### Video Transcript

Consider the function π of π₯
equals cos of π₯. Find the Taylor series expansion of
π of π₯ equals cos of π₯ at π₯ equals π and write the Taylor series expansion of
π of π₯ in sigma notation.

And we recall that the Taylor
series expansion of a function about π₯ equals π is given by π of π₯ equals π of
π plus π prime of π over one factorial times π₯ minus π plus π double prime of
π over two factorial times π₯ minus π all squared and so on. In this question, our function is
cos of π₯. And weβre looking to find the
Taylor series expansion at π₯ equals π. So weβre going to let π be equal
to π. And so we see that π of π₯ here is
equal to π of π plus π prime of π over one factorial times π₯ minus π and so
on. Now, we can quite easily evaluate
π of π. We would substitute π into the
function cos of π₯. But what about π prime of π, π
double prime of π, and so on?

Well, weβre going to differentiate
our function with respect to π₯. We might recall the first
derivative of cos of π₯ to be negative sin of π₯. Then to get the second derivative,
we differentiate negative sin π₯ with respect to π₯ and we get negative cos π₯. Differentiating once more, we find
that π triple prime of π₯ is sin π₯. And in fact, weβll repeat this
process one more time because when we differentiate sin π₯, we get cos π₯ again. And you might notice, we have a
cycle. The fifth derivative of π will be
negative sin π₯, the sixth derivative will be negative cos π₯, and so on. And letβs use all of these to
evaluate π of π, π prime of π, π double prime of π, and so on and look for a
pattern. π of π is cos π, which is
negative one. π prime of π is negative sin π,
which is zero. π double prime of π is negative
cos π, which is one. And π triple prime of π is sin
π, which is once again zero.

It, of course, follows that the
fourth derivative of π will be cos of π again, which is negative one. Letβs substitute all of these back
into our Taylor series expansion. Now of course, every other term is
going to be equal to zero as weβve seen. And so we find the Taylor series
expansion to be negative one plus a half times π₯ minus π all squared minus one
over 24 times π₯ minus π to the fourth power plus one over 720 times π₯ minus π to
the sixth power and so on. The second part of this question
asks us to write the Taylor series expansion of π of π₯ in sigma notation. And so letβs see if thereβs a way
we can find a pattern. We saw that π prime of π was
equal to zero, while π triple prime of π is equal to zero. And extending the pattern, we would
see that the fifth derivative, the seventh derivative, and so on would also be equal
to zero. And so our derivative alternate
between negative one, one, negative one, and so on.

Similarly, our denominators are
factorials of ascending even numbers each time this even number is the same as the
exponent. So letβs define that even number as
two π. Then the denominator is two π
factorial and the exponent of π₯ minus π is two π. To achieve alternating powers of
negative one, we write negative one times π plus one. That means when π is zero,
negative one to the power of one is negative one, when π is one, negative one to
the power of two gives us one, and so on. And so we find that π of π₯ equals
the sum of negative one to the power of π plus one times π₯ minus π to the power
of two π over two π factorial for values of π between zero and infinity.