### Video Transcript

In this video, we will learn how to
find Taylor polynomials and use them to approximate a function. Weβll then see some examples of
Taylor polynomials of some functions. This is the general form for a
Taylor polynomial up to the πth derivative for a function π centered at the point
π.

But to understand where this comes
from and why we use Taylor polynomials, think about a power series centered at
π. We know that if we can write a
function as a power series, we find that this is much easier to differentiate and
integrate. So being able to write a function
as a power series is a really useful tool.

Taylor polynomials give us a
general method for writing a power series representation near a point. Weβre going to try and rewrite
these coefficients in terms of π and its derivatives. And we firstly assume that π does
have a power series representation about the point π₯ equals π and π has
derivatives of every order. And that the absolute value of π₯
minus π is less than π
, where π
is the radius of convergence.

Centering our approximation at π,
so substituting π₯ equals π, we find for π of π we have all the terms canceling
except π zero. So π of π equals π zero. Then letβs differentiate π of π₯
to find π prime of π₯. π zero is a constant. So this differentiates to zero. π one multiplied by π₯ minus π
will just differentiate to π one. Then π two π₯ minus π squared
differentiates to two π two π₯ minus π. And we continue in the same
way. We can then evaluate this first
derivative at π₯ equals π. And we find that all the terms
except π one actually disappear.

Now letβs find the second
derivative of π. And we do this by differentiating
the first derivative of π. And then we make the substitution
π₯ equals π. And we find that all the terms
except two π two disappear. Now remember that our goal is to
rewrite these coefficients in terms of π and its derivatives. So we rearrange this for π
two. And we find that π two equals π
double prime of π over two.

We then continue and find the third
derivative of π. Evaluating this at the point π
gives us six π three, which rearranges to give us that π three equals π triple
prime of π over six. Now if we look at what we found for
π zero, π one, π two, and π three, we can put together a more general rule. Each π π equals the πth
derivative of π at π over π factorial. And remember that π factorial is
the product of π and all the integers below it up to one. For example, three factorial is
three multiplied by two multiplied by one, which is six. So this leads us to the following
definition.

The Taylor polynomial of π
centered at π up to the πth derivative is π of π add π prime of π multiplied
by π₯ minus π add π double prime of π over two factorial. Multiplied by π₯ minus π squared
add π triple prime of π over three factorial multiplied by π₯ minus π cubed. And this continues up to the πth
derivative of π at π over π factorial multiplied by π₯ minus π to the πth
power.

You might sometimes see this
written as the sum from π equals zero to π of the πth derivative of π at π over
π factorial multiplied by π₯ minus π to the πth power. There are a couple of things to
note about this formula. Sometimes we see π prime of π
written as π prime of π over one factorial. But as one factorial is just one,
this is exactly the same.

Also note that even though this
first term looks a little bit different, it still follows this general rule. If we apply the formula when π
equals zero, then this is just going to be π of π. Also, π₯ minus π to the zero power
is just going to be one because anything raised to the zero power is one. And finally, zero factorial is
defined as one. So we can say that this is just
going to give us π of π.

Letβs now see a visual
representation of how Taylor polynomials help us to approximate a function.

If we take the graph of π of π₯
equals π to the π₯ power and find the Taylor polynomials for this function centered
at π₯ equals two. You can see here the graph of π to
the π₯. And we remind ourselves of the
Taylor polynomials formula. If we try and approximate this with
π equals zero, our Taylor polynomial is π squared. And that just gives us a straight
line, which is not a very good approximation for our function.

But if we now find our Taylor
polynomials up to the first degree, even though this still isnβt a great
approximation, we can see itβs a little bit better. But if we then take Taylor
polynomials up to the second degree, we get an even closer approximation. And here are the graphs for π
equals three and π equals four, the Taylor polynomial up to the third degree and
the Taylor polynomial up to the fourth degree. We can see that each time the
approximation gets a little bit better. So the higher the value of π, the
better the approximation.

Letβs now see some examples of
Taylor polynomials to approximate a function.

Find the Taylor polynomials of
degree two approximating the function π of π₯ equals π₯ cubed add two π₯ minus
three at the point π equals two.

Letβs start by writing out the
general form of a Taylor polynomial up to degree π for a function π. And for this question, weβre
centering our approximation at the point π equals two. So we make the substitution π
equals two. Also note that weβve only gone up
to the degree two power as thatβs what weβve been asked to do in the question.

So now we need to evaluate π of
two, π prime of two, and π double prime of two. Well, we know that π of π₯ equals
π₯ cubed add two π₯ minus three. And so π of two equals two cubed
add two multiplied by two minus three. So π of two equals nine.

To find π prime of two, we need
to, first of all, find π prime of π₯. To do this, we differentiate π of
π₯, which we do term by term using the power rule, to get π prime of π₯ equals
three π₯ squared add two. Note that negative three
differentiates to zero because itβs a constant. And so π prime of two equals three
multiplied by two squared add two, which gives us 14.

To find π double prime of two, we
need to, first of all, find π double prime of π₯, which we find by differentiating
the first derivative, to get six π₯. And so π double prime of two
equals six multiplied by two, which is 12.

We then substitute these values
into our workings. From here, we can evaluate these
factorials. We know that the factorial of a
number is the product of that number and all the integers below it to one. So one factorial is one multiplied
by one, which is one. And two factorial is two multiplied
by one, which is two. From here, we just need to do some
simplification. 14 over one is just 14, and 12 over
two is just six. And so that gives us our final
answer: nine add 14 multiplied by π₯ minus two add six multiplied by π₯ minus two
squared.

Find the Taylor polynomials of the
third degree approximating the function π of π₯ equals the square root of π₯ at the
point π equals nine.

Letβs start this question by
writing out the general form for Taylor polynomials up to degree π. And letβs begin by substituting in
the point where weβre centering our approximation, at π equals nine. Also note that weβve also been
asked to find the Taylor polynomials up to the third degree. From here, we can see that weβre
going to need to find π of nine, π prime of nine, π double prime of nine, and π
triple prime of nine. To do this, weβre going to need to
find π prime of π₯, π double prime of π₯, and π triple prime of π₯.

Well, we know from the question
that π of π₯ equals the square root of π₯, which we know is the same as π₯ to the
half power. Writing it this way is going to
help us differentiate it to find the first derivative. Using the power rule, which tells
us to multiply by the power and then subtract one from the power, gives us the first
derivative. π prime of π₯ equals one over two
π₯ to the power of negative one-half.

And we then differentiate this to
get the second derivative of π. Applying the power rule again gives
us that the second derivative of π is negative one over two multiplied by one over
two π₯ to the power of negative three over two. But we can just simplify this to
negative one over four π₯ to the power of negative three over two.

And we differentiate once more to
get the third derivative of π. This gives us negative three over
two multiplied by negative one over four π₯ to the power of negative five over
two. But again, we can simplify this to
give us three over eight π₯ to the power of negative five over two.

But what we actually want here is
π of nine, π prime of nine, π double prime of nine, and π triple prime of
nine. So we substitute nine into each of
these functions. And then we substitute the values
that we found back into our working. And from here, we just need to do
some simplification to achieve our final answer.

If we start here with one over six
over one factorial and we recall that π factorial is the product of π and all the
integers below it down to one. Then one factorial is just one. So one over six over one factorial
is just one over six over one, which is just one over six.

Now letβs simplify negative one
over 108 divided by two factorial. Well, two factorial is just two
multiplied by one, which is two. So this is just negative one over
108 over two. And we know from fraction laws that
this is just negative one over 108 multiplied by two, which is negative one over
216. And finally, we need to simplify
one over 648 divided by three factorial. Well, three factorial is three
multiplied by two multiplied by one, which is six. So this is just one over 648
multiplied by six, which gives us one over 3888. So that gives us our final
answer. The Taylor polynomials of the third
degree approximating the function π of π₯ equals the square root of π₯ at the point
π equals nine.

Find the fourth-degree polynomial
of the function π of π₯ equals sin of π₯ at the point π equals π over two.

Letβs start by writing out the
general form for a Taylor polynomial which approximates a function π of π₯ at the
point π₯ equals π. Weβve been asked to find the
fourth-degree Taylor polynomial of this function at the point π equals π over
two. So letβs write out our general form
for a Taylor polynomial up to the fourth degree and substitute in π equals π over
two.

We can see here that weβre going to
need to substitute in values for π of π over two, the first derivative of π at π
over two, the second derivative of π at π over two, the third derivative of π at
π over two, and the fourth derivative of π at π over two.

So letβs start by finding the
derivatives that we need. π of π₯ is the function sin of
π₯. At this point, we recall this
useful cycle, which shows us what each of these functions differentiates to. So we see that sin of π₯
differentiates to cos of π₯. So π prime of π₯ equals cos of
π₯. Then we differentiate the first
derivative to get the second derivative. And we see that this is going to be
negative sin of π₯. We then differentiate again to get
the third derivative, which we see is negative cos of π₯. And we differentiate once more to
get the fourth derivative of π. And we get sin of π₯.

But what we actually need to do is
evaluate each of these functions at π over two. So we substitute in π over two
into each of these functions. We can work these out either on a
calculator or by using the graphs. So we can see that sin of π over
two is going to give us one. So negative sin of π over two will
be negative one. And we can see that cos of π over
two is zero. So negative cos of π over two will
also be zero.

So now we can substitute these
values back into our working. When we do this, we find that two
of our terms are actually zero. From here, we just need to simplify
our answer. Remembering that the factorial of a
number is the product of that number and the integers below it down to one. So two factorial is two multiplied
by one, which is two. And four factorial is four
multiplied by three multiplied by two multiplied by one, which is 24. So substituting those values gives
us our final answer.

So to summarize, Taylor polynomials
allow us to approximate functions. And here is the general form of the
Taylor polynomial which approximates the function π centered at the point π. A good place to start with these
questions is by finding the necessary derivatives and then evaluating these at the
point π.