### Video Transcript

For π₯ is greater than zero, which
of the following does the power series π₯ minus π₯ cubed over two factorial plus π₯
to the fifth power over four factorial minus π₯ to the seventh power divided by six
factorial and then we add an infinite number of these terms together where the πth
term is negative one to the πth power times π₯ to the power of two π plus one
divided by two π factorial converge to? Option (a) the sin of π₯, option
(b) π₯ times the sin of π₯, option (c) π₯ times the cos of π₯, or option (d) the sin
of π₯ divided by π₯.

The question gives us a power
series. And weβre told that this power
series will converge for any value of π₯ greater than zero. We need to determine which function
this power series converges to. Weβre given four options. If we look at the power series
weβre given in the question, we can see each term only includes π₯. This should remind us of our
Maclaurin series or alternatively our Taylor series centered at zero. And we can see that our answers
only include the sin of π₯, the cos of π₯, and π₯. So letβs start by recalling the
Maclaurin series for the sin of π₯ and the cos of π₯.

First, the Maclaurin series for the
sin of π₯ is π₯ minus π₯ cubed over three factorial plus π₯ to the fifth power over
five factorial and we add infinitely many terms of this form where the πth term is
negative one to the πth power times π₯ to the power of two π plus one divided by
two π plus one factorial. Second, the Maclaurin series for
the cos of π₯ is one minus π₯ squared over two factorial plus π₯ to the fourth power
over four factorial and we add infinitely many terms of this form where the πth
term is negative one to the πth power times π₯ to the power of two π divided by
two π factorial. And we know both of these Maclaurin
series are valid for all real values of π₯.

We can use these two Maclaurin
series to find Maclaurin series for all four of the answers given to us in the
question. For example, in option (a), we want
to find the Maclaurin series for the sin of π₯. Weβve already done this. So we can move on to option (b) π₯
times the sin of π₯. To find the Maclaurin series for π₯
times the sin of π₯, weβre going to use our Maclaurin series for the sin of π₯.

So weβll substitute this into our
expression. This gives us π₯ times the sin of
π₯ is equal to the following expression. Remember, we know the Maclaurin
series for the sin of π₯ is valid for all real values of π₯. So we can just multiply each term
by π₯. It will still be valid. To multiply each term by π₯, we
just need to increase the exponent of π₯ by one. This gives us the following
expression for π₯ times the sin of π₯. And we can see this is not equal to
the power series given to us in the question. For example, our power series here
has a term for π₯ squared. However, the power series given to
us in the question does not have a term for π₯ squared.

And although not strictly necessary
to answer this question, itβs worth pointing out that the power series of a function
centered at π₯ is equal to π is unique. It must be equal to the Taylor
series of that function centered at π₯ is equal to π. So far, weβve found power series
for option (a) and option (b). Letβs now find one for option (c)
π₯ times the cos of π₯.

Weβll do this in the same way we
did for π₯ times the sin of π₯. Weβll substitute our Maclaurin
series for the cos of π₯ into this expression. Substituting in our Maclaurin
series for the cos of π₯, we get the following expression for π₯ times the cos of
π₯. Again, we know the Maclaurin series
for the cos of π₯ is convergent for all real values of π₯. So to multiply this by π₯, weβll
just multiply each term by π₯. And to do this, we just need to add
one to our exponent of π₯ on each term. This gives us π₯ minus π₯ cubed
over two factorial plus π₯ to the fifth power over four factorial minus π₯ to the
seventh power over six factorial and we add an infinite number of terms of this form
where the πth term is negative one to the πth power times π₯ to the power of two
π plus one divided by two π factorial.

And we can see this is exactly the
same as the power series given to us in the question. So the power series given to us in
the question must converge to the function π₯ times the cos of π₯, which was our
option (c).

Itβs also worth noting we couldβve
done the same for option (d). We wouldβve got the sin of π₯
divided by π₯ is equal to the sum from π equals zero to β of negative one to the
πth power times π₯ to the power of two π divided by two π plus one factorial. Of course, this wouldnβt be valid
when π₯ was equal to zero because the sin of π₯ divided by π₯ is not defined when π₯
is equal to zero. But this wasnβt the correct answer
either. Therefore, we were able to show for
values of π₯ greater than zero, the power series given to us in the question
converges to π₯ times the cos of π₯, which was option (c).