We want to represent the function
f of x equals two x
times natural log of the quantity one plus x
as a power series
in this form here,
then we'll compute the first few
coefficients of the power series
where c sub two
would be the coefficient of the degree two term,
c sub three would be the coefficient
of the degree three term, and so on.
We also want to find the radius of convergence.
Notice how the given function does not resemble
the formula used here to find the infinite sum
of a convergent geometric series.
But if we focus on just part of this function first,
for example, we want to focus on
natural log of the quantity one plus x,
since we know the integral
of one divided by x integrated with respect to x
is equal to natural log of the absolute value
of x plus c,
then it follows that the integral
of the quantity one divided by the quantity one plus x,
integrated with respect to x,
would equal the natural log of the absolute value
of the quantity one plus x
plus c.
So notice how the integrand of this integral
does resemble the formula
for the infinite sum of a geometric series.
So we can use this to help us write the power series
for our function f of x.
But, again, we'll first focus on
the natural log part of the function,
then if we can write that as a power series,
we can just multiply that power series by two x.
So from this equation, it follows that natural log
of the quantity one plus x
is equal to the integral of one
divided by the quantity one plus x,
integrated with respect to x,
plus c.
And now let's rewrite this integrand
so it fits the formula here
for the sum of an infinite geometric series.
So now we'll write natural log of the quantity one plus x
equals the integral of one divided by,
because we want this to be a difference,
we'll write one minus negative x.
And now we can write this integrand as a power series,
where notice that a
is equal to one
and the common ratio r
would be equal to negative x.
So this would give us natural log of the quantity one plus x
equals the integral of
the summation from n equals zero to infinity
of a, which is one,
times negative x raised to the power of n.
Let's go ahead and change the form of this slightly
and write this as natural log of the quantity one plus x
equals the integral of
the summation from n equals zero to infinity.
Let's write this as negative one
raised to the power of n
times x raised to the power of n.
And now because our function is f of x equals two x times
natural log of the quantity one plus x,
we can represent our function
by multiplying by two x.
Let's do this on the next slide.
So, again, because our function f of x
equals two x times
natural log of the quantity one plus x,
this would be equal to two x
times the summation
from n equals zero to infinity
of negative one raised to the power of n
times x raised to the power of n,
integrated with respect to x,
plus c.
Now, we'll integrate this power series with respect to x.
So this would give us two x
times the summation from n equals zero to infinity of,
we still have negative one raised to the power n,
but then we'd add one to the exponent of x,
that would give us x
raised to the power of n plus one
divided by n plus one.
And, of course, we still have the plus c.
And now we'll bring the factor of two x
into the power series.
So I can write this as the summation
from n equals zero to infinity of,
notice how we're bringing in a factor of two,
so we'd have two times negative one
raised to the power of n.
We're bringing in an extra factor of x,
so we'd have x raised to the power of n plus two
divided by the quantity n plus one
plus c.
So, again, all this is still equal to two x
times natural log of the quantity one plus x.
And now from here, notice we let x equals zero.
The left side of this equation would be zero
and the power series would also be equal to zero,
and, therefore, c equals zero.
So now we know that our function f of x,
which equals two x times
natural log of the quantity one plus x,
is equal to the power series
where we have the summation
from n equals zero to infinity of
two times negative one raised to the power of n
times x raised to the power of n plus two
divided by the quantity n plus one.
And now we'll generate the first several terms
of our power series
to find the coefficients for our question.
So notice when n is zero,
we'd have two times one times x to the second,
or two x to the second
divided by one.
When n is one, notice the term is negative.
So we have minus two
times x to the third
divided by two
plus,
when n is two, we have two times one,
that's two x to the fourth,
divided by three.
So far we have [mumbles] n equals zero,
n equals one,
and n equals two.
Notice how this is degree four term.
And going back to our question,
we need to find the coefficients
up until the degree six term.
So we have to continue to find the terms
for n equals three and n equals four.
Notice when n is three, the term is negative.
So we have minus two times x to the fifth
divided by four.
And then finally, when n is four,
term is positive so we have plus two
times x to the sixth
divided by five.
And, of course, this continues.
Let's go and simplify this.
We have two x to the second
minus
x to the third,
plus 2/3
x to the fourth,
minus 2/4 or 1/2
x to the fifth,
and then finally, plus 2/5
x to the sixth.
Which means c sub two would be two,
c sub three
would be negative one,
c sub four is 2/3,
c sub five is negative 1/2,
and c sub six is 2/5.
Let's go ahead and record this.
Again, c sub two is equal to two,
c sub three is equal to negative one,
c sub four is equal to 2/3,
c sub five is equal to negative 1/2,
and c sub six is equal to 2/5.
Now, we still have to find the radius of convergence.
But looking at the integral here,
notice how the common ratio r is equal to negative x,
which means the power series will converge
when the absolute value of negative x
is less than one.
If it's helpful, we can think of this as
the absolute value of negative one x
less than one.
And since the absolute value of negative one is one,
this is equivalent to the absolute value of x
is less than one.
And, therefore, the radius of convergence is one.
This also tells us that the interval of convergence
would be the open interval from
negative one to positive one.
Now, before we go,
let's compare the graph of the given function
and the graph of the function
using the first five terms of this power series
which would be called the Maclaurin polynomial
since it is centered at x equals zero.
The original function is graphed here in blue
and the polynomial function
using the terms from the power series
is graphed here in red.
Notice around x equals zero
where the power series is centered,
the red function is an excellent representation
of the given blue function.
I hope you found this helpful.