Video: Finding the Maclaurin Series of an Inverse Trigonometric Function

Find the Maclaurin series of arctan 5π‘₯.

08:25

Video Transcript

Find the Maclaurin series of arctan of five π‘₯.

We need to find the Maclaurin series for the function the arctan of five π‘₯, and there’s a few different ways we could approach this. We might be tempted to try and do this directly from the definition of a Maclaurin series. And we recall the Maclaurin series for the function 𝑓 of π‘₯ is the sum from 𝑛 equals zero to ∞ of the 𝑛th derivative of 𝑓 of π‘₯ evaluated at zero divided by 𝑛 factorial times π‘₯ to the 𝑛th power. So to try and find the Maclaurin series directly from the definition of the arctan of five of π‘₯, we need to start differentiating the arctan of five π‘₯.

So if we were to attempt to do this, we would start by setting the function 𝑓 of π‘₯ to be equal to the arctan of five π‘₯. We would then need to differentiate this. In fact, we could do this by using one of our standard inverse trigonometric derivative results. We know for any real constant π‘˜, the derivative of the arctan of π‘˜π‘₯ with respect to π‘₯ is equal to π‘˜ divided by one plus π‘˜ squared π‘₯ squared. In our case, the value of π‘˜ is equal to five, so we get that 𝑓 prime of π‘₯ is equal to five divided by one plus 25π‘₯ squared.

If we then wanted to carry on finding the Maclaurin series using this method, we would then need to differentiate five divided by one plus 25π‘₯ squared with respect to π‘₯. And this means we would even need to use the quotient rule, the chain rule, or the general power rule. So while this is possible, we can see that this will get more and more complicated. In fact, there’s an easier method. We can find a power series representation of five divided by one plus 25π‘₯ squared by using what we know about infinite geometric series. But then this power series will be equal to 𝑓 prime of π‘₯. So if we integrate this power series, we’ll then get a power series representation for the arctan of five π‘₯.

So let’s discuss how we would do this. We know for real constants π‘Ž and π‘Ÿ where π‘Ž is not equal to zero, if the absolute value of π‘Ÿ is less than one, then the sum from 𝑛 equals zero to ∞ of π‘Ž times π‘Ÿ to the 𝑛th power is convergent and is equal to π‘Ž divided by one minus π‘Ÿ. We also know if the absolute value of π‘Ÿ is greater than one, then this series will be divergent. We want to use this to find a power series representation of our function 𝑓 prime of π‘₯. And we can see that 𝑓 prime of π‘₯ is already in this form. We just need to set our value of π‘Ž equal to five and our value of π‘Ÿ equal to negative 25π‘₯ squared. Then, we’ve rewritten 𝑓 prime of π‘₯ to be equal to π‘Ž divided by one minus π‘Ÿ.

Now, by using what we know about infinite geometric series, we can rewrite 𝑓 prime of π‘₯ as the sum from 𝑛 equals zero to ∞ of five times negative 25π‘₯ squared raised to the 𝑛th power. And it’s also worth pointing out since this was a geometric series, we know it will converge when the absolute value of the ratio is less than one and diverge when the absolute value of the ratio is greater than one. So we could in fact use this to find the radius of convergence of our power series. However, we’re not required to do this in this question, so we could ignore this part of our working. We just need to be mindful that this power series will only be valid for values of π‘₯ inside of our radius of convergence.

Remember, we’re going to want to integrate this power series, so we should rewrite our summand into a form which is easy to integrate. We’ll start by distributing the exponent over our parentheses. This gives us the sum from 𝑛 equals zero to ∞ of five times negative 25 raised to the 𝑛th power multiplied by π‘₯ to the power of two 𝑛. Next, we’ll simplify negative 25 all raised to the 𝑛th power to be negative one raised to the 𝑛th power times five to the power of two 𝑛. We can do this because negative 25 is negative five squared. This gives us the sum from 𝑛 equals zero to ∞ of five times negative one to the 𝑛th power multiplied by five to the power of two 𝑛 times π‘₯ to the power of two 𝑛.

Finally, we know that five multiplied by five to the power of two 𝑛 is five to the power of two 𝑛 plus one. So we’ve shown that our function 𝑓 prime of π‘₯ is equal to the sum from 𝑛 equals zero to ∞ of negative one to the 𝑛th power times five to the power of two 𝑛 plus one multiplied by π‘₯ to the power of two 𝑛, provided that this power series converges. And there’s also one more thing worth pointing out about this power series. It must be the Maclaurin series for our function 𝑓 prime of π‘₯. This is because any power series representation centered at a point for a function must be equal to its Taylor series centered at this point. And this power series is centered at zero, so it must be a Maclaurin series. We’re now ready to use this to find the Maclaurin series for the arctan of five π‘₯.

To start, remember that 𝑓 prime of π‘₯ is the derivative of the arctan of five π‘₯ with respect to π‘₯. So the arctan of five π‘₯ must be equal to the integral of 𝑓 prime of π‘₯ with respect to π‘₯. Now, instead of integrating 𝑓 prime of π‘₯, we’ll integrate the Maclaurin series expansion we found for 𝑓 prime of π‘₯. This gives us the integral of the sum from 𝑛 equals zero to ∞ of negative one to the 𝑛th power times five to the power of two 𝑛 plus one multiplied by π‘₯ to the power of two 𝑛 with respect to π‘₯. And now we need to use rules which we know about finding the integral of a power series.

We know that instead of integrating a power series, we can integrate this term by term as long as we’re within the radius of convergence. In other words, we have the sum from 𝑛 equals zero to ∞ of the integral of negative one to the 𝑛th power times five to the power of two 𝑛 plus one multiplied by π‘₯ to the power of two 𝑛 with respect to π‘₯. And this looks rather complicated. However, negative one to the 𝑛th power times five to the power of two 𝑛 plus one is a constant with respect to π‘₯; if won’t vary as π‘₯ varies. So the only part of this expression varying as π‘₯ varies is π‘₯ to the power of two 𝑛. And we know how to integrate this by using the power rule for integration.

But just before we use the power rule for integration, it’s worth checking we don’t have π‘₯ to the power of negative one. Well, our values of 𝑛 start at 𝑛 is equal to zero. So our exponents of π‘₯ are greater than or equal to zero, so we can use the power rule for integration for all of these terms. So we add one to our exponents of π‘₯ and then divide by this new exponent. Our exponent of π‘₯ was two 𝑛. We add one to this to get two 𝑛 plus one, so we need to divide by two 𝑛 plus one. This gives us the sum from 𝑛 equals zero to ∞ of negative one to the 𝑛th power times five to the power of two 𝑛 plus one divided by two 𝑛 plus one all multiplied by π‘₯ to the power of two 𝑛 plus one.

And remember, we would need to add a constant of integration for each of these terms. However, we can combine all of these into one constant of integration we add outside of our series. So we now have a power series representation for the arctan of five π‘₯ centered at π‘₯ is equal to zero. We know this has to be the Maclaurin series for our function. The only thing left to do is find the value of our constant of integration, 𝐢. To do this, we’ll substitute π‘₯ is equal to zero because we know our power series must converge for π‘₯ is equal to zero.

Substituting π‘₯ is equal to zero into this expression, we get the arctan of five times zero is equal to the sum from 𝑛 equals zero to ∞ of negative one to the 𝑛th power times five to the power of two 𝑛 plus one divided by two 𝑛 plus one multiplied by zero to the power of two 𝑛 plus one. And then we add onto this series our constant of integration, 𝐢. Well, the arctan of five times zero is the arctan of zero. And we know the tan of zero is equal to zero, so this is just equal to zero.

Next, to evaluate our series, we notice every term in our series has a factor of zero. So, in fact, every term in this series is just equal to zero. So, in fact, the right-hand side of this equation simplifies to give us 𝐢. So we’ve shown that 𝐢 must be equal to zero. So, by using the fact that 𝐢 is equal to zero and rearranging our summand, we’ve shown the arctan of five π‘₯ is equal to the sum from 𝑛 equals zero to ∞ of negative one to the 𝑛th power times five to the power of two 𝑛 plus one multiplied by π‘₯ to the power of two 𝑛 plus one divided by two 𝑛 plus one. And this is of course only true for values of π‘₯ where this power series converges.

There’s one more thing we’ll do to simplify this power series. We see that both five and π‘₯ have the same exponent, so we’ll combine this into five π‘₯ all raised to the power of two 𝑛 plus one. And this gives us our final answer. We were able to show the Maclaurin series of the arctan of five π‘₯ is equal to the sum from 𝑛 equals zero to ∞ of negative one to the 𝑛th power times five π‘₯ all raised to the power of two 𝑛 plus one all divided by two 𝑛 plus one.

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