Video: Finding the Maclaurin Series of a Rational Function

Find the Maclaurin series for 𝑓(π‘₯) = 1/(10π‘₯ + 1).

02:46

Video Transcript

Find the Maclaurin series for 𝑓 of π‘₯ is equal to one divided by 10π‘₯ plus one.

We need to find the Maclaurin series for the function 𝑓 of π‘₯. There’s a few different ways of doing this. We might be tempted to use the definition of a Maclaurin series. And we recall the Maclaurin series for 𝑓 of π‘₯ is the sum from 𝑛 equals zero to ∞ of the 𝑛th derivative of 𝑓 of π‘₯ with respect to π‘₯ evaluated at zero divided by 𝑛 factorial multiplied by π‘₯ to the 𝑛th power. And this will only be valid for values of π‘₯ where our series is convergent. And we see, to find this power series, we need to find the 𝑛th derivative of 𝑓 of π‘₯ with respect to π‘₯ evaluated at zero.

In other words, if we wanted to do this directly from our definition, we would need to differentiate our function 𝑓 of π‘₯ 𝑛 times. And in fact, this would work. We could do this by using the quotient rule, the chain rule, or the general power rule. But that’s quite a complicated method; it’s very easy to make a mistake. In fact, there’s an easier method we could use. Instead, remember, we can find a power series representation of 𝑓 of π‘₯ in this form by using geometric series. To do this, recall the formula for the infinite sum of a geometric series. If the absolute value of our ratio 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 equal to π‘Ž divided by one minus π‘Ÿ.

The expression for the infinite sum of our series is very similar to our function 𝑓 of π‘₯. One way of seeing this is to rewrite our function 𝑓 of π‘₯ as one divided by one minus negative 10π‘₯. Now, we just need to set our value of π‘Ž equal to one and our value of π‘Ÿ equal to negative 10π‘₯. This will then give us a power series representation for 𝑓 of π‘₯. So by using our formula for the infinite sum of a geometric series with π‘Ž equal to one and π‘Ÿ equal to negative 10π‘₯, we get that our function 𝑓 of π‘₯ is equal to the sum from 𝑛 equals zero to ∞ of one times negative 10π‘₯ raised to the power of 𝑛. And of course, we can simplify this. Multiplying by one doesn’t change our value.

And by using our laws of exponents, we can distribute the exponent of 𝑛 over our parentheses. This gives us the sum from 𝑛 equals zero to ∞ of negative 10 raised to the 𝑛th power times π‘₯ to the 𝑛th power. And in fact, we know this has to be a Maclaurin series for our function 𝑓 of π‘₯. And the reason for this is, any power series representation of a function 𝑓 of π‘₯ centered at a value must be equal to its Taylor series represented at this value.

This means that this series must be our Maclaurin series for 𝑓 of π‘₯. Therefore, we were able to show the Maclaurin series for the function 𝑓 of π‘₯ is equal to one divided by 10π‘₯ plus one is the sum from 𝑛 equals zero to ∞ of negative 10 to the 𝑛th power times π‘₯ to the 𝑛th power.

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