Finding Taylor or Maclaurin series for a function
Geometric series interval of convergence
So as we talked about in the last video, we've seen many examples of starting with a geometric series expanded out, and then assuming that its common ratio, that the absolute value of the common ratio is less than 1, finding what the sum of that might be. We've proven with this formula in previous videos. But now let's go the other way around. Let's try to take some function-- let's say h of x being equal to 1 over 3 plus x squared-- and let's try to put it in this form. And then once we put it in that form, we can think about what a and our common ratio is. And then try to represent it as an actual geometric series. So I encourage you to pause the video and try to do that right now. So let's see, the first thing that you might notice is we have a 1 here instead of a 3. So let's try to factor out a 3. So this is equal to 1 over 3 times 1 plus x squared over 3. And now, since we don't want that 3 in the denominator, we can think about this as 1 over 3. So we could say this is 1/3 over-- let me do it in that purple color. 1/3 over 1. And we don't want to just add something, we want to subtract our common ratio. So 1 minus-- and let me write our common ratio here in yellow. 1 minus negative x squared over 3. So now we've written this in that form. And so now we could say that the sum-- let me write it here in-- let me do it in a new color. So let me do it in blue. So now we could say that the sum from n equals 0 to infinity of-- let's see, our first term is 1/3. 1/3 times our common ratio to the n-th power. Common ratio is negative x squared over 3. And if we wanted to expand this out, this would be equal to-- so the first term is 1/3 times all of this to the 0-th power. So it's just going to be 1/3. And so each successive term is just going to be the previous term times our common ratio. So 1/3 times negative x squared over 3 is going to be negative 1/9 x squared. To go from that to that, you have to multiply by-- let's see, 1/3 to negative 1/3, you have to multiply it by negative 1/3. And we multiplied by x squared as well. Now in our next term, we're going to multiply by negative x squared over 3 again. So it's going to be plus-- a negative times a negative is a positive-- plus 1/27 x to the fourth. x squared times x squared, x to the fourth power. And we just keep going on and on and on. And when this converges, so over the interval of convergence, this is going to converge to h of x. Now, what is the interval of convergence here? And I encourage you to pause the video and think about it. Well, the interval of convergence is the interval over which your common ratio, the absolute value of your common ratio, is less than 1. So let me write this right over here. So our absolute value of negative x squared over 3 has to be less than 1. Well, the absolute value, this is going to be a negative number. This is the same thing as saying-- let me scroll down a little bit. This is the same thing as saying that the absolute value of x squared over 3 has to be less than 1. And this is another way of saying-- well, one thing that might jump out at you is that x squared, this is going to be positive no matter what. Or I guess I should say, this is going to be non-negative no matter what. So this is another way of saying that x squared over 3 has to be less than 1. Right? I don't want to confuse you in this step right over here. But the absolute value of x squared over 3 is just going to be x squared over 3, because this is never going to take on a negative value. And so we can multiply both sides by 3. I'll go up here now to do it. Multiply both sides by 3 to say that x squared needs to be less than 3. And so that means that the absolute value of x needs to be less than the square root of 3. Or we could say that x is greater than the negative square root of 3, and it is less than the square root of 3. So this is the interval of convergence. This is the interval of convergence for this series, for this power series. It's a geometric series, which is a special case of a power series. And over the interval of convergence, that is going to be equal to 1 over 3 plus x squared. So as long as x is in this interval, it's going to take on the same values as our original function, which is a pretty neat idea.
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