Calculus Two: Sequences and Series is an introduction to sequences, infinite series, convergence tests, and Taylor series. The course emphasizes not just getting answers, but asking the question "why is this true?"

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From the course by The Ohio State University

Calculus Two: Sequences and Series

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Calculus Two: Sequences and Series is an introduction to sequences, infinite series, convergence tests, and Taylor series. The course emphasizes not just getting answers, but asking the question "why is this true?"

From the lesson

Power Series

In this fifth module, we study power series. Up until now, we had been considering series one at a time; with power series, we are considering a whole family of series which depend on a parameter x. They are like polynomials, so they are easy to work with. And yet, lots of functions we care about, like e^x, can be represented as power series, so power series bring the relaxed atmosphere of polynomials to the trickier realm of functions like e^x.

- Jim Fowler, PhDProfessor

Mathematics

Evidence for e to the x.

[SOUND]

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.

[SOUND].

[SOUND]

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