### Video Transcript

The following table shows the value
of function π and some of its derivatives at π₯ equals negative two. Write the first five terms of the
Taylor series of π.

So, we have a table here with some
values. And at this point, itβs worth me
pointing out that this notation means the second, third, and fourth derivatives of
π. Weβve been asked to write the first
five terms of the Taylor series of π, so letβs remind ourselves of the Taylor
series expansion. This allows us to express a
function as a sum of its derivatives about a point.

The Taylor series of a function π
about the point π is the sum from π equal zero to infinity of the πth derivative
of π at the point π over π factorial multiplied by π₯ minus π to the power
π. Recall that π factorial is the
product of all the integers less than or equal to π but greater than or equal to
one. We can equivalently express this
sum as π of π add π dash of π over one factorial multiplied by π₯ minus π add
the second derivative of π at π over two factorial multiplied by π₯ minus π
squared, etcetera.

So, letβs apply this to our
question. Weβve been given the derivatives at
the point π₯ equals negative two. So, for our question, π equals
negative two. Letβs go ahead and substitute π
equals negative two into our formula. From here, all we need to do is
substitute in the values that we know and simplify. In these brackets, we have π₯ minus
negative two. And because weβre subtracting a
negative, we can write this as π₯ add two. We can also calculate these
factorials.

One factorial is one. Two factorial is two. Three factorial is six. And four factorial is 24. Letβs now substitute in the
derivatives weβve been given in the question in our new line of working. π of negative two is three. π dash of negative two is negative
one. The second derivative of π at
negative two is five. The third derivative of π at
negative two is eight. And the fourth derivative of π at
negative two is negative six.

From here, we just need to
simplify. Negative one over one is just
negative one. So, we just need to write a minus
before the bracket. Eight over six simplifies to four
over three. And negative six over 24 simplifies
to negative one over four. And so, that gives us our final
answer. The first five terms of the Taylor
series of π are three minus π₯ add two add five over two add two squared add four
over three π₯ add two cubed minus one over four π₯ add two to the power of four.