Welcome to a video on Differentiating
and Integrating with Power Series.
By now we should all be fairly familiar with
the most common power series as we see listed here.
And, we can use these power series to approximate
the derivative and anti-derivatives
of a variety of functions.
For example, we can use our series to approximate
the derivative of f of x equals sine x.
Now, of course we know the derivative of sine x
is equal to cosine x.
Let's go ahead and approximate this derivative
using a power series.
There's a couple ways of doing this.
We can either differentiate this summation here
or we can differentiate the right side
by differentiating these individual terms.
Let's start by finding the derivatives
of the first several terms of the expansion
of the power series.
So, we'll find the derivative of x
minus x cubed divided by three factorial
plus x to the fifth divided by five factorial
minus x to the seventh divided by seven factorial
and so on.
And, these derivatives should give us
the first several terms of the power series for cosine x.
So, this would be one minus three x squared
divided by three factorial
plus five x to the fourth divided by five factorial.
Minus seven x to the sixth
divided by seven factorial and so on.
Let's go ahead an simplify these terms.
The three and the three factorial would simplify
to two factorial.
So, we'd have minus x squared over two factorial.
Plus, the five and the five factorial would simplify.
We'd have a denominator of four factorial.
And again, the seven and the seven factorial simplify
like so.
This should be the first several terms
of the power series for cosine x.
Before we go and take a look at this
let's go ahead and find the derivative
of this summation with respects to x.
So, the negative one and the power of n
will not be affected by finding the derivative.
And, to find the derivative of this term
with respects to x we would have two n plus one
times x to the two n plus one minus one
or x to the two n and we still have a denominator
of two n plus one factorial.
But, remember that would be two n plus one
times two n factorial.
And, so we can simplify a factor of two n plus one.
So, the derivative would be the summation
from n equals zero to infinity.
Negative one to the n
times x to the two n divided by two n factorial.
Now, let's go ahead and go back to our formulas
and verify that both of these would be the power series
for cosine x.
Well, here's cosine x.
And, this does match the derivative that we found
using the terms of the power series for sine x.
And, the formula for the summation
is the derivative of the power series
for sine x as we see here.
Now, let's take a look at how we can use
power series to integrate.
We want to integrate sine x divided by x
with respects to x on the interval from zero to 0.5
Now, because of this enneagram we don't have
a nice way to integrate this.
We can use power series to approximate this
depth in integral.
This would be equal to the integral from zero
to 0.5 of the power series
for sine x divided by x.
Well, here's the power series for sine x.
We want to divide this by x, which is the same
as multiplying by one over x.
So, we'll have a factor of x here.
Let's go ahead and simplify this and then we can integrate.
Well, here we have two n plus one factors of x.
And, here we have one factor of x.
So, we can simply this to just x to the power of two n.
Now what we're going to do is find
the first several terms of this power series,
but when we start to evaluate this
we'll keep finding terms until we find the term
that is less than 0.001.
So, when n is equal to zero we're going to have one.
When n is equal to one we're going to have minus
this will be x to the second divided by
three factorial.
When n is two we'll have plus x to the fourth
divided by five factorial,
minus and so on.
I think we have enough terms now
lets go ahead and find the anti-derivative
and then evaluate it at the upper and lower
limits of integration.
So, the anti-derivative of one in respects to x would be x.
Here, we have minus x to the third
divided by three times three factorial.
Plus, this will be x to the fifth
divided by five times five factorial and so on.
So, go ahead and evaluate this at 0.5
and then also at zero.
So, when x is .5 we'll have 0.5 minus
0.5 to the third,
divided by three times three factorial,
plus 0.5 to the fifth
divided by five times five factorial and so on.
And then, when x is equal to zero
these will all be zero.
And, since we want to approximate this
with an error that's less than 0.001
as soon as we find the term that's less that this
we have found the approximation that we're looking for.
So, this is 0.5
minus, let's see what the value of this fraction would be.
.5 cubed
divided by three
and three factorial would be six.
So, you have approximately .0069.
And, if this fraction here is less than 0.001
this value here will be our approximation.
Let's go ahead and check it.
Divided by five times five factorial.
And, this number here is less than .001
and therefore our approximation will be
this difference here.
So, our approximation is approximately .493.
Okay, that's going to do it for this video
thank you for watching.