### Video Transcript

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.