I never really considered the quo chen lu to be its own rule, namely because it can very easily be derived from the prada lu and the chen lu. Sure, you could derive all the other rules using just the chen lu and the derivative of e^x, but the prada lu and the power rule come up frequently enough. However, the quo chen lu doesn’t really come up in a lot of contexts where it isn’t just as convenient to use the prada lu instead. To me, it just seems like an unnecessary formula to memorize - but maybe that’s just because my way of learning is weird. If I may ask, what are the experiences of your students with the quo chen lu? Does learning it facilitate their journey through calculus?
Wow! My professor went totally off the deep end with his explanation. Used some version of the trigonometric half angles…in short, he confused everybody in class, and then told us that we would have to memorize it for the test. I am going to submit the method you used tomorrow in class and ask if it would be acceptable. Thank you so much!!!
blackpenredpen can you do a video on integrating this formula to get a Quo Chen Lu version of Integration by Parts? My friend and I figured it out but I wanted to know if it had any uses.
For the past several years, I’ve been stuck on this one math puzzle: Imagine with a circle, with radius R, and draw a circular sector with an angle theta. Next, draw a chord, thus creating a triangle, and a circular segment. What angle does theta need to be so that the triangle and circular segment are equal in area? I cheated a bit, so I already know the angle is (approximately) 1.895~ radians. I’m not, to be completely honest, interested in the value of theta; I’m interested, instead, in an expression that defines theta exactly (kinda like how Pi can be expressed as an infinite sum.)
That’s as far as I got, setting the two areas equal to each other, and simplifying to 2sin(theta)=theta. I’m hoping to find a different route... kinda like a “back door” approach?
Using some formulas you get this equation: 2sinx - x = 0 It has 3 solutions: 0, A and -A. The bad thing is you can't write A (~1.895) using other numbers, you can just aproximate it. That's because A is a trancendental number (like pi) :D
Sir We can also have : g = e^ (ln g ), f= e^(ln f) then f/g = e^ (lnf -- lng ) then we can proceed to differentiate as usual to get ( f/g ) ' = e^ (lnf -- lng ) . ( f' / f -- g' / g = f/g ( f' /f -- g' / g ) = (gf' -- fg' )/ g^2
Could you do a video showing how to prove the 2πr circumference formula by taking the limit as n approaches infinity of the perimeter of an n-sided polygon inscribed in a circle? That would be awesome! 😀
The chain rule is unnecessary here: One begins wih the product rule, (hg)' = h'g + hg' (hg)' - hg' = h'g [(hg)' - hg']/g = h' [g(hg)' - hgg']/g² = h' Now, let h=f/g (therefore hg = f). By replacing we get: [gf' - fg']/g² = (f/g)'. There we have it.
Ferdinand Grein ln(a + b) = ln(a)ln(b) => a + b = b^ln(a). Now let c = ln(a), so e^c + b = b^c. Then (e/b)^c + b^(1 - c) = 1 = e^[dc] + e^[ln(b)(1 - c)] = [e^c]^d + [b][e^c]^[-ln(b)]. This equation is unsolvable, in that e^c cannot be isolated by applying any finite set of operations. As such, one cannot find all the solutions a & b analytically.
More clearly, simply let x = ln(a) and y = ln(b), so ln(e^x + e^y) = xy. This implies e^x + e^y = e^xy, which implies (e^x)^y - (e^x) - e^y = 0, but this equation is unsolvable. Let e^x = p and e^y = q. Then the equation is p^y - p - q = 0, and now it is evident that one cannot obtain p as a relation of q. This was proved in the Abel-Ruffini theorem, where it was discovered that the equation in general is unsolvable for y = 5 and y > 5.
Mmm.. let's say: 1. ln(a) and ln(b) are defined and so: neither a nor b are zero. 2. So the quotient of a and b is always defined, let's label that u, such that u=b/a and b=au 3. ln(a+b) = ln(a) * ln(b) ln(a(1+u))=ln(a) * ln(au) = ln(a) + ln(1+u) = ln(a) * (ln(a) + ln(u)) 4. Let's label ln(a) as x 5. [3] /\ [4] => x^2 + x*(ln(u)-1) - ln(1+u) = 0 = x^2 + x*ln(u/e) - ln(1+u) = (x + ln(u/e)/2)^2 - ((ln(u/e))^2)/4 - ln(1+u) => 6. x + ln(u/e)/2 = +/- sqrt( ln(1+u) + ((ln(u/e))^2)/4 ) x = ln(e/u)/2 +/- sqrt( ln(1+u) + ((ln(u/e))^2)/4 ) 7. Choose a value of u from the positive real domain and [6] gives two relevant solution values for x, and the relevant a and b via a=exp(x) [4] and b=au [2] Note: u>0 => 1+u>1 => ln(1+u)>0 u>0 => u/e>0 => ln(u/e) is defined and real => (ln(u/e))^2 is greater than or equal to zero. So the square-root in [6] is always from a positive real number.
I don't think it's possible to solve . I'm working on an approach in 3D , the couples solution are the intersection of the surfaces of respective equations : z = ln(a+b) & z = ln(a)*ln(b) (where a & b are the x & y axes ) . At least to see how it looks like ...
lol Funny thing is, I remember when I used to study these back in college (and HS), I don't remember they gave us names to any of these "rules" at all. We are just taught the methods in these cases (multiplication, division, powers...etc). I wish if you do any videos, if possible, about contour integrals. I still, after all these years, can't figure out what's so special about them and how different they are from regular integrals?
@@angelmendez-rivera351 done them long time ago in my mathematical physics classes and maybe calc C classes. Thats why im asking coz ive totally mixed their meanings with regular integrals
@@nimmira They're integrals of a function of a complex variable. Since going from 0 to i could be different depending on what path you take (in particular, which way you go around poles), you need to specify the path as well as the endpoints. Then you find an antiderivative and choose branch cuts so you aren't crossing any. The most common use for them is to use residue calculus to get the value, then write the integral as the sum of some easy integrals and a definite integral you can't otherwise solve.
I was learning 20 pages of theorems on continuous functions and derivative... I was going to go asleep. Thanks for the little push 😍😍 (love ur vidz btw😝) Oh and could you make a video on how to write properly a demonstration, i’ m having troubles writing them clearly...
QUO
CHEN
LU?!
I'm dead.
The best Street Fighter character of all time
Beautiful Intro! 🥰 And of course: Chen Lu!!! 😄
I don't think it matters how many languages you can speak if you can speak mathematics.
We also have to deal with books not using factoreo wtf
I never really considered the quo chen lu to be its own rule, namely because it can very easily be derived from the prada lu and the chen lu. Sure, you could derive all the other rules using just the chen lu and the derivative of e^x, but the prada lu and the power rule come up frequently enough. However, the quo chen lu doesn’t really come up in a lot of contexts where it isn’t just as convenient to use the prada lu instead. To me, it just seems like an unnecessary formula to memorize - but maybe that’s just because my way of learning is weird. If I may ask, what are the experiences of your students with the quo chen lu? Does learning it facilitate their journey through calculus?
Own lu*
As usual, such a nice video, well explained and easy to understand. Well done. Thanks
Wow! My professor went totally off the deep end with his explanation. Used some version of the trigonometric half angles…in short, he confused everybody in class, and then told us that we would have to memorize it for the test. I am going to submit the method you used tomorrow in class and ask if it would be acceptable. Thank you so much!!!
Quo Chen Lu... that made my day!
blackpenredpen can you do a video on integrating this formula to get a Quo Chen Lu version of Integration by Parts? My friend and I figured it out but I wanted to know if it had any uses.
The way this was taught to me was using this:
Low d high plus high d low, draw a line and square below
For the past several years, I’ve been stuck on this one math puzzle:
Imagine with a circle, with radius R, and draw a circular sector with an angle theta. Next, draw a chord, thus creating a triangle, and a circular segment. What angle does theta need to be so that the triangle and circular segment are equal in area?
I cheated a bit, so I already know the angle is (approximately) 1.895~ radians. I’m not, to be completely honest, interested in the value of theta; I’m interested, instead, in an expression that defines theta exactly (kinda like how Pi can be expressed as an infinite sum.)
That’s as far as I got, setting the two areas equal to each other, and simplifying to 2sin(theta)=theta.
I’m hoping to find a different route... kinda like a “back door” approach?
Using some formulas you get this equation:
2sinx - x = 0
It has 3 solutions: 0, A and -A. The bad thing is you can't write A (~1.895) using other numbers, you can just aproximate it. That's because A is a trancendental number (like pi) :D
I did some more math here and I found out that A is the solution to this following sum (n goes from 0 to inf):
Sum of (x^2n)/((2n+1)!) = 1/2
Sir We can also have : g = e^ (ln g ), f= e^(ln f) then f/g = e^ (lnf -- lng ) then we can proceed to differentiate as usual to get ( f/g ) ' = e^ (lnf -- lng ) . ( f' / f -- g' / g = f/g ( f' /f -- g' / g ) = (gf' -- fg' )/ g^2
2:20 d/DX(1/y) = (dy/DX) * (d(1/y) /dy )
How about dancing with Carl Cool Lass?
But....
But it's *Koshen Lu*
Steve? Your secret identity came out of the blue .. but this is only your English name. One day, we will learn your real Chinese first name
He is... Chen Lu!
integrate of cos x/x^4 please if is it
Could you do a video showing how to prove the 2πr circumference formula by taking the limit as n approaches infinity of the perimeter of an n-sided polygon inscribed in a circle? That would be awesome! 😀
So, what would be the definition of π, if not π
You can, of course, derive it from the basic definition of a derivative without bothering the the other rules.
Disclaimer ploddud dule means product rule and davvededd rule means division rule 😁😁😁😁
C est vraiment super ces vidéos. Vous avez une très bonne pédagogie, de l humour. Vous êtes tous les deux géniaux !. 👍👍
Do the Trippa Prada Lu
VERY WELL AND PREECISE EXPLANATION THANK YOU VEERY VEERY MUCH
powa lu by induction and prada lu
Chen Lu and Quo Chen Lu... Is that like Sign and Quo Sign?
Finally it has come on my birthday!!! 😁😁😁😁
Best crossover ever
integrate 1/(1+x^5) please
Can you make a video on integration by parts?
The chain rule is unnecessary here:
One begins wih the product rule, (hg)' = h'g + hg'
(hg)' - hg' = h'g
[(hg)' - hg']/g = h'
[g(hg)' - hgg']/g² = h'
Now, let h=f/g (therefore hg = f). By replacing we get:
[gf' - fg']/g² = (f/g)'. There we have it.
Wan lu to lu them all!
Integration of sin x/sin x + cos x dx ?
saving my life here
Is there a 1/2 Lu theorem?
So that is chen lu! Wow
Isn't it guo shen lu
Au fait l étudiant russe qui avait fait une démonstration sur les suites et leurs sommations, qu' est ce qu' il est devenu ?
LMAO why do you call them that, it took me half a minute to realize the names were the words!! prada lu and chen lu LMAO
Plz help me to integrate e^(sinx) dx
新年快乐！
Black pen redpen keep up the good work can you please make a video about sec^3x tan^3x
quo chen lu := quotient rule #aha #mathforfun
I literally googled who tf is Quo Chen Lu😒
Next time use 律 instead of „Lu“ 😉
(Yes, I know the pronunciation differs slightly but at least it means „rule“.)
Great!
Happy lunar year %
Fake ln rule:
Find a and b, so that ln(a+b) = ln(a) * ln(b) .
Ferdinand Grein ln(a + b) = ln(a)ln(b) => a + b = b^ln(a). Now let c = ln(a), so e^c + b = b^c. Then (e/b)^c + b^(1 - c) = 1 = e^[dc] + e^[ln(b)(1 - c)] = [e^c]^d + [b][e^c]^[-ln(b)]. This equation is unsolvable, in that e^c cannot be isolated by applying any finite set of operations. As such, one cannot find all the solutions a & b analytically.
More clearly, simply let x = ln(a) and y = ln(b), so ln(e^x + e^y) = xy. This implies e^x + e^y = e^xy, which implies (e^x)^y - (e^x) - e^y = 0, but this equation is unsolvable. Let e^x = p and e^y = q. Then the equation is p^y - p - q = 0, and now it is evident that one cannot obtain p as a relation of q. This was proved in the Abel-Ruffini theorem, where it was discovered that the equation in general is unsolvable for y = 5 and y > 5.
Mmm.. let's say:
1. ln(a) and ln(b) are defined and so: neither a nor b are zero.
2. So the quotient of a and b is always defined, let's label that u, such that u=b/a and b=au
3. ln(a+b) = ln(a) * ln(b) ln(a(1+u))=ln(a) * ln(au) = ln(a) + ln(1+u) = ln(a) * (ln(a) + ln(u))
4. Let's label ln(a) as x
5. [3] /\ [4] => x^2 + x*(ln(u)-1) - ln(1+u) = 0 = x^2 + x*ln(u/e) - ln(1+u) = (x + ln(u/e)/2)^2 - ((ln(u/e))^2)/4 - ln(1+u) =>
6. x + ln(u/e)/2 = +/- sqrt( ln(1+u) + ((ln(u/e))^2)/4 ) x = ln(e/u)/2 +/- sqrt( ln(1+u) + ((ln(u/e))^2)/4 )
7. Choose a value of u from the positive real domain and [6] gives two relevant solution values for x, and the relevant a and b via a=exp(x) [4] and b=au [2]
Note: u>0 => 1+u>1 => ln(1+u)>0
u>0 => u/e>0 => ln(u/e) is defined and real => (ln(u/e))^2 is greater than or equal to zero.
So the square-root in [6] is always from a positive real number.
I don't think it's possible to solve . I'm working on an approach in 3D , the couples solution are the intersection of the surfaces of respective equations : z = ln(a+b) & z = ln(a)*ln(b) (where a & b are the x & y axes ) . At least to see how it looks like ...
🐷你新年快乐！
Thank you!!
Lu family
Back when I learned it long ago, we called it the hoho rule. That is, d(hi/ho) = hodhi minus hidho over hoho.
今天我们用普拉达陆和陈路的方法证明郭晨露😂
沒錯！
括晨陆
擴沈路！
How is the power rule related to everyone?
(I don’t know how to say power rule in terms of lu)
Maybe Pao Wa Lu?
Pawa Lu!!!
Pawa (Paoa) Lu
lol
Funny thing is, I remember when I used to study these back in college (and HS), I don't remember they gave us names to any of these "rules" at all. We are just taught the methods in these cases (multiplication, division, powers...etc).
I wish if you do any videos, if possible, about contour integrals. I still, after all these years, can't figure out what's so special about them and how different they are from regular integrals?
nimmira Do you know anything about contour integrals at all to begin with?
@@angelmendez-rivera351 done them long time ago in my mathematical physics classes and maybe calc C classes. Thats why im asking coz ive totally mixed their meanings with regular integrals
@@nimmira They're integrals of a function of a complex variable. Since going from 0 to i could be different depending on what path you take (in particular, which way you go around poles), you need to specify the path as well as the endpoints. Then you find an antiderivative and choose branch cuts so you aren't crossing any. The most common use for them is to use residue calculus to get the value, then write the integral as the sum of some easy integrals and a definite integral you can't otherwise solve.
@@iabervon Thanks!
Pro Chen Lu 😂😂
Lmao
I was learning 20 pages of theorems on continuous functions and derivative... I was going to go asleep. Thanks for the little push 😍😍 (love ur vidz btw😝)
Oh and could you make a video on how to write properly a demonstration, i’ m having troubles writing them clearly...
Your g looks like 9
nedisxx It really does not
First!