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

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

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

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

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

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Remember to wrap math in `$` ... `$` or `\[` ... `\]` to ensure proper formatting.
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Comments

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.

Otto Bretscher - 6 years, 1 month 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.

Lucas Tell Marchi - 6 years, 1 month 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.

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

@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 $e^{ix},$ and Euler's formula is his answer.

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

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

@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 – What motivated you to consider the differential equation $a^{ix}=f'(x)+if(x)$ ?

@Otto Bretscher – Boredom at my electrodynamics class.

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
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`2^{34}`	$2^{34}$
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`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$