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

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

## Comments

Sort by:

TopNewestThis 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 · 1 year, 10 months ago

Log in to reply

– Lucas Tell Marchi · 1 year, 10 months 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.Log in to reply

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 · 1 year, 10 months ago

Log in to reply

– Lucas Tell Marchi · 1 year, 10 months ago

But why did you choose the base \(a = e\)?Log in to reply

– Otto Bretscher · 1 year, 10 months 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.Log in to reply

– Lucas Tell Marchi · 1 year, 10 months ago

That's what I'm saying. My original differential equation has nothing to do with Euler's.Log in to reply

– Otto Bretscher · 1 year, 10 months ago

The point of your note seems to be to present a proof of Euler's formula?!Log in to reply

– Lucas Tell Marchi · 1 year, 10 months 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.Log in to reply

– Otto Bretscher · 1 year, 10 months ago

What motivated you to consider the differential equation \(a^{ix}=f'(x)+if(x)\)?Log in to reply

– Lucas Tell Marchi · 1 year, 10 months ago

Boredom at my electrodynamics class.Log in to reply