140 lines
7 KiB
Markdown
Executable file
140 lines
7 KiB
Markdown
Executable file
---
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draft: true
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title: Hypercomplex Interest
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summary: You know what, fuck you _rotates your interest rates 90°_
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---
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# Compound Interest
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Y'all remember this from like 4th grade right, I barely do so here's a refresh
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$$ z = Pe^{rt} $$
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- \(P\) is your principal, or how much money you initially put in or took out
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- \(r\) is the interest rate, you want this to be low if you're borrowing and high if you're lending
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- \(t\) is time, unless you have a TARDIS, this one is pretty out of your control
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- \(z\) is how much money you owe/are owed, I know this isn't the standard variable name but bear with me
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Now let's make it spicy
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# Complex Interest
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Who says you need to have a real number for your interest rate? The Fed moves interest rates up and down all the time, when will those cowards start moving it left and right.
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But first, what does it mean to move money to the left? \[Vsauce Sting]
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Let's keep things simple for now and just rotate it around 0 like it's a circle
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We can represent this as a complex number in the form \(z=Pe^{i\theta}\), where \(\theta\) is some angle.
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Hold on a second this looks kinda like the the interest rate formula from earlier. Let's add in time as a factor to get: \(z=Pe^{i\theta t}\).
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Okay so we can see that this is basically just compound interest with an interest rate of \(\theta\).
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What does this even mean, can I use this to get out of student loans? Do I owe MOHELA imaginary money?
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Well I left out an important detail, \(Pe^{i\theta}=P\cos(t\theta)+iP\sin(t\theta)\)
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The proof is trivial and would be left as an exercise to the reader but unfortunately I need to use it later on so here's why this works:
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> [!TODO] Explain Euler's Identity here
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So let's say I owe 1,000 USD at a 5% interest rate (I wish lmao), that would look like:
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> [!TODO] Todo
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That makes sense, but now let's look at what happens if we do \(5i\%\)
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Well the real part ends up being
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$$\Re(z)=P\cos(\frac\pi 2rt)$$
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And the imaginary part ends up being
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$$\Im(z)=P\sin(\frac\pi 2rt)$$
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So if I wait about 30 years after my initial loan, I can owe none money with left bread, if I wait 60 years, MOHELA needs to double my money for free.
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Honestly I think the stock market is a better investment strategy. But that seems a bit too based either way, so no wonder I can't get my loans refinanced like this
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Hm.
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_Morpheus voice_ what if I told you there's more than one way to rotate a cow?
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> [!TIP] Okay listen closely because this next part is how I got my partner to fall for me
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# Split-Complex Interest
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Normally when you rotate something you move it in a circle, which is cool and all, but what if you can rotate it in the opposite of a circle?
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Introducing: the Hyperbola
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So with the unit circle we had an equation like \(x^2+y^2=1\), well there's a unit Hyperbola too and the equation for that is \(x^2-y^2=1\).
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There's actually a lot more stuff that circles have that have a hyperbolic equivalent.
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Remember \(\sin\) and \(\cos\)? There's also a \(\sinh\) and \(\cosh\), don't ask me to explain these because we didn't cover them at all in high school.
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Here's the big one that we care about though, we know multiplying something by \(i\) rotates it around a circle, there's actually a \(j\) that rotates something around a hyperbola.
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But what exactly is this mysterious \(j\)?. It's shrimple as really, \(j=\sqrt{1}\).
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> [!QUOTE] Isn't that just 1?
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No. don't think about it too hard
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> [!QUOTE] So what does \(e^{\theta j}\) break down into, if anything?
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I wouldn't be asking this rhetorical if there wasn't a semi-interesting answer
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It's actually \(Pe^{j\theta t} = P\cosh(t\theta)+jP\sinh(t\theta)\)
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# Dual Interest rates
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Okay this one isn't as interesting but I want to include it for completeness,
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We have \(i=\sqrt{-1},\ j=\sqrt{1}\), now get ready for \(\varepsilon=\sqrt{0}\). I guess was taken when they invented this one?
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> [!Todo] todo
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while we're on the topic of dual numbers, lets try shoving \(x+\varepsilon\) into some functions for shits and giggles:
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$$(x+\varepsilon)^2=x^2+2x\varepsilon+\varepsilon^2=2x\varepsilon$$
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> [!quote] Okay that was a waste of time, was that supposed to be interesting?
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Yes. I'm getting there
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$$\sin(x+\varepsilon)=(x+\varepsilon)-\frac{(x+\varepsilon)^3}{3!}+\frac{(x+\varepsilon)^5}{5!}\ldots$$
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Okay now lets expand that out
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$$=(x+\varepsilon)-\frac{x^3+3x^2\varepsilon+3x\varepsilon^2+\varepsilon^3}{3!}+\frac{x^5+5x^4\varepsilon+10x^3\epsilon^2+10x^2\varepsilon^3+5x\varepsilon^4+\varepsilon^5}{5!}\ldots$$
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of course \(\varepsilon\) to powers higher than 1 just ends up being zero so we can simplify this monstrosity down to
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$$=x+\varepsilon-\frac{x^3+3x^2\varepsilon}{3!}+\frac{x^5+5x^4\varepsilon}{5!}$$
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...except this still doesn't mean too much, lets try factoring \(\varepsilon\) out
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$$(x-\frac{x^3}{3!}+\frac{x^5}{5!}+\ldots)+\varepsilon(1-\frac{x^2}{2!}+\frac{x^4}{4!})$$
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$$=\sin(x)+-\cos(x)\varepsilon$$
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> [!QUOTE] Hold on, what the shit? we the 'imaginary' part is the derivative?
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Yeah actually. If you remember back to 8th grade,
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$$f'(x)=\frac{d}{dx}f(x)=\lim_{h\to0}\frac{f(x+h)-f(x)}{h}$$
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We can kinda replace \(h\to 0\) here with \(\varepsilon\) since it's basically an infinitely tiny number that's not zero or negative, which gives us
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$$f'(x)=\frac{f(x+\varepsilon)-f(x)}{\varepsilon}$$
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> [!note] Dividing by is sus as fuck, but just bear with me here
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$$\varepsilon f'(x)=f(x+\varepsilon)-f(x)$$
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$$f(x+\varepsilon)=f(x)=\varepsilon f'(x)$$
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There you go, typesetting this was a bitch.
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# Errata
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## Matrix Representations
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Fun fact, you can also represent \(i,j,\varepsilon\) as matrices too
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Its already kinda standard to use \(\begin{bmatrix}1&0\\0&1\end{bmatrix}\) to cast real numbers into 2D matrices, but there's two zeros in there doing nothing, maybe we can yoink those for real estate.
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Introducing:
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$$i=\begin{bmatrix}0&1\\-1&0\end{bmatrix},\ j=\begin{bmatrix}0&1\\1&0\end{bmatrix},\ \varepsilon=\begin{bmatrix}0&1\\0&0\end{bmatrix}$$
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These versions actually obey the same multiplication rules as our original derivations of the hypercomplex numbers. Now we can turn something like \(6+3j\) into \(\begin{bmatrix}6&3\\3&6\end{bmatrix}\)
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## Colors
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As somewhat of a convention, I use red, green, and blue in this post to differentiate between graphs relating to complex numbers, split-complex numbers, and dual numbers respectively
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## Honey, I broke the concept of division!
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Yeah... division doesn't really work right when you bring \(j\) and \(\varepsilon\) into the mix since you can multiply non-zero stuff and get zero out
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$$\frac{1}{1+j}\cdot\frac{1}{1-j}=\frac{1}{1-j^2}=\frac{1}{1-1}=\frac10$$
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$$\frac 1\varepsilon \cdot \frac 1\varepsilon=\frac 1{\varepsilon^2}=\frac 10$$
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