Video: Geometric Interpretation of Multiplication by 𝑖

Let us consider 𝑔(π‘₯) = π‘₯/(3 βˆ’ π‘₯). Find a power series representation for 𝑔(π‘₯). Find the radius of convergence.

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Video Transcript

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

The question gives us a rational function 𝑔 of π‘₯. It wants us to find the power series representation of the function 𝑔 of π‘₯. 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 π‘Ž multiplied by π‘Ÿ to the 𝑛th power is equal to π‘Ž divided by one minus π‘Ÿ, when the absolute value of our ratio π‘Ÿ is less than one.

So if we can write our rational function 𝑔 of π‘₯ in the form π‘Ž divided by one minus π‘Ÿ, 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 π‘Ÿ is less than one. We’ll start by noticing that we can factor a coefficient of π‘₯ outside of our rational function. This gives us π‘₯ multiplied by one over three minus π‘₯. We want our denominator to be in the form one minus the ratio π‘Ÿ. However, we have three minus π‘₯ as our denominator. To have a denominator in the form one minus π‘Ÿ, we’re going to factor out a coefficient of three in our denominator. This gives us three multiplied by one minus π‘₯ 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 π‘Ž equal to one and π‘Ÿ equal to π‘₯ 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 π‘₯ is equal to π‘₯ divided by three multiplied by the sum from 𝑛 equals zero to ∞ of one multiplied by π‘₯ over three to the 𝑛th power, when the absolute value of our ratio π‘₯ 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 π‘₯ divided by three inside of our sum to give us π‘₯ over three multiplied by π‘₯ over three to the 𝑛th power. However, we can simplify this further to just be π‘₯ over three all raised to the power of 𝑛 plus one. This gives us that our function 𝑔 of π‘₯ is equal to the sum from 𝑛 equals zero to ∞ of π‘₯ over three to the power of 𝑛 plus one, when the absolute value of π‘₯ 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 π‘Ÿ the radius of convergence of a power series if the power series converges when the absolute value of π‘₯ is less than π‘Ÿ and diverges when the absolute value of π‘₯ is greater than π‘Ÿ. Unless the power series converges for all values of π‘₯, in which case we call the radius of convergence ∞. We recall that for a geometric series, the sum from 𝑛 equals zero to ∞ of π‘Ž multiplied by π‘Ÿ to the 𝑛th power will diverge whenever the absolute value of our ratio π‘Ÿ is greater than one. In our case, our ratio π‘Ÿ is π‘₯ divided by three. So our power series converges when the absolute value of π‘₯ over three is less than one and diverges when the absolute value of π‘₯ over three is greater than one.

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

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