Euler's Formula

#redirect Euler's formula

Euler's formula

''This article is about the '''Euler's formula in complex analysis'''. For Euler's formula in algebraic topology, see Euler characteristic.'' ---- '''Euler's formula''', named after Leonhard Euler, is a mathematics formula in the subfield of complex analysis that shows a deep relationship between the trigonometric functions and the exponential function. (Euler's identity is a special case of the Euler formula). Euler's formula states that, for any real number ''x'', :

e^{ix} = \cos x + i\;\sin x

where :''e'' is the e (mathematical constant) :''i'' is the imaginary unit :sin and cos are trigonometric functions. ==History== Euler's formula was proved (in an obscured form) for the first time by Roger Cotes in 1714, then rediscovered and popularized by Euler in 1748. It is interesting to note that neither of these men saw the geometrical interpretation of the formula: the view of complex numbers as points in the plane arose only some 50 years later (see Caspar Wessel). ==Notes== This formula can be interpreted as saying that the function ''e''^''ix'' traces out the unit circle in the complex number as ''x'' ranges through the real numbers. Here, ''x'' is the angle that a line connecting the origin with a point on the unit circle makes with the positive real axis, measured counter clockwise and in radians. The formula is valid only if sin and cos take their arguments in radians rather than in degrees. The proof is based on the Taylor series expansions of the exponential function ''e''^''z'' (where ''z'' is a complex number) and of sin ''x'' and cos ''x'' for real numbers ''x'' (see below). In fact, the same proof shows that Euler's formula is even valid for all ''complex'' numbers ''x''. The formula provides a powerful connection between mathematical analysis and trigonometry. It is used to represent complex numbers in coordinates (elementary mathematics) and allows the definition of the logarithm for complex arguments. By using the exponential laws :

e^{a + b} = e^a \cdot e^{b}

and :

(e^a)^b = e^{a b} \,

(which are valid for all complex numbers ''a'' and ''b''), one can also readily derive several trigonometric identity as well as de Moivres formula from it. Euler's formula also allows one to interpret the sine and cosine functions as mere variations of the exponential function: :

\cos x = {e^{ix} + e^{-ix} \over 2}

\sin x = {e^{ix} - e^{-ix} \over 2i}

These formulas can even serve as the definition of the trigonometric functions for complex arguments ''x''. You can derive the two equations above simply by adding or subtracting Euler's formulas: :

e^{ix} = \cos x + i \sin x \;

e^{-ix} = \cos x - i \sin x \;

and solving for either cosine or sine. The formulae above can also be used to relate the hyperbolic sine and hyperbolic cosine functions to the usual trigonometric functions. In differential equations, the function ''e''^''ix'' is often used to simplify derivations, even if the final answer is a real function involving sine and cosine. Euler's identity is an easy consequence of Euler's formula. In electrical engineering and other fields, signals that vary periodically over time are often described as a combination of sine and cosine functions (see Fourier analysis), and these are more conveniently expressed as the real part of exponential functions with imaginary number exponents, using Euler's formula. ==Proofs== ===Using Taylor series=== Here is a proof of Euler's formula using Taylor series expansions as well as basic facts about the powers of ''i'': :

i^0=1 \,

i^1=i \,

i^2=-1 \,

i^3=-i \,

i^4=1 \,

and so on. The functions ''e''^''x'', cos(''x'') and sin(''x'') (assuming ''x'' is real number) can be written as: :

e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots

\cos x = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \cdots

\sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \cdots

and for complex ''z'' we ''define'' each of these function by the above series, replacing ''x'' with ''iz''. This is possible because the radius of convergence of each series is infinite. We then find that :

e^{iz} = 1 + iz + \frac{(iz)^2}{2!} + \frac{(iz)^3}{3!} + \frac{(iz)^4}{4!} + \frac{(iz)^5}{5!} + \frac{(iz)^6}{6!} + \frac{(iz)^7}{7!} + \frac{(iz)^8}{8!} + \cdots

= 1 + iz - \frac{z^2}{2!} - \frac{iz^3}{3!} + \frac{z^4}{4!} + \frac{iz^5}{5!} - \frac{z^6}{6!} - \frac{iz^7}{7!} + \frac{z^8}{8!} + \cdots

= \left( 1 - \frac{z^2}{2!} + \frac{z^4}{4!} - \frac{z^6}{6!} + \frac{z^8}{8!} + \cdots \right) + i\left( z - \frac{z^3}{3!} + \frac{z^5}{5!} - \frac{z^7}{7!} + \cdots \right)

= \cos (z) + i\sin (z) \,

The rearrangement of terms is justified because each series is absolutely convergent. Taking ''z'' = ''x'' to be a real number, gives the original identity as Euler discovered it. Q.E.D. ===Using calculus=== Define the complex number

z

such that :

z=\cos x + i\sin x \,

Differentiating

z

with respect to

x

: :

\frac{dz}{dx}=-\sin x + i\cos x

Using the fact that ''i''² = -1: :

\frac{dz}{dx}=i^2\sin x + i\cos x=i(\cos x + i\sin x)=zi

Separating variables and integrating both sides: :

\int\frac{1}{z}\,dz=\int i\,dx

\ln z=xi + C\,

where :

C

is the constant of integration. To finish the proof we have to argue that it is zero. This is easily done by substituting

x=0

. :

\ln z = C\,

But

z

is just equal to: :

z = \cos x + i\sin x = \cos 0 + i \sin 0 = 1 \,

thus :

\ln 1 = C \,

C = 0 \,

So now we just exponentiate :

\ln z = xi \,

e^{\ln z} = e^{xi} \,

z = e^{xi} \,

e^{xi} = \cos x + i\sin x \,

''Q.E.D.'' ==External links== *[http://agutie.homestead.com/files/Eulerformula.htm Euler and his beautiful and extraordinary formula] by Antonio Gutierrez from Geometry Step by Step from the Land of the Incas. *[http://agutie.homestead.com/files/Puzzle_EulerFormula.htm Euler's Formula - Puzzle: 55 pieces in a six star style of piece] by Antonio Gutierrez from Geometry Step by Step from the Land of the Incas. Complex analysisTheorems vi:C�ng thức Euler

Euler's formula

Can you show a proof of Euler's equation? There is another way of demonstrating the formula which I find to be more beautiful: Let z = cos t + i sin t then dz = (-sin t + i cos t) dt = i (cos t + i sin t) dt = i z dt. Integrating: int dz/z = int i dt or ln z = i t. Exponentiating: z = exp i t. :let

\ln z = it + C_1

z = e^{it +C_1} = e^{C_1}\cdot e^{it}

C = e^{C_1} \rightarrow z = Ce^{it}

---- The proof using Taylor series is silly! If one is allowed to assume the Taylor expansions of ''exp(x)'', ''sin(x)'' and ''cos(x)'', then just add the series for ''cos x + i sin x'' and note that it is the same as the series for ''exp(i x)''. --User:Zero0000 09:38, 12 Oct 2003 (UTC) You have an error anyway in your proof: i(-sin t + i cos t) = - (cos t + i sin t) = -z. I don't think you can differentiate like you're doing in any case since ''z'' is a complex variable (I could be wrong, I haven't done any complex analysis stuff for a while). User:Dysprosia 10:03, 12 Oct 2003 (UTC) No, that part of the proof is fine. The only problematic step is the integration, since it really gives ln z = i t + C for a constant C. One then has to find an argument that C=0. --User:Zero0000 12:46, 12 Oct 2003 (UTC) The argument that C=0 can be easily found by substituting t=0 and evaluating. --Komp, 10th Sept 2004. == Taylor Series for e^x == I'm a little confused about one thing for the e^ix = cosx + (i)sinx derivation. It looks like the Taylor Series of e^ix is exanded around the point a = 0. Wouldn't that mean the proof is only valid near x = 0? The series is valid for all x. User:Charles Matthews 09:42, 18 Dec 2003 (UTC) : Radius of convergence of exp x is infinite, btw. User:Dysprosia 09:48, 18 Dec 2003 (UTC) That explains it, thanks a lot! :You could expand it about any point, and as long as you took all (an infinite number of) the elements, it would still work. If you're only going to use a few terms you should expand it about whatever local operating point you're using. User:Moink 05:12, 13 Jan 2004 (UTC) ---- I would like to suggest moving the complex analysis to the top, above the other one. In my experience it's much more common. User:Moink 05:12, 13 Jan 2004 (UTC) How about more -- split this into 2 articles; the two results have almost nothing to do with each other. They don't belong together. ---- I propose replacing the e^ix = cosx + (i)sinx derivation by the following simplified version. It is known that exp(''x''), sin(''x''), and cos(''x'') have Talyor series which converge for all complex ''x'': :

\exp(x) = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots

\sin(x) =  x - \frac{x^3}{3!} + \frac{x^5}{5!} - \cdots

\cos(x) = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} + \cdots

Adding the series for cos(''x'') to ''i'' times the series for sin(''x'') gives the series for exp(''ix''). :It misses some parts of the proof though; the periodicy where i^2 = -1 ; i^3 = -i ; i^4 = 1. User:Sverdrup 14:52, 6 May 2004 (UTC) ::Sverdup is right, but I think the notation in the current proof is more a hindrance than a help. It's much easier to visually see what's going on by writing "dot-dot-dot"'s and collecting terms than by using a jillion sigma notations. :I suggest we use the proof on top of this talk page to motivate the formula, and keep the current taylor series proof as ''the'' proof. We need to be accurate, and we are also elegant if the math is done right with summation etc. User:Sverdrup 22:27, 6 May 2004 (UTC) ::I'm not sure what you mean by "the" proof -- most results have multiple proofs, and this is no exception. The proof using "dz/z = it dt" is good motivation, yes, but it's also a completely rigorous proof, as well, so by including it as a "real" proof, we would not lose any accuracy. I still maintain that the Taylor series proof is much easier to understand without sigma notation, without losing any rigor -- "dot-dot-dot's" are fully rigorous, as long as it's obvious what is intended, which is the case here if enough terms are spelled out. Having four different summations with "4n", "4n+1", "4n+2", and "4n+3" is only going to confuse people who aren't used to the notation for partitioning integers into congruence classes -- they will have to spell out what the sums say for themselves, so why not do it for them? (BTW, in case you wonder why the "dz/z = it dt" proof is rigorous, it comes down to this. We are basically dealing with the analytic continuation of the real exponential to the entire complex plane -- this is known to exist because the Taylor series at z = 0, say, has infinite radius of convergence. So, we can define the exponential as exp(z) = Taylor series. It's pretty trivial to show that d/dz(exp(z)) = exp(z) for all z, everything's abs/unif converg, etc. By the chain rule, d/dz(exp(iz)) = i*exp(iz), i.e. exp(iz) satisfies the diff eq w' = iw. Now, note that if w = cos(z) + isin(z), then this w also satisfies the equation; this means w = C*exp(iz) for some constant C; z = 0 gives 1 = w = C, so w = exp(iz) = cos(z) + isin(z). Now, just take z = x to be real. This is basically what is going on with the shorthand notation "dz/z = it dt". The shorthand notation proof is somewhat glossy over a couple of these details, but then again, a lot of proofs at wikipedia are really just "sketches of a proof".) Let me put a copy of what I would have as my Taylor series proof here, so people can see it and compare. ---- Here is my proposal to replace the current Taylor series proof: ===Derivation=== Here is a derivation of Euler's formula using Taylor series expansions as well as basic facts about the powers of ''i'': :

i^0 = 1, \ i^1 = i, \ i^2 = -1, \ i^3 = -i, \ i^4 = 1, \ i^5 = i, \ i^6 = -1, \ i^7 = -i, \ i^8 = 1, \dots

The functions ''e''^''x'', cos(''x'') and sin(''x'') (assuming ''x'' is a real) can be written as: :

e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots

\cos x = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \cdots

\sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \cdots

and for complex ''z'' we ''define'' each of these function by its series. This is possible because the radius of convergence of each series is infinite. Now, take ''z'' = ''ix'', where ''x'' is real, and note that :

e^{ix} = 1 + ix + \frac{(ix)^2}{2!} + \frac{(ix)^3}{3!} + \frac{(ix)^4}{4!} + \frac{(ix)^5}{5!} + \frac{(ix)^6}{6!} + \frac{(ix)^7}{7!} + \frac{(ix)^8}{8!} + \cdots

= 1 + ix - \frac{x^2}{2!} - \frac{ix^3}{3!} + \frac{x^4}{4!} + \frac{ix^5}{5!} - \frac{x^6}{6!} - \frac{ix^7}{7!} + \frac{x^8}{8!} + \cdots

= \left( 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \frac{x^8}{8!} + \cdots \right) + i\left( x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \cdots \right)

= \cos (x) + i\sin (x)

The rearrangement of terms is justified because each series is absolutely convergent. QED I think people will find it much easier to follow this proof. : One problem is that you wrote "''z'' = ''ix''" but ''z'' is not defined and it does not appear elsewhere. Also, the proof works for all complex ''x'' but you limited it to real ''x''. --User:Zero0000 09:27, 7 May 2004 (UTC) ::I see how might not be entirely clear -- actually, I did say what z is, when I said, "for complex z, we define each of these functions by these series". Also, yes, the proof works for all complex z, but "Euler's formula" is usually taken to mean when z is purely imaginary, primarily for historical reasons (Euler "derived" it for ix, not z); also, when most people say "Euler's formula", they're usually intimating at the periodic nature of exp around the unit circle. But, it's certainly true for any z, and I can add this. :: If ''z'' = ''a'' + ''bi''; e^''z'' = e^''a''e^''bi'', so there is no problem with complex ''x''. User:Sverdrup 13:12, 7 May 2004 (UTC) :I'm convinced, this looks very good. User:Sverdrup 13:12, 7 May 2004 (UTC) :The "dz/z = i dt" argument can be made even more legit for most folks by taking the 2nd order linear diff eq, w'' = -w, gotten by iterating the 1st order one, then everything is real and you don't have to think about analytic continuation, etc. :It would be interesting to note how Euler actually "discovered" this. The way he "proved" it is completely backwards from how it is usually presented in modern form -- he assumed DeMoivre's identity, and did some clever fooling around with i's (infinitely small numbers) and w's (infinitely large numbers), treating them as ordinary numbers. Of course, not rigorous at all, but historically very interesting, providing insight into Euler's brain circuitry. ==Moved Euler characteristic material== I moved the material about the euler characteristic to the Euler_characteristic page. My reasons were *elimination of duplicated material *article should only focus on one topic and not on two totally unrelated topics. For people who are looking for the euler characteristic on this page I have put a short note at the top. I have tried to fix all broken links but I probably forgot some.User:MathMartin 22:18, 2 Aug 2004 (UTC) == The "by calculus" proof == ...is wrong because it ignores the constant of integration. Please fix it! --User:Zero0000 03:23, 5 Dec 2004 (UTC)

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