### Video Transcript

Write the first four terms of the power series of the function π of π₯ is equal to two divided by two minus five π₯.

Weβre given a rational function π of π₯, and weβre asked to write the first four terms of the power series representation of this function π of π₯. And thereβs a few different ways of doing this. For example, we could use directly the definition of the Maclaurin series or the Taylor series to find the expression for this power series. However, both of these will involve differentiating our rational function three times. Instead, we can use the following power series representation, which we got by using what we know about infinite geometric series.

Specifically, if we set our initial value of π equal to one and our ratio of successive terms π equal to π₯, then the infinite sum of a geometric series with π equal to one and π equal to π₯ is given by one divided by one minus π₯, so long as the absolute value of our ratio of successive terms π₯ is less than one. And this is a useful result worth committing to memory, since we can manipulate this to find power series representations for different rational functions. We want to manipulate this so that one divided by one minus π₯ is equal to two divided by two minus five π₯.

Since we wanted two in the numerator, the first thing weβll do is multiply both our numerator and our denominator through by two. Of course, multiplying both our numerator and our denominator through by two doesnβt change our value. This gives us two divided by two minus two π₯ is equal to the sum from π equals zero to β of π₯ to the πth power as long as the absolute value of π₯ is less than one. And this is now almost in the form we want it. However, we want negative five π₯ instead of negative two π₯. And thereβs a few different ways of thinking about this. For example, we could have initially called our variable π instead of π₯ for the ratio of successive terms.

This just gives us the following equivalent statement. Then we want negative two π to be equal to negative five π₯. And if we want negative two π to be equal to negative five π₯, we can divide both sides of this equation through by negative two, meaning that we set π equal to five over two π₯. So all we have to do is substitute π is equal to five over two π₯ into this expression. So by substituting π is equal to five over two π₯ into every instance of π in this equation gives us that two divided by two minus two times five over two π₯ is equal to the sum from π equals zero to β of five over two π₯ all raised to the πth power as long as the absolute value of five over two π₯ is less than one.

And itβs worth pointing out because this comes from an infinite geometric series, we can guarantee our power series converges when the absolute value of five over two π₯ is less than one. And we can guarantee it diverges when the absolute value of five over two π₯ is greater than one. However, in this question, weβre not actually asked to find any intervals of convergences, so we can skip this step out. Next, we can simplify slightly. In the denominator of our rational function, we have negative two multiplied by five over two, which is, of course, equal to negative five. So this gives us the rational function two divided by two minus five π₯, which is, of course, equal to our function π of π₯.

Now, remember, the question is only asking us to find the first four terms of this power series. So all we need to do is find the first four terms of this series. So we just need to write this series out term by term. Weβll start with the first term. Thatβs when π is equal to zero. This gives us five over two π₯ all raised to the zeroth power. And of course, raising a number to the zeroth power is just equal to one. So the first term in this series expansion is one.

Letβs now move on to the second term. Thatβs when π is equal to one. So we get five over two π₯ all raised to the first power. And of course, raising a number to the first power doesnβt change its value. So our second term is just five over two times π₯. We can do the same to find the third term in our series. Thatβs when π is equal to two. We get five over two π₯ all squared. To simplify this, we need to distribute the square over our parentheses. This gives us five over two all squared times π₯ squared. And then we can just calculate five over two all squared to be equal to 25 divided by four.

So the third term in our series expansion is 25 over four times π₯ squared. And we find this in exactly the same way. We substitute π is equal to three into our summand giving us five over two π₯ all cubed. And then we evaluate this to get 125 over eight times π₯ cubed. And so the fourth term in our power series representation is 125 over eight times π₯ cubed. Of course, thereβs going to be an infinite number of terms of this power series.

However, the question only asks us to find the first four terms of this power series. So at this point, weβre done. Therefore, we were able to show the first four terms of the power series representation of the function π of π₯ is equal to two divided by two minus five π₯ is equal to one plus five over two π₯ plus 25 over four π₯ squared plus 125 over eight π₯ cubed.