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Consider the function 𝑓(π‘₯) = 1/(2 + π‘₯), Find the power series for 𝑓(π‘₯). Identify its interval of convergence.

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Video Transcript

Consider the function 𝑓 of π‘₯ is equal to one divided by two plus π‘₯, find the power series for 𝑓 of π‘₯. Identify its interval of convergence.

The question gives us a rational function 𝑓 of π‘₯. It wants us to find the power series for the function 𝑓 of π‘₯ and then to find the interval of convergence for this power series. We recall a fact about geometric series. The sum from 𝑛 equals zero to ∞ of π‘Ž multiplied by π‘Ÿ to 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. And we also recall that this power series will diverge when the absolute value of π‘Ÿ is greater than one. If we could rewrite our function 𝑓 of π‘₯ in the form π‘Ž divided by one minus π‘Ÿ, then we could use this fact about geometric series to rewrite our function as a power series.

We want the denominator of our fraction to be in the form one minus the ratio π‘Ÿ. However, in our fraction for 𝑓 of π‘₯, we have a two in the denominator. Since we want our denominator to be one minus the ratio π‘Ÿ, we’ll take out our factor two in our denominator. We have that two multiplied by one gives us two and then two multiplied by π‘₯ divided by two gives us π‘₯. We can then take the constant of a half outside of our fraction. This gives us a half multiplied by one divided by one plus π‘₯ over two. This is now almost in the form that we need it. We have the constant of one in the numerator. But instead of one minus π‘Ÿ in the denominator, we have one plus π‘₯ over two.

We can do this by using a little bit of algebraic manipulation. Instead of adding π‘₯ over two, we can subtract negative π‘₯ over two. Now, if we set the numerator of our function to be equal to π‘Ž and the denominator of our fraction to be one minus π‘Ÿ, so that setting π‘Ÿ to be equal to negative π‘₯ over two. Then, we’ve shown that our fraction is in the form of the infinite sum of a geometric series with first term π‘Ž equal to one and ratio π‘Ÿ equal to negative π‘₯ over two.

So, using our fact about geometric series, we can write this as a half multiplied by the sum from 𝑛 equals zero to ∞ of one multiplied by negative π‘₯ over two raised to the 𝑛th power. And in particular, we’ll know that this is definitely true, when the absolute value of our ratio π‘Ÿ is less than one. And we know that it will diverge when the absolute value of our ratio π‘Ÿ is greater than one. In our case, we have that π‘Ÿ is equal to negative π‘₯ divided by two. So this power series must converge when the absolute value of negative π‘₯ divided by two is less than one.

We can simplify our power series slightly by noticing that negative π‘₯ over two is equal to a half multiplied by negative π‘₯. We can then distribute the exponent over the parenthesis. This gives us a half multiplied by the sum from 𝑛 equals zero to ∞ of a half to the 𝑛th power multiplied by negative π‘₯ to the 𝑛th power. Finally, we can bring the coefficient of a half inside of our summand. But then, a half multiplied by a half to the 𝑛th power is just equal to a half raised to the power of 𝑛 plus one. So the power series representation for our function 𝑓 of π‘₯ is the sum from 𝑛 equals zero to ∞ of a half raised to the power of 𝑛 plus one multiplied by negative π‘₯ to the 𝑛th power.

The second part of our question wants us to identify the interval of convergence of this power series. We recall that the interval of convergence is all of the values of π‘₯ where our series will converge. We’ve already shown that our series has to converge when the absolute value of our ratio π‘Ÿ is less than one. And it has to diverge when the absolute value of our ratio π‘Ÿ is greater than one. In our case, our ratio π‘Ÿ is equal to negative π‘₯ over two. So our power series must converge when the absolute value of negative π‘₯ divided by two is less than one. The absolute value of negative π‘₯ over two is just equal to the absolute value of π‘₯ over two.

Next, solving the absolute value of π‘₯ over two is less than one is the same as solving negative one is less than π‘₯ over two is less than one. We can then multiply this inequality through by two. This gives us that negative two is less than π‘₯ is less than two. This is our radius of convergence. Our power series must converge for all values of π‘₯ between negative two and two. And we know it will diverge when π‘₯ is greater than two or when π‘₯ is less than negative two. However, we don’t know what will happen when π‘₯ is equal to two or when π‘₯ is equal to negative two.

To find out what happens at the bounds of our radius of convergence, let’s substitute in π‘₯ is equal to two and π‘₯ is equal to negative two into our power series. Substituting π‘₯ is equal to two gives us the sum from 𝑛 equals zero to ∞ of a half raised to the power of 𝑛 plus one multiplied by negative two to the power of 𝑛. There are several different ways of evaluating this series. For example, we could use the 𝑛th term divergence test. However, we’re going to look term by term at the expansion of our series.

Our first term when 𝑛 is equal to zero gives us a half raised to the power of zero plus one, which is a half. And then, we multiply this by negative two raised to the power of zero, which is one. We can do the same of our second term. When 𝑛 is equal to one, we get a half squared multiplied by negative two, which we can work out to be negative one-half. Our third term when 𝑛 is equal to two gives us a half cubed multiplied by negative two squared, which is plus a half. In fact, this series continues. We add a half. Then, we subtract a half. Then, we add a half and we subtract a half et cetera. We can then consider what the partial sums will look like.

Our first partial sum will be equal to a half. Our second partial sum, the sum of the first two terms of our series, will be a half minus a half which is zero. Our third partial sum, the sum of the first three terms of our series, will be a half. And our fourth partial sum, the sum of the first four terms of our series, will be zero. And we can see that this pattern will continue. In fact, this means that our partial sums won’t converge. They will fluctuate between a half and zero. Since the sequence of partial sums doesn’t converge, we can conclude that our power series does not converge when π‘₯ is equal to two. So π‘₯ is equal to two is not in our interval of convergence.

We can do the same when π‘₯ is equal to negative two. We substitute π‘₯ is equal to negative two into our power series and then calculate the series term by term. We see that every single term in our series is equal to a half. Therefore, our 𝑛th partial sum of this series will be a half plus a half plus a half 𝑛 times or 𝑛 divided by two. Therefore, since our 𝑛th partial sum is unbounded, we can conclude that the power series does not converge when π‘₯ is equal to negative two. Therefore, since we’ve shown that neither of the end points of our radius of convergence are in our interval of convergence, we can conclude that the interval of convergence for the power series of the function one divided by two plus π‘₯ is the open interval from negative two to two.

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