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visit MIT OpenCourseWare at ocw.mit.edu. Today we’re going to start

with a simple problem that many of you may

have already encountered. For example, when you

got a student loan or if your family took a

loan out on your house, you know, a home mortgage

loan, and it’s the problem of pricing an annuity. An annuity is a

financial instrument that pays a fixed

amount of money every year for some

number of years, and it has a value

associated with it. For example, with

a student loan, the value is the amount of

money they gave you to pay MIT, then later in life– every

year or every month– you’re going to send them a check. And you want to sort of equate

those two things to find out, are you getting enough

money for the money you’re going to pay back

monthly sometime in the future? So let’s define this, and

there’s a lot of variations on annuities, but we’ll

start with one that’s called an n-year $m-dollar

payment annuity, and it works by paying m dollars

at the start of each year, and it lasts for n years. Now, usually n is

finite, but not always, and in a few minutes

we’ll talk about the case when it’s infinite. But this includes

home mortgages, where you pay every

month for 30 years. Megabucks, the lottery. They don’t actually give

you the million dollars. They give you $50,000

per year for 20 years, and call it a million bucks. Retirement plans. You pay in every

year, and then you get some big lump sum later. Life insurance benefits,

you know, and so forth. Now, if you go to Wall

Street, this is a big deal. A lot of the stuff that

happens on Wall Street involves annuities in

one form or another, packaging them all up

and, in fact, we’ll look at later when

we do probability. It was how these things

were packaged and sold that led to the sub-prime

mortgage disaster, and we’ll see how some confusion

over independents, when you look at random variables,

led to the global recession, a real disaster. Of course, some people

understood how all that worked, made hundreds of billions

of dollars at the same time. So it was sort of money went

from one place to another. So it’s pretty

important to understand how much this is worth. What is this

instrument– a piece of paper that says

it will pay you m dollars at the beginning of

each year for n years, what is that worth today? For example, say I

gave you a choice– the Megabucks choice–

$50,000 a year for 20 years or a

million dollars today. How many people would prefer

the $50,000 a year for 20 years? A few of you. How many would prefer the

million bucks right up front? Much better. OK. Always better to have

the cash in hand, because there’s things

like inflation– pretty low now– interest. You can put the

money in the bank or invest it and make

some money hopefully. So the million dollars

today is a lot better, which is why the State pays you

50 grand a year for 20 years. It’s better for them, and

they call it a million bucks. So that was pretty

clear, but what if I gave you this option– 700

grand today or 50 grand a year for 20 years? How many people want the cash

upfront– 700 grand only. A few. How many people want 50

grand a year for 20 years? All right, we’re almost–

that’s pretty close to half way. How about 500 grand today versus

50 grand a year for 20 years? How many want half

a million today? A lot of people like the cash. You know, it’s this kind of a

time, you have the recession, it’s a disaster on Wall Street. You know, Street Wall

Street didn’t like the cash. How many want, instead

of a half a million up front, 50 grand

a year for 20 years? All right, now

we’re about halfway. All right, well

that’s pretty good. So we’re going to find

out what you should pay, or at least one way

of estimating that. Now, to do that, we’ve

got to figure out what $1 today is worth in a year. And to do that, we

make an assumption, and the assumption

is that there’s a fixed– we’ll call

it an interest rate. It’s sort of the devaluation

of the money per year. And we’re going to call it p. Later we’ll plug

in values for p, but you can think of

it as like 6%, 1%. You know, it’s the money,

if you put money in a bank, they’ll give you some

percent back every year. And, of course, the fact

that different people have different ideas

of what this would be, allows people to make

money on Wall Street. As we’ll see, a

slight difference in p can make big differences in

what the annuity is worth. So for example, $1 today

is going to equal 1 plus p dollars in one year. Similarly, $1 today–

how much is that going to be worth in two years? Say that p stays fixed,

the same over all time. One plus p squared,

because every year you multiply what you got by

1 plus p, because that’s the interest you’re getting. All right, we’ll think of

it in terms of interest. In three years, $1 today

is worth 1 plus p cubed in three years and so forth. All right. Now, what we really care about

is what’s $1, or m dollars, worth today if you’re

getting it next year? So we need to sort of flip

this back the other way. So what is $1 in a year

worth today in terms of p? So if I’m going to

be paid $1 in a year, what would be the equivalent

amount to be paid today? One over 1 plus p, because

what’s happening here is, as you go forward in a year,

you just multiply by 1 plus p. So 1 over 1 plus p turns

into $1 in a year– being paid in a year. All right. What is $1 a year in

two years worth today? One over 1 plus p squared. So $1 in two years is

worth this much today. Well, now we can use

this to go figure out the current value

of that annuity. We just figure out what every

payment is worth today and then add it up. So we’ll put the

payments over here, and we’ll compute

the current value of every payment on this side. So with the annuity,

the way we’ve set it up is it pays n dollars

at the start of every year, so the first payment is now. So the first of the

n payments is now, and since it’s being paid

now, that’s worth m dollars. There’s no devaluation. The next payment is m

dollars in one year, and so that’s going to be

worth m over 1 plus p today. And the next payment is

m dollars in two years. That’s worth m over

1 plus p squared, and we keep on going

until the last payment. It’s the n-th payment, so it’s

m dollars in n minus 1 years. And so that’s going to

be worth m over 1 plus p to the n minus 1. All right. So we can compute the current

value of all those payments, then the annuity is computed–

the value is computed just by adding these up, of

all the current values. So the total current value is

the sum i equals 0 to n minus 1 of m over 1 plus p to the i. And that is the

total current value. That’s what you should

pay today for the annuity. Any questions? What we did here? All right. Well, of course, what we’d like

is a closed form expression here. Something that’s simple

so we could actually get a feel without having

to add up all those terms, and that’s not hard to get. In fact, let’s put

this sum in a form that might be more familiar. This equals– we’ll pull

the m out in front– and let’s use x to be 1

over 1 plus p to the i. And so x equals 1 over 1 plus

p, and I wrote it this way because this might be familiar. Does everybody remember

that from– I think it was the second recitation? Anybody remember the formula? What this evaluates to? The sum of x to the i, where

i goes from 0 to n minus 1? Remember that? One minus x to the n. Remember 1 minus x. In the second

recitation, I think, we proved that this equals that. What was the proof

technique we used? Induction. OK? So, in fact, there’s

a theorem here. For all n bigger and equal

to 1 and x not equal to 1, we proved the sum from i equals

0 to n minus 1 x to the i equals 1 minus x to

the n over 1 minus x. And so this is a

nice, closed form. No sum any more, just

that, which is nice. Now, induction proved

it was the right answer. Once you knew it– we gave

it to you– using induction to prove that

theorem wasn’t hard. What we’re going to look at

doing this week and next week is figuring out how

to figure out this was the answer in

the first place. Methods for doing that– to

evaluate the sum– and there’s a lot of ways that you can

do that particular sum. Probably the easiest is known

as the perturbation method. This sometimes works. Certainly with sums like

that, it often works. The idea is as follows. We’re trying to compute

the sum S, which is 1 plus x plus x squared

plus x to the n minus 1, and what we’re going to do

is perturb it a little bit and then subtract to

get big cancellation. In this case, it’s

pretty simple. We multiply the sum by x to get

x plus x squared plus– I’ve defined S to be that,

x times S– well, I get x plus x squared and

so forth, up to x to the n, and now I can subtract one from

the other and almost everything cancels. So I get 1 minus x

times S equals 1. These cancel, cancel, cancel. Minus x to the n. And therefore S equals 1 minus

x to the n over 1 minus x. So that’s a vague method. This gets used all the time

in applied mathematics, and they call it the

perturbation method. Take your sum, wiggle

it around a little bit, get something that looks

close, subtract it, everything cancels,

life is nice, and all of a sudden you’ve

figured out the answer. So getting back to

our annuity problem, we can plug that

formula back in here. So the value of the annuity

is m times 1 minus x to the n over 1 minus x. We’ll plug in x equals

1 over 1 plus p, and we get m 1 minus

1 over 1 plus p to the n over 1 minus 1 over

1 plus p, just plugging in. And now to simplify this, I’ll

multiply the top and bottom by 1 plus p, and I’ll get 1

plus p minus 1 on the bottom. Just gives me a p on the bottom. I have a 1 plus p on

the top minus 1 over 1 plus p to the n

minus 1, all over p. So now we have a formula–

closed form expression formula– for the

value of the annuity. All’s we’ve got to plug in

is m, the payment every year, n, the number of years, and

then p, the interest rate. And so, for example, if

we made m be $50,000, as in the lottery. We made n be 20 years, and

say we took 6% interest, which is actually very good these

days, and I plug those in here, the value is going

to be $607,906. All right. So those of you that preferred

700 grand– if you assume 6% interest– you’re right. Those of you who

preferred 500 grand, no, you’re better off waiting and

getting your 50 grand a year. Now, of course, if the

interest rate is lower, well, that changes things. That shifts it even more. The annuity is worth even more

if the interest rate is lower– if p is smaller. In fact, say p was 0. Say the interest rate is 0, so

$1 today equals $1 tomorrow, then what is the lottery worth? A million dollars. And the bigger p gets, the

less your payment is worth. Any questions about that? OK. What if you were paid $50,000 a

year forever– you live forever or it goes to your

estate and your heirs, $50,000 a year forever or

a million dollars today. How many people want the

million dollars today? How many want 50

grand a year forever? Sounds good. You know, that’s an infinite

amount of money, sort of. It’s not as good as it sounds. Let’s see why. So this is a case where

n equals infinity, and so I’ll claim that

if n equals infinity, then the value of this

annuity is just m times 1 plus p over p. Let’s see why that’s the case. You know, it sounds

hard to evaluate, because it’s an infinite

number of payments, but what happens here

when n goes to infinity? What happens to this thing? That goes to 0 as

n goes to infinity, as long as p is bigger than 0. So we’re going to

assume 6% interest. So that goes away, so

the annuity is worth just that, m times 1 plus p over

p, because the limit as n goes to infinity of 1 over

1 plus p to the n minus 1, that’s going to 0. So the value for m is $50,000,

and at 6% V is only $883,000. So you’re better off

taking a million dollars today than $50,000

a year forever. Now, if you think about

it, and think about it as an interest

rate, why should it be obvious that you’re better

off with a million dollars today than 50 grand

a year forever? Think about what you could

do with that million dollars if you had it today

at 6% interest. What would you do with it to

make more money than 50 grand a year forever? Yeah. AUDIENCE: You could have like–

you could make $50,000 per year just off of [INAUDIBLE]. PROFESSOR: Yeah. In this model, if the

interest rate is 6%, you can put in the bank. It makes 6% every year, that’s

60 grand a year forever. Better than 50 grand

a year forever. So maybe it’s– even

without doing the math, you can tell which

way it’s going to go, but this tells you

exactly what it’s worth. Any questions? OK. Yeah. AUDIENCE: [INAUDIBLE]

current value– that’s how much it’s

worth to you right now. PROFESSOR: Yeah. AUDIENCE: So and, like, if

you have $1 in the future– PROFESSOR: Yeah. AUDIENCE: –where S right

now is m over 1 plus p– PROFESSOR: Yeah. AUDIENCE: –is it worth less to

you [INAUDIBLE] then later on, it’s going to be

worth more to you. PROFESSOR: Ah, so

if you move yourself forward in time, $1 a year, in

a year it’ll worth $1 to you– AUDIENCE: Yeah. PROFESSOR: –but today it’s

worth less than $1 to you, because you could take the

dollar today and invest in the bank, and it’s worth more

in a year, because the money grows in value, as a

way to think of it, because you can

earn interest on it. What’s that? AUDIENCE: You could spend it. PROFESSOR: Yeah. Yeah, if you just

spend it, well, then at least you

had the use of what you spent it on for the year. So there’s some other

kind of value, right? Maybe you bought a

house or something that– maybe something that

even appreciated in value. OK. But these things

get sort of squishy, and that is where

a lot of people make money on Wall Street, is

because different companies have different needs for money. They have different views of

what the interest rates are going to be, and you

can play in the middle and make a lot of

money that way. So more generally, there’s

a corollary to the theorem, and that is that if the absolute

value of x is less than 1, then the sum i equals 0

to infinity x to the i is just 1 over 1 minus x. We didn’t prove this back

in the second recitation, because there’s no

n to induct on here, but the proof is simple

from the theorem. And it’s simply because

if x is less than 1, an absolute value, as n goes

to infinity, that goes to 0, and so you’re just left

with 1 over 1 minus x. So, for example,

what’s this sum? This one you all know, I’m sure. Out to infinity. What’s that sum to? To 2. Yeah. It’s 1 over 1 minus

1/2, which is 2. What about this sum

out to infinity? What does that sum to? AUDIENCE: [INAUDIBLE] PROFESSOR: Yeah, 3/2, 1

over 1 minus 1/3 is 3/2. So, easy corollaries. These are all examples

of geometric series. That’s what a definition

of a geometric series is. Something that’s going

down by a fixed– each term goes down by the same

fixed amount every time. And geometric

series, generally sum to something that is very

close to the largest term. In this case, it’s 1. Very common, because

of that formula. It’s 1 over 1 minus x. Any questions about this

or geometric series? All right. Well, those are

straight geometric sums. Sometimes you run into things

that are a little bit more complicated. For example, say I have this

kind of a sum, i equals 1 to n, i times x to the i. Now, those are adding up x

plus 2x squared plus 3x cubed, and so forth, up

to n x to the n. You know, that’s a

little more complicated. The terms are

getting– decreasing by a factor of x, increasing by

1 in terms of the coefficient every time. A little trickier. So say we wanted to get a

closed form expression for that? There are several

ways we can do it. The first would be to try to use

perturbation– the perturbation method. Let’s try that. So we write S equals x plus 2x

squared plus 3x cubed plus nx to the n, and let’s try

the same perturbation. Multiply by x, I

get that x squared plus 2x cubed plus n minus 1x to

the n, plus nx to the n plus 1. And then I subtract to try

to get all the cancellation. So then I do that. I get 1 minus x times S. Well, I

didn’t quite cancel everything. X plus x squared

plus x cubed plus x to the n plus n– or

minus nx to the n plus 1. Ah, it didn’t quite work. Anybody see a way that

I can fix this up? What about this piece? That’s still a mess here. Can I simplify that? Yeah? AUDIENCE: [INAUDIBLE] PROFESSOR: Yeah. There’s a simpler way. Yeah. AUDIENCE: That’s a

geometric series. PROFESSOR: That’s

a geometric series. We just got the formula

for it, so that’s easy. This equals 1 minus x to the

n over 1 minus x minus the 1, because I’m missing the 1 here. So we can rewrite this

whole thing over here. Oops. Yikes. Got attacked. What’s that? AUDIENCE: Would it be 1

minus x should be n plus 1? PROFESSOR: Yes it would. That’s right. I’ve got to add 1 to there. OK. So that’s good. So that says that

1 minus x times S equals 1 minus x to the n

plus 1 over 1 minus x minus 1, and then I’ve got to remember

to subtract that term too. Minus nx to the n plus 1. That means now I

just divide through, and I simplify– divide through

by 1 minus x– and simplify, and I get the following formula. I won’t go through all the

details, but it’s not hard. Let’s see if I got that right. Yeah, that looks right. OK, so that is the

closed form expression for that sum, which we can

get from the perturbation method and the fact

that we’d already done the geometric series. There’s another way to

compute these kinds of sums, which I want to show you,

because it can be useful. So we’re going to do the same

sum and derive the formula a different way. This method is called

the derivative method, and the idea is to start

with a geometric series which it’s close to and then

take a derivative. So for x not equal to 1, we

already know from the theorem that i equals 0 to n, x to the

i equals 1 minus x to the n plus 1 over 1 minus x. That was the theorem. We already know that. Now, I can take the

derivative of both sides, and let’s see what we get

by taking the derivative. Well, here I get the sum. The derivative of x to the

i is just i times x to the i minus 1. The derivative over here

is a little messier. I’ve got to have–

well, I take 1 minus x times a derivative of that. Derivative of this is now

minus n plus 1, x to the n. Then I take this times

the derivative of that is now minus minus 1, 1

minus x to the n plus 1, and then I divide

by that squared. Now, when we compute all that

out, we get this, 1 minus n plus 1, x to the n plus

nx to the n plus 1 over 1 minus x squared. I won’t drag you

through the algebra, but it’s not hard to

go from there to there. This is pretty close

to what we wanted. We’re trying to figure out

this, and we almost got there. What do I do to finish it up? AUDIENCE: Multiply by x. PROFESSOR: Multiply by x. Good. So if I take this and multiply

by x, i equals zero to n, ix to the i equals x minus

n plus 1, x to the n plus 1, plus nx to the n plus 2,

all over 1 minus x squared. OK? Which should be the same–

yeah– the same formula we had up there. So that’s called the

derivative method. You can start manipulating–

you treat these things as polynomials– these sums–

and you start manipulating them like you would polynomials. In fact, there’s a whole

branch of mathematics called generating functions

that we won’t have time to do in this class that’s

in chapter 12 of the text. But you do things

like that to get sums. Any questions about

what we did there? You can also do a version where

you take integrals of this if you want, and

then you get the i’s in the denominator instead

of those coefficients. For homework, I

think we’ve given you the sum of i squared x to the i. How do you think you’re

going to do that? Any thoughts about how

you’re going to solve that? Get the sum, a closed

form for the sum of i squared x to the i? AUDIENCE: Do the

derivative method twice. PROFESSOR: Yeah, do it twice. Take this, which now you know. Take the derivative again. Won’t be too hard. You can also take the version of

this where n goes to infinity. Let’s do that. If the absolute value of x

is less than 1, the sum i equals 1 to infinity of ix to

the i, what does that equal? This one, you can see

it easier up here. What happens when

n goes to infinity? What does this do? X is less than 1,

an absolute value. What happens to this

as n goes to infinity? This term. Goes to 0, right? This gets big, but this

gets smaller faster. What happens to this term

as n goes to infinity? Same thing, 0. All I’m left with is x

over 1 minus x squared. Now, this formula

is useful if you’re trying to, say, get

the value of a company, and the company is growing. Every year the company grows

its bottom line by m dollars. So the first year, the

company generates m dollars, the next year it generates

two m dollars in profit, the next year is

three m dollars. So you’ve got an

entity that every year is growing by a fixed amount. It’s not doubling every

year, but every year adds in m dollars

more of profit. What would you pay

to buy that company? What is that worth? So you can think of this

as, again, an annuity. Here the annuity pays im

dollars, in this case, at the end, not the beginning,

say, of the year i forever. This company is– or this

annuity– is worth, well, we just plug into the formula. Instead of $1 each year, it’s

m so there’s an m out front. x is 1 over 1 plus p, and

then we have 1 minus 1 over 1 plus p squared. And if we multiply the top and

bottom by 1 plus p squared, we get m 1 plus

p over p squared. So it’s possible with

a very simple formula to figure out how

much you should spend to buy this company,

what its value is today. So, for example, say the company

was adding $50,000 a year in profit. The interest rate was 6%,

the value of this company is $14 million– $14.7

million, just plugging into that formula. So people that buy

companies and stuff, they use formulas like this

to figure out what it’s worth. Of course, you’ve

got to make sure it’s really going to keep paying

the $50,000 more every year and that this is the

right interest rate to be thinking about. You know, and the

guys on Wall Street– the bankers on Wall

Street– they all have their estimations for

what these things are– the value of p they would

put into these formulas. Any questions about that? Yeah. AUDIENCE: [INAUDIBLE] PROFESSOR: OK. Good. So this one is OK? OK. I plugged x equals

1 over 1 plus p, like we did before–

remember for the annuity– because every year

you’re degrading it, devaluing by 1 over 1 plus p. So that’s the x term, and it’s

paying– in the i-th year, it’s paying im dollars. All right? So the first year it pays n

dollars, the next year 2 m, the next year 3 m,

the next year 4 m, but every year you’re

knocking it down by 1 plus 1 over p to the number of years. So what you get– the sum

you’ve really got here– is i equals 1 to infinity,

im dollars are paid, but those dollars are

worth 1 over 1 plus p to the i today,

the current value. That’s a good question. I should’ve said that. That’s a great question. So that’s how we

connected this up, because you’re getting paid

this much in our years, and that’s worth

that much degradation or that much devaluation

today, and now we add up a total current value. So even a company that is

paying you more and more every year still has a finite

value, because the extra– the payments are increasing

but only linearly. The value today is

decreasing geometrically, and the geometric decrease

wipes out the value of the company in the future. Yeah. AUDIENCE: Are you [? squaring ?]

quantity of [INAUDIBLE]. PROFESSOR: What did I do? Oh, wait, wait, wait. I screwed up here too. Is that what you’re

asking about? Yeah. That’s what I should

have done, right? Because I got 1 minus x is

1 minus 1 over 1 plus p. That gets squared, and now when

I multiply 1 plus p squared, it’s multiplying

this by 1 plus p. It’s 1 plus p minus 1 is p. So I have p squared. All right, so this part’s OK. That part I wrote wrong. That’s good. Any other questions? So let’s do a simple example. What is this sum? A 1/2 plus 2/4 plus

3/8 plus 4/16 forever. What’s that sum equal? You can plug that

in the formula. AUDIENCE: [INAUDIBLE] PROFESSOR: Yeah. That’s right. Good. These details. Otherwise that

sum would be what? If I didn’t put

the negative here, it’s going to be infinity,

and that’s not so interesting. The negative makes it more

interesting, so I got 1 over 2 to the i in there. What’s it worth then? Well, I can plug in the formula. What’s x? One half. So I get 1/2 over 1 minus

1/2 squared is 1/2 over 1/4, and that’s 2. Any questions on that formula? It’s amazing how useful

these things get to be later. So that’s sort of the

geometric kinds of things. Next I want to talk about more

of the arithmetic kinds of sums and what you do there. In fact, we’ve already

seen one that we’ve done. If I sum i equals 1 to n of

i– I think we’ve already done this one– that’s just

n times n plus 1 over 2, and probably most

of you even learned that formula back

in middle school, I’m guessing– maybe before. How many people know

the answer for this sum? The sum of the squares–

the first n squares. Somebody knows. What is it? AUDIENCE: n times n plus 1

times 2n plus 1, all over 6. PROFESSOR: Very good. That is correct. Most people don’t

remember that one. It’s a little harder to derive. How would you prove

this by induction? Unfortunately, induction

doesn’t tell you how to remember what

the formula was, and there’s a couple of

ways you can go about that. One is, you can

remember or guess that the answer is

a polynomial in n. In fact, because

you’re summing squares, you might guess that it’s

a cubic polynomial in n, and if you remember just that

or guess just that, then you could actually plug in

values and get the answer. And this is– you

know, a common method of solving these sums

is you sort of guess the form of the solution. In this case you might

guess that for all n, the sum i equals 1 to n of

i squared equals a cubic. And then what you would do is

plug in the value n equals 1, n equals 2, maybe

even– we’ll make it n equals 0– make

it simple and start getting some constraints

on the coefficients. If you would plug in n

equals 0, the sum is 0. The polynomial evaluates to d. That tells you what d has got

to be right away. n equals 1. The sum is 1, and when you

plug into the polynomial, you get a plus b plus c plus d. n equals 2. Well, that’s 1 plus

4 is 5, 2 cubed is 8, so you have 8a plus

4b plus 2c plus d, and you’ll need one more since

you’ve got four variables. Let’s see, 1 plus

4 plus 9 is 14. I’ve now got 3 cubed is

27a plus 9b plus 3c plus d. So now I’ve got four

equations and four variables and, with any luck, I can

solve that system of equations and get the answer. And, in fact, you can. When you solve this system, you

get a equals 1/3, b equals 1/2, c equals 1/6, and d equals 0. And that’s exactly what

you get in that formula. So that’s a way to reproduce

the formula if you forgot it. Now, this method–

really to be sure you got the right answer–

you’ve got to go prove it by induction, because I

derived the answer– if it was a polynomial, I would

have gotten it right, but I might be

wrong in my guess. And to make sure

your guess is right, you’ve got to go back

and use induction to prove it for this approach. Yeah. AUDIENCE: How do you know that

it would be [INAUDIBLE] and not some higher power? PROFESSOR: Well, it turns

out that anytime you’re summing powers, the answer

is a polynomial to one higher degree. So if you just remembered that

fact, or you guessed that fact. Another way to sort of

imagine that might be true is that I’m getting

n of them, so I might be multiplying–

the top one is n squared, so I’m going to have n

of them about n squared. Might be something like n cubed. That’s another way you could

think of it, to guess that. Any other questions on this? So far, all these sums had nice

closed forms, and a lot of them do that you’ll encounter

later on, but not all, and sometimes you get sums that

don’t have a nice closed form– at least nobody has

ever figured out one, and probably doesn’t

always exist. For example, what if I

want to sum the first n square roots of integers? Let’s write that down. So say I want a closed

form for this guy. Nobody knows an answer

for that, but there are ways of getting very

good, close bounds on it that are closed form, and

these are very important, and we’re going to use

this the rest of today and the rest of next time. And they’re based on replacing

the sum with an integral, and the integral is very

close to the right answer, and then we can see what

the error terms are. So let’s first look

at the case when we’ve got a sum where the

terms are increasing as i grows, and we’ll call

these integration bounds, and a general sum will look

like this– i equals 1 to n of f of i, and the first case is

when f is a positive increasing function, increasing in i. Integration bounds, and so

we’re increasing function. So let me draw a picture

that will hopefully make the bounds that we’re

going to get pretty easy. So let’s draw the

sum here as follows. I’ve got 0, 1, 2, 3, n

minus 2, n minus 1, n, and draw the values of f here. Here’s f of 1, f of 2–

it’s increasing– f of 3, f of n minus 1, and f of n. Then I’ll draw the

rectangles here. So this has area of f of

1, this has area f of 2, this has area f of 3,

and we keep on going. Let’s see, this will

be f of n minus 2 on this one– I’ll just do f of

n minus 1, draw this guy here. So its unit width, its

height is f of n minus 1, so its area is f of n

minus 1, and then f of n. And let me also–

so the sum of f of i is the areas in the rectangles. That’s what the

sum is, and I want to get bounds on this

sum using the integral, because integrals are

easier to compute. So let’s draw the function

f of x from 1 to n. All right, so this is

f of x as a function. Now I claim that the sum

i equals 1 to n of f of i is at least f of 1 plus

the integral from 1 to n, f of x, dx. Now, the integral

from 1 to n of f of x is this stuff, the

stuff under the curve. It comes down here,

starts at 1, and it’s the stuff under the curve. And what I’m saying

here is that if you take that stuff under

the curve and add f of 1, which is

this piece, that’s a lower bound on our sum. The sum’s bigger than that. So what I’m saying is the

area in the rectangles is at least as big as the

area in the first rectangle plus the area under the curve. Does everybody see why that is? I’m saying the sum is the

area in the rectangles, right? That’s pretty clear. And that is at least as big

as the first rectangle f of 1 plus the stuff under the

curve, which is the integral, and I’ve left– I’ve

chopped off these guys. That’s extra. OK? Is that all right? So lower bound. Any questions on

the lower bound? This is a picture proof, which

we always tell you not to do, but we’re going to do one here. And, of course, it totally hides

why did I need f is increasing, but we’ll see that in a minute. The proof would not work

unless it is increasing here. Any questions,

because now going I’m going to do the other

bound, the other side. I also claim– this will be

a little trickier to see– that the sum is at most f of n

plus the integral from 1 to n. So this is the lower

bound add in f of 1, the upper bound

just add in f of n. So let’s see why that’s true. Now, to see that,

this is– I’m not going to be able to draw it. I want you to imagine taking

this curve and the area under it, down to here, and

sliding it left one unit. sliding it left to here, sliding

it left one unit over to here. Now, when I slide it left one

unit, did the area under it change? No. It’s the same area

under it, just where it sits on the picture

is now out here. It’s this area under this

guy, but it’s the same thing, its the same integral. And you can see that

it’s more than what’s in these rectangles, because

I got all this stuff. And, of course, I didn’t

even include this, so now I add the f n. So if I take the area under the

curve, which is the integral, shift it left one,

so it only goes up to here now, and then add in

this rectangle, that dominates the area in the rectangles. Bigger than. Do you see that? I could do a lot of

equations on the board but, for sure, that would

be hopeless to follow. Any questions about this? Yeah. AUDIENCE: I guess I understand

the lower amounts because we’re cutting off the triangles. PROFESSOR: Yeah. But is there– is it

a lot of hand waving, or am I just missing something,

that it’s always going to be the f of n is what we– PROFESSOR: Yeah. There’s a little

hand waving going on, but I do believe it is true. With equations you

can make it precise. Let’s look at it again,

do this one more time. AUDIENCE: [INAUDIBLE] PROFESSOR: Yeah, I’m

waving hands a little bit, but it also– hopefully

the intuition comes across, because when I

shift left by one, this point becomes this

point, and that point becomes that point, and

I’m catching sort of the cusp of these rectangles,

and now I take this curve here, shifting the whole

curve left one unit, and the area doesn’t change. It would be equivalent of

taking the integral from 0 to n minus 1. You notice what I’m

doing– of f of x plus 1. There’s another way to

look at it mathematically, maybe– probably the

formal way to do it– but the idea is that

this now contains all these first n

minus 1 rectangles, are contained underneath it. And then I just add this

in, and I’m good to go. I’ve contained all

the rectangles now for the upper bounds. All right,

mathematically what this is, this curve is

0 to n minus 1, fx plus 1, which is the

same thing as what that is. Any questions for that? But now I have good

bounds on the sum. We know that the sum is at

least this and at most that, and those two bounds only differ

by a single term in the sum, so they’re very close. These actually are good

formulas to write down for your crib sheet. That is worth doing on the test,

and we’ll use them more today, and we’ll also use

them on Thursday. So now we can actually

get close bounds on the sum of the square roots. So let’s see how

this works for f of i equal to the square root

of i, which is increasing. So I compute the integral from

1 to n of square root of x dx, and that’s just x to the 3/2

over 3/2, evaluated at n and 1, and that just equals 2/3

n to the 3/2 minus 1. And now I can compute

the bounds on the sum of the first n square roots. So I know that square root

i equals 1 to n– well, the upper bound is f

n plus the integral, and the lower bound is f

of 1 plus the integral. What is f of n? That’s the square root of n. What is f of 1? The square root

of 1, which is 1. So I’ll plug that in here. I get 2/3 n to the 3/2 plus

1 minus 2/3 is plus 1/3, and here I get 2/3 n to the

3/2 plus the square root of n minus 2/3. So now I have pretty good

bounds on the sum of the first n square roots using this method. So for example,

take n equals 100. This number evaluates to 667. This number evaluates to

676, and the difference is 9, which is the square

root of 100 minus 1. This is the square

root of n, this is 1, so the gap here, square root

of n minus 1, and n equals 100. Square root of 100 minus

1 is 9, so I shouldn’t be surprised my gap here is nine. So I didn’t get exactly

the right answer, but I’m pretty close here. Now, as n grows, what happens to

the gap between the upper bound and the lower bound? What does it do? It gets bigger. So my gap gets bigger. That’s not so nice. Doesn’t always stay 9

forever, but somehow though this is still pretty

good, because the gap grows but the gap only grows

as square root of n, where the answer– the bounds–

are growing as n to the 3/2. In other words, my error

is somewhere around here, and that gets smaller

compared to my answer, which is somewhere around there. And there’s a special

notation that people use. In fact, let’s write that

down and then do the notation. Another way of writing this is

that the sum i equals 1 to n of square root of i. The leading term here

is 2/3 n to the 3/2, and then there’s some error

term– delta term here. We’ll call that

delta n, and we know that the error term is at

least 1/3 and, at most, the square root of n minus 2/3. So this delta term is bound

by the square root of n. That’s getting

bigger as n gets big, but this value compared to

your answer is getting small. That’s nice, and so the way that

gets represented is as follows. We would say– it’s using

that tilde notation. We write tilde 2/3

times n to the 3/2, and now we’ve gotten

rid of the delta, because this tilde is

telling us that everything else out here gets small

compared to this as n gets big. And the formal definition–

Let’s write out the formal definition for it. Now, a lot of times

you’ll see people use this symbol to mean about. That’s informal. When I’m using it here,

it’s a very formal meaning mathematically. A function g of x is

tilde, a function h of x means that the limit as x goes

to infinity of g over h is 1. In other words, that

as x goes to infinity– as x gets big– the ratio

of these guys becomes 1. And let’s see if

that’s true here. Well, square root

of i equals this, so I need to show the limit

of this over that is 1. So let’s check that. A limit as n goes to

infinity of 2/3 n to the 3/2 plus that delta term

over 2/3 n to the 3/2. Well, that equals– divide out

by 2/3 n to the 3/2, I get a 1. Did I– should I

have subtracted? No, that’s OK. One plus delta n over

2/3 n to the 3/2. If I can pull the 1 out front,

that’s 1 plus the limit, and this is now delta n is

at most square root of n, so I get square root of

n over 2/3 n to the 3/2. Square root of n over n to

the 3/2, that goes to 0. So this equals 1. So this limit is 1,

and so therefore I can say that the sum of

the first n square roots is tilde 2/3 n to the 3/2. Any questions about this? We’re going to do a lot of this

kind of notation next time. Yeah. AUDIENCE: When you

took the integral and got 2/3 into

the 3/2 minus 1, why did the minus 1 not become

part of the actual solution and become part of the delta? PROFESSOR: OK. So you brought it

up to here, right? OK. So then I plug that

into the integral that appears on both sides here,

and here I add the f 1, here I add f n, and now I

have the lower bound here and the upper bound here. Are you good with those? All right. Now some judgment takes

place, and what I’m really trying to do here is figure out

what are the important terms in these bounds as n gets big? How big is this

growing as n gets big? Well, as n gets big, the

1/3 is not doing much. As n gets big, the

square root of n grows, but it’s nothing

like what’s happening here. If you had to

describe to somebody what’s going on in this bound,

would you start here or here? No. You’d start here. This is the action, and there’s

a little bit– the rest is just in the slop, in the air. And so now I’ve used judgement

to say that delta is somewhere in this stuff here. What’s really happening, and

the nice thing is they match. The lower bound and the upper

bound match on that term. So what I do is I write

it equals this term plus something that’s

smaller and, in particular, it’s between 1/3 and

square root of n minus 2/3. All right. So what I’m trying

to capture here is just the guts of what’s

happening to this function as n grows, and the

guts of it is this. It’s not exactly equal to

2/3 times n to the 3/2, but it’s close

and, in fact, if I take the limit of this over

that, that limit goes to 1. It’s a way of saying they’re

approximately the same that’s called asymptotically

the same, and we’ll talk a lot more about

asymptotic notation next time. We’ll give you five more symbols

besides tilde that people use. Any questions about– maybe

start with the bounds. Any question on the

bounds that we got? That’s the integration method,

first getting the bounds. You take the integral, you add

f of 1 for the lower bound, you add f of n for

the upper bound, and it’s in between,

somewhere in there. Questions there? All right. Then we plugged

it in, and now we look at this tilde notation

that says– well, first we’d write it like this. The sum is this value plus an

error term and, lo and behold, that error term is small. If I take the limit of the whole

thing divided by the big term, I get 1, which means this

thing is really not important. So I write this. Questions on that? All right. There’s one more

case to consider, and now we’re going to go back

to the integration bounds, and that is when f is

a decreasing function, and we’re going to do

the analysis for that, and then we’ll be all done. So we’re going to look

at integration bounds when f is decreasing

and positive still. The example here might

be, for example, the sum i equals 1 to n, 1 over

the square root of i. Say you had to get some idea

of how fast that function is growing as a function of n. I’m summing the first n

inverse as the square roots. What is that

roughly going to be? How fast does that grow

as a function of n? So let’s do that. And, of course, 1

over square root of i decreases as i gets bigger. So let’s do the general picture

again and see what happens. So we have 0, 1, 2, 3, n minus

2, n minus 1, n, and now f of n is the small term and f of n

minus 1, f of 3 here, f of 1. Now, I’m going to

draw the rectangle. This has area f of 1. This one has area f of 2. This one has area f of

3, and then this one has area f of n minus 1 here,

and then, lastly, area f of n. So the sum is the area in the

rectangles, just like before, except now the rectangles

are getting smaller. Let’s draw the integral

like we did before. The integral is the area

under this curve, f of x here, just like before. So this is f of x,

only it’s decreasing. Now, let’s take the area under

this curve, and add f of 1 to it. If I take the area under

this curve, all the way down to here and then add

f of 1, what do I get? Upper bound on my sum. OK. So the sum i equals

1 to n f of i is upper bounded by that

guy, which is f of 1, plus my integral, which is

the area under the curve. The integral is the area under

that curve, starting here, and that contains

all these rectangles, and then I just add in f

of 1 to get an upper bound. Now, for the lower bound,

think about shifting the whole curve left by one. That goes to there,

this goes to here, that goes to here, that goes to

there, and that goes to there. The area under the

curve did not change when I shifted it left by one. This is now my area

under the curve. Stops here. What do I get when I take that

area and add in this last box, f of n? A lower bound,

because it’s contained in all the rectangles. Now, what’s really weird

about these formulas, do they look familiar? Yeah. Yeah. I switched them. Yeah. They’re the same formulas

we had over here, except we switched the direction

on the less than and greater than signs. Well, I swapped f of 1

and f of n, however you want to think about it. The lower bound

here in that case became the upper

bound in this case. Is that possible that the lower

bound became the upper bound? Yeah. Yeah, because what

really happened here– which is the big term

in this case, fn or f1? f1 is the big term

because it’s decreasing, so it’s totally symmetric. All right? The proof was very

similar, so the nice thing is you’ve only got to remember

the bounds are now simple for any sum as long as an

increasing or decreasing, it’s the same as the integral. The lower bound is the smaller

of the first and last term, and the upper bounds are larger

of the first and last term. Very easy to remember. Probably don’t even

need the crib sheet for it, although to be safe,

want to write that down. So now it’s easy to compute

good bounds on the sum of the inverse square roots. Any questions there

before I go do it? So let’s take the case

where we’re summing 1 over square root of i. So we compute the integral of

1 over square root of x dx. That equals the square root of x

over 1/2, evaluated at n and 1. That equals 2 square

root of n minus 1, or two square root of n minus

2, and now we can bound the sum. The upper bound is f

of 1 plus the integral. The lower bound is f

of n plus the integral. What is f of 1? One? One over the square

root of 1 is 1. What is f of n? One over the square root of n. Small. So these bounds are

pretty close here, right? In fact, this gets really

tiny as n gets big, so I’m just going

to replace this with 2 square root of n minus

2 and make it a strict lower bound, and this–

cancel there– I get 2 square root of n minus 1. Wow, these bounds are great. They’re within one for all n. That’s really good. So we can rewrite this in

terms of what really matters. What really matters

in these bounds? How fast is this

function growing? AUDIENCE: 2 root n. PROFESSOR: 2 root n. That’s what really matters,

so let’s write that down. So this says that the

sum i equals 1 to n of 1 over square root i

equals 2 square root of n– we have a minus delta n, where

delta is between 1 and 2. And so if I use the tilde

notation, what would I write down here for the tilde? Past the tilde? I don’t want to mess– AUDIENCE: Tilde. PROFESSOR: –I don’t

want to keep track of all the delta stuff as n gets big. AUDIENCE: 2 root n. PROFESSOR: 2 root

n, because this term over that goes to 0 as

n gets large, so let’s just check that. So we take the limit as n

goes to infinity of 2 root n minus delta n over 2 root n. I’m just checking

the definition now. That’s what the

definition would be. Equals 1 minus the

limit as n goes to infinity of 2 over 2 root n. This is 0. So it equals 1. And so now you know that the

sum of the first n inverse square roots grows as 2 root

n, which is the integral. Yeah. AUDIENCE: [INAUDIBLE] dropped

off the lower bound that f of n was 1 over root n? PROFESSOR: Yeah, I dropped it

off, because it was so tiny and going to zero, I just

made a strict less than. In fact, yes. I don’t hurt myself

by dropping it off. In fact, the lower

bound was a little bit– I made a little

weaker lower bound. So this is still true. I just– it wasn’t as

tight as it used to be, so I could keep it around. Yeah, it doesn’t hurt

to keep it around, then it’s a less

than or equal there. And now this would be

something like that. So I could keep it around,

but I’m going to get rid of it anyway, because I’m going

to go to the tilde notation, and as n gets big,

this is really tiny. So in this case, the

bounds are great. You can nail it

pretty much right on. Yeah. AUDIENCE: You said one

over n still there, the number is bigger than

it would normally be, so when you take it

out, it becomes smaller, so how could you go to

a less than [INAUDIBLE]? PROFESSOR: Well, you’re

saying you don’t like the fact I dropped it here? AUDIENCE: Yes. When you drop it, why do you

go to a less than instead of [INAUDIBLE]? PROFESSOR: Oh, because I’ve got

a bigger bound that I made less when I dropped it. I took something away, so I

know I could never equal this, because I know it’s

bigger than this. I know that the real answer

has to be at least this big, and so it has to be bigger

than something smaller. That’s why I did it. Any other questions? We’ll get more practice tomorrow

and next time with this stuff.