is this definition of the natural log VALID?

[@dogandonmez5274]

Second derivative zero means function is at most first degree polynomial 🙂

[@jaimeduncan6167]

Yep

[@keremcanknay2277]

Yeah that's true but since we've been said that a is equal to 0 that doesn't change anything. Also I think if we'd written y equals to ax^3 + bx^2 + cx + d, we can find out a and b are both equal to 0 by the same technique we have used before to find a equals 0.

[@tolberthobson2610]

woah, your right. That really does seem like a pretty big mistake on Michaels part.

[@Happy_Abe]

Was just about to comment this thanks

[@RexxSchneider]

@@keremcanknay2277 Yes, but the reason we find the coefficients of all polynomial terms of degree greater than one are zero is that _it's a first degree polynomial_ as the OP said. We really should just write y = ax + b and save a lot of messing around.

[@filbranden]

If the second derivative is zero, then the function is linear, not quadratic

[@harrymattah418]

Fortunately enough, he found a=0.

[@RexxSchneider]

@@harrymattah418 I think the word you're looking for is "inevitably", not "fortunately". There's nothing fortunate about about finding that the coefficient of x^2 in a first degree polynomial is zero.

[@harrymattah418]

@@RexxSchneider I think the word you don't understand is humour.

[@RexxSchneider]

@@harrymattah418 Sure, every time you make a mistake, pretend it was a joke. Then tell everybody who points out your mistake that they don't understand humour. A great recipe for learning. Not.

[@harrymattah418]

@@RexxSchneider I think another word you don't understand is ridiculous.

[@francescomussin]

y' = y/x could also be solved another way: since we are looking for a differentiable function y(x) which only takes positive values for x, the differential equation y' = y/x could also be written as y'x -y = 0, which in turn is the same as (y'x-y)/x^2 = 0. Here we can recognize the quotient rule and realize that our differential equation is really just (y/x)' = 0, whose only solution is y = cx for some constant c.

[@harrymattah418]

Elegant

[@joaovictormacedosales2520]

Wonderful!

[@harrymattah418]

@@Nolord_ I am not sure we are interested by non positive values of x since we are talking of a logarithm.

[@shiboliu1736]

Good. I also got this idea. Moreover, to find c we simply observe that y(1) = 1.

[@JM-us3fr]

Very slick

[@jplikesmaths]

I watched an 11 min video of Michael taking everything I already know and making them interesting again

[@Mitschy2007]

Nice! A similar video by defining the arctan as integral of 1/(1+t^2) might be interesting.

[@manucitomx]

This is straightforward (almost) and fun. I shall use it in teaching. Thank you, professor.

[@robshaw2639]

This integral is exactly how the Apostol text introduces the log function, and it is motivated by the fact that we want log ab to be log a + log b. It isn't even introduced as an inverse - the exp function actually comes later in the text (motivated by exp (a+b) = exp(a) exp(b)), and then he shows their realationship. The Apostol texts are a great calculus course.

[@stephenbeck7222]

Any calculus textbook labeled as ‘late transcendentals’ does the same thing. Larson’s Calculus for example is published in both early trans and late trans editions.

[@iooooooo1]

100% agreed. Apostol is a really good book. Spivak (another good book) also defines log() this way and defines exp() as its inverse.

[@GBloxers]

My courses in university use the apostol texts & they're great!

[@abstractnonsense3253]

Apostol's "Calculus" is a great 2 volume textbook in university mathematics, containing calculus 1, 2, and 3, linear algebra, and introduction to differential equations. It's clear, concise, rigorous, and accessible. And it has a ton of exercises.

[@robshaw2639]

@@GBloxers glad to hear they are still being taught! which university?

[@yoyoyogames9527]

CHALK WHATVE YOU DONE CHALK

[@robertmonroe9728]

7:04 why second order poly? Just ax+b. And then show that a is 1, b is 0

[@ArmedStevos]

What can we use to show the integral at @7:30 diverges to negative infinity (w/o using ln(x))?

[@sigmaoctantis5083]

For any natural number n we can consider the partition 0 < 1/n < 2/n < ... < 1. The lower Darboux sum of f(t)=1/t with respect to that partition is the n-th partial sum of the harmonic series. Since this series diverges, the lower Darboux sums are not bounded above, and hence the integral diverges to infinity.

[@AyoubSaab7]

In an algorithms course, I learned a nice explanation of log(x) in base 2: it's the number of times need to divide x by 2 to reach 1 or lower. Ever since, this definition has never left my mind due to how practical it is (akin to the compound interest origins of e). Naturally, it applies to any valid base and can be used -- as a starting point -- to explain what the natural log(x) means in practice: it's the number of times we must divide x by around 2.718 to reach 1 or lower. I believe it's important to emphasize the physical interpretations when possible, because rigorous definitions are difficult to remember in general. Thank you for the video!

[@adayah2933]

This is just the definition of log...

[@AyoubSaab7]

​@@adayah2933 Yes of course. I am just saying this one feels easier to remember/intuit than alternative definitions.

[@MarcoMate87]

Michael saved the day at 7:06 saying that y is a quadratic polynomial AT THE MOST, which is indeed a correct sentence. The only problem is that he prolonged the computations for this reason. Here is an example of how he could have continued if he had set y = mx + q. From the differential equation y' = y/x we get m = m + q/x, which implies q = 0; thus we have y = mx. From the definition of y, we compute y(1) = 1, which implies that m = 1; thus we have y = x.

[@GreenMeansGOF]

Fun fact: The natural logarithm was discovered before e.

[@jaafarmejri3361]

This is how it was defined as I was in school. We just needed to show that f(ab)=f(a)+f(b) which satisfies the functional equation for Ln + a bit of initial conditions

[@CM63_France]

Hi, For me it's not at all a "backward direction", that was precisly that way we defined these functions (in France in the 70's). First of all you define the log as the anti-derivative of 1/x that is null for x=1. Then you see that ln(x) and ln(ax) have the same derivative, so they differ by a constant. Doing x=1 tells you that constant is equal to ln(a). So you just showed that for all x,y ln(x)+ln(y)=ln(xy) . Then, the ln function is continuous and monotonous , so you can define the inverse (reciprocal). Let's call it exp(x), but we don't know if it is any number to any power yet. But we know that it has the property : exp(x+y)=exp(x)*exp(y) Then you can easily prove that, for a rationnal number y, exp(xy)=(exp(x))^y . Then, let e be the value of exp(x) for x=1, you can write exp(x)=e^x and that defines the exponentiation for real numbers.

[@JoelRosenfeld]

It’s a lot faster to show that the inverse function of the natural log defined in the manner of this video satisfies y’ = y with y(0)=1. Then the exponential follows really quickly. Going the rational power route also requires additional steps to get real values through the least upper bound property, which can be a pain.

[@TheEternalVortex42]

It's more natural to define ln(x) as the inverse of exp(x) since you encounter that earlier than you do integrals. Of course once you have all the machinery of analysis then you can start defining things differently...

[@Anokosciant]

en prépa, c'est toujours la manière de définir le logarithme naturel

[@Achill101]

Our high school textbook in Germany defined ln and exp the same way, and it worked well for the real numbers. Do you know where this way of introducing them started? . . . For complex numbers, however, you need power series.

[@JoelRosenfeld]

@@TheEternalVortex42 historically the definition of e from compound interest and the natural logarithm as an integral were discovered in parallel. The logarithm in the integral form arises naturally from Archimedes’ study of the area under a hyperbola.

[@Quasarbooster]

Very cool exploration! Thanks for going through this

[@goodplacetostop2973]

10:43

[@rodrigoappendino]

3:06 I'm trying to understand this part there you differentiate the integral. How did you do?

[@tomkerruish2982]

This is how we defined it in when I took calculus. We used Apostol.

[@terryendicott2939]

Same here - different book - Johnson and Kiokemeister 3rd edition.

[@harrymattah418]

I intensively hope so, as it is the very way I was introduced to the natural log! Please save my math world!! I remember my teacher having us finding primitives of 1/x^n for different values of n, therefore having us failing for n=1. He would then define the natural log to be the primitive of 1/x that is zero at x=1. We would then check all the natural logs properties, notably ln(ab)=ln(a)+ln(b) and the limits of the natural log with this integral definition. Good exercise.

[@jursamaj]

"Primitives"?

[@harrymattah418]

@@jursamaj primitive function. A function we know the derivative of. If you know one primitive, then all other primitives are equal up to an additive constant.

[@jursamaj]

@@harrymattah418 Oh, the indefinite integral.

[@harrymattah418]

@@jursamaj Indeed

[@Nolord_]

​@@jursamaj We say Primitive in french and Antiderivative in english.

[@eduardochappa4761]

if the second derivative is equal to zero the function is linear. If y' = y/x, then the derivative of y/x is 0 by the quotient rule. The function g composed with f is not defined at 0 because 0 is not in the domain of f, so you cannot evaluate y = g composed with f at 0. Instead you can take limit when x approaches 0 and that will give you the value of C.

[@sebastian8870]

That's the definition in Michael Spivak's

[@whyyat3470]

In most of Michael Penn's videos, there is a WTF? moment for me. This one was at @7:05. Second derivative of 0 indicates a quadratic. Um, what?

[@tcmxiyw]

My problem with using the ln as the inverse of exponentiation base e is the problem of exponentiation for irrational numbers. When the exponent is rational, exponentiation is straightforward. If the exponent is irrational, say a=sqrt(2), what is, say, 2^a? It’s okay to say approximate a with a rational number b and then 2^b approximates 2^a. The problem of saying there is a unique 2^a which can be so approximated involves at the Calculus I level a little hand waving. This problem goes away if we use the FTOC to define ln(x). Start off by acknowledging that exponentiation is defined only for rational exponents. By FTOC, ln(x)=int(1,x,1/t,dt) we have a function whose derivative is 1/x and whose domain is the positive real numbers (rational and irrational). Now develop the properties of ln: its range is all of R, it is increasing, it has the logarithmic properties, and it is one-to-one. Therefore ln has an inverse, exp, that is defined for all real numbers. The difference between exp(x) and e^x is that the latter is defined only for rational exponents. Now develop the properties of exp, including that it follows for all real numbers the laws of exponents that we are familiar with. Finally, show that for rational exponents exp(x)=e^x. It is now natural to extend the definition of e^x to include irrational exponents by defining e^x=exp(x). Exponentiation with other bases can be extended to irrational exponents by defining a^x=e^(x ln(a)).

[@tcmxiyw]

My second Calculus I teacher (college vs. high school) showed us this development. She started of by asking some questions: what is 2^3? What is 2^(9/7)? What is 2^pi? The last one generated a discussion that led into this application of the FTOC.

[@Bermatematika]

In my university we define the natural log first through the definite integral and then define e^x as the inverse of ln x. How do you define e^x without having ln x. What exponentiation a^x even mean?

[@thiantromp6607]

You can define it as its power series after you've shown that converges everywhere.

[@Bermatematika]

@@thiantromp6607 Ok. But usually power series comes later in Calc 2.

[@coldservings]

My first "serious" introduction to calculus was a book "Calculus the Easy Way". The "Calculus" I had in high school wasn't in any way serious--just things like "here's the power rule" and the like with no attempt to define why those rules apply. I mean, how can you have a calculus class where you don't even go into limits? Anyway, one of the chapters introduced the integral of 1/x, demonstrated that the result was a logarithm, worked out an approximation of the base, and declared it was a "natural logarithm" "because it arose naturally from our work." That was something like 40+ years ago. I still have fond memories of that book, whatever its weaknesses, because it gave me a big head start when I returned to school to study physics some years later.

[@byronwatkins2565]

All curves with zero curvature are lines, y=ax+b. The only complication is a second constant of integration multiplying x.

[@robshaw2639]

Today's Mathologer's vid (Anti-shapeshifters) shows some more really nice math going on here

[@JoelRosenfeld]

I like to show that ln(x) being defined via the integral gives the differential equation for e^x directly. We know already that ln(x) is continuous and strictly monotonic, which means that the inverse is also differentiable (via a theorem in Rudin). Call g its inverse, then ln( g(x) ) = x, and d/dx ln ( g(x) ) = 1/(g(x)) g'(x) = 1. Hence, g'(x) = g(x). Finally, since we know that ln(1) = 0 (by the integral definition), this means g'(x) = g(x) with g(0) = 1. The Picard theorem says there is only one function that satisfies that, and thus g(x) = e^x.

[@bobbybobsen123]

The best definition is the following: First, define any “log” to be a function from positive reals to the reals, which satisfies log(xy) = log(x) + log(y) for all x,y>0. Then ln(x) is the unique “log” function, such that ln(x) ≤ x-1 for all x>0. You can then define exp as the inverse to ln.

[@dogandonmez5274]

I just wanted to help Michael 🙂. What he has done is also correct of course!

[@not_vinkami]

If y"=0 shouldn't y be a linear function ax+b at most?

[@scottmiller2591]

As noted by others, the 2nd 1/2 of the proof is more complicated than it needs to be. But still, a pretty result. It got me thinking about how any function h(x) = 1/t + sum(delta(t,t_i)) could also be used as the definition of the ln - that is, you could introduce a finite number (probably even infinite number, but with some constraints) of finite deviations from 1/t, and still get a valid definition of ln(x). This, of course, is not invertible, but it might have some interesting applications to Poisson or other counting-type processes.

[@coolfreaks68]

Yes. You are right. I just checked. *log (2) = integral(@(p) 1/p, 1, 2, "ArrayValued",true ) = 0.6931*

[@MattMcIrvin]

I always thought of this as slightly mysterious, because the integral of x^(n-1) was always proportional to x^n but there didn't seem to be any sense in which ln(x) corresponded to "x^0". But you can actually get there that way. Use your choice of a constant of integration to take the integral of x^(n-1) that is equal to 0 at x=1, that is, f(x) = (x^n - 1)/n. ln(x) is now the limit of that as n approaches 0, from either direction. For positive real x, at least... x^0 is a constant, x^0/0 is nonsense, but you can think of ln(x) as the limit of what's left over if you subtract a constant from x^n and scale it up, in the limit of n going to 0. There's more! Suppose we try to invert that function and write x in terms of f(x). Now we have (1 + n f(x))^(1/n) = x Make the substitution f(x) -> y x -> g(y) n -> 1/m and we have (1 + y/m)^m = g(y) Now if we take the limit of g(y) as m goes to infinity, that's the famous "compound interest" definition of the exponential function. So, yes, the functional inverse is e^x.

[@MattMcIrvin]

(it actually all works for complex numbers too)

[@APaleDot]

Amazing comment. I've been missing this connection for so long.

[@raymondhusser2915]

Odd question here (a couple of them actually). If you do integration by parts letting u=1/t and dv=dt → v=t, du=-dx/t², you get ln(x)=1+integral (dt/t) meaning ln(x) = 1 + ln(x) or after another iteration 1 + 1 + ln(x)... Second question: where did the absolute value go and is this the reason we've got a problem?

[@bilalabbad7954]

6:57 that should be a linear function

[@adamwho9801]

I could swear that Micheal was in my graduate real analysis class.... but I know he didn't go to my school.

[@tsawy6]

This was the route my real analysis course took!

[@tsawy6]

My processor felt it strange to start with power series, after all, they start with a list of all the derivatives, so we used this for exp and the line integral of the half circle for cos

[@MasterHigure]

It is possible to define logarithms (in general) as the inverse of exponentials, show that 1/x has an antiderivative that is a logarithm, and then define e to be the base of said logarithm. I have seen authors do it this way.

[@admink8662]

You kinda make it hard with the quadratic, hehe.

[@ashotdjrbashian9606]

I watch almost all of Michael's videos just because they are interesting and he presents it very well. One unfortunate thing which is almost constant is that he makes mistakes in virtually every video. Yes, they're mostly "typos" but sometimes more serious mistakes happen. This video is one of those. Lots of people already pointed out that vanishing second derivative means the function is linear, not quadratic. I think in this particular case it would be only fair if he edits this video completely and removes the error.

[@beeble2003]

I don't think that's a _serious_ mistake. He claims it's quadratic and finds that the coefficient of the squared term is zero, which means the function is linear. That doesn't lead to any error -- it's just wasted time. But I agree, there always seem to be mistakes in these videos.

[@ashotdjrbashian9606]

@@beeble2003 This is not a serious mistake for a professional's point of view, but keep in mind, these videos are mainly for students. I still think he should edit it considerably.

[@seneca983]

@@ashotdjrbashian9606 Strictly speaking, he didn't say anything incorrect. What he said was that the function is at most quadratic which is stille true albeit less specific than what he could have said. He was still technically correct, the best kind of correct.

[@MrDynamite110]

These descriptions are killing me xD

[@rv706]

Fun fact: *the integral function of 1/x is not a primitive of 1/x.* (In fact, the integral function of 1/x is not even _defined_ on the connected component of R-{0} that doesn't contain the lower integration extreme because 1/x is not locally integrable around 0. Same thing happens with 1/x^2, cause it's not locally summable around 0) To get a primitive of 1/x on all of its domain, you have to define it by two different integral functions on the two connected components of R-{0}, one with negative lower integration extreme and the other with positive lower integration extreme.

[@cparks1000000]

What is a "noninverse counter-part"?

[@StuartSimon]

Larson, Hostetler, and Edwards define it this way.

[@GKinWor]

sheeeeeeesh thats valid af no cap 💯 💯

[@Bodyknock]

As a fun side note: Set f(x) = ∫1/t dt from t=1 to x f'(x) = 1/x Let g(x) = eˣ . Then 1/f'(g(x)) = 1/ (1/eˣ) = eˣ But recall (or prove as a lemma) that for any invertible, differentiable function F, that (F⁻¹)'(x) = 1 / F'(F⁻¹(x)). So if we substitute in f for F, we see 1/f'(g(x)) = g'(x) (since g(x) = g'(x) = eˣ) Which means that g(x) = eˣ satisfies the same equation as when you look at the derivative of the inverse of f(x).

[@MyOneFiftiethOfADollar]

The result you mentioned is easy to show using f(f^(-1)(x))=x along with with chain rule where it is simple to isolate inverse function algebraically. Fun exercise is to use the result to find derivative of sqrt(x) differently from the mindless exponent rule.

[@jamiewalker329]

More succinctly : Taking the integral definition above d/dx(lnx) = 1/x via fundamental theorem of calculus. Also, if f(x) = e^x then f'(x) = e^x = f(x). If we define g as the inverse of f, then we have g'(x)=1/f'(g(x)) = 1/f(g(x)) = 1/x, where we have used the property that for any function and it's inverse f(g(x)) = x. Hence d/dx(Lnx-g(x))=1/x-1/x = 0 And thus lnx - g(x) = constant. Since f(0)=1, we have g(1)=0, and according to the integral definition, Ln1 = 0 as the limits are both 1. Thus the constant is 0, and we must have ln(x) = g(x) as required

[@HagenvonEitzen]

In the first part, you showed that f(g(x))=x for all x in R. Then already g(f(x))=x for all x in the image of g (namely, if x=g(z), then g(f(x))= g(f(g(z)))=g(z)=x). So the second part could be replaced to showing that g is onto (0,infty). But that quickly follows from (1) g(x+y)=g(x)g(y), (2) g not constant, (3) g continuous

[@IsYitzach]

I saw this and said to myself, "I don't see why not, but why would someone want to do it?" The answer at the end being, "because we can and highlight methods."

[@carultch]

Historically, that is the way it was defined. Natural log was discovered before exponentials and the number e. The function e^x used to be called antilog(x), instead of exp(x) as we call it today. Today, we see e^x as the forward function and ln(x) as the inverse function, but that is only because there are a lot more applications where it is more intuitive to set it up that way, and it's more intuitive to introduce it to students that way.

[@Jack_Callcott_AU]

The Prof has shown that the the two definitions are equivalent‼

[@qillerdaemon9331]

Mathematicians really do live in a differently defined universe... 😯

[@tommyrjensen]

In high school we defined ln x as the integral from 1 to ∞ of dx/x, and e^x as the inverse of ln x. I have not watched the video in which he defined the exponential function. If he defined it in the same way, then it should take less work to prove what he wants.

[@EdenGranot]

This is actually the first definition of e that I saw, one can take e to be the unique root of ln(x)-1, to show that ln(ab)=ln(a)+ln(b) Then to deduce ln(e^n)=n For all integers n. Then for all integers m,n where m!=0 n=ln(e^n)=ln((e^(n/m))^m)=m*ln(e^(n/m)) So ln(e^q)=q For all rationals. By continuity ln(e^r)=r for all real numbers. By the chain rule we may conclude that (e^x)'=e^x And then to do the rest as in the video!

[@Achill101]

Our high school textbook defined ln(x) as the Integral over 1/x from 0 to x and defined exp(x) as the inverse of that logarithm: it worked well in the real numbers. For complex numbers, however, we introduced power series.

[@nikita_x44]

integral from 0 to x dt/t is ln(x)-1, not ln(x)

[@adayah2933]

@@nikita_x44 It's divergent you guys

[@Achill101]

@@adayah2933 - yes, you're right. I meant, ofc, the integral from 1 to x.

[@tommyrjensen]

Actually, even if I don't know what he thinks of as being the definition of the exponential function, it does not really matter. It is quite easy to show that f(x) (his integral) has the properties f(1)=0, f(xy)=f(x)+f(y) for all x,y > 0, and f'(x) = 1/x, from which it follows again very easily that the inverse function g satisfies g(0)=1, g(x+y)=g(x)g(y) , and g'(x)=g(x) for all x,y. If you already know that these are exactly the properties of e^x, then this is all a lot easier than what he does.

[@zakiabg845]

What's the value of "t" and its derivetive son we can calcul ln(X)?

[@douglasmagowan2709]

If y''=0 then y is quadratic? Anyway, when I learned this stuff, my prof defined the logarithm in terms of the given integral, showed that the result had the properties of a logarithm, and used that result to introduce the exponential function and its invariability under differentiation.

[@seneca983]

He needn't have included a quadratic term.

[@carultch]

Well really just a line. y" = constant implies that it is a quadratic. A line is a degenerate quadratic with the a-term equaling 0, which it did in this case.

[@JamesLewis2]

I was expecting a recap of the usual approach, showing that that particular antiderivative of 1/x has the properties of a logarithm, its inverse has the properties of an exponential, and their common base is a certain number between 2⅔ and 2¾; also, at the end, it would have been much easier if you had noticed that a had to be zero for the second derivative of y to be 0, or in other words that the form of a second antiderivative of 0 is bx+c.

[@rolfs5854]

Is the following proof possible: f(x) = e^x, g(y) = f^-1(y), d f(x) / dx = f(x) hence d g(y) /dy = 1 / f(x) = 1/y Hence the integral of 1/y is the inverse of exponential function because the if the derivatives of two functions are the same, the two functions are the same.

[@felipemagalhaes3678]

For me it's not clear that the integral from 1 to 0 diverges. Well, I know it's true because it's ln(x), but how it was trivial here?

[@micayahoo500]

The integral IS the definition of Ln (x).

[@Ligatmarping]

It's true he messed up with the quadratic polynomial, but is really need to point it out dozens of times? It isn't that bad, just makes it a little more complicated.

[@JorgeLuis-ts6qp]

Spivak I Love you

[@shimi994]

If y double prime is zero y is a linear function, not quadratic function.

[@martincohen8991]

I like the definition from 😮the functional equation f(xy)=f(x)+f(y). From this, assuming differential ability, it can be shown that f'(x)=f'(1)/x. The natural log is a solution with f'(1)=1.

[@bridgeon7502]

I wonder if we taught it like that, people would treat it like Ei(x) and Li(x)

[@mtaur4113]

Both take some theory to establish as well-defined on the positive reals. And of course more theory to show the two to be equivalent definitions.

[@savonliquide7677]

I think the second derivative part is not necessary : by y'/y= ln(y)' (witch follows from our local* definition of ln and does NOT involve the e fonction) we have at most one constant to deal with...) *however in my undergraduates studies (french system, my later studies were not about calculus fondations, so that I'm not aware of how other undergraduate student of other system were teached) i've never been teached that the définition of ln is the inverse of e, but the contrary! So that the " local" définition was presented to us as the main one. It is indeed more natural to me : what would be the définition of e? the only good candidate (if not the inverse of ln) is the Taylor développement, but ln(1+x) has a Taylor dev too so no reason for e " before " ln, from that point of view. We could define e^x as a particular a^x, where a is a real constant, that would be defined from the rationnal case but then why the constant e? the number e comes from the intégral definition of ln, althought it can be defined alternativally by anecdoctical séries or séquences. I m pretty sure that from any relèvent way that you take the problem it is at least shorter to define ln before e then the contrary!

[@matthewparker9276]

e is defined by the Taylor series, or more accurately the solution to the set of eqns f'=f f(0)=1 Evaluated at 1. But either approach is well motivated and equivalent.

[@Channel_98.6]

I think once you have shown f(g(x)) = x and both f and g are continuous and monotonic and that the domains/ranges are correct, a general argument from these facts would show that f and g are inverses of each other. I really enjoy this channel. But I rarely comment bc I watch from a TV.

[@edmundwoolliams1240]

I tried to do this proof years ago just using the Maclaurin expansion definition of e^x... Needless to say I got nowhere with it. I didn't realise the result would just pop out so easily just by taking derivatives!

[@kappasphere]

Doesn't the fundamental theorem of calculus just give you this equation (with extra confirmation of the constant C=0) if you know d/dx ln(x) = 1/x, which seems to be a given in this video? Just integrate both sides and you get ln(x) = int 1/x dx + C

[@adayah2933]

If ln(x) means the inverse of exponential, Michael is not using d/dx ln(x) = 1/x anywhere. He's defining a new function L(x) = int(1 to x) 1/t dt and then doing all the work to show that L(x) = ln(x).

[@FirstnameLastname-hg5gt]

I felt angry when I heard Michael at the 1.14 mark calling the definite integral "Antiderivative".

[@ayushmanchakraborty5744]

Second derivative of a polynomial is zero means the first derivative is C And the polynomial is x + C I think that is a mistake Michael Penn.

[@henryD9363]

Ummm, I don't know what compose or composed of means. I'm in a fog about it.

[@RexxSchneider]

In simple terms, it just means that the left-most function is taking the output of the function to its right as its input. Crudely, you can think of it as f( g(x) ).

[@videolome]

Respectfully. You seem to over complicate everything. The step of why the integral at zero is - infinity is not necessary and it is not justified. You don’t need to use the second derivative, and you don’t need to determine three constants. OMG Prove: (y/x)’=0. From this, you get y/x= c. Substitute x=1 and you get c=1.

[@Noam_.Menashe]

How did you get the quadratic?

[@TheOiseau]

He actually made a mistake (well not really, it still works out to a=0 in the end). If the second derivative is zero, then the first derivative is a constant, so the original function was a line, not a parabola.

[@MyOneFiftiethOfADollar]

If NOT valid, thousands of calculus textbooks, including Salas, are wrong. Salas rigorously showed integral definition of the logarithm satisfied the defining axioms of log functions.

[@MarsAnonymous]

I wonder if you could derive the value(s) of ln(-1) using that...

[@rshawty]

no because the integral would be improper at x=0 and you can prove the integral diverges

[@MattMcIrvin]

You can, but you have to make it a contour integral in the complex plane, and mind the branch cut on the negative real axis. Say, go in a unit semicircle from z=1 to -1. I think that will give you ln(-1) = i*pi if you do it right.

[@MattMcIrvin]

@@rshawty That's why you don't do it along the real axis.

[@rshawty]

@@MattMcIrvin oh okay nice, i didn’t know because I haven’t learnt contour integration yet

[@edmundwoolliams1240]

@@MattMcIrvin Which would only give you the Cauchy principal value

[@edoardodemichelis9319]

In order to say that 1/t integrated between 1 and 0 diverges, aren’t you using the fact that its antiderivative is log x? Is there another way of showing that it diverges?

[@beeble2003]

No, it's basically the same as showing that the sum of 1/n diverges as n->infinity. For t in (1/2,1], 1/t >=1, so the area under the curve there is at least 1/2. For t in (1/4,1/2], 1/t >= 2, so the area under that part of the curve is also at least 1/2. For t in (1/8,1/4], 1/2 >= 4, so we have another area 1/2 under the curve, etc.

[@juhisteri]

The LHS and RHS agree for x=1. Their derivatives agree for all x>0. They start from same place and move with equal speed. Done?

[@m.lebedev1493]

If you can show that exp is bijective then the 2nd part is unnecessary (i.e if f bijective and f(g)= id then g(f) = id)

[@Blackmuhahah]

7:00 y''=0 => y= ax+b, y(1)=a+b=1; y'=y/x=a=a+b/x=>(x=1) a=a+b =>b=0; => a+0=1 => y=x

[@audience2]

Here's an excellent video on the history of the natural logarithm in terms of it being the area under the hyperbola from 1 to x. kzbin.info/www/bejne/npLFeX5srLGen8U

[@Etothe2iPi]

"acheive" again! Maybe you should use the word "get" instead.

[@General12th]

Hi Dr. Penn! This explanation makes much more sense than Angel's. (I don't why he overcomplicates things.)

[@MarcelCox1]

You made the last part unnecessary complicated. In fact, y"=0 means y=bx+c. With your formula, y" would be 2a and not 0.

[@pranjalchaudhary4206]

Hey penn, you should try jee advanced paper from India. We 12 graders solve more tougher questions than in your videos in that paper.

[@joyboricua3721]

Alle Wige kommt nach log(x)

[@rayniac211]

2:00 Could you elaborate why the domain is only the positive real numbers? Is it because the integral cannot be evaluated over an interval that contains zero? While I wholeheartedly agree that this is a fun excercise and the mathematics introduced in this video is truly fascinating and educational, I can not for the life of me figure out why the natural logarithm is, (as many viewers have already pointed out) ACTUALLY DEFINED THIS WAY IN SCHOOLS! This kind of abuse on our young students should never be allowed to happen as it is a crime against humanity. This convention does a disservice to students by failing to promote a deep understanding of mathematical concepts. This kind of definition is something that you can come up with only AFTER you have studied the exponential function and the natural logarithm (defined as an inverse exponential!) and discovered that the derivative of ln(x) is 1/x. Honestly I think some mathematicians go through the trouble of coming up with these "cool definitions" because they don't want anyone else to understand. /rant

[@JamesLewis2]

It's done that way so it will be easier to explain arbitrary real exponents; FWIW, some textbooks start with exponentiation by rationals, then define real exponents as limits of exponentials of rationals, then use the definition of a derivative to show that if a^x has a derivative at x, it's a^x*lim((a^h--1)/h,h,0) and after showing that limit is an increasing function of a, define e as the value of a that makes that limit equal to 1. Generally, the value of "that limit" turns out to be ln(a), and that can be shown by proving it has the same properties as a logarithm. It's the reverse of the usual way, where a particular antiderivative of 1/x is shown to have the properties of a logarithm and then its inverse is shown to have the properties of an exponential.

[@rayniac211]

@@JamesLewis2 I get what you're saying. I still think it's better to start from the exponential function. This way it will not feel like something is handed down to you from high, and you can build your understanding from the ground up in a natural way instead of "Oh if you try this it just works". I can show you what I mean: You start out by accepting that there is no good definition (yet) for raising things to irrational powers. You then hope to extend the initial definition by some way, and just as you said, it will lead you to lim h->0 (e^h - 1)/h = 1. (which, by the way, naturally gives you: d/dx e^x = e^x) But then you can manipulate this expression to get to the famous limit: lim n-> ∞ (1+ 1/n)^n = e. (another "definition" that mathbooks love to throw at you!) You can then further show that raising this whole limit to the x power is actually the same as the limit: lim n-> ∞ (1 + x/n)^n. (just assume that x is rational for now) You can then start by expanding the binomial in the parenthesis and (after a lot of steps that I'm skipping here) show that the limit is actually: x^0/0! + x^1/1! + x^2/2! + x^3/3! ... Boom. You now have THE exponential function. Defined for all real numbers. (also for complex numbers later down the road) And you didn't even need to know anything about Taylor series! If you really want to, you could now prove that this infinite polynomial obeys all the old rules for exponentiation ( exp(x+y) = exp x * exp y, and (exp x)^y = exp (xy) ). I'm not going to bother because, after all, this all started from e^x, so I think it's fairly obvious this way. Since you now have a well defined expression for exponentiation, you can call ln(x) the inverse of this and call it a day. Later on in your calculus journey you can learn how to take the derivative of the inverse of a function. It's quite straightforward and you will find that because the derivative of exp(x) is itself, that will neatly lead to d/dx ln(x) = 1/x. I really think this is the more natural way because honestly ask yourself this: What do you care more about? Exponentiation and its inverse or calculating the area under 1/x? I'm not saying it's less work this way. Actually I think if you really do this carefully and rigorously, it's a lot more work. But it's a path that starts out from what you know and every step of the way you build all those "definitions" by yourself instead of receiving them from high.

[@JamesLewis2]

@@rayniac211, that is an excellent approach, but the way I was thinking about it was less on-high and more about the rabbit-hole that begins when you notice that the graph of 1/x has a well-defined signed area, and yet the power rule does not work for that case. An observation that is similar to yours that I would like to see made more often is about what the integral from 1 to x of t^r is for other real values of r, and what the limit as r→−1 looks like. (That is, it's (x^(r+1)−1)/(r+1), or after re-indexing, the question is about the limit of (x^r−1)/r as r→0.) It would help put the natural logarithm in context, rather than just looking like that one weird exception to the power rule; after all, the usual canonical antiderivatives for x^r, namely x^(r+1)/(r+1), do *not* approach a limit as r→−1. (On the other hand, that observation could be misleading, because it could imply that pointwise convergence implies convergence in measure, which it does not.)

[@rayniac211]

@@JamesLewis2 That limit is a completely novel way for me to look at things and it's interesting. Plugging in r = -1 for the canonical antiderivative gives you 1/0, which is bad :D However plugging in r = 0 to the limit of (x^r−1)/r as r→0.) gives you a 0/0-type indeterminate form, suggesting that an answer is somewhere out there. You just have to do more work to find out what it is. That stuff you said about pointwise convergence is something I know very little of. The wikipedia definition was... confusing :D

[@SGKdi]

5:58 y'=(1/x)*y ==> (x*y'-y)=0 ==> (x*y'-y)/x^2=0 ==> (y/x)'=0 ==> y/x = cste From limit conditions cste=1 ==> y=x

[@johnpaterson6112]

Your integral definition defines a function which you show to be equivalent to the natural logarithm, ie logarithm to base e. That is fine. But to say that that is an alternative definition is an abuse of language. Will you proceed to redefine the general "logarithm" in terms of your integral? It can indeed be done, but you will soon realise how silly it is. The redefinition trick should be confined to functions where the range of inputs is usefully expanded (eg factorials).

[@mathunt1130]

It's not only valid, but it's the definition!!!

[@jackychanmaths]

y"=0 => y=ax+b

[@primenumberbuster404]

Hahahahahahaha

[@jaimeduncan6167]

Have never seen it done before, pretty compact the only homework was the undefined was the exponential of the indefinite integral. Also as @dogandonmez5274 says the second derivative condition is stronger than Michael Penn assumed.

[@hra1465]

There is an easier way to show that g∘f (x) = x when f is injective and g∘f (x) = x: f [g ∘ f (x)] = f∘g∘f (x) = f∘g [f (x)] =f(x) => g ∘ f (x) = x.