# Power Series in X & the Interval of Convergence

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• 0:04 Power Series in X
• 1:42 Interval of Convergence
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Lesson Transcript
Instructor: Gerald Lemay

Gerald has taught engineering, math and science and has a doctorate in electrical engineering.

A power series is often used in physics and engineering to approximate functions. In this lesson, we'll explore the power series in x and show how to find the interval of convergence.

## Power Series in X

My favorite part of the hardware store is the power tool section, especially the battery-powered tools. Speaking of power and tools, there's this mathematical tool called the power series in x.

A power series in x is an infinite sum of terms where each term is a factor multiplying an x plus a constant raised to a power.

### Building a Power Series

Infinity is a lot of terms to add! What if we start with the first 4 terms? The sum of those four terms could look like:

We let x take on a value. Then, we add the four terms and get a result. What if we let x have values from -5 to -1 ? Then we could substitute these values for x, sum the four terms, and save the results. Plotting these results as a function of x:

Can you predict what the next term in this series looks like? If you said (x + 3)^4 divided by 5, then you are correct! What if we sum the first 10 terms of the series? Let's see, the tenth term will be (x + 3)^9 divided by 10. Again, substitute x values from -5 to -1, sum the ten terms, and save the results. If these results for 10 terms are plotted with the results for 4 terms we see:

There seems to be some values for x where the results have changed a lot when we went from 4 terms to 10 terms. The values to the left of -4 and the values to right of -2 are quite different. The sum is staying pretty much the same between x = -4 and x = -2. This is getting interesting.

While our battery continues to charge, let's build on these ideas.

## Interval of Convergence

What's the general term we have been summing? Is it (x + 3)^n divided by n + 1? Let's check this out. If n = 2, we have (x + 3)^2 divided by 3. Looking back when we summed four terms, this was our third term.

What about the first term, which is a 1? No problem. Letting n = 0 gives us (x + 3)^0 divided by 0 + 1. The (x + 3)^0 is 1 and 0 + 1 is 1. So we have 1 divided by 1, which is 1.

To express the infinite sum starting from n = 0, we write:

The capital sigma means to take the sum from n = 0 to n = infinity of the general term. We're going to use the symbol An for the general term.

The ratio test says a series converges if

We know the An general term:

To get the An+1 expression, we replace the n with n + 1:

Let's work out the absolute value part first:

Since (n + 1)/(n + 2) is always greater than 0, we can take it outside of the absolute value:

Now, we deal with the limit part of the ratio test:

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