Welcome to Calculus. I'm Professor Ghrist. We're about to begin Lecture 56,
on Approximation and Error. We've seen the big picture
of how Taylor series and polynomial approximations
to functions work. In this lesson,
we're going to change focus and detail exactly what happens
when we perform approximations. By now you know that the sum of
one of our N squared converges. Since it is a P series
with P equal to two. But to what does it converge? We've claimed in the past that this
converges to pi squared over 6. That is a deep result. We can't get that easily,
so let's approximate. The true value of pi squared over 6
is in decimal form, 1.6449 et cetera. How many terms would we have
to sum up in this series? To get within a certain
amount of that true value. Well, let's fire up the computer, and see what happens if we add up let's say,
the first 10 terms. Then we get an answer that is,
well, it's within a neighborhood. It's not exactly really close. So let's add up the first 20 terms and
see how close we get. Now we're doing a fair bit better. If we take a little bit more time and
effort, compute the first 100 terms, then it seems as though we're definitely
within 1% of the true answer. And if we added up the first 1,000 terms,
well now we're getting something that is really,
fairly close to pi squared over six. But how close? Well in general, you're never going
to be able to get to the truth. The true answer involves a limit. And that just takes a lot of work. In general, what you can do is
come up with an approximation. Let's say by summing up terms up to and
including a sub capital N now what's left over is the error
that we'll denote E sub capital N. You're not gonna be able to
compute that error exactly, since then you would know the truth, but
you can control or bound that error. Let's see how that works in
the context of an alternating series. Let's consider a series that satisfies the
criteria of the alternating series test. That is, the sum of -1 to the n a sub
n where the a sub ns are positive, decreasing, and limiting to 0. Then remember how this
convergence happens. As you take the partial sums, you're
always jumping over the true answer. To the right, and then back to the left,
because of the alternating nature. Then in this case,
it's easy to get an upper bound on the error E sub N,
in absolute value. It is precisely a sub n+1,
the next term in this series, because you're always overshooting. When you have an alternating series,
this result is simple and useful. Let's consider the approximation
of one over square root of e, with the goal of getting
within one one-thousandth. Well, if we use our familiar expansion for e to the X where X equals
negative one-half, then we see this is really an alternating
series with the ace of n term being equal to one over n
factorial times two to the n. Now, if our goal is to add up
a finite number of these terms and get the error less than
one one thousandth. And the alternating series bound says
that we need to find a capital N so that a sub capital N plus one
is less than one one-thousandth. Well, what does a sub capital N plus one,
it is capital N plus one quantity factorial times
two to the capital N plus one. Getting that less than one one
thousandth is not so hard. That's going to work. Whenever (n+1) factorial
times 2 to the N+1 > 1000. That's true whenever N is bigger than or
equal to 4. So it doesn't take many terms to
approximate, 1 over square root of e. But, let's consider what it
would take to approximate log of 2 using the alternating
harmonic series. In this case, a sub n is 1 over n. In order to get the error less than
one one thousandth, what do we need? Well again,
by the alternating series bound, when need a sub N plus one
less than one one thousandth. That's then same thing as saying N
is greater than or equal to 1000, and that is a lot of terms to get within the same error amount
that we used for an exponential. What happens if you don't
have an alternating series? Well, you need a different error bound. There is one associated
to the integral test. Let's say that you have continued
your series a sub n to a function to a of x and have shown convergence by
means of integrating this function. Then one can see that the tail,
the E sub N term has a natural lower bound in
terms of the integral of a of x. Specifically, if one
integrates x goes from N+1. To infinity, then that is
a strict lower bound for E sub N. Now that's not optimal. We tend to want an upper bound for
the error. But note what we can do if we
slide everything over by one unit. Then the appropriate upper bound for
the error is the integral of [INAUDIBLE] as x goes from N to infinity. This is remarkable,
in that we get a bound in both directions. With this in mind, let's see what
it would take to get close to pi squared over 6 when we sum
up terms from 1 / n squared. Now we know the value of pi squared / 6,
and let's say that we wanna get within 0.001. Well, we know that this P series
converges by the integral test. Using a continuous function,
a of x equals 1 over x squared. By the integral test bound
E sub capital N is less than the integral from capital
N to infinity of this a sub x. We've done the integral of 1
sub x squared enough times so that you'll believe me when I say that
this integral comes to one over capital N. Now, if we want that to be
less than one one thousandth, that's really saying that N
has to be larger than 1,000. And if you'll recall when we
did some of our computations, that is exactly what we saw. When we summed up the first 1,000 terms. The integral test gives
very precise bounds. If you don't have an integral test and you don't have an alternating series,
what can you do? Well there is one last error bound that
involves only Taylor expansion but we are going to pay for
that generality in terms of complexity for the following result is deep and
difficult to grasp. For that reason, we'll keep it simple by
looking at what happens at f of x for x close to 0. Assume that f is smooth,
then Taylor expand f, about X equals zero, keeping only
terms up to and including order n. Now of course,
f is not equal to this Taylor polynomial, it's just an approximation. So there's some error, but
here the error term E sub N is a function of x and not a constant. So what can we say about
that error function? Well the first thing that we can say
is that E sub N of x is in big O of x to the N plus 1. That's not too surprising. Everything else is in higher order terms. On the other hand, this is kind of
a weak result in that in big O, you're only find out what
happens up to a constant and in the limit as x goes to zero. What we'd really like is
a more explicit bound that we can use to get numerical results. Well, there is a strong form
of this theorem that says that the error is bounded
in absolute value. I some constant c times x to the n + 1, over (N + 1)!. Where this constant C serves as an upper
bound to the N + first derivative of f, at all values of t between zero and
x inclusive. This is a much stronger
version of the arrow bound and it tells you that it's
really the n plus first term in the Taylor expansion that is
giving you control over the error. In fact if you wanna get really strong we
can replace the constant big C i exactly. The n + first derivative of f at
some t that is between zero and x and this is a very remarkable result
in that you're not bounding the error, you're saying exactly
what the error equals. As a function of x,
what I'm not telling you is what t you have to choose in order to evaluate. That n plus first derivative. Now, I'll let you work out what this
would be if you replace 0 with a. Let's see how this bound
works in an example. Let's approximate the square root
of E within 10 to the negative 10 using the familiar expansion for
e to the x and evaluating at x equals one
half then what do we get? We have some E sub N. Where by the Taylor Theorem,
E sub N is less than some constant C over N+1 factorial
times x to the n plus one. In this case, x equals one half and
c is some constant that bounds the n plus first
derivative of e to the x for all values of x between zero and one half. Now fortunately, derivatives of E to the X are easy
to compute, that's just E to the X. So, what is a good upper bound for
E to the X? Well, since E to the X is increasing. Then, a good upper bound would
be the right hand end point. E to the 1/2. Well, that number is maybe not so
easy to work with. So let's just say 2 because I
know that 2 is a reasonable upper bound for e to the 1/2. Therefore, we get that N is
less than 1 / (N+1)!, 2N. That's may be not the best bound we can
come up with but it will get the job done. Because if we tell you up and
plus one factorial times two to the N for various values of N. We see without too much effort that having
N bigger than or equal to ten will work. Well that went so well. Let's do it again. This time to estimate ARCSIN of one
tenth within ten to the negative ten. Now, I won't go through the details of
the taylor expansion for ARCSIN of x. The terms are a little bit complicated,
but not too bad if you assume
that they're given. What matters is the Taylor error bound
that E sub N is less than a constant C over (N+1) factorial times x to the N+1,
where x equals one-tenth. Now, this constant C is
the critical piece of information. It's an upper bound for
the n plus first derivative of ARCSIN(x) for all x between 0 and 1/10. Now, who remembers the formula for
the n plus first derivative of ARCSIN(x)? Anybody? I don't remember it either. And this is the difficult part
of using the Taylor bound. You don't necessarily know a good
bound for the N plus first derivative. How are we going to solve this? Well, if the Taylor theorem is not gonna
work and it's not an alternating series, and I don't think I wanna integrate
this function, then what do we do? Well, we're just going to have to think. But, if we think, well this is not so bad. Look at the terms in this series. We have one tenth, and
then something times one tenth cubed, plus something times one
tenth to the fifth, etc. It seems as though every step
where we go from n to n+2 we're picking up an extra
one tenth squared. Okay, so that's 1/100, but if we look
at the coefficients the 2n+1 and the product of odds over
the products of evens, then we're picking up another
factor of 10 in the denominator. And I claim that a,
n+2 the next term in the series is less than the previous term
a sub n divided by 1,000. What this means really is that
you're picking up three decimal places of accuracy,
with each subsequent term. And that means that if we want
to get within 10 to the -10th, it's going to suffice to choose
N bigger than or equal to 7. So the first four terms we
have represented on this slide suffice to approximate. ARCSIN one-tenth within
ten to the negative ten. Never forget to think even if
a Taylor bound doesn't work. In general, bounding errors is just hard. There's no getting around it. If you're fortunate enough to have an
alternating series, then it's not so bad. If you've got something that works
with an integral test, you're great. If not, you're either going to have
to resort to the Taylor theorem or use your head. That brings to a close all of
the main topics of this course. We're not quite done, though. We still have a few things to say to
wrap up this chapter, this course. To give you the big picture. And to point ahead to some of the broader
ideas you may encounter in calculus and in the world of mathematics.