If it's not (and it's not), you know you need to stretch more. The goal isn't the splits, just some self-determined level.
My math goals are similar. I don't need expert proficiency, just a "touch your toes" understanding for the topics I care about.
Here's an internal dialog I might have to verify my understand of exponents.
Gutcheck: Roughly speaking, what's 2^{100}?
It's a large, even, positive number. (This intuition should appear almost instantly. If it takes 10 seconds of thinking to realize it's large, even, or positive, exponents aren't natural.)
Gutcheck: Roughly speaking, what's 2^{-100}?
It's a tiny, almost undetectable positive decimal. Intuition: It's like going "back in time" by 100 doublings.
Gutcheck: Roughly speaking, what's 2^{i}?
Uh oh. Imaginary exponents! With enough intuition, you realize: "It's on the unit circle, at about ln(2) ~ .693 radians."
There's a few gutchecks here. The first is that an imaginary exponent puts you on the unit circle (no matter the base). The next level is a rough "important constant" gutcheck, where you remember ln(2) ~ .693. (Not as important, but good to remember. It helps with things like the Rule of 72)
Gutcheck: Roughly speaking, what's i^{i}?
Oh, here's a tricky one. Remember how we blurted out that 2^{100} was large, even, and positive? How proud we were of our quick thinking? Well, what can you say about i^{i}, hotshot?
Yikes. Realizing I couldn't instantly rattle of any properties of i^{i} meant my intuition for exponents wasn't complete. After getting an intuition for imaginary exponents, the thought becomes:
i^{i} starts as growth pointing sideways, whose direction is rotated again. It's a positive real number less than 1.0.
Phew. If I truly understand exponents, the gutcheck for 2^{100} and i^{i} should be similar in speed and detail. A painful stretch means I need more understanding.
The gutcheck process doesn't quite translate to text. These internal back-and-forths happen pre-verbally: I think of a question and quickly feel/visualize/remember an analogy. (It's a gutcheck, not a think-aloud-for-minutes check.)
Here's a few examples I run through from time to time:
Imaginary numbers: What's the cube root of -1?
Fourier Transform: What's the transform of [1 0 0 0]
?
[.25 .25 .25 .25]
(using the notation in the Fourier Transform article).Trigonometry: What's the connection between the 6 major trig functions?
Thought: I think "dome, wall, ceiling" and visualize this trigonometry diagram:
Calculus: Explain the derivative of x^{3}
Bayes Theorem: What's the plain-English description?
Exponents: What does discrete vs. compound exponential growth look like?
The key element is speed: an intuitive response should bubble as you hear the question. Struggling for an hour to touch my toes, though admirable, still means I'm not flexible enough.
The goal isn't learning minutia, it's a working understanding of an idea, enough to solve a problem without tremendous effort. It's a diagnostic, not a value judgement. If I struggle, I simply need a better intuition.
Strangely enough, not everyone wants to keep math insights top-of-mind. But pick something that's important to you and occasionally try a 5 second gutcheck on the essentials.
Happy math.
]]>I think of imaginary multiplication as turning your map 90 degrees. East becomes North; no matter how long you drove East, now you're going North.
An imaginary exponent is like turning just the steering wheel. Where you will end up? Depends how long you drive!
That's the intuition, let's work through the details.
The math is straightforward when multiplying by i. We perform a 90-degree rotation around the origin, so 1 becomes 1i, 2 becomes 2i, and so on:
This is Multiplication World, where numbers are plopped on the number line then slid around (added), stretched (multiplied), shrunk (divided), and so on. Rotating a number is new, but not overly strange.
Unfortunately, the Multiplication World perspective isn't great for exponents. If we see exponents as "repeated multiplication" we're stuck when we try to count i times. It's the wrong model.
Nope -- to use i in the exponent, we need to enter Exponent World.
Here, numbers are grown, not simply plopped down on the number line. Every number starts at 1.0, then we run an exponential growth engine at 100% for some period of time:
In Exponent World, a familiar number like "2" is just 1.0, grown for .693 seconds at a 100% continuous interest rate. In other words:
And in general:
e^{x} is an rocketship that pushes our numbers ever further from our starting point of 1. At t=3 we're around 20, and at t=10 we're over 20,000.
So, what happens if we drop an imaginary number into the exponent (e^{1} rightarrow e^{1i})? We keep the same amount of fuel, but rotate our engine sideways:
With regular exponential growth, we expect to speed along the real dimension. With our sideways engine, we'll need to compute what will happen.
Euler's Formula gives us the answer: constant force in a perpendicular direction creates an orbit:
and in general:
A few notes/gotchas:
We always start at 1.0. When seeing e = 2.718..., it's tempting to think growth starts from 2.718. But no -- when we write e^{x} we're still begin the growth process at 1.0 (e^{0} = 1 implies no change from 1.0).
I try to remember we can swap e for its official definition at any time:
The only starting point we see in the definition is 1.
Every number orbits at a radius of 1.0. In exponent world, every number is grown from 1.0, just with varying amounts of fuel. When we put the engine sideways, the orbits are at 1.0, for varying distances around the circle.
The orbit doesn't get faster. Regular exponential growth has runaway compounding because our changes accumulate in the same direction. With sideways growth, changes don't accumulate (always in a new direction) and we spin at a constant speed. In other words, e^{10} is thousands of times larger than e^{1}, but e^{10i} is only ten times around the circle compared to e^{i}.
So, a setup like 2^{i} tells us to use .693 units of fuel in a sideways direction:
To get the coordinates for our final position, we see how far ln(2) = .693 units of fuel takes us around the unit circle:
Phew! Working out 2^{i} (rotated exponential growth) is much trickier than 2i (a simple rotation).
Now let's get tricky. What is i^{i}?
Remember, we're in Exponent World, and even i is something we had to grow to! In other words, we start at 1.0 and orbit a quarter of the way around the circle (90 degrees, or frac(π)(2) radians).
Whoa. Don't like how i appears in its own exponential definition? It must also bother you that every word in the dictionary is defined by other words.
Coming back to i^{i}, we have two operations.
Ah. The result is
And we end up with a real number. Intuitively, we can roughly predict this because we start at 1.0 and point the engine backwards on the real axis. Once you can mentally estimate the direction i^{i} goes, in seconds, you've got a sold intuition on imaginary exponents.
Happy math.
Just for fun, what about i · i^{i}?
We know that i^{i} is a purely real number smaller than 1.0. The first i (doing the multiplication) will just rotate us, so now we have a purely imaginary number smaller than 1i.
Not bad! Work through the exponents, then rotate the final position.
From a physics perspective, if f(x) = e^{ix} is our position, then f'(x) is our velocity. Working this out we get:
In other words, our velocity is perpendicular to our position. Taking the derivative of e^{ix} might seem weird, but treat i like any other constant: frac(d)(dx) e^{ax} = ae^{ax}.
]]>First off, e was discovered, not chosen. Think of the speed of light, c. It wasn't originally decided to be 299,792,458 m/s -- we did experiments and realized under ideal, universal conditions (a vacuum), this was the fastest light could move^{1}.
Let's look at growth and ask under ideal, universal conditions, what's the fastest something can possibly grow? Ideal, universal assumptions would be:
Turning these assumptions into a formula, we get:
If we actually use the formula (using large values of n for more accuracy) we get e = 2.718281828459...
Objection: But 13.74^{x} can model exponential growth just like e^{x} can!
Sure. But what assumptions did you make to get 13.74? They probably weren't "unit rate, unit time, perfectly compounded". (You can pick k as the speed of light through Kool Aid too -- but why?)
Arguably, 2^{n} is also universal ("the discrete e"), because you have zero compounding (n is an integer like 0, 1, 2, 3). Instead of perfectly continuous, it's perfectly non-continuous (discrete), and we take growth step-by-step.
So, I'd say either 2^{n} (in discrete systems) or e^{x} (in continuous systems) are "universal".
Objection: But things can grow faster than e^{x}, which is just 2.71828^{x} -- what about 13.74^{x}?
What is it with you and 13.74? Yes, you can beat e^{x} in an exponential footrace, if you use a rate more than 100%. 13.74^{x} is really [e^{ln(13.74})]^{x}. Because ln(13.74) ~ 2.6, you are assuming a 260% continuous interest rate, more than the 100% e^{x} uses. (Alternatively, you can grow for 260% of the unit time period that e^{x} uses.)
Related:
Funny enough, in 1983 c was decreed to be 299,792,458 m/s by redefining the length of a meter. Similarly, you could decide that e is a clean "10" in your base-e number system. ↩
What goes through our head the first time we see this?
Ok, y is usually the final value in an equation. It's based on two parts... mx and b. If x goes up by 1 let's say, we only change y by... m.
Hrm, let's try an example. If m=.3, and x goes from 0 to 10, we go up by 10 * .3 = 3. Ok. Got that part.
Now what does b do? It adds something on top of whatever we just figured out. If x = 0, I guess we just have y = b.
There's a lot of Math <> English translation. In math jargon, we say the slope (m) determines the direct rate of change between x and y, and the y-intercept (b) adds a starting value.
Ugh. It's abstract and doesn't resonate. Here's a more natural wording:
Makes sense, right? Start somewhere, make a change, get to the new location.
Zooming in, we realize a change has two parts: how fast are we changing, and for how long?
Ah. I can see the role of each of the pieces. Turning this into succint math notation:
"Start at b, add the rate of change (m) times the duration (x), and you end up at y."
We aren't forced to graph the equation and describe b as the "y intercept". Who thinks (or talks) like that?
In plain English:
Boom. Later, if we decide to graph our line (fun fun), then b is the height on the y-axis when x=0. But thinking in terms of our starting value (the bias) just clicks better.
The graphed line is a list of all possible outcomes under some set of circumstances. We might not wait at all (x =0), or wait for a bit (x=1), wait for a long time (x = 100), or look back in time (x = -3). The line shows what would have happened in each of those scenarios.
People claim they don't think in terms of math. Sure, you might not describe things as slope/intercept form, but I'd guess "original + change = new" is pretty familiar to you. Even the simplest equations can use an intuitive wording.
Q: Find the equation for a line between (3, 4) and (7, 12)
Let's think about this intuitively. We know ultimately we want an equation of the form:
position = start + change
For the two points we have, let's look at the change:
Between those points, x changed by 4 and y changed by 8. The rate was 8/4 = 2, so the equation should look like:
position = start + x * 2
We don't know where we started, but whenever x changes, y changes by 2 times that amount.
Ok. Now what was the starting position, when x = 0? Let's see the change between our default value of (0, ?) to (3, 4).
Well, if x changed by 3, and the rate of change is 2, then y changed by 3*2 = 6. The original value was therefore 4 - 6 = -2.
The final equation is then:
position = start + change
position = start + x * 2
position = -2 + x * 2
and written in more standard y = mx + b notation,
y = 2x - 2
Let's double check. We start at (0, -2) and go 3 units on the x coordinate. Our final position is 2*3 - 2 = 6 -2 = 4
, which matches up with (3, 4).
This approach is slower than a plug-and-chug shortcut, but the idea is to figure out exactly what you're doing:
Happy math.
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