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

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

No vote yet

1 vote

×

Problem Loading...

Note Loading...

Set Loading...

Easy Math Editor

`*italics*`

or`_italics_`

italics`**bold**`

or`__bold__`

boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

paragraph 2

`[example link](https://brilliant.org)`

`> This is a quote`

Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.`2 \times 3`

`2^{34}`

`a_{i-1}`

`\frac{2}{3}`

`\sqrt{2}`

`\sum_{i=1}^3`

`\sin \theta`

`\boxed{123}`

## 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.

Log in to reply

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

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.

Log in to reply

Log in to reply

Log in to reply

Log in to reply

Log in to reply

Log in to reply

Log in to reply

Log in to reply