### Video Transcript

One way to think about the function π to the π‘ 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 inputting zero returns one, itβs actually the
only function with this property. And you can illustrate what this means with a physical model.

If π to the π‘ 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 zero you are, the faster you move. So even before knowing how to compute π to the π‘ 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 youβll 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 π to the two times π‘, 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 π to the two π‘ 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. Youβd go the other direction, slowing down in an exponential decay towards zero.

But what about if that constant was π, the square root of negative one? If your position was always π to the ππ‘, how would you move as the time π‘ takes
forward? Well, now the derivative of your position will always be π times itself and
multiplying by π 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 π to the π times π‘. 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 π‘ equals zero, π to the ππ‘ will be one. Thatβs our initial condition. And there is only one trajectory starting from that position where your velocity is
always matching the vector that itβs passing through. A 90-degree rotation of the position. Itβs when you go around the circle of radius one at a speed of one unit per
second. So after π seconds, youβve traced a distance of π around. So π to the π times π should be negative one. After π seconds, youβve gone full circle. π to the π times π equals one. And more generally, π to the π times π‘ equals a number thatβs π‘ 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 π to the π‘ is a bit of a notational disaster, giving the number π
and the idea of repeated multiplication way more emphasis than they deserve. But my time is up, so Iβll spare you the full rant until the next video.