Video: Represent an Algebraic Fraction as a Power Series

Use a power series to represent π‘₯/(1+ π‘₯^2).

03:20

Video Transcript

Use a power series to represent the function π‘₯ divided by one plus π‘₯ squared.

The question wants us to represent this function as a power series. And we recall that one way of doing this is by using a property of geometric series which says, if the absolute value of the ratio π‘Ÿ is less than one, then the sum from 𝑛 equals zero to ∞ of π‘Ž multiplied by π‘Ÿ to the 𝑛th power is equal to π‘Ž divided by one minus π‘Ÿ. On the left-hand side of our equation, we have something which can be described as a power series in terms of π‘Ÿ. And on the right-hand side of our equation, we have something which looks very similar to the function given to us in the question.

At this point, it would be tempting to set π‘Ž equal to π‘₯ and π‘Ÿ equal to negative π‘₯ squared. This would give us that π‘Ž divided by one minus π‘Ÿ is equal to our function π‘₯ divided by one plus π‘₯ squared. In fact, this actually works. Substituting π‘Ž is equal to π‘₯ and π‘Ÿ is equal to negative π‘₯ squared into our geometric series formula gives us π‘₯ divided by one plus π‘₯ squared is equal to the sum from 𝑛 equals zero to ∞ of π‘₯ multiplied by negative π‘₯ squared to the 𝑛th power when the absolute value of π‘₯ squared is less than one.

Then, we can simplify this by distributing our 𝑛th power of our parentheses to get negative one to the 𝑛th power multiplied by π‘₯ to the power of two 𝑛. Then, we can multiply π‘₯ by π‘₯ the power two 𝑛, giving us the sum from 𝑛 equals zero to ∞ of negative one to the 𝑛th power multiplied by π‘₯ to the power of two 𝑛 plus one. However, it’s worth noting that our step of setting π‘Ž to be a function of π‘₯, in this case we just had π‘Ž equal to π‘₯, will not work in general. To see why, we need to recall the definition of a power series.

We recall that we call the sum from 𝑛 equals zero to ∞ of 𝐢 𝑛 multiplied by π‘₯ to the 𝑛th power, where 𝐢 𝑛 a constant real number’s a power series in π‘₯. The part we’re interested in is that the coefficients of π‘₯ to the 𝑛th power need to be constant real numbers. Since we want the values of 𝐢 𝑛 to be constant real numbers, it’s best not to set π‘Ž to be a function of π‘₯, since we will not always get a power series of π‘₯ if we do this. There are several methods we should know if we’re given a function of π‘₯ in the numerator of our fraction.

One of these is partial fractions. But a simpler method which works here is factorization. We could’ve set π‘₯ divided by one plus π‘₯ squared to be equal to π‘₯ multiplied by one divided by one plus π‘₯ squared. Then, by saying π‘Ž equal to one and π‘Ÿ equal to negative π‘₯ squared, we can use our property of geometric series. Substituting these values into our formula gives us π‘₯ multiplied by the sum from 𝑛 equals zero to ∞ of negative π‘₯ squared raised to the 𝑛th power. We can then bring the coefficient of π‘₯ inside of our summation. And we can see that this is exactly the same as the summation we had before.

So, we can do exactly what we did before. We can distribute the 𝑛th power over our parentheses and then simplify to get the sum from 𝑛 equal zero to ∞ of negative one to the 𝑛th power multiplied by π‘₯ to the power of two 𝑛 plus one. The key difference being that we did not set π‘Ž to be a function of π‘₯. Otherwise we would’ve risked our answer ending up to be not a power series of π‘₯. In conclusion, what we’ve shown is that π‘₯ divided by one plus π‘₯ squared could be represented by the power series the sum from 𝑛 equal zero to ∞ of negative one to the 𝑛th power multiplied by π‘₯ to the power of two 𝑛 plus one.

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