One way to think about the function E to the T is to ask what properties does it have.
Probably the most important one, and from some points of view, the defining property,
is that it is its own derivative. Together with the added condition that in putting 0 returns
one, it's actually the only function with this property. And you can illustrate what this means
with a physical model. If E to the T describes your position on a number line as a function of time,
then you start at the number one. And what this equation is saying is that your velocity,
the derivative of position is always equal to that position. The farther away from 0 you are,
the faster you move. So even before knowing how to compute E to the T exactly,
going from a specific time to a specific position, this ability to associate each position with a
velocity paints a very strong intuitive picture of how the function must grow. You know that you'll
be accelerating and at an accelerating rate with an all-around feeling of things getting out of hand quickly.
And if you add a constant to that exponent, like E to the two times T, the chain rule tells
us that the derivative is now two times itself. So at every point on the number line,
rather than attaching a vector corresponding to the number itself, first double the magnitude
of the position, then attach it. Moving so that your position is always E to the T,
is the same thing as moving in such a way that your velocity is always twice your position.
The implication of that two is that our runaway growth feels all the more out of control.
If that constant was negative, say negative 0.5, then your velocity vector is always negative 0.5
times your position vector, meaning you flip it around 180 degrees and scale its length by a half.
Moving in such a way that your velocity always matches this flipped and squished copy of your
position vector, you'd go the other direction, slowing down in an exponential decay towards 0.
But what about if that constant was I, the square root of negative 1? If your position was
always E to the I T, how would you move as the time T takes forward? Well, now the derivative
of your position will always be I times itself and multiplying by I has the effect of rotating
numbers 90 degrees. So as you might expect, things only make sense here if we start thinking
beyond the number line and in the complex plane. So even before you know how to compute E to the I times
you know that for any position this might give for some value of time, the velocity at that time
will be a 90 degree rotation of that position. Drawing this for all possible positions you might
come across, you get a vector field, where as usual with vector fields, you shrink things down
to avoid clutter. At time T equals 0, E to the I T will be 1, that's our initial condition,
and there's only one trajectory starting from that position where your velocity is always
matching the vector that it's passing through, a 90 degree rotation of the position. It's when
you go around a circle of radius 1 at a speed of 1 unit per second. So after pi seconds you've
traced a distance of pi around, so E to the I times pi should be negative 1. After tau seconds
you've gone full circle. E to the I times tau equals 1, and more generally E to the I times T
equals a number that's T radians around this unit circle in the complex plane.
Nevertheless something might still feel immoral about putting an imaginary number up in that
exponent, and you would be right to question that. What we write as E to the T is a bit of a
notational disaster, giving the number E and the idea of repeated multiplication, we more emphasis
than they deserve. But my time is up, so I'll spare you the full rant until the next video.