Question Video: Find the Rational Function Given by a Power Series | Nagwa Question Video: Find the Rational Function Given by a Power Series | Nagwa

Question Video: Find the Rational Function Given by a Power Series Mathematics

Consider the power series ∑_(𝑛 = 0)^(∞) 3^(𝑛) 𝑥^(𝑛). Find the function 𝑓 represented by this series. Determine the interval of convergence of the series.

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Video Transcript

Consider the power series the sum from 𝑛 equals zero to ∞ of three to the 𝑛th power multiplied by 𝑥 to the 𝑛th power. Find the function 𝑓 represented by this series. Determine the interval of convergence of this series.

The first thing we want to do is recall a fact about geometric series. Which is that for any constant 𝑎, if the absolute value of our ratio 𝑟 is less than one, then the sum from 𝑛 equals zero to ∞ of 𝑎 multiplied by 𝑟 to the 𝑛th power is just equal to 𝑎 divided by one minus 𝑟. And if our constant 𝑎 is not equal to zero and the absolute value of our ratio 𝑟 is greater than or equal to one, then the sum from 𝑛 equals zero to ∞ of 𝑎 multiplied by 𝑟 to the 𝑛th power does not converge.

To utilize this, we need to rewrite the power series given to us in the question as a geometric series. We notice that our summand three to the 𝑛th power multiplied by 𝑥 to the 𝑛th power is just equal to three 𝑥 all to the 𝑛th power. And we see that this gives us the sum from 𝑛 equals zero to ∞ of three 𝑥 all to the 𝑛th power. Which is a geometric series with an initial term of one and a ratio of three 𝑥.

So using our rule about geometric series, we must have that this is equal to 𝑎 divided by one minus 𝑟. If the absolute value of 𝑟 is less than one, we substitute 𝑛 𝑎 is equal to one. And then we substitute 𝑟 is equal to three 𝑥. Giving us one divided by one minus three 𝑥, if the absolute value of three 𝑥 is less than one. Therefore, we have shown that the function 𝑓, which represents this power series, is equal to one divided by one minus three 𝑥.

To determine the interval of convergence of this series, it’s worth noting that our initial term 𝑎 is equal to one. So we must have that the absolute value of three 𝑥 is greater than or equal to one. Then our series does not converge. Therefore, since the series does not converge when the absolute value of three 𝑥 is greater than or equal to one. We must have that the interval of convergence is what we had before, those values of 𝑥 which satisfy the absolute value of three 𝑥 is less than one.

There are several different ways of solving inequalities like this. One such way is if we know the absolute value of three 𝑥 is less than one. Then we know that three 𝑥 must be between negative one and one. We can then divide all parts of this inequality by three to get negative a third is less than 𝑥 is less than one-third. And we know that this is the same as saying that 𝑥 is in the open interval from negative one-third to a third.

What we have shown is that if access in the open interval from negative a third to a third, then the power series, the sum from 𝑛 equals zero to ∞ of three to the 𝑛th power multiplied by 𝑥 to the 𝑛th power, is equal to the function one divided by one minus three 𝑥.

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