### Video Transcript

Consider the function π of π₯
equals π to the power of two π₯. Find the Taylor series
representation of π about π₯ equals three. And in fact, there are two further
parts of this question that weβll consider in a moment.

We begin by recalling that the
Taylor series of a function π about π is given by the sum of the πth derivative
of π evaluated at π over π factorial times π₯ minus π to πth power values of π
between zero and infinity. And so what do we know about our
function? Well, itβs given by π of π₯ equals
π to the power of two π₯. We want to find the Taylor series
about π₯ equals three. So we let π be equal to three. Then we see that π of π₯ is equal
to the sum of the πth derivative of π evaluated at three of π factorial times π₯
minus three to the πth power for values of π between zero and infinity. This is, of course, equal to π of
three plus π prime of three over one factorial times π₯ minus three plus π double
prime of three over two factorial times π₯ minus three all squared and so on.

Itβs quite clear that weβre going
to need to work out π of three, π prime of three, π double prime of three, and
possibly look for a pattern. π of three is quite
straightforward. Itβs π to the power of two times
three which is, of course, π to the power of six. So letβs next look for the
derivative of π of π₯. The derivative of π to the power
of two π₯ is two π to the power of two π₯. So π prime of three must be two π
to the power of two times three, which is two π to the power of six. Weβll differentiate once again. This time we get four π to the
power of two π₯. And π double prime of three is for
π to the power of two times three, which is four π to the sixth power.

Now actually, we can see that the
πth derivative of π of π₯ is equal to two to the πth power times π to the power
of two π₯. And this means the πth derivative
of π evaluated at three is two to the πth power times π to the sixth power. Our Taylor series expansion using
sigma notation is, therefore, the sum of two to the πth power times π to the sixth
power times π₯ minus three to the πth power all over π factorial for values of π
between zero and infinity.

The second part of this question
asks us to find the interval of convergence of the Taylor series representation of
π about π₯ equals three.

We say that the intervals of all
π₯s for which the power series converges is called the interval of convergence of
the series. Now actually, we know that the
power series will converge for π₯ equals π. So here π₯ equals three. But thatβs all we know right
now. Weβll use the ratio test to
determine the rest of the π₯s for which the power series will converge. This says suppose that we have some
series, the sum of ππ, and we defined π to be equal to the limits as π
approaches infinity of ππ plus one over ππ. If π is less than one, the series
is absolutely convergent and hence convergent. This is the scenario weβre
interested in. So our π must be the limit as π
approaches infinity of the absolute value of two to the power of π plus one times
π to the sixth power times π₯ minus three to the power of π plus one over π plus
one factorial divided by two to the πth power times π to the sixth power times π₯
minus three to the πth power over π factorial.

Now of course, when we divide by a
fraction, itβs the same as multiplying by the reciprocal of that fraction. So we can rewrite our limit as
shown. Then we notice we can perform some
simplification. We can cancel out π to the power
of six and two to the power of π. Similarly, we can cancel π₯ minus
three to the power of π. And if we, finally, recall that π
plus one factorial is the same as π plus one times π factorial, we see that we can
cancel by dividing through by π factorial. Well, this all simplifies really
nicely. We see that π is equal to the
limit as π approaches infinity of the absolute value of two times π₯ minus three
over π plus one.

Now in fact, the expression two
times π₯ minus three is independent of π. So we can factor this, remembering
that we must write the absolute value of two times π₯ minus three. And so π is equal to the absolute
value of two times π₯ minus three times the limit as π approaches infinity of the
absolute value of one over π plus one. Now, as π gets larger, one over π
plus one gets smaller. So the limit approaches zero. This means π is equal to zero. This is of course less than
one. And it doesnβt change no matter the
value of π₯. And so we say our series is
absolutely convergent for all values of π₯. Its interval of convergence is
negative infinity to infinity.

The third part of this question
asks, what is the radius of convergence of the Taylor series representation of π
about π₯ equals three?

Recall that there exist a number
capital π
, which we call the radius of convergence, so that the power series will
converge for absolute values of π₯ minus π less than π
and will diverge for
absolute values of π₯ minus π greater than π
. Well here, weβve seen that our
power series is absolutely convergent. And thereβs a special case for the
radius of convergence. In cases such as these, we say that
the radius of convergence is infinity. Capital π
is, therefore, equal to
infinity.