### Video Transcript

For the given function π of π₯ is
equal to the natural logarithm of one plus π₯ squared, find a power series
representation of π by integrating the power series for π prime.

The question wants us to find a
power series representation of our function π, which is the natural logarithm of
one plus π₯ squared. And it tells us to do this by
integrating the power series representation of our function π prime. And this leaves us with the
question, why can we integrate a power series of π prime of π₯ to find a power
series for π of π₯?

We recall if we have a function π
of π₯, which is given by a power series, when the absolute value of π₯ is less than
our radius of convergence π
. Then integrating our function π of
π₯ with respect to π₯ is the same as integrating each term of our power series with
respect to π₯ when the absolute value of π₯ is less than our radius of convergence
π
. The question tells us to use this
on the function π prime. So, weβll set π equal to π
prime. Applying this to our function π
prime of π₯ tells us the integral of π prime of π₯ with respect to π₯ is the same
as integrating each term of our power series for π prime of π₯ when the absolute
value of π₯ is less than the radius of convergence π
.

And we can simplify the left-hand
side of this expression. The integral of the derivative
function π prime of π₯ with respect to π₯ will be just equal to π of π₯ plus the
constant of integration πΆ. And we can combine the constant of
integration we get on the left-hand side of the equation with the ones we will get
on the right-hand side of the equation. This will then give us a power
series for our function π of π₯ which will be true for all values of π₯ whose
absolute value is less than the radius of convergence π
.

So, to find our power series for a
function π prime of π₯, letβs first find out what the function π prime of π₯ is by
differentiating the natural logarithm of one plus π₯ squared. We see that the natural logarithm
of one plus π₯ squared is a composite function. So, weβll differentiate this by
using the chain rule. We see if we set π’ equal to the
inner function of one plus π₯ squared, then our function π of π₯ is a function of
π’. Itβs equal to the natural logarithm
of π’. And π’ is a function of π₯; itβs
one plus π₯ squared.

And the chain rule tells us that if
our function π is a function of π’ and π’ is a function of π₯. Then we can calculate the
derivative of π with respect to π₯ by calculating the derivative of π with respect
to π’ and then multiplying this by the derivative of π’ with respect to π₯. So, by using the chain rule, we
have that π prime of π₯ is equal to the derivative of π with respect to π’, which
is the derivative of the natural logarithm of π’ with respect to π’.

And we recall the derivative of the
natural logarithm of π’ with respect to π’ is just equal to one over π’. Then we multiply this by the
derivative of π’ with respect to π₯. Thatβs the derivative of one plus
π₯ squared with respect to π₯. And we can differentiate this using
the power rule for differentiation. It gives us two π₯. Finally, we substitute π’ is equal
to one plus π₯ squared to see that our function π prime of π₯ is equal to two π₯
divided by one plus π₯ squared. So, we found our function π prime
of π₯. We now need to turn this into a
power series.

To do this, weβre going to use one
of our standard results of power series. One divided by one minus π₯ is
equal to the sum from π equals zero to β of π₯ to the πth power for all values of
π₯, whose absolute value is less than one. However, we want a power series for
π prime of π₯; thatβs two π₯ divided by one plus π₯ squared. So, instead of our negative π₯ in
the denominator, we want a positive π₯ squared. So, if we rewrite our denominator
of one plus π₯ squared as one minus negative π₯ squared, then we can see if we
rewrite all instances of π₯ with negative π₯ squared in our power series
representation for one divided by one minus π₯. Then weβll have a power series
representation for a function with one plus π₯ squared in the denominator.

This gives us one divided by one
minus negative π₯ squared is equal to the sum from π equals zero to β of negative
π₯ squared all raised to the πth power when the absolute value of negative π₯
squared is less than one. And we can simplify the denominator
of one minus negative π₯ squared to be one plus π₯ squared. We almost have the power series for
our function π prime of π₯. All we need is a numerator of two
π₯. Weβll achieve this by multiplying
both sides of our equation by two π₯.

This gives us that two π₯ divided
by one plus π₯ squared is equal to two π₯ times the sum from π equals zero to β of
negative π₯ squared all raised to the πth power. We can simplify this further by
distributing the exponent over the parentheses inside of our summand. This gives us negative one to the
πth power multiplied by π₯ squared all raised to the πth power which, by our laws
of exponents, gives us π₯ to the power of two π.

Next, weβll bring the two π₯ inside
of our sum. And we can simplify this by
grouping our factors of π₯. Doing this and then rearranging
gives us that our derivative function π prime of π₯ is equal to the sum from π
equals zero to β of two. Times negative one to the πth
power times π₯ to the power of two π plus one when the absolute value of negative
π₯ squared is less than one. Weβre now ready to integrate this
power series to find the power series for our function π of π₯.

We have π of π₯ is equal to the
integral of π prime of π₯ with respect to π₯ up to our constant of integration. And we know this is the same as
integrating each term of our power series when the absolute value of negative π₯
squared is less than one. And we can integrate all of these
using the power rule for integration. Since two times negative one to the
πth power is a constant and we can see our exponent of two π plus one from π is
equal to zero up to β is never equal to negative one.

So, we add one to our exponent and
then divide by this new exponent. This gives us the sum from π
equals zero to β of two times negative one to the πth power times π₯ to the power
of two π plus two all divided by two π plus two. And then, weβll combine all of our
constants of integration into one constant of integration outside of our sum, which
we will call πΆ. We can also cancel the shared
factor of two in our numerator and our denominator.

We now almost have our power series
representation for our function π of π₯. We just need to find what our
constant of integration πΆ is. To start, we see that this power
series representation is representing our function π of π₯, which is the natural
logarithm of one plus π₯ squared. And we know that this power series
representation must be true whenever the absolute value of negative π₯ squared is
less than one. So, what we could do is substitute
π₯ is equal to zero. Then, we can evaluate the power
series on the right-hand side. And we can evaluate the natural
logarithm of one plus zero squared on the left.

And of course, this power series
representation is valid when π₯ is equal to zero. Substituting π₯ is equal to zero
gives us the natural logarithm of one plus zero squared is equal to the sum from π
is equal to zero to β of negative one to the πth power times zero to the power of
two π plus two all divided by π plus one. And then we add our constant of
integration πΆ. On the left-hand side of our
equation, we have one plus zero squared is equal to one, and then the natural
logarithm of one is just equal to zero. And on the right-hand side of our
equation, we see that every term in our sum has a factor of zero. Therefore, our sum itself is just
equal to zero. This gives us the equation zero is
equal to zero plus πΆ. So, πΆ is also equal to zero.

Therefore, by integrating a power
series representation of the derivative of the natural logarithm of one plus π₯
squared. Weβve shown that our function π of
π₯ can be represented by the power series the sum from π equals zero to β of
negative one to the πth power times π₯ to the power of two π plus two divided by
π plus one.