10:50 Why the inequality method doesn't work? Can't we solve it like this? For t between 0 to 1: e^t is smaller than or equal e, so 1/e^t is greater than or equal 1/e so 1/(te^t) is greater than or equal 1/(et) so the integral from 0 to 1 of 1/te^t diverge
OMG! You are right! I forgot for that bound is from 0 to 1... This is what happens when I have to do two different types of improper integral back to back...
@@alexwang982 using gamma function (-1)! diverges, and you cant even say that you can arrange it infinitely many ways, limit of x! as x->-1 doesnt exist
@@SEBithehiper945 How to actually calculate the negative number factorial without the intervention of gamma function plot. I want to plot (-1/3)!,(-2/3)!,(-5/3)!,... etc. i tried to solve by gamma integral. But didn't ended up in answer
As I said in your poll, this is a definition issue. There are of course well defined ways to extend the factorial function beyond the nonnegative integers. But the exclamation mark is reserved for that original, integral definition. It's unfortunate that the Gamma function is "off by one" or it would be easy to just use that and call it a day.
I prefer the second method, from 16:00 onwards - it is much more intuitively appealing If you insist on using the PI function we can still do the same Having already shown the relationship PI (n) = n.PI (n-1) in an earlier video, we can simply apply this result instead of repeating the Laplace integral again.
Yes, cause e^t in the interval [0,1] is always less than e, so if you replace e^t with e the value gets smaller. and cause e is a constant it can be ignored entirely.
Why Gamma function? check out this page math.stackexchange.com/questions/1537/why-is-eulers-gamma-function-the-best-extension-of-the-factorial-function-to
Great this reminded me of the old questions we got back in elementary school where they asked things like: 2_2_2_2 = and you had to put signs in to make it equal as many numbers as you could usually like from 0 to 10 but now knowing whats -0.5! there is a cool question you can ask your friends 2_2_2_2=π and see if they can solve it! my solution is (-2^(-2)*2)!^2=π
I remember that in my Calculus exam my teacher put a question with that divergent integral... It took me like all the exam time to realize that cannot be solved. XD
You could also look at (-1)! as a sequence. Every time you subtract 1 you multiply by a larger value (in terms of absolute value) and change the sign. Roughly speaking it looks (vaguely) like the graph of x*sin(x) in that it approaches both infinity AND negative infinity which is why saying (-1)! = infinity is incorrect.
If we treat the number line from negative infinity to positive infinity as a infinite circle then the undefined part is when both ends meet at infinity, I think they showed that each undefined part has its own infinity, I think there's some mathematical theory that uses that.
Not necessarily. Remember, Infinity is a concept, so if you want, you can treat infinity like a number, but not exactly like one. Like the idea of ∞+n=∞ and 1/0≠+∞ or -∞, but instead is 1/0=±∞
@@orngng did you even look at the last part :| Listen, (-1)! gets you a vertical asymtope as you can literally see in the graph of Π(x), and a vertical asymtope has the value of 1/0, which could possibly be ±∞. The original comment literally described a vertical asymtope is and why it's a problem to 1/0. Do YOU understand what the comment is even saying?
Yes, GAMMA in complex analysis is a meromorphic function, it has poles with residues at negative integers, and you can compute the integral in all the positive complex domain (Re(z)>0 otherwise the integral in undefined). Beware the real part of z being negative though since you need the mirror to compute the analytic conitnuation, example , GAMMA (-3.15) = pi/sin(pi*(-3.15_)/GAMMA(4.15), same thing goes for any complex value with negative real part, you need to mirror into the original domain, GAMMA(z) = pi/sin(pi*z)/GAMMA(1-z)
(-1)! is undefined, but 1/(-1)! = 0 just fine. You can show this without resorting to the gamma function by considering the number of ways to write n symbols in a list of length k. That is given by n! / (n-k)!. First: how many ways are there to write the list when it has length n? n!, obviously, but that requires 0! = 1. Similarly, if k = 0 the formula gives 1, but that is also 0!. Put another way, there is one way to write an empty list, just put the grouping symbols (that avoids the philosophical worry over how to arrange 0 things). Second: how many ways are there to write the list when k > n? Zero ways, because you can never successfully write such a list, but that requires n!/(n-k)! = 0 for k > n.
instead of saying (-1)! is undefined or infinity, I think there is a need to put a strict and new definition to something like 1=0*infinity .... maybe something new symbol that is very super and like complex number i that avoid explain what is sqrt(-1)
The Pi function has a singularity at each negative integer, but because those singularities are poles, not essential singularities, it is reasonable (so long as you take appropriate care) to say the value at those points is projective infinity in much the same way that other intuitive processes (like splitting up dy/dx and working with the dy and dx as individual values, or pretending the dirac delta is a function even though it isn't) are not only reasonable, but helpful, so long as you properly account for the caveats. That said, it is safer, if you're not confident of your ability to properly handle the caveats, to just say the value is undefined.
Hi, I really enjoy your videos. Could you show something about the wau(or digamn) number. I saw it, and got curious. Thanks for your amazing videos here.
3:48 actually i made a comment similar to this in a peyam video. as a polynomial of degree 6 differentiated 7 times should get 0, if we differentiate 5.5 times to get a degree of 1/2, differentiate again for a power of -1/2, and finally half-differentiate then the 7th derivative of x^6 ~ 1/x
Hey blackpenredpen. May I clarify something about your students? According to me, I study at the top 2 (or 1) university of Ukraine, and our students are so lazy that about 60% of all of them at my specialty do not pass calculus exam ('cause we have a strict tutor=)). So the question is: how many students pass your exams in average?
This method can work because 0!=0*(0-1)! which gives you 1=0*(-1)! which if you do in another way 0-1=-1 then you can say 1=1/-1*(-1)! then you get (-1)!=1/1*(-1)! which you tern the 1/1=1 then you put the factorial in front of the positive one and multiply it by (-1) which would look something like (-1)!=1!*(-1) which then 1!=1 so then you get (-1)!=1*(-1) which then equals to (-1)!=-1 because 1 multiplied by -1 = -1 so this means this works and (-1)! does exist and equals -1
4! 24 We divide 4 and 3! 6 We divide 3 and 2! 2 We divide 2 and 1! 1 We divide 1 and 0! 1 We divide "0" and -1! 1/0 nondefined We divide -1 and -2! -1/0 nondefined And for other negatife numbers x/0
i have a question. pi(x) is a good function for factorials. But pi(x)*cos(2*pi*x) it's also a correct function for factorials. Why do use one and not the other one
f(t) = 1/e^t is absolutely continuous over any closed interval and it has a max and min in the [O, 1] so it is easy to compare the initial integral with that of 1/t multiplied by some certain constant which is Devergent.
@@Fokalopoka Didn't the suggestion imply that somebody of a high mathematical level, such as a serious mathematician, do it, or at least somebody who thinks that they can produce an answer? But if it is the approximation of an infinite summation, then perhaps it is impossible to do it symbolically, until somebody sees some insight as to another way to do it. But we could write the infinite summation as the answer?
If I remember correctly, there's a neat trick where you can "extract" the divergences/poles (on the negative integers) by using the by-parts expansion of Γ. This gives Γ(x)=Γ(x-n)/(x(x-1)...(x-n)) or something along those lines (it has been a while since I did complex analysis so my memory is a bit hazy) where you end up with the first n poles along the negative integers in the denominator.
Also, putting in any n+1 to the gamma function... You'll find that it gives the same integral as the pi function by subtracting 1! The gamma function has a very useful property that gamma (x+1)=x gamma (x) which flirts very closely with the Reimann Zeta Function and a ton of other series in higher math
@@tracyh5751 yep! Gamma distributions are useful for modeling continuous random variable distributions that are positively skewed. Also, other distributions like the chi squared distribution, and the exponinetial distribution are really special cases of the gamma distribution.
The gamma function is indeed very convenient for the riemann zeta function, but what i really don't get is people using it for calculating simple factorials, WHY??? You are doing more work when you could be using the capital pi function which is simpler.
Have you thought about bringing your accessible approach to explaining derangements, subfactorials and the partial gamma function? When I first researched this, I find the appearance of e unexpected and delightful and the appearance of the nearest integer function completely counter intuitive. Now I've started looking at the analytic continuation of subfactorials and I find it counter intuitive in two ways. First that, as far as I can tell, it's defined everywhere, including negative integers and second that it maps real numbers into the complex plain.
Behavior of factorial in the vicinity of a negative integer: When n is a positive integer, and ε is an infinitesimal quantity, (-n + ε)! ~ (-1)¹⁻ⁿ·n!/ε An interesting plot to show this, is y = 1/x! It oscillates for x < 0, crossing the x-axis for each negative integer; the amplitude increases "factorially" as x becomes more negative. For x > 0, y > 0, and goes asymptotically to 0 as x increases toward ∞. y has a local maximum for x between 0 and 1. Fred
@Gerben van Straaten Agreed: The value is undefined, because you reach different limits if you approach from left vs right. In fact, my comment shows that approaching -1 from above (i.e., x --> -1+), it approaches infinity.
Like I have said many times before, It is *_undefined_* due to *_definition issues._* A common solution used by the math community is ±∞ as it's own value instead of +∞ or -∞.
So when I look up the gamma function on wikipedia, why is it defined for negative non-integers? This integral diverges for all exponents of t less than negative 1, right? So this integral can't be used to find those values. Is it just analytic continuation using the identity x!=(x+1)!/(x+1)?
To blackpenredpen You analytically continue the integral so maybe maybe not using the integral for one partial gamma and sum for the other help get -1!. That's how mathematicians convergent for all values. Solve for f(n) being 1/(n!) and the solution for -1 is 0 thus defined. Also 1/0 the solution is unsigned ∞ and technically greater than ∞ so calling it ∞ is inaccurate. It's rather unsigned 1/0. Take the sum equation for for example sin and instead use it with sum replaced by integral from -∞ to ∞, averaging all multiple solutions in complex math of each integrand, and no dt. Now the negative coefficient multiply by f(n)=0 so zero out so 1/0 for result for factorials is valid.
Riemann Zeta function's integral expression involves gamma function. We know that Riemann Zeta function is defined everywhere in the complex plane (by analytic continuation), except for the line where Re(z)=1. Thus Zeta(-n) is defined (where n a positive integer). But when (-n) is plugged into the integral expression,the zeta function (LHS) is defined whereas RHS is undefined because gamma (-n) is undefined. How to resolve this problem????
himanshu mallick zeta integral is only convergent for Re(z)>1 the same thing for GAMMA function (Re(z)>0). For any negative complex number I mean the real part you need to use the mirror for the analytic continuation (functional equation)
Hey! Maybe if factorial of (-1/2) i.e. -0.5! is sqrt(pi) then maybe if there is some rule as one factorial solution can be written as summ of other factorial solutions, then ! (-1) can be calculated and finally be defined from that?
Set tan(u)=2x, then sec^2(u)du=2dx. After that you end up with half of the integral of sec^3(u). You can find the solution to that on his channel. Then use the fact that u=arctan(2x).
Simply set 2x = sh(u), then 2dx = ch(u) du, dx = ch(u)/2 du. Putting everything in your integral and knowing that 1 + sh^2(u) = ch^2(u) and ch(u) is always possible, you get ch(u) * ch(u)/2 du = 1/2 (ch(u))^2 du. After that use fact ch(u) = 1/2 (e^u + e^(-u)) and you'll get pretty simple integral with exponents.
(-1)! it’s easy: you can just use (-1)!=1/0 if you put it on the “recall” part it goes fine: 0!= 0 * (-1)! 1=0*(-1)! 1=0*1/0 (0 and 0 cancels out) 1=1 so 1/0 is a solution
Hello! Can you make a video explaining this optimization word problem? I would really appreciate it! Love your videos btw! A woman in a rowboat 3 miles from the nearest point on a straight shore line wishes to reach the dock which is 4 miles farther down the shore. If she can sail at a rate of 6 miles per hour and run at a rate of 4 miles per hour, how should she proceed in order to reach the dock in the shortest amount of time? I can't figure this out! Thanks
Trick question. If she sails -- although _rowing_ would be more consistent with her stated mode of transportation -- faster than she runs, and the shortest path is only rowing, then obviously taking the shortest path is not only the path of least distance, but also the path of least time. Rowing the *sqrt((3^2) + (2 ^ 2)) = 5 miles* (by Pythagoras' Theorem) at *6 miles / hour* would take her 50 minutes. All other paths are slower than that. There is a much more interesting type of problem that's similar than this, but it only works if the speed in the water -- or whatever travel medium the starting point is in -- is actually _slower_ than the one on the sand (or whatever type of medium the end point is in). It also only works if the end point is _not_ on the line that is the transition from one medium to the other (the shore in this case). Instead, it must be at least marginally "land-inwards", so to speak If you're interested, watch this video by VSauce. ua-cam.com/video/skvnj67YGmw/v-deo.html The whole thing is brilliant and I definitely recommend watching the whole thing, but the type of problem I was talking is given an example at around the 6:25 mark (or maybe a few seconds after that -- it's the one with the mud and the road).
hey I looking for same though like sum of factorial solutions can be some factorial, then some summ of (-1/2)! can be put as solution to -1!. Have you find something?
Here's how I defined it: we can invent a new system of numbers. Let's call this j. 1j is the result of 1/0. By multiplying xj by 0, we get x by definition. We can't have a lonesome j and here's why: what is 0*j? Well 0*j could be viewed as 0*1j and that equals 1. On the other hand, 0*j could be 0j and that equals 0. Contradiction! So to continue the sequence, remember how -1! = 0!/0 = 1/0? Well that can be 1j. To continue, -1!/-1 = -1j, -2!/-2 = 1j/2, etc.
I think it's a logical leap to say the pi function is equivalent to the factorial function. Just because the pi function happens to intersect with positive integers for the factorial function does not mean it *is* the factorial function. The factorial function is defined by using integers.
What about 1/n factorial f.e 1/3? I will try it on my own but I will probably fail. It may be e^t^3 but then I don't know what to do next, maybe it is related to higher dimensions.
wo997 we don't usually use factorial but GAMMA function. And yes you can compute GAMMA(1/3) and by Euler reflection the other ones. This number is transcendental (PROVEN!) I guess what you mean is 1/3! = GAMMA 4/3
![Lp. Сердце Вселенной #60 РОЖДЕНИЕ ЛОЛОЛОШКИ [Финал] • Майнкрафт](http://i.ytimg.com/vi/YoR0pAV9FVQ/mqdefault.jpg)
10:50 Why the inequality method doesn't work?
Can't we solve it like this?
For t between 0 to 1:
e^t is smaller than or equal e,
so 1/e^t is greater than or equal 1/e
so 1/(te^t) is greater than or equal 1/(et)
so the integral from 0 to 1 of 1/te^t diverge
OMG! You are right!
I forgot for that bound is from 0 to 1...
This is what happens when I have to do two different types of improper integral back to back...
Are you argentinian?
@@Prxwler ttrrrryyyrfi
What is the Integration of {x+1/x}½ ?
Please solve this problem.🙏🙏🙏
"How many ways can you arrange negative 1 apples?"
...
R.
1
@@alexwang982
Oh really? Then n! for n
@@yosefmacgruber1920 gamma function, mate
and we meet again!
@@alexwang982 using gamma function (-1)! diverges, and you cant even say that you can arrange it infinitely many ways, limit of x! as x->-1 doesnt exist
also gamma of -1 is 0!, remember your definitions
*someone:* how many ways can you arrange negative half of a quarter
*me:* square root of pi ways.
You are too uneducated mathematically for this channel
Robin Sailo I think he meant -1/2 of a quarter (coin)
@@arnavanand8037 Ohh and you're too educated for a joke?
@@arnavanand8037 get over yourself buddy
@@arnavanand8037 you too sit, have a nice day.
try to do a complex factorial
You can do it like this: When doing Π(z), plug in complex integration.
💀
@@SEBithehiper945 How to actually calculate the negative number factorial without the intervention of gamma function plot. I want to plot (-1/3)!,(-2/3)!,(-5/3)!,... etc. i tried to solve by gamma integral. But didn't ended up in answer
As I said in your poll, this is a definition issue. There are of course well defined ways to extend the factorial function beyond the nonnegative integers. But the exclamation mark is reserved for that original, integral definition. It's unfortunate that the Gamma function is "off by one" or it would be easy to just use that and call it a day.
The Π function (mentioned in the video) is what you’re looking for, it’s the Γ function, but displaced by 1 unit: Π(x) = Γ(x+1).
@@GRBtutorials thank you!
I prefer the second method, from 16:00 onwards - it is much more intuitively appealing
If you insist on using the PI function we can still do the same
Having already shown the relationship
PI (n) = n.PI (n-1)
in an earlier video, we can simply apply this result instead of repeating the Laplace integral again.
Would ∏(n) = n • ∏(n-1) be an improvement upon your syntax, or did I do it wrong in some way?
But can you do dis?
Yes, cause e^t in the interval [0,1] is always less than e, so if you replace e^t with e the value gets smaller. and cause e is a constant it can be ignored entirely.
Chvocht - thats how i found this video aswell, dont even know why i clicked on it
Peter Auto r/wooosh
Yes *Leans Chair Backwards*
@@PeterAuto1 I know this was 4 years ago but woooooosh dude
Thank you! This is very advanced for me , but I am so glad I can find answers to my math questions! Awesome!
Why Gamma function? check out this page math.stackexchange.com/questions/1537/why-is-eulers-gamma-function-the-best-extension-of-the-factorial-function-to
blackpenredpen so what's 0.5! ?
factOREO!
Great this reminded me of the old questions we got back in elementary school where they asked things like:
2_2_2_2 =
and you had to put signs in to make it equal as many numbers as you could usually like from 0 to 10
but now knowing whats -0.5! there is a cool question you can ask your friends
2_2_2_2=π
and see if they can solve it!
my solution is (-2^(-2)*2)!^2=π
Amazing
I remember that in my Calculus exam my teacher put a question with that divergent integral... It took me like all the exam time to realize that cannot be solved. XD
Jorge Eduardo Pérez Tasso in which exam bro
In one of my College's exam @Hritik Rastogi
Me: Can we have (-1)! at home?
Mom: We have (-1)! at home.
(-1)! at home: Undefined
You could also look at (-1)! as a sequence. Every time you subtract 1 you multiply by a larger value (in terms of absolute value) and change the sign. Roughly speaking it looks (vaguely) like the graph of x*sin(x) in that it approaches both infinity AND negative infinity which is why saying (-1)! = infinity is incorrect.
If we treat the number line from negative infinity to positive infinity as a infinite circle then the undefined part is when both ends meet at infinity, I think they showed that each undefined part has its own infinity, I think there's some mathematical theory that uses that.
Not necessarily.
Remember, Infinity is a concept, so if you want, you can treat infinity like a number, but not exactly like one.
Like the idea of ∞+n=∞ and 1/0≠+∞ or -∞, but instead is 1/0=±∞
@@VenThusiaist What do you mean "not necessarily", only to reply with something else that doesn't follow up on the original comment
@@orngng
did you even look at the last part :|
Listen, (-1)! gets you a vertical asymtope as you can literally see in the graph of Π(x), and a vertical asymtope has the value of 1/0, which could possibly be ±∞.
The original comment literally described a vertical asymtope is and why it's a problem to 1/0.
Do YOU understand what the comment is even saying?
@@pon1 That is called "Wheel Algebra", my friend.
14:11
“That I want my students to show...”
OMG !! YOU HAVE STUDENTS !!!
Nicholas Leclerc
Yes
Can you show us the graph of the Pi function?
Take the Gamma function's graph and shift to the left by 1, because Pi(x) = Gamma(x+1).
Can you take the factorial of complex numbers, like i or 1+i? Or even quaternions like 1+i+j+k?
That would be cool just try and plug it in and see what happens
@@bonkuto7679 I don't think there is a meaningful or useful notion of what it means to raise a number to a quaternion exponent power
@@wraithlordkoto maybe not in today’s conditions of math and science
@@The-Devils-Advocate I dont remember what it means, but quaternion exponents are a thing actually
@@wraithlordkoto I meant that they might not be useful today, but later they could be, like imaginary numbers
Ok ok i can do neg factorials...
BUT CAN YOU DO THIS?
Complex factorials possible?
Yes, GAMMA in complex analysis is a meromorphic function, it has poles with residues at negative integers, and you can compute the integral in all the positive complex domain (Re(z)>0 otherwise the integral in undefined). Beware the real part of z being negative though since you need the mirror to compute the analytic conitnuation, example , GAMMA (-3.15) = pi/sin(pi*(-3.15_)/GAMMA(4.15), same thing goes for any complex value with negative real part, you need to mirror into the original domain,
GAMMA(z) = pi/sin(pi*z)/GAMMA(1-z)
materiasacra
Yes, since Re(z)=1 >0, the integral is convergent, GAMMA(1+i) = .4980156681-.1549498284*I
Whatever the hell a "factorial" of a non-ordered field means. :-D
reddit.com
(-1)! is undefined, but 1/(-1)! = 0 just fine. You can show this without resorting to the gamma function by considering the number of ways to write n symbols in a list of length k. That is given by n! / (n-k)!. First: how many ways are there to write the list when it has length n? n!, obviously, but that requires 0! = 1. Similarly, if k = 0 the formula gives 1, but that is also 0!. Put another way, there is one way to write an empty list, just put the grouping symbols (that avoids the philosophical worry over how to arrange 0 things). Second: how many ways are there to write the list when k > n? Zero ways, because you can never successfully write such a list, but that requires n!/(n-k)! = 0 for k > n.
instead of saying (-1)! is undefined or infinity, I think there is a need to put a strict and new definition to something like 1=0*infinity .... maybe something new symbol that is very super and like complex number i that avoid explain what is sqrt(-1)
That was a good clickbait tittle, i stoped immediately what i was doing.
hehehe
I don't know why but hat scream at the very end just scared me so freaking much.
sorry.... I think I forgot to lower the volume on that..
Extremely nice vid bprp
I don't understand all of this but it's fun to watch him get going on math
The Pi function has a singularity at each negative integer, but because those singularities are poles, not essential singularities, it is reasonable (so long as you take appropriate care) to say the value at those points is projective infinity in much the same way that other intuitive processes (like splitting up dy/dx and working with the dy and dx as individual values, or pretending the dirac delta is a function even though it isn't) are not only reasonable, but helpful, so long as you properly account for the caveats. That said, it is safer, if you're not confident of your ability to properly handle the caveats, to just say the value is undefined.
Hi, I really enjoy your videos. Could you show something about the wau(or digamn) number. I saw it, and got curious. Thanks for your amazing videos here.
It's all one.
Check the date of the "wau" video.
So this means that the factorial is undefined for all negative integers right?
Was wondering if you can make a video on the analytical continuation / poles of the gamma function? That'd be interesting.
3:48 actually i made a comment similar to this in a peyam video.
as a polynomial of degree 6 differentiated 7 times should get 0, if we differentiate 5.5 times to get a degree of 1/2, differentiate again for a power of -1/2, and finally half-differentiate then the 7th derivative of x^6 ~ 1/x
You are literally bringing those questions which i always thought about 👍 thanks 😊
Hey blackpenredpen. May I clarify something about your students? According to me, I study at the top 2 (or 1) university of Ukraine, and our students are so lazy that about 60% of all of them at my specialty do not pass calculus exam ('cause we have a strict tutor=)). So the question is: how many students pass your exams in average?
This method can work because
0!=0*(0-1)! which gives you 1=0*(-1)! which if you do in another way 0-1=-1 then you can say 1=1/-1*(-1)! then you get (-1)!=1/1*(-1)! which you tern the 1/1=1 then you put the factorial in front of the positive one and multiply it by (-1) which would look something like (-1)!=1!*(-1) which then 1!=1 so then you get (-1)!=1*(-1) which then equals to (-1)!=-1 because 1 multiplied by -1 = -1 so this means this works and (-1)! does exist and equals -1
4! 24
We divide 4 and
3! 6
We divide 3 and
2! 2
We divide 2 and
1! 1
We divide 1 and
0! 1
We divide "0" and
-1! 1/0 nondefined
We divide -1 and
-2! -1/0 nondefined
And for other negatife numbers x/0
i have a question. pi(x) is a good function for factorials. But pi(x)*cos(2*pi*x) it's also a correct function for factorials. Why do use one and not the other one
So can we conclude that f : x => x! is defined on R/Z-* ?
James Grime (on numberphile) extended the function and it didn't work... i mean it kinda worked... i guess...
-1! = 1÷0
Well the result is right...you still end up with infinity :)
so which for negative numbers are indetermined the factorial function?
Wait can't we integrate e^(-t)/t by Feynman/Leibnitz rule?
Eventi with that technic it doesn't converge
f(t) = 1/e^t is absolutely continuous over any closed interval and it has a max and min in the [O, 1] so it is easy to compare the initial integral with that of 1/t multiplied by some certain constant which is Devergent.
Keep doing what you're doing!
Do you live in ישראל (Israel)?
Could you please calculate (e)! and (pi)!
?
(e)! = 4.2608204741
(pi)! = 7.1880827328
@@Cjnw
How about calculating it in symbolic form, rather than decimal approximation?
@@yosefmacgruber1920 goodluck dealing with x^(pi) in a integral, if its doable, then its way over calc 2 level
@@Fokalopoka
Didn't the suggestion imply that somebody of a high mathematical level, such as a serious mathematician, do it, or at least somebody who thinks that they can produce an answer? But if it is the approximation of an infinite summation, then perhaps it is impossible to do it symbolically, until somebody sees some insight as to another way to do it. But we could write the infinite summation as the answer?
@@yosefmacgruber1920 im pretty sure it doesnt have a nice series, because of x^π, by nics series, i mean a series that will help at integration
If I remember correctly, there's a neat trick where you can "extract" the divergences/poles (on the negative integers) by using the by-parts expansion of Γ. This gives
Γ(x)=Γ(x-n)/(x(x-1)...(x-n)) or something along those lines (it has been a while since I did complex analysis so my memory is a bit hazy) where you end up with the first n poles along the negative integers in the denominator.
Let g be a continuous function which is not differentiable at 0 and let g(0) = 8. If
f(x) = x.g(x), then f(0)?
A) 0 B) 4 C) 2 D) 8.
Gamma upsets me. The pi function is much more logical! What gives?!
Also, putting in any n+1 to the gamma function... You'll find that it gives the same integral as the pi function by subtracting 1! The gamma function has a very useful property that gamma (x+1)=x gamma (x) which flirts very closely with the Reimann Zeta Function and a ton of other series in higher math
Gamma function is usually much more convenient when studying the Riemann Zeta function
gamma function also arises in statistics very naturally.
@@tracyh5751 yep! Gamma distributions are useful for modeling continuous random variable distributions that are positively skewed. Also, other distributions like the chi squared distribution, and the exponinetial distribution are really special cases of the gamma distribution.
The gamma function is indeed very convenient for the riemann zeta function, but what i really don't get is people using it for calculating simple factorials, WHY??? You are doing more work when you could be using the capital pi function which is simpler.
Please do a video on complex numbers factorials
I will try!
@@blackpenredpen
And what about quaternion factorials? Is there any such thing? Are quaternions the ultimate numbers?
Have you thought about bringing your accessible approach to explaining derangements, subfactorials and the partial gamma function?
When I first researched this, I find the appearance of e unexpected and delightful and the appearance of the nearest integer function completely counter intuitive.
Now I've started looking at the analytic continuation of subfactorials and I find it counter intuitive in two ways. First that, as far as I can tell, it's defined everywhere, including negative integers and second that it maps real numbers into the complex plain.
*plane. Oops.
Great! So, as 0! = 1! = 1 but n!
ふぇええ こうやって拡張できるのがガンマ関数の面白いところですねえ
そして(-1)!がこれまた面白い
Behavior of factorial in the vicinity of a negative integer:
When n is a positive integer, and ε is an infinitesimal quantity,
(-n + ε)! ~ (-1)¹⁻ⁿ·n!/ε
An interesting plot to show this, is y = 1/x!
It oscillates for x < 0, crossing the x-axis for each negative integer; the amplitude increases "factorially" as x becomes more negative.
For x > 0, y > 0, and goes asymptotically to 0 as x increases toward ∞.
y has a local maximum for x between 0 and 1.
Fred
While (-1)! is undefined, it seems that you should be able to show that lim(x --> -1+, x!) approaches +inf.
Yes, but the same limit approached from the left approaches -inf, hence the "undefined".
Agreed: There's a nice plot of the Gamma function (not the Pi function) at Wikipedia: en.wikipedia.org/wiki/Gamma_function
@Gerben van Straaten Agreed: The value is undefined, because you reach different limits if you approach from left vs right. In fact, my comment shows that approaching -1 from above (i.e., x --> -1+), it approaches infinity.
Very ingenious. Congratulation.
DEAR SIR, I REQUEST YOU TO POST VIDEOS ON MULTIPLE INTEGRALS
3:48 that voice crack though
3! = 1*2*3 = 6
2! = 3! /3 = 2
So that means
(n-1)! = n!/n
(-1)! = 0!/0= 1/0 = undefined
It's worth noting that as you approch (-1)! from the negative side, then it diverges to negative infinity too.
Like I have said many times before,
It is *_undefined_* due to *_definition issues._*
A common solution used by the math community is ±∞ as it's own value instead of +∞ or -∞.
So when I look up the gamma function on wikipedia, why is it defined for negative non-integers? This integral diverges for all exponents of t less than negative 1, right? So this integral can't be used to find those values. Is it just analytic continuation using the identity x!=(x+1)!/(x+1)?
Yes, it is analytic continuation using that identity.
To blackpenredpen
You analytically continue the integral so maybe maybe not using the integral for one partial gamma and sum for the other help get -1!. That's how mathematicians convergent for all values. Solve for f(n) being 1/(n!) and the solution for -1 is 0 thus defined. Also 1/0 the solution is unsigned ∞ and technically greater than ∞ so calling it ∞ is inaccurate. It's rather unsigned 1/0.
Take the sum equation for for example sin and instead use it with sum replaced by integral from -∞ to ∞, averaging all multiple solutions in complex math of each integrand, and no dt.
Now the negative coefficient multiply by f(n)=0 so zero out so 1/0 for result for factorials is valid.
So what is the domain of the factorial function?
Can you use the squeeze theorem to find a value?
In general sir....
Tell me that ...
Can we find ( R )!
Where R is any real number...
Yay! This makes so much sense!
Riemann Zeta function's integral expression involves gamma function.
We know that Riemann Zeta function is defined everywhere in the complex plane (by analytic continuation), except for the line where Re(z)=1. Thus Zeta(-n) is defined (where n a positive integer). But when (-n) is plugged into the integral expression,the zeta function (LHS) is defined whereas RHS is undefined because gamma (-n) is undefined. How to resolve this problem????
himanshu mallick zeta integral is only convergent for Re(z)>1 the same thing for GAMMA function (Re(z)>0). For any negative complex number I mean the real part you need to use the mirror for the analytic continuation (functional equation)
Me: hmmm lets open youtupe because i am tired of studying
This man:
why did you multiple by 2
Where do I find the graph?
I used feynman's technique although its undefined its a pleasure to use that technique like its soo gud yk
Hey! Maybe if factorial of (-1/2) i.e. -0.5! is sqrt(pi) then maybe if there is some rule as one factorial solution can be written as summ of other factorial solutions, then ! (-1) can be calculated and finally be defined from that?
Steve, could you integrate sqrt(1+4x²)? I'd appreciate it so much
Set tan(u)=2x, then sec^2(u)du=2dx. After that you end up with half of the integral of sec^3(u). You can find the solution to that on his channel. Then use the fact that u=arctan(2x).
kaszimidaczi Oh thank you! I haven't though it could be possible with tangent. Thanks :D
Simply set 2x = sh(u), then 2dx = ch(u) du, dx = ch(u)/2 du. Putting everything in your integral and knowing that 1 + sh^2(u) = ch^2(u) and ch(u) is always possible, you get ch(u) * ch(u)/2 du = 1/2 (ch(u))^2 du. After that use fact ch(u) = 1/2 (e^u + e^(-u)) and you'll get pretty simple integral with exponents.
Rafa xD No problem :)
His name is Steve?!
(-1)! it’s easy:
you can just use (-1)!=1/0
if you put it on the “recall” part it goes fine:
0!= 0 * (-1)!
1=0*(-1)!
1=0*1/0
(0 and 0 cancels out)
1=1 so 1/0 is a solution
can you do Gamma(n+1/2) and Gamma(-n+1/2) formula? pls
Hello! Can you make a video explaining this optimization word problem? I would really appreciate it! Love your videos btw!
A woman in a rowboat 3 miles from the nearest point on a straight shore line wishes to reach the dock which is 4 miles farther down the shore. If she can sail at a rate of 6 miles per hour and run at a rate of 4 miles per hour, how should she proceed in order to reach the dock in the shortest amount of time?
I can't figure this out!
Thanks
Trick question. If she sails -- although _rowing_ would be more consistent with her stated mode of transportation -- faster than she runs, and the shortest path is only rowing, then obviously taking the shortest path is not only the path of least distance, but also the path of least time. Rowing the *sqrt((3^2) + (2 ^ 2)) = 5 miles* (by Pythagoras' Theorem) at *6 miles / hour* would take her 50 minutes. All other paths are slower than that.
There is a much more interesting type of problem that's similar than this, but it only works if the speed in the water -- or whatever travel medium the starting point is in -- is actually _slower_ than the one on the sand (or whatever type of medium the end point is in). It also only works if the end point is _not_ on the line that is the transition from one medium to the other (the shore in this case). Instead, it must be at least marginally "land-inwards", so to speak
If you're interested, watch this video by VSauce. ua-cam.com/video/skvnj67YGmw/v-deo.html The whole thing is brilliant and I definitely recommend watching the whole thing, but the type of problem I was talking is given an example at around the 6:25 mark (or maybe a few seconds after that -- it's the one with the mud and the road).
Is it allowed to say (-1)!=1/0
Can u get a value of -1! If you get -.5! And the find -1.5! And find theoretical value that’s between those two?
hey I looking for same though like sum of factorial solutions can be some factorial, then some summ of (-1/2)! can be put as solution to -1!. Have you find something?
can you also view it as the series of sum ,and easy compare to known harmonic series and 1/n^2 series
Here's how I defined it: we can invent a new system of numbers. Let's call this j. 1j is the result of 1/0. By multiplying xj by 0, we get x by definition. We can't have a lonesome j and here's why: what is 0*j? Well 0*j could be viewed as 0*1j and that equals 1. On the other hand, 0*j could be 0j and that equals 0. Contradiction! So to continue the sequence, remember how -1! = 0!/0 = 1/0? Well that can be 1j. To continue, -1!/-1 = -1j, -2!/-2 = 1j/2, etc.
Is it only the antiderivative of e^-1 that is a logarithm?
At 2:11 you cut the 2u and not just one u. You are dividing for only one u.
16:03
I don't undestand, because, 0 factorial is 1, but using the formula we have other values!!!
How do we find i!
I think it's a logical leap to say the pi function is equivalent to the factorial function. Just because the pi function happens to intersect with positive integers for the factorial function does not mean it *is* the factorial function. The factorial function is defined by using integers.
Why don't you need to put ± in front of u for (-½)!, since you've taken the root of t?
Yet sqrt -1 is Real, no -1! is UNREAL, Euler had Whiskey on Weekends.
isnt it possible to invent a new type of complex number that satisfies 0*x = 1 ?
Derive math with it
But the 0 powers follow separate rules as complex as calculous
Define j to be 1/0
Then is 5*0*j eqaul to (5*0)j=1 or 5(0*j)=5?
couldn't you just say that?
(1/2)! = (1/2 -1)! *1/2 = (-1/2)!/2
2*(1/2)! = (-1/2)!
sqrt(PI) = (-1/2)!
QED
it still gets to the same answer
nvm
^Lul
Yes this is completely right
He has to make you watch his videos longer 😏😏😏😏
Sir can we say 1/(-1)! = 0 ??
What about 1/n factorial f.e 1/3? I will try it on my own but I will probably fail. It may be e^t^3 but then I don't know what to do next, maybe it is related to higher dimensions.
wo997 we don't usually use factorial but GAMMA function. And yes you can compute GAMMA(1/3) and by Euler reflection the other ones. This number is transcendental (PROVEN!)
I guess what you mean is 1/3! = GAMMA 4/3
I can barely understand anything but its so satysfying to watch lol
integral e- x3 zero to infinity ??
2iπ=what. plese tell me
15:58 i definitely heard "Ладно я шучу"
Sir U r great,👍👍👍💥💥💥💥
Loved video
Can you try to do "(1/3)!"?
Maybe make the next video about the derivative of factorial?
Adam Kangoroo the derivative of the GAMMA function is called the digamma function and it is also meromorphic with the same poles
But can you do this?!
Its a meme you dip
Thank you
What is the minimum of the gamma function of x for all positive real numbers x?
andyct1982 it is the only zero of the digamma function in the positive numbers
Exactly. I want him to do a video on that.
andyct1982 I can tell you the minimum if you need it
I know it. It's 1.461632144968.... I just want him to do a video on it.
Can we calculate factorial of a complex number?
And if so, whatever would it mean?
Well if we know half! = root pi /2, then by a property of the gamma function (half - 1)! = half! /half = 2*half! = root pi.