Question Video: Differentiating a Power Series Term by Term to Help Evaluate a Function Mathematics • Higher Education

Consider the series 𝑓(π‘₯) = 1/(1 βˆ’ π‘₯)Β² = βˆ‘_(𝑛 = 0)^(∞) (𝑛 + 1)π‘₯^(𝑛). Differentiate the given series expansion of 𝑓 term by term to find the corresponding series expansion for the derivative of 𝑓. Use the result of the first part to evaluate the sum of the series βˆ‘_(𝑛 = 0)^(∞) ((𝑛 + 1)(𝑛 + 2))/4^(𝑛).

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Video Transcript

Consider the series 𝑓 of π‘₯ is equal to one divided by one minus π‘₯ all squared is equal to the sum from 𝑛 equals zero to ∞ of 𝑛 plus one times π‘₯ to the 𝑛th power. Differentiate the given series expansion of 𝑓 term by term to find the corresponding series expansion for the derivative of 𝑓. Use the result of the first part to evaluate the sum of the series the sum from 𝑛 equals zero to ∞ of 𝑛 plus one times 𝑛 plus two divided by four to the 𝑛th power.

We’re given a power series representation of the function 𝑓 of π‘₯. The first thing we need to do is differentiate the power series expansion of 𝑓 term by term to help us find a corresponding power series expansion for the derivative of our function 𝑓. There’s a few different ways of doing this. We’re going to write our series out term by term.

Let’s start with the first term in our series. That’s when 𝑛 is equal to zero. We get zero plus one times π‘₯ to the zeroth power. Of course, we can simplify this. Zero plus one simplifies to give us one. And π‘₯ to the zeroth power is just equal to one. So the first term in our series expansion for 𝑓 of π‘₯ is equal to one.

Let’s now find the second term in this series. That’s when 𝑛 is equal to one. We get one plus one times π‘₯ to the first power. Raising a number to the first power doesn’t change anything. So we can rewrite this as π‘₯. And one plus one simplifies to give us two. So our second term simplifies to give us two π‘₯.

We can do the same to find the third term in this series expansion. When 𝑛 is equal to two, we get two plus one multiplied by π‘₯ squared. And we can simplify this to give us three π‘₯ squared. And we could keep doing this. However, we can also notice a pattern. Our coefficient is always one higher than our exponent of π‘₯. So this term will be four π‘₯ cubed, the next one five π‘₯ to the fourth power, and this pattern will continue. And we know this pattern will continue because we’re given the power series for 𝑓 of π‘₯ in the question.

Now, we’ve written our power series for 𝑓 of π‘₯ out term by term. It’s time to differentiate this term by term. And it’s worth pointing out we know we can do this provided our value of π‘₯ is within the radius of convergence of our power series. We want to differentiate each term with respect to π‘₯. We know the derivative of our first constant term will be equal to zero. To differentiate two π‘₯, we’ll write this as two π‘₯ to the first power and then use the power rule for differentiation. This gives us one times two times π‘₯ to the zeroth power.

Let’s keep going. To differentiate our next term, we’ll once again use the power rule for differentiation. We multiply by our exponent of π‘₯ and then reduce this exponent by one. This gives us two times three times π‘₯ to the first power. We can do the same to differentiate four π‘₯ cubed. We get three times four times π‘₯ squared. And if we differentiate five π‘₯ to the fourth power with respect to π‘₯, we get four times five times π‘₯ cubed. And we could differentiate the rest of the terms in the same manner.

But the question wants us to write this as a power series. To do this, we’ll start by removing our term of zero. We can see that the exponents of π‘₯ increase by one each term. So our power series will include π‘₯ to the 𝑛th power. And we can see that our exponents of π‘₯ start at zero and go on forever. So our series should sum 𝑛 from zero to ∞.

We just need to find an expression for our coefficients. Let’s start with the coefficient in our first term. That’s when 𝑛 is equal to zero. We get one multiplied by two. We can also see when 𝑛 is equal to one, we get two multiplied by three. And we can already see a pattern. We’re multiplying two numbers together, and each of these numbers increases by one on each term. So we just need to find expressions for each of the factors of our coefficient.

Let’s start with the first coefficient in each pair. In our first term, the value of one is zero plus one. In the second term, two is one plus one. And this pattern continues. The first factor in our coefficient is our exponent plus one. But remember, our exponent is equal to 𝑛. So the first factor in the coefficient will be 𝑛 plus one. And of course we know the second number is just one bigger than the first number. So we can just write this as 𝑛 plus two.

Therefore, we’ve shown 𝑓 prime of π‘₯ is equal to the sum from 𝑛 equals zero to ∞ of 𝑛 plus one times one plus two all multiplied by π‘₯ to the 𝑛th power provided this power series is convergent for that value of π‘₯.

But we’re not done. The next part of this question tells us to use the answer to the first part of this question to evaluate the following sum: the sum from 𝑛 equals zero to ∞ of 𝑛 plus one times 𝑛 plus two divided by four to the 𝑛th power. We can see that this series is very similar to the series we got in our answer. Instead of having π‘₯ to the 𝑛th power, we’re dividing by four to the 𝑛th power. In other words, the series we want to evaluate is our series where we substitute π‘₯ is equal to one-quarter.

This is a complicated series to evaluate directly. But remember, our power series is equal to 𝑓 prime of π‘₯. So if our power series is convergent, it must be equal to 𝑓 prime evaluated at one-quarter. So we need to show that our power series is convergent when π‘₯ is equal to one-quarter. Then we can evaluate this series by just evaluating 𝑓 prime of one-quarter.

To prove that this series is convergent when π‘₯ is equal to one-quarter, we’ll prove it by using the ratio test. We recall one small part of the ratio test. This tells us the sum from 𝑛 equals zero to ∞ of π‘Žπ‘› will be convergent if the limit as 𝑛 approaches ∞ of the absolute value of π‘Žπ‘› plus one divided by π‘Žπ‘› is less than one. In actual fact, this would make our series absolutely convergent. But we don’t need this. We just need to evaluate this limit for our series where π‘₯ is equal to one-quarter.

So let’s evaluate this limit with our series where π‘₯ is equal to one-quarter. And we’ll write one-quarter as 0.25. We get the limit as 𝑛 approaches ∞ of the absolute value of 𝑛 plus two times 𝑛 plus three multiplied by 0.25 raised to the power of 𝑛 plus one divided by 𝑛 plus one times 𝑛 plus two multiplied by 0.25 raised to the power of 𝑛. And we can simplify this expression. First, we’ll cancel the shared factor of 𝑛 plus two in our numerator and our dominator. Next, we’ll cancel 𝑛 of the shared factors of 0.25 in our numerator and our denominator. This leaves us with the limit as 𝑛 approaches ∞ of the absolute value of 0.25 times 𝑛 plus three all divided by 𝑛 plus one.

To help us evaluate this limit, we’ll now take the constant 0.25 outside of our limit. We’ll write this again as one-quarter. Remember, since we were taking the absolute value of this, we also need to take the absolute value of one-quarter. But this is still just equal to one-quarter. So we now have one-quarter times the limit as 𝑛 approaches ∞ of the absolute value of 𝑛 plus three divided by 𝑛 plus one.

There’s a few different ways of evaluating this limit. One way is to divide our numerator and our denominator through by 𝑛. Doing this, we get one-quarter times the limit as 𝑛 approaches ∞ of the absolute value of one plus three over 𝑛 divided by one plus one over 𝑛. Now, we can just evaluate this limit directly. As 𝑛 approaches ∞, the constants of one don’t change as 𝑛 changes. However, three divided by 𝑛 and one divided by 𝑛 both approach zero as 𝑛 approaches ∞. So this approaches the absolute value of one divided by one, which is equal to one.

We then multiply this by one-quarter to get one-quarter. And this is of course less than one. So our power series converges when π‘₯ is equal to one-quarter. This means we’ve justified our use of substituting π‘₯ is equal to one-quarter into 𝑓 prime of π‘₯ to evaluate our series. So let’s clear some space and do this.

However, we run into a problem. We don’t actually have an expression for 𝑓 prime of π‘₯. Using our laws for exponents, we’ll write 𝑓 of π‘₯ as one minus π‘₯ all raised to the power of negative two. This means we know 𝑓 prime of π‘₯ is the derivative of one minus π‘₯ all raised to the power of negative two with respect to π‘₯. And we could differentiate this by using substitution or the chain rule. However, we can also use the general power rule.

The general power rule tells us, for any real constant π‘˜ and differentiable function 𝑔 of π‘₯, the derivative of 𝑔 of π‘₯ all raised to the power of π‘˜ is equal to π‘˜ times 𝑔 prime of π‘₯ multiplied by 𝑔 of π‘₯ all raised to the power of π‘˜ minus one. In our case, the value of our exponent π‘˜ is negative two and our function 𝑔 is one minus π‘₯. And of course, to apply the general power rule, we need to find 𝑔 prime of π‘₯.

Well, 𝑔 of π‘₯ is just a linear function. So 𝑔 prime of π‘₯ will just be the coefficient of π‘₯, which in our case is negative one. So by using the general power rule, we get 𝑓 prime of π‘₯ is equal to negative two times negative one multiplied by one minus π‘₯ all raised to the power of negative three. And of course we can simplify this to give us two divided by one minus π‘₯ all cubed.

Now that we’ve found an expression for 𝑓 prime of π‘₯, let’s clear some space. We’re now ready to evaluate our series by substituting π‘₯ is equal to one-quarter into our function 𝑓 prime of π‘₯. Remember, we can do this because we proved that our series is convergent when π‘₯ is equal to one-quarter. Substituting π‘₯ is equal to one-quarter into our expression for 𝑓 prime of π‘₯, we get 𝑓 prime of one-quarter is equal to two divided by one minus one-quarter all cubed. And if we evaluate this expression, we get 128 divided by 27.

In this question, we were able to differentiate the power series for the function 𝑓 of π‘₯ is equal to one divided by one minus π‘₯ all squared term by term. We then wrote this in sigma notation to give us a power series representation for the derivative of 𝑓. We then showed that this power series must be convergent when π‘₯ is equal to one-quarter and then used this to show that the sum from 𝑛 equals zero to ∞ of 𝑛 plus one times 𝑛 plus two divided by four to the 𝑛th power must be equal to 128 divided by 27 by substituting π‘₯ is equal to one-quarter into our expression for 𝑓 prime of π‘₯.

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