# Question Video: Find the Power Series Representation of a Composition between the Natural Logarithm Function and a Polynomial Mathematics • Higher Education

For the given function 𝑓(𝑥) = ln (1 + 𝑥²), find a power series representation of 𝑓 by integrating the power series of 𝑓′.

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### Video Transcript

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

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

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

And we can simplify the left-hand side of this expression. The integral of the derivative function 𝑓 prime of 𝑥 with respect to 𝑥 will be just equal to 𝑓 of 𝑥 plus the constant of integration 𝐶. 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 𝑓 of 𝑥 which will be true for all values of 𝑥 whose absolute value is less than the radius of convergence 𝑅.

So, to find our power series for a function 𝑓 prime of 𝑥, let’s first find out what the function 𝑓 prime of 𝑥 is by differentiating the natural logarithm of one plus 𝑥 squared. We see that the natural logarithm of one plus 𝑥 squared is a composite function. So, we’ll differentiate this by using the chain rule. We see if we set 𝑢 equal to the inner function of one plus 𝑥 squared, then our function 𝑓 of 𝑥 is a function of 𝑢. It’s equal to the natural logarithm of 𝑢. And 𝑢 is a function of 𝑥; it’s one plus 𝑥 squared.

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

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

To do this, we’re going to use one of our standard results of power series. One divided by one minus 𝑥 is equal to the sum from 𝑛 equals zero to ∞ of 𝑥 to the 𝑛th power for all values of 𝑥, whose absolute value is less than one. However, we want a power series for 𝑓 prime of 𝑥; that’s two 𝑥 divided by one plus 𝑥 squared. So, instead of our negative 𝑥 in the denominator, we want a positive 𝑥 squared. So, if we rewrite our denominator of one plus 𝑥 squared as one minus negative 𝑥 squared, then we can see if we rewrite all instances of 𝑥 with negative 𝑥 squared in our power series representation for one divided by one minus 𝑥. Then we’ll have a power series representation for a function with one plus 𝑥 squared in the denominator.

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

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

Next, we’ll bring the two 𝑥 inside of our sum. And we can simplify this by grouping our factors of 𝑥. Doing this and then rearranging gives us that our derivative function 𝑓 prime of 𝑥 is equal to the sum from 𝑛 equals zero to ∞ of two. Times negative one to the 𝑛th power times 𝑥 to the power of two 𝑛 plus one when the absolute value of negative 𝑥 squared is less than one. We’re now ready to integrate this power series to find the power series for our function 𝑓 of 𝑥.

We have 𝑓 of 𝑥 is equal to the integral of 𝑓 prime of 𝑥 with respect to 𝑥 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 𝑥 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 𝑛th power is a constant and we can see our exponent of two 𝑛 plus one from 𝑛 is equal to zero up to ∞ 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 𝑛 equals zero to ∞ of two times negative one to the 𝑛th power times 𝑥 to the power of two 𝑛 plus two all divided by two 𝑛 plus two. And then, we’ll combine all of our constants of integration into one constant of integration outside of our sum, which we will call 𝐶. 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 𝑓 of 𝑥. We just need to find what our constant of integration 𝐶 is. To start, we see that this power series representation is representing our function 𝑓 of 𝑥, which is the natural logarithm of one plus 𝑥 squared. And we know that this power series representation must be true whenever the absolute value of negative 𝑥 squared is less than one. So, what we could do is substitute 𝑥 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 𝑥 is equal to zero. Substituting 𝑥 is equal to zero gives us the natural logarithm of one plus zero squared is equal to the sum from 𝑛 is equal to zero to ∞ of negative one to the 𝑛th power times zero to the power of two 𝑛 plus two all divided by 𝑛 plus one. And then we add our constant of integration 𝐶. 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 𝐶. So, 𝐶 is also equal to zero.

Therefore, by integrating a power series representation of the derivative of the natural logarithm of one plus 𝑥 squared. We’ve shown that our function 𝑓 of 𝑥 can be represented by the power series the sum from 𝑛 equals zero to ∞ of negative one to the 𝑛th power times 𝑥 to the power of two 𝑛 plus two divided by 𝑛 plus one.