Video Transcript
Consider the function 𝑓 of 𝑥 is equal to 15 times the inverse tan of two 𝑥 over five. Find the power series expansion for 𝑓 by integrating the power series of 𝑓 prime.
We’re given a function 𝑓 of 𝑥, and we want to find the power series expansion for 𝑓 of 𝑥. Now we could do this directly. However, there’s actually an easier method, and we’re told to do this in the question. We’re going to find the power series expansion for 𝑓 by integrating the power series expansion for 𝑓 prime. So the first thing we’re going to need to do is find a power series expansion for 𝑓 prime. To do this, we’re going to need to find an expression for 𝑓 prime. 𝑓 prime of 𝑥 will be the derivative of 15 times the inverse tan of two 𝑥 over five with respect to 𝑥.
There’s a few different ways we could evaluate this derivative. For example, we could do this by using the chain rule. However, we can also recall the following derivative result for inverse trigonometric functions. For any real constant 𝑎, the derivative of the inverse tan of 𝑎𝑥 with respect to 𝑥 is equal to 𝑎 divided by one plus 𝑎𝑥 all squared. In our case, the value of the constant 𝑎 is two over five. So by setting our value of 𝑎 equal to two over five and remembering we’re multiplying this entire expression by 15, we get that 𝑓 prime of 𝑥 is equal to 15 times two over five all divided by one plus two 𝑥 over five all squared.
And we can simplify this expression. First, in our numerator, 15 divided by five is equal to three, but then we have three times two which is equal to six. Next, in our denominator, we want to distribute the square over our parentheses. This means we’ll need to square everything inside of our parentheses. This will be four 𝑥 squared over 25. And at this point, we might be tempted to continue simplifying. But remember, we’re trying to find the power series representation for 𝑓 prime. And we can see that this representation for 𝑓 prime is in a special form. It’s in the form 𝑎 divided by one minus 𝑟, so we can find the power series representation for 𝑓 prime by using what we know about the sum of infinite geometric series.
So let’s recall what we know about the infinite sum of a geometric series. We know if the absolute value of our ratio 𝑟 is less than one and 𝑎 is not equal to zero, then 𝑎 over one minus 𝑟 will be equal to the sum from 𝑛 equals zero to ∞ of 𝑎 times 𝑟 to the 𝑛th power. And we can see our expression for 𝑓 prime is in this form. First, our value of 𝑎 will be six. Next, our value of 𝑟 will be negative four 𝑥 squared over 25. This means we can substitute 𝑎 is equal to six and 𝑟 is equal to negative four 𝑥 squared over 25 into our infinite sum for a geometric series formula. This will give us a power series representation for 𝑓 prime. We get 𝑓 prime of 𝑥 will be equal to the sum from 𝑛 equals zero to ∞ of six times negative four 𝑥 squared over 25 all raised to the 𝑛th power.
And it’s worth pointing out we can guarantee this converges as long as the absolute value of 𝑟, which is negative four 𝑥 squared over 25, is less than one. Finally, the question tells us to integrate this power series representation of 𝑓 prime of 𝑥 to find a power series expansion for 𝑓 of 𝑥. So let’s discuss how we’ll do this. First, remember, 𝑓 of 𝑥 is an antiderivative of 𝑓 prime of 𝑥. In other words, 𝑓 of 𝑥 is the integral of 𝑓 prime of 𝑥 with respect to 𝑥 up to a constant of integration 𝐶. And we’ll need to worry about this constant of integration at the end.
First, we’ll write our power series representation for 𝑓 prime of 𝑥 into this expression. This gives us the integral of the sum from 𝑛 equals zero to ∞ of six times negative four 𝑥 squared over 25 all raised to the 𝑛th power with respect to 𝑥. And to evaluate this, we need to recall we can integrate each term of our power series separately. And this would be guaranteed to be true within the radius of convergence. So we’ll integrate each term of our power series separately. This gives us the sum from 𝑛 equals zero to ∞ of the integral of six times negative four 𝑥 squared over 25 all raised to the 𝑛th power with respect to 𝑥.
To evaluate this integral, we’re going to first need to simplify. We’re going to distribute the exponent of 𝑛 over our parentheses. And when we do this, we’ll only raise negative four over 25 to the 𝑛th power and 𝑥 squared to the 𝑛th power. So by doing this, we get negative four over 25 all raised to the 𝑛th power multiplied by 𝑥 squared all raised to the 𝑛th power. Of course, we can simplify 𝑥 squared all raised to the 𝑛th power to just be 𝑥 to the power of two 𝑛. And at this point, we’re ready to start evaluating our integral term by term. First, six times negative four over 25 all raised to the 𝑛th power is not varying as 𝑥 varies. Therefore, it’s constant with respect to our integral.
Next, we know how to integrate 𝑥 to the power of two 𝑛 by using the power rule for integration. However, remember, we do need to be careful. We need to check that we never have an exponent of negative one. Otherwise, we would need to use our laws for integrating the reciprocal function. In this case, we don’t need to worry, though, our values of 𝑛 start at zero. So our exponent will never be negative one. So we can integrate every single term in this power series by using the power rule for integration. We need to add one to our exponent and then divide by this new exponent.
We add one to our exponent of two 𝑛, giving us a new exponent of two 𝑛 plus one, and then divide by the new exponent of two 𝑛 plus one. This gives us the sum from 𝑛 equals zero to ∞ of six times negative four over 25 all raised to the 𝑛th power times 𝑥 to the power of two 𝑛 plus one divided by two 𝑛 plus one.
And remember, every time we integrate, we will get a constant of integration. However, it’s easier to combine all of these constants into one constant of integration outside of our sum we’ll call 𝐶. Remember though, this is supposed to be the power series representation of 𝑓 of 𝑥, which is 15 times the inverse tan of two 𝑥 over five. So we can’t just have an unknown constant 𝐶 in this expression; we need to find a value for 𝐶. And usually, the easiest way to do this is to substitute a value into our power series which will give us zero.
In this case, we can substitute 𝑥 is equal to zero into this expression. We know our power series will be convergent because it would just be the start of zeros. And this will be useful because we can also substitute 𝑥 is equal to zero into our expression for 𝑓 of 𝑥. So we substitute 𝑥 is equal to zero into our equation. This gives us 15 times the inverse tan of two times zero over five is equal to the sum from 𝑛 equals zero to ∞ of zero plus the constant of integration 𝐶. And we can just solve this equation. First, the inverse tan of zero is just equal to zero. Next, the infinite sum of zero is also going to be equal to zero. So 𝐶 is just equal to zero. And if 𝐶 is equal to zero, we can just remove this from our expression for 𝑓 of 𝑥. And this just gives us our final answer.
Therefore, for the function 𝑓 of 𝑥 is equal to 15 times the inverse tan of two 𝑥 over five, we were able to find the power series expansion for 𝑓 by integrating the power series for 𝑓 prime. We got the power series the sum from 𝑛 equals zero to ∞ of six times negative four over 25 all raised to the 𝑛th power multiplied by 𝑥 to the power of two 𝑛 plus one divided by two 𝑛 plus one.
