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>> In this chapter, and indeed in this course, we've seen several instances of

Â turning functions into sequences through some sampling process.

Â We now turn to the problem of reversing that machine,

Â of converting sequences into functions.

Â One mechanism for doing so is the power series.

Â A power series in x is simply a series that has

Â the variable x inside of it as a monomial term.

Â Something of the from sum and goes from 0 to infinity.

Â Of a sub n, x to the n, where now a sub n is considered a coefficient.

Â We're going to think of the notion of a power series as a machine or

Â an operator that converts a sequence,

Â a sub n, into a function, f of x.

Â Let's begin with some simple examples.

Â If we have a sequence, all of whose terms are 0,

Â except for one in the nth slot,

Â then the power series is simply the monomial x to the n.

Â Now if we have a finite sequence,

Â in the sense that only finitely many of the terms are non 0,

Â then this will have a power series that is a sum of monomials with

Â coefficients in front of them that eventually terminates.

Â We call that a polynomial.

Â Now, if we have something that does not eventually end in 0s,

Â we might have to be a bit more careful, but this isn't so mysterious.

Â If we have the sequence of all 1s, we recognize that the power series for

Â that sequence is simply the geometric series.

Â And this gives us the function 1 over 1-x,

Â at least when x is less than 1, an absolute value.

Â Other sequences can be easily seen to correspond to interesting functions,

Â though one may have to do a bit of algebra to see it.

Â If we look at certain sequences, we'll recognize some old friends.

Â Log of one+x, or e to the x.

Â These correspond to certain power series.

Â Now, what if we have a sequence that does not go to 0.

Â Say, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, etc.

Â Will that converge to anything?

Â Well, of course, this isn't so bad.

Â This one corresponds to the function 1 over quantity 1+x squared,

Â as you can see from squaring the geometric series.

Â In fact, we can have sequences that grow rather large but

Â which have power series that converge to something very simple.

Â Now, what if we just take any old interesting sequence of coefficients.

Â What will that converge to?

Â Sometimes it's a bit of a mystery.

Â Let's look at the Fibonacci sequence, and

Â let us denote by script F the power series whose coefficients are F sub n,

Â the terms in the Fibonacci sequence.

Â Now, if we were to multiply this power series by x what would happen?

Â It would give us a new power series associated to a shift,

Â a right shift, of the Fibonacci Sequence.

Â If we were to do it again, then we would shift over to the right yet once more.

Â And the reason for doing this is to line up the coefficient to

Â match the recursion relation that the Fibonacci Sequence satisfies.

Â Fn+2 = Fn + 1 + Fn.

Â 4:46

Now if we take a look at the coefficients in front of the monomial term,

Â x to the n, for these three power series that we have written down.

Â Then we see from the recursion relation the relationship between them.

Â The first is the sum of the latter two.

Â And so if we were to subtract x times script F and

Â x squared times script F from script F, what would we obtain?

Â Most of the coefficients of the monomial terms are zero,

Â because of the recursion relation.

Â When we add up these power series, the only thing that

Â is non-zero is in the constant term or we get F naught and

Â the first order term where we get F1x- F naught x.

Â Now because we know the exact values of F naught and

Â F1, we see that, on the left, we have F- xF- x squared F.

Â On the right, we have simply, x.

Â Solving for script F gives us the power series

Â formula that script F, the power series,

Â the Fibonacci sequence is equal to x over 1- x- x squared.

Â Now, that's what it is, but what does it mean?

Â What do we do with this?

Â These functions that generate the power series have within them,

Â all sorts of interesting questions.

Â They tie together recursive properties, enumerative properties,

Â approximation properties and convergence.

Â We're going to focus on issues of convergence.

Â Consider a power series F of x, sum of a sub N, x to the n.

Â The following theorem holds.

Â For some capital R between 0 and infinity,

Â perhaps achieving either of those values f of x has

Â the following convergence behavior.

Â It converges absolutely, if x is less than R in absolute value,

Â it diverges if x is bigger than R in absolute value.

Â Now this R is a very special number associated to the power series.

Â It's called the radius of convergence.

Â Let's see, how would we prove this theorem?

Â What tool would we turn to, to determine convergence?

Â Well, let's start with our friend, the ratio test.

Â Here, what we need to do is compute rho.

Â The limit is n goes to infinity of the ratio,

Â the n plus first term to the nth term.

Â Now, what are the terms?

Â Well, here they are a sub n times x to the n.

Â So taking the ratio of incident terms will tell us what we need.

Â Now, we have to be careful to take the absolute values

Â since the ratio test requires a positive sequence.

Â Okay, so doing that what do we see?

Â We get some cancellation with the xs and we can rewrite this as

Â the limit as n goes to infinity of the absolute value of a sub n + 1 over

Â the absolute value of a sub n times the absolute value of x.

Â Now, if that is less than 1, we have absolute convergence.

Â If that is bigger than 1, we have divergence.

Â So where is this capital R coming from?

Â Well, it is precisely the reciprocal of this limiting coefficient.

Â Since if we divide by this on both sides,

Â we get the appropriate answer.

Â When rho is less than 1, that's the same thing as

Â saying that x is strictly less than R in absolute value.

Â That means we have absolute convergence.

Â When rho is bigger than 1 that is the same thing as saying x is

Â greater than R in absolute value, and that means divergence.

Â So what matters is this radius of convergence capital R.

Â The limit as n goes to infinity of absolute a sub n over a sub n+1.

Â Be sure to remember that, and be sure to remember that it's the nth

Â term over the n plus first term, not the other way.

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Now, this Radius of convergence is called a radius of convergence because

Â it's telling you the distance away from 0 past

Â which your power series is going to diverge.

Â But within that, you're fine.

Â Now, you do have to be careful, we've said nothing about the end points of

Â the domain, because of course, the ratio test tells us nothing in that case.

Â So you're going to have to be careful and check explicitly at those end points.

Â Let's see a simple example if we consider f of x.

Â The sum over n -1 to the n, square root of n, quantity 2x to the n.

Â Then our coefficient a sub n is -1 to the n,

Â square root of n, 2 to the n.

Â To compute the radius of convergence, we take the limit of

Â the a sub n divided by a sub n + 1 all in absolute value.

Â That gives us the limit as n goes to infinity of square root of n,

Â 2 to the n, over square root of n + 1, 2 to the n + 1.

Â That is, as you can easily see, one-half.

Â Therefore, within one-half of 0, we have absolute convergence.

Â Outside of that, we have divergence.

Â What happens at the end points?

Â Well, we need to check what happens at x = -one-half if we plug

Â in a value of -one-half in 4x, what do we get?

Â We get the sum over n of square root of n.

Â Since the -1 to the ns cancel.

Â That is a divergent series.

Â So at the left hand end point we have divergence.

Â At the right hand end point, at x= one-half,

Â you can easily see that we get the same sort of series that,

Â too, is going to diverge by the nth term test.

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Now other series that we've seen throughout this course have similar

Â behavior.

Â If we look at the geometric series,

Â this is the power series where all the coefficients are equal to one.

Â The radius of convergence in this case equals 1, and

Â this diverges at both endpoints at plus and minus 1.

Â If on the other hand we consider the power series with coefficient 1 over n,

Â then goes from 1 to infinity.

Â Then what do we see?

Â This, too, has a radius of convergence equal to 1, and

Â at the right hand end point we obtain the harmonic series, which certainly diverges.

Â But at the left-hand end point, that x = -1,

Â we get the alternating harmonic series.

Â That converges, but it does so conditionally,

Â not absolutely.

Â If we consider the power series x to the n over n squared,

Â where the coefficient is 1 over n squared,

Â this also has radius of convergence equal to 1.

Â But it converges at both end points and

Â does so, therefore, absolutely.

Â 13:23

Some power series come to us in a slightly different form,

Â that of a shifted power series, something of the form sum over n

Â of a sub n times quantity x- c to the nth.

Â This of course is just a power series shifted over by c units.

Â In this case, if we compute the exact same radius of convergence,

Â then what we will find is that this shifted power series converges

Â when x-c is less than R in absolute value.

Â That is when x is within R of c.

Â R is thus a distance to the center, c, of the shifter power series.

Â For an example of a shifted power series,

Â consider the sum of 3x-2 to the n over n times 4 to the n.

Â Now this is not in the form of x-c to the n, so

Â we're going to have to do a little bit of algebra to get it into that form.

Â Claim that by factoring out a three-quarters to the n, and

Â absorbing that 3 to the n term from the denominator into the numerator.

Â We get the equivalent sum of 1 over n times

Â three-quarters to the n times x -two-thirds to the n.

Â That means that this shifted power series is centered, and x = two-thirds.

Â Now to determine the radius of convergence,

Â we isolate the coefficient a sub n as 1 over n times three-quarters to the n.

Â We compute the radius R is the limit then goes to infinity

Â of a sub n over a sub n + 1.

Â That yields, after some algebra, the limit of n + 1 over n times

Â four-thirds, which is precisely four-thirds.

Â Therefore, the domain of absolute convergence is all x

Â within four-thirds of two-thirds the center.

Â Now, what happens at the boundary points?

Â Well, on the left, we have x = -two-thirds.

Â That is, two-thirds minus four-thirds.

Â Substituting in x = -two-thirds, we obtain,

Â after a little bit of cancellation, the alternating harmonic series.

Â That, of course, converges.

Â When we substitute in the right hand end point, that is,

Â x = two-thirds + four-thirds or 2.

Â Then we obtain, substituting in a series which is the harmonic series.

Â This, of course, diverges, so

Â that we have conditional convergence on the left, divergence on the right.

Â