Video Transcript
Consider the function lowercase 𝑓 of 𝑥 is equal to 18 divided by one minus three 𝑥 all squared. Find a power series expansion for lowercase 𝑓 by differentiating the power series for capital 𝐹.
We’re given a function lowercase 𝑓 of 𝑥, which is a rational function. And we’re asked to find a power series representation for this function lowercase 𝑓 of 𝑥 by differentiating the power series for capital 𝐹. First, we should recall what we mean by capital 𝐹. It will be an antiderivative of lowercase 𝑓 of 𝑥. In other words, capital 𝐹 of 𝑥 will be equal to the integral of lowercase 𝑓 of 𝑥 with respect to 𝑥. Remember, this will give us a constant of integration. However, it won’t mess up what we choose for our constant of integration.
Since we want to find a power series representation for capital 𝐹 of 𝑥, we’re going to need to find an expression for capital 𝐹 of 𝑥. And to do this, we’re going to need to integrate the lowercase 𝑓 of 𝑥. So we’ll set capital 𝐹 of 𝑥 to be the integral of 18 divided by one minus three 𝑥 all squared with respect to 𝑥. And the easiest way to evaluate this integral is to use a substitution.
We’ll use the substitution 𝑢 is equal to one minus three 𝑥. So we want to integrate this by substitution. We’re going to need to find an expression for d𝑢 by d𝑥. That’s the derivative of one minus three 𝑥 with respect to 𝑥. This is a linear function. So its derivative will be the coefficient of 𝑥, which is negative three. Then we know that d𝑢 by d𝑥 is not a fraction. However, when we’re using integration by substitution, we can treat it a little bit like a fraction. This gives us the equivalent statement in terms of differentials d𝑢 is equal to negative three d𝑥.
Now, to make our substitution slightly easier, we can see we have 18 in our numerator. However, we could write 18 as negative six multiplied by negative three. Doing this gives us negative three multiplied by d𝑥, which we know is equal to d𝑢. We’re now ready to evaluate this integral by using our substitution. First, we change 18 d𝑥 with negative six d𝑢. Then, in our denominator, we substitute one minus three 𝑥 with 𝑢. This gives us the integral of negative six over 𝑢 squared with respect to 𝑢.
And now we’re almost ready to evaluate this integral. However, we’ll rearrange our integrand slightly. We’ll write this as negative six times 𝑢 to the power of negative two. And the reason we do this is now our expression is easier to integrate by using the power rule for integration. We add one to our exponent of 𝑢 and then divide by this new exponent. This gives us negative six 𝑢 to the power of negative one all divided by negative one. And of course we can simplify this. Negative six divided by negative one is equal to six. And of course don’t forget this is an indefinite integral. So we need to add a constant of integration 𝐶. The last thing we’ll do is use our laws of exponents to move 𝑢 into our denominator and then use our substitution 𝑢 is equal to one minus three 𝑥.
Therefore, we were able to show that capital 𝐹 of 𝑥 will be equal to six divided by one minus three 𝑥 plus the constant of integration 𝐶. And this is our general antiderivative of the function lowercase 𝑓 of 𝑥. We might be worried about having a constant of integration in this expression. However, once we find a power series representation of this function, we’re going to differentiate it. And that will remove our constant term.
So in actual fact, we could pick a value for 𝐶 or we could just leave it as it is. It won’t make a difference because we’re going to differentiate this anyway. So to make things easier, we’ll choose the value of 𝐶 is equal to zero. So now we need to find a power series representation of our specific antiderivative. Capital 𝐹 of 𝑥 is equal to six divided by one minus three 𝑥. We could do this directly from the definition of a Maclaurin series. Or we could use what we know about the sum of infinite geometric series.
However, in this case, we’re going to use the fact that one divided by one minus 𝑥 is equal to the sum from 𝑛 equals zero to ∞ of 𝑥 to the 𝑛th power. This is a useful result that’s worth committing to memory. In fact, this is always valid provided the absolute value of 𝑥 is less than one.
We want to use this power series to find the power series for our function capital 𝐹 of 𝑥. In our numerator, we have six. So we’ll need to multiply through by six. Multiplying both sides of this equation through by six, we get six over one minus 𝑥 is equal to six times the sum from 𝑛 equals zero to ∞ of 𝑥 to the 𝑛th power. And we can simplify this slightly. We know six is a constant, so we can bring it inside of our series. Then we can rearrange the two factors of our summand.
So now we have a power series representation for six divided by one minus 𝑥. However, in our function capital 𝐹 of 𝑥, we have three 𝑥 instead of 𝑥. So we need to replace 𝑥 with three 𝑥. So by replacing 𝑥 with three 𝑥, we have six divided by one minus three 𝑥 is equal to the sum from 𝑛 equals zero to ∞ of six times three 𝑥 all raised to the 𝑛th power.
And there is something worth pointing out here. Since our original power series was only valid at the absolute value of 𝑥 was less than one, multiplying through by six won’t change the convergence. However, replacing 𝑥 with three 𝑥 will change the convergence.
We now have the absolute value of three 𝑥 must be less than one. However, in this case, we’re not asked to find the values of 𝑥 for which our power series is valid. So we don’t actually need to worry about this step. We’re now almost ready to use this to find a power series representation for lowercase 𝑓 of 𝑥. There’s one more thing we’ll do. We’re going to distribute the exponent of 𝑛 over our parentheses. Doing this gives us the sum from 𝑛 equals zero to ∞ of six times three to the 𝑛th power multiplied by 𝑥 to the 𝑛th power.
We’re now ready to find our power series representation for lowercase 𝑓 of 𝑥 by differentiating the power series for capital 𝐹. First, recall capital 𝐹 is an antiderivative of lowercase 𝑓 of 𝑥. So lowercase 𝑓 of 𝑥 is equal to the derivative of capital 𝐹 of 𝑥 with respect to 𝑥. We now want to replace capital 𝐹 of 𝑥 with our power series representation for capital 𝐹 of 𝑥. Doing this gives us the derivative of the sum from 𝑛 equals zero to ∞ of six times three to the 𝑛th power multiplied by 𝑥 to the 𝑛th power with respect to 𝑥.
Next, we need to recall we’re allowed to evaluate the derivative of a power series term by term provided it’s for a value of 𝑥 which our power series is convergent. So by instead differentiating our power series term by term, we now have the sum from 𝑛 equals one to ∞ of the derivative of six times three to the 𝑛th power multiplied by 𝑥 to the 𝑛th power with respect to 𝑥. And it’s worth pointing out here our sum now goes from 𝑛 is equal to one to ∞. And this is because when 𝑛 is equal to zero, we will have a constant term. And we know the derivative of a constant with respect to 𝑥 is equal to zero. And this will always be true when we’re differentiating a power series. So we can always do this step.
Now, we need to evaluate this derivative. For any value of 𝑛, six times three to the 𝑛th power will be a constant. The only thing varying as 𝑥 varies is 𝑥 to the 𝑛th power. So we can differentiate all of these by using our power rule for differentiation. We want to multiply by our exponent of 𝑥, which in this case is 𝑛, and then reduce this exponent by one. This gives us six times three to the 𝑛th power multiplied by 𝑛 times 𝑥 to the power of 𝑛 minus one. This then gives us the sum from 𝑛 equals one to ∞ of six times three to the 𝑛th power multiplied by 𝑛 times 𝑥 to the power of 𝑛 minus one.
And we could leave our answer like this. However, as it currently stands, our series starts when 𝑛 is equal to one. We instead want our series to start when 𝑛 is equal to zero. To do this, we’re going to need to add one to all of the values of 𝑛 in our summand. By doing this and simplifying, we get the sum from 𝑛 equals zero to ∞ of six times three to the power of 𝑛 plus one times 𝑛 plus one multiplied by 𝑥 to the 𝑛th power, which is our final answer.
Therefore, for the function lowercase 𝑓 of 𝑥 is equal to 18 divided by one minus three 𝑥 all squared, we were able to integrate this by using a substitution, find a power series representation for its antiderivative capital 𝐹 of 𝑥, and then differentiate this power series term by term to find a power series representation for lowercase 𝑓 of 𝑥. We found the power series representation for lowercase 𝑓 of 𝑥 was the sum from 𝑛 equals zero to ∞ of six times three to the power of 𝑛 plus one multiplied by 𝑛 plus one times 𝑥 to the 𝑛th power.
