.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)
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
The solution is not complete by the end of the video - it evaluated the absolute value of the factorial of i, not the factorial of i. I'm a bit disappointed.
@@NotBroihon Sure - that was the original scope of the problem. But that is just an artificial constraint of the problem. The problem is evaluating the factorial of i arises naturally in the course of solving the original problem. The video clearly should have addressed it.
@@davidchung1697 Considering the fact that - to my knowledge - there's no closed form for i! the video would kinda suck if you just took the estimate for i! and then calculated the absolute value. This shows that you can get the absolute value of i! without knowing what i! actually is. Why there's no closed form for i! is a completely different (and definitely interesting) topic and not relevant to this video.
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".
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 =)
@@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.
Me: "Why and how did you come up with imaginary numbers? What kind of drugs were u on?" Guy who came up with imaginary numbers: "My motives are complex"
@@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.
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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
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
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.
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!
Amazing video, the relationship between the Gamma function and pi is just incredible.
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.
scandalous for engineers
I really don't know how this guy doesn't have at least 1m sub for his good explanation. HE MADE LOVE MATH, THX!
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
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....
Beautiful! That was awesome!. Excellent presentation !.
I love these weird types of functions;
they are always interesting to take a look at.
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
So many loopholes in the solution. This video is perfect example of using incomplete solution that luckily leads to the right answer.
|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.
This class is nothing short of breathtaking! Your ability to make complex concepts easily understandable is truly remarkable.
I understand Euler's identity but this one's alluded me. I'll come back to it thanks.
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!))
@@Woah9394 the factorial function is not analytical
please go into crazy depth sometime, i'd love that!
At 3:38 I Visualized 4D By Accident
A bit above my level but got the gist of some of it. Thanks.
The video starts with wondering what value i! has but ends up with calculating |i!| instead
Quickly into the weeds. More power to you
Well that just totally blew me away. Have really looked at this stuff from my college days and now I remember why. 😃
Awesome explanation , the gamma function is great function
UA-cam is getting scary I thought about i factorial in my head this morning and then I get recommended this video
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)
You only showed the distance i! lies from the origin, that's not enough to know it's exact value
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
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.
Awesome video. Though you do often confuse i! with |i!|. Saying that i! is something when really you mean |i!|
Beautiful! That was awesome!
you said click the video hard but the video isn't showing can you please add it too the description
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 😂
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
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.
Such a beautiful result
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
But what does it mean? And how do these complex factorials behave? And what are they useful for?
These are cleanest animations I've seen on any math channel! Thanks for the quality content
You’ve never watched 3Blue1Brown, I suppose
I am too new to complex numbers to know how we got to hyperbolic space, as in your mention of hyperbolic sin. Help.
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
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?
Excellent presentation !
(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".
You were right. I don't believe the outcome.
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 factorial function is simpler than the Gamma function. It's the same integral but with z instead of z - 1.
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)!
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.
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 ^^
What about de argument/fase of the complex number i! ?
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.
3:23 Your wrong because if you do absolute value of one its positive one so you are still adding
3:29 where did you get absolutes here?
animations are really cool!
Me, a high school student who recently learnt of i and has no idea what anything in this video means: mmm interesting
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.
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
(Quizzical look) i is an integer.
Very nice!
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 :(
yeah that's just beautiful
In decimal, it's about 0.52
Maybe I Should Stop Reading Comments Before Watching The Entire Thing
Brilliant.
I understood some of those words.
Hyperbolic is where Goku trained to fight Cell.
Besides real arguments, can the gamma function be calculated for any point in a non-numeric way?
The integral gets so much harder.
In which world is absolute value good enough??????
The question was |i!| no more, no less.
@@adiaphoros6842 He made the question... And reasoning why exactly this question was "good enough" which I disagree with
(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".
my brain says zero times zero zero times is zero
i ! is a complex number , you give the answer only for absolute value .However, I like your presentation .
UA-cam's getting real comfortable with these 90 second double unskippable ads...
(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.”
Bro assembled the infinity gauntlet of math concepts
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: ua-cam.com/video/a9l1E-KhXC4/v-deo.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)
You know what isn't brilliant? Going from a pleasing black background video to a white background sponsor segment.
|amazing|
So it's about 12/23. That's almost Christmas! :D
I used factorials today at my job in the military industrial complex. Weird.
btw it looks fun to write
|i!|
sorry, i lack understanding, i do not see the sense of the matter, empty formalisms - whatever should be the sense of e powered by pi?
You only gave the magnitude of the complex answer. What is the phase?
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
Now I understand why e^iπ + 1 =0
Sqrt[pi/sinh(pi)]=sqrt[pi×csch(pi)]
The solution is not complete by the end of the video - it evaluated the absolute value of the factorial of i, not the factorial of i. I'm a bit disappointed.
Well, the point was to evaluate the absolute value and not the factorial. Nothing incomplete here.
@@NotBroihon Sure - that was the original scope of the problem. But that is just an artificial constraint of the problem.
The problem is evaluating the factorial of i arises naturally in the course of solving the original problem. The video clearly should have addressed it.
@@davidchung1697 Considering the fact that - to my knowledge - there's no closed form for i! the video would kinda suck if you just took the estimate for i! and then calculated the absolute value. This shows that you can get the absolute value of i! without knowing what i! actually is.
Why there's no closed form for i! is a completely different (and definitely interesting) topic and not relevant to this video.
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".
matrix factorials please
Brilliant
Next the value of !|i| 😉
BlackpenRedpen did this 4 years ago
the brilliant ad seared my eyeballs when the screen went white lol
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.
This is nice. But this is modulus of i!. So what is i!?
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
And what's the argument of this number?
Me: "Why and how did you come up with imaginary numbers? What kind of drugs were u on?"
Guy who came up with imaginary numbers: "My motives are complex"
It's been widely observe that the name _"imaginary"_ was poorly chosen and not accurately evocative.
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
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)