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Evidence for e to the x.

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[SOUND]

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I've already mentioned this

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remarkable result.
Let's define a function f of x which is

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equal to the value of this power series,
the sum n goes from zero to infinity.

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Of x to the n over n factorial.

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Now, earlier, we've seen that the radius
of convergence of this power series

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is infinite.

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So, this power series converges regardless
of what value I choose for X.

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So this is a function whose domain is all
the real numbers.

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Now the big result here, is that this
function, f of x, ends

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up being equal to the more familiar
function, just e to the x.

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Let me now try to convince you that this
is true.

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Well, first of all, f of 0 is what?

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Well, what happens when I plug in 0 for x?

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Then all of these terms vanish, except for
the n equals 0 term.

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Because by convention, 0 to the 0 is 1.

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So that's 1 over 0 factorial, that's just
1.

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So the value of this power series when x
equals 0 is 1.

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And note that that's also e to the 0th
power, that's also equal to 1.

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Not only do these

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functions agree at a single value, but

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they also satisfied the same differential
equation.

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Well, let's see.

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So, let's look at e to the x.
What's the derivative of e to the x?

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Well, that's itself, right?

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So e to the x is a function, which is its
own derivative.

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Now let's take a look at the derivative of
the power series, the sum n goes from

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0 to infinity of x to the n over n
factorial, right.

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I'm really just asking, what's the
derivative

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of this function that I'm calling f.

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Well I'm differentiating a power series,
and I can do that term by term.

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So the derivative of this power series is
the sum n goes from 1 to infinity, I can

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throw away the n=0 term because that's a
constant

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term and the derivative of a constant is
zero.

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So,

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it's the sum n goes to 1 to infinity

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of the derivative of x to the n over
n-factorial.

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So now I just have to differentiate e to

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the term separately and add all of those
up.

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So this is the sum n goes from 1

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to infinity, well what's the derivative of
x to

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the n that's n times x to the n minus 1,
and then the constant is just n factorial.

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But I can simplify

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this, this is n times x to the n minus 1
over n factorial.

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That is the same as the sum n

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goes from 1 to infinity of n over n
factorial, that's just n minus

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1 factorial in the denominator, and the
numerator is x to the n minus 1.

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But now, if you think about what this
series is, when

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I plug in n equals 1, that's x to the 0

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over 0 factorial.

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When I plug in n equals 2, that's x to the
1 over 1 factorial.

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When I plug in n equals, 3, that's just x
to the 2, over 2 factorial.

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When I plug in n equals 4, that's x to the
3rd, over 3 factorial.

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This is actually the same as just the sum,
n goes

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from 0 to infinity, of x to the n, over n
factorial.

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So what I've shown is that

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f is a function which is also its own

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derivative just like e to the x, because
if

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I differentiate f, if I differentiate this
power series,

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well what I get back is just f again.

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Two functions, both alike in dignity.

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These two star-crossed functions agree at
a single point, and they're changing

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in the same way, and consequently, they
must be the same function.

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And therefore,

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f of x is equal to e to the x, right?
What I'm saying is that e to

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the x is the sum, n goes from zero to
infinity of x to the n over n factorial.

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[SOUND].

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[SOUND]