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Personally, my favourite method is as follows: The diff equation y'' = -y has cosine and sine as obvious solutions, and obviously any linear combination of them as well. But we could also have taken exp(c•x), whose second derivative is c²exp(cx), with c some constant. If c²=-1, i.e. if c=i, this is also a solution. But how do we compute exp(ix)? Luckily, we don't need to, because by Picard-Lindelöf we know there is a two parameter family of solutions for this equation, already covered by the sines and cosines from before, so e^ix must be some combination of the two. Adjusting constants and noticing e^i0 = 1 and its derivative at zero is i, we reach the famous result. I love this method, because it is both intuitive and really highlights the connection between the trigonometric functions and the exponential. In fact, if you're more knowledgable in complex analysis, you'll know the relationship is actually backwards, and sine and cosine are actually combinations of complex exponentials and not the other way around
My favourite as well. Also, works off the way I define e^x: e^x is the unique function such that it is equal to it's derivative, and it''s value when x=0 is 1.
That last statement is debatable because in order to use the series for sine and cosine you need their derivatives and thus standard limits. Now where do they come from? I don't see a way to prove the limit sinx/x without basic sohcahtoa.
@@mathisnotforthefaintofheartIm not following you, I did not use the series for any function, neither exponentials nor trig functions. I did use their derivative rules, and both can be easily obtained with elementary math. The exponential is easy (that one in fact is easier if you do define it as its series, but it's not necessary), and for sin you need to do it through inequalities and the squeeze theorem. That being said, it does not come up for this proof, they could easily be taken as given without any circular reasoning The elementary proof would be coefficient comparison from their Taylor series of course, in fact I wouldnt be surprised if that comes up in the proof of Picard Lindelof, but to me the elegance of this method is that it gives a useful "why" for why theyre equal, it was not chance their taylor coefficients ended lining up like that
@@ClaraDeLemonMaybe I understood you wrong. Apologies. ok, here is what happened. When I did a second MA in math, my instructor DEFINED the sine and cosine as a difference / sum of complex e-powers. Form there he so called "proved" Euler's identity. And from there he proved the series. But when I asked where these definitions were coming from, he said, "they are just definitions". So when I asked why don't you define the sine and cosine vice versa, he said that it didn't have any purpose in the subsequent theory. When I told him that you didn't have any other choice to define them that way in order to make the theory "work", he got mad. This was the same teacher who claimed that L'Hospital Rule could be used to prove the limit sinx/x=1, even though I believe that is also circular reasoning. So we agreed to disagree. Now Euler may not always have been rock solid in his mathematical proofs, but the way how he came to Euler's formula is well accepted in modern days
Fun video. 🙂 One small thing, in method 3 there's a substitution of x=tan(θ/2) , which is fine but note that there's a restricted domain on it since tan(π/2) is undefined. So unlike the first two methods you might need to be a little careful handling the domain.
The domain is still R. Because, domain is defined as the values of x (independent variable) for which y(dependent variable) is defined, *without solving the equation, like it's raw!*
@@rajneeshmishra6969 Whenever you substitute one variable for another in an equation you have to also specify the bounds of the new variable. For instance, when you take an integral from x=0 to infinity, and then substitute in x=ln(t) in the equation, you need to recognize that the new bounds range from you need to note that that new bounds are from t=1 to infinity. Similarly in this case, by making the replacement x=tan(θ/2), you may need to recognize that θ is going to range from (-π to π) (or something similar but shifted). Otherwise you can end up working with resulting equations using θ and accidentally apply θ=π, for example, without realizing it can result in an undefined value there.
here's another way: set f(θ) = cos(θ) + i sin(θ) and take the derivative f'(θ) = -sin(θ) + i cos(θ) = i (cos(θ) + i sin(θ)) = i f(θ) which is a differential equation whose solution (with initial condition f(0) = 1) is f(θ) = exp(iθ)
well actually when we differentiating maybe e^iθ with respect to θ , we get ie^iθ when i is a constant , for sure i=root -1 is a constant but is it strict? since it involved with complex number
The problems of these proofs is how do you define each functions? What is assumed about each functions? For example, in the second proof, you are supposed to know that the derivative of exp(iθ) wrt θ is equal to iexp(iθ), but you first need to define the exponential of a complex number.
That's true, all of these proofs are based on assumptions about certain properties of the involved functions (in the complex plane!). Here's my take on this: Before working with those functions, you need to extend the definition of e.g. exp(x) from the real numbers to the complex numbers. Whatever way you do this, you want this extension to be natural, i.e. you want to retain as much of their essential properties as possible. For the exponential function, this could be the functional equation exp(x+y) = exp(x)*exp(y), or it could be the differential equation y' = y, or it could be the series expansion, or it could be Euler's limit definition for e^x. So you could take one of these properties to be the defining property of the complex extension and then use this property to prove Euler's identity without circular reasoning. Of course, you would still have to show that all the other useful properties still hold for the complex numbers and follow from your extended definition.
So good to see the return of the magic chalk squares as you progress through the problem. It's what got me into your videos in the first place. Long may they continue...
There is a way with the definition through the limit. e^(ix)=lim (1+ix/n)^n=lim [sqrt(1+x^2/n^2)] ^n * [cos(atg(x/n))+i sin (atg(x/n))]^n=lim [sqrt(1+x^2/n^2)]^n * [cos(n atg(x/n))+i sin (n atg(x/n))] =1 [cos(x)+i sin(x)]
The derivations using calculus are probably circular. That is, the derivative of e raised to an imaginary power is undefined until and unless Euler’s formula is derived. For example, taking the derivative of both sides of Euler’s formula yields Euler’s formula. But that only works if you already know that the derivative of e^i*theta = i*e^i*theta. The power series for the sine, cosine and exponential function are really fundamental to proving the formula without any circularity. I think that’s how Euler discovered his famous formula.
@@Alex_Deam Possibly. Recall that complex numbers arose from efforts to find the roots of polynomials. Polynomials only involve addition and multiplication operations. The Taylor power series for the exponential function involves only derivatives over the real numbers. After you find the power series for the exponential function over the reals it's only valid for reals. The next step is expand the argument to the complex numbers which is now possible because a power series is also a polynomial. But that step is really a definition of e^z since you only proved the power series is valid over the reals using the Taylor expansion. But then you also need to show the series converges for complex arguments to show the definition is valid.
@@Alex_Deam Weierstrass M-test applied to the Taylor series uses the absolute value of the terms or alternatively, the magnitudes of each term in the case of substituting a complex variable z in place of the real variable x. If the Taylor series converges x it also must converge for z. The problem with defining e^z by its derivative is that you first have to define complex differentiation. That’s not required in this definition because the Taylor series for e^x was created by taking the derivatives with respect to the real variable x. Complex differentiation is not required. Then after you have the Taylor series for e^x where x is real you replace x with z to define the meaning of e^z. And convergence is not an issue because it depends only on the magnitude of z, which is a real number.
15:44 we cannot set theta = 0 as , in one of the revious steps we had 2-2cos(theta) in denominator , so while reducing the fraction we had to do a assumption that theta is not = 2 n pie
z = cos(x) + i sin(x) = cos(xN/N) + i sin(xN/N) = (cos(x/N) + 1 sin(x/N))^N by de Moivre's formula which is proved by indiction and intuitively clear from the fact that the angles of complex numbers add up when multiplied. Letting N -> infinity: cos(x)+i sin(x) = (1 + ix/N)^N = e^(ix).
Try using rotation matrices: Look at R(theta) = R(theta/n)^n in the limit as n->inf. Show that in this limit you get the limit of (I+J*theta/n)^n =exp(J*theta). Where Z=xI+yJ is the matrix representation of the complex numbers. (I = diag(1,1) J = ((0,-1),(1,0))^T. R(theta)= ((c,-s),(s,c))^T c =cos(theta), s = sin(theta). [I'm using column of rows to represent square matrix, and column = row^T ... 'cause, YT comments don't allow LaTex input. ]
I love the complex logarithm way. You can derive the formula from the harmonic oscillator equation. The eigenvalues are complex and when you diagonalize you end up with a rotation matrix. This proves Euler formula.
I derived it in a similar way as the last one, but without making any initial assumptions about what the polar form should be. Take some complex number 𝑧 = 𝑎 + 𝑖∙𝑏. From the interpretation that a complex number is a point in the complex plane, we can verify that 𝑧 can be equivalently defined as the point that is at a distance 𝑟 from the origin 𝑂 and which has a line segment connecting it to the origin forming an angle 𝜃 with the positive real axis measured in the counter-clockwise direction. 𝑟 and 𝜃 will be called the polar coordinates of 𝑧 and it follows that 𝑧 = 𝑟(cos 𝜃 + 𝑖∙sin 𝜃) from trigonometry. Similarly, we can take a second complex number 𝑤 = 𝑠(cos 𝜑 + 𝑖∙sin 𝜑), where 𝑠 and 𝜑 are the polar coordinates of 𝑤. Then we can calculate the product 𝑧𝑤 = 𝑟𝑠[cos 𝜃 cos 𝜑 - sin 𝜃 sin 𝜑 + 𝑖∙(cos 𝜃 sin 𝜑 + sin 𝜃 cos 𝜑)] = 𝑟𝑠[cos (𝜃 + 𝜑) + 𝑖∙sin(𝜃 + 𝜑)], by using trigonometric identities, which has polar coordinates 𝑟𝑠 and 𝜃 + 𝜑. This fact motivates us to represent a complex number in polar form as 𝑧 = 𝑟∙𝑢^𝜃, where 𝑢 is some yet to be determined complex number, such that 𝑧𝑤 = 𝑟𝑠∙𝑢^(𝜃 + 𝜑). To determine 𝑢, we can take the derivative of 𝑧 w.r.t. 𝜃 in both the rectangular and polar forms: ∂𝑧/∂𝜃 = 𝑟(-sin 𝜃 + 𝑖∙cos 𝜃) = 𝑟∙𝑖∙(cos 𝜃 + 𝑖∙sin 𝜃) = 𝑖∙𝑧; ∂𝑧/∂𝜃 = 𝑟∙ln 𝑢∙𝑢^𝜃 = ln 𝑢∙𝑧. Therefore, ln 𝑢 = 𝑖 ⟺ 𝑢 = 𝑒^𝑖. In conclusion, it was shown that a complex number 𝑎 + 𝑖∙𝑏 can be equivalently represented in polar form as 𝑟∙𝑒^(𝑖∙𝜃) and Euler's formula was automatically proven along the way.
Yes I do have another method, which I discovered before I had ever heard of Euler. It is so unique that it was published in a Mathematical Journal because I had discovered Euler's equation before I had learned of its existence. Actually I was searching for any meaning to the logarithms of negative numbers. After discovering the formula, I kept quiet about it because it seemed so weird that I did not understand what it meant. However, after I had learned of Euler, I sent it to the Journal as a unique way to arrived at his equation. I would happily reproduce it here, except that I have not the ability to type Maths, but it can be found on the internet by searching M500 and scrolling to page 24. (It is important that you scroll to page 24, because the fast track way to get to it is 2 pages out of phase.)
Here's a method I learned that is very similar to your METHOD 2: Let f_1(x) = e^{ix} and f_2(x) = \cos(x) + i\sin(x). Both of these functions satisfy the differential equation: F' = iF, and the initial condition F(0) = 1. Therefore, by the existence-uniqueness theorem of first-order, homogeneous, linear ordinary differential equations, the two functions must be identical.
I know of a 5th way! It uses the close approximation of theta and sine (and by extension tangent) and the limit form of e^(r). Pretty lengthy but decent enough for a PreCalculus class
From De Moivre’s Theorem (cosx + isinx)^n=cos(nx) +isin(nx), let t=nx, then (cos(t/n)+isin(t/n)^n = cos(t) + isin(t) Now let n be big… approximate cos(t/n) by 1+(t/2n)^2 which goes to 1 as n gets big and sin(t/n) by t/n… Then cos(t) + isin(t) = limit as n goes to infinity of (1+it/n)^n which is, of course e^(it).
Consider z = cos θ + i sin θ, so that dz/dθ = -sin θ + i cos θ dz/dθ = i(i sin + cos θ) dz/dθ = zi dz/z = i dθ By integrating both side we get ln z = θi + C z = e^(θi + C) z = Ce^(θi) cos θ + i sin θ = Ce^(θi) Substitue θ = 0 to get rid of the C cos 0 + i sin 0 = C 1 = C Therefore cos θ + i sin θ = e^(θi)
I like to do it by showing the following: Let f(x) = exp(i*x) Let g(x) = cos(x) + i sin(x) Proceed to show that f'''' = f, f''''' = f', f'''''' = f'', f''''''' = f''. Show that the same holds for g. Then show that f(0) = g(0), f'(0) = g'(0), f''(0) = g''(0), and f'''(0) = g'''(0). Then observe that this means that f(x) and g(x) are not only equal at x=0, but all of their infinite derivatives are also equal. Thus f and g must be equal everywhere.
A simple way of showing that Euler's equation is correct is to define a function y(x) = cosx+i*sinx. If you differentiate y, you end up with a differential equation dy/dx = i*y. If you separate variables and integrate, you get ln(y) = i*x+C1. If you exponentiate both sides, you get y(x) = C2*exp(i*x), where C2=exp(C1). If you evaluate both forms of y(x) at x = 0, you find C2=1, and Euler's equation is obtained.
Hi, I've been thinking about this problem for quite a long time, but it seems to me that there is another way of doing that. If the Euler's formula is true, then you have to proof that ix is the logarithm of cos x + i sin x. The whole problem is to define the complex logarithm without using Euler's formula. Why not use the definition for a positive real, and extend it for a complex number? For a positive real number : ln x = int_1^x {dt/t) . Let's say that, for a complex number : ln z = int_1^z {dw/w) , where w is the dummy complex integration variable. As the function 1/z as only a pole in z=0, the result of the integration will not depend upon the path used, as long as you don't cross the "cut direction" y=0 and x0, and I found the expected result: ln (|z|) + i arctan (y/x) . Once again : without using the Euler formula, only by integration of 1/z . And for x
Another way: Let z=cosθ+isinθ Then dz/dθ=-sinθ+icosθ=iz So dz/dθ=iz Obviously z=C×e^(iθ) And by plugging 0 into the original function for z we get that C=1 Therefore cosθ+isinθ=e^(iθ)
Foundation of these proof assume we have d(e^i theta)/ d theta = i e^i theta But behavior of i in exponentiation is undefined here. Can you share why do we have that?
I've always struggled to understand how this formula makes any sense but when you mentioned that exp(i*theta) is a complex number so it has to have a polar form... suddenly it was so obvious!
@@bluelemon243 Perhaps the nicest way of defining it is with power series. You define e^x = sum_{n>=0} x^n/n!, then using Hadarmard's theorem, you find that this has an infinite radius of convergence, then you know that for analytic functions, their derivative is the component-wise derivative, so you get that d/dx e^x = e^x. This is how Cartan defines the exponential function on C in his complex analysis book. You then just define cos and sin in terms of e^ix and show that they really are what you think they are.
My proof is pretty short. Define a function f(t)=cos(t)+i*sin(t). Take the derivative of this function by taking the derivative of real and imaginary part separately:: f'(t)=-sin(t)+i*cos(t) By definition i*i=-1, thus f'(t)=i*(i*sin(t)+cos(t))=i*f(t). But this a linear differential equation, and these can only have solutions consisting of sums of exponential functions, so in this case f(t) must be of the form f(t)=C*exp(k*t), where C and k are constants. Plugging this into the differentiial equation you'll find that k=i, and because of the border condition f(0) = cos(0)+i*sin(0) = 1, therefore C=1. QED. In case you're not convinced you can explicitly solve the differential equation by separating the variables and integrating both sides: df/dt=i*f => ∫df/f = ∫i*dt => ln(f) = i*t+c => f(t)=exp(i*t+c) Because f(0)=1, c=0.
How do we know that i behaves like a real constant wrt complex derivatives and integrals. We all know it does, but isn't that an assumption that needs to be proved?
Because -1 is also a constant. There are only two numbers, that when squared, equal -1, which are +i and -i respectively. Since i cannot abruptly jump to -i for a continuous function, this means that i is a constant.
The derivative operator D is a linear transformation on the vector space of analytic functions, and both cos, sin are analytic. Therefore, if f(x) = cosx + i sinx, then Df(x) =D(cosx + isinx) = Dcosx + i Dsinx
Notice also that we can show that the differential operator D is linear without any calculus. By defining analytic functions as their corresponding power series, we can define D via the power rule on formal power series.
The simplest and most obvious way to demonstrate that e^ix = cos x + Isin x is to just take the derivative of the polar form of a complex number. For example: a+bi = rcos x + irsin x D(a+bi) = -rsin x + ircos x D(a+bi) = (i^2)rsin x + ircos x D(a+bi) = i( rcos x + irsin x) D(a+bi) = i(itself) So the derivative of a complex number is itself times a constant (in this case is the imaginary unit). The only function that does this is e^ix, as D(e^ix) = i(itself)
Is it correct to assume that derivation on complex numbers works identical compared to the real numbers? Michael did this too in this video, I was wondering the same then. Seems to me, for a 100%-perfect proof you have to show this as well.
The way my teacher did was to use maclaurin series for sin, cos and e^(ix) to show how the series for e^(ix) has the series for sin and cos inside of it
That is a very good entry level method. The only caveat is: Did the teacher show that the resulting series (when replacing x by ix) is still convergent for all x"values? It is though, but that should be proven, which is not so hard
cos(phi) is variable term, we are supposed to compare the coefficient of the term only. So, we get a(phi)'=a➡️(phi)'=1 Note that we can divide a here because r=a is not equal to zero! If r=0, we have e^(iθ) =rcos(phi)+risin(phi)=0 But we know e^(iθ) is never zero
For the second method, how are we sure that the rules of derivation with functions involving complex numbers are the same which hold in the real field?
Cool video but how did u define e to the complex power? In my degree I learn this formula as a definition of it wich is a simple way but also a boring one.
Using complex logarithm ! (*Screams in Terror*) f(x)=log(cos(x)+isin(x)) f'(x)=(-sin(x)+icos(x))/(cos(x)+isin(x)) f'(x)=i (cos(x)+isin(x))/(cos(x)+isin(x)) f'(x)=i => f(x)=ix+ C f(0)=log(1)=0 =>C=0 So ix=log(cos(x)+isin(x)) So e^(ix)=cos(x)+isin(x)
My calc book (Thomas/Finney 7th ed) says the following after deriving Euler's formula using the Taylor Series expansions, "It would not be accurate to say that the calculations just completed have proved [Euler's formula]. Rather, we shall adopt the point of view that [Euler's formula] is the definition of e^i*theta." Can anyone comment on what is meant by that?
that you need to define the exponential to complex numbers and that the way Euler did it was by pretending it was just to apply a complex number at the Taylor Series.
Euler start from cosθ+ i sin θ proves to be equal to e^iθ , not the other way around. The beginning starts with the most basic trig identity cos θ^2 + sin θ^2 = 1 . that is (cos θ + i sin θ)(cos θ - isin θ)=1 .
There might be some limitation of rigour in reverse application of Taylor expansions. As Michael was noting, everything here is periodic as well so multiple functions may equivalently map to the same taylor expansion along their periods. I'm sure there's probably a way to be quite precise about it in a way that takes care of every case, but that hardly feels necessary for a first year college course. Your book may have also been noting that Sine and Cosine are a little bit weird in their definition. There are such a variety of (usually geometric) ways to derive Sine and Cosine that results in their taylor expansion being that way. But maybe it would be neater to say that actually we might as well just call Sine and Cosine "the two formulas that go in the real and imaginary part of euler's formula." Such a definition gives us all the qualities we want out of the trig functions.
It cannot be a proof because we don't know what e^(iθ) means. More specifically, we don't know that it's valid to plug in the imaginary number iθ for the real variable x in the Maclaurin series for e^x derived using the real number system. Instead, this should be thought of an argument of the form "if we expect e^(iθ) to behave the same way as e^x does, what should e^(iθ) be?" From there, you can follow this sort of argument and reach the conclusion that if you want to define e^(iθ) in such a way that it behaves the way we expect based on how e^x behaves for real numbers x, then we should define e^(iθ) as cos(θ)+isin(θ). The Maclaurin expansion argument can be made rigorous by first developing the theory of convergence in the complex plane, defining e^z as the series ∑z^n/n!, showing that ∑z^n/n! is absolutely convergent everywhere in the complex plane, and then performing the computation as in the video.
@@backyard282 as i said, assuming you know polar coordinates, then exp(iθ) is a complex number with radius 1 and argument θ. you can draw that in the plane and geometrically determine the real and imaginary part
With all these ways of proving Euler's formula, one must make sure you don't assume Euler's Formula and get a circular proof. None of these solutions seem to implicitly use Euler's Formula, from what I can see.
I always feel strange about the proofs of this formula. The formula is more like a notation to me, through which we could define powers with complex number exponents.
The second ‘derivation’ is just a check. You overlook that f(theta)=1. Now take the derivative of that, too, and you’re not having to integrate and find C. You’ve made things harder than they need to be, I would say in order to make things more involved, then claim a derivation. The clear and obvious difference between the first two is that you start with the very thing you want to derive, in the second. That’s not the case with the Taylor series; there, I don’t see the result until the end! Honestly, you’re a Ph.D. and all you do is confuse people, make math harder than it really is? On the other hand, that would certify you as having a Ph.D. in math. Crazy smart? More like practically devious for having outsized respect for your achievement.
I think method two should be problematic to many people because it involves using trigonometric idientities whose demonstrations are usually only taught by using Euler's formula. Therefore, circular logic.
Given a complex number with a magnitude r, and an angle θ, the log of this number is: ln(r) + i*(θ + 2*π*k) where k is any arbitrary integer. The log of the magnitude tells us the real value, and the angle tells us the imaginary value. We also add any arbitrary number of full rotations to the imaginary part, which is where 2*π*k comes from.
@@carultch You must be joking? "THE" in english means it should be something DEFINITE, or in other words it should be ONE THING. That nonsens you wrote, means that every one of the complex numbers which can be expressed as ln(r) + i*(θ + 2*π*k) is THE logarithm of the number r*exp(θ*i) !! Are you making an idiot of me or of yourself ?? The answer is clear.
@@adamkolany1668 Well I'm sorry you asked a question that has a multiplicity of answers. There's nothing I can do about the mathematical fact that there are an infinite number of solutions to the log of any given complex number.
@@carultch Simply do not answer questions that go beyond your competencies. Your response suggests that you're not very proficient when it comes to mathematics.
@@adamkolany1668 Then what did you want for the answer to your question, if you are going to split hairs on trivial issues like which article I used?" I've given you a mathematically accurate answer to your question, what more do you want?
In the last proof Mike states that Φ is a function of θ. He then differentiates with respect to θ, but how can he know the this function: Φ(θ) is necessarily differentiable❔
in the last method, you didn't need to consider the case of a=-1 because it stands for the value of r - the magnitude in polar coords, defined to be positive.
I once had a professor suggest there were alternative ways to derive Euler's Formula there, but that gave different, logically consistent meanings to e^itheta. Is there an alternative definition? I haven't encountered one. Having 4 methods to prove the usual relationship seems to settle it, there are no alternatives.
Euler used 1st method + Trig identity, but not in the same order from the video . a good read for you is "HEDI-2007-08.pdf" . google it .On the side note . Only the first method is a construction way to do, you only start from e^iθ . without need to know cosθ+i sinθ
With the third method i believe there is circular reasoning. When both sides are exponentiated you are using the fact that the exponential is the inverse of the logarithm. Since the logarithm is complex the associated exponetial must be though as the complex one, therefore including euler´s formula as a special case and hence the circular reasoning.
it is literally not u simply dont know what i means there is literally nothing special about e^iθ u plug it into ur stupid power series, u get the answer and u can differentiate or integrate e^iθ and u also get the answer ALL THOSE METHOD LITERALLY DIDN'T INVOLVE POWER SERIES NOR e^iθ=cosθ+isinθ in the first place, they all arrived at cosθ+isinθ also, power series IS NOT the definition of sine and cosine or whatever you are saying, it is a representation, the definition is literally triangles and even if it is, it is still not circular reasoning, it simply proved the power series, it didn't use it
it literally doesnt change anything nobody said u cant i is still a number using the vector, axis or just simple sqrt(-1) definition all of these allow the operation just as u can on a vector
Just because you can derive something from something doesn’t make it mentionable! I see math geeks do this all the time. Ask yourself, Why would anyone think to start there? The only derivation you do that begins in a rational sense is the power series. The others merely show that, having made yourself lost in a forest, you can find your way out. But getting lost on purpose is dumb is the point. Honestly, most of this video amounts to grandstanding, not teaching. You are far from alone in this, however.
Which mathematical formula or theorem has the MOST known valid proofs? I guess we'd have to agree on what constitutes an 'atomic' sub proof and where we'd be getting into 'reductio ad absurdam '
@@xizar0rg My second sentence gets at what can you take as a sub proof. Otherwise you could go all the way down to basic axioms and therefore the number of different ways of working back up the proof chain would increase considerably
Euler formula is the epitome of ridiculeness. Imaginary numbers are fake invented math. The very fact that you apply series that are valid for countable real numbers to non countable imaginary numbers says it all, but I guess you don't grasp that.
yep, one of the best ways to do it tbh. All of the proofs assumed a definition of the exponential of complex numbers. The first one assumed we could extend the Taylor Series to complex numbers, the second one assumed the derivative of the exponential of complex numbers, the third assumed we could use the logarithm as if "i" was a real constant... not much different than just defining it to be the case.
A lot of these are just so obscure or lengthy, it is like you are deliberately trying to find it, and that would be because you know the identity in the first place, for the differential equation, it is a proof sure, but deriving? Let us assume nobody knew the identity in the first place, who the hell would write that exact equation to derive it, how would they know to write that? Power series or Taylor series is like the standard proof but make stuff more complex, we need to define extra stuff like their representation in infinite series which is also abstract since you can't actually solve it because it involves infinity and such. Whose idea was it to make this the "official" proof? The simplest proof is the best, the polar form is great, you just know it must be something a+ib which is rsinθ+ircosθ because vector. A simpler and more straightforward method is to just use Laplace Transform, it doesn't use the identity, no circular reasoning. Straight up just evaluate L{e^iθ} and you get the answer. Some people are very weird and asks why can you evaluate e^iθ or complex exponential or blablah without defining the identity, like bruh this isn't even using the identity, this is simply using the definition of i=√-1 or that i is another axis, essentially a vector. An operator and a number must result in another number, √-1 must be a NUMBER which is CONSTANT even if we don't know, i must be a constant and can be manipulated as such. With vector definition, we can also manipulate it just as we would with vector, there is literally nothing here that says we cant manipulate or evaluate it like normal, we can integrate e^iθ all the same. It's not even about e^iθ, these people probably don't even understand i in the first place......
I'm one of those weird people. The Laplace Transform is *_defined in terms of_* e^(complex number), so you need to already know what e^(complex number) means - how it's defined - before you can use the Laplace Transform to derive what e^(complex number) means. That's as circular as it gets.
@@mathboy8188 you literally don't, you simply use the standard definition for e and the definition for imaginary number You know how to manipulate e and you know i is a constant, e^iθ is not even special, why do you even need to specifically define e^iθ, you already have the definition for i, tell me which of these symbol is undefined? e is defined, i is defined, θ is defined, why does it suddenly not work when you put i in e? u literally know i is constant and the operation for e, just do it, it is not special! Even power series uses this, straight up plug iθ into the power series like normal e^x, even power series doesn't define it, it plugs the complex number into existing operation or manipulation for standard e. e^iθ is literally defined the moment you define i also, laplace transform is literally defined as the continuous analog of the power series, at this point you might as well say the power series uses circular reasoning as it assumed or included e^(complex number) in the first place do you know how to integrate and differentiate e^x? do you know how to integrate and differentiate a constant? then where s the problem? listen here, we didn't define e^iθ, we simply defined i and it just works
@@joaogoncalves-tz2uj where in the world do u need to define e^iθ to find the derivative, let us assume we don't know the identity, like we are Newton or somebody. Do you think we need to define e^iθ? Hell no, people just differentiate, they don't need to even know e^iθ. No we don't need the identity to know how to differentiate it, we simply need the definition for i to know how to differentiate it. e^iθ is derived not defined.
The argument here isn't that e^iθ is defined by Laplace Transform. Heck, it was never defined, it was simply derived. The argument here is that e^iθ is defined by i, or the definition of the imaginary number. The power series didn't define it, it is simply one of the many methods that proved or DERIVED it, even the power series take i as a constant and plug it in. (The power series isn't a definition, it is a representation) e^iθ was never defined, it was derived or proved. e^iθ is defined the moment i was defined. It was literally defined in the first place, that's why there are so many methods to derive it. As long as it doesn't involve the identity itself, which is e^iθ=cosθ+isinθ, it is not circular reasoning. Do you get it? i is not special, it isn't some special function or variable or anything, IT IS A NUMBER, A CONSTANT, because it is literally defined as that. A complex number is still a constant, following the definition of i. You don't need to know or define e^iθ=cosθ+isinθ to know how to manipulate e^iθ or anything complex. It is not about whether you understand e^iθ, it is literally about whether you understand the definition of imaginary number, whether you understand i......
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Personally, my favourite method is as follows:
The diff equation y'' = -y has cosine and sine as obvious solutions, and obviously any linear combination of them as well. But we could also have taken exp(c•x), whose second derivative is c²exp(cx), with c some constant. If c²=-1, i.e. if c=i, this is also a solution. But how do we compute exp(ix)? Luckily, we don't need to, because by Picard-Lindelöf we know there is a two parameter family of solutions for this equation, already covered by the sines and cosines from before, so e^ix must be some combination of the two. Adjusting constants and noticing e^i0 = 1 and its derivative at zero is i, we reach the famous result.
I love this method, because it is both intuitive and really highlights the connection between the trigonometric functions and the exponential. In fact, if you're more knowledgable in complex analysis, you'll know the relationship is actually backwards, and sine and cosine are actually combinations of complex exponentials and not the other way around
My favourite as well. Also, works off the way I define e^x: e^x is the unique function such that it is equal to it's derivative, and it''s value when x=0 is 1.
I completly agree, that is "my" method too.
That last statement is debatable because in order to use the series for sine and cosine you need their derivatives and thus standard limits. Now where do they come from? I don't see a way to prove the limit sinx/x without basic sohcahtoa.
@@mathisnotforthefaintofheartIm not following you, I did not use the series for any function, neither exponentials nor trig functions. I did use their derivative rules, and both can be easily obtained with elementary math. The exponential is easy (that one in fact is easier if you do define it as its series, but it's not necessary), and for sin you need to do it through inequalities and the squeeze theorem. That being said, it does not come up for this proof, they could easily be taken as given without any circular reasoning
The elementary proof would be coefficient comparison from their Taylor series of course, in fact I wouldnt be surprised if that comes up in the proof of Picard Lindelof, but to me the elegance of this method is that it gives a useful "why" for why theyre equal, it was not chance their taylor coefficients ended lining up like that
@@ClaraDeLemonMaybe I understood you wrong. Apologies. ok, here is what happened. When I did a second MA in math, my instructor DEFINED the sine and cosine as a difference / sum of complex e-powers. Form there he so called "proved" Euler's identity. And from there he proved the series. But when I asked where these definitions were coming from, he said, "they are just definitions". So when I asked why don't you define the sine and cosine vice versa, he said that it didn't have any purpose in the subsequent theory. When I told him that you didn't have any other choice to define them that way in order to make the theory "work", he got mad. This was the same teacher who claimed that L'Hospital Rule could be used to prove the limit sinx/x=1, even though I believe that is also circular reasoning. So we agreed to disagree. Now Euler may not always have been rock solid in his mathematical proofs, but the way how he came to Euler's formula is well accepted in modern days
Fun video. 🙂 One small thing, in method 3 there's a substitution of x=tan(θ/2) , which is fine but note that there's a restricted domain on it since tan(π/2) is undefined. So unlike the first two methods you might need to be a little careful handling the domain.
The domain is still R. Because, domain is defined as the values of x (independent variable) for which y(dependent variable) is defined, *without solving the equation, like it's raw!*
@@rajneeshmishra6969 Whenever you substitute one variable for another in an equation you have to also specify the bounds of the new variable. For instance, when you take an integral from x=0 to infinity, and then substitute in x=ln(t) in the equation, you need to recognize that the new bounds range from you need to note that that new bounds are from t=1 to infinity.
Similarly in this case, by making the replacement x=tan(θ/2), you may need to recognize that θ is going to range from (-π to π) (or something similar but shifted). Otherwise you can end up working with resulting equations using θ and accidentally apply θ=π, for example, without realizing it can result in an undefined value there.
here's another way: set f(θ) = cos(θ) + i sin(θ) and take the derivative
f'(θ) = -sin(θ) + i cos(θ) = i (cos(θ) + i sin(θ)) = i f(θ)
which is a differential equation whose solution (with initial condition f(0) = 1) is f(θ) = exp(iθ)
That's my method too. And you can take a second derivative too and eliminate "i" from the resulting DFQ
well actually when we differentiating maybe e^iθ with respect to θ , we get ie^iθ when i is a constant , for sure i=root -1 is a constant but is it strict? since it involved with complex number
That's my favorite way, too.
The problems of these proofs is how do you define each functions? What is assumed about each functions? For example, in the second proof, you are supposed to know that the derivative of exp(iθ) wrt θ is equal to iexp(iθ), but you first need to define the exponential of a complex number.
That's true, all of these proofs are based on assumptions about certain properties of the involved functions (in the complex plane!). Here's my take on this: Before working with those functions, you need to extend the definition of e.g. exp(x) from the real numbers to the complex numbers. Whatever way you do this, you want this extension to be natural, i.e. you want to retain as much of their essential properties as possible. For the exponential function, this could be the functional equation exp(x+y) = exp(x)*exp(y), or it could be the differential equation y' = y, or it could be the series expansion, or it could be Euler's limit definition for e^x. So you could take one of these properties to be the defining property of the complex extension and then use this property to prove Euler's identity without circular reasoning. Of course, you would still have to show that all the other useful properties still hold for the complex numbers and follow from your extended definition.
So good to see the return of the magic chalk squares as you progress through the problem. It's what got me into your videos in the first place. Long may they continue...
There is a way with the definition through the limit. e^(ix)=lim (1+ix/n)^n=lim [sqrt(1+x^2/n^2)] ^n * [cos(atg(x/n))+i sin (atg(x/n))]^n=lim [sqrt(1+x^2/n^2)]^n * [cos(n atg(x/n))+i sin (n atg(x/n))] =1 [cos(x)+i sin(x)]
Amazing as usual. Thanks for this weekend mental refreshment.
@20:42 cos(φ) *First equation + sin(φ)* second equation
The derivations using calculus are probably circular. That is, the derivative of e raised to an imaginary power is undefined until and unless Euler’s formula is derived. For example, taking the derivative of both sides of Euler’s formula yields Euler’s formula. But that only works if you already know that the derivative of e^i*theta = i*e^i*theta. The power series for the sine, cosine and exponential function are really fundamental to proving the formula without any circularity. I think that’s how Euler discovered his famous formula.
Can't you just *define* the function e^z by its derivative to avoid circularity?
@@Alex_Deam Possibly. Recall that complex numbers arose from efforts to find the roots of polynomials. Polynomials only involve addition and multiplication operations. The Taylor power series for the exponential function involves only derivatives over the real numbers. After you find the power series for the exponential function over the reals it's only valid for reals. The next step is expand the argument to the complex numbers which is now possible because a power series is also a polynomial. But that step is really a definition of e^z since you only proved the power series is valid over the reals using the Taylor expansion. But then you also need to show the series converges for complex arguments to show the definition is valid.
@@dougrife8827 Can't we use the Weierstrass M-test for convergence?
@@Alex_Deam Weierstrass M-test applied to the Taylor series uses the absolute value of the terms or alternatively, the magnitudes of each term in the case of substituting a complex variable z in place of the real variable x. If the Taylor series converges x it also must converge for z. The problem with defining e^z by its derivative is that you first have to define complex differentiation. That’s not required in this definition because the Taylor series for e^x was created by taking the derivatives with respect to the real variable x. Complex differentiation is not required. Then after you have the Taylor series for e^x where x is real you replace x with z to define the meaning of e^z. And convergence is not an issue because it depends only on the magnitude of z, which is a real number.
15:44 we cannot set theta = 0 as , in one of the revious steps we had 2-2cos(theta) in denominator , so while reducing the fraction we had to do a assumption that theta is not = 2 n pie
z = cos(x) + i sin(x) = cos(xN/N) + i sin(xN/N) = (cos(x/N) + 1 sin(x/N))^N by de Moivre's formula which is proved by indiction and intuitively clear from the fact that the angles of complex numbers add up when multiplied. Letting N -> infinity: cos(x)+i sin(x) = (1 + ix/N)^N = e^(ix).
Try using rotation matrices: Look at R(theta) = R(theta/n)^n in the limit as n->inf. Show that in this limit you get the limit of (I+J*theta/n)^n =exp(J*theta). Where Z=xI+yJ is the matrix representation of the complex numbers. (I = diag(1,1) J = ((0,-1),(1,0))^T. R(theta)= ((c,-s),(s,c))^T c =cos(theta), s = sin(theta). [I'm using column of rows to represent square matrix, and column = row^T ... 'cause, YT comments don't allow LaTex input. ]
Looks like the relationship between the groups U(1) and SO(2)
I love the complex logarithm way. You can derive the formula from the harmonic oscillator equation. The eigenvalues are complex and when you diagonalize you end up with a rotation matrix. This proves Euler formula.
I'm a math teacher here in France, and I love your videos! Very well explained, a pure pleasure! Thanks 💜
I derived it in a similar way as the last one, but without making any initial assumptions about what the polar form should be.
Take some complex number 𝑧 = 𝑎 + 𝑖∙𝑏. From the interpretation that a complex number is a point in the complex plane, we can verify that 𝑧 can be equivalently defined as the point that is at a distance 𝑟 from the origin 𝑂 and which has a line segment connecting it to the origin forming an angle 𝜃 with the positive real axis measured in the counter-clockwise direction. 𝑟 and 𝜃 will be called the polar coordinates of 𝑧 and it follows that 𝑧 = 𝑟(cos 𝜃 + 𝑖∙sin 𝜃) from trigonometry. Similarly, we can take a second complex number 𝑤 = 𝑠(cos 𝜑 + 𝑖∙sin 𝜑), where 𝑠 and 𝜑 are the polar coordinates of 𝑤.
Then we can calculate the product 𝑧𝑤 = 𝑟𝑠[cos 𝜃 cos 𝜑 - sin 𝜃 sin 𝜑 + 𝑖∙(cos 𝜃 sin 𝜑 + sin 𝜃 cos 𝜑)] = 𝑟𝑠[cos (𝜃 + 𝜑) + 𝑖∙sin(𝜃 + 𝜑)], by using trigonometric identities, which has polar coordinates 𝑟𝑠 and 𝜃 + 𝜑. This fact motivates us to represent a complex number in polar form as 𝑧 = 𝑟∙𝑢^𝜃, where 𝑢 is some yet to be determined complex number, such that 𝑧𝑤 = 𝑟𝑠∙𝑢^(𝜃 + 𝜑).
To determine 𝑢, we can take the derivative of 𝑧 w.r.t. 𝜃 in both the rectangular and polar forms:
∂𝑧/∂𝜃 = 𝑟(-sin 𝜃 + 𝑖∙cos 𝜃) = 𝑟∙𝑖∙(cos 𝜃 + 𝑖∙sin 𝜃) = 𝑖∙𝑧;
∂𝑧/∂𝜃 = 𝑟∙ln 𝑢∙𝑢^𝜃 = ln 𝑢∙𝑧.
Therefore, ln 𝑢 = 𝑖 ⟺ 𝑢 = 𝑒^𝑖.
In conclusion, it was shown that a complex number 𝑎 + 𝑖∙𝑏 can be equivalently represented in polar form as 𝑟∙𝑒^(𝑖∙𝜃) and Euler's formula was automatically proven along the way.
Yes I do have another method, which I discovered before I had ever heard of Euler. It is so unique that it was published in a Mathematical Journal because I had discovered Euler's equation before I had learned of its existence. Actually I was searching for any meaning to the logarithms of negative numbers.
After discovering the formula, I kept quiet about it because it seemed so weird that I did not understand what it meant. However, after I had learned of Euler, I sent it to the Journal as a unique way to arrived at his equation. I would happily reproduce it here, except that I have not the ability to type Maths, but it can be found on the internet by searching M500 and scrolling to page 24. (It is important that you scroll to page 24, because the fast track way to get to it is 2 pages out of phase.)
Here's a method I learned that is very similar to your METHOD 2: Let f_1(x) = e^{ix} and f_2(x) = \cos(x) + i\sin(x). Both of these functions satisfy the differential equation: F' = iF, and the initial condition F(0) = 1. Therefore, by the existence-uniqueness theorem of first-order, homogeneous, linear ordinary differential equations, the two functions must be identical.
The Beatles of math formulas. Nicely done! 🙂
Too bad he didn't show us the Best derivation.
I know of a 5th way!
It uses the close approximation of theta and sine (and by extension tangent) and the limit form of e^(r). Pretty lengthy but decent enough for a PreCalculus class
Fantastic. Thanks for your efforts 🙏
From De Moivre’s Theorem (cosx + isinx)^n=cos(nx) +isin(nx), let t=nx, then (cos(t/n)+isin(t/n)^n = cos(t) + isin(t)
Now let n be big… approximate cos(t/n) by 1+(t/2n)^2 which goes to 1 as n gets big and sin(t/n) by t/n…
Then cos(t) + isin(t) = limit as n goes to infinity of (1+it/n)^n which is, of course e^(it).
If I recall correctly, this is pretty much the same as Euler's original demonstration. It's interesting how no one presents it this way anymore.
Brilliant as usual. Thanks Michael
I feel the 2nd one is the best one! Thanks 😀
The last one may be complet in logic. Thanks
Consider z = cos θ + i sin θ, so that
dz/dθ = -sin θ + i cos θ
dz/dθ = i(i sin + cos θ)
dz/dθ = zi
dz/z = i dθ
By integrating both side we get
ln z = θi + C
z = e^(θi + C)
z = Ce^(θi)
cos θ + i sin θ = Ce^(θi)
Substitue θ = 0 to get rid of the C
cos 0 + i sin 0 = C
1 = C
Therefore
cos θ + i sin θ = e^(θi)
I like to do it by showing the following:
Let f(x) = exp(i*x)
Let g(x) = cos(x) + i sin(x)
Proceed to show that f'''' = f, f''''' = f', f'''''' = f'', f''''''' = f''.
Show that the same holds for g.
Then show that f(0) = g(0), f'(0) = g'(0), f''(0) = g''(0), and f'''(0) = g'''(0).
Then observe that this means that f(x) and g(x) are not only equal at x=0, but all of their infinite derivatives are also equal. Thus f and g must be equal everywhere.
A simple way of showing that Euler's equation is correct is to define a function y(x) = cosx+i*sinx. If you differentiate y, you end up with a differential equation dy/dx = i*y. If you separate variables and integrate, you get ln(y) = i*x+C1. If you exponentiate both sides, you get y(x) = C2*exp(i*x), where C2=exp(C1). If you evaluate both forms of y(x) at x = 0, you find C2=1, and Euler's equation is obtained.
I love all patterns that bring together exponential and trig functions.
You can also define sine and cosine to be imaginary and real parts of e^ix and then proof that those functions are indeed normal trygonometric ones
Hi,
I've been thinking about this problem for quite a long time, but it seems to me that there is another way of doing that.
If the Euler's formula is true, then you have to proof that ix is the logarithm of cos x + i sin x. The whole problem is to define the complex logarithm without using Euler's formula.
Why not use the definition for a positive real, and extend it for a complex number? For a positive real number : ln x = int_1^x {dt/t) . Let's say that, for a complex number :
ln z = int_1^z {dw/w) , where w is the dummy complex integration variable.
As the function 1/z as only a pole in z=0, the result of the integration will not depend upon the path used, as long as you don't cross the "cut direction" y=0 and x0, and I found the expected result: ln (|z|) + i arctan (y/x) . Once again : without using the Euler formula, only by integration of 1/z .
And for x
An awesome video - well done !
20:40 isn't this supposed to be cos φ instead of cos θ? Because I don't think we know at this point that φ=θ
yes, you're right. Should be φ at this stage.
Thank you 💖
Another way:
Let z=cosθ+isinθ
Then dz/dθ=-sinθ+icosθ=iz
So dz/dθ=iz
Obviously z=C×e^(iθ)
And by plugging 0 into the original function for z we get that C=1
Therefore cosθ+isinθ=e^(iθ)
Fantastic!!!...& very nice!..Thank you.
Foundation of these proof assume we have d(e^i theta)/ d theta = i e^i theta
But behavior of i in exponentiation is undefined here. Can you share why do we have that?
I've always struggled to understand how this formula makes any sense but when you mentioned that exp(i*theta) is a complex number so it has to have a polar form... suddenly it was so obvious!
how did he define exp(i*theta*), to start? (what does it mean? why is it a complex number?)
@@frentz7the limit defenition for exp lim n->infinity (1+theta/n)^n
Work in complex number too
@@bluelemon243 Perhaps the nicest way of defining it is with power series. You define e^x = sum_{n>=0} x^n/n!, then using Hadarmard's theorem, you find that this has an infinite radius of convergence, then you know that for analytic functions, their derivative is the component-wise derivative, so you get that d/dx e^x = e^x. This is how Cartan defines the exponential function on C in his complex analysis book. You then just define cos and sin in terms of e^ix and show that they really are what you think they are.
On of your best videos
My proof is pretty short.
Define a function f(t)=cos(t)+i*sin(t). Take the derivative of this function by taking the derivative of real and imaginary part separately:: f'(t)=-sin(t)+i*cos(t)
By definition i*i=-1, thus f'(t)=i*(i*sin(t)+cos(t))=i*f(t).
But this a linear differential equation, and these can only have solutions consisting of sums of exponential functions, so in this case f(t) must be of the form f(t)=C*exp(k*t), where C and k are constants. Plugging this into the differentiial equation you'll find that k=i, and because of the border condition f(0) = cos(0)+i*sin(0) = 1, therefore C=1. QED.
In case you're not convinced you can explicitly solve the differential equation by separating the variables and integrating both sides: df/dt=i*f => ∫df/f = ∫i*dt => ln(f) = i*t+c => f(t)=exp(i*t+c)
Because f(0)=1, c=0.
Your videos are always nice :)
IMHO, the second proof is the best...it's the most elegant and simple.
Is there a part in the video where z^w is defined, for complex numbers z, w?
I personally think it is because of this formula that the powers with complex number exponents can be defined.
How do we know that i behaves like a real constant wrt complex derivatives and integrals. We all know it does, but isn't that an assumption that needs to be proved?
Because -1 is also a constant. There are only two numbers, that when squared, equal -1, which are +i and -i respectively. Since i cannot abruptly jump to -i for a continuous function, this means that i is a constant.
The derivative operator D is a linear transformation on the vector space of analytic functions, and both cos, sin are analytic. Therefore, if f(x) = cosx + i sinx, then Df(x) =D(cosx + isinx) = Dcosx + i Dsinx
Notice also that we can show that the differential operator D is linear without any calculus. By defining analytic functions as their corresponding power series, we can define D via the power rule on formal power series.
The simplest and most obvious way to demonstrate that e^ix = cos x + Isin x is to just take the derivative of the polar form of a complex number. For example:
a+bi = rcos x + irsin x
D(a+bi) = -rsin x + ircos x
D(a+bi) = (i^2)rsin x + ircos x
D(a+bi) = i( rcos x + irsin x)
D(a+bi) = i(itself)
So the derivative of a complex number is itself times a constant (in this case is the imaginary unit). The only function that does this is e^ix, as D(e^ix) = i(itself)
Is it correct to assume that derivation on complex numbers works identical compared to the real numbers? Michael did this too in this video, I was wondering the same then. Seems to me, for a 100%-perfect proof you have to show this as well.
The way my teacher did was to use maclaurin series for sin, cos and e^(ix) to show how the series for e^(ix) has the series for sin and cos inside of it
That is a very good entry level method. The only caveat is: Did the teacher show that the resulting series (when replacing x by ix) is still convergent for all x"values? It is though, but that should be proven, which is not so hard
That Nintendo switch click ❤
@22:00 we assume a and cos(phi) are non-zero by dividing by them. Why are we allowed to do that?
cos(phi) is variable term, we are supposed to compare the coefficient of the term only. So, we get a(phi)'=a➡️(phi)'=1
Note that we can divide a here because r=a is not equal to zero! If r=0, we have e^(iθ) =rcos(phi)+risin(phi)=0
But we know e^(iθ) is never zero
@@khoozu7802thanks that makes sense comparing coefficients of variable terms.
What’s the arrow emoji mean here?
@@Happy_Abe u can treat it as implication but actually it is for saving space only
@@Happy_Abe TH-cam do not have limit for how many words but twitter does have
@@khoozu7802 ah gotcha thanks!
I'm curious, for problem 3, where di the fraction 2i/(x^2+1) come from? Does that fraction come from somewhere?
For the second method, how are we sure that the rules of derivation with functions involving complex numbers are the same which hold in the real field?
Cool video but how did u define e to the complex power? In my degree I learn this formula as a definition of it wich is a simple way but also a boring one.
Take your favorite defention of exp(x) except the e^x (fox example the limit one or the series one) and change every x to z lol
Using complex logarithm ! (*Screams in Terror*)
f(x)=log(cos(x)+isin(x))
f'(x)=(-sin(x)+icos(x))/(cos(x)+isin(x))
f'(x)=i (cos(x)+isin(x))/(cos(x)+isin(x))
f'(x)=i => f(x)=ix+ C
f(0)=log(1)=0 =>C=0
So ix=log(cos(x)+isin(x))
So e^(ix)=cos(x)+isin(x)
This one looks like the a perfect "Animation vs Math" video explanation 😀
25:31
🤔
Nice. I love all of the methods.
Method 2 is wonderful, but makes me laugh as it's clearly a strategy come up with in hindsight.
I would have shown that both exp(iθ) and cos(θ) + i sin(θ) are solutions of the differential problem y' = iy , y(0) = 1.
My calc book (Thomas/Finney 7th ed) says the following after deriving Euler's formula using the Taylor Series expansions, "It would not be accurate to say that the calculations just completed have proved [Euler's formula]. Rather, we shall adopt the point of view that [Euler's formula] is the definition of e^i*theta." Can anyone comment on what is meant by that?
that you need to define the exponential to complex numbers and that the way Euler did it was by pretending it was just to apply a complex number at the Taylor Series.
Euler start from cosθ+ i sin θ proves to be equal to e^iθ , not the other way around. The beginning starts with the most basic trig identity cos θ^2 + sin θ^2 = 1 . that is (cos θ + i sin θ)(cos θ - isin θ)=1 .
There might be some limitation of rigour in reverse application of Taylor expansions. As Michael was noting, everything here is periodic as well so multiple functions may equivalently map to the same taylor expansion along their periods. I'm sure there's probably a way to be quite precise about it in a way that takes care of every case, but that hardly feels necessary for a first year college course. Your book may have also been noting that Sine and Cosine are a little bit weird in their definition. There are such a variety of (usually geometric) ways to derive Sine and Cosine that results in their taylor expansion being that way. But maybe it would be neater to say that actually we might as well just call Sine and Cosine "the two formulas that go in the real and imaginary part of euler's formula." Such a definition gives us all the qualities we want out of the trig functions.
It cannot be a proof because we don't know what e^(iθ) means. More specifically, we don't know that it's valid to plug in the imaginary number iθ for the real variable x in the Maclaurin series for e^x derived using the real number system.
Instead, this should be thought of an argument of the form "if we expect e^(iθ) to behave the same way as e^x does, what should e^(iθ) be?" From there, you can follow this sort of argument and reach the conclusion that if you want to define e^(iθ) in such a way that it behaves the way we expect based on how e^x behaves for real numbers x, then we should define e^(iθ) as cos(θ)+isin(θ).
The Maclaurin expansion argument can be made rigorous by first developing the theory of convergence in the complex plane, defining e^z as the series ∑z^n/n!, showing that ∑z^n/n! is absolutely convergent everywhere in the complex plane, and then performing the computation as in the video.
if you assume polar coordinates are known, you can literally just convert exp(iθ) into cartesian coordinates graphically.
How could you convert it graphically given that you apriori don't know Euler's formula
@@backyard282 as i said, assuming you know polar coordinates, then exp(iθ) is a complex number with radius 1 and argument θ. you can draw that in the plane and geometrically determine the real and imaginary part
Circular reasoning @@demenion3521
15:50 how can we have C=-1 when at 10:15 C was defined as e^c?
Because c was complex
With all these ways of proving Euler's formula, one must make sure you don't assume Euler's Formula and get a circular proof. None of these solutions seem to implicitly use Euler's Formula, from what I can see.
I like how only the first proof is correct and complete. The rest three are either incomplete or incorrect (perhaps both)
16:35 wait, bro you cannot assume that.
0:29 aren’t the Taylor expansions only made for real numbers ? I believe it’s not right to plug in iθ here
isn't the last one circular reasoning?
How so?
I always feel strange about the proofs of this formula. The formula is more like a notation to me, through which we could define powers with complex number exponents.
The second ‘derivation’ is just a check. You overlook that f(theta)=1. Now take the derivative of that, too, and you’re not having to integrate and find C. You’ve made things harder than they need to be, I would say in order to make things more involved, then claim a derivation. The clear and obvious difference between the first two is that you start with the very thing you want to derive, in the second. That’s not the case with the Taylor series; there, I don’t see the result until the end!
Honestly, you’re a Ph.D. and all you do is confuse people, make math harder than it really is? On the other hand, that would certify you as having a Ph.D. in math. Crazy smart? More like practically devious for having outsized respect for your achievement.
Shut up
In the next video, please try to prove de Moivre's formula.
I think method two should be problematic to many people because it involves using trigonometric idientities whose demonstrations are usually only taught by using Euler's formula. Therefore, circular logic.
U can get de moivre theorem without eulers identity
@@peamutbubber I know. You could use rotation matrices for example. But I said they're usually only taugh through Euler's identity and that's true.
what is a log of a complex number??
Given a complex number with a magnitude r, and an angle θ, the log of this number is:
ln(r) + i*(θ + 2*π*k)
where k is any arbitrary integer.
The log of the magnitude tells us the real value, and the angle tells us the imaginary value. We also add any arbitrary number of full rotations to the imaginary part, which is where 2*π*k comes from.
@@carultch
You must be joking?
"THE" in english means it should be something DEFINITE, or in other words it should be ONE THING. That nonsens you wrote, means that every one of the complex numbers which can be expressed as ln(r) + i*(θ + 2*π*k) is THE logarithm of the number r*exp(θ*i) !!
Are you making an idiot of me or of yourself ??
The answer is clear.
@@adamkolany1668 Well I'm sorry you asked a question that has a multiplicity of answers. There's nothing I can do about the mathematical fact that there are an infinite number of solutions to the log of any given complex number.
@@carultch Simply do not answer questions that go beyond your competencies. Your response suggests that you're not very proficient when it comes to
mathematics.
@@adamkolany1668 Then what did you want for the answer to your question, if you are going to split hairs on trivial issues like which article I used?"
I've given you a mathematically accurate answer to your question, what more do you want?
In the last proof Mike states that Φ is a function of θ. He then differentiates with respect to θ, but how can he know the this function: Φ(θ) is necessarily differentiable❔
Are there any ways to do this where one only needs precalculus skills? Anyone?!
in the last method, you didn't need to consider the case of a=-1 because it stands for the value of r - the magnitude in polar coords, defined to be positive.
Negative magnitude is accepted in polar coordinates.
@@jursamaj huh? since when?
John McEnroe, mathematician
I once had a professor suggest there were alternative ways to derive Euler's Formula there, but that gave different, logically consistent meanings to e^itheta. Is there an alternative definition? I haven't encountered one. Having 4 methods to prove the usual relationship seems to settle it, there are no alternatives.
im at a loss what you are talking about
The nintendo switch sound was kinda jarring
Um - How did Euler do it?
Euler used 1st method + Trig identity, but not in the same order from the video . a good read for you is "HEDI-2007-08.pdf" . google it
.On the side note . Only the first method is a construction way to do, you only start from e^iθ . without need to know cosθ+i sinθ
@@bosorot ty
With the third method i believe there is circular reasoning. When both sides are exponentiated you are using the fact that the exponential is the inverse of the logarithm. Since the logarithm is complex the associated exponetial must be though as the complex one, therefore including euler´s formula as a special case and hence the circular reasoning.
Zero rizz
You had to replace teta by x to know that it's variable.
Surely all but the first method are circular reasoning as the power series expansion is the definition of sine and cosine.
it is literally not
u simply dont know what i means
there is literally nothing special about e^iθ
u plug it into ur stupid power series, u get the answer
and u can differentiate or integrate e^iθ and u also get the answer
ALL THOSE METHOD LITERALLY DIDN'T INVOLVE POWER SERIES NOR e^iθ=cosθ+isinθ in the first place, they all arrived at cosθ+isinθ
also, power series IS NOT the definition of sine and cosine or whatever you are saying, it is a representation, the definition is literally triangles
and even if it is, it is still not circular reasoning, it simply proved the power series, it didn't use it
Sorry… that is 1+(1/2)(t/n)^2
The log identities shouldn't be applied just like that because these are complex numbers.
it literally doesnt change anything
nobody said u cant
i is still a number
using the vector, axis
or just simple sqrt(-1) definition
all of these allow the operation just as u can on a vector
@@urnoob5528 0 != ln(-1)+ln(-1) = ln(1) = 0
doraemon is correct.
@@tracyh5751 u r making no damn sense and not explaining anything
@@tracyh5751 can we correct method 3 to make it work?
Just because you can derive something from something doesn’t make it mentionable! I see math geeks do this all the time. Ask yourself, Why would anyone think to start there? The only derivation you do that begins in a rational sense is the power series. The others merely show that, having made yourself lost in a forest, you can find your way out. But getting lost on purpose is dumb is the point.
Honestly, most of this video amounts to grandstanding, not teaching. You are far from alone in this, however.
cool
Last
Which mathematical formula or theorem has the MOST known valid proofs? I guess we'd have to agree on what constitutes an 'atomic' sub proof and where we'd be getting into 'reductio ad absurdam '
Pythagoras' theorem probably has the most proofs.
I'm not sure where you're going with your second sentence as it seems a non sequitur.
@@xizar0rg My second sentence gets at what can you take as a sub proof. Otherwise you could go all the way down to basic axioms and therefore the number of different ways of working back up the proof chain would increase considerably
Euler formula is the epitome of ridiculeness. Imaginary numbers are fake invented math. The very fact that you apply series that are valid for countable real numbers to non countable imaginary numbers says it all, but I guess you don't grasp that.
But you aren’t deriving the formula so much as you are defining equivalences.
If your name is Hilary Priestly, then you simply DEFINE it to be the case...
Who?
@@CTJ2619 Introduction to Complex Analysis by H. A. Priestly.
yep, one of the best ways to do it tbh. All of the proofs assumed a definition of the exponential of complex numbers. The first one assumed we could extend the Taylor Series to complex numbers, the second one assumed the derivative of the exponential of complex numbers, the third assumed we could use the logarithm as if "i" was a real constant... not much different than just defining it to be the case.
@20:56, why do those two terms cancel?
A lot of these are just so obscure or lengthy, it is like you are deliberately trying to find it, and that would be because you know the identity in the first place, for the differential equation, it is a proof sure, but deriving? Let us assume nobody knew the identity in the first place, who the hell would write that exact equation to derive it, how would they know to write that?
Power series or Taylor series is like the standard proof but make stuff more complex, we need to define extra stuff like their representation in infinite series which is also abstract since you can't actually solve it because it involves infinity and such. Whose idea was it to make this the "official" proof?
The simplest proof is the best, the polar form is great, you just know it must be something a+ib which is rsinθ+ircosθ because vector.
A simpler and more straightforward method is to just use Laplace Transform, it doesn't use the identity, no circular reasoning. Straight up just evaluate L{e^iθ} and you get the answer.
Some people are very weird and asks why can you evaluate e^iθ or complex exponential or blablah without defining the identity, like bruh this isn't even using the identity, this is simply using the definition of i=√-1 or that i is another axis, essentially a vector. An operator and a number must result in another number, √-1 must be a NUMBER which is CONSTANT even if we don't know, i must be a constant and can be manipulated as such. With vector definition, we can also manipulate it just as we would with vector, there is literally nothing here that says we cant manipulate or evaluate it like normal, we can integrate e^iθ all the same. It's not even about e^iθ, these people probably don't even understand i in the first place......
I'm one of those weird people.
The Laplace Transform is *_defined in terms of_* e^(complex number), so you need to already know what e^(complex number) means - how it's defined - before you can use the Laplace Transform to derive what e^(complex number) means. That's as circular as it gets.
@@mathboy8188 correct. Even the method 4 makes use of the derivative of exp(ix)
@@mathboy8188 you literally don't, you simply use the standard definition for e
and the definition for imaginary number
You know how to manipulate e and you know i is a constant, e^iθ is not even special, why do you even need to specifically define e^iθ, you already have the definition for i, tell me which of these symbol is undefined? e is defined, i is defined, θ is defined, why does it suddenly not work when you put i in e? u literally know i is constant and the operation for e, just do it, it is not special! Even power series uses this, straight up plug iθ into the power series like normal e^x, even power series doesn't define it, it plugs the complex number into existing operation or manipulation for standard e.
e^iθ is literally defined the moment you define i
also, laplace transform is literally defined as the continuous analog of the power series, at this point you might as well say the power series uses circular reasoning as it assumed or included e^(complex number) in the first place
do you know how to integrate and differentiate e^x? do you know how to integrate and differentiate a constant? then where s the problem? listen here, we didn't define e^iθ, we simply defined i and it just works
@@joaogoncalves-tz2uj where in the world do u need to define e^iθ to find the derivative, let us assume we don't know the identity, like we are Newton or somebody. Do you think we need to define e^iθ? Hell no, people just differentiate, they don't need to even know e^iθ. No we don't need the identity to know how to differentiate it, we simply need the definition for i to know how to differentiate it.
e^iθ is derived not defined.
The argument here isn't that e^iθ is defined by Laplace Transform. Heck, it was never defined, it was simply derived.
The argument here is that e^iθ is defined by i, or the definition of the imaginary number.
The power series didn't define it, it is simply one of the many methods that proved or DERIVED it, even the power series take i as a constant and plug it in. (The power series isn't a definition, it is a representation)
e^iθ was never defined, it was derived or proved.
e^iθ is defined the moment i was defined.
It was literally defined in the first place, that's why there are so many methods to derive it. As long as it doesn't involve the identity itself, which is e^iθ=cosθ+isinθ, it is not circular reasoning.
Do you get it? i is not special, it isn't some special function or variable or anything, IT IS A NUMBER, A CONSTANT, because it is literally defined as that. A complex number is still a constant, following the definition of i. You don't need to know or define e^iθ=cosθ+isinθ to know how to manipulate e^iθ or anything complex.
It is not about whether you understand e^iθ, it is literally about whether you understand the definition of imaginary number, whether you understand i......