Video: Integrating a Power Series Representation of a Function

For the given function 𝑓(π‘₯) = tan⁻¹ (2π‘₯), find a power series representation for 𝑓 by integrating the power series for 𝑓′.

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Video Transcript

For the given function 𝑓 of π‘₯ is equal to the inverse tan of two π‘₯, find a power series representation of 𝑓 by integrating the power series for 𝑓 prime.

The question is asking us to integrate the power series for the function 𝑓 prime so that we can find the power series representation of our function 𝑓 of π‘₯ is equal to the inverse tan of two π‘₯. So let’s start by calculating what our function 𝑓 prime of π‘₯ is. We know that the derivative with respect to π‘₯ of the inverse tan of π‘Žπ‘₯ is just equal to π‘Ž divided by one plus π‘Ž squared π‘₯ squared. We can use this to calculate that the function 𝑓 prime of π‘₯ is equal to two divided by one plus two squared π‘₯ squared. And then, we can just simplify two squared π‘₯ squared to just be four π‘₯ squared.

Next, the question tells us that we should find the power series for our function 𝑓 prime of π‘₯. And we can do this by using what we know about geometric series. We know that if the absolute value of our ratio π‘Ÿ is less than one, then the sum from 𝑛 equals zero to ∞ of π‘Ž multiplied by π‘Ÿ to the 𝑛th power is just equal to π‘Ž divided by one minus π‘Ÿ. Next, we can notice that our function 𝑓 prime of π‘₯ is already in the form π‘Ž divided by one minus π‘Ÿ if we set π‘Ž to equal two and π‘Ÿ to equal negative four π‘₯ squared.

So if we substitute this into what we know about geometric series, we get that if the absolute value of negative four π‘₯ squared is less than one, then we know that the sum from 𝑛 equals zero to ∞ of two multiplied by negative four π‘₯ squared all raised the 𝑛th power is equal to two divided by one minus negative four π‘₯ squared, where we can notice that our denominator, one minus negative four π‘₯ squared, is just one plus four π‘₯ squared. So this is equal to two divided by one plus four π‘₯ squared, which is equal to 𝑓 prime of π‘₯.

So we’ve now found the power series for our function 𝑓 prime. That is, it’s the sum from 𝑛 equals zero to ∞ of two multiplied by negative four π‘₯ squared raised to the 𝑛th power. So we can write the power series for our function 𝑓 prime of π‘₯ into our top equation. Now, we’re ready to simplify this power series. Let’s start by distributing the exponent 𝑛 over our parentheses. This gives us negative four to the 𝑛th power multiplied by π‘₯ squared to the 𝑛th power.

Next, we can rewrite negative four as negative one multiplied by two squared. We can then distribute the exponent 𝑛 over our first set of parentheses to get negative one to the 𝑛th power multiplied by two squared to the 𝑛th power. Finally, we can evaluate the exponents in our second and third set of parentheses, giving us a final answer of negative one to the 𝑛th power multiplied by two to the power of two 𝑛 multiplied by π‘₯ to the power of two 𝑛. This gives us a new sum from 𝑛 equals zero to ∞ of two multiplied by negative one to the 𝑛th power multiplied by two to the power of two 𝑛 multiplied by π‘₯ to the power of two 𝑛.

Now, we can simplify this further by noticing that two multiplied by two to the power of two 𝑛 is just equal to two to the power of two 𝑛 plus one, giving us the sum from 𝑛 equals zero to ∞ of two to the power of two 𝑛 plus one multiplied by negative one to the 𝑛th power multiplied by π‘₯ to the power of two 𝑛. We now recall that the integral of our derivative function 𝑓 prime of π‘₯ with respect to π‘₯ is equal to the function 𝑓 of π‘₯ up to a constant of integration 𝐢.

At this point, we substitute our power series representation for our derivative function 𝑓 prime of π‘₯. At this point, we’re going to move the integral symbol inside our sum. And we know we can do this because we’re integrating a power series. At this point, we can notice that two to the power of two 𝑛 plus one and negative one to the 𝑛th power are a constant with respect to π‘₯. So integrating the term inside our summand is equivalent to just integrating π‘Ž multiplied by π‘₯ to the power of two 𝑛 with respect to π‘₯, which we know is equal to π‘Ž multiplied by π‘₯ to the power of two 𝑛 plus one divided by two 𝑛 plus one plus our constant of integration 𝐢.

Instead of adding a constant of integration for every single term in our sum, we’re just going to add one right at the end. Evaluating the integral inside our sum gives us 𝐢 plus the sum from 𝑛 equals zero to ∞ of two to the power of two 𝑛 plus one multiplied by negative one to the 𝑛th power multiplied by π‘₯ to the power of two 𝑛 plus one divided by two 𝑛 plus one.

We now have a power series representation for our function 𝑓 of π‘₯. However, it still has an unknown variable which we’ve labelled 𝐢. To eliminate this unknown variable, we recall that this is a power series for our function 𝑓 of π‘₯, which is equal to the inverse tan of two π‘₯. Therefore, if we were to substitute, say, π‘₯ is equal to zero into our function, we would get the inverse tan of zero is equal to 𝑓 evaluated at zero is equal to our power series where we substitute π‘₯ equals zero.

So 𝐢 plus the sum from 𝑛 equals zero to ∞ of two to the power of two 𝑛 plus one multiplied by negative one to the 𝑛th power multiplied by zero to the power of two 𝑛 plus one divided by two 𝑛 plus one. And we know we can justify using the substitution since it’s true if the size of negative four π‘₯ squared is less than one, which is true for a substitution π‘₯ is equal to zero. We see that, in this infinite sum, every term is being multiplied by zero. So this is just a sum of zeros and we can conclude is equal to zero.

Therefore, we can conclude that the inverse tan of zero is equal to 𝐢. And of course, we know the inverse tan of zero evaluates to just equal to zero. Therefore, what we have shown is a power series representation of the inverse tan of two π‘₯ is equal to the sum from 𝑛 equals zero to ∞ of two to the power of two 𝑛 plus one multiplied by negative one to the 𝑛th power multiplied by π‘₯ to the power of two 𝑛 plus one divided by two 𝑛 plus one.

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