I had this thought at lunch today. It's not a sophisticated treatment of the subject, but I found it amusing.

We know that when you differentiate a sinusoid twice, you get back a scaled and negated version of the original. Here the scaling factor is unity.

\[ y = sin(t) \\ \ddot{y} = -sin(t) = -y\]

We also know that double-differentiating an exponential gets us back a scaled (but not negated) version of the original. Here again, the scaling factor is unity.

\[ y = e^t \\ \ddot{y} = e^t = y\]

These two behaviors are tantalizingly similar. So how might we get the exponential to behave like the sinusoid with respect to double-differentiation? Maybe we could throw in the square root of negative one.

\[ y = e^{j t} \\ \ddot{y} = j^2 e^{j t} = -e^{j t} = -y\]

Making the exponent complex makes the exponential behave like a sinusoid with respect to double-differentiation. Hence, we've stumbled onto something like Euler's equation (shown below for reference).

\[ e^{j t } = cos \, t + j \, sin \, t\]

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:

TopNewestGood Lord. Thanks sir for posting these.

Log in to reply

Glad you liked it

Log in to reply