Evidence for e to the x.
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I've already mentioned this
remarkable result.
Let's define a function f of x which is
equal to the value of this power series,
the sum n goes from zero to infinity.
Of x to the n over n factorial.
Now, earlier, we've seen that the radius
of convergence of this power series
is infinite.
So, this power series converges regardless
of what value I choose for X.
So this is a function whose domain is all
the real numbers.
Now the big result here, is that this
function, f of x, ends
up being equal to the more familiar
function, just e to the x.
Let me now try to convince you that this
is true.
Well, first of all, f of 0 is what?
Well, what happens when I plug in 0 for x?
Then all of these terms vanish, except for
the n equals 0 term.
Because by convention, 0 to the 0 is 1.
So that's 1 over 0 factorial, that's just
1.
So the value of this power series when x
equals 0 is 1.
And note that that's also e to the 0th
power, that's also equal to 1.
Not only do these
functions agree at a single value, but
they also satisfied the same differential
equation.
Well, let's see.
So, let's look at e to the x.
What's the derivative of e to the x?
Well, that's itself, right?
So e to the x is a function, which is its
own derivative.
Now let's take a look at the derivative of
the power series, the sum n goes from
0 to infinity of x to the n over n
factorial, right.
I'm really just asking, what's the
derivative
of this function that I'm calling f.
Well I'm differentiating a power series,
and I can do that term by term.
So the derivative of this power series is
the sum n goes from 1 to infinity, I can
throw away the n=0 term because that's a
constant
term and the derivative of a constant is
zero.
So,
it's the sum n goes to 1 to infinity
of the derivative of x to the n over
n-factorial.
So now I just have to differentiate e to
the term separately and add all of those
up.
So this is the sum n goes from 1
to infinity, well what's the derivative of
x to
the n that's n times x to the n minus 1,
and then the constant is just n factorial.
But I can simplify
this, this is n times x to the n minus 1
over n factorial.
That is the same as the sum n
goes from 1 to infinity of n over n
factorial, that's just n minus
1 factorial in the denominator, and the
numerator is x to the n minus 1.
But now, if you think about what this
series is, when
I plug in n equals 1, that's x to the 0
over 0 factorial.
When I plug in n equals 2, that's x to the
1 over 1 factorial.
When I plug in n equals, 3, that's just x
to the 2, over 2 factorial.
When I plug in n equals 4, that's x to the
3rd, over 3 factorial.
This is actually the same as just the sum,
n goes
from 0 to infinity, of x to the n, over n
factorial.
So what I've shown is that
f is a function which is also its own
derivative just like e to the x, because
if
I differentiate f, if I differentiate this
power series,
well what I get back is just f again.
Two functions, both alike in dignity.
These two star-crossed functions agree at
a single point, and they're changing
in the same way, and consequently, they
must be the same function.
And therefore,
f of x is equal to e to the x, right?
What I'm saying is that e to
the x is the sum, n goes from zero to
infinity of x to the n over n factorial.
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