A better product rule?

[@davidgolden1607]

When I was sophomore in college taking ODEs, our professor gave us a pretty tough problem to work on as homework. When we turned in the HW, she asked if anyone had a solution. No one answered. She kept asking and finally I raised my hand, and told my answer which she said was correct. Then of course, she asked how I did it. I was very embarrassed to say that none of the techniques I knew worked, so I kind of guessed until I got something that worked. And our normally very calm professor got very excited and said 'Yes!! That's how mathematicians actually work. Guessing is an important part of math.' I can't tell you how much that impacted me to this day (many decades later). It was not only reaffirming, but gave me great insight into how to be a researcher. Prof. Penn, kudos to you for making that such an integral part of your teaching. You are modeling for your students how scientists and mathematicians do a lot of our work. I love these videos and applaud you for your efforts and approach to pedagogy.

[@akshatvats7992]

You sound like a software engineer.

[@az0rs]

that pun is nice

[@aqeel6842]

What was the problem?

[@zachbowles4516]

Pun intended?

[@xl000]

Nice story. But did it actually happen?

[@willyh.r.1216]

Indeed, Math should be taught through exploration by soliciting student's curiosity. Rules come later. Thank you Michael Penn.

[@feitan8033]

not really, rules are much easier to remember as a student

[@EpicMathTime]

Have you read Lockhart's Lament? Touches on this beautifully.

[@meccamiles7816]

Well put.

[@alvinlepik5265]

So, the process of exploration is something most of the class won't understand. They're explained only later what all of it meant? I can't get behind that idea at all. Sure, we can give examples of applications, but even then, often understanding the application requires knowledge of theory, as well. E.g how to solve very big systems of linear equations using Banach fixpoint theorem. You won't even understand what it means without theory. On a cynical note, I am getting quite tired of this mantra of "best way of learning something without putting in any serious effort". There is none.

[@nexusclarum8000]

@@feitan8033 It's not *purely* about remembering. It's about understanding. And I find the understanding of concepts is lacking.

[@sk8erJG95]

It's very refreshing to see another mathematician with such an exploratory style to their teaching. It's the best way to transfer your enthusiasm to the students!

[@paulkohl9267]

7:57 explanation of bottom up math research using examples that build intuition that lead to proofs, hell yes. Thank you Michael Penn for explaining that idea to the audience at large! Most have no idea what research is like, so they think it is like how it is in the textbooks with top-down axioms elegantly solving everything from the start; when instead one has blind alleyways of hypothesis testing that occasionally get rewarded with the exciting possibility of proof.

[@wafiklotfallah9951]

We could have also written the original equation as: (ln f)' + (ln g)' = (ln f)'(ln g)' Or ((ln f)' - 1)((ln g)' - 1) = 1 So take any piecewise continuous h(x) not equal to 0, and get: f = exp(int(1+h)dx) and g = exp(int(1+1/h)dx).

[@user-en5vj6vr2u]

You mean h(t) and integral dt right

[@wafiklotfallah9951]

Thanks for your comment. I had typos and corrected my comment above.

[@priteshsrivastava5850]

This channel is not just about finding answers but building an approach towards it . Great to learn🙏🙏

[@GalaxyGal-]

It’s so weird seeing my professors channel blow up like this.

[@GalaxyGal-]

Adam Romanov Both. I’ve seen him at the rock climbing gym with his kids before. I’ve also heard things about his pizza. I just graduated as a physics BS,l from Randolph College, so I had him for calc 2, differential equations, and linear algebra. He likes to do calisthenics in class and he ripped his jeans one time in an 8AM DE class.

[@adenpower249]

@@GalaxyGal- From my perspective you a very lucky to have such a professor.

[@mariomuysensual]

@@GalaxyGal- for real? You are lucky to have him as Professor

[@xanderabbey8529]

@@GalaxyGal- Oh my God, I feel so bad! He does sound like a legendary professor though. Wish I had him as my prof.

[@GalaxyGal-]

Aden Power He’s awesome. He has great chemistry with the other math professors from what I can tell. He got me roped into hagoromo too. It’s a shame the school’s social media don’t talk about his channel that much.

[@TheOneThreeSeven]

It's easy to assume everything is "nice" when working on a new mathematical idea, the hard part is to figure out the right idea to work on. In my experience the best way to find good idea's is to have a playful attitude and focus on learning something new instead of trying to prove something amazing.

[@easymathematik]

In this case "everything is nice" is okay. All you need is "differentiability" and "not divide by zero". So it´s okay.

[@TranquilSeaOfMath]

It is pleasant to see a video with an approachable differential equation in the discussion. I really like the last example. I enjoyed this video.

[@spideramazon5032]

Very beautiful math problem. Thanks for the class!

[@jeremyredd4232]

This is very functional calculusy! You should do some functional calculus videos. I'm enjoying your differential forms videos.

[@rileyjeffries1282]

I kinda cheesed this one. I let f=g and it was easy to discover that f and g could equal e^(2x) for the rule to be satisfied.

[@tracyh5751]

It might be fun to take your solutions for g=x^r and turn them into local power series solutions for analytic g.

[@armanrasouli2779]

I'm loving your channel dude...

[@jblaskovich8675]

Wonderful example. Thank you so much for posting.

[@LucaIlarioCarbonini]

I hope a video about what "nice" means might follow. Awesome video indeed!

[@duckymomo7935]

it's a very copout term in math. It is most likely referring to a smooth function (infinitely differentiable). In other contexts, it might be further restricted to analytic. A lot of functions that we by default work with happen to be smooth and analytic: polynomials, trig, exponential

[@LucaIlarioCarbonini]

I see @Adriano Andrade , the solution for g(x)=e^ax have a-1 at the denominator so a can't be 1

[@appleslover]

İt's a mathematician's orgasm

[@LucaIlarioCarbonini]

Playing with g(x)=ln(x) made my day, next try: g(x)=log_b(x)

[@easymathematik]

I don´t get you. It is obvious, what "nice" means in this context.

[@easymathematik]

This kind of product rule holds for all "nice" functions, if someone define the derivative * via f*(x) = lim h -> 0 (f(x + h) / f(x) )^(1/h) The motivation (for people who are interessted in such things) behind that is a "exponential slope". And f* can be written with the "classic" derivative ' via f*(x) = exp( (ln o |f|)´(x) ) With this definition we have: (fg)* = f* g* And what is "nice"? f is "classic" diff f is *-diff. And so on.

[@iang0th]

What is that type of derivative called, so I can look it up?

[@abdelazizm.7729]

@@iang0th Yes you can.. It is called the multiplicative derivative.. Every bijection or more generally a homeomorphism gives rise to a pair of calculi.. In this case, the bijection is the exponential function and it gives rise to multiplicative(or geometric) derivative and bigeometric derivative and respective integrals as well.. The idea to is to convert the usual four algebraic operations to a subset of the real numbers using a bijection then use the converted operations to construct these calculi and much more .. A wider scope is Differential Geometry .. However, you can look this up from papers and a book called Non-Newtonian Calculi by Michael Grossman and Robert Katz.. Several good papers were also published on this topic by Agamirza Bashirov.

[@jonaskoelker]

I'm conjecturing that the rule "(f + g)*(x) = f*(x) + g*(x)" doesn't hold in general. Win some lose some I guess. Also (e^x)* = lim [h -> 0] of (e^{x+h}/e^x)^(1/h). This expression equals (e^h)^(1/h) which equals e^{h * (1/h} = e which is independent of x and h. So (e^x)* = e for all x. This seems like the most expected thing. Conjectured corollary: if f*(x) = c for some c then x is an exponential function, hopefully even f(x) = c^x. Proof left as an exercise for you ;-)

[@xaxuser5033]

when u divide by g(x)-g'(x) then u automatically say that a fonction that satisfy g(x)=g'(x) is not a choice for your g's however u have either proof that is impossible to have such a g as solution or discuss the cases before dividing by g(x)=g'(x) to be more rigorous

[@R3lay0]

g=g' => f'g+fg' = g'(f+f') = g'f' => f+f'=f' => f=0. So g' can only be equal to g if f=0. And f can only be f' when g=0 for symetry reasons. I'm not very good at this so please tell me if I made a mistake.

[@R3lay0]

I'll also have to add that g=g'=0 is also possible (since I divided by g')

[@camilonaranjo9926]

I started to watch your videos last week, your channel is amazing dude :)

[@brooksbryant2478]

Me too

[@LorenzoClemente]

it is true for the Hadamard product of power series (term-wise product)

[@zackalexander4178]

Playing around and setting g(x)=sin(x) gives you a very similar result to when g(x)=cos(x). Both of these yield some really crazy looking graphs and Im wondering if you can explain why they look the way that they do, and how this ties into when the functions are "nice"

[@aljuvialle]

By division to (g' - g) you lost at least 2 special cases from g' = g. This kinda shown with g=exp(ax), but e^x is an impossible function here.it'd be worth a while to have these classes shown in this video. This gives at least g = 0 and g = g' != 0, in first case f can be any, and in the second we get f=f' (any form). WIthout even mentioning of this case, I'll get 0 points for a task because of division by 0

[@michaelpersiyanov6131]

I really wish my math class at the university was this fun!

[@danodet]

The differential operator u*=exp(u’/u) satisfy the identity (uv)*=u* v*.

[@joshrobo1000]

Superb video sir. Thanks for the entertainment.

[@rateeshk8175]

Constant functions are a simple solution to such hard differential equations

[@mesplin3]

Great video. One critique though. Affirming the consequent: If P then Q Q is true Therefore P I wished we checked the examples to verify that they actually do what they were designed to do.

[@mesplin3]

@@angelmendez-rivera351 let g(x) = exp(x)

[@mesplin3]

@VeryEvilPettingZoo that's fair. The algebra checks out. I think it's a good idea to just verify where the rule works. Like f and g need to be differential-able g' != g etc.

[@aryangarg3200]

"Freshman's dream" epic 😂😂😂

[@raphaeljacobs3518]

awesome video, michael!

[@VK1710]

Try taking f=sin x+cos x. May be a little nicer fn may occur

[@kbluedye]

"... and we know how to integrate the number one" Lol!

[@pullingrabbitsouttaahat]

Great. When I first saw this, I had a hunch that exponential was in it. I was right.

[@Grassmpl]

Did you integrate a hyperbola without inserting an abs value inside the ln?

[@3manthing]

Awsome video, as always. I study math at colledge, so it is really nice to hear some of the theorems again. It is really nice to see that your videos cover large variety of mathemathics. Now to be clear, i'm not asking for it, as it is a subject i dislike most in the field of mathemathics, but can we expect a video of more topological nature? If there are some already, excuse me for asking.☺

[@indocesare14]

I love Neil Patrick Harris teaching math. Really good teacher

[@davidbizzozero3458]

The simplest non-trivial solution I could come up with was the symmetric solution: f(x) = g(x) = exp(2*x) because 2 + 2 = 2 * 2.

[@KS-lb9uc]

Functions of the form C*e^(2x) also work, right ? It's weird to see how two different paths can lead to two different answers...

[@KS-lb9uc]

I mean, the equation is satisfied when both f and g are of this form.

[@VladYakovlev-ix8zi]

С*exp(-2*x) also fits

[@chaosredefined3834]

There is a case where the formula breaks, and we should check if it can be avoided. Consider g(x) = exp(x). This means g'(x) = exp(x), and our formula now involves dividing by zero. So, let's try finding the matching f(x) from scratch. f(x) g'(x) + f'(x) g(x) = f'(x) g'(x) f(x) exp(x) + f'(x) exp(x) = f'(x) exp(x) f(x) exp(x) = 0 f(x) = 0 Not a very insightful approach, but it is a result that wasn't covered in the original.

[@jamesrockybullin5250]

Is this part of series? This is incomprehensible to me, who has taken A-level maths at 18 years old. Please could you say where an introduction to this topic and notation is?

[@tracyh5751]

You will want to learn differential calculus, integral calculus, and differential equations.

[@maciejkozowski6063]

That's amazing.

[@Subro_Plays]

I really liked your content it really motivates me to know more about math😊😊😊 thank you for motivating me.

[@LorenzoClemente]

so much dedication to menial calculations

[@anshum1675]

8:53 Who else noticed that he said r times r to the x-1?

[@jeffburdette2025]

Nice video thank you 😊

[@mendezbcs20]

Is there a way to extend this "Freshman's Dream" to multivariate vector functions? I am trying to solve a matrix calculus problem and I was wondering if such an approach could be taken; that would actually solve my problem. Although my intuition says this is not possible due to incompatible dimensions and things getting pretty messy with the Kronecker product, I do not want to give up that easily :'(

[@rc210397]

Those typed brackets at 10:46 show how meticulous you are :D

[@Patapom3]

Amazing!

[@Patapom3]

Don't you have a patreon or something?

[@blackloop1861]

i already know the answer But watched the video tell the end nice videos

[@jultoflerber]

The anti-derivative of f'/f should be ln(|f|). The constant of integration c can be any real value, C=exp(c) is a positive constant. At the end you should find f(x)=+/- C•exp(integral of g'/(g'-g)) which is the same as having a real constant C... I know it's a technicallity, but it's a common redaction mistake. Great video anyway ! (Pardon my spelling, I'm from France)

[@SellusionStar]

This was a nice demonstration. I liked it a lot! Just one question: don't you need second parentheses on logarithms when putting outer factory into the exponents? Like ln((....)^1/2) and not ln(....)^1/2

[@ps200306]

I thought this too.

[@steffahn]

Not necessarily if you allow to apply functions like ln or sin without any parentheses. Which is actually done in the video, notice the sin 𝑥 everywhere. To elaborate, this would mean you interpret something like sin 𝑥² as sin(𝑥²). For squaring the result instead, you’d write (sin 𝑥)² or some people write sin² 𝑥 (but I don’t like this notation). This means that parentheses like in ln(𝑟−𝑥) are still needed, but only since ln 𝑟−𝑥 would mean ln(𝑟)−x, not because ln always needs some parentheses. Then ln(𝑟−𝑥)⁰·⁵ would be similar to the sin 𝑥² above and it would mean ln((𝑟−𝑥)⁰·⁵). But don’t forget, notation is not universal, everyone can do whatever they like. For example you can now debate whether to allow products inside the sin or ln without parentheses like sin 𝑥𝑦, or if that’s gonna mean (sin 𝑥)𝑦 instead. Also consider sin 𝑥 sin 𝑦. My personal take on these would be to disallow too confusing notations entirely and always use parentheses to disambiguate in such cases, for example in the last case either with (sin 𝑥)(sin 𝑦) or with sin(𝑥) sin(𝑦). If you think: Nested function application with sin, like interpreting sin 𝑥 sin 𝑦 as sin(𝑥 sin 𝑦), i.e. sin(𝑥 ⋅ sin(𝑦)), wouldn’t make much sense, yeah probably true, but at least with ln it does. The function ln  ln 𝑥 is something you actually need sometimes (I’ve seen it written in the form log  log 𝑛 in computer science [admittedly that’s not base 𝑒 anymore]), and it’s pretty nice to not be in need for all the extra parentheses of ln(ln(𝑥)).

[@SellusionStar]

@@steffahn you got a good point. I think here the problem is -as you said- not that it's a different notation, but that it is handwritten. Thus there is no such thing like consistent spacings or no spacing. sin sinxy might be sin x y and without parentheses here, you can't know for sure what is meant. I mean yes, in the context it is mostly clear what is meant, but the fact that it's handwritten makes this topic debatable in my opinion.

[@papapap2]

Im confused about what the value of this product rule. It seems to be circularly defined as in, it only holds if you choose f and g such that it holds.

[@jfwrg3461]

There is no direct value to my knowledge. The basic premise of the video is that many first year calculus students would suppose that the derivative of the product of two functions would simply be the product of their derivatives only to find it that there is a bit more to it than that! This video just examines what functions would work for that naive assumption.

[@TJStellmach]

It's only referred to as a "rule" by analogy to the actual product rule. The exercise was simply to find functions f and g where the statement holds true, knowing that it does not do so in the general case.

[@geosalatast5715]

Hahaha thats great!! I miss the old days when I was at school and and I wad playing with derivatives and integrals.

[@DrWizardMother]

Using a 'sloppier' but much shorter route, I find that any set of functions such that f = C(1/g)^x will give you this relationship. Basically, I rewrote the integrand of the integral he didn't solve in terms of g, dg, and dx. I get the integrand to be dx*dg/(dg-g), which is approximately dx*dg/(-g). This is a double integral. A solution is nominally x*ln(1/g) + C (there may be a more general class of functions also). Set this equal to the left side which is ln(f). Solve for f in terms of x and g. Then check to see if f'g +fg' = f'g'. It does.

[@chaosend3815]

This is awesome!

[@carrotfacts]

Why did you assume the anti derivative of 1/u is ln(u) and not the natural log of the absolute value of u? How would having the more truthful ln|u| affect the answer?

[@dehnsurgeon]

it's not more truthful - they both differentiate to the same thing; for x

[@ryanfreitas5703]

Ganhou um inscrito brasileiro Won a Brazilian subscriber

[@pb6270]

Are there any cool implications to this freshman's dream identity being satisfied? Even something geometric?

[@matron9936]

Nice simple video:)

[@alexanderandonov4683]

WONERFUL! I calculated for f(x)=(sin x)^2; => g(x)= C*exp(0,8x)/[(2cos x - sin x)^0,4] ;

[@ChanAlex]

integrating factor?

[@gamedepths4792]

This is soo awesome !!!!

[@aymanalgeria7302]

Great video

[@Varde1234]

Taking differential eqns right now, this is a lot of fun

[@klementhajrullaj1222]

And for (f/g)'=f'/g', how will proced?!

[@davidbrisbane7206]

Very good. Nice cadence.

[@omaralhafez5014]

Pretty nice, thank you

[@PauxloE]

I'm somehow missing after the examples the check whether those actually fulfill the freshman's dream product rule.

[@hocineslamene9135]

You must add two conditions: g(x) # 0 and g'(x) # g(x)

[@zunaidparker]

g(x)=0 satisfies the original problem trivially so is a valid solution. If g=g', then f(x)=0 is a valid solution to the original problem. Neither condition is a restriction of the problem originally posed, they're only restrictions on the specific class of solutions his derivation can find.

[@hindigente]

Neat exploratory video.

[@thomasfink2385]

At 10:24 he saved some chalk and a pair of parenthesies. It should have been ln( (r-x)^-r)

[@af9466]

Yes, but, by the way, the interesting thing is, that in our country such a "parenthesesless" form of logarithm, along with the other functions like trigonometric, is pretty much common. The only case we use parentheses here is when disambiguation is needed: ln(2x • y + 5z)

[@الرياضياتالعظيمة-ز6ك]

There's some thing messing in 4:55 the primitive of f'/f is Ln|f| with absolute value in logarithm.

[@vicktorioalhakim3666]

Here is an idea without integration. Suppose f and g are both differentiable and their derivatives are nonzero on all of their domains. Then: (fg)' = f'g + fg' = f'g' => f'g' = f'(g + (1/f')g') => g' = g + (1/f')g' => g' = [1/(1 - f'^-1)]g. Assuming that 1/(1 - f'^-1) = C is constant, then g(t) = e^{Ct}. Now, we solve for f: 1/(1 - f'^-1) = C => 1 = C - Cf'^-1 => f'(C-1) = Cf => f' = [C/(C-1)]f => f = e^{Dt} where D = C/(C-1). Both functions are nonzero on their domain and differentiable, so they satisfy the equation and it is easy to verify that (gf)'(t) = (e^{Ct}e^{Dt})' = Ce^{Ct}e^{Dt} + De^{Ct}e^{Dt} = e^{Ct}e^{Dt} [C + C/(C-1)] = e^{Ct}e^{Dt} [C^2/(C-1)] = Ce^{Ct}De^{Dt} = g'(t)f'(t) It would now be interesting how this can be extended by letting C be other functions of t, such as a polynomial, which has been brilliantly described in the video. It all amounts to integrating C(t) and C(t)/(C(t) - 1).

[@bigdhav]

Chalkboards are SO much better than the squeaky whiteboards we have now. My old maths teacher insisted on keeping it when the school changed them because he could draw near perfect circles on them!

[@onderozenc4470]

You could have equated both sides in f and g to a constant.

[@ingobojak5666]

Try g(x) = x, which also corresponds to r = 1 in your first example. Mild problems ensue...

[@matthewjames7513]

I had a very different dream as a freshman... (x+a)(x+b) = x^2 + ab

[@mathieugouttenoire9665]

3:40 you can't just divide by f(x) or g'(x)-g(x) ...

[@easymathematik]

On the board is the note: assuming everything is "nice".

[@pi4meterftw2]

It just means that what is derived is a source of examples, but not exhaustive. There could be other examples. One example not captured here by the final equation, for instance, is f = g = 0.

[@rccalytrix]

the "nice" idea- is it inspired by the "wouldn't it be nice?" videos by blackpenredpen?

[@benalkan8559]

"nice" here means that you assume that your function is sufficiently smooth on the domain of interest so that you can differentiate your function, and you're not worrying about g'(x)≈g(x). Not very nice functions f(x)=Weierstrass function (continuous yet non differentiable). Not differentiable anywhere. f(x)={ 1 if x is odd, -1 if x is even, 0 otherwise}. Not differentiable at those discrete jumps. f(x)=|x| near x=0. Not differentiable at x=0

[@omargaber3122]

Wonderfull board ,congeratilation

[@tryingtopredict134]

Satisfied with your videos Michael. I would like to point out something which you can improve that is just look towards camera while recording

[@bendeoliveira2768]

The title was kinda misleading though

[@bobh6728]

I think it is hilarious that he will pause to explain how he multiplied and moved terms to the other side of the equation (Algebra 1), but blow by things like obviously by some rule from Calculus 3, the result is .... Still great videos.

[@naimulhaq9626]

Interesting.

[@elai3147]

so basically freshmen can keep dreaming

[@leastmachine8693]

I'd just like to see some follow up proofs. I tried going through the algebra on the second example and, unless the a/(a-1) just disappears into the C, it doesn't pan out.

[@matthiasbergner8911]

Your formula for f(x) does not work when g' = g everywhere, i.e. g(x) = c exp(x). P.S. I do consider the exponential function to be "nice".

[@imacds]

Well yeah, in order to make g(x) = c exp(x) you would need f(x) = c exp(x/0)... I don't consider 0 in the denominator to be "nice". :P

[@bacon8979]

I suspect this is why he chose g(x) = e^(ax); he did note that they were carefully picked

[@hectormartinpenapollastri4557]

It's easy to see that f = constant is the only solution in that case. Not a big loss. You miss a trivial solution only.

[@maypiatt3766]

This was a Putnam question!!

[@maypiatt3766]

I don’t remember the question but it was from a long time ago 😊

[@photonicsauce7729]

@@maypiatt3766 Great!!

[@photonicsauce7729]

Would i get some free marks if i wrote the answer as: F(x) =0 or g(x) =0 or both

[@massimogirola5375]

It would be interesting to define a new kind of derivative that has this property, a bit like you do with the Levi-Civita connection to define the covariant derivative (ex. in General Relativity). Is it possible to introduce this kind of new derivative?

[@alexanderandonov4683]

this is not new kind of derivative; this holds just for special couples of funcs; it is like a differential equation;

[@TDSONLINEMATHS]

Nice

[@stearin1978]

V.I. Arnold said that Leibnitz actually defined the rule of product differentiation in that way. Possibly Arnold was joking as usually...

[@نعمللوحدة]

Great

[@gourabjitbiswas]

f(x)=c1 g(x)=c2 😁

[@malawigw]

d(you) divided by you

[@dustinbachstein3729]

Unfortunately, none of the three pairs of functions have a form such that anyone would use the product rule to take the derivative... :D

[@a.osethkin55]

These are private examples.. huh.. not prove((

[@JohnWick-xd5zu]

Ok good 👌

[@SillySussySally]

Saw 999 likes. Hit dat spicy button. Now 1k.

[@Grassmpl]

Are you using "C" for 2 different purposes? Not cool. You should change the name to "k" or something.

[@TJStellmach]

He changed "c" to "C," which I'll admit could have been clearer.

[@diarandor]

Dropping the absolute values in logarithms from integrals is just wrong. This had to be said.

[@skulliam4]

It's okay in this scenario because the plus/minus becomes part of C

[@거미남자_spidy]

if g(x)=BesselJ(alpha,x) Your brain will be overloaded :)

[@photonicsauce7729]

Whats bessel?

[@NarutoSSj6]

@@photonicsauce7729 i think its a herb

[@lionel0353]

Same

[@Flammewar]

PhotonicSauce Bessel functions are special kind function which are solving the Bessel differential equations.

[@거미남자_spidy]

ya. Diff equa x^2*y''(x)+x*y'(x)+(x^2-a^2)*y(x)=0 Solution is C1*J(a,x)+C2*Y(a,x)