Video: Geometric Interpretation of Multiplication by 𝑖

Video Transcript

Let us consider the function 𝑔 of 𝑥 is equal to 𝑥 divided by three minus 𝑥. Find a power series representation for the function 𝑔 of 𝑥. Find the radius of convergence.

The question gives us a rational function 𝑔 of 𝑥. It wants us to find the power series representation of the function 𝑔 of 𝑥. And it wants us to then find the radius of convergence for this power series. To do this, we recall a fact about the sum of an infinite geometric series. The sum from 𝑛 equal zero to ∞ of 𝑎 multiplied by 𝑟 to the 𝑛th power is equal to 𝑎 divided by one minus 𝑟, when the absolute value of our ratio 𝑟 is less than one.

So if we can write our rational function 𝑔 of 𝑥 in the form 𝑎 divided by one minus 𝑟, then we can use this fact about geometric series to write it as a power series with a radius of convergence the absolute value of 𝑟 is less than one. We’ll start by noticing that we can factor a coefficient of 𝑥 outside of our rational function. This gives us 𝑥 multiplied by one over three minus 𝑥. We want our denominator to be in the form one minus the ratio 𝑟. However, we have three minus 𝑥 as our denominator. To have a denominator in the form one minus 𝑟, we’re going to factor out a coefficient of three in our denominator. This gives us three multiplied by one minus 𝑥 divided by three. We can then simplify this by taking the coefficient of three in our denominator and putting it outside of our rational function.

If we then set 𝑎 equal to one and 𝑟 equal to 𝑥 divided by three, we can use our infinite geometric series to rewrite our expression. Using our infinite geometric series formula, we have that 𝑔 of 𝑥 is equal to 𝑥 divided by three multiplied by the sum from 𝑛 equals zero to ∞ of one multiplied by 𝑥 over three to the 𝑛th power, when the absolute value of our ratio 𝑥 divided by three is less than one. We can simplify this further.

First, we can remove the multiplication by one. Next, we can take our coefficient of 𝑥 divided by three inside of our sum to give us 𝑥 over three multiplied by 𝑥 over three to the 𝑛th power. However, we can simplify this further to just be 𝑥 over three all raised to the power of 𝑛 plus one. This gives us that our function 𝑔 of 𝑥 is equal to the sum from 𝑛 equals zero to ∞ of 𝑥 over three to the power of 𝑛 plus one, when the absolute value of 𝑥 over three is less than one.

Next, the question wants us to find the radius of convergence of this power series. We recall that we called 𝑟 the radius of convergence of a power series if the power series converges when the absolute value of 𝑥 is less than 𝑟 and diverges when the absolute value of 𝑥 is greater than 𝑟. Unless the power series converges for all values of 𝑥, in which case we call the radius of convergence ∞. We recall that for a geometric series, the sum from 𝑛 equals zero to ∞ of 𝑎 multiplied by 𝑟 to the 𝑛th power will diverge whenever the absolute value of our ratio 𝑟 is greater than one. In our case, our ratio 𝑟 is 𝑥 divided by three. So our power series converges when the absolute value of 𝑥 over three is less than one and diverges when the absolute value of 𝑥 over three is greater than one.

We can then rearrange both of these expressions. The absolute value of 𝑥 over three being less than one is the same as saying the absolute value of 𝑥 is less than three. And if the absolute value of 𝑥 over three is greater than three, this is the same as saying the absolute value of 𝑥 is greater than three. So our power series converges when the absolute value of 𝑥 is less than three and diverges when the absolute value of 𝑥 is greater than three. Therefore, our radius of convergence 𝑟 is equal to three. And our power series for 𝑔 of 𝑥 will converge whenever the absolute value of 𝑥 is less than three.