5) seems very roundabout since you already proved ln(xy) = ln(x) + ln(y). Use that to say ln(x (1/y)) = ln(x) + ln(y^-1). Then from 2) you have = ln(x) - ln(y)
Smart, because the power rule for non-rational values of r can only be derived via properties of the ln. If we define the ln however not by the integral shown in the video but as the inverse of e^x, the latter we might define as: A. the function equal to it's own derivative, as well as equal to 1 for the input value 0. B. of what according to A it's Taylor-series around x=0 are. C. the limit of (1+x/n)^n as n goes to positive infinite. .. then the argument would no longer be circular, I think.
@@Apollorion Yeah! It works, but I think it would be way more difficult to prove these properties. At least for me, the definition by the area under the hyperbola it's more intuitive. But for this we need another argument.
Next on the list, let exp be the inverse function to ln. ln(exp(x)) = x, then differentiate both sides and use the chain rule and algebra to see that exp' = exp.
It's probably best to define it with the Taylor Series, 1) because that's what's used in practice, but also 2) it will have students work with infinite series to prove exp' = exp to provide more variety
@@fotnite_ Often the first time you see it is with compound interest, which is at best an important limit that comes up a lot, but power series and exp' = exp is much more satisfying and fundamental overall. Some textbooks do ln before exp because the number of things you have to prove to formally define it is smaller, and many beginners have a very hard time understanding the idea of an infinite sum and summation notation, for whatever reason.
Once you have the logarithm of a product formula it is actually pretty intuitive to get the power formula. (Cuz a number raised to a power its just a number multiplying itself).
For step 5 wouldn’t it be simpler to rewrite it to be ln(x*y^(-1)) and then use a combination of the results in step 4 and step 2 in order to find the answer: ln(x/y) = ln( x*y^(-1) ) =(step 4)= = ln(x) + ln(y^(-1)) =(step 2)= = ln(x) +( -1*ln(y) ) = ln(x) - ln(y)
Sam Harper Actually, the Chen Lu, as Steve Chow/blackpenredpen calls it, is called the Chain Rule. They’re similar, but he calls it the Chen Lu because it’s funny that way. Just deal with it. ; )
To prove #5, you can just use the product rule (for natural logs) and the power rule (for natural logs), since ln(x/y)=ln(x*1/y)=ln(xy^(-1))=lnx+ln(y^(-1))=lnx+(-1)lny=lnx-lny.
Skylar Deslypere It is the point of the video, though. He never used the definition to actually prove property 2. He used property 1, which itself was proven fron the definition. You only need to prove some properties from the definition, the rest can be proven from the other properties. That is how the video works. This means he overly complicated the video.
I miss a piece of the puzzle for full comprehension. I never understood integration by substitution. Is there a video that explain how this thing works ?
@blackpenredpen have you or anyone you else you know done a video on the integral of (cos(t)-1)/t from zero to x? I would be interested, it involves special integrals and the Euler-Mascheroni Constant. I know Dr. Peyam has done videos on the EMC in the past, but when I searched could not find this one. Thank you! Very gracious for your hard work!!
2:35 I guess I will be the guy with the comment about formalism in mathematics... XD. I guess there would be no problem if r is rational, but if r is irrational then it would be necessary to have a previous definition for x^r. A common definition for x^r is if x>0, x^r=:e^(r*ln(x)). From that it follows that if x>0 for any real number r then (x^r)'=rx^(r-1)... However, that result was used in the proof so it would be circular thought... I liked the proofs for 4 and 5. Using induction, 4 and 5 it is possible to prove 2 for any rational r. Then that result would motivate a definition for x^r for any real r.
That's typically how you do it. But then you used the inverse of the e-function to define it. Bprp put the kettle on the head. He used the integral as the definition. And let's be real it is not absurd to do that. It after all is non-elementary so defining a function to solve it was only not necessary because we happened to know it before we had a definition of integrals. (Areas under curves were studied for a long time. And many people concluded the area under 1/x to be some strange logarithm). And i would not be surprised to see some textbooks define it like that. I literally saw sine and cosine defined by their Taylor series. But if you take a different definition, you need to check for that definition to maintain all properties of the function you are defining. And that's where exercises like this come in. And in fact I'm missing one. We proved our integral is a logarithm without any doubt. But we did not prove it's base e. But truth be told this might be overkill for a youtube video xD
I don't like the way he proved 2. He should have done a substitution of t=u^r, and you would have got that out easily. Property 5 follows from properties 2 and 4.
Hey I know this is super old but here's the answer: define function f(n) = integral from 0 to 1 of (x^n)*(e^x) dx. Using integration by parts (differentiate x^n and integrate e^x) we clearly see that f(n) = e - n*f(n-1) for n > 0. Since we are integrating a finite function over a finite interval, we are able to take the limit as n->+inf of f(n) by moving the limit inside of the integral. Notice that e^x is uneffected but x^n -> 0 for all numbers in the interval. Essentially the function x^inf * e^x is equal to 0 on the interval (0,1). There are ways to make this intuition rigorous but I won't do it here. Here's what we have so far: f(n) = e - n*f(n-1) for n > 0 lim as n->+inf f(n) = 0 Your limit is just lim n->+inf of (n+2)*f(n+1). Consider f(n+2), using the recursion we derived we know that f(n+2) = e - (n+2)*f(n+1) and therefore (n+2)*f(n+1) = e - f(n+2). Taking the limit of both sides we get that your desired limit is e.
I love it. 2 things though: isn't property six trivial if you know property 5 and 2, and can you prove that the integral definition of ln is the same as the "normal" definition.
This is a very interesting and pedagogical exercise. P.S.: in (4) you are assuming y > 1, right? Otherwise, you cannot split the integral like that (?)
Dani Borrajo Gutiérrez You can still split the integral. It’s like the following integral: Integral from a to b of f(x) dx = Integral from a to c of f(x) dx + Integral from c to b of f(x) dx for c>b, since the second Integral is negative the extra part from b to c in the first Integral cancels out. In the video, xy represents b and x represents c. c>b if y
Knowing (ln(x))' we also know (ln(f(x)))'. If we define e^x as the function such that it is equal to its derivative and equal to 1 for x=0 using f = e^x we finished
Hi
blackpenredpen Hey! I just want to tell you this: Thank you so much for approving the captions! : D Someday, I may include Chinese captions. ; )
Hello
I would like to translate to Spanish and Portuguese! How could I do it?
hey
Hello
Could just write x/y as x*(1/y), use property 4 to write ln(x/y) as ln(x) + ln(1/y), then ln(1/y) = ln(y⁻¹) = -ln(y) through property 2.
Oh yea!!! I forgot I had that already lol
Can not could!.
Yeah, that's what I came up with. I just didn't convert ln(1/y) to -ln(y).
I was going to comment about that.
@@hasanzia1613
I can see that several people suggested that as well. Looks like I'm getting all the credit though :)
Sleep: its 01:23, go to bed
New bprp video: *appears*
Sleep: nonono
Karol Chojnacki So true! XD
Holy shit! I read this comment at exactly 01:32 am!
Can you do the differentiation and integration of x! It would be fun
It is e^(sum(1/i), i=x to oo)
Derivative of x! is ln^x(1/t)
@@tsndiffoperayo im late but that diverges
Change of base property: "I guess I'm just an illusion"
Me : AM 8:40, I'm late for school!
BPRP : *Properties of logarithm function with Integral*
Isn't it holiday?
@@andreimiga8101 Not everywhere is America
5) seems very roundabout since you already proved ln(xy) = ln(x) + ln(y). Use that to say ln(x (1/y)) = ln(x) + ln(y^-1). Then from 2) you have = ln(x) - ln(y)
An explanation of the elliptic functions sn, tn, cn, dn, and so on, from a geometric standpoint, would be a very good video to make.
At 2:18:
How do we know that [x^r]'=r*x^(r-1)? It seems a circular argument, because we don't even know what x^r means without ln.
Smart, because the power rule for non-rational values of r can only be derived via properties of the ln.
If we define the ln however not by the integral shown in the video but as the inverse of e^x, the latter we might define as:
A. the function equal to it's own derivative, as well as equal to 1 for the input value 0.
B. of what according to A it's Taylor-series around x=0 are.
C. the limit of (1+x/n)^n as n goes to positive infinite.
.. then the argument would no longer be circular, I think.
@@Apollorion Yeah! It works, but I think it would be way more difficult to prove these properties.
At least for me, the definition by the area under the hyperbola it's more intuitive. But for this we need another argument.
Next on the list, let exp be the inverse function to ln. ln(exp(x)) = x, then differentiate both sides and use the chain rule and algebra to see that exp' = exp.
It's probably best to define it with the Taylor Series, 1) because that's what's used in practice, but also 2) it will have students work with infinite series to prove exp' = exp to provide more variety
@@fotnite_ Often the first time you see it is with compound interest, which is at best an important limit that comes up a lot, but power series and exp' = exp is much more satisfying and fundamental overall. Some textbooks do ln before exp because the number of things you have to prove to formally define it is smaller, and many beginners have a very hard time understanding the idea of an infinite sum and summation notation, for whatever reason.
*prove integrals from log properties*
hahaha!
epic
Once you have the logarithm of a product formula it is actually pretty intuitive to get the power formula. (Cuz a number raised to a power its just a number multiplying itself).
thats my favorite kind of videos
Proofs?
You are the best one in integrals my brother thank's soo much
For step 5 wouldn’t it be simpler to rewrite it to be ln(x*y^(-1)) and then use a combination of the results in step 4 and step 2 in order to find the answer:
ln(x/y) = ln( x*y^(-1) ) =(step 4)=
= ln(x) + ln(y^(-1)) =(step 2)=
= ln(x) +( -1*ln(y) ) = ln(x) - ln(y)
Prove that a function s.t. f(xy) = f(x) + f(y) has e^(x/f'(1)) as its inverse.
You can't it's not enough info, you need continuity as well
Great video!
So what about proving ln(x) is the inverse of e^x? Since that is not how you chose to define ln(x)
d/dx (ln(e^x))= 1/(e^x)*e^x=1 by chain rule. integrate to get ln(e^x)=x
0:50 *cries in Chinese* DON'T TALK ABOUT CHEN LIKE DAT
Sam Harper Actually, the Chen Lu, as Steve Chow/blackpenredpen calls it, is called the Chain Rule. They’re similar, but he calls it the Chen Lu because it’s funny that way. Just deal with it. ; )
@@einsteingonzalez4336 It's from Dr. P : )
blackpenredpen Of course I’ve heard of Dr. Peyam! : )
Yea. He is the originator. But I guess people associate me and Chen lu more ..
@@einsteingonzalez4336 omg. EVERYBODY KNOWS CHEN LU = CHAIN RULE IS A CHANNEL MEME. Jesus. The name Einstein really has lost its luster.
To prove #5, you can just use the product rule (for natural logs) and the power rule (for natural logs), since ln(x/y)=ln(x*1/y)=ln(xy^(-1))=lnx+ln(y^(-1))=lnx+(-1)lny=lnx-lny.
But that's not the point of the video is it, Gabriel? You can prove all of them without the integrals jut that's not the point. :)
Skylar Deslypere It is the point of the video, though. He never used the definition to actually prove property 2. He used property 1, which itself was proven fron the definition. You only need to prove some properties from the definition, the rest can be proven from the other properties. That is how the video works. This means he overly complicated the video.
THANK YOU
I miss a piece of the puzzle for full comprehension. I never understood integration by substitution. Is there a video that explain how this thing works ?
@blackpenredpen have you or anyone you else you know done a video on the integral of (cos(t)-1)/t from zero to x? I would be interested, it involves special integrals and the Euler-Mascheroni Constant.
I know Dr. Peyam has done videos on the EMC in the past, but when I searched could not find this one.
Thank you! Very gracious for your hard work!!
You could have concluded 5 from 2 and 4
In another video you used ln(x)-properties to prove that 1/x is an indefinite integral of ln(x), right? is math sometimes cyclic?
I remember commenting about this somewhere and then also on Reddit, but I'm not sure if it was on a video of yours
You mean the problem?
Excellent.
But, why to choose a tortuous path?
Very gooooood Golden boy.
2:35 I guess I will be the guy with the comment about formalism in mathematics... XD. I guess there would be no problem if r is rational, but if r is irrational then it would be necessary to have a previous definition for x^r. A common definition for x^r is if x>0, x^r=:e^(r*ln(x)). From that it follows that if x>0 for any real number r then (x^r)'=rx^(r-1)... However, that result was used in the proof so it would be circular thought... I liked the proofs for 4 and 5. Using induction, 4 and 5 it is possible to prove 2 for any rational r. Then that result would motivate a definition for x^r for any real r.
Yes okay that’s what I needed , but how could you prove that lnx is the reciprocal of e^x ??
ln(e^x)=x*ln(e) so you just have to prove that ln(e)=1, using substitution t=e^u, dt=e^u*du you get ln(e)=int(1/t,t=1..e)=int(e^u/e^u,u=0..1)=1
Can you prove ln(e)=1?
That would be a definition of the number e.
Amazing!
The first property is how we derived the integral in the first place
That's typically how you do it. But then you used the inverse of the e-function to define it. Bprp put the kettle on the head. He used the integral as the definition. And let's be real it is not absurd to do that. It after all is non-elementary so defining a function to solve it was only not necessary because we happened to know it before we had a definition of integrals. (Areas under curves were studied for a long time. And many people concluded the area under 1/x to be some strange logarithm).
And i would not be surprised to see some textbooks define it like that. I literally saw sine and cosine defined by their Taylor series. But if you take a different definition, you need to check for that definition to maintain all properties of the function you are defining.
And that's where exercises like this come in. And in fact I'm missing one. We proved our integral is a logarithm without any doubt. But we did not prove it's base e. But truth be told this might be overkill for a youtube video xD
No thumbnail!
😮
Kind of funny! XD
Math is so beautiful. Great video :)
8:45 U have to be Y
Ok but you still didn’t explain what I have to be and why
This is so fun to watch. Thank you bprp! We will teach integration class first then log class in future 😂
Thanks!!
How to integrate e^(-x^2)
Use TSeries. (Taylor s.)
@@alessiorusso1508 ??
You can't, really. It does not have an elementary anti derivative
I don't like the way he proved 2. He should have done a substitution of t=u^r, and you would have got that out easily. Property 5 follows from properties 2 and 4.
For number 2, you could just integrate the r/x result from the first derivative to get the r ln x, which would justify where the r ln x came from.
What is frobenius's theorem?
i love ur blackpenredpen
Hi! Can you help me with this math problem?
lim n->+inf of (n+2)*(integral from 0 to 1 of x^(n+1)*e^x dx)
Hey I know this is super old but here's the answer:
define function f(n) = integral from 0 to 1 of (x^n)*(e^x) dx. Using integration by parts (differentiate x^n and integrate e^x) we clearly see that f(n) = e - n*f(n-1) for n > 0. Since we are integrating a finite function over a finite interval, we are able to take the limit as n->+inf of f(n) by moving the limit inside of the integral. Notice that e^x is uneffected but x^n -> 0 for all numbers in the interval. Essentially the function x^inf * e^x is equal to 0 on the interval (0,1). There are ways to make this intuition rigorous but I won't do it here.
Here's what we have so far:
f(n) = e - n*f(n-1) for n > 0
lim as n->+inf f(n) = 0
Your limit is just lim n->+inf of (n+2)*f(n+1). Consider f(n+2), using the recursion we derived we know that
f(n+2) = e - (n+2)*f(n+1) and therefore (n+2)*f(n+1) = e - f(n+2). Taking the limit of both sides we get that your desired limit is e.
also where did you get the problem, I was surprised when solving it to find that it actually had a pretty nice solution.
Why did you start with ln(x^r). Why not start with ln(xy)? You could have then moved on to ln(x^2), ln(x^3) and so on...
we need fractional powers
I love it. 2 things though: isn't property six trivial if you know property 5 and 2, and can you prove that the integral definition of ln is the same as the "normal" definition.
Love your content!
I was bored until 4. But I got to say: that substitution is beautiful.
Saying 'u has..' sounds funny even in non-native english speakers!
should be starting my geometry baseline review but this is better
Love BPRP always with the hyped drip 👑
Where‘s the plus C? Joke xD
WHY DO YOU USE A HAND MIC?
Ishaan Agarwal Because it helps him use his left hand. Basically, he prefers to keep both of his arms and hands moving. : )
He already explained in this video ruclips.net/video/x1BXVUSZ6Ug/видео.html
But where are the red and white chalks? #Supreme
Hahahaha I like this comment!
I will guess it’s bc of sin(z)=2
@@blackpenredpen When complex numbers were an *integral* part of the content.
yes but noone is gonna start by defining ln that way
This is a very interesting and pedagogical exercise.
P.S.: in (4) you are assuming y > 1, right? Otherwise, you cannot split the integral like that (?)
Dani Borrajo Gutiérrez You can still split the integral. It’s like the following integral:
Integral from a to b of f(x) dx
= Integral from a to c of f(x) dx + Integral from c to b of f(x) dx for c>b, since the second Integral is negative the extra part from b to c in the first Integral cancels out. In the video, xy represents b and x represents c. c>b if y
@@D-Bar but of course! thank you so much and sorry for the lapsus! 😅
Request: Show the proof -from- using both definitions of ln and e that ln(x) is the inverse of e^x.
Knowing (ln(x))' we also know (ln(f(x)))'. If we define e^x as the function such that it is equal to its derivative and equal to 1 for x=0 using f = e^x we finished
神是最伟大的几何学家😊He made Earth a sphere so that it has the greatest surface area for a fixed volume.
nice
good job
SO GOOD!!! The things they don’t teach you in school!!!
Could you not just do ln(x/y)=
ln(x)+ln(1/y)
Of course you can, if you need to
Hi
Now make a video proving these integral properties
For x>0, 1∫x(1/t)dt = (ln|t|)1|x = ln|x| - ln|1| = ln(x), hence ln(x) = 1∫x(1/t)dt. I believe everything else can be worked out as shown.
Didnt know that ln((-1)*(-1))=ln(-1)+ln(-1)
Good good
'properites' :thonk:
27th to comment!
Amazing!
How to integrate e^(-x^2)
Hi