I love how you always find an intersting topic and go down a deep rabbit hole of making maybe 10 videos about that topic. Truly shows your passion for mathematics and the true desire to learn more. Never stop learning
I honestly enjoy seeing your enthusiasm for mathematics , you have way more passion and better teacher then all the math profs I had in my university haha
It's a great choice of a problem for students to build an understanding of what's going on. I can see how you put a lot of thought into example selection, and your subsequent delivery for an audience is something to be admired.
Well, it's useful to have a sufficiently appropriate "coarse feeling" of the value. The integral at the end is not straightaway self-explanatory, so lets make sense of it! Maybe for the "coarse feeling" of the derivative, we don't need to take the exact value of the gamma function. By taking the difference over a full 1 in x, then taking the "appropriate" average... difference one up: (x+1)! - x! == x!*(x+1) - x! == x! * ((x+1)-1) == x! * x difference one down: x! - (x-1)! = (x-1)! * (x-1) Since the series is growing by multiplication (by a rather constant factor, since the difference between x and x+1 for the growths is the smaller the bigger x becomes), it is appropriate to take the geometric average from the difference up und down to get a pretty good fitting approximation of the value for the difference at spot x: average (one up, one down) = sqrt( x! * x * (x-1)! * (x-1) ) == sqrt( x!^2 * (x-1) ) == x! * sqrt(x-1) The "-1" in the sqrt we can qietly ignore since the whole thing goes about a "coarse feeling" anyways, thus we land at: derivative (x!) ≈ x! * sqrt(x) That's a very easy to remember (but very coarse) approximation for practical usage. Check with Wolfram Alpha yields that this is actually better approximated by: derivative ((x-1)!) ≈ x! / (sqrt(x) * ln(sqrt(x))) The "-1" on the LHS because the Gamma function is one of against the factorial function. To rectify that for easier use: derivative (x!) ≈ x! * sqrt(x) / ln(sqrt(x)) That is sufficiently easy to remember and to calculate and in the range of a few percentage off the exact value. And it gives a good "feeling" for the look of that derivative function.
Thanks for the extra insight and explanation! I think for a coarse approximation you could also differentiate Stirling's factorial formula. I'm curious if that'd look anything similar to the approximation you explained.
these are the most wholesome advanced calculus videos ive ever seen in my life. i say advanced calculus only because my high school calculus teacher was a devoutly religious, elderly vietnamese woman who stood 4'11"
Good job. You can actually represent the derivative of the gamma function using the definition of the digamma function and its series representation. Keep up the good work!
I think you are very very ... passionate about mathematics. The 10s of videos you make about the same topic in different ways show this. And I like your way of explanation that is different from other YT people. I hope you do more videos like this
There is another maybe shorter way to show that the partial derivative with respect to x of t^x is ln(x)t^x . We know that t is considered as a constant . The derivative with respect to x of y=e^(ax) is ae^(ax) . Start with t = e^(lnt) ( where t and also lnt are constants ) and substitute this into t^x = (e^(lnt))^x = e^[(lnt)x}] . Now the derivative with respect to x of this last expression is lnxe^[(lnt)x} . But , in this last equation we know that e^[(lnt)x} = t^x ; therefore , the partial derivative with respect to x of t^x is (lnt)t^x .
Saw an interesting definition of the gamma function: lim (n goes to infinity) n! * u^n / Product (other Pi function) ( v as v goes from 0 to n) of (u+v) u > 0 In an old 1960's Finite Differences textbook.
I did not understand much of it without delving into it more. But the beginning is interesting by making the x factorial as pi of x. I think you can do that with any irrational number, so why not chose square root of 2? Or another irrational number.
All I could think of is that the derivative would be huge, quickly. Factorials grow fast 😅 I am going to have to rewatch this to really get my head around it.
Thankyou for a fun and useful result! 😄 In the early pages of Bleistein & Handelsman "Asymptotic Expansions of Integrals", they talk about: \limits_{N \to \infty } \left[ {{{\left( { - 1} ight)}^N}N!x{e^x}\int\limits_x^\infty {\frac{{{e^{ - t}}}}{{{t^{N + 1}}}}dt} } I have been wrestling with this for some time; thanks to your videos combining the Leibnitz rule, l'Hopital, second FTC etc with limits, I am (slowly! haha) gaining some traction. Much appreciated!
Hace mucho que no practico matemáticas, pero me parece, solo me parece, que hay un grave error en cambiar una función que solo es continua en puntos concretos y aislados en una función continua en todo el intervalo. Lo considero un error, aunque puedo estar equivocado
Hi! I am curious, why is there no way? At the end of the video you had the intention to replace t^x e^-t with x! ? You didnt do it because it would be abusive notation or im missing the smth?
Since the function on the left always increasing , there can be maximum one solution. One may guess that it's x=1 , but I am afraid , you have to use numerical ways to solve it, like Newton's method.
Abuse of notation is pretty common in math (as long as it is clear from the context what a given notation mean). After all, there are not so many math symbols to denote the variety of similar concepts.
You can go further. That derivative you speak can be obtained in terms of what is called the digamma function (Psi) . en.wikipedia.org/wiki/Digamma_function i.e. Int(t^x*ln(t)*exp(-t), t = 0 .. infinity) = Psi(x+1)*GAMMA(x+1)
Correct; however, he differentiated the Pi function, which is a popular extension of the factorial function to all reals except negative integers, essentially making a continuous factorial function
I love how you always find an intersting topic and go down a deep rabbit hole of making maybe 10 videos about that topic. Truly shows your passion for mathematics and the true desire to learn more. Never stop learning
I honestly enjoy seeing your enthusiasm for mathematics , you have way more passion and better teacher then all the math profs I had in my university haha
Thank you
You, Sir, are the epitome of what teaching with passion is all about.
It's a great choice of a problem for students to build an understanding of what's going on. I can see how you put a lot of thought into example selection, and your subsequent delivery for an audience is something to be admired.
Wow! You're one of the most fantastic instructors I have ever seen! Great video!
6:42 John 1:4? Amen
Thanks for the tutorial ❤
Well, it's useful to have a sufficiently appropriate "coarse feeling" of the value. The integral at the end is not straightaway self-explanatory, so lets make sense of it!
Maybe for the "coarse feeling" of the derivative, we don't need to take the exact value of the gamma function. By taking the difference over a full 1 in x, then taking the "appropriate" average...
difference one up: (x+1)! - x! == x!*(x+1) - x! == x! * ((x+1)-1) == x! * x
difference one down: x! - (x-1)! = (x-1)! * (x-1)
Since the series is growing by multiplication (by a rather constant factor, since the difference between x and x+1 for the growths is the smaller the bigger x becomes), it is appropriate to take the geometric average from the difference up und down to get a pretty good fitting approximation of the value for the difference at spot x:
average (one up, one down) = sqrt( x! * x * (x-1)! * (x-1) ) == sqrt( x!^2 * (x-1) ) == x! * sqrt(x-1)
The "-1" in the sqrt we can qietly ignore since the whole thing goes about a "coarse feeling" anyways, thus we land at:
derivative (x!) ≈ x! * sqrt(x)
That's a very easy to remember (but very coarse) approximation for practical usage.
Check with Wolfram Alpha yields that this is actually better approximated by:
derivative ((x-1)!) ≈ x! / (sqrt(x) * ln(sqrt(x)))
The "-1" on the LHS because the Gamma function is one of against the factorial function.
To rectify that for easier use:
derivative (x!) ≈ x! * sqrt(x) / ln(sqrt(x))
That is sufficiently easy to remember and to calculate and in the range of a few percentage off the exact value. And it gives a good "feeling" for the look of that derivative function.
Thanks for the extra insight and explanation!
I think for a coarse approximation you could also differentiate Stirling's factorial formula. I'm curious if that'd look anything similar to the approximation you explained.
I was just wondering about this a few days ago, can't stop living!
I like it when you smile. Love the videos ❤️
these are the most wholesome advanced calculus videos ive ever seen in my life. i say advanced calculus only because my high school calculus teacher was a devoutly religious, elderly vietnamese woman who stood 4'11"
Great delivery and informative.
Good job.
You can actually represent the derivative of the gamma function using the definition of the digamma function and its series representation. Keep up the good work!
Sir your videos helps me a lot..
From Iloilo Philippines ❤❤❤
They are some clean as hell blackboards you got there. 😜 You do good work man, love the pace and energy
Very clearly explained...thanks
Great videos! Love the scripture at the end.
I think you are very very ... passionate about mathematics. The 10s of videos you make about the same topic in different ways show this. And I like your way of explanation that is different from other YT people. I hope you do more videos like this
There is another maybe shorter way to show that the partial derivative with respect to x of t^x is ln(x)t^x . We know that t is considered as a constant . The derivative with respect to x of y=e^(ax) is ae^(ax) . Start with t = e^(lnt) ( where t and also lnt are constants ) and substitute this into t^x = (e^(lnt))^x = e^[(lnt)x}] . Now the derivative with respect to x of this last expression is lnxe^[(lnt)x} . But , in this last equation we know that e^[(lnt)x} = t^x ; therefore , the partial derivative with respect to x of t^x is (lnt)t^x .
You, Michael Penn & Papa Flammy all make me miss *real* chalkboards.
Very nice lecture
Thank you, Sir!
❤️🙏
I was waiting for this
nicely done!
Now it's time for integral x factorial
Very nice writing.
Thanks a lot 😊
Saw an interesting definition of the gamma function:
lim (n goes to infinity) n! * u^n / Product (other Pi function) ( v as v goes from 0 to n) of (u+v)
u > 0
In an old 1960's Finite Differences textbook.
Love the shirt! Where did you get it?
YOU ARE AMAZING
Mister I think leibniz rule hold for proper integrals. How would you justify using it for the improper integral here?
Thank you.
I would like to enroll in your class this year!!
You can simplify using the digamma function, though (if you can really call that a simplification).
He is awesome
Hey sir, a doubt is can't we write ln(t) t^x as ln(t)^(t^x) which would give x?
Interesting one there. Good video.
I did not understand much of it without delving into it more. But the beginning is interesting by making the x factorial as pi of x. I think you can do that with any irrational number, so why not chose square root of 2? Or another irrational number.
All I could think of is that the derivative would be huge, quickly. Factorials grow fast 😅
I am going to have to rewatch this to really get my head around it.
Wouldnt it be easier to use stirlings aproximation
splendid
Is this channel for postgraduates?
you are so close to discovering the di-gamma function
This is the gamma function
Sir, it seems to me that you could use Lambert Function to continue the last result.
Thankyou for a fun and useful result! 😄
In the early pages of Bleistein & Handelsman "Asymptotic Expansions of Integrals", they talk about:
\limits_{N \to \infty } \left[ {{{\left( { - 1}
ight)}^N}N!x{e^x}\int\limits_x^\infty {\frac{{{e^{ - t}}}}{{{t^{N + 1}}}}dt} }
I have been wrestling with this for some time; thanks to your videos combining the Leibnitz rule, l'Hopital, second FTC etc with limits, I am (slowly! haha) gaining some traction. Much appreciated!
Can you upload videos about complex geometrical problems(drawing graphs), like polygons? That would be great to see.
Sounds like something I don't know yet
@@PrimeNewtons never stop learning, those who stop learning ! stops living
@@paraskumar9850 He never said he wasn't willing to figure it out ... Looks to me like a way for him to sustain life!
Hace mucho que no practico matemáticas, pero me parece, solo me parece, que hay un grave error en cambiar una función que solo es continua en puntos concretos y aislados en una función continua en todo el intervalo. Lo considero un error, aunque puedo estar equivocado
But i dont know the define of x! if x in R
the bounds are in terms of t or x?
t
sir, for the power of t, shouldn't it be x-1? Because the y = x!, not y = (x-1)! hence it should be gamma of x, so t's power has to be (x-1)
I used the π function
❤️❤️
You could just differentiate Stirling formula.
i love your videos!
i have a question that's unrelated to the video but still mathematical
i can put it in the replies of this question if you'd like
An email with be better. Primenewtons@gmail.com
Учитывая, что Гамма функция- это интеграл, найти от неё производную не так уж сложно
i ❤ Mathematics
Oops derivative of a factorial function 🥶
Hi! I am curious, why is there no way? At the end of the video you had the intention to replace t^x e^-t with x! ? You didnt do it because it would be abusive notation or im missing the smth?
You can not replace t**x*exp(-t) with x! because integral(t**x*exp(-t)) from 0 to inf equals x!, not function inside.
Dy/Dx = X! [ Sum from {i = 0 to x-1} (1/(X-i))]
Isn't it ?
This fuction is not continu how could it be derivable ???
why don t you ask if this fuction is derivable before anything
Not zero x/ y if y=0 that mean not knowing
I guess the next derivative would square the logarithm.
Please try to solve this equation
(X+1/x)^x=2
Please
Please
x = 1 (I just guessed)
x = 1 is a trivial solution
Since the function on the left always increasing , there can be maximum one solution. One may guess that it's x=1 , but I am afraid , you have to use numerical ways to solve it, like Newton's method.
Γ(x) = (x-1)! → x! = Γ(x+1) (Gamma function)
Γ'(x) = Γ(x)(ψ(x)) → Γ'(x+1) = Γ(x+1)(ψ(x+1))
d(x!)/dx = x!(ψ(x+1)) → ψ(x+1) = ψ(x) + 1/x (Digamma function)
ψ(x) = Hₓ₋₁ - γ (Harmonic number & Euler's constant)
d(x!)/dx = x!(Hₓ₋₁ - γ + 1/x)
Factorial is part of N, not R.
Abuse of notation is pretty common in math (as long as it is clear from the context what a given notation mean). After all, there are not so many math symbols to denote the variety of similar concepts.
Well in that case the derivative of x! is (x+1)! - x! = x!(x+1 - 1) = x!x. But technically that’s a difference not a derivative.
Taking the derivative under an integral requires some justification.
The FTOC.
You can go further. That derivative you speak can be obtained in terms of what is called the digamma function (Psi) . en.wikipedia.org/wiki/Digamma_function i.e. Int(t^x*ln(t)*exp(-t), t = 0 .. infinity) = Psi(x+1)*GAMMA(x+1)
X! is not continuous, so has no derivative.
Correct; however, he differentiated the Pi function, which is a popular extension of the factorial function to all reals except negative integers, essentially making a continuous factorial function
anyone from India ( JEE aspirant) here
Thé chaîne de zéro is unknowing
( x-1)!...?
X Munier multiply by zero the result zero
Derivatives an anti derivative
Ugh.
Me totally forgot gamma function