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## Finding Taylor or Maclaurin series for a function

# Geometric series interval of convergence

## Video transcript

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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