At 11:00, there is the claim that e^(m/n) = e(m/n) for all ratoinal numbers; the proof until then only shows this for positive rational numbers. This is sufficient for the rest of the definition of e since 1 is a positive rational number, but to make it complete, the additivity rule gives us 1= e(m/n - m/n) = e(m/n)*e(-m/n) = e^(m/n)*e(-m/n), so e(-m/n) = 1/e^(m/n) = e^(-m/n) for positive integers m and n.
I've proposed this definition to other people. Considering that the reason we care about e is because the derivative of e^x is e^x, we should define it by that. Sure, it was originally discovered some other way, but that's just an argument from tradition.
I love to see how this is done rigorously. When teaching about the exponential function in school, I actually start with that differential equation as well because, at least for me, it is WHY this function is so important. Then we go on a bit of a goose chase what kind of function it could be, build the taylor polynomial and I typically act like "nah that's not right, we're going on forever and ever." And kids nod. Then we simply use a graphical idea to figure out it must be AN exponential (because let's be real nothing skyrockets like those) leaving us with the problem, what's the base? And that is where we use the "failed" Taylor series to "get an idea". Thank god for Geogebra xD you can set up 30 terms in 5 seconds and just plug in 1. Then just to see if we maybe are on the right track, square the result of T(1) and compare to T(2) with 30 terms you end up reasonably close. Of course I show them a proper introduction to Euler's number after. The classic instantaneously compounding interest. Aka "you think modern day credit sharks are bad?"
@jagatiello6900 : The problem with that is there already is a "[Ham] Sandwich Theory" in, I believe, topology (it definitely has strong applications in topological subjects), which posits that: For any shape, in any dimension, there is at least one unique "cut" that can be performed upon it that will divide said shape exactly in half Numberphile (if you watch that math-themed channel here on RUclips) has a video on it
I like this definition of e! I think this is conceptually easier than the usual definition in terms of the limit you prove here. Since understanding limits already implies some knowledge of calculus, your definition of e isn't really any more complicated than the usual one, which you were able to derive in any case. Good job!
I remember I showed the limit definition was equivalent to the differential equation definition back in 10th grade by using Euler's method to solve the differential equation for an arbitrary step size and then taking the limit as the step size goes to zero. I used to keep it on the wall above my desk along with other cool stuff I derived back in highschool lol.
I once had a math encyclopedia that defined pi as the smallest positive root of the unique solution f(x) to the IVP given by the second-order differential equation f''(x) = -f(x) with initial conditions f(0) = 0 and f'(0) = 1, which seems like your approach to defining e here. Of course, the solution to given IVP is f(x) = sin(x), and we all know that pi is the smallest positive root of the sine function, but this definition requires no previous knowledge of trig functions or even of circles! In any case, since the usual definition of pi involving circles is so simple, I was quite surprised that my math encyclopedia defined it this way instead!
That backflip (which he's also used in other videos) is amazing. Somehow it erases the part of the blackboard he wants erased and nothing else. Incredible! I don't know how he does it. 🙂🙃🙂 The only thing slightly off in this video is that there are too many "e"s. I lost count of how many times even Michael mixed them up and said, "e to the..." when he should have been saying "e of..." or "e evaluated at...". It would have been less confusing if he had picked a different function name.
This is how I always define Euler's constant and define exponentiation by this differential equation, and then define ln by the inverse of exponential. I know that is the opposite of the way the two functions were historically discovered, but it seems more natural to get the specific value of e as the solution to this initial value problem than it does to either define ln (n) as the integral of 1/x from 1 to n or to define it by the historical definition of property of logarithm F(xy) = F(x) + F(y), but then e is sot of hard to get out of that. But a function proportional to its slope makes every sense as something to define with real world applications, and then it's pretty clear that there would be a unique such function by setting y(0) to be 1, with all other such solutions of the differential equation being a multiple of that one.
Surely it's a whole lot simpler just to take f(x) = exp(x) - e(x) where e(x) is the solution of your DE. Then this has derivative 0 everywhere so must be a constant function. Since they have the same value at 0 they are equal everywhere. Your supposed proof has the problem that in the reals (as opposed to the complex numbers) there is no guarantee that the Taylor series converges to the original function. exp(-1/x^2) is infinitely differentiable at 0 and has every differential equal to 0. But the function is not the zero function.
Yeah, to complete the proof he would need to show that e(x) is analytic, but well, there are other ways to show that e(x) is always positive so it's not that big of a deal.
exp(-1/x^2) is undefined at 0 and has no derivatives there. You need to say {f(x) = 0 for x ≤ 0; exp(-1/x^2) for x > 0} It is fairly easy to show that e^x is analytic, so the proof is salvageable. But you are right it happens to be missing that step.
@@honourabledoctoredwinmoria3126 Define a function which is exp(-1/x^2) everywhere except x=0 where we define the function value to be 0. Then this function is continuous and differentiable at x=0 and the derivative at that point is 0. The derivative is continuous and differentiable everywhere and the second derivative is 0 at x=0. This continues for each successive derivative because the exp(-1/x^2) term dominates the value.
@@normanstevens4924 Yep. I see what you are saying. Michael Penn indeed missed a step. So what Michael Penn had to do to show that was not the case here is use Taylor's Theorem. Just knowing y'(x) = y(x), you can show the series converges everywhere to its value, while having no idea of said value, and since it has a nonzero value somewhere, it must not identically converge to 0.
In order to prove the solution is positive, one can consider the initial value problem g'=g , g(a)= 0 with unique solution g(t) = 0 and which has to coincide with y(t) in a neighbourhood of proving y(t) = 0
I have a problem for you all. The electric circuit represents a connected undirected graph without multiple edges, where edges (numbered N) are wires, and vertices are either light bulbs or the single power source. Each edge has a relay. A light bulb is lit if there exists a path connecting it to the power source, along which all relays are in the "on" position. It is known that exactly one of the relays is defective and never allows current to pass. You can toggle relays (and see if bulbs are lit). Initially, all relays are in the "on" position. Describe a method for finding the defective relay in O(N) toggle operations.
I've been using this definition of e for teaching for a while now. Although I couldn't go into this much detail for high school students. I'd usually start out with general exponential functions, and show that their derivatives are the same except for a factor. And with a few examples, it becomes evident that there has to be one exp function where the derivative is an exact reproduction.
e^(-1/x^2) has infinite radius of convergence centered at 0, but its taylor series is not equal to the function. I believe the argument at 2:30 is not quite correct.
You can prove that e(x) > 0 for all x without using the Taylor series. Here’s one way of doing it. Define f(x) = e(x)e(-x). Then f is differentiable everywhere, and by the product rule f’(x) = e’(x)e(-x) - e(x)e’(-x) = e(x)e(-x) - e(x)e(-x) = 0 where in the second equality I used the fact that e’ = e. So f’ is always 0, and thus f is constant. But f(0) = e(0)e(0) = 1. So f(x) = 1 for all x. Suppose there was an x such that e(x) = 0. Then 1 = e(x)e(-x) = 0e(-x) = 0, a contradiction. So e(x) is never 0. By the intermediate value theorem, it can never be negative either.
isn't there like a specific example of a function that has infinite derivatives which are all zero (edit: at some point) and so has a constant Taylor series but isn't constant also how exactly do negative numbers work with your proof? as in, i don't think you looked at the case of negative rationals
He did talk about negative numbers. If y(x) = y'(x), then either y(x) is 0 everywhere, or it is by necessity only defined for positive 0. The Taylor expansion he talked about is one way to see this, and another is to define the properties of exponentials that pop out of the differential equation and see that you get a contradiction in terms if you have y(x) ≤ 0. Note however that x itself can absolutely be negative. Just that if x is a positive number, y(-x) = 1/y(x), which is greater than 0.
@@honourabledoctoredwinmoria3126 i was talking about negative values of x. i see it's simple now (1=e(0)=e(x+(-x))=e(x)e(-x))but i think that should've been mentioned in the proof
@@bjornfeuerbacher5514 The Taylor series is identically 0 so has an infinite radius of convergence. However it does not converge to the initial function except at x=0.
So, I'm someone who went no higher than college-level algebra in terms of the upper limit of my accumulated math(s) skills, but after watching a lot of math content here on AdTube over the years, I now have some anecdotal familiarity and understanding of more-advanced math subjects. I've been thinking about this "e" constant recently, and a thought literally occurred to me the other day where I wondered that: If 0 and 1 are the additive and multiplicitive identities, respectively, might e be the exponential identity? This seems to indicate that it indeed is that
Re: proof of e(a + b) = e(a).e(b), can't we just differentiate f (x) = e (x + b) / e (b) Leads to f' (x) = e' (x + b) /e (b) = e (x + b) / e (b) = f (x). We also have f (0) = 1 Whence, by uniqueness, f (x) = e (x); i.e., e (x + a) / e (b) = e (x) => e (x + b) = e (x).e (b) => e (a + b) = e (a).e (b)
I like to think of limit (1+1/y)^y→e as y→∞. On my pocket calculator (1+1/1000000000)^1000000000=2.7182818271... is about as close to e as it can get without giving an error. Of course, I'm not a mathematician, so what do I know?
Dr. Penn is quite athletic. He's shown off his backflipping abilities in a few other videos. Also... AI? How? What program could possibly take in videos of Dr. Penn moving around and output a clip of him backflipping?
He's starting with the differential equation and then showing the solution has the same properties as e^x. This is why showing that e(x)-e^x=0 is so important.
Wow, the mathematician who flips himself like a coin, hoping to never land on head! Mike, you'd better quit doing that backflip, or one day you'll land on your head and ruin that great mathematical brain that you have. Before that happens, you'll probably make a video where you start with a backflip, then either calculate the probability that you would land on your head, or how much time it would take you to land on your head after repeated backflips. No, no, no, stop the madness!
07:15 a good thing to stop. If I want to see gymnastics, there are channels for that. Those channels showcase young women. I am okay with them showing curves without math.
🌟🌟To try everything Brilliant has to offer-free-for a full 30 days, visit brilliant.org/michaelpenn.🌟🌟
I tried to repeat the demonstration at home and ended up hitting my head on the floor.
eeeeee, that’s terrible!
7:17 Backflip
16:53 Nice Place To Be
22:37 Good Place To Stop
BACKFLIP!
Nice place to stop
Good place to be
DID HE JUST BACKFLIP?!?!??
WHHHHHATTT
Nope! He rotated within the xz plane :-)
@@jaafarmejri3361yz*
actually yz to us, xz to him
welcome to the real world, younglings
In French high schools, this is the definition of the exponential function. Existence is omitted, uniqueness is proven.
That backflip came out of nowhere.
At 11:00, there is the claim that e^(m/n) = e(m/n) for all ratoinal numbers; the proof until then only shows this for positive rational numbers.
This is sufficient for the rest of the definition of e since 1 is a positive rational number, but to make it complete, the additivity rule gives us 1= e(m/n - m/n) = e(m/n)*e(-m/n) = e^(m/n)*e(-m/n), so
e(-m/n) = 1/e^(m/n) = e^(-m/n) for positive integers m and n.
I've proposed this definition to other people. Considering that the reason we care about e is because the derivative of e^x is e^x, we should define it by that. Sure, it was originally discovered some other way, but that's just an argument from tradition.
This is the way the function exp is defined in French high school.
I love to see how this is done rigorously. When teaching about the exponential function in school, I actually start with that differential equation as well because, at least for me, it is WHY this function is so important.
Then we go on a bit of a goose chase what kind of function it could be, build the taylor polynomial and I typically act like "nah that's not right, we're going on forever and ever." And kids nod.
Then we simply use a graphical idea to figure out it must be AN exponential (because let's be real nothing skyrockets like those) leaving us with the problem, what's the base?
And that is where we use the "failed" Taylor series to "get an idea".
Thank god for Geogebra xD you can set up 30 terms in 5 seconds and just plug in 1. Then just to see if we maybe are on the right track, square the result of T(1) and compare to T(2) with 30 terms you end up reasonably close.
Of course I show them a proper introduction to Euler's number after. The classic instantaneously compounding interest. Aka "you think modern day credit sharks are bad?"
I love the squeeze theorem!
Here in Argentina we call it teorema del sandwich.
@@jagatiello6900 Way better name!
@@honourabledoctoredwinmoria3126 only if you speak Spanish!
@jagatiello6900 : The problem with that is there already is a "[Ham] Sandwich Theory" in, I believe, topology (it definitely has strong applications in topological subjects), which posits that: For any shape, in any dimension, there is at least one unique "cut" that can be performed upon it that will divide said shape exactly in half
Numberphile (if you watch that math-themed channel here on RUclips) has a video on it
@@shruggzdastr8-facedclown I don't know much about topology. I'll check up that vid. Thanks for the heads-up.
Backflip tutorial next?
michael penn never fails to do hard math
michael penn is the definition for hard math.
GothamChess menes never fail to invade other channels
I like this definition of e! I think this is conceptually easier than the usual definition in terms of the limit you prove here. Since understanding limits already implies some knowledge of calculus, your definition of e isn't really any more complicated than the usual one, which you were able to derive in any case. Good job!
OMG HE DID A BACKFLIP
7:17 - I was not expecting that. ^.^
I remember I showed the limit definition was equivalent to the differential equation definition back in 10th grade by using Euler's method to solve the differential equation for an arbitrary step size and then taking the limit as the step size goes to zero. I used to keep it on the wall above my desk along with other cool stuff I derived back in highschool lol.
I once had a math encyclopedia that defined pi as the smallest positive root of the unique solution f(x) to the IVP given by the second-order differential equation f''(x) = -f(x) with initial conditions f(0) = 0 and f'(0) = 1, which seems like your approach to defining e here. Of course, the solution to given IVP is f(x) = sin(x), and we all know that pi is the smallest positive root of the sine function, but this definition requires no previous knowledge of trig functions or even of circles! In any case, since the usual definition of pi involving circles is so simple, I was quite surprised that my math encyclopedia defined it this way instead!
That backflip (which he's also used in other videos) is amazing. Somehow it erases the part of the blackboard he wants erased and nothing else. Incredible! I don't know how he does it. 🙂🙃🙂
The only thing slightly off in this video is that there are too many "e"s. I lost count of how many times even Michael mixed them up and said, "e to the..." when he should have been saying "e of..." or "e evaluated at...". It would have been less confusing if he had picked a different function name.
This is how I always define Euler's constant and define exponentiation by this differential equation, and then define ln by the inverse of exponential. I know that is the opposite of the way the two functions were historically discovered, but it seems more natural to get the specific value of e as the solution to this initial value problem than it does to either define ln (n) as the integral of 1/x from 1 to n or to define it by the historical definition of property of logarithm F(xy) = F(x) + F(y), but then e is sot of hard to get out of that.
But a function proportional to its slope makes every sense as something to define with real world applications, and then it's pretty clear that there would be a unique such function by setting y(0) to be 1, with all other such solutions of the differential equation being a multiple of that one.
Black flip was siiiiiick.
From what I remember , that's how exponential function was first introduced to me in France !
If you want e^x be zero and negative, just use an analytic continuation….
that backflip transition caught me off guard
Surely it's a whole lot simpler just to take f(x) = exp(x) - e(x) where e(x) is the solution of your DE. Then this has derivative 0 everywhere so must be a constant function. Since they have the same value at 0 they are equal everywhere.
Your supposed proof has the problem that in the reals (as opposed to the complex numbers) there is no guarantee that the Taylor series converges to the original function.
exp(-1/x^2) is infinitely differentiable at 0 and has every differential equal to 0. But the function is not the zero function.
Sure, that's simpler but the whole point of the video is that we're defining exp(x), so we use it in this way to prove it
Yeah, to complete the proof he would need to show that e(x) is analytic, but well, there are other ways to show that e(x) is always positive so it's not that big of a deal.
exp(-1/x^2) is undefined at 0 and has no derivatives there. You need to say {f(x) = 0 for x ≤ 0; exp(-1/x^2) for x > 0} It is fairly easy to show that e^x is analytic, so the proof is salvageable. But you are right it happens to be missing that step.
@@honourabledoctoredwinmoria3126 Define a function which is exp(-1/x^2) everywhere except x=0 where we define the function value to be 0. Then this function is continuous and differentiable at x=0 and the derivative at that point is 0. The derivative is continuous and differentiable everywhere and the second derivative is 0 at x=0. This continues for each successive derivative because the exp(-1/x^2) term dominates the value.
@@normanstevens4924 Yep. I see what you are saying. Michael Penn indeed missed a step.
So what Michael Penn had to do to show that was not the case here is use Taylor's Theorem. Just knowing y'(x) = y(x), you can show the series converges everywhere to its value, while having no idea of said value, and since it has a nonzero value somewhere, it must not identically converge to 0.
In order to prove the solution is positive, one can consider the initial value problem g'=g , g(a)= 0 with unique solution g(t) = 0 and which has to coincide with y(t) in a neighbourhood of proving y(t) = 0
I have a problem for you all.
The electric circuit represents a connected undirected graph without multiple edges, where edges (numbered N) are wires, and vertices are either light bulbs or the single power source. Each edge has a relay. A light bulb is lit if there exists a path connecting it to the power source, along which all relays are in the "on" position. It is known that exactly one of the relays is defective and never allows current to pass. You can toggle relays (and see if bulbs are lit). Initially, all relays are in the "on" position. Describe a method for finding the defective relay in O(N) toggle operations.
I've been using this definition of e for teaching for a while now. Although I couldn't go into this much detail for high school students.
I'd usually start out with general exponential functions, and show that their derivatives are the same except for a factor. And with a few examples, it becomes evident that there has to be one exp function where the derivative is an exact reproduction.
Cool video! Can you do a version of this where you construct ln(x) from the integral of 1/x definition?
e^(-1/x^2) has infinite radius of convergence centered at 0, but its taylor series is not equal to the function. I believe the argument at 2:30 is not quite correct.
yeah i was in the comments exactly to see if someone solved that issue in the argument, tell me if u find a solution to that
You can prove that e(x) > 0 for all x without using the Taylor series. Here’s one way of doing it.
Define f(x) = e(x)e(-x). Then f is differentiable everywhere, and by the product rule
f’(x) = e’(x)e(-x) - e(x)e’(-x) = e(x)e(-x) - e(x)e(-x) = 0
where in the second equality I used the fact that e’ = e.
So f’ is always 0, and thus f is constant. But f(0) = e(0)e(0) = 1. So f(x) = 1 for all x.
Suppose there was an x such that e(x) = 0. Then 1 = e(x)e(-x) = 0e(-x) = 0, a contradiction. So e(x) is never 0. By the intermediate value theorem, it can never be negative either.
isn't there like a specific example of a function that has infinite derivatives which are all zero (edit: at some point) and so has a constant Taylor series but isn't constant
also how exactly do negative numbers work with your proof? as in, i don't think you looked at the case of negative rationals
exp (-1/x^2) if I remember correctly.
He did talk about negative numbers. If y(x) = y'(x), then either y(x) is 0 everywhere, or it is by necessity only defined for positive 0. The Taylor expansion he talked about is one way to see this, and another is to define the properties of exponentials that pop out of the differential equation and see that you get a contradiction in terms if you have y(x) ≤ 0. Note however that x itself can absolutely be negative. Just that if x is a positive number, y(-x) = 1/y(x), which is greater than 0.
@@honourabledoctoredwinmoria3126 i was talking about negative values of x. i see it's simple now (1=e(0)=e(x+(-x))=e(x)e(-x))but i think that should've been mentioned in the proof
For that function with infinite derivatives with value zero, the radius of convergence of the Taylor series is zero, if I remember correctly.
@@bjornfeuerbacher5514 The Taylor series is identically 0 so has an infinite radius of convergence. However it does not converge to the initial function except at x=0.
BACKFLIP 😮✌️👍
This is how we were taught about e (and exponential function) in first maths class at University. I was majoring Engineering Physics.
Michael Penn is now backflipping again, just like he used to... awesomesauce.
The easier version of the Sum for e should start at n=1 right ? Because otherwise you wold have to define 1/0! 😳
Dear Michael i have a question
Sum(k=1,Infinity){(Zeta(k+1)(-1)^k/(k+2)}
So, I'm someone who went no higher than college-level algebra in terms of the upper limit of my accumulated math(s) skills, but after watching a lot of math content here on AdTube over the years, I now have some anecdotal familiarity and understanding of more-advanced math subjects. I've been thinking about this "e" constant recently, and a thought literally occurred to me the other day where I wondered that: If 0 and 1 are the additive and multiplicitive identities, respectively, might e be the exponential identity? This seems to indicate that it indeed is that
1 is also the exponential identity, since x^1 = x. That said, e is an identity in a lot of other ways.
Re: proof of e(a + b) = e(a).e(b), can't we just differentiate f (x) = e (x + b) / e (b)
Leads to f' (x) = e' (x + b) /e (b) = e (x + b) / e (b) = f (x). We also have f (0) = 1
Whence, by uniqueness, f (x) = e (x); i.e., e (x + a) / e (b) = e (x)
=> e (x + b) = e (x).e (b)
=> e (a + b) = e (a).e (b)
Never mind the backflip... Did you notice the magically appearing right parenthesis at around 19:37?
I like to think of limit (1+1/y)^y→e as y→∞. On my pocket calculator (1+1/1000000000)^1000000000=2.7182818271... is about as close to e as it can get without giving an error. Of course, I'm not a mathematician, so what do I know?
backflip like
Do You not need to add either continuity or monotonicity to the conditions to ensure uniqueness?
Well, I guess you do by requiring diffentiability everywhere ☺️
Differentiability gives continuity and f'(x) > 0 ensure it's monotonic increasing.
If you take taylor series for granted, can't you just take the taylor series at e(0) and get e(x)=exp(x), with e=exp(1) being our familiar definition?
:Real Analysis Done Right.
Been watching this channel for years where tf did that backfill come from and why tf did he do it lmao. That must be AI right?
Dr. Penn is quite athletic. He's shown off his backflipping abilities in a few other videos.
Also... AI? How? What program could possibly take in videos of Dr. Penn moving around and output a clip of him backflipping?
@ 14:48 you mean c_n
It seems to me that everything is true because of the first observation and because of the claim.
Picasso
Nice
So the assumptions are: you know of e^x solves a derivative equation, now let me derive e^x from the solution??
He's starting with the differential equation and then showing the solution has the same properties as e^x. This is why showing that e(x)-e^x=0 is so important.
The taylor step isn't 100% legit
You have to prove the solution must be analytic
@@anshumanagrawal346 The function satisfies f'(x)=f(x) so it's infinitely differentiable. You don't have to prove it because it's assumed.
@@abebuckingham8198 Infinitely differentiable and analytic aren't the same thing. (At least for functions from R to R)
Wow, the mathematician who flips himself like a coin, hoping to never land on head! Mike, you'd better quit doing that backflip, or one day you'll land on your head and ruin that great mathematical brain that you have. Before that happens, you'll probably make a video where you start with a backflip, then either calculate the probability that you would land on your head, or how much time it would take you to land on your head after repeated backflips. No, no, no, stop the madness!
07:15 a good thing to stop. If I want to see gymnastics, there are channels for that. Those channels showcase young women. I am okay with them showing curves without math.