### Video Transcript

Write the first four terms for the
Taylor series expansion of π of π₯ is equal to 11 times π to the power of two π₯
in ascending powers of π₯ minus two.

We need to find the first four
terms of the Taylor series expansion for the function π of π₯ is equal to 11 times
π to the power of two π₯ in ascending powers of π₯ minus two. Remember, ascending powers means
the exponents of π₯ minus two will be increasing. And the fact that weβre told to use
ascending powers of π₯ minus two tells us that we need to center our Taylor
expansion at two.

Letβs recall what we mean by a
Taylor series expansion for a function π of π₯. The Taylor series expansion for a
function π of π₯ centered at π is given by the following power series: the sum
from π equals zero to β of the πth derivative of π of π₯ with respect to π₯
evaluated at π divided by π factorial multiplied by π₯ minus π raised to the πth
power.

We call π the center of our Taylor
series. In our case, weβre told to use π
is equal to two since then our series will be given in terms of powers of π₯ minus
two. So to find the first four terms of
this series, weβre going to need to start differentiating our function π of π₯.

Our series starts from π is equal
to zero. So letβs find the first term in
this series. Thatβs when π is equal to
zero. To do this, weβre going to need to
find the zeroth derivative of π of π₯ with respect to π₯ evaluated at two. And we need to remember what this
means. The zeroth derivative of a function
is just equal to that function itself. In other words, we just need to
find π evaluated at two. So we just substitute π₯ is equal
to two into our function π of π₯. This gives us 11 times π to the
power of two times two, which we can evaluate is equal to 11π to the fourth
power.

To find the next term in our
series, weβre going to want to find the first derivative of π of π₯ with respect to
π₯ evaluated at two. But we canβt do this directly. So instead, weβre going to need to
differentiate our function π of π₯. Thatβs the derivative of 11π to
the power of two π₯ with respect to π₯.

And to differentiate this, we need
to recall our rules for differentiating exponential functions. We know for any constant π, the
derivative of π to the power of ππ₯ with respect to π₯ is equal to π times π to
the power of ππ₯. In our case, our value of π is
equal to two. So differentiating π of π₯, we get
11 times two π to the power of two π₯, which of course we can simplify to give us
22π to the power of two π₯.

Weβre now ready to find our first
derivative of π of π₯ with respect to π₯ evaluated at two. We just substitute π₯ is equal to
two into our expression for π prime of π₯. We get 22π to the power of two
times two. And we can just simplify this to
get 22π to the fourth power.

To find the third term in our
Taylor series expansion, weβre going to need to find the second derivative of π of
π₯ with respect to π₯ evaluated at two. Again, we canβt do this
directly. So weβre going to need to
differentiate our function π prime of π₯ with respect to π₯. Our second derivative of π of π₯
with respect to π₯ is equal to the derivative of 22π to the power of two π₯ with
respect to π₯.

Again, we can do this by using our
derivative rule for exponential functions. This time, our value of π is also
equal to two. So by using this rule, we get the
second derivative of π of π₯ with respect to π₯ is equal to 22 times two π to the
power of two π₯. And of course, we can simplify this
to get 44π to the power of two π₯.

We can now use this to find the
second derivative of π of π₯ with respect to π₯ evaluated at two. We just substitute π₯ is equal to
two into π double prime of π₯. We get 44π to the power of two
times two, which we can simplify to give us 44π to the fourth power.

We just need to do this one more
time to find the fourth term in our Taylor series expansion. We need to find the third
derivative of π of π₯ with respect to π₯ evaluated at two. Just as we did before, we canβt do
this directly. So we need to differentiate our
second derivative of π of π₯ with respect to π₯. Just as we did before, weβll
differentiate this by using our rules for differentiating exponential functions. And our value of π is equal to
two.

This gives us our third derivative
of π of π₯ with respect to π₯ is equal to 44 times two π to the power of two
π₯. And we can simplify this to get
88π to the power of two π₯.

And now we can use this expression
to find the third derivative of π of π₯ with respect to π₯ evaluated at two. We just substitute π₯ is equal to
two into this expression. We get 88π to the power of two
times two, which simplifies to give us 88π to the fourth power.

We now need to substitute these
four values into our Maclaurin series. And remember, weβre only writing
out the first four terms of this series. Our first term when π is equal to
zero is 11π to the fourth power divided by zero factorial multiplied by π₯ minus
two to the zeroth power. Of course, we can simplify
this. We know zero factorial is just
equal to one. And π₯ minus two all raised to the
zeroth power is also equal to one. So our first term just simplifies
to give us 11π to the fourth power.

Letβs now find the second term in
our series. Remember, we substitute π is equal
to one and π is equal to two into the summand in our Taylor series. And we already showed that the
first derivative of π of π₯ with respect to π₯ at two is 22π to the fourth
power. So using this, our second term is
22π to the fourth power divided by one factorial multiplied by π₯ minus two raised
to the first power. And once again, we can simplify
this.

First, π₯ minus two raised to the
first power is just π₯ minus two. Next, one factorial is just equal
to one. And dividing by one doesnβt change
our value. So we just get 22π to the fourth
power times π₯ minus two.

The third term in our series will
be when π is equal to two. And we already showed the second
derivative of π of π₯ with respect to π₯ at two is 44π to the fourth power. So substituting these values and π
is equal to two into the summand in our Taylor series, we get the third term is 44π
to the fourth power divided by two factorial multiplied by π₯ minus two squared.

And once again, we can
simplify. First, two factorial is just equal
to two. And then 44 divided by two can be
simplified to just give us 22. So our third term simplifies to
give us 22π to the fourth power times π₯ minus two all squared.

We just need to do this one more
time to find the fourth term in our series. Thatβs when π is equal to
three. This time, by substituting these
values into the summand of our Taylor series, we get 88π to the fourth power
divided by three factorial times π₯ minus two all cubed. And we can simplify this term. First, three factorial is equal to
three times two times one. And now we can cancel the shared
factor of two in our numerator and our denominator. This gives us 44 divided by
three. So we were able to simplify our
fourth term to 44π to the fourth power divided by three times π₯ minus two all
cubed. And this is our final answer.

Therefore, we were able to show the
first four terms in the Taylor series expansion of π of π₯ is equal to 11π to the
power of two π₯ in ascending powers of π₯ minus two is given by 11π to the fourth
power plus 22π to the fourth power times π₯ minus two plus 22π to the fourth power
multiplied by π₯ minus two all squared plus 44π to the fourth power over three
multiplied by π₯ minus two all cubed.