### Video Transcript

Consider the function π of π₯
equals cos of π₯. What are the first four derivatives
of π with respect to π₯? Write the general form for the πth
derivative of π with respect to π₯. And hence, derive the Maclaurin
series for cos of π₯.

There is a fourth part to this
question, which asks us to find the radius π
of convergence of the Maclaurin series
for cos of π₯. And weβll consider that at the very
end. So we begin by finding the first
four derivatives of π with respect to π₯. Weβre told that π of π₯ is equal
to cos of π₯. And so we quote the general result
for the first derivative of cos of π₯ is negative sin π₯. And therefore, π prime of π₯, the
first derivative of π with respect to π₯, is negative sin π₯.

Next, we recall the general result
for the derivative of sin π₯. Itβs cos π₯. And this means the derivative of
negative sin π₯ must be negative cos of π₯. By returning to the first general
result we quoted, weβll be able to differentiate negative cos of π₯. Itβs negative, negative sin of π₯,
which is, of course, simply sin of π₯. So the third derivative is sin of
π₯. And finally, we obtain the fourth
derivative to be equal to cos of π₯. And so we have the first four
derivatives. They are negative sin π₯, negative
cos π₯, sin π₯, and cos π₯.

The second part of this question
asks us to find a general form for the πth derivative of π with respect to π₯. And so we need to spot a pattern
with our derivatives. Firstly, we should notice that if
we differentiate cos of π₯, we go back to negative sin π₯. And this cycle will continue. Next, we recall that sine and
cosine are horizontal translations of one another. And so if we sketch the curve of π¦
equals negative sin of π₯, that, of course, is a reflection on the π₯-axis for the
graph of π¦ equals sin π₯. We see that we can write π¦ equals
negative sin π₯ as π¦ equals cos of π₯ plus π by two. Itβs a horizontal translation of
the graph of π¦ equals cos of π₯, left by π by two radians. So π prime of π₯ is equal to cos
of π₯ plus π by two.

Weβll repeat this for the graph of
π¦ equals negative cos of π₯. Itβs a reflection again in the
π₯-axis of the graph π¦ equals cos of π₯. And it, therefore, also can be
represented by a horizontal translation of the graph of π¦ equals cos of π₯ left by
π radians. So we can write π double prime of
π₯, which was negative cos of π₯, as cos of π₯ plus π. By applying a similar thought
process, we find that the third derivative, sin of π₯, can be written as cos of π₯
plus three π by two. And the fourth derivative can be
written as cos of π₯ plus two π.

Remember, cos itself is periodic
with a period of two π radians. So cos of π₯ plus two π is exactly
the same as cos of π₯. And we might be starting to spot a
pattern. In fact, if we write the second
derivative, cos of π₯ plus π, as cos of π₯ plus two π over two and the fourth
derivative as cos of π₯ plus four π over two, we see we can write the πth
derivative as cos of π₯ plus ππ over two. And so the general form for the
πth derivative of π with respect to π₯ is cos of π₯ plus ππ over two.

The third part of this question
asks us to derive the Maclaurin series for cos of π₯. And so we recall this is the sum
from π equals zero to β of the πth derivative of π evaluated at zero over π
factorial times π₯ to the πth power. Well, in this case, the πth
derivative of π evaluated at zero is cos of zero plus ππ over two. When π is zero, we have π of
zero, which is cos of zero which is one. When π is one, we have π prime of
zero, which is cos of π by two, which is zero. When π is two, we have cos of π,
which is negative one. And when π is equal to three, we
have cos of three by two, which is once again zero. And of course, as we saw in the
second part of this question, this cycle continues.

And this means the first few terms
will be one minus π₯ squared over two factorial plus π₯ to fourth power over four
factorial minus π₯ to the six over six factorial, which we can write as the sum from
π equals zero to β of negative one to the πth power. Remember, this part will just give
us the alternating signs. We go from positive to negative and
back to positive again. This is all over two π
factorial. This bit gives us the even
factorials and the denominator times π₯ to the power of two π. And so weβve derived the Maclaurin
series for cos of π₯. Itβs the sum from π equals zero to
β of negative one to the πth power over two π factorial times π₯ to the power of
two π. And, of course, if we chose to
alternatively write this as shown, that would be absolutely fine too.

Weβre now going to clear some space
and consider the fourth and final part of this question. Weβll keep the Maclaurin series for
cos of π₯ on screen as weβre going to use that in a moment. And this question says, what is the
radius π
of convergence of the Maclaurin series for cos of π₯? We recall that, in general, thereβs
an open interval from negative π
to π
, in which a power series converges. And that number π
is called the
radius of convergence. And we can use the ratio test to
find this value. The part of the test weβre
interested in says that, given a series of sum of π π, if the limit as π
approaches β of the absolute value of π π plus one over π π is less than one,
then the series is absolutely convergent and hence convergent.

In our case, we let π π be equal
to negative one to the πth power times π₯ to the power of two π over two π
factorial. Then, π π plus one is negative
one to the power of π plus one times π₯ to the power of two times π plus one over
two times π plus one factorial. Weβll distribute the parentheses to
make the next step easier. Two times π plus one is two π
plus two. According to the ratio test, we
want to find where the limit as π approaches β of the absolute value of the
quotient of these is less than one.

We know that when dividing by a
fraction, we simply multiply by the reciprocal of that fraction. So we can say that we want the
limit as π approaches β of the absolute value of negative one to the power of π
plus one times π₯ to the power of two π plus two over two π plus two factorial
times two π factorial over negative one to the πth power times π₯ to the power of
two π. And then, we recall that when
dividing two numbers with equal basis, we simply subtract their exponents so that π₯
to the power of π divided by π₯ to the power of π is π₯ to the power of π minus
π. Weβll use this to simplify negative
one to the power of π plus one divided by negative one to the power of π. Itβs simply negative one. Similarly, weβll be able to
simplify π₯ to the power of two π plus two divided by π₯ to the power of two
π. Itβs π₯ squared.

And what about these
factorials? Well, we know that two π plus two
factorial is the same as two π plus two times two π plus one times two π times
two π minus one and so on. Alternatively, thatβs the same as
two π plus two times two π plus one times two π factorial. And that allows us to divide
through by two π factorial. When we do, weβre left with two π
plus two times two π plus one on the denominator of our fraction. So we want the limit as π
approaches β of the absolute value of this fraction to be less than one. We spot that negative one times π₯
squared is independent event. And so we can take the absolute
value of negative π₯ squared outside of the limit. And we have the absolute value of
negative π₯ squared times the limit as π approaches β of one over two π plus two
times two π plus one.

Weβre now ready to evaluate this
limit. As π grows larger, one over two π
plus two times two π plus one approaches zero. And so, in fact, weβre looking to
find the values of π₯ such that the absolute value of negative π₯ squared times zero
is less than one. Well, when we multiply the absolute
value of negative π₯ squared by zero, weβll always get zero. And this means our Maclaurin series
converges for all values of π₯. And we can therefore say that π
equals β or positive β.