### Video Transcript

In this video, we will learn how to
find the Maclaurin series representation of common functions such as exponential and
trigonometric functions and binomial expansion. Weβll then see some important uses
of Maclaurin expansions, such as using the Maclaurin series expansion to estimate
certain values of functions, find Maclaurin expansions of different forms of common
functions, and to estimate integrals.

We know that this is the Taylor
series of a function π at π. But when we center our
approximation at π equals zero, we get this series which has a special name. We call this the Maclaurin
series. We use this to approximate
functions by firstly finding the first, second, third, and so on derivatives of π,
and then evaluating these at π₯ equals zero. We then substitute these values
back into the Maclaurin series and simplify. One application of the Maclaurin
series is that we can use our approximation to estimate our function for different
values of π₯. Letβs see an example.

Consider π of π₯ equals π to the
π₯ power. Thereβs two parts to this
question. The first part says, find the
Maclaurin series of π of π₯. And the second part says, use the
first three terms of this series to find an approximate value of π to the power of
0.4 to two decimal places.

Letβs start by writing out the
Maclaurin series expansion for a general function π of π₯. So, letβs begin by finding some of
the derivatives of our function π of π₯. π of π₯ equals π to the π₯ power,
so π of zero equals π to the zero power, which is one. We differentiate π to the π₯ to
get the first derivative. So, we recall the fact that the
derivative of π to the power of π₯ with respect to π₯ is just π to the power of
π₯. So, π prime of π₯ equals π to the
π₯ power. And so, it follows that π prime of
zero is one. And in fact, we see that all the
derivatives of π of π₯ are π to the power of π₯.

So, now letβs make the
substitutions. As our function is defined as π of
π₯, we can replace π with π and begin to make some substitutions. We found that π of zero is
one. And we found that π prime of zero
is one. And we found that π double prime
of zero is one. And π triple prime of zero is
one. And this will continue in this way
for all the future terms. We can bring the π₯, the π₯
squared, and the π₯ cubed, and so on on to the top of the fraction. And then, we can spot that we can
actually write this as the sum from π equals zero to β of π₯ to the πth power over
π factorial. So, this gives us the Maclaurin
series expansion of π to the π₯ power.

The second part of this question
asks us to use the first three terms of this series to find an approximate value of
π to the 0.4 power to two decimal places. Here are the first three terms of
our series. And we want to approximate π to
the π₯ when π₯ equals 0.4. So, we replace π₯ with 0.4, and we
find that this is one plus 0.4 over one factorial. But we know that one factorial is
just one. So, this is 0.4. And 0.4 squared is 0.16. And thatβs over two factorial,
which is two multiplied by one, which is two. But 0.16 over two is just 0.08. So, adding these up, we find that
our approximation is 1.48. We could then use a calculator to
check how good our approximation is. Using a calculator gives us the π
to the power of 0.4 is 1.4918 to four decimal places. So, we can see that even though we
only use the first three terms of our series, our approximation is actually quite
good.

Being able to find the Maclaurin
series expansion of common functions is really useful because this opens the door to
finding Maclaurin series of more difficult forms of common functions. For instance, in the last example,
we saw that the Maclaurin series of π to the π₯ power is the sum from π equals
zero to β of π₯ to the πth power over π factorial. So, if we want to find the
Maclaurin expansion of π to the two π₯ power, rather than start from scratch to
find this expansion, we can replace π₯ in the series with two π₯. We must be careful here because,
remember, weβve got to raise both of the terms in the parentheses to the power. And we then see that we can write
the series as the sum from π equals zero to β of two to the πth power multiplied
by π₯ to the πth power over π factorial. Letβs now see another example.

Find the Maclaurin series of the
hyperbolic sin of three π₯ equals π to the three π₯ power minus π to the negative
three π₯ power over two.

Letβs start by writing out the
general form for the Maclaurin series expansion for a function π. To make things easier for
ourselves, letβs start by finding the Maclaurin series expansion for the hyperbolic
sin of π₯. Weβll then be able to substitute π₯
with three π₯ to get the Maclaurin series of the hyperbolic sin of three π₯. So, weβre going to need to take our
function, π of π₯, to be the hyperbolic sin of π₯, and weβre going to need to
evaluate this at zero and its derivatives at zero. So, weβll also have to remember
that the derivative with respect to π₯ of the hyperbolic sin of π₯ is the hyperbolic
cos of π₯. And the derivative of the
hyperbolic cos of π₯ with respect to π₯ is the hyperbolic sin of π₯.

Letβs create a table for the
derivatives of π of π₯ equals the hyperbolic sin of π₯. So, when π is zero, we just have
π of π₯, which is the hyperbolic sin of π₯. So, to evaluate this when π₯ is
zero, we use the fact that the hyperbolic sin of π₯ is equal to π to the π₯ power
minus π to the negative π₯ power over two. So, the hyperbolic sin of zero is
π to the zero power minus π to the negative zero power over two. But as π to the zero power is just
one, this is one minus one over two, which is zero. When π is equal to one, we have
the first derivative of π, which weβve seen already is the hyperbolic cos of
π₯.

In order to evaluate this at zero,
we recall that the hyperbolic cos of π₯ is equal to π to the π₯ power add π to the
negative π₯ power over two. So, the hyperbolic cos of zero is
equal to π to the zero power add π to the negative zero power over two. But we know that π to the zero
power is just one. So, this is one add one over two,
which is just one. When π is two, weβre looking for
the second derivative of π of π₯. We can get this by differentiating
the first derivative, which was the hyperbolic cos of π₯. And because the hyperbolic cos of
π₯ differentiates to give us the hyperbolic sin of π₯, this is the hyperbolic sin of
π₯. We know what this is; when we
evaluate it at zero, it just gives us zero.

And when π is three, we want to
find the third derivative of π, which is the derivative of the second derivative,
which we found to be the hyperbolic sin. So, this is the hyperbolic cos of
π₯, which we know when π₯ is zero, this is one. And We can start to see a bit of a
pattern here. Each derivative of π evaluated at
zero alternates between zero and one. So, weβre going to use this table
to write out the Maclaurin series for the hyperbolic sin of π₯. When we substitute in our values,
we see that every other term is zero. Weβre only left with the odd powers
of π₯. We want to write this as a series,
but we need to make sure we only end up with those odd powers of π₯.

So, our series is the sum from π
equals zero to β of π₯ to the power of two π add one over two π add one
factorial. We can use this now to find the
Maclaurin series for the hyperbolic sin of three π₯. We do this by replacing π₯ with
three π₯. And we can see that this gives us
three π₯ over one factorial add three π₯ cubed over three factorial add three π₯ to
the fifth power over five factorial, and so on. And we can see that we can write
this as the sum from π equals zero to β of three π₯ raised to the power of two π
add one over two π add one factorial.

One use of Maclaurin series is that
we can integrate functions which would normally be hard to integrate much more
easily if we can find the Maclaurin series expansion for the function. We can compute both definite and
indefinite integrals. If we do have limits of
integration, we can use the Maclaurin series expansion to estimate the value of the
integral. Letβs see an example.

Approximate the integral between
zero and one of sin of π₯ squared with respect to π₯ using the first two terms of an
appropriate series.

Letβs begin by writing out the
Maclaurin series expansion for a function π. Now, the question asks us to use an
appropriate series. But what would an appropriate
series be? Well, if we can find the Maclaurin
series expansion for sin of π₯ squared, we could then integrate from there. To find the Maclaurin series
expansion of a function, we need π and a few derivatives of π. Sin of π₯ squared could get a
little bit complicated trying to find the derivatives. So, letβs start by finding the
Maclaurin series expansion of sin of π₯. Weβll let π of π₯ equal sin of
π₯. Weβll find the first five
derivatives of π of π₯. To do this, we recall that the
derivative with respect to π₯ of sin of π₯ is cos of π₯. And the derivative with respect to
π₯ of cos of π₯ is negative sin of π₯.

So, we differentiate sin of π₯ to
get the first derivative, which is cos of π₯. We differentiate the first
derivative to get the second derivative, which is negative sin of π₯. And we differentiate again to get
the third derivative, and so on. But what we actually need to do is
evaluate each of these at zero. Well, sin of zero is zero, and cos
of zero is one. So, negative cos of zero is
negative one. So, now we can write out the first
few terms of the Maclaurin series expansion of sin of π₯. We use the general form of
Maclaurin series expansion and replace π and its derivatives with the ones that we
found. And we find that every other term
is zero. In fact, all the terms with
positive powers of π₯ disappear here. So, weβre left with only the odd
powers of π₯. And also notice how the terms
alternate between positive and negative.

So, this is the Maclaurin expansion
for sin of π₯. Weβll leave it in this form for now
and use it to find the Maclaurin series for sin of π₯ squared. So, letβs clear some space to do
this. We use the series expansion that we
found for sin of π₯ in order to find the Maclaurin series expansion for sin of π₯
squared. All we need to do is replace π₯
with π₯ squared. We then remember the index law that
tells us that π₯ raised to the power of π raised to the power of π is equal to π₯
raised to the power of π multiplied by π. So, π₯ squared and then raised to
the power of three is π₯ raised to the sixth power. π₯ squared and then raised to the
fifth power is just π₯ raised to the 10th power.

So, we can rewrite the integral
between zero and one of sin of π₯ squared with respect to π₯ as the integral between
zero and one of π₯ squared minus π₯ to the sixth power over three factorial add π₯
to the 10th power over five factorial and so on with respect to π₯. A few things to notice at this
point is that I rewrote π₯ squared over one factorial as π₯ squared because one
factorial is just one and dividing by one gives us that same number. Also note that itβs okay that weβve
only written out three terms here. Remember that in the question,
weβre asked to approximate this integral using only the first two terms. So, weβre actually only going to
need the first two terms here.

From here, Itβs just a case of
integrating using integration rules that we know. We use the power rule for
integration, which tells us that we add one to the power and then divide by the new
power. So, π₯ squared integrates to π₯
cubed over three. π₯ to the sixth power over three
factorial just integrates to π₯ to the seventh power over seven multiplied by three
factorial. And π₯ to the 10th power over five
factorial integrates to π₯ to the 11th power over 11 multiplied by five factorial,
and so on. And we need to evaluate this for
our limits. We firstly substitute in our upper
limit, which is one. And then, we subtract our function
with zero substituted in.

But notice that if we substitute
zero, all of these terms will just be zero. So, weβd just be subtracting
zero. So, from here, letβs just simplify
what weβve got. One raised to anything will just
give us one. So, all these numerators will be
one. We can also calculate our
denominators by remembering that the factorial of a number is the product of that
number and all the integers below it to one. And remember, weβve been asked to
approximate this integral using only the first two terms. So, our approximation for this
integral is one over three minus one over 42, which we calculate to be 13 over
42.

Thereβs another really important
result that we get from the Maclaurin series expansion, and it comes from the
binomial theorem. We know that the binomial theorem
gives us a way of expanding π plus π to the πth power, where π and π are real
numbers and π is a positive integer, where these are called the binomial
coefficients. For example, π plus π raised to
the power of two is π squared plus two ππ plus π squared. Or, π plus π raised to the third
power is π cubed plus three π squared π plus three ππ squared plus π
cubed. So, this expands a polynomial when
π is a positive integer with a finite sum.

But thereβs a special case of the
binomial theorem which comes from the case when π equals one and π equals π₯. So, this is one plus π₯ to the πth
power. In 1665, Isaac Newton generalize
this theorem for powers which are fractional or negative or both. However, when π is not positive,
one plus π₯ to the πth power is no longer a polynomial, so we canβt find a finite
sum of terms that equals π of π₯. So, when π is fractional or
negative, this is not a finite sum anymore. Itβs an infinite series. To find the series, we compute the
Maclaurin series of one plus π₯ to the πth power. To do this, we need to find π and
derivatives of π evaluated at zero. π of π₯ is one plus π₯ to the πth
power. We differentiate this to get the
first derivative.

Using the chain rule, this is π
multiplied by one plus π₯ raised to the power of π minus one. And we continue in this way to find
the second, third, and so on derivatives. But we need to evaluate each of
these at zero, which we do by substituting in π₯ equals zero. And then, we substitute these
values into our general form for Maclaurin series expansion, which gives us the
Maclaurin series expansion of one plus π₯ to the πth power. This gives us the important series
known as the binomial series. This can be shown to converge if
the absolute value of π₯ is less than one and diverge if the absolute value of π₯ is
greater than one. Therefore, the binomial expansion
is only valid when the absolute value of π₯ is less than one.

Letβs summarize the main points
from this lesson. Here is the Maclaurin series
expansion for a function π of π₯. We can use the Maclaurin series
expansion of common functions such as the exponential function or trig functions to
approximate these functions for different values of π₯. We can use the Maclaurin series
expansion of common functions to find the Maclaurin series expansion of different
forms of the function with a simple substitution and manipulation. Maclaurin series expansions allow
us to approximate integrals which would normally be tricky to compute. And finally, Maclaurin series
expansion gives us the binomial series derived from the binomial theorem with π
equal to one and π equal to π₯ and works for π being negative or fractional or
both.