Euler's formula and no sequences

Let f:ABf:A\rightarrow B be a function of xx only such that ARA\subseteq \mathbb{R} and BRB\subseteq \mathbb{R}. Let us, now, investigate the following differential equation:

aix=Df(x)+if(x);     [1] a^{ix} = Df(x) + if(x) ;\;\;\ \left [ 1 \right ]

Where Df(x)Df(x) stands for the derivative of ff in relation to xx. Now, the relation above gives us iDf(x)=iaix+f(x)iDf(x) = ia^{ix} + f(x) simply by multiplying the relation by ii. Now, differentiating equation [1]\left [ 1 \right ] will give us

ilna  aix=D2f(x)+iDf(x) i \ln{a} \; a^{ix} = D^{2}f(x) + iDf(x)

and using the relation iDf(x)=iaix+f(x)iDf(x) = ia^{ix} + f(x) that we just proved will result in

iaix(lna1)=(D2+1)f(x) ia^{ix}\left ( \ln{a}-1 \right ) = \left ( D^{2}+1 \right )f(x) i(lna1)Df(x)(lna1)f(x)=(D2+1)f(x);     [2] i\left ( \ln{a}-1 \right )Df(x) - \left ( \ln{a}-1 \right )f(x) = \left ( D^{2}+1 \right )f(x) ;\;\;\ \left [ 2 \right ]

Now, as f(x)Rf(x) \in \mathbb{R} for any xx on its domain, it follows that

i(lna1)Df(x)=0 i\left ( \ln{a}-1 \right )Df(x) = 0

As we don't want the trivial solution f(x)=cf(x) = c with cc a constant, we need to take a=ea = e. Now let us find the solution to the rest of the differential equation [2] \left [ 2 \right ] . The equation is

(lna1)f(x)=(D2+1)f(x) - \left ( \ln{a}-1 \right )f(x) = \left ( D^{2}+1 \right )f(x)

or

(D2+lne)f(x)=0\left ( D^{2} + \ln{e} \right )f(x) = 0 (D+i)(Di)=0\left ( D+i \right )\left ( D-i \right ) = 0

And the solutions is obviosly

f(x)=Asinx+Bcosxf(x) = A\sin{x} + B\cos{x}

Feeding it back into [1]\left [ 1 \right ] turns into

eix=(A+iB)cosx+(iAB)sinxe^{ix} = \left ( A+iB \right ) \cos{x} + \left ( iA-B \right ) \sin{x}

as we want e0=1e^{0} = 1 it turns out that A=1iBA = 1 - iB which is only possible if B=0B = 0, because AA and BB must be real once they are part of f(x)f(x), and also A=1A = 1 because A=1iB=10=1A = 1 - iB = 1 - 0 = 1. Then the equation becomes

eix=cosx+isinx e^{ix} = \cos{x} + i\sin{x}

Note by Lucas Tell Marchi
4 years, 2 months ago

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This looks like a correct but pretty roundabout way to derive Euler's formula. If you know the solutions to the differential equation f(x)=f(x)f''(x)=-f(x), as you assume we do, then you can get the result by writing eix=f(x)+ig(x)e^{ix}=f(x) + ig(x), taking the first and second derivative, and using the initial values.

Otto Bretscher - 4 years, 2 months ago

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But you see, I just wanted to find the solutions to the differential equation [1], and I didn't even set a=ea = e; what I did here is: if ff assumes only real values, then differential equation [1] is Euler's formula.

Lucas Tell Marchi - 4 years, 2 months ago

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You assumed that we know that the real solutions of f(x)=f(x)f''(x)=-f(x) are of the form f(x)=asin(x)+bcos(x)f(x)=a\sin(x)+b\cos(x).

Now if we write eix=f(x)+ig(x)e^{ix}=f(x)+ig(x), then we have, taking derivatives, ieix=f(x)+ig(x)ie^{ix}=f'(x)+ig'(x) and eix=f(x)+ig(x)-e^{ix}=f''(x)+ig''(x). The second derivative tells us that f(x)f(x) and g(x)g(x) are solutions of our Diff Equ, and the initial values (f(0)=1,f(0)=0f(0)=1, f'(0)=0 etc) tell us that f(x)=cosxf(x)=\cos{x} and g(x)=sinxg(x)=\sin{x}... done.

Otto Bretscher - 4 years, 2 months ago

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@Otto Bretscher But why did you choose the base a=ea = e?

Lucas Tell Marchi - 4 years, 2 months ago

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@Lucas Tell Marchi I didn't choose it... my country man Euler chose it (hey, it's his number): He asked for the real and imaginary parts of eix,e^{ix}, and Euler's formula is his answer.

Otto Bretscher - 4 years, 2 months ago

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@Otto Bretscher That's what I'm saying. My original differential equation has nothing to do with Euler's.

Lucas Tell Marchi - 4 years, 2 months ago

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@Lucas Tell Marchi The point of your note seems to be to present a proof of Euler's formula?!

Otto Bretscher - 4 years, 2 months ago

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@Otto Bretscher It turned out to have something to do with Euler's formula, but at first it didn't have anything to do with it.

Lucas Tell Marchi - 4 years, 2 months ago

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@Lucas Tell Marchi What motivated you to consider the differential equation aix=f(x)+if(x)a^{ix}=f'(x)+if(x)?

Otto Bretscher - 4 years, 2 months ago

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@Otto Bretscher Boredom at my electrodynamics class.

Lucas Tell Marchi - 4 years, 2 months ago

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