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

Consider the power series βˆ‘_(𝑛 = 0)^(∞) 3^(𝑛) π‘₯^(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 π‘₯ 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 π‘Ž, if the absolute value of our 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 just equal to π‘Ž divided by one minus π‘Ÿ. And if our constant π‘Ž is not equal to zero and the absolute value of our ratio π‘Ÿ is greater than or equal to one, then the sum from 𝑛 equals zero to ∞ of π‘Ž multiplied by π‘Ÿ 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 π‘₯ to the 𝑛th power is just equal to three π‘₯ all to the 𝑛th power. And we see that this gives us the sum from 𝑛 equals zero to ∞ of three π‘₯ all to the 𝑛th power. Which is a geometric series with an initial term of one and a ratio of three π‘₯.

So using our rule about geometric series, we must have that this is equal to π‘Ž divided by one minus π‘Ÿ. If the absolute value of π‘Ÿ is less than one, we substitute 𝑛 π‘Ž is equal to one. And then we substitute π‘Ÿ is equal to three π‘₯. Giving us one divided by one minus three π‘₯, if the absolute value of three π‘₯ 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 π‘₯.

To determine the interval of convergence of this series, it’s worth noting that our initial term π‘Ž is equal to one. So we must have that the absolute value of three π‘₯ 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 π‘₯ is greater than or equal to one. We must have that the interval of convergence is what we had before, those values of π‘₯ which satisfy the absolute value of three π‘₯ 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 π‘₯ is less than one. Then we know that three π‘₯ 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 π‘₯ is less than one-third. And we know that this is the same as saying that π‘₯ 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 π‘₯ to the 𝑛th power, is equal to the function one divided by one minus three π‘₯.

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