First, note that this formula is a DEFINITION in complex analysis, so this argument is not a “proof”. It is, however, a very nice explanation of why this definition makes sense. Note that the step at 2:24 isn’t exactly rigorous as it requires integration over a complex domain, which may not be well-defined.
@@thinkers6616 Because c is a constant, it will be the same value no matter what value any of the variables has. Putting in 0 for theta just makes the whole process easier, because you can easily get the value of c; inputting 0 eliminates many other parts of the expression. As an example of an unfavorable outcome, if you input pi, you get: z = cos(pi) + isin(pi) z = -1 + i*0 z = -1 -1 = c * e^(i * pi) Here, you can’t proceed to calculate c because you don’t know what e^(i * pi) is, as we haven’t proven the formula yet.
This is not a proof . You can't take an equation and prove it by it's equal . A linguistic example to more easily understand it is this : Q:What is the definition of Shame? A:The Shame you feel when someone shames you . This is false . You can't define a word by using the word because then it doesn't define anything it just says I exist because I exist
As pointed out by someone else, this is not really a proof, because you are assuming things about complex integration and complex functions that require justification. So what this really shows is the consistency of various operations that are familiar with real variables and functions. For instance, how do you define the natural logarithm ln for a complex argument z? (In fact, it is a multivalued function with infinitely many branches.) Similarly, how do you define the exponential function for a complex argument, and show that (locally, in some domain) it is the inverse of the logarithm? One way to do this is by solving the ODE dz/dt = z in the complex domain (for complex argument t). Then by integrating along a curve in the t plane you can prove that the solution is a holomorphic function of t, hence it has a Taylor series (which you can compute from the ODE); and after that you can set t = i theta and justify the rest of your argument.
Glad someone is still looking at this. Complex numbers aside, I think the integration needs to be ln abs z, because z = - 1 for theta = pi. I can’t see the justification for then dropping abs, to then have exactly what you want to show. Your thoughts?
@@212ntruesdale z is a complex number, and you really do need to integrate w.r.t. z (not the absolute value |z| which is real and non-negative) to get the complex logarithm, which is log z = ln |z| + i arg z ; the multivaluedness comes from the argument, which is only defined up to integer multiples of 2 pi. The Euler identity specifically concerns the case |z| =1, where the real part of the log is zero.
The only problem here is using logarithm in the complex setting which is a bit tricky, it’s better to just use the characteristic equation of the ODE to show that this is the solution and we know that such a linear ODE has a unique solution
I shared this with my high school students to lay the foundation for answering the question: "Is there a complex number z such that a nonzero real number raised to the power of z will equal zero?" Thank you for such a wonderful argument!
@@thelaststraw1467 I don't think I'm expert enough to give a hint without giving away my full argument. If you would like my full argument for why there is no complex exponent that will take a real number to zero, by all means, read on: Since any complex number can be expressed as a+bi, and exponents add under multiplication, e^(a+bi) = (e^a)(e^bi). e^a, with a being a real number, can never equal 0, so the question becomes: Is there a real number b such that e^bi = 0? If there is, then (e^a)(e^bi) will equal 0, regardless of what e^a is. Well, this is equivalent to solving the equation e^i(theta) = 0, with theta being some real number. Since e^i(theta) = cos(theta) + isin(theta), we now need to find a real number theta whose cosine *and* sine are both zero. That is the only way that cos(theta) + isin(theta) could equal zero, since i is a nonzero number. A quick check of the graphs of the cosine and sine functions, along with an understanding of their periodic nature, will verify that there is no real number whose cosine and sine are both zero. Therefore, there is no real number b, and consequently no complex number a+bi, such that e^(a+bi) = 0. I'm comfortable with assuming that, since no complex number a+bi exists for which e^(a+bi) = 0, then no complex number a+bi exists for which r^(a+bi) = 0, with r being any nonzero real number. This is not a rigorous proof, but in my opinion, it's good enough for a high school math class. Cheers!
If you're implying that it exists, then *you have taught your high school students something wrong!* There is no nonzero complex number w and complex number z such that w^z=0. [Here, as usual, we're defining w^z as exp(z ln(w)) for a fixed determination of the complex natural logarithm] This trivially follows from the fact that the complex exponential never attains the value 0.
@@rv706 Fear not! The answer to the question posed in my original comment was a conclusive "No!" Thank you for the elegant proof. My class was unfamiliar with the range of the natural exponential and logarithmic functions over the complex domain, so we were unaware of that fact.
To anyone confused about taking theta=0: If I give you the plot of a line y=x+C and you want to find C, you can just pick any point on the line to find C. But typically you'd choose the y-intercept (x=0) since it gives you the answer directly. Picking any other point would give the same answer, since it's the same line.
Nice, I'm only a little bit worried of complex integrals, because they usually behave very differently from real ones. And ln(z) can exist in many domains, the usual one is C\{Re(z)
Yeah this is not a proof. The manipulation with differentials is very informal, even if one interprets that equation as differential forms I don't believe it is correct as one side of the equation is a complex differential form and the other a real form. I don't see a way in which this argument can be mended, I think it should be thought more like an interesting algebraic manipulation that is not very correct but it leads you to a correct formula.
I'd say this is about as rigorous as the taylor series argument, but is shorter and is physically motivated from the context of solving ODEs of various physical systems. Nice work
This how my friend and I managed to convince ourselves it was true when we first saw the formula. This is the first time I've ever seen someone prove it that way and it makes me happy that we were right back then.
The whole thing is backward or circular. The problem arises because cos and sin are ill-defined in high school and at the beginning of calculus. To properly define sin and cos, you need to define them by Taylor series or by the exponential function or by solutions to differential equations. Here is how to do this properly and in the most elementary way. You define ln(x) = \int 1/t dt first. Then you define exp(x) as the inverse function of ln(x). At this point, you introduce the complex numbers and define cos(x) := (exp(ix) + exp(-ix))/2 and sin(x) := (exp(ix) - exp(-ix))/(2i). The Euler formula then comes out almost as the definition. There are alternatives, but they all come down to this: These functions are all solutions to the 4-th order differential equation (d^4/dx^4) f(x) = f(x). exp(i x), cos(x) and sin(x) are all solutions to this equation. The Euler formula is just a linear dependence relation of these three solutions over the complex numbers.
This doesn't strike me as a particularly rigorous proof. 1) the way the differential equation was "solved" by separation of variables would need more clarification, especially since it involves the complex differential dz/z. 2) the natural logarithm is only single-valued on a simply connected domain not containing the origin. Which domain do you choose? And what determination of the log? 3) the integration on the left-hand side is a different type of integration than the one appearing on the right-hand side. The first is the integration of a complex differential form along a path; the second is just the indefinite integral of a real differential. The dθ can be thought of as a (non-holomorphic) complex differential form too, but that shouldn't be skipped over so fast.
A very neat demonstration indeed. The problem that you have here though is that you still need to define what e^(it) is (using t instead of theta for writing simplicity in this comment), as an exponential with imaginary argument is not an obvious entity, and why you can assume it is the inverse operation of the complex natural logarithm. Instead, by defining e^x as the infinite polinomial 1 + x + x^2/2! + x^3/3! +... , proving Euler's Formula becomes the simple act of evaluating the polynomial when its argument is on the imaginary axis. Sure one has to rearrange the terms of an infinite sum, and there are rules to do that, but I'm not sure why you consider this less rigorous when these rules are satisfied. I would call it maybe more "technical" (even though there are hidden technicalities in your chosen path as well) but there are no issues about its mathematical validity
Direct proof with no calculus: Begin with e=(1+1/n)^n as n->infinity, from which follows that e^x=(1+x/n)^n as n->infinity. From this, first compute the modulus of e^(ix). |(1+ix/n)^n|^2=(1+x^2/n^2)^n->e^(x^2/n)->1 as n->infinity. This shows that the modulus |e^(ix)|=1. As for the argument of e^(ix), let u=arg(1+ix/n), then tan(u)=x/n, and hence, arg((1+ix/n)^n)=n*arg(1+ix/n)=nu=xu/tan(u), and since u/tan(u)->1 as n->infinity, it follows that arg(e^(ix))=x. This shows that e^(ix) is a complex number of modulus 1 and argument x. There is only one such number, cos(x)+isin(x).
I really like this! Wouldn't it be more rigorous to use the generalised solution for such a differential equation, where dz/dθ - iz = 0 z = Ce^(iθ) ? Or would that not work in the complex plane ?
I have 2 Math degrees & I'd have NEVER guessed this method exists as an alternative Euler's proof! U have amazing mathematical insight my friend....love the few adv Math tutorials u got here!!
The way I have seen it done before (using x for theta) is to divide both sides by e^ix. This gives you 1 = e^(-ix)*(cosx + isinx). Now just show that the right side is equal to one by taking its derivative, noting that the derivative is zero so the original expression is a constant, and doing the substitution x=0 to find what that constant is. There is the issue that you need to know that e^(ix) is never zero. That's true, but a more rigorous examination might want to show it.
If you have 2 math degrees then you should know this is a definition in complex analysis, not a proof, and that it lacks way to many details that are being glossed over
@@rv706 In practice I'm not. Work through the proof sketch I gave if you're having trouble understanding it. If you really want to be a stickler, yes, we have to know what functions we're talking about. However, the method I described is not intended to be a proof from first principles (or even a complete proof) and is understandable for anyone who has enough calculus to apply the chain rule and take derivatives of trigonometric and exponential functions with the understanding that i is a constant. It doesn't go so deep as to appeal directly to the definitions of those functions. Note that sin and cos in Euler's formula have real arguments, so a definition of those functions that supports non-real complex arguments is not required in any case.
I am curious, why do your have two math degrees?, I mean, did you study the same degree at two different universities or did you study two different programs but both math based?
I think you can’t integrate 1/z dz to ln z, because it’s not true for complex numbers. Even with the complex log it doesn’t work, I.e. at the branch cut, whereas 1/z integrated is holomorph on C without 0.
So how did you find that the indefinite integral of 1/z is ln(z)? How do you even define what integrating through the complex numbers means? Especially when you don't specify what curve you are integrating over?
As a civil engineer and reading safety navigation yournal for constructing elements in see/ocean crutial role was in this kind of Euler's formula. All fascilitates of the trigonometric form are now integrate with the Fourier transform which means it couldn't express wave forms of periodici transformation as simple super position of su wave form. This is exp. how many of us CAN APPLY ALL DICIPLINES TO GREAT PATH FOR FUTURE GENERATIONS. 👏
Can we really use natural logarithms to prove it? Shouldn't we first derive how e is related to ln(z) from first principles before using that fact to prove it?
And how do you apply that formula in designing Microwave circuits and other electronic design calculations, can you provide and example how it is used in the real word both in 'radians' and 'degrees'?
Thats great and everything but what are you drawing lines and numbers around? Can you extrapolate each function and explain it? Or will you just give me more numbers to gawk at?
The proof starts with the derivatives: but are these derivatives (deriv of cos is -sin; deriv of sin is cos) not proven by Taylor series we wanted to avoid ? I mean, are the Taylor series not hiding behind the formulas for derivatives of cos and sin?
What was the motivation for taking the first derivative of Z? You just did it without saying why. Taking the first derivative, then taking the integral later on, seems like some circular reasoning.
not easy to justify the switch of z and dtheta, especialy for "lower grades". I have used quite a similar approach at the begining but using the second derivative => z'' = -z So exp(i thetha) is a particular solution (in C) of this differential equation. General solution is z = Acos(theta) + Bsin(theta). Then use limit condition z(0) = 1 and z' = iz and you easily find A and B.
If I give you the plot of a line y=x+C and I want you to find C, you can just pick any point on the line to find C. But typically you'd choose the y-intercept (x=0) since it gives you the answer directly.
The euler formula can't be right. The left side, exponential equation, gives values from 1 to infinity by definition. The right side is an addition of 2 trig functions which give values from -1 to 1. Do you see the problem?
nice proof! altho during rearranging at 2:05 you divide both parts by z, so you have to check that z ≠ 0, which is actually true, cause |z| = 1 since it's a unit circle
i guess my next question is: what is the motivation for putting a number in the form cosθ +isinθ? i understand that polar coordinates and the complex plane are useful, but i still dont see where someone would come up with this form to represent something just by complete chance as a result of liking to use the complex plane.
The thing that I am curious about is not the proof. I mean it is easier to prove an identity to be correct when you know that it IS correct. How did euler come up with the identity is what I'm in for. Euler probably didn't randomly come up with an equation and went hey e^ix =cosx +isinx looks pretty correct, let's prove. Can you show me how euler came up with it?
Amazing ! But i feel a bit weird about taking the ln of a complex number... It's probably well defined and e^ln(z) = z being true as well, but I don't know much about it ? How does this work ?
log(z) is by definition, a complex number w, such that e^w = z. But you are right. This argument is not rigorous. We do not define what it means to differentiate or integrate over complex variables, which requires some complex analysis. It is a hand-wavy sketch aimed to give an intuition as to why this statement is true.
@@indigoselinger1640 You can't actually say that with certainty unless you have first performed some elementry complex analysis. It's good to hope that though as indeed is the hope behind this elegant intuative demonstration of what could be proven without Taylor's ugliness :-)
@@alphalunamare Yeah, idk why I said that. I do have experience in complex analysis and I should know that a lot of functions aren't 1-1 on complex numbers. This definitely wasn't before I did complex analysis... lol
How to prove Euler really depends on how the exponential function is defined. In this video it is defined as the solution of a differential equation, or more precisely, the unique solution to an initial value problem. Implicitly it takes the preparation of the existence and uniqueness of a (linear) ODE to justify this method. The proof using Taylor series also requires substantial preparation in Differential Calculus. It seems that at the level of high school math, proving Euler has a fundamental obstacle which is the very definition of the exponential function e^x. I am curious if there is a way to get around this, i.e., a proof without "advanced" mathematics.
It would be great to pair this with the visualisation that explains why d/dx(sin x) = (cos x) etc, by reference to a point moving around the unit circle.
So eventually, Taylor is unavoidable. Nevertheless, exp is the function equal to its derivative and such that the image of zero is one. If you accept that this function's domain can be extended to complex numbers, you have a good way to help students accept this strange and powerful definition, before power series are defined. As a teacher, I can't help introducing it with the idea of power series of exponential. But when I introduce complex functions, I point out this consistency between e^ix and the derivative.
1:04 how do we know the derivative of i? I know it’s a constant and that checks out algebraically but how exactly do you define the derivative of imaginary numbers?
Is just finding the slope at any point of the tangent line on an argand diagram, so I don’t think complex numbers mean anything. You have to imagine them as normal numbers otherwise they don’t make any actual sense. I might have said something stupid idk
Derivative of a complex number or any number for that matter doesn't make sense , as long as it's constant you can just put it out of integration or differentiation, differentiation will ork similar to other constants
Sure, but how would one know to do this if we didn't already know what we're supposed to obtain? The nice thing about the Taylor series method is that it moves forward without knowing what to expect...yielding the wonderful and surprising result!
You could start differently : Define e^(i x) as cosx + i sinx . If you differentiate cos x+ i sinx to get -sin x+i cos x = i(cos x + i sinx ) ,then you see that it makes sense to define e^(i x) this way.
But z is not necessarily real. There are places need to be justified, such as why the line integral of 1/z is lnz, and they are not justified. It is very hard to say the proof is rigorous.
This is quite definitely not a "proof", but it is a beautiful heuristic argument. Before we consider this identity, we need to know what complex exponentiation means. Traditionally we define exp(z) to be the Maclaurin series for eˣ extended to complex numbers z (since it converges for all complex z). Then we find that exp(z) satisfies some of the laws of exponents such as exp(z+w)=exp(z)exp(w), and say, let's define eᶻ=exp(z) for complex z. At that point, the Euler's Formula is meaningful and can be proved. Here you are essentially trying to replace the Maclaurin series definition of complex exponentiation by the differential equation definition. dy/dx=y with y(0)=1 is definitely a legitimate alternative definition of exp(x) for real x, and one can then prove that exp(x+y)=exp(x)exp(y) etc and that exp(x)=eˣ where e=exp(0), and I expect that this approach can be extended to complex analysis.
Yes it’s really a hand wavy argument and says essentially nothing. It’s certainly not a proof, but perhaps another reason why the definition of the complex exponential is what it is
The old became new. It's not proof. It's circular around itself. I did that in my homework, and my math professor gave me 0. But it's a good way to PROVE to some special people that "Euler's identity is right".
If you are do to it rigorously, put f(x)=exp(ix) and show it is solution of ODE f’(x)-if(x)=0 and then show g(x)=cos(x)+isin(x) does the same and leverage Cauchy unicity theorem.
"Proof" fails when doing the integral: until that point, you can say ok e^iθ is seen as a function from the reals to a two-dimensional real vector space (say, the Clifford algebra cl(0,1)) and we're just using knowledge from real analysis. However, the integration can only be a true complex integration as no integration domain is specified. Thus we can't conclude that the complex logarithm solves the integral, we would have to prove that without using Euler's formula (good luck). You could try to salvage by choosing a real path integral from 1 to z (both members of the 2D space), but it's not trivial as it has to be path independent, so you can't just pick your favourite path, you have to do it for an arbitrary path.
They would have different constants, so if you imagine the left integral having the constant K and the right integral hanging the constant R then C = R - K, do you see?
i was always annoyed with using the taylor series since its based on real values, but this really solidifies in my mind that this is without doubt a valid expansion of the exponential function. Amazing
Seems good, but have a step that without a good definition have no sense, and is when you take primitive to 1/z you say is ln(z) wich is not true, unless you define ln with a complex domine-codomine cause the equation lnx= ia € C; just have no solution for x defining ln(.) in the usual way.
The big problem with all the Euler's formula proofs is that this equation has very little to do with Calculus, even less with the Taylor series. It will stand even if Calculus was never invented. Unfortunately, almost nobody is aware of it.
Depends on how you define the exponential function in C. You can define it as the unique function whose derivative is itself and initial condition at 0 is 1, by power series, or by the limit definition. All of these definitions use calculus
I am not sure what you mean. The exponential function with a complex argument cannot be defined without the processes of calculus, or rather analysis (using limits of some kind). Even for real numbers this is true. For example, try defining 2^x for real values x that are irrational. It is impossible without limits: really this is the function exp(x ln(2) ). So the simplest way is to define the function exp and its inverse function ln (otherwise you have to mess around with limits of powers of rational numbers). Once you have done that, then exponents of all real numbers are defined, including the number e=exp(1). But you can't define the number e without a power series or a limit. (Without a power series, another way is to calculate the limit of (1 + x/n)^n as n tends to infinity, which gives exp(x) ; so this can be taken as a definition once it has been shown that the limit exists, and then it can be related to the other properties of exp.)
The integration is not so problematic, but rather the derivative done in the second line is doubtful since it requires the definition by the Taylor series.
my fav proof is: consider g(θ) and f(θ) with g(θ) = cos (θ) + i sen (θ) and f(θ) = e^iθ, initial conditions g(0) = 1 and f(0) = 1, deriving g'(θ) = i (cos (θ) + i sen (θ)), f'(θ) = i e^iθ we can see that g'(θ) = i g(θ) and f'(θ) = i f(θ) so i g(θ) = i f(θ), therefore g(θ) = f(θ) and e^iθ = cos (θ) + i sen (θ)
"dz/dθ" is a symbol or single notation and can't just be split for no reason. The only split is z from "d/dθ". If you want to apply a chain rule, then I think this should be made more obvious in this case.
First, note that this formula is a DEFINITION in complex analysis, so this argument is not a “proof”. It is, however, a very nice explanation of why this definition makes sense. Note that the step at 2:24 isn’t exactly rigorous as it requires integration over a complex domain, which may not be well-defined.
I think it's okay because the same formula is in the Wikipedia
@@krx3070 It’s most definitely okay, but it requires more justification than just adding an integral sign.
@@willassad8670 ya ik
@@willassad8670 why do we need to equate theta to zero? what will happen if theta is geater than zero? thanks
@@thinkers6616 Because c is a constant, it will be the same value no matter what value any of the variables has. Putting in 0 for theta just makes the whole process easier, because you can easily get the value of c; inputting 0 eliminates many other parts of the expression. As an example of an unfavorable outcome, if you input pi, you get:
z = cos(pi) + isin(pi)
z = -1 + i*0
z = -1
-1 = c * e^(i * pi)
Here, you can’t proceed to calculate c because you don’t know what e^(i * pi) is, as we haven’t proven the formula yet.
My mind has been blown. I had never seen this proof before and yet it's so elegant and simple.
@@judesalles arguments?
Rip your blown mind. We almost had a great mathematician 🥀😭
It's from Complex Analysis.
This is not a proof . You can't take an equation and prove it by it's equal . A linguistic example to more easily understand it is this :
Q:What is the definition of Shame?
A:The Shame you feel when someone shames you .
This is false . You can't define a word by using the word because then it doesn't define anything it just says I exist because I exist
Also we need to determine complex integral and complex functions like Ln and exp
As pointed out by someone else, this is not really a proof, because you are assuming things about complex integration and complex functions that require justification. So what this really shows is the consistency of various operations that are familiar with real variables and functions. For instance, how do you define the natural logarithm ln for a complex argument z? (In fact, it is a multivalued function with infinitely many branches.) Similarly, how do you define the exponential function for a complex argument, and show that (locally, in some domain) it is the inverse of the logarithm? One way to do this is by solving the ODE dz/dt = z in the complex domain (for complex argument t). Then by integrating along a curve in the t plane you can prove that the solution is a holomorphic function of t, hence it has a Taylor series (which you can compute from the ODE); and after that you can set t = i theta and justify the rest of your argument.
Shouldn't you only need to answer such questions for this exercise if your definition of real-value multiplication doesn't handle constants?
Sorry, not sure what you mean: this is about complex integration, not real multiplication.
Glad someone is still looking at this. Complex numbers aside, I think the integration needs to be ln abs z, because z = - 1 for theta = pi. I can’t see the justification for then dropping abs, to then have exactly what you want to show. Your thoughts?
@@212ntruesdale z is a complex number, and you really do need to integrate w.r.t. z (not the absolute value |z| which is real and non-negative) to get the complex logarithm, which is log z = ln |z| + i arg z ; the multivaluedness comes from the argument, which is only defined up to integer multiples of 2 pi. The Euler identity specifically concerns the case |z| =1, where the real part of the log is zero.
Lots of thanks
The only problem here is using logarithm in the complex setting which is a bit tricky, it’s better to just use the characteristic equation of the ODE to show that this is the solution and we know that such a linear ODE has a unique solution
Can you explain how you show it only has a unique solultion?
@@pqb0Picard's theorem on the existance and uniqueness of solutions of initial value problems
I shared this with my high school students to lay the foundation for answering the question: "Is there a complex number z such that a nonzero real number raised to the power of z will equal zero?"
Thank you for such a wonderful argument!
umm is there? cant seem to figure it out. any hints?
@@thelaststraw1467 I don't think I'm expert enough to give a hint without giving away my full argument. If you would like my full argument for why there is no complex exponent that will take a real number to zero, by all means, read on:
Since any complex number can be expressed as a+bi, and exponents add under multiplication, e^(a+bi) = (e^a)(e^bi). e^a, with a being a real number, can never equal 0, so the question becomes: Is there a real number b such that e^bi = 0? If there is, then (e^a)(e^bi) will equal 0, regardless of what e^a is. Well, this is equivalent to solving the equation e^i(theta) = 0, with theta being some real number.
Since e^i(theta) = cos(theta) + isin(theta), we now need to find a real number theta whose cosine *and* sine are both zero. That is the only way that cos(theta) + isin(theta) could equal zero, since i is a nonzero number. A quick check of the graphs of the cosine and sine functions, along with an understanding of their periodic nature, will verify that there is no real number whose cosine and sine are both zero.
Therefore, there is no real number b, and consequently no complex number a+bi, such that e^(a+bi) = 0. I'm comfortable with assuming that, since no complex number a+bi exists for which e^(a+bi) = 0, then no complex number a+bi exists for which r^(a+bi) = 0, with r being any nonzero real number. This is not a rigorous proof, but in my opinion, it's good enough for a high school math class. Cheers!
If you're implying that it exists, then *you have taught your high school students something wrong!*
There is no nonzero complex number w and complex number z such that w^z=0.
[Here, as usual, we're defining w^z as exp(z ln(w)) for a fixed determination of the complex natural logarithm]
This trivially follows from the fact that the complex exponential never attains the value 0.
@@rv706ok did you even read the explanation?
@@rv706 Fear not! The answer to the question posed in my original comment was a conclusive "No!"
Thank you for the elegant proof. My class was unfamiliar with the range of the natural exponential and logarithmic functions over the complex domain, so we were unaware of that fact.
To anyone confused about taking theta=0:
If I give you the plot of a line y=x+C and you want to find C, you can just pick any point on the line to find C. But typically you'd choose the y-intercept (x=0) since it gives you the answer directly. Picking any other point would give the same answer, since it's the same line.
Thank you very much.
lol how would this be this video's point of confusion
Because he didn't explain it as well as here, which is not confusing.
This is such a nice proof ! :D Thank you for explaining it so clearly and briefly :)
That’s such an elegant proof. Very good video!
Nice, I'm only a little bit worried of complex integrals, because they usually behave very differently from real ones. And ln(z) can exist in many domains, the usual one is C\{Re(z)
Yeah this is not a proof. The manipulation with differentials is very informal, even if one interprets that equation as differential forms I don't believe it is correct as one side of the equation is a complex differential form and the other a real form. I don't see a way in which this argument can be mended, I think it should be thought more like an interesting algebraic manipulation that is not very correct but it leads you to a correct formula.
I'd say this is about as rigorous as the taylor series argument, but is shorter and is physically motivated from the context of solving ODEs of various physical systems. Nice work
This how my friend and I managed to convince ourselves it was true when we first saw the formula. This is the first time I've ever seen someone prove it that way and it makes me happy that we were right back then.
The whole thing is backward or circular. The problem arises because cos and sin are ill-defined in high school and at the beginning of calculus. To properly define sin and cos, you need to define them by Taylor series or by the exponential function or by solutions to differential equations.
Here is how to do this properly and in the most elementary way. You define ln(x) = \int 1/t dt first. Then you define exp(x) as the inverse function of ln(x). At this point, you introduce the complex numbers and define cos(x) := (exp(ix) + exp(-ix))/2 and sin(x) := (exp(ix) - exp(-ix))/(2i). The Euler formula then comes out almost as the definition.
There are alternatives, but they all come down to this: These functions are all solutions to the 4-th order differential equation (d^4/dx^4) f(x) = f(x). exp(i x), cos(x) and sin(x) are all solutions to this equation. The Euler formula is just a linear dependence relation of these three solutions over the complex numbers.
This doesn't strike me as a particularly rigorous proof.
1) the way the differential equation was "solved" by separation of variables would need more clarification, especially since it involves the complex differential dz/z.
2) the natural logarithm is only single-valued on a simply connected domain not containing the origin. Which domain do you choose? And what determination of the log?
3) the integration on the left-hand side is a different type of integration than the one appearing on the right-hand side. The first is the integration of a complex differential form along a path; the second is just the indefinite integral of a real differential. The dθ can be thought of as a (non-holomorphic) complex differential form too, but that shouldn't be skipped over so fast.
Oh my god the most elegant beautiful démonstration it really chocked me
how do even calcute the integral of dθ dz
You can't take integral w.r.t. z and integral w.r.t. θ on both sides.... can you?
A very neat demonstration indeed. The problem that you have here though is that you still need to define what e^(it) is (using t instead of theta for writing simplicity in this comment), as an exponential with imaginary argument is not an obvious entity, and why you can assume it is the inverse operation of the complex natural logarithm. Instead, by defining e^x as the infinite polinomial 1 + x + x^2/2! + x^3/3! +... , proving Euler's Formula becomes the simple act of evaluating the polynomial when its argument is on the imaginary axis. Sure one has to rearrange the terms of an infinite sum, and there are rules to do that, but I'm not sure why you consider this less rigorous when these rules are satisfied. I would call it maybe more "technical" (even though there are hidden technicalities in your chosen path as well) but there are no issues about its mathematical validity
Direct proof with no calculus: Begin with e=(1+1/n)^n as n->infinity, from which follows that e^x=(1+x/n)^n as n->infinity. From this, first compute the modulus of e^(ix).
|(1+ix/n)^n|^2=(1+x^2/n^2)^n->e^(x^2/n)->1 as n->infinity. This shows that the modulus |e^(ix)|=1. As for the argument of e^(ix), let u=arg(1+ix/n), then tan(u)=x/n, and hence, arg((1+ix/n)^n)=n*arg(1+ix/n)=nu=xu/tan(u), and since u/tan(u)->1 as n->infinity, it follows that arg(e^(ix))=x. This shows that e^(ix) is a complex number of modulus 1 and argument x. There is only one such number, cos(x)+isin(x).
The second equation needs the first one, that you take for garanted. (actually it is an arbitrary definition). So it's a circular proof.
Very explanatory and concise video, good job!
Really nice! I am impressed how this proof came together!
Don’t use the expression “Taylor series approximation” . The Taylor series IS the function.
☹️ can you give me some tips ij studying differentiatial equations ?. I am struggling currently
I really like this! Wouldn't it be more rigorous to use the generalised solution for such a differential equation, where dz/dθ - iz = 0 z = Ce^(iθ) ? Or would that not work in the complex plane ?
omg that was awesome!! and your handwriting >>
this made more sense to me than eulers actual proof
I have 2 Math degrees & I'd have NEVER guessed this method exists as an alternative Euler's proof!
U have amazing mathematical insight my friend....love the few adv Math tutorials u got here!!
The way I have seen it done before (using x for theta) is to divide both sides by e^ix. This gives you 1 = e^(-ix)*(cosx + isinx). Now just show that the right side is equal to one by taking its derivative, noting that the derivative is zero so the original expression is a constant, and doing the substitution x=0 to find what that constant is. There is the issue that you need to know that e^(ix) is never zero. That's true, but a more rigorous examination might want to show it.
If you have 2 math degrees then you should know this is a definition in complex analysis, not a proof, and that it lacks way to many details that are being glossed over
@@bobbun9630: What definitions of exp(z), sin(z) and cos(z) are you using?
@@rv706 In practice I'm not. Work through the proof sketch I gave if you're having trouble understanding it. If you really want to be a stickler, yes, we have to know what functions we're talking about. However, the method I described is not intended to be a proof from first principles (or even a complete proof) and is understandable for anyone who has enough calculus to apply the chain rule and take derivatives of trigonometric and exponential functions with the understanding that i is a constant. It doesn't go so deep as to appeal directly to the definitions of those functions. Note that sin and cos in Euler's formula have real arguments, so a definition of those functions that supports non-real complex arguments is not required in any case.
I am curious, why do your have two math degrees?, I mean, did you study the same degree at two different universities or did you study two different programs but both math based?
I think you can’t integrate 1/z dz to ln z, because it’s not true for complex numbers. Even with the complex log it doesn’t work, I.e. at the branch cut, whereas 1/z integrated is holomorph on C without 0.
So how did you find that the indefinite integral of 1/z is ln(z)?
How do you even define what integrating through the complex numbers means? Especially when you don't specify what curve you are integrating over?
I had the same doubt
As a civil engineer and reading safety navigation yournal for constructing elements in see/ocean crutial role was in this kind of Euler's formula. All fascilitates of the trigonometric form are now integrate with the Fourier transform which means it couldn't express wave forms of periodici transformation as simple super position of su wave form.
This is exp. how many of us CAN APPLY ALL DICIPLINES TO GREAT PATH FOR FUTURE GENERATIONS. 👏
Can we really use natural logarithms to prove it? Shouldn't we first derive how e is related to ln(z) from first principles before using that fact to prove it?
Yes. We should. We have assumed it to be a given in the above video.
A lot of elegant proofs use the properties of ODEs thank you so much for showing me another one
I love it. I think it brilliant for bringing it home after the Taylor expansion series.
And how do you apply that formula in designing Microwave circuits and other electronic design calculations, can you provide and example how it is used in the real word both in 'radians' and 'degrees'?
Pl ease note that in your derivation, e^C is not equal to C. This could be misleading
If e^C was substituted by another arbitrary constant like A, it would be ok. But yes e^C is not C in itself
Yeah, our diff eq teacher always used K or some other variable during these integrals, and just made note that K = e^C
This is definitely the best way to prove and explain eulers identity
Thats great and everything but what are you drawing lines and numbers around? Can you extrapolate each function and explain it? Or will you just give me more numbers to gawk at?
Congratulatios!I never thought of that demonstration. Anyway is a proof that needs a higher mathematics.
2:28 shouldn't it be ln|z|, which results in 2 possible solutions for u, u=±e^(iθ)
Eullor proved his formula using sequences and series , and you proved the validity of his formula in a genius way. Well d
The proof starts with the derivatives: but are these derivatives (deriv of cos is -sin; deriv of sin is cos) not proven by Taylor series we wanted to avoid ? I mean, are the Taylor series not hiding behind the formulas for derivatives of cos and sin?
What was the motivation for taking the first derivative of Z? You just did it without saying why. Taking the first derivative, then taking the integral later on, seems like some circular reasoning.
Read the pinned comment. You cannot prove this formula in the first place- it is a definition. All such “proofs” are motivations for the definition
Your pinned comment is spot on. Thanks for the heads up.
from which book you found this explanation
not easy to justify the switch of z and dtheta, especialy for "lower grades".
I have used quite a similar approach at the begining but using the second derivative => z'' = -z
So exp(i thetha) is a particular solution (in C) of this differential equation.
General solution is z = Acos(theta) + Bsin(theta).
Then use limit condition z(0) = 1 and z' = iz and you easily find A and B.
3:08 why it has to be tetha equal 0? can we subsitute other number such as tetha equal 45?
You can
Because e^i0=e^0=1 and cos0+isin0=1+i0=1
We don’t know what e^i45° equals as we have not finished the proof
If I give you the plot of a line y=x+C and I want you to find C, you can just pick any point on the line to find C. But typically you'd choose the y-intercept (x=0) since it gives you the answer directly.
The euler formula can't be right. The left side, exponential equation, gives values from 1 to infinity by definition. The right side is an addition of 2 trig functions which give values from -1 to 1. Do you see the problem?
nice proof! altho during rearranging at 2:05 you divide both parts by z, so you have to check that z ≠ 0, which is actually true, cause |z| = 1 since it's a unit circle
i guess my next question is: what is the motivation for putting a number in the form cosθ +isinθ? i understand that polar coordinates and the complex plane are useful, but i still dont see where someone would come up with this form to represent something just by complete chance as a result of liking to use the complex plane.
The thing that I am curious about is not the proof. I mean it is easier to prove an identity to be correct when you know that it IS correct. How did euler come up with the identity is what I'm in for. Euler probably didn't randomly come up with an equation and went hey e^ix =cosx +isinx looks pretty correct, let's prove. Can you show me how euler came up with it?
Amazing ! But i feel a bit weird about taking the ln of a complex number... It's probably well defined and e^ln(z) = z being true as well, but I don't know much about it ? How does this work ?
log(z) is by definition, a complex number w, such that e^w = z. But you are right. This argument is not rigorous. We do not define what it means to differentiate or integrate over complex variables, which requires some complex analysis. It is a hand-wavy sketch aimed to give an intuition as to why this statement is true.
Since ln and exponents are one-to-one, inverse functions work on complex numbers without that problem.
@@indigoselinger1640 You can't actually say that with certainty unless you have first performed some elementry complex analysis. It's good to hope that though as indeed is the hope behind this elegant intuative demonstration of what could be proven without Taylor's ugliness :-)
@@indigoselinger1640 in the complex numbers, ln is in fact not 1 to 1
@@alphalunamare Yeah, idk why I said that. I do have experience in complex analysis and I should know that a lot of functions aren't 1-1 on complex numbers. This definitely wasn't before I did complex analysis... lol
How to prove Euler really depends on how the exponential function is defined. In this video it is defined as the solution of a differential equation, or more precisely, the unique solution to an initial value problem. Implicitly it takes the preparation of the existence and uniqueness of a (linear) ODE to justify this method. The proof using Taylor series also requires substantial preparation in Differential Calculus. It seems that at the level of high school math, proving Euler has a fundamental obstacle which is the very definition of the exponential function e^x. I am curious if there is a way to get around this, i.e., a proof without "advanced" mathematics.
Without « ln », we can say that e^z = e^itheta+C and C=1 because z(0) = 1
Amazing video, finally somebody not using the taylor series. Plus only in 4 minutes🙏🙏
Nice piece of self-defining circular logic, which is excellent Teaching practice, when you look at it holistically.
It would be great to pair this with the visualisation that explains why d/dx(sin x) = (cos x) etc, by reference to a point moving around the unit circle.
De Moivre’s before Euler knew expression z^n = r^n( cos(nθ) + isin(nθ) ). Euler much improved equation by introducing e exponential.
Mayn I thought this is gonna be a complex one, it's nice and elegant. Glad that I found your videos, love your content 👍, Jesus bless.
How do you define ln(z) for a complex number z?
I completely agree! This proof is circular. You would have to define the logarithm of a complex number, which you would do with Taylor series!
So eventually, Taylor is unavoidable.
Nevertheless, exp is the function equal to its derivative and such that the image of zero is one. If you accept that this function's domain can be extended to complex numbers, you have a good way to help students accept this strange and powerful definition, before power series are defined.
As a teacher, I can't help introducing it with the idea of power series of exponential. But when I introduce complex functions, I point out this consistency between e^ix and the derivative.
@@YannCoganwhat do you mean with the image of zero is one, I didn't get that.
@@wolfvash22 exp(0)=1
ln(z) = ln |z|+ i arg(z) where arg(z) = arctan( Imag(z) / Real(z)) over all domain in [0, 2pi]
1:04 how do we know the derivative of i? I know it’s a constant and that checks out algebraically but how exactly do you define the derivative of imaginary numbers?
Is just finding the slope at any point of the tangent line on an argand diagram, so I don’t think complex numbers mean anything. You have to imagine them as normal numbers otherwise they don’t make any actual sense. I might have said something stupid idk
Derivative of a complex number or any number for that matter doesn't make sense , as long as it's constant you can just put it out of integration or differentiation, differentiation will ork similar to other constants
Sure, but how would one know to do this if we didn't already know what we're supposed to obtain? The nice thing about the Taylor series method is that it moves forward without knowing what to expect...yielding the wonderful and surprising result!
beautiful and elegant proof.
You could start differently : Define e^(i x) as cosx + i sinx . If you differentiate cos x+ i sinx to get -sin x+i cos x = i(cos x + i sinx ) ,then you see that it makes sense to define e^(i x) this way.
It is good . But if we do not know that relation from the beginning?
But the proof using Taylor series starts from one side and reaches to the other
That was really awesome and faster than the other method
Proofs of function usaully involves ODE. Another example is: what f(x) subjects to f(a+b)=(f(a)+f(b))/(1-f(a)f(b))? The answer is tan(x).
In your 3rd line you replaced the minus sign with i^2, but couldn't you have equally replaced it with (-i)^2 and got a different result?
No. If you want to factor +2 from (-2)^2
Do you want to write it like (+2)(-2)?
Or 🐤
Do you want to write it like (+2)(+2)?
Very smart - Job well done, Will.
The proof is rigurous since theta is real. Thank you very much!
But z is not necessarily real. There are places need to be justified, such as why the line integral of 1/z is lnz, and they are not justified. It is very hard to say the proof is rigorous.
This is quite definitely not a "proof", but it is a beautiful heuristic argument.
Before we consider this identity, we need to know what complex exponentiation means.
Traditionally we define exp(z) to be the Maclaurin series for eˣ extended to complex numbers z (since it converges for all complex z). Then we find that exp(z) satisfies some of the laws of exponents such as exp(z+w)=exp(z)exp(w), and say, let's define eᶻ=exp(z) for complex z. At that point, the Euler's Formula is meaningful and can be proved.
Here you are essentially trying to replace the Maclaurin series definition of complex exponentiation by the differential equation definition. dy/dx=y with y(0)=1 is definitely a legitimate alternative definition of exp(x) for real x, and one can then prove that exp(x+y)=exp(x)exp(y) etc and that exp(x)=eˣ where e=exp(0), and I expect that this approach can be extended to complex analysis.
This argument reminded me of the chicken and egg dialectic.
Yes it’s really a hand wavy argument and says essentially nothing. It’s certainly not a proof, but perhaps another reason why the definition of the complex exponential is what it is
Won't there be different domain and definition for complex integral of Z
Super demonstration! Bravo!
wowww..... so easy , nicely explained. THANKS....
The old became new. It's not proof. It's circular around itself. I did that in my homework, and my math professor gave me 0. But it's a good way to PROVE to some special people that "Euler's identity is right".
This really helped Thank you :D
i did not understand why we did not take the absolute value of Z after integration of 1/Z
That was elegant. Thank you sir!
This should have been the first day of differential equations if not earlier. Sigh.... Well better learned later than never.
How do you know i is a constant term while doing derivative?
i^2 = -1 bro
Wow. Nice approach. Love and respect from Bangladesh 🇧🇩
If you are do to it rigorously, put f(x)=exp(ix) and show it is solution of ODE f’(x)-if(x)=0 and then show g(x)=cos(x)+isin(x) does the same and leverage Cauchy unicity theorem.
"Proof" fails when doing the integral: until that point, you can say ok e^iθ is seen as a function from the reals to a two-dimensional real vector space (say, the Clifford algebra cl(0,1)) and we're just using knowledge from real analysis. However, the integration can only be a true complex integration as no integration domain is specified. Thus we can't conclude that the complex logarithm solves the integral, we would have to prove that without using Euler's formula (good luck). You could try to salvage by choosing a real path integral from 1 to z (both members of the 2D space), but it's not trivial as it has to be path independent, so you can't just pick your favourite path, you have to do it for an arbitrary path.
wouldn't the integration of 1/z *dz produce a constant as well? so the 2 constants cancel each other out? help my math is terrible.
They would have different constants, so if you imagine the left integral having the constant K and the right integral hanging the constant R then C = R - K, do you see?
My favourite intuitive proof of Euler s theorem is that a curve with derivative that is always perpendicular is a circle
Can you find sin(1)
I am not able to find
Solution to cubic equation using cardano
Wonderful, so elegant, thanks a lot
i was always annoyed with using the taylor series since its based on real values, but this really solidifies in my mind that this is without doubt a valid expansion of the exponential function. Amazing
Simple and lovely proof.
Seems good, but have a step that without a good definition have no sense, and is when you take primitive to 1/z you say is ln(z) wich is not true, unless you define ln with a complex domine-codomine cause the equation lnx= ia € C; just have no solution for x defining ln(.) in the usual way.
The big problem with all the Euler's formula proofs is that this equation has very little to do with Calculus, even less with the Taylor series. It will stand even if Calculus was never invented. Unfortunately, almost nobody is aware of it.
Depends on how you define the exponential function in C. You can define it as the unique function whose derivative is itself and initial condition at 0 is 1, by power series, or by the limit definition. All of these definitions use calculus
I am not sure what you mean. The exponential function with a complex argument cannot be defined without the processes of calculus, or rather analysis (using limits of some kind). Even for real numbers this is true. For example, try defining 2^x for real values x that are irrational. It is impossible without limits: really this is the function exp(x ln(2) ). So the simplest way is to define the function exp and its inverse function ln (otherwise you have to mess around with limits of powers of rational numbers). Once you have done that, then exponents of all real numbers are defined, including the number e=exp(1). But you can't define the number e without a power series or a limit. (Without a power series, another way is to calculate the limit of (1 + x/n)^n as n tends to infinity, which gives exp(x) ; so this can be taken as a definition once it has been shown that the limit exists, and then it can be related to the other properties of exp.)
what a nice proof!! it helped me alot!!!
isnt the integration of dz/z = z+C so wouldnt it be e^(z+C1)=e^(iθ+C2)?
yes but he can join both C1 and C2 into C anyways, as he did turn e^C into C
ah alright, thanks
The integration is not so problematic, but rather the derivative done in the second line is doubtful since it requires the definition by the Taylor series.
can’t you just stop at line 2, factor out -i and then get dz/dx = iz and solve the DE to get e^ix?
Marvelous, Simply Marvelous.
Beautiful proof 🔝
i remember learning this in middle school… good times :)
my fav proof is: consider g(θ) and f(θ) with g(θ) = cos (θ) + i sen (θ) and f(θ) = e^iθ, initial conditions g(0) = 1 and f(0) = 1, deriving g'(θ) = i (cos (θ) + i sen (θ)), f'(θ) = i e^iθ we can see that g'(θ) = i g(θ) and f'(θ) = i f(θ) so i g(θ) = i f(θ), therefore g(θ) = f(θ) and e^iθ = cos (θ) + i sen (θ)
"dz/dθ" is a symbol or single notation and can't just be split for no reason. The only split is z from "d/dθ". If you want to apply a chain rule, then I think this should be made more obvious in this case.
This is called a separable first order differential equation. This is a valid method for solving them, at least in the first order case.
@@carultch It's a non-standard use which abuses the notation, some kind of explanation should be made even if it leads to a correct solution
This approach only proves that for any two functions f(x) and g(x) such that df/dx = -g and dg/do = f that e^(i×x) = f(x) !+ i×g(x). Scott McCaughrin
PHENOMINAL!
Very nice. Simple and direct