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Function as a geometric series | Series | AP Calculus BC | Khan Academy


3m read
·Nov 11, 2024

We're asked to find a power series for f, and they've given us f of x is equal to 6 over 1 + x to the 3 power. Now, since they're letting us pick which power series, you might say, "Well, let me just find the McLaurin series," because the McLaurin series tends to be the simplest to find, centered at zero.

You might immediately go out and say, "All right, well, let me evaluate this function at 0." Evaluating the first derivative at zero is pretty straightforward, but then once you start taking the second and third derivatives, it gets very hard very fast. You could do a simplification where you could say, "Well, let me find the McLaurin series for f of u is equal to 6 over 1 + u," where u is equal to x cubed.

So you find this McLaurin expansion in terms of u and then substitute for x cubed. Actually, that makes it a good bit simpler. So, that is another way to approach it, but the simplest way to approach it is to say, "Hey, you know what? This form right over here, this rational expression, looks similar; it looks like the sum of a geometric series."

Let's just remind ourselves what the sum of a geometric series looks like. If I have a plus a * r, so a is my first term, r is my common ratio, plus I'm going to multiply times r again, plus a * r squared, plus a * r cubed, and I keep going on and on and on forever. We know that this is going to be equal to a over 1 minus r, where a is the first term and r is the common ratio. This comes from the sum of a geometric series.

Notice that what we have here, our f of x, our definition of f of x, and the sum of a geometric series look very, very similar. If we say that this right over here is a, so a is equal to 6, and if negative r is equal to x cubed, or we could say, "Let me rewrite this," I could write this denominator as 1 minus x cubed.

Now, you can say, "Okay, well, r could be equal to x cubed," and just like that, we can expand it out! Well, if a is equal to 6 and r is equal to x cubed, then we can just write this out as a geometric series, which is very straightforward.

So, let's do that, and I will do this in this nice pink color. The first term would be 6, plus 6 times our common ratio, which is -6x cubed. Actually, let me just write that as -6x cubed.

Then we're going to multiply by x cubed again, so that's going to be... If I multiply this by - x cubed, that's going to be positive 6 * x to the 6th power. Then, I'm going to multiply by x cubed again, so it's going to be - 6 * x to the 9th power, and I'm going to go on and on and on.

I could keep going. If I multiply times x cubed, I will get 6 * x to the 12th power, and we can go on and on and on and on forever. The key here was... and this is the McLaurin series expansion for our f of x, but the key is to not have to go through all of this business and just to recognize that, "Hey, the way this function was defined looks a lot like the sum of a geometric series."

It can be considered the sum of a geometric series, and we can use that to find the power series expansion for our function. This is a very, very, very useful trick.

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