0:25

By now you know that the sum of one of our N squared converges.

Â Since it is a P series with P equal to two.

Â But to what does it converge?

Â We've claimed in the past that this converges to pi squared over 6.

Â That is a deep result.

Â We can't get that easily, so let's approximate.

Â The true value of pi squared over 6 is in decimal form, 1.6449 et cetera.

Â How many terms would we have to sum up in this series?

Â To get within a certain amount of that true value.

Â Well, let's fire up the computer, and

Â see what happens if we add up let's say, the first 10 terms.

Â Then we get an answer that is, well, it's within a neighborhood.

Â It's not exactly really close.

Â So let's add up the first 20 terms and see how close we get.

Â Now we're doing a fair bit better.

Â If we take a little bit more time and effort, compute the first 100 terms,

Â then it seems as though we're definitely within 1% of the true answer.

Â And if we added up the first 1,000 terms, well now we're getting something

Â that is really, fairly close to pi squared over six.

Â But how close?

Â 1:52

Well in general, you're never going to be able to get to the truth.

Â The true answer involves a limit.

Â And that just takes a lot of work.

Â In general, what you can do is come up with an approximation.

Â Let's say by summing up terms up to and including a sub capital N

Â now what's left over is the error that we'll denote E sub capital N.

Â You're not gonna be able to compute that error exactly,

Â since then you would know the truth, but you can control or bound that error.

Â Let's see how that works in the context of an alternating series.

Â Let's consider a series that satisfies the criteria of the alternating series test.

Â That is, the sum of -1 to the n a sub n where the a sub ns are positive,

Â decreasing, and limiting to 0.

Â Then remember how this convergence happens.

Â As you take the partial sums, you're always jumping over the true answer.

Â To the right, and then back to the left, because of the alternating nature.

Â Then in this case, it's easy to get an upper

Â bound on the error E sub N, in absolute value.

Â It is precisely a sub n+1, the next term in this series,

Â because you're always overshooting.

Â When you have an alternating series, this result is simple and useful.

Â Let's consider the approximation of one over square root of e,

Â with the goal of getting within one one-thousandth.

Â Well, if we use our familiar expansion for

Â e to the X where X equals negative one-half,

Â then we see this is really an alternating series with the ace of n term

Â being equal to one over n factorial times two to the n.

Â Now, if our goal is to add up a finite number of these terms and

Â get the error less than one one thousandth.

Â And the alternating series bound says that we need to find a capital N so

Â that a sub capital N plus one is less than one one-thousandth.

Â Well, what does a sub capital N plus one, it is capital N plus

Â one quantity factorial times two to the capital N plus one.

Â 4:59

But, let's consider what it would take to approximate

Â log of 2 using the alternating harmonic series.

Â In this case, a sub n is 1 over n.

Â In order to get the error less than one one thousandth, what do we need?

Â Well again, by the alternating series bound,

Â when need a sub N plus one less than one one thousandth.

Â That's then same thing as saying N is greater than or equal to 1000,

Â and that is a lot of terms

Â to get within the same error amount that we used for an exponential.

Â 5:43

What happens if you don't have an alternating series?

Â Well, you need a different error bound.

Â There is one associated to the integral test.

Â Let's say that you have continued your series a sub n to a function to

Â a of x and have shown convergence by means of integrating this function.

Â Then one can see that the tail, the E sub N term

Â has a natural lower bound in terms of the integral of a of x.

Â Specifically, if one integrates x goes from N+1.

Â To infinity, then that is a strict lower bound for E sub N.

Â 6:58

With this in mind, let's see what it would take to get close to

Â pi squared over 6 when we sum up terms from 1 / n squared.

Â Now we know the value of pi squared / 6, and

Â let's say that we wanna get within 0.001.

Â Well, we know that this P series converges by the integral test.

Â Using a continuous function, a of x equals 1 over x squared.

Â By the integral test bound E sub capital N is less

Â than the integral from capital N to infinity of this a sub x.

Â We've done the integral of 1 sub x squared enough times so

Â that you'll believe me when I say that this integral comes to one over capital N.

Â Now, if we want that to be less than one one thousandth,

Â that's really saying that N has to be larger than 1,000.

Â And if you'll recall when we did some of our computations,

Â that is exactly what we saw.

Â When we summed up the first 1,000 terms.

Â The integral test gives very precise bounds.

Â 8:14

If you don't have an integral test and

Â you don't have an alternating series, what can you do?

Â Well there is one last error bound that involves only Taylor expansion but

Â we are going to pay for that generality in terms of complexity for

Â the following result is deep and difficult to grasp.

Â For that reason, we'll keep it simple by looking at what happens at f of x for

Â x close to 0.

Â Assume that f is smooth, then Taylor expand f,

Â about X equals zero, keeping only terms up to and including order n.

Â 8:58

Now of course, f is not equal to this Taylor polynomial,

Â it's just an approximation.

Â So there's some error, but here the error term E sub N

Â is a function of x and not a constant.

Â So what can we say about that error function?

Â Well the first thing that we can say is that E sub N of x is in big O

Â of x to the N plus 1.

Â That's not too surprising.

Â Everything else is in higher order terms.

Â On the other hand, this is kind of a weak result in that in big O,

Â you're only find out what happens up to a constant and

Â in the limit as x goes to zero.

Â What we'd really like is a more explicit bound

Â that we can use to get numerical results.

Â Well, there is a strong form of this theorem that says

Â that the error is bounded in absolute value.

Â I some constant c times x to the n + 1,

Â over (N + 1)!.

Â Where this constant C serves as an upper bound to the N + first derivative of f,

Â at all values of t between zero and x inclusive.

Â This is a much stronger version of the arrow bound and

Â it tells you that it's really the n plus first term

Â in the Taylor expansion that is giving you control over the error.

Â In fact if you wanna get really strong we can replace the constant big C i exactly.

Â The n + first derivative of f at some t that is between zero and

Â x and this is a very remarkable result in that you're not bounding the error,

Â you're saying exactly what the error equals.

Â As a function of x, what I'm not telling you is what

Â t you have to choose in order to evaluate.

Â That n plus first derivative.

Â Now, I'll let you work out what this would be if you replace 0 with a.

Â 11:21

Let's see how this bound works in an example.

Â Let's approximate the square root of E within 10 to the negative

Â 10 using the familiar expansion for e to the x and

Â evaluating at x equals one half then what do we get?

Â We have some E sub N.

Â Where by the Taylor Theorem, E sub N is less than

Â some constant C over N+1 factorial times x to the n plus one.

Â In this case, x equals one half and c is some constant

Â that bounds the n plus first derivative of e to the x for

Â all values of x between zero and one half.

Â Now fortunately,

Â derivatives of E to the X are easy to compute, that's just E to the X.

Â So, what is a good upper bound for E to the X?

Â Well, since E to the X is increasing.

Â Then, a good upper bound would be the right hand end point.

Â E to the 1/2.

Â Well, that number is maybe not so easy to work with.

Â So let's just say 2 because I know that 2 is a reasonable

Â upper bound for e to the 1/2.

Â Therefore, we get that N is less than 1 / (N+1)!, 2N.

Â That's may be not the best bound we can come up with but it will get the job done.

Â Because if we tell you up and plus one factorial times two to the N for

Â various values of N.

Â We see without too much effort that having N bigger than or equal to ten will work.

Â 13:12

Well that went so well.

Â Let's do it again.

Â This time to estimate ARCSIN of one tenth within ten to the negative ten.

Â Now, I won't go through the details of the taylor expansion for ARCSIN of x.

Â The terms are a little bit complicated, but

Â not too bad if you assume that they're given.

Â What matters is the Taylor error bound that E sub N is less than a constant C

Â over (N+1) factorial times x to the N+1, where x equals one-tenth.

Â Now, this constant C is the critical piece of information.

Â It's an upper bound for the n plus first derivative

Â of ARCSIN(x) for all x between 0 and 1/10.

Â Now, who remembers the formula for the n plus first derivative of ARCSIN(x)?

Â Anybody?

Â I don't remember it either.

Â And this is the difficult part of using the Taylor bound.

Â You don't necessarily know a good bound for the N plus first derivative.

Â How are we going to solve this?

Â Well, if the Taylor theorem is not gonna work and it's not an alternating series,

Â and I don't think I wanna integrate this function, then what do we do?

Â Well, we're just going to have to think.

Â But, if we think, well this is not so bad.

Â Look at the terms in this series.

Â We have one tenth, and then something times one tenth cubed,

Â plus something times one tenth to the fifth, etc.

Â It seems as though every step where we go from n to n+2

Â we're picking up an extra one tenth squared.

Â Okay, so that's 1/100, but if we look at the coefficients the 2n+1 and

Â the product of odds over the products of evens,

Â then we're picking up another factor of 10 in the denominator.

Â And I claim that a, n+2 the next term in the series is less

Â than the previous term a sub n divided by 1,000.

Â What this means really is that you're picking up three decimal

Â places of accuracy, with each subsequent term.

Â And that means that if we want to get within 10 to the -10th,

Â it's going to suffice to choose N bigger than or equal to 7.

Â So the first four terms we have represented on this

Â slide suffice to approximate.

Â ARCSIN one-tenth within ten to the negative ten.

Â Never forget to think even if a Taylor bound doesn't work.

Â In general, bounding errors is just hard.

Â There's no getting around it.

Â If you're fortunate enough to have an alternating series, then it's not so bad.

Â If you've got something that works with an integral test, you're great.

Â If not, you're either going to have to resort to the Taylor theorem or

Â use your head.

Â