### Video Transcript

For the function π, π evaluated
at three is equal to two. The first derivative of π
evaluated at three is equal to seven. And the πth derivative of π
evaluated at three is equal to negative a half multiplied by π multiplied by the π
minus oneth derivative of π evaluated at three for π is greater than or equal to
two. Find the first five terms of the
Taylor series representation of π at π₯ is equal to three.

We recall that we call the sum from
π equals zero to infinity of the πth derivative of π evaluated at π divided by
π factorial multiplied by π₯ minus π raised to the πth power the Taylor series
representation of the function π at π. The question asks us for the first
five terms of this Taylor series representation. And weβre told that this is at π₯
is equal to three. So we set π equal to three.

So weβll substitute π equals three
into our Taylor series representation. And weβll only do a partial sum up
to four to find the first five terms. For our first term, when π is
equal to zero, we get the zeroth derivative of π evaluated at three divided by zero
factorial multiplied by π₯ minus three to the zeroth power.

We can simplify this term by
noticing that zero factorial is equal to one and π₯ minus three raised to the zeroth
power is also equal to one. Then we know that the zeroth
derivative of π is just equal to our function π. So our entire first term just
simplifies to π evaluated at three.

Now letβs do the same of our second
term when π is equal to one. This gives us the first derivative
of π evaluated at three divided by one factorial multiplied by π₯ minus three to
the first power. We have that one factorial is just
equal to one, and π₯ minus three to the first power is just equal to itself, π₯
minus three. And we can rewrite the first
derivative of π to just be equal to π prime. So this gives us that our second
term can be written as π prime evaluated at three multiplied by π₯ minus three.

Weβll then just write out our
third, fourth, and fifth term by substituting in π is equal to two, three, and
four, respectively. The question tells us that π
evaluated at three is equal to two. So we can change our first term to
just be equal to two. The question also tells us that π
prime evaluated at three is equal to seven. So we can change the second term in
our series to be equal to seven multiplied by π₯ minus three.

The later terms in our series all
use higher-order derivatives of our function π. Luckily, the question gives us the
formula for calculating the πth derivative of π evaluated at three when π is
greater than or equal to two. Substituting π is equal to two
into this formula gives us the second derivative of π evaluated at three is equal
to negative a half multiplied by two multiplied by π. But we take two minus one
derivatives and evaluate this at three.

We can simplify this by noticing
that negative a half multiplied by two is equal to negative one. And taking two minus one
derivatives of the function π is the same as just taking one derivative. This gives us that the second
derivative of π evaluated at three is equal to negative π prime evaluated at
three, which is equal to negative seven.

We can now substitute π equals
three into this formula to find that the third derivative of π evaluated at three
is equal to negative a half multiplied by three multiplied by π. But we take three minus one
derivatives and evaluate at three. Taking three minus one derivatives
of π and evaluating at three is evaluating the second derivative of π at three,
which we calculated above to be negative seven. So the third derivative of π
evaluated at three is equal to negative a half multiplied by three multiplied by
negative seven, which is equal to 21 divided by two.

Finally, we can do the same by
substituting π equals four into our formula. We can get an expression for the
fourth derivative of π evaluated at three. Using our answer from before that
the third derivative of π evaluated at three is 21 divided by two. We get that the fourth derivative
of π evaluated at three is equal to negative a half multiplied by four multiplied
by 21 divided by two, which is equal to negative 21.

We can now substitute these values
in to find the coefficients of terms three, four, and five in our Taylor series
representation. Substituting these values in and
evaluating our factorials gives us the coefficient of the third term as negative
seven divided by two, the coefficient of the fourth term as 21 divided by two all
divided by six, and the coefficient of the fifth term as negative 21 divided by
24. We can evaluate negative seven
divided by two to be equal to negative seven over two.

If we calculate 21 divided by two
all divided by six, weβll get seven divided by four. And we can calculate negative 21
divided by 24 to be equal to negative seven divided by eight. Giving us that the first five terms
of the Taylor series representation of the function π to find in the question about
π₯ is equal to three are. Two plus seven multiplied by π₯
minus three minus seven over two multiplied by π₯ minus three squared plus seven
over four multiplied by π₯ minus three cubed minus seven over eight multiplied by π₯
minus three to the fourth power.