### Video Transcript

Find the Maclaurin series for the
function the hyperbolic cos of two π₯ which is equal to π to the power of two π₯
plus π to the power of negative two π₯ all divided by two.

We recall that for an infinitely
differentiable function π of π₯, the Maclaurin series of π of π₯ is equal to π
evaluated at zero plus the first derivative of π evaluated at zero multiplied by π₯
plus the second derivative of π evaluated at zero divided by two factorial
multiplied by π₯ squared. And this sum goes on infinitely,
giving us the sum from π equals zero to β of the πth derivative of π evaluated at
zero divided by π factorial multiplied by π₯ to the πth power.

We might be tempted at this point
to define π of π₯ to be the cosh of two π₯ and then differentiate this function and
try to find a pattern for the πth derivative evaluated at zero. And this would work. However, a slightly simpler method
would be to use the cosh of π₯ and then substitute in to π₯ at the end. So letβs set our function π of π₯
to be equal to the cosh of π₯, which is equal to π to the power of π₯ plus π to
the power of negative π₯ all divided by two.

To find the first derivative of our
function π of π₯, we differentiate each term separately. First, we notice that the
derivative of π to the power of π₯ all divided by two is just equal to itself, π
to the power of π₯ all divided by two. Similarly, the derivative of π to
the power of negative π₯ all divided by two is negative π to the power of negative
π₯ all divided by two. We sometimes call this function the
hyperbolic sin of π₯ or sin π₯.

Now, to find the second derivative
of our function π of π₯, we take the derivative of our first derivative. So thatβs the derivative of our
function, the hyperbolic sign of π₯. The derivative of π to the power
of π₯ all divided by two is just equal to itself. And when we differentiate our
second term, which has an π to the power of negative π₯, weβll get two negative
ones, giving us plus π to the power of negative π₯ all divided by two. So we have our second derivative of
our function π of π₯ with respect to π₯ is equal to the hyperbolic cos of π₯.

Therefore, what we have shown is if
we differentiate our function π of π₯ an odd number of times, we will get the
hyperbolic sin of π₯. And if we differentiate our
function π of π₯ an even number of times, we will get the hyperbolic cos of π₯. Weβre now ready to start evaluating
these functions at zero. We have that our function π
evaluated at zero is equal to the hyperbolic cos evaluated at zero is equal to π to
the zeroth power plus π to the negative zeroth power all divided by two. And since π to the zeroth power
and π to the negative zeroth power are both just equal to one, we just have one
plus one all divided by two which is equal to one.

We can then do the same to find the
value of our first derivative function evaluated at zero, which is equal to
hyperbolic sin evaluated at zero, which is just equal to zero. Now, using the fact that if we
differentiate our function π of π₯ an odd number of times with respect to π₯, we
get the hyperbolic sin of π₯. Then if we evaluate this π₯ is
equal to zero, we will just get zero. Similarly, since weβve shown if we
differentiate our function π an even number of times, weβll get the hyperbolic cos
of π₯. If we evaluate this when π₯ is
equal to zero, weβll just get one.

In particular, what weβve shown
here is that for odd π in our Maclaurin series we would just get zero in the
series. So we can leave these out. So using our definition of
Maclaurin series. We have the hyperbolic cos of π₯ is
equal to the sum from π equal zero to β of the πth derivative evaluated at zero
divided by π factorial multiplied by π₯ to the πth power. We can split this series into two
series, the first giving us all of the odd values of π and the second one giving us
all the even values of π.

As we discussed previously, every
coefficient in our first series is equal to zero. So we can just remove the
series. And in our second series, for our
even values of π, we have that the evaluations of our function at zero is always
equal to one. Giving us the Maclaurin series for
the hyperbolic cos of π₯ is equal to the sum from π equals zero to β of π₯ to the
power of two π divided by two π factorial.

Now, we want to use our Maclaurin
series for the hyperbolic cos of π₯ to find a Maclaurin series for the hyperbolic
cos of two π₯. And we can do this by just
substituting two π₯ into our series. This gives us that the hyperbolic
cos of two π₯ is equal to the sum from π equals zero to β of two π₯ raised to the
power of two π all divided by two π factorial. Therefore, we have shown that the
Maclaurin series for the hyperbolic cos of two π₯ is equal to the sum from π equals
zero to β of two π₯ raised to the power of two π all divided by two π
factorial.