# 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
6 years, 2 months ago

This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.

When posting on Brilliant:

• Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused .
• Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting "I don't understand!" doesn't help anyone.
• Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge.

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold
- bulleted- list
• bulleted
• list
1. numbered2. list
1. numbered
2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in $$ ... $$ or $ ... $ to ensure proper formatting.
2 \times 3 $2 \times 3$
2^{34} $2^{34}$
a_{i-1} $a_{i-1}$
\frac{2}{3} $\frac{2}{3}$
\sqrt{2} $\sqrt{2}$
\sum_{i=1}^3 $\sum_{i=1}^3$
\sin \theta $\sin \theta$
\boxed{123} $\boxed{123}$

Sort by:

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.

- 6 years, 2 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.

- 6 years, 2 months 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.

- 6 years, 2 months ago

But why did you choose the base $a = e$?

- 6 years, 2 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.

- 6 years, 2 months ago

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

- 6 years, 2 months ago

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

- 6 years, 2 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.

- 6 years, 2 months ago

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

- 6 years, 2 months ago

Boredom at my electrodynamics class.

- 6 years, 2 months ago