Today in this video we're going to be
looking at analyzing a function that's
defined in terms of power series, in this
case f of x is sigma kx^k sum
from 1 to infinity. We sum from 0 to
infinity of course but the 0 term will
be 0 anyway, so the first thing we want
to know is where does this series
converge for which x does this
actually make a sensible definition of a
function. In the second half of the video
we're going to look at what the function
actually is in a closed form. So Part A:
Where does the series converge? Now you
may have a formula for the radius of
convergence of a power series. I tend not
to use it. I prefer to go straight to the
ratio test but it's important that you
remember that the power series will
always converge in some interval
converging from a-r to a+r
for sum a and sum r but I always
find these things go much more smoothly
if you just forget that for the moment and just rely
on the ratio test. That's all you do in
the end anyway and you're also going to
be testing for absolute convergence not
just convergence. So what you do is you
consider the absolute value of the k+1
in our case k+1 x^k
divided by kx^k.
Well as is usual
most of these terms cancel off that
should be an x+1 there, k+1
there of course we're going to get
k+1/k that's positive so I
don't need the absolute value signs on
that k+1/k and I'm going to be
left with an x, mod x in fact. Taking
absolute values.
What we want to know is where is this going to tend to a
limit that's less than 1. What's
obvious enough that is k goes to
infinity this thing just tends to have
mod x and the ratio test tells me that
the series will be convergent if this
ratio is strictly less than 1 and that's
the thing that you want for a power
series because we want interval of
convergence so f of x. Well we can say f
of x converges
the series that converges if mod x is
less than 1 i.e. in the open interval
-1 to 1. Because this leaves open
the question of what happens at the end
point we only consider the endpoints of
these intervals of convergence of power
series in our higher first-tier courses
but in this case it's not difficult to
see what happens when x equals 1
with sigma k, 1 + 2 + 3
clearly doesn't converge and when you
put x is -1 your sum becomes well
it becomes -1 + 2 - 3 + 4.
That's not going to converge either so
in fact this series wouldn't converge at
either endpoint. Typically when you do a
convergence at the endpoints one of them
does and one doesn't but that's not
guaranteed. It's only the sort of
questions you tend to get asked.
That's where the series converges.
The next thing is can we actually find a
closed form for this particular function
can we actually write this function down
in something that's more familiar to us.
Normally you can't. Normally you write
down a power series it defines a brand
new function that doesn't have any other
we've been defined, but in this case you
can actually do that, so that's what
I'm gonna do in the second half the
video. So now we know where this function
is well defined is defined from -1
to 1. That's where the series converges.
You want to know what function is it. Can
we ever find something that's a bit more
understandable? something we can actually
deal with well? In this case we can. The
series itself looks a lot like the
geometric progression if the key word
would be sigma x to the k which
we know is just the geometric series
-x and something similar can be
done here because what we're going to be
relying on is the fact that if you take
a power series you can differentiate
term by term. You can take polynomials,
more or less everything works and that's
what we're going to do here. So we are
given that the sum from k is 0 to
infinity x^k is 1/1-x
if mod x one particular function exists,
because the series is convergent then we
can differentiate both sides and we get
-x^2.
Again it's usually a good idea to keep track of where these series are
convergent, well that's not quite what we
wanted because we've got k*x^k-1. Here we wanted kx^k.
We could just multiply through by
x
and of course the k = 0 term
forget about because it's
just going to be a 0, so we get the
sum of k is 1 to infinity. Okay
x^k is x/1-x^2 if mod x is less than 1.
