The Student Room Group

Euler's formula "from scratch"

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(edited 2 years ago)
Oooooooooo awesome.
Reply 2
I thought the proof comes from Maclaurin's series. I watched it on Khan Academy recently.
Reply 3
This is very cool :colondollar:
A somewhat similar, but shorter argument.

ddxcosx+isinx=icosxsinx=i(cosx+isinx)\dfrac{d}{dx} \cos x + i \sin x = i \cos x - sin x = i(\cos x + i \sin x)

So if z=cosx+isinxz = \cos x + i \sin x, then dzdx=iz\dfrac{dz}{dx} = i z. If we assume separation of variables works(*) we go 1zdz=idx\displaystyle \int \dfrac{1}{z} \, dz = \int i \, dx, so lnz=ix+C\ln z = ix + C. Setting x = 0 we see C = 0 and so z=eixz = e^{ix}.

(*) If you don't want to assume this works, you can use the integrating factor idea from FM to realise that ddxzeix\dfrac{d}{dx} ze^{-ix} is constant. Unless I'm missing something, the only unjustified step with this approach is the assumption that ddxeix=ieix\dfrac{d}{dx} e^{ix} = i e^{ix}.
Original post by DFranklin
A somewhat similar, but shorter argument.

ddxcosx+isinx=icosxsinx=i(cosx+isinx)\dfrac{d}{dx} \cos x + i \sin x = i \cos x - sin x = i(\cos x + i \sin x)

So if z=cosx+isinxz = \cos x + i \sin x, then dzdx=iz\dfrac{dz}{dx} = i z. If we assume separation of variables works(*) we go 1zdz=idx\displaystyle \int \dfrac{1}{z} \, dz = \int i \, dx, so lnz=ix+C\ln z = ix + C. Setting x = 0 we see C = 0 and so z=eixz = e^{ix}.

(*) If you don't want to assume this works, you can use the integrating factor idea from FM to realise that ddxzeix\dfrac{d}{dx} ze^{-ix} is constant. Unless I'm missing something, the only unjustified step with this approach is the assumption that ddxeix=ieix\dfrac{d}{dx} e^{ix} = i e^{ix}.


This is exactly how I first saw Euler's formula derived. Very interesting although I still haven't seen proofs/derivations of certain aspects of it like 1z=lnz\int\frac{1}{z}=lnz etc.
Original post by arkanm
This assumes that their is something special about cosx+isinx.
Well, yes, but there is something special about cos x + i sin x. It's a point on the unit circle. And "as any fule kno", when an object moves in a circle, its velocity (derivative) is at right angles (multiplied by +/- i) to the radius vector. In other words, the realization that cos x + i sin x has an "interesting" derivative is directly motivated by geometrical considerations.

In my proof (although I don't call it one :redface: ) you discover the result sponteneously, starting with the common pythagorean trig formula and deducing stuff, until your are led naturally to the result.
It's subjective, but I find my argument a lot more natural than yours. (Natural enough that I did actually derive this on the fly starting from thinking about complex numbers on rotation - although after about half a line i realised "I've seen this somewhere before").
Reply 7
Original post by arkanm

So we have cosx+isinx=ax\cos x+i\sin x=a^{x}. Differentiating both sides gives sinx+icosx=axlna-\sin x+i\cos x=a^x\ln a,


Original post by DFranklin

ddxcosx+isinx=icosxsinx=i(cosx+isinx)\dfrac{d}{dx} \cos x + i \sin x = i \cos x - sin x = i(\cos x + i \sin x)


Sorry, stupid question here, but why is ddxcosx+isinx=icosxsinx\dfrac{d}{dx} \cos x + i \sin x = i \cos x - sin x? I don't understand that part. :smile:
Reply 8
Original post by arkanm
d/dx(cosx+isinx)=d/dx(cosx)+d/dx(isinx)=d/dx(cosx)+i×d/dx(sinx)=...d/dx(\cos x+i\sin x)=d/dx(\cos x)+d/dx(i\sin x)=d/dx(\cos x)+i\times d/dx(\sin x)=...
(you have to know how to differentiate trig functions)


Oh my god! :eek: Massive brain failure. I assumed the cosx\cos x got turned into icosxi \cos x Sorry, sorry, Understand it fully now. Gosh, I need to get some sleep. :colondollar:
Original post by arkanm
d/dx(cosx+isinx)=d/dx(cosx)+d/dx(isinx)=d/dx(cosx)+i×d/dx(sinx)=...d/dx(\cos x+i\sin x)=d/dx(\cos x)+d/dx(i\sin x)=d/dx(\cos x)+i\times d/dx(\sin x)=...
(you have to know how to differentiate trig functions)



I suspect it may be the 1\sqrt{-1} that is confusing him.

It is just a multiple and is not to be differentiated so can be taken out to the front.

Just like d(2sinx)dx=2cosx\frac{d(2sinx)}{dx}=2cosx.

Good explanation though.
Original post by arkanm
Method 1: n+1n=(n+1n)×n+1+nn+1+n=1n+1+n12n\sqrt{n+1}-\sqrt{n}=(\sqrt{n+1}-\sqrt{n})\times \frac{\sqrt{n+1}+\sqrt{n}}{\sqrt{n+1}+\sqrt{n}}=\frac{1}{\sqrt{n+1}+\sqrt{n}}\approx \frac{1}{2\sqrt{n}}

Method 2: Let x=n+1x=\sqrt{n+1} and y=ny=\sqrt{n}. Naturally squaring both equations we get x2=n+1x^2=n+1 and y2=ny^2=n, so eliminating nn we obtain x2y2=1x^2-y^2=1, then (xy)(x+y)=1(x-y)(x+y)=1, and finally xy=1x+yx-y=\frac{1}{x+y}. Back substituting for x,y we get n+1n=1/(n+1+n)\sqrt{n+1}-\sqrt{n}=1/(\sqrt{n+1}+\sqrt{n}), and then we continue as before.

This removes the rabbit element, essentially breaking down a "flash of insight" into easily understood pieces.
I don't think this is a great example: any experienced mathmo is instantly going to at least think about what happens when you multiply by the conjugate, and this leads you to method 1.

You might say the conjugate is an unbroken down "flash of insight", but if you break it down, the idea is much more "if x^2 and y^2 are nice and xy isn't, then (x-y)(x+y) is likely to be worth considering" than anything in your reasoning.

Your mileage may vary.

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