Question Video: Finding the Taylor Series Expansion of a Rational Function Mathematics • Higher Education

Find the Taylor series of the function 𝑓(π‘₯) = (1 βˆ’ π‘₯)⁻² about π‘₯ = βˆ’5.

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Video Transcript

Find the Taylor series of the function 𝑓 of π‘₯ is equal to one minus π‘₯ all raised to the power of negative two about π‘₯ is equal to negative five.

We’re given a function 𝑓 of π‘₯, and we’re asked to find the Taylor series expansion of this function 𝑓 of π‘₯ about π‘₯ is equal to negative five. The first thing we’re going to need to do is recall what we mean by the Taylor series of a function 𝑓 of π‘₯. The Taylor series of a function 𝑓 of π‘₯ centered at a value of π‘Ž is equal to the sum from 𝑛 equals zero to ∞ of the 𝑛th derivative of 𝑓 evaluated at π‘Ž divided by 𝑛 factorial multiplied by π‘₯ minus π‘Ž all raised to the 𝑛th power. And we sometimes call the value of π‘Ž the center. In this case, we’re told to use a value of π‘Ž equal to negative five. So we can set the value of π‘Ž equal to negative five in our Taylor series.

In fact, we see we can then simplify this. π‘₯ minus negative five is equal to π‘₯ plus five. Now by looking at our definition of the Taylor series, we see we’re going to need to find the 𝑛th derivative of 𝑓 evaluated at negative five. To do this, we’re going to need to differentiate our function 𝑓 of π‘₯ and see if we can spot a pattern. So let’s start with our function 𝑓 of π‘₯. We want to differentiate this with respect to π‘₯. In this case, we can see there’s a couple of options. Since 𝑓 of π‘₯ is the composition of two functions, we could do this by using the chain rule. However, because our outer function is a power function, we’ll do this by using the general power rule.

We recall the general power rule tells us, for differentiable function 𝑔 of π‘₯ and real constant π‘š, the derivative of 𝑔 of π‘₯ all raised to the 𝑛th power with respect to π‘₯ is equal to 𝑛 times 𝑔 prime of π‘₯ multiplied by 𝑔 of π‘₯ all raised to the power of π‘š minus one. We want to apply this to 𝑓 of π‘₯. We need to set our exponent value of π‘š equal to negative two and our inner function 𝑔 of π‘₯ equal to one minus π‘₯. We then see to apply the general power rule, we need to find an expression for 𝑔 prime of π‘₯. That’s the derivative of one minus π‘₯ with respect to π‘₯. This is a linear function, so its derivative will be the coefficient of π‘₯, which is negative one.

Substituting in our values for π‘š and 𝑔 prime of π‘₯ and our expression for 𝑔 of π‘₯, 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 two minus one. And we can simplify this expression. Negative two multiplied by negative one is equal to two. And our exponent negative two minus one is equal to negative three. This gives us 𝑓 prime of π‘₯ is two times one minus π‘₯ all raised to the power of negative three. We could then find an expression for the second derivative of 𝑓 of π‘₯ with respect to π‘₯ by using the general power rule.

We want to differentiate our expression for 𝑓 prime of π‘₯. So we’ll set our value of π‘š equal to the exponent of negative three, and 𝑔 of π‘₯ will be our inner function one minus π‘₯. However, we already found an expression for 𝑔 prime of π‘₯. This will stay the same at negative one. And of course, since we’re multiplying this by two, we also need to multiply our derivative by two. So by using the general power rule, we get 𝑓 double prime of π‘₯ will be equal to two times negative three multiplied by negative one times one minus π‘₯ all raised to the power of negative three minus one.

We got this by substituting π‘š is equal to negative three, 𝑔 prime of π‘₯ is negative one, and 𝑔 of π‘₯ is one minus π‘₯ into our general power rule and then multiplying through by two. And we can simplify this. We get 𝑓 double prime of π‘₯ is equal to six times one minus π‘₯ all raised to the power of negative four. If we were to try and differentiate this again, we would once again use the general power rule. And this lets us spot a pattern. To find the next derivative of our function, we multiply by negative one, we multiply by our exponent, and then reduce the exponent by one.

So if we were to apply this, we would get 𝑓 triple prime of π‘₯ is equal to four times six times one minus π‘₯ all raised to the power of negative five. Well, it’s worth pointing out we simplified our negative times our negative already to give us a positive. And now we need to notice one more thing before we can start finding our Taylor series. In each step in our derivative process, our coefficient is multiplied by one higher integer number. We can see this starts at one. And at 𝑓 triple prime of π‘₯, we have four times three times two times one. And we know this is equal to four factorial. So we can write 𝑓 triple prime of π‘₯ as four factorial times one minus π‘₯ all raised to the power of negative five.

And of course, we could do something very similar with the rest of our derivatives. We could write six as three factorial, two as two factorial, and the coefficient in front of 𝑓 of π‘₯ which is one as one factorial. We’re now ready to find an expression for the 𝑛th derivative of 𝑓 of π‘₯ with respect to π‘₯. When we took one derivative, we got two factorial as our coefficient. When we took two derivatives, we got three factorial as our coefficient. And when we took three derivatives, we got four factorial as our coefficient. So our coefficient will be the factorial of one higher than the number of derivatives we’ve taken. In other words, in this case, our coefficient will be 𝑛 plus one factorial.

Similarly, we know we’re reducing our exponent by one at each step, and our exponent starts at negative two. So after 𝑛 derivatives, we’ll reduce negative two 𝑛 times. This gives us a new exponent of negative two minus 𝑛. Now that we’ve found our expression for the 𝑛th derivative of 𝑓 of π‘₯ with respect to π‘₯, we’ll clear some space and find our Taylor series expansion for 𝑓 of π‘₯ about π‘₯ is equal to negative five. Now that we’ve found an expression for the 𝑛th derivative of 𝑓 of π‘₯ with respect to π‘₯, we’re ready to start finding our Taylor series expansion.

First, we need to substitute in an expression for the 𝑛th derivative of 𝑓 of π‘₯ evaluated at π‘₯ is equal to negative five into our power series. And we can find this by substituting π‘₯ is equal to negative five into the expression we found earlier. Doing this, we get the following expression. And now we want to start simplifying. First, we know one minus negative five is equal for six. Next, we need to recall that 𝑛 plus one factorial will be equal to 𝑛 plus one multiplied by 𝑛 multiplied by all the integers down to one. However, we could write this as 𝑛 plus one times 𝑛 factorial. And then we see our numerator and our denominator share a factor of 𝑛 factorial. So we can cancel these out.

So with these two pieces of simplification and a small bit of rearranging, we get the sum from 𝑛 equals zero to ∞ of six to the power of negative two minus 𝑛 times 𝑛 plus one multiplied by π‘₯ plus five all raised to the 𝑛th power. And this is our final answer. Therefore, by using the general power rule, we were able to find the Taylor series expansion for the function 𝑓 of π‘₯ is equal to one minus π‘₯ all raised to the power of negative two about π‘₯ is equal to negative five. We found that this Taylor series was given by the sum from 𝑛 equals zero to ∞ of six to the power of negative two minus 𝑛 times 𝑛 plus one multiplied by π‘₯ plus five all raised to the 𝑛th power.

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