The goals of this video are to determine a
Taylor or Maclaurin series that represents a function
and also to determine the interval of convergence.
We've already talked about Taylor and Maclaurin polynomials
so Taylor and Maclaurin series is just an extension of this.
The Taylor series is a representation of a function
as an infinite series calculated by the values
of its derivatives at a single point.
If f of x is continuously differentiable at x equals c
then the function is equal to the Taylor series
defined as the sum, from n equals zero to infinity
of the nth derivative of the function
evaluated at c, divided by n factorial
times the quantity x minus c raised to the nth
where the Taylor series will be centered at c.
A couple things to mention here.
When n is equal to zero
we'd have the zero derivative of the function
which is actually just the original function
evaluated at c, as we see here
and zero factorial equals one
and x minus c to the zero power is also equal to one
so the first term will just be f of c.
Then, when n is equal to one
we'll have the first derivative evaluated at c
divided by one factorial, which is one
and then we have x minus c to the first
and then the pattern continues from there.
Now, if the series is centered at zero
the series is called the Maclaurin series
so it's the same formula as the Taylor series
except now, c is equal to zero
so we'll always have the nth derivative evaluated at zero
and instead of having x minus c to the nth power
we'll just have x.
Let's go ahead and take a look at a couple of examples.
Here, we want to determine the Maclaurin series
for f of x equals sine x
so again, here's the formula.
What we're going to have to do is
start with the original function
and then find the first four or five derivatives
of this function, evaluate the derivatives at x equals zero
since this is a Maclaurin series
and then see if we can write a formula for the pattern
we see as we find more and more terms.
So f of x equals sine x, so the first derivative
will equal cosine x.
The second derivative will equal negative sine x.
The third derivative would be negative cosine x.
The fourth derivative would be positive sine x
and let's find one more.
The fifth derivative would be equal to cosine x.
Now, we'll go back and evaluate all of these
at x equals zero.
So f of zero would be sine zero, which is zero.
F prime of zero would be equal to cosine zero or one.
F double prime of zero
would be zero again.
The third derivative at zero would be negative one
and then the fourth derivative at zero would be zero
and the fifth derivative at zero would be equal to one again
Now, let's go ahead and find the first six terms
of this Maclaurin series.
So, the first term is just f of zero, which is zero
plus the first derivative evaluated at zero times x.
That will be one times x
plus now we'll have the second derivative
evaluated at zero, that would be zero
divided by n factorial, which is two factorial
times x to the second.
This term is equal to zero
but it is nice to see the pattern.
Plus the third derivative evaluated at zero
that would be negative one
divided by n factorial, that's three factorial
times x to the third
plus the fourth derivative at zero is zero
divided by four factorial times x to the fourth
and we'll go ahead and find one more term.
That'll be plus the fifth derivative at zero will be one
divided by five factorial times x to the fifth
and of course, this continues.
Let's go ahead and simplify this and see if we can
discover the pattern to write the formula
for the Maclaurin series.
Well, this term is zero, this term is zero
and this term is zero
so we have sine x is equal to x
and then, instead of writing plus a negative one
over three factorial, we'll write minus one
over three factorial, x to the third
plus one over five factorial
times x to the fifth.
Here, you can probably see the pattern.
Notice that we have odd exponents on x
and we also have odd factorials in the denominator
as well as alternating signs
so the next term would be minus one over seven factorial
x to the seventh plus one over nine factorial
x to the ninth, and so on.
So, we have sine x equals the summation
from n equals zero to infinity of
Now, notice we have alternating signs
which means we'll have a factor of negative one to the nth.
Our denominator is one factorial
three factorial, five factorial
so we'll have two n plus one factorial
and then we also have x raised to an odd power
so we have x to the two n plus one.
Now, we know this is true when x is equal to zero
because that's where it's centered.
Now let's take a look at the interval of convergence.
Because of the factorials and the exponentials
we'll go ahead and apply the ratio test
so we'll take the limit as n approaches infinity
of a sub n plus one, so we'll replace n with n plus one
so we'll have x to the power of two times n plus one
plus one, all over two times n plus one plus one factorial.
Instead of dividing by a sub n
we're going to multiply by the reciprocal of a sub n
so we'll have two n plus one factorial
all over x to the two n plus one.
Let's go ahead and take a look at this exponent here.
We have two n plus two plus one
so this exponent is really two n plus three
and this is also two n plus three factorial.
Now let's take a look at the factors of x.
This has two n plus one factors of x
and this has two n plus three factors of x.
There are two more factors of x in the numerator
and then two n plus three factorial would be
two n plus three times two n plus two
times two n plus one factorial
so we'll have a two n plus three
and a two n plus two in the denominator.
Now, as n approaches infinity
the denominator increases without bound
so this tells us this limit will always equal zero
for any value of x
therefore the interval of convergence will be from
negative infinity to positive infinity.
Let's go ahead and take a look at this
Maclaurin series graphically.
Here we have the graph of f of x equals cosine x in blue
and then in red, we're going to start adding terms
to the Maclaurin series, and since the interval
of convergence is from negative infinity
to positive infinity, as we increase the number of terms
we're going to see
the red function match perfectly with the blue function.
This is a nice way to show that the interval of convergence
looks like it will be from negative infinity
to positive infinity.
Let's go ahead and take a look at another example.
We want to determine the Taylor series
for f of x equals e to the x
but now it's going to be centered at x equals negative one
so this tells us that c will equal negative one.
Therefore, we'll have a binomial factor of x plus one.
Another nice thing about f of x equals e to the x
all of the derivatives of the function
are also equal to e to the x
so with this information, we can go ahead
and start finding the first five or six terms
of the Taylor series and then hopefully discover
the pattern to write the formula for the infinite series
so we're going to have e to the x.
It's going to equal the function value
when x is equal to negative one.
That's going to be e to the negative one
plus the first derivative evaluated at negative one
times x plus one.
Well, the derivative is the same as the original function
so we'll have e to the negative one
times x plus one to the first
plus now we'll have the second derivative
evaluated at negative one.
That'll still be e to the negative one
divided by two factorial times x plus one to the second
plus the third derivative evaluated at negative one
is going to be e to the negative one again
divided by three factorial
times x plus one to the third, and so on.
I think we have enough information here to see a pattern.
Again, remember, this was divided by zero factorial
and this was divided by one factorial
if that helps recognize the pattern.
The first thing we should notice is that
all of these terms have a factor of e to the negative one
so we could factor that out of the summation
and remember, e to the negative one
is the same as one over e
so I have one over e times the summation
from n equals zero to infinity.
Then we have these factors of x plus one
so we'll have x plus one to the nth.
This should be x plus one to the zero
x plus one to the first, x plus one to the second
and so on, and then our denominator would be n factorial.
Now, this is true when x is equal to negative one
so let's now determine the interval of convergence.
To determine the interval of convergence of this
we'll go ahead and apply the ratio test
so we'll have x plus one to the n plus one power
divided by n plus one factorial
and then, instead of dividing by a sub n
we'll multiply by the reciprocal again
so we'll have n factorial over x plus one to the nth
and this simplifies a little bit easier
than the last example.
Notice we have one more factor
of x plus one in the numerator
and n plus one factorial would be n plus one
times n factorial, so we'll have
n plus one in the denominator.
Again, as n approaches infinity
our denominator increases without bound
and therefore this limit is going to equal zero
for any values of x, which tells us
that the interval of convergence will be
from negative infinity to positive infinity.
Okay, at this point, usually the textbooks provide
a nice list of Taylor and Maclaurin series
or power series, that would look something like this
and so, in the next video, what we're going to do is
use this information to determine
additional Maclaurin or Taylor series.
I hope you found this video helpful.
Thank you for watching.