### Video Transcript

In this video, weβll learn how to
find a Taylor representation of an arbitrary function by using the Taylor series
expansion. Weβll consider how we can find the
radius of convergence of these series as well as intervals of convergence, primarily
using the ratio test for convergence. We begin by supposing that π is a
function that can be represented by a power series π of π₯ equals π zero plus π
one π₯ minus π plus π two π₯ minus π all squared and so on, where the absolute
value of π₯ minus π is less than some value capital π
. We can begin to work out what ππ,
the coefficients, must be in terms of π. For example, if we let π₯ be equal
to π, we see that π₯ minus π becomes π minus π, which is zero. And then we find that π of π
equals π zero.

Okay, well, thatβs great. But thereβs nothing else we can do
with this function as it stands to help us represent the other coefficients in terms
of π. So instead, we differentiate our
function π with respect to π₯, remembering that we can do so term by term. And π zero is a constant so its
derivative is zero. Then the derivative of π one times
π₯ minus π is just π one. And we could achieve this result by
using the product rule or distributing our parentheses and differentiating as
normal. Similarly, we find the derivative
of π two times π₯ minus π all squared to be two times π two times π₯ minus
π. The derivative of π three times π₯
minus π all cubed is three π three times π₯ minus π all squared and so on.

Once again, we let π₯ be equal to
π. And this time we obtain π prime of
π to be equal to π one. Now, it follows that we can repeat
this process. We differentiate our function once
again with respect to π₯. π double prime of π₯ is two π two
plus two times π three times π₯ minus π plus another term, three times four π
four times π₯ minus π all squared and so on. And this time when we let π₯ be
equal to π we find that π double prime of π is equal to two π two. Weβll repeat this one more time to
help us establish a pattern. The third derivative is two times
three π three plus two times three times four π four times π₯ minus π plus three
times four times five π five times π₯ minus π all squared and so on. And letting π₯ be equal to π now,
we see that π triple prime of π, the third derivative of our function with respect
to π₯, evaluated at π₯ is equal to π is two times three times π three, which we
could write alternatively as three factorial π three.

Now, if we were to continue to
different and let π₯ be equal to π, we would find that the πth derivative of our
function evaluated at π is two times three times four times five, all the way up to
π lots of ππ. In other words, itβs π factorial
times ππ. Weβre going to solve the equation
the πth derivative of π evaluated at π equals π factorial ππ for ππ. When
we do, we find that ππ is equal to the πth derivative of π evaluated at π over
π factorial. Itβs useful to remember that this
formula does remain valid even if π is equal to zero, since zero factorial is just
one and the zeroth derivative of π is simply π. So we obtain our first theorem. If
π has a power series representation at π β β that is, if π of π₯ is equal to the
sum of ππ times π₯ minus π to the πth power for values of π between zero and
infinity β β where the modules of π₯ minus π is less than sum capital π
. Then its coefficients are given by
the formula ππ equals the πth derivative of π evaluated at π over π
factorial.

Weβre going to substitute this back
into our original series. We find that if π has a power
series expansion at π, then the Taylor series of the function π at π or about π
is the sum of the πth derivative of π evaluated at π over π factorial times π₯
minus π to the πth power for values of π between zero and infinity. And this can alternatively be
written as π of π plus π prime of π over one factorial times π₯ minus π plus π
double prime of π over two factorial times π₯ minus π all squared and so on. Itβs useful to know that if π is
equal to zero, we do have a special case that has its own name. Itβs called the Maclaurin
series. And we wonβt be investigating its
properties in this video. Letβs now have a look at a couple
of examples of how to apply the formula for the Taylor series.

What are the first four terms of
the Taylor series of the function π of π₯ equals the square root of π₯ about π₯
equals four?

We recall that the Taylor series of
a function π about π is given by the sum of the πth derivative of π evaluated at
π of π factorial times π₯ minus π to the πth power for values of π between zero
and infinity. So what do we know about our
function? Itβs given as π of π₯ equals the
square root of π₯, which we might alternatively say is π₯ to the power of
one-half. We want to find the Taylor series
about π₯ equals four, so weβre going to let π be equal to four. Using the second form of the Taylor
series, we find that π of π₯ is equal to π of four plus π prime of four over one
factorial times π₯ minus four plus π double prime of four over two factorial times
π₯ minus four squared plus π triple prime of four over three factorial times π₯
minus four cubed.

We can work out π of four fairly
easily. Weβll substitute four into our
original function. But to be able to work out π prime
of four, π double prime of four, and π triple prime of four, weβre going to need
to differentiate our function π of π₯ with respect to π₯ three times. We said that π of π₯ was equal to
π₯ to the power of one-half and we recall that to differentiate a polynomial term,
we multiply the entire term by the exponent and then reduce that exponent by
one. This means π prime of π₯, the
first derivative of π is a half π₯ to the power of negative one-half. π double prime of π₯ is negative a
half times a half π₯ to the power of negative three over two, which simplifies to
negative one-quarter π₯ to the power of negative three over two. Finally, π triple prime of π₯ is
negative three over two times negative a quarter π₯ to the power of negative five
over two, which is three-eighths π₯ to the power of negative five over two.

Weβre now going to substitute π₯
equals four into our functions for π of π₯, π prime of π₯, π double prime of π₯,
and π triple prime of π₯. π of four is the square root of
four; itβs simply two. π prime of four is a half times
four to the power of negative one-half which is a quarter. π double prime of four is negative
one-quarter times four to the power of negative three over two, which is negative
one over 32. And π triple prime of four is
three-eighths times four to the power of negative five over two, which is three over
256. Our final job is simply to
substitute these into our expansion. When we do, we find the first four
terms of the Taylor series of the function π of π₯ equals the square root of π₯
about π₯ equals four to be two plus a quarter times π₯ minus four minus one over 64
times π₯ minus four squared plus one over 512 times π₯ minus four cubed.

Weβll have a look at another
example of this type.

Consider the function π of π₯
equals cos of π₯. Find the Taylor series expansion of
π of π₯ equals cos of π₯ at π₯ equals π and write the Taylor series expansion of
π of π₯ in sigma notation.

And we recall that the Taylor
series expansion of a function about π₯ equals π is given by π of π₯ equals π of
π plus π prime of π over one factorial times π₯ minus π plus π double prime of
π over two factorial times π₯ minus π all squared and so on. In this question, our function is
cos of π₯. And weβre looking to find the
Taylor series expansion at π₯ equals π. So weβre going to let π be equal
to π. And so we see that π of π₯ here is
equal to π of π plus π prime of π over one factorial times π₯ minus π and so
on. Now, we can quite easily evaluate
π of π. We would substitute π into the
function cos of π₯. But what about π prime of π, π
double prime of π, and so on?

Well, weβre going to differentiate
our function with respect to π₯. We might recall the first
derivative of cos of π₯ to be negative sin of π₯. Then to get the second derivative,
we differentiate negative sin π₯ with respect to π₯ and we get negative cos π₯. Differentiating once more, we find
that π triple prime of π₯ is sin π₯. And in fact, weβll repeat this
process one more time because when we differentiate sin π₯, we get cos π₯ again. And you might notice, we have a
cycle. The fifth derivative of π will be
negative sin π₯, the sixth derivative will be negative cos π₯, and so on. And letβs use all of these to
evaluate π of π, π prime of π, π double prime of π, and so on and look for a
pattern. π of π is cos π, which is
negative one. π prime of π is negative sin π,
which is zero. π double prime of π is negative
cos π, which is one. And π triple prime of π is sin
π, which is once again zero.

It, of course, follows that the
fourth derivative of π will be cos of π again, which is negative one. Letβs substitute all of these back
into our Taylor series expansion. Now of course, every other term is
going to be equal to zero as weβve seen. And so we find the Taylor series
expansion to be negative one plus a half times π₯ minus π all squared minus one
over 24 times π₯ minus π to the fourth power plus one over 720 times π₯ minus π to
the sixth power and so on. The second part of this question
asks us to write the Taylor series expansion of π of π₯ in sigma notation. And so letβs see if thereβs a way
we can find a pattern. We saw that π prime of π was
equal to zero, while π triple prime of π is equal to zero. And extending the pattern, we would
see that the fifth derivative, the seventh derivative, and so on would also be equal
to zero. And so our derivative alternate
between negative one, one, negative one, and so on.

Similarly, our denominators are
factorials of ascending even numbers each time this even number is the same as the
exponent. So letβs define that even number as
two π. Then the denominator is two π
factorial and the exponent of π₯ minus π is two π. To achieve alternating powers of
negative one, we write negative one times π plus one. That means when π is zero,
negative one to the power of one is negative one, when π is one, negative one to
the power of two gives us one, and so on. And so we find that π of π₯ equals
the sum of negative one to the power of π plus one times π₯ minus π to the power
of two π over two π factorial for values of π between zero and infinity.

Weβre also able to find the radius
of convergence of a Taylor series. We say that there exists the number
capital π
such that the power series will converge for absolute values of π₯ minus
π less than π
and will diverge for absolute values of π₯ minus π greater than
π
. And this number capital π
is
called the radius of convergence of the series. And note that the series may or may
not converge, if the absolute value of π₯ minus π is equal to π
. We then say that the interval of
convergence is an interval containing all the values of π₯, for which the series
converges. Now, this may or may not include
the end points of the interval and even be the whole set of real numbers. And we say that if we know the
radius of convergence of a power series is π
, then for values of π₯ greater than π
minus π
and less than π plus π
, the power series converges. For values of π₯ less than π minus
π
and greater than π plus π
, the power series diverges. It also follows that the power
series will always converge for π₯ is equal to π. Letβs have a look at an application
of this theory.

Consider the function π of π₯
equals π to the power of two π₯. Find the Taylor series
representation of π about π₯ equals three. And in fact, there are two further
parts of this question that weβll consider in a moment.

We begin by recalling that the
Taylor series of a function π about π is given by the sum of the πth derivative
of π evaluated at π over π factorial times π₯ minus π to πth power values of π
between zero and infinity. And so what do we know about our
function? Well, itβs given by π of π₯ equals
π to the power of two π₯. We want to find the Taylor series
about π₯ equals three. So we let π be equal to three. Then we see that π of π₯ is equal
to the sum of the πth derivative of π evaluated at three of π factorial times π₯
minus three to the πth power for values of π between zero and infinity. This is, of course, equal to π of
three plus π prime of three over one factorial times π₯ minus three plus π double
prime of three over two factorial times π₯ minus three all squared and so on.

Itβs quite clear that weβre going
to need to work out π of three, π prime of three, π double prime of three, and
possibly look for a pattern. π of three is quite
straightforward. Itβs π to the power of two times
three which is, of course, π to the power of six. So letβs next look for the
derivative of π of π₯. The derivative of π to the power
of two π₯ is two π to the power of two π₯. So π prime of three must be two π
to the power of two times three, which is two π to the power of six. Weβll differentiate once again. This time we get four π to the
power of two π₯. And π double prime of three is for
π to the power of two times three, which is four π to the sixth power.

Now actually, we can see that the
πth derivative of π of π₯ is equal to two to the πth power times π to the power
of two π₯. And this means the πth derivative
of π evaluated at three is two to the πth power times π to the sixth power. Our Taylor series expansion using
sigma notation is, therefore, the sum of two to the πth power times π to the sixth
power times π₯ minus three to the πth power all over π factorial for values of π
between zero and infinity.

The second part of this question
asks us to find the interval of convergence of the Taylor series representation of
π about π₯ equals three.

We say that the intervals of all
π₯s for which the power series converges is called the interval of convergence of
the series. Now actually, we know that the
power series will converge for π₯ equals π. So here π₯ equals three. But thatβs all we know right
now. Weβll use the ratio test to
determine the rest of the π₯s for which the power series will converge. This says suppose that we have some
series, the sum of ππ, and we defined π to be equal to the limits as π
approaches infinity of ππ plus one over ππ. If π is less than one, the series
is absolutely convergent and hence convergent. This is the scenario weβre
interested in. So our π must be the limit as π
approaches infinity of the absolute value of two to the power of π plus one times
π to the sixth power times π₯ minus three to the power of π plus one over π plus
one factorial divided by two to the πth power times π to the sixth power times π₯
minus three to the πth power over π factorial.

Now of course, when we divide by a
fraction, itβs the same as multiplying by the reciprocal of that fraction. So we can rewrite our limit as
shown. Then we notice we can perform some
simplification. We can cancel out π to the power
of six and two to the power of π. Similarly, we can cancel π₯ minus
three to the power of π. And if we, finally, recall that π
plus one factorial is the same as π plus one times π factorial, we see that we can
cancel by dividing through by π factorial. Well, this all simplifies really
nicely. We see that π is equal to the
limit as π approaches infinity of the absolute value of two times π₯ minus three
over π plus one.

Now in fact, the expression two
times π₯ minus three is independent of π. So we can factor this, remembering
that we must write the absolute value of two times π₯ minus three. And so π is equal to the absolute
value of two times π₯ minus three times the limit as π approaches infinity of the
absolute value of one over π plus one. Now, as π gets larger, one over π
plus one gets smaller. So the limit approaches zero. This means π is equal to zero. This is of course less than
one. And it doesnβt change no matter the
value of π₯. And so we say our series is
absolutely convergent for all values of π₯. Its interval of convergence is
negative infinity to infinity.

The third part of this question
asks, what is the radius of convergence of the Taylor series representation of π
about π₯ equals three?

Recall that there exist a number
capital π
, which we call the radius of convergence, so that the power series will
converge for absolute values of π₯ minus π less than π
and will diverge for
absolute values of π₯ minus π greater than π
. Well here, weβve seen that our
power series is absolutely convergent. And thereβs a special case for the
radius of convergence. In cases such as these, we say that
the radius of convergence is infinity. Capital π
is, therefore, equal to
infinity.

In this video, weβve seen that if
π has a power series expansion at π, then the Taylor series of the function π at
π or about π is the sum of the πth derivative of π evaluated at π over π
factorial times π₯ minus π to the πth power for values of π between zero and
infinity. We saw that capital π
represents
the radius of convergence such that the power series will converge for absolute
values of π₯ minus π less than π
and will diverge for absolute values of π₯ minus
π greater than π
. And finally, we learned that we can
use the ratio test or other suitable tests to find the intervals of convergence. The power series converges for
values of π₯ greater than π minus π
and less than π plus π
. And it follows that the power
series will always converge when π₯ is equal to π.