Video Transcript
Find the Maclaurin series of three divided by one plus 𝑥. Write your answer in Σ notation.
The question wants us to find the Maclaurin series of the function three divided by one plus 𝑥. We need to give our answer in Σ notation. Let’s start by recalling what the Maclaurin series of a function 𝑓 of 𝑥 is. The Maclaurin series for a function 𝑓 of 𝑥 is the following power series. The sum from 𝑛 equals zero to ∞ of the 𝑛th derivative of 𝑓 with respect to 𝑥 evaluated at zero divided by 𝑛 factorial times 𝑥 to the 𝑛th power. And if this power series converges for a certain value of 𝑥, then it’s equal to 𝑓 of 𝑥.
To find the Maclaurin series of a function 𝑓 of 𝑥, we need to evaluate each of its 𝑛th derivatives at 𝑥 is equal to zero. So the first thing we’ll do is call three divided by one plus 𝑥 𝑓 of 𝑥. We need to find the terms in this series. Our series starts when 𝑛 is equal to zero. This means we need to find the 𝑛th derivative of 𝑓 of 𝑥 with respect to 𝑥 when 𝑥 is equal to zero. To do this, we need to recall when we say the zeroth derivative of a function, we just mean that function. So this is just 𝑓 evaluated at zero. We can find this by substituting 𝑥 is equal to zero into our function. We get three divided by one plus zero which is equal to three.
To find the next term in our series, we need to find the first derivative of 𝑓 of 𝑥 with respect to 𝑥 evaluated at zero. Alternatively, this is 𝑓 prime of zero. Of course, we can’t do this directly. We first need to find an expression for 𝑓 prime of 𝑥. We might be tempted to try using the quotient rule, since 𝑓 of 𝑥 has given us the quotient of two functions. Instead, however, we’ll rewrite 𝑓 of 𝑥 as three times one plus 𝑥 all raised to the power of negative one. We can now differentiate this by using the chain rule or, equivalently, the general power rule.
Recall, the general power rule tells us for a differentiable function 𝑔 of 𝑥 and real constant 𝑘, the derivative of 𝑔 of 𝑥 raised to the 𝑘th power with respect to 𝑥 is equal to 𝑘 times 𝑔 prime of 𝑥 multiplied by 𝑔 of 𝑥 raised to the power of 𝑘 minus one. In fact, this is just an application of the chain rule. In our case, our function 𝑔 of 𝑥 is the linear function 𝑥 plus one and our value of 𝑘 is negative one. So to use the general power rule, we need to find an expression for 𝑔 prime of 𝑥. That’s the derivative of one plus 𝑥 with respect to 𝑥. Of course, this is a linear function, so its derivative will just be the coefficient of 𝑥 which is one.
So by the general power rule, we get that 𝑓 prime of 𝑥 is equal to three times negative one multiplied by one times one plus 𝑥 raised to the power of negative one minus one. And of course we can simplify this. We get negative three times one plus 𝑥 to the power of negative two. Remember, we found this expression for 𝑓 prime of 𝑥 to find the first derivative of 𝑓 of 𝑥 with respect to 𝑥 evaluated at 𝑥 is equal to zero. So we substitute 𝑥 is equal to zero into our expression for 𝑓 prime of 𝑥. This gives us negative three times one plus zero raised to the power of negative two. And one plus zero raised to the power of negative two is just equal to one. So this simplifies to just give us negative three.
Now that we’ve evaluated this, let’s clear some space and find the next term in our Maclaurin series. To do this, we’re going to need to find the second derivative of 𝑓 of 𝑥 with respect to 𝑥 evaluated at zero. And to find the second derivative, we can once again use the general power rule. We have 𝑓 double prime of 𝑥 will be equal to the derivative of 𝑓 prime of 𝑥 with respect to 𝑥. Doing this by using the general power rule, this time our value of 𝑘 will be equal to negative two and 𝑔 of 𝑥 will once again be equal to one plus 𝑥. And of course, this means that 𝑔 prime of 𝑥 will just be equal to one.
Now, by applying the general power rule, we get 𝑓 double prime of 𝑥 is equal to negative three times negative two multiplied by one times one plus 𝑥 raised to the power of negative two minus one. And we can simplify this to get six times one plus 𝑥 all raised to the power of negative three. And now, if we substitute 𝑥 is equal to zero into this expression, we’ll just get six. So our second derivative of 𝑓 with respect to 𝑥 at 𝑥 is equal to zero is six. But now we can start to see a pattern. Each time we differentiate 𝑓, we multiply by our exponent and then reduce this exponent by one. It doesn’t matter how many times we reduce this exponent. If we substitute 𝑥 is equal to zero into this part, we will always get one.
So when we’re substituting 𝑥 is equal to zero, we just need to find a formula for this coefficient. To really show this point, let’s now find the third derivative of 𝑓 of 𝑥 with respect to 𝑥. And we know the easiest way to do this is by using the general power rule. Doing this, we would get 𝑓 triple prime of 𝑥 is equal to six times negative three multiplied by 𝑥 plus one raised to the power of negative four. The coefficient of negative three comes from the exponent of the function we were taking before. It’s also worth pointing out this is also equal to negative the value of the derivative we’re taking. It’s negative three and we’re taking the third derivative. And this will always be the case, since we always reduce our exponent by one.
So our third derivative of 𝑓 of 𝑥 with respect to 𝑥 when 𝑥 is equal to zero is equal to negative 18. So, by analyzing our differentiation process, we notice the pattern. To get from 𝑓 of zero to 𝑓 prime of zero, we multiplied by negative one. To get from 𝑓 prime of zero to 𝑓 double prime of zero, we multiplied by negative two. And then to get from 𝑓 double prime of zero to 𝑓 triple prime of zero, we multiplied by negative three. So what does this mean for our 𝑛th derivative of 𝑓 evaluated at zero? Well, we have one times two times three et cetera. We know this is equal to 𝑛 factorial. Of course, we also have negative one multiplied by itself 𝑛 times, so we get negative one to the 𝑛th power. And finally, we started with 𝑓 of zero being equal to three, so we need a factor of three in this expression.
This means we’ve shown the 𝑛th derivative of 𝑓 with respect to 𝑥 evaluated at zero is equal to three times negative one raised to the 𝑛th power multiplied by 𝑛 factorial. All we need to do now is write this into our Maclaurin series. Substituting in this expression into our formula for the Maclaurin series, we get three divided by one plus 𝑥 is equal to the sum from 𝑛 equals zero to ∞ of three times negative one to the 𝑛th power times 𝑛 factorial divided by 𝑛 factorial all multiplied by 𝑥 to the 𝑛th power.
And of course, we can simplify this. We can cancel the shared factor of 𝑛 factorial in the numerator and the denominator. And we’ll do one more thing. We’ll take our factor of three outside of our series. And this gives us our final answer, three times the sum from 𝑛 equals zero to ∞ of negative one to the 𝑛th power times 𝑥 to the 𝑛th power.
It’s worth pointing out though, there is a simpler method to find this answer. Recall that we can represent three divided by one plus 𝑥 as a geometric series. To do this, recall if the absolute value of our ratio 𝑟 is less than one, then the sum from 𝑛 equals zero to ∞ of eight times 𝑟 to the 𝑛th power is equal to 𝑎 divided by one minus 𝑟. And this is very similar to our function given to us in the question, with our value of 𝑎 equal to three and 𝑟 equal to negative 𝑥. So we could just substitute these values into our series. Doing this, we get three divided by one plus 𝑥 is equal to the sum from 𝑛 equals zero to ∞ of three times negative 𝑥 raised to the 𝑛th power provided the absolute value of negative 𝑥 is less than one.
And we can see we could rearrange this to get the exact same series. This is another way of getting our Maclaurin series. And this has to be our Maclaurin series. We know this because a power series of a function centered at a point must be equal to its Taylor series. Therefore, we found two different ways of showing the Maclaurin series of three divided by one plus 𝑥 is equal to three times the sum from 𝑛 equals zero to ∞ of negative one to the 𝑛th power times 𝑥 to the 𝑛th power.
