Video: Estimating Products of Decimals and Whole Numbers

Suppose βˆ‘_(𝑛 = 0)^(∞) π‘Ž_(𝑛) π‘₯^(𝑛) is a power series whose interval of convergence is (βˆ’3, 3) and that βˆ‘_(𝑛 = 0)^(∞) 𝑏_(𝑛) π‘₯^(𝑛) is a power series whose interval of convergence is (βˆ’5, 5). Find the interval of convergence of the series βˆ‘_(𝑛 = 0)^(∞) ((π‘Ž_(𝑛) π‘₯^(𝑛)) βˆ’ (𝑏_(𝑛) π‘₯^(𝑛)). Find the interval of convergence of the series βˆ‘_(𝑛 = 0)^(∞) (𝑏_(𝑛) 2^(𝑛) π‘₯^(𝑛)).

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Video Transcript

Suppose the sum from 𝑛 equals zero to ∞ of π‘Ž 𝑛 multiplied by π‘₯ to the 𝑛th power is a power series whose interval of convergence is the open interval from negative three to three. And the sum from 𝑛 equals zero to ∞ of 𝑏 𝑛 multiplied by π‘₯ to the 𝑛th power is a power series whose interval of convergence is the open interval from negative five to five. Find the interval of convergence of the series the sum from 𝑛 equals zero to ∞ of π‘Ž 𝑛 π‘₯ to the 𝑛th power minus 𝑏 𝑛 π‘₯ to the 𝑛th power. And find the [interval] of convergence of the series the sum from 𝑛 equals zero to ∞ of 𝑏 𝑛 two the 𝑛th power π‘₯ to the 𝑛th power.

The question wants us to find the interval of convergence for two different power series. We recall that an interval of convergence is an interval which contains all values of π‘₯ such that our series converges. If both power series converges so the absolute value of π‘₯ is less than the smaller of three or five. Therefore, since we’re adding together two convergent power series, we can combine the summands just as we would in a regular series. This gives us the sum from 𝑛 equals zero to ∞ of π‘Ž 𝑛 π‘₯ to the 𝑛th power minus 𝑏 𝑛 π‘₯ to the 𝑛th power will converge when the absolute value of π‘₯ is less than the smaller of three and five.

So we’ve shown that the power series converges when the absolute value of π‘₯ is less than three. But what about when it’s equal to three? What about when it’s greater than three? Let’s consider the case when the absolute value of π‘₯ is equal to three. And let’s assume that our series converges. Then, since the absolute value of π‘₯ is equal to three, the sum from 𝑛 equals zero to ∞ of 𝑏 𝑛 π‘₯ to the 𝑛th power must also converge because its radius of convergence is five. Now, what would happen if we tried to add these two series together? Well, we’ve assumed that our first power series will converge when the absolute value of π‘₯ is equal to three. And if the absolute value of π‘₯ is equal to three, this is less than five. So our second power series must also converge.

This means we’re trying to add together two power series which both converge. So we can just add the coefficients of π‘₯ to the 𝑛th power together. We see that negative 𝑏 𝑛 and 𝑏 𝑛 cancel. So this simplifies to give us the sum from 𝑛 equals zero to ∞ of π‘Ž 𝑛 multiplied by π‘₯ to the 𝑛th power. However, the sum from 𝑛 equals zero to ∞ of π‘Ž 𝑛 π‘₯ to the 𝑛th power will only converge when the absolute value of π‘₯ is less than three. So this series must diverge. The only way this could’ve happened is if the sum over 𝑛 of π‘Ž 𝑛 π‘₯ to the 𝑛th power minus 𝑏 𝑛 π‘₯ to the 𝑏 𝑛 π‘₯ to the 𝑛th power was divergent when the absolute value of π‘₯ was equal to three.

Therefore, since neither three nor negative three are in our interval of convergence, we’ve shown that the sum from 𝑛 equals zero to ∞ of π‘Ž 𝑛 π‘₯ to the 𝑛th power minus 𝑏 𝑛 π‘₯ to the 𝑛th power has the interval of convergence the open interval from negative three to three. So let’s clear some space and work on the second case of finding our interval of convergence.

We’ll let the function 𝑓 of π‘₯ be equal to the sum from 𝑛 equals zero to ∞ of 𝑏 𝑛 π‘₯ to the 𝑛th power. And this will be true whenever the absolute value of π‘₯ is less than five. We want to make this power series look like the one given to us in the question. We’ll start by multiplying and dividing our summands by two to the 𝑛th power. Next, we can use our laws of exponents to notice that a half to the 𝑛th power multiplied by π‘₯ to the 𝑛th power is equal to π‘₯ over two all raised to the 𝑛th power. We want our summands to have π‘₯ to the 𝑛th power instead of π‘₯ over two to the 𝑛th power. So we’ll let 𝑒 be equal to π‘₯ divided by two. This gives us that the sum from 𝑛 equals zero to ∞ of 𝑏 𝑛 two to the 𝑛th power multiplied by 𝑒 to the 𝑛th power will converge when the absolute value of two 𝑒 is less than five.

If the absolute value of two 𝑒 is less than five, this is the same as saying the absolute value of 𝑒 is less than five over two. So we’ve shown that the radius of convergence of the sum from 𝑛 equals zero to ∞ of 𝑏 𝑛 multiplied by two to the 𝑛th power multiplied by 𝑒 to the 𝑛th power is five over two. Since it doesn’t matter if we call our variable 𝑒 or π‘₯, we’ve shown that our second power series, the sum from 𝑛 equals zero to ∞ of 𝑏 𝑛 two to the 𝑛th power π‘₯ to the 𝑛th power, will have the interval of convergence the open interval from negative five over two to five over two.

Let’s now take a look at what would happen if we tried to multiply two power series together.

So let’s say we wanted to multiply two power series together, how would we do this? Well, let’s start by writing each out by term by term. Writing these out term by term gives us the following expression. Well, we remember that both of these expressions go on infinitely. If we look at the two expressions we have, these just seem like two polynomials. And we know how to multiply two polynomials together. To find the constant term, we multiply both of our constant terms together. This gives us π‘Ž nought multiplied by 𝑏 nought.

Now, to find all our terms of π‘₯, we multiply the constant terms by the coefficients with π‘₯. This gives us π‘Ž nought multiplied by 𝑏 one plus π‘Ž one multiplied by 𝑏 nought. We can then do the same for the coefficient of π‘₯ squared. To get our terms of π‘₯ squared in the product, we’ll be multiplying the constant terms by the terms of π‘₯ squared. And we’ll also be multiplying the terms of a single π‘₯ in them together. And we could continue doing this process indefinitely.

We can now notice when we were choosing our coefficient to multiply to get the term π‘₯ squared in our product, we were choosing them so that when we sum their indexes, we get two. And this makes sense because the index of our coefficient is exactly the same as the power of π‘₯. Let’s now take a closer look at our second term. We could write this as the sum from 𝑗 equals zero to one of π‘Ž 𝑗 multiplied by 𝑏 one minus 𝑗 all multiplied by π‘₯ to the first power. And again, this is because we wanted the sum of our indexes to be equal to the power of π‘₯. We could do exactly the same for the coefficient of π‘₯ squared. In fact, we could make a similar sum as the coefficient of any power of π‘₯. We can finally write this as a sum over our different powers of π‘₯. Where each coefficient of our power of π‘₯ will be equal to one of the coefficients we found earlier.

As we discussed earlier, to find the coefficient of π‘₯ to the 𝑛th power, we need to add together the products of all terms whose indexes add to 𝑛. And this gives us a motivation of how to multiply two power series together. If we’re given two power series which converge for a certain value of π‘₯, then the product of these two power series is equal to. We sum over 𝑛 to find the coefficient of π‘₯ to the 𝑛th power. And this coefficient is the sum over 𝑗 of π‘Ž 𝑗 multiplied by 𝑏 𝑛 minus 𝑗.

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