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# Euler's formula and no sequences

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

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

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

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

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

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

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

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

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

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

or

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

And the solutions is obviosly

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

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

$e^{ix} = \left ( A+iB \right ) \cos{x} + \left ( iA-B \right ) \sin{x}$

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

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

Note by Lucas Tell Marchi
2 years 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 $$f''(x)=-f(x)$$, as you assume we do, then you can get the result by writing $$e^{ix}=f(x) + ig(x)$$, taking the first and second derivative, and using the initial values. · 2 years ago

But you see, I just wanted to find the solutions to the differential equation [1], and I didn't even set $$a = e$$; what I did here is: if $$f$$ assumes only real values, then differential equation [1] is Euler's formula. · 2 years ago

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

Now if we write $$e^{ix}=f(x)+ig(x)$$, then we have, taking derivatives, $$ie^{ix}=f'(x)+ig'(x)$$ and $$-e^{ix}=f''(x)+ig''(x)$$. The second derivative tells us that $$f(x)$$ and $$g(x)$$ are solutions of our Diff Equ, and the initial values ($$f(0)=1, f'(0)=0$$ etc) tell us that $$f(x)=\cos{x}$$ and $$g(x)=\sin{x}$$... done. · 2 years ago

But why did you choose the base $$a = e$$? · 2 years ago

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 $$e^{ix},$$ and Euler's formula is his answer. · 2 years ago

That's what I'm saying. My original differential equation has nothing to do with Euler's. · 2 years ago

The point of your note seems to be to present a proof of Euler's formula?! · 2 years ago

It turned out to have something to do with Euler's formula, but at first it didn't have anything to do with it. · 2 years ago

What motivated you to consider the differential equation $$a^{ix}=f'(x)+if(x)$$? · 2 years ago