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One way to think about the function E to the T is to ask what properties does it have.

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Probably the most important one, and from some points of view, the defining property,

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is that it is its own derivative. Together with the added condition that in putting 0 returns

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one, it's actually the only function with this property. And you can illustrate what this means

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with a physical model. If E to the T describes your position on a number line as a function of time,

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then you start at the number one. And what this equation is saying is that your velocity,

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the derivative of position is always equal to that position. The farther away from 0 you are,

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the faster you move. So even before knowing how to compute E to the T exactly,

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going from a specific time to a specific position, this ability to associate each position with a

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velocity paints a very strong intuitive picture of how the function must grow. You know that you'll

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be accelerating and at an accelerating rate with an all-around feeling of things getting out of hand quickly.

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And if you add a constant to that exponent, like E to the two times T, the chain rule tells

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us that the derivative is now two times itself. So at every point on the number line,

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rather than attaching a vector corresponding to the number itself, first double the magnitude

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of the position, then attach it. Moving so that your position is always E to the T,

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is the same thing as moving in such a way that your velocity is always twice your position.

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The implication of that two is that our runaway growth feels all the more out of control.

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If that constant was negative, say negative 0.5, then your velocity vector is always negative 0.5

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times your position vector, meaning you flip it around 180 degrees and scale its length by a half.

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Moving in such a way that your velocity always matches this flipped and squished copy of your

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position vector, you'd go the other direction, slowing down in an exponential decay towards 0.

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But what about if that constant was I, the square root of negative 1? If your position was

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always E to the I T, how would you move as the time T takes forward? Well, now the derivative

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of your position will always be I times itself and multiplying by I has the effect of rotating

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numbers 90 degrees. So as you might expect, things only make sense here if we start thinking

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beyond the number line and in the complex plane. So even before you know how to compute E to the I times

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you know that for any position this might give for some value of time, the velocity at that time

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will be a 90 degree rotation of that position. Drawing this for all possible positions you might

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come across, you get a vector field, where as usual with vector fields, you shrink things down

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to avoid clutter. At time T equals 0, E to the I T will be 1, that's our initial condition,

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and there's only one trajectory starting from that position where your velocity is always

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matching the vector that it's passing through, a 90 degree rotation of the position. It's when

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you go around a circle of radius 1 at a speed of 1 unit per second. So after pi seconds you've

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traced a distance of pi around, so E to the I times pi should be negative 1. After tau seconds

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you've gone full circle. E to the I times tau equals 1, and more generally E to the I times T

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equals a number that's T radians around this unit circle in the complex plane.

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Nevertheless something might still feel immoral about putting an imaginary number up in that

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exponent, and you would be right to question that. What we write as E to the T is a bit of a

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notational disaster, giving the number E and the idea of repeated multiplication, we more emphasis

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than they deserve. But my time is up, so I'll spare you the full rant until the next video.