Russell, Ch.39 of Principles of Mathematics, ‘The Infinitesimal Calculus’, summary notes

 

by Corry Shores
[Search Blog Here. Index-tags are found on the bottom of the left column.]

[Central Entry Directory]

[Bertrand Russell, entry directory]

[Other entries in the Russell Principles of Mathematics, series]

[The following is summary and quotation. All boldface, underlining, and bracketed commentary are mine. Please see the original text, as I did not follow it closely. Proofreading is incomplete, so mistakes are still present.]

 

 

Bertrand Russell

Principles of Mathematics

Part 5: Infinity and Continuity

Ch.39: The Infinitesimal Calculus

Brief Summary

Leibniz’ infinitesimal calculus used a concept of the infinitely small (the infinitesimal quantity). Russell explains how differential and integral calculus now function with the concept of limit rather than the concept of infinitesimal.

 

Summary

 

§303

‘Infinitesimal calculus’ refers to differential and integral calculus, however “there is no allusion to, or implication of, the infinitesimal in any part of this branch of mathematics.” [330]

Leibniz was its inventor, but he considered it to be more practically applicably than metaphysically truthful. “He appears to have held that, if metaphysical subtleties are left aside, the Calculus is only approximate, but is justified practically by the fact that the errors to which it gives rise are less than those of observation”. [330]

But because Leibniz believed in the actual infinitesimal, he was unable to see that calculus rests on the doctrine of limits. Newton’s fluxions are closer to this truer foundation.

When he was thinking of Dynamics, his belief in the actual infinitesimal hindered him from discovering that the Calculus rests on the doctrine of limits, and made him regard his dx and dy as neither zero, nor finite, nor mathematical fictions, but as really representing the units to which, in his philosophy, infinite division was supposed to lead. And in his mathematical expositions of the subject, he avoided giving careful proofs, contenting himself with the enumeration of rules. At other times, it is true, he definitely rejects infinitesimals as philosophically valid; but he failed to show how, without the use of infinitesimals, the results obtained by means of the Calculus could yet be exact, and | not approximate. In this respect, Newton is preferable to Leibniz: his Lemmas give the true foundation of the Calculus in the doctrine of limits, and, assuming the continuity of space and time in Cantor’s sense, they give valid proofs of its rules so far as spatio-temporal magnitudes are concerned. [330-331]

Leibniz’ error has misled philosophers and mathematicians from his time to Weierstrass.

it is at any rate certain that, in his first published account of the Calculus, he defined the differential coefficient by means of the tangent to a curve. And by his emphasis on the infinitesimal, he gave a wrong direction to speculation as to the Calculus, which misled all mathematicians before Weierstrass (with the exception, perhaps, of De Morgan), and all philosophers down to the present day. It is only in the last thirty or forty years that mathematicians have provided the requisite mathematical foundations for a philosophy of the Calculus. [331]

§304

“The differential coefficient depends essentially upon the notion of a continuous function of a continuous variable”. [331]

In §254 we noted that a function relates the elements of one set to those of another, usually in an order-preserving way. [Also see Edwards and Penney’s account of functions here.]
And in §277 we defined a continuum as  a dense series of terms whose values are all definable by means of limits contained with in it. [Russell says we examined continuous variables in this chapter. There is no use of this term, just continuum and continuous series, which might therefore be equivalent.] If the function is one-valued and ordered correlatively with a continuous variable, then the function is continuous. [332d]. But if the function has an order indepent of correlation, it could possible be that the series obtained in its correlation is not continuous. When the correlation does produce a continuous series in some interval, then the function is continuous in that interval. [From the informal and formal definitions, it seems that a function is continuous at a point when the limits to either side of that point have equal value. If we were thinking in infinitesimal terms, if we were to move infinitesimally to the right or left of the point, it would be the same value (except for the infinitesimal, inassignable difference). But it is discontinuous when the side-limits are different. As  Mr. Flatcher writes “The graph of a continuous function has no holes, jumps, or gaps. Think of a continuous function as one that you can graph without ever lifting your pencil.” (Mr. Flatcher / Flatchermatics) Consider for example this function.

At x = 1, the limits on either of its sides should be 0. However, as we can see, the y values are much different from zero.

(Image and function from ‘Classification of Discontinuities’, wikipedia)

As you can see, at the limit right before x = 1, the y value is greater than zero, and at the limit right after, the y value is even greater than that. Below we have an animation showing a transition from continuity to discontinuity. The caption reads: “A sequence of continuous functions f_n(x) whose (pointwise) limit function f(x) is discontinuous. The convergence is not uniform.”

(Animated diagram and above caption from ‘Continuous function’, wikipedia)

]

If the function is one-valued, and is only ordered by correlation with the variable, then, when the variable is continuous, there is no sense in asking whether the function is continuous; for such a series by correlation is always ordinally similar to its | prototype. But when, as where the variable and the field of the function are both classes of numbers, the function has an order independent of correlation, it may or may not happen that the values of the function, in the order obtained by correlation, form a continuous series in the independent order. When they do so in any interval, the function is said to be continuous in that interval. The precise definitions of continuous and discontinuous functions, where both x and f(x) are numerical, are given by Dini as follows. The independent variable x is considered to consist of the real numbers, or of all the real numbers in a certain interval; f(x), in the interval considered, is to be one-valued, even at the end-points of the interval, and is to be also composed of real numbers. We then have the following definitions, the function being defined for the interval between α and β, and ɑ being some real number in this interval.

“We call f(x) continuous for x = ɑ, or in the point ɑ, in which it has the value f(ɑ), if for every positive number σ, different from 0, but as small as we please, there exists a positive number ε, different from 0, such that, for all values of δ which are numerically less than ε, the difference f(ɑ + δ) − f(ɑ) is numerically less than σ. In other words, f(x) is continuous in the point x = ɑ, where it has the value f(ɑ), if the limit of its values to the right and left of a is the same, and equal to f(ɑ).”

“Again, f(x) is discontinuous for x = ɑ, if, for any positive value of σ, there is no corresponding positive value of ε such that, for all values of δ which are numerically less than ε, f(ɑ + δ) − f(ɑ) is always less than σ; in other words, f(x) is discontinuous for x = ɑ, when the values f(a + h) of f(x) to the right of a, and the values f(ɑ − h) of f(x) to the left of ɑ, the one and the other, have no determinate limits, or, if they have such, these are different on the two sides of ɑ; or, if they are the same, they differ from the value f(ɑ), which the function has in the point ɑ.” [331-332]

But the limit of a function is slightly different than the limit in general (of series) that we have discussed so far. Russell defines the limit in this way. [Put in simple terms, it seems that the limit is the value most immediate to the point, were we using infinitesimal terms.]

A function of a perfectly general kind will have no limit as it approaches any given point. In order that it should have a limit as x approaches a from the left, it is necessary and sufficient that, if any number ε be mentioned, any two values of f(x), when x is sufficiently near to a, but | less than a, will differ by less than ε; in popular language, the value of the function does not make any sudden jumps as x approaches a from the left. Under similar circumstances, f(x) will have a limit as it approaches a from the right. But these two limits, even when both exist, need not be equal either to each other or to f(ɑ), the value of the function when x = ɑ. The precise condition for a determinate finite limit may be thus stated:

“In order that the values of y to the right or left of a finite number a (for instance to the right) should have a determinate and finite limit, it is necessary and sufficient that, for every arbitrarily small positive number σ, there should be a positive number ε, such that the difference y_{ɑ + ε} − y_{ɑ+ δ} between the value y_{ɑ + ε} of y for x = a + ε, and the value yɑ + δ, which corresponds to the value a + δ of x, should be numerically less than σ, for every δ which is greater than 0 and less than ε.” It is possible, instead of thus defining the limit of a function, and then discussing whether it exists, to define generally a whole class of limits. In this method, a number z belongs to the class of limits of y for x = ɑ, if, within any interval containing ɑ, however small, y will approach nearer to z than by any given difference. Thus, for example, sin 1/x, as x approaches zero, will take every value from −1 to +1 (both inclusive) in every finite interval containing zero, however small. Thus the interval from −1 to +1 forms, in this case, the class of limits for x = 0. This method has the advantage that the class of limits always exists. It is then easy to define the limit as the only member of the class of limits, in case this class should happen to have only one member. This method seems at once simpler and more general.
[332-333]

§305

Russell will now discuss the derivative or differential coefficient of the function. [To better grasp Russell’s example, we will draw from our summary of one of David Jerrison’s class lectures on the differential. We will find an equivalent formulation to Russell’s so that we can make what he is saying more concrete. So first let’s understand the formulation. Consider a curve with point P.

The line has a different slope (tendency of variation) at each point. We will ask, what is its slope at x₀, or point P? We determine the y value on the basis of the function f(x). And since P = (x,y), then P = (x0,(fx0)). Our calculation will involve looking at the change in x to the change in y (or change in f we might say).

The slope is found at the limit as Δx goes to zero.

We see the coordinates given here:

Slope is rise-over-run, or Δy / Δx. So

m = (y₂ – y₁) / (x₂ – x₁)

or in our case,

m = (f₂ – f₁) / (x₂ – x₁)

Notice in the above diagram that the y values in P and Q are: f(x₀) and (f(x₀ + Δx). For (x₂ – x₁) we only need Δx. So if we substitute these values into the slope formula, we have:

(f(x0 + Δx) – f(x0)) / Δx

The derivative we will denote as f’(x₀). Thus

We call that formulation “the difference quotient”.

Now let’s use a specific function.

f(x) = 1 / x

We want to find the derivative at x₀, and the dotted line is the tangent whose slope we seek. So we need to find Δf / Δx. The formula for Δf was (f(x₀ + Δx) – f(x₀)). When we plug in our function, we (multiplicatively) invert the x values, that is, put a one over them, hence we obtain [18]:

When we remove the embedded fractions (by dropping the top couched-denominators to the entire bottom denominator), we get:

As you can see, 1 / Δx is common to both parts. When we factor it out, we get:

The subtracted parts need a common denominator for us to simplify them. To give them both a common denominator, we multiply each by the other’s denominate set over itself (thus equaling 1).

Let’s combine these figures to get:

We can now subtract the terms:

We then distribute the negative in the right side of the numerator to make x₀ – x₀ –Δx, thereby leaving –Δx; hence:

Since both sides share Δx inversely, we can cancel them, leaving us with 1/1 on the left side, which can as well be eliminated, and remaining on the left is:

The last step is to take the limit, as delta tends to zero, and substitute zero for Δx. We can do this now, because before the numerator and denominator gave us number divided by zero, which is undefined. But through algebraic operations, we were able to make the Δx negate-out of the equation without leaving a zero in the denomenator. Thus we now substitute-in the limit, that is, make Δx equal zero, leaving us with:

Let’s put all of this together into one large formulation:

We then compare with our chart. This is negative, and likewise the slope is negative. Also, as x_o goes to infinity, so as x moves to the right, it becomes less steep:

As we will see, Russell uses a very similar formulation, except with δ instead of our Δx]

 

If f(x) be a function which is finite and continuous at the point x, then it may happen that the fraction

{f(x + δ) − f(x)}/δ

has a definite limit as δ approaches to zero. If this does happen, the limit is denoted by f '(x), and is called the derivative or differential of f(x) in the point x. If, that is to say, there be some number z such that, given any number ε however small, if δ be any number less than some number η, but positive, then {f(x ± δ) − f(x)}/ ± δ differs from z by less than ε, then z is the derivative of f(x) in the point x. If the limit in question does not exist, then f(x) has no derivative at the point x. If f(x) be not continuous at this point, the limit does not exist; if f(x) be continuous, the limit may or may not exist.
[334]

§306

[Russell will say that the notion of the infinitesimal was not used in this definition. This probably results from the notion of the limit as standing outside the series, and the values approaching can only ever get closer and closer with no final value. I challenge this view, because it implies the value of the interval between the limit and the series that approaches it is finite. Let’s take this idea that between any two values is a middle value, infinitely. So a series of diminishing values perhaps could be something like 1/1, 1/2, 1/4, 1/8. What is important regarding Cantor’s infinity is that the cardinal value for infinity  α₀ is not among the natural numbers. It is the limit to which their law of genesis implicitly strives toward but does not precisely attain. But so long as the interval is finite, which Russell insists it must be, then does it fulfill the definition for “given any number ε however small”? It seems here the idea is not “the interval is so small it is infinitely small and thus continuous with zero” but rather “the interval is very small but still finite, yet it is close enough to zero that we can substitute one for the other.” Perhaps this is where the term “arbitrarily” small comes from. Is it strange that this ‘fudging’ sort of operation where we exchange a finite value for zero is considered more precise than when we think of this value as infinitely small, especially since in both cases we assume that there are an infinity of subdivisions? If intervals really are infinitely sudividable, why is it so hard to conceive of the intervals between them as being infinitely small? If there were not infinitely small, then they would be finitely small, and an infinity of them would compose an infinitely large interval. But an infinity of infinitely small values could conceivably compose a finite interval (for we would multiply infinity times one over infinity, equaling a finite unit after the infinities cancel.)] Russell emphasizes that the fact he has defined the derivative using limits and not infinitesimals is philosophically the most important part of his treatment on calculus. [The philosophical implication of this might be that the law of continuity does not hold, and thus change is not a matter paradoxically co-given contrary states. Also, change or motion would be ‘at-at’; infinitesimal intervals suggest ‘between-between’ or ‘at-at plus at-at’. I think Russell insists on this philosophical point for logical reasons. The problem with the infinitesimal calculus and its law of continuity is that it is a dialetheia, a true contradiction, and Russell will not allow exceptions to his rigid logic of perfect self-consistency. Because of the advances in dialetheic logic, it is no longer illogical to say that change is inherently paradoxical. And on account of the invention of non-standard analysis, it is no longer conceptually sloppy to use the notion of the infinitesimal in calculus. So here too is equally my greatest point of emphasis and the purpose for all these mathematical technicalities: despite Russell’s insistence, we can have a dialetheic between-between theory of motion, and it will not have such oddities like we find in Russell’s account, such as an infinity of finite intervals composing a finite interval and not an infinite one, and a moving object always being in no state other than rest, just rest in different places at different times.]

The only point which it is important to notice at present is, that there is no implication of the infinitesimal in this definition. The number δ is always finite, and in the definition of the limit there is nothing to imply the contrary. In fact, {f(x + δ) − f(x)}/δ, regarded as a function of δ, is wholly indeterminate when δ = 0. The limit of a function for a given value of the independent variable is, as we have seen, an entirely different notion from its value for the said value of the independent variable, and the two may or may not be the same number. In the present case, the limit may be definite, but the value for δ = 0 can have no meaning. Thus it is the doctrine of limits that underlies the Calculus, and not any pretended use of the infinitesimal. This is the only point of philosophic importance in the present subject, and it is only to elicit this point that I have dragged the reader through so much mathematics. [383]

[In the above, it seems Russell is absolutely clear that the small interval does not equal 0, but it also does not equal an infinitesimally small value. Rather, it equals a finite value that is as small as you want it to be.]

§307

[In Russell’s description of the definite integral, we divide the interval up into n portions. We find their ‘areas’ or products. Then we want the sum of all such interval areas or products. As we increase n, the sum tends toward a definite limit, which gives us the integral sum.  For more on this operation, see David Jerison’s definite integral class or Edwards & Penney’s section on Riemann sums.]

Just as the derivative of a function is the limit of a fraction, so the definite integral is the limit of a sum. The definite integral may be defined as follows: Let f(x) be a function which is one-valued and finite in the interval α to β (both inclusive). Divide this interval into any n portions by means of the | (n − 1) points x₁, x₂, . . . x_n − 1, and denote by δ₁, δ₂, . . . δ_n the n intervals x₁ − α, x₁ − x₂, . . . β − x_{n − 1} . In each of these intervals, δ_s, take any one of the values, say f(ζ_s), which f(x) assumes in this interval, and multiply this value by the interval δ_s. Now form the sum

This sum will always be finite. If now, as n increases, this sum tends to one definite limit, however f(ζs) may be chosen in its interval, and however the intervals be chosen (provided only that all are less than any assigned number for sufficiently great values of n)—then this one limit is called the definite integral of f(x) from α to β . If there is no such limit, f(x) is not integrable from α to β.
[384-385]

§308

 

Russell now explains that neither the concept of the infinitesimal nor of infinity were used in this account of integral calculus. In fact, it is not even a sum [which might be impossible were the terms infinite in number]. Rather, this value is the limit of a sum [to boundary value to which its summing is tending]. But it never reaches that value. [If it did, then that could only be by means of an infinitesimally small increment between the limit and the next value near it. But so long as it never gets there but instead continually gets nearer without arriving, then it is made of many finite units. Imagine a segment that is built by adding 1/2 + 1/4 + 1/8 + etc., it is tending toward the total 1. See the divided square diagram on this page for an illustration (found near the end, under “geometric series that halves each time.”) We don’t need infinitely many to know it is tending to that limit. So we do not need the concepts of infinitesimal and infinity. But if we think that it does actually reach that limit, then it could only do so with an infinity of divisions with the smallest being infinitesimal.]

As in the case of the derivative, there is only one important remark to make about this definition. The definite integral involves neither the infinite nor the infinitesimal, and is itself not a sum, but only and strictly the limit of a sum. All the terms which occur in the sum whose limit is the definite integral are finite, and the sum itself is finite. If we were to suppose the limit actually attained, it is true, the number of intervals would be infinite, and the magnitude of each would be infinitesimal; but in this case, the sum becomes meaningless. Thus the sum must not be regarded as actually attaining its limit. But this is a respect in which series in general agree. Any series which always ascends or always descends and has no last term cannot reach its limit; other infinite series may have a term equal to their limit, but if so, this is a mere accident. The general rule is, that the limit does not belong to the series which it limits; and in the definition of the derivative and the definite integral we have merely another instance of this fact. The so-called infinitesimal calculus, therefore, has nothing to do with the infinitesimal, and has only indirectly to do with the infinite—its connection with the infinite being, that it involves limits, and only infinite series have limits.
[386]

 

 

Sources [unless otherwise notes, all bracket page citations are from]:

Bertrand Russell. Principles of Mathematics. London/New York: Routledge, 2010 [1st published 1903].

 

Otherwise:

Mr. Flatcher. Continuity and Differentiability.
http://fletchmatics.weebly.com/continuity-and-differentiability.html

Wikipedia. ‘Classification of Discontinuities’.
http://en.wikipedia.org/wiki/Classification_of_discontinuities

Wikipedia. ‘Continuous function’.
http://en.wikipedia.org/wiki/Continuous_function

 

 

Jerison. ‘Definite Integrals’, MIT 18.01, class 18

by Corry Shores
[Search Blog Here. Index-tags are found on the bottom of the left column.]

[Central Entry Directory]
[Mathematics, Calculus, Geometry, Entry Directory]
[Calculus Entry Directory]

The following summarizes a class by MIT math professor David Jerison. (Bibliography information given at the end.) Unless otherwise noted, all images are from either screen shots of the video or from the class pdf handout. My own commentary is found between brackets of some kind.

David Jerison

Single Variable Calculus (Course at MIT)

Class 18: Definite Integrals

Brief summary:

The definite integral allows us to calculate the area under curves and the cumulative sums of functions.

Summary

 

This is an introduction to integral calculus, and this lesson’s topic is definite integrals. The idea of definite integrals can be presented in a number of ways, but Jerison will begin with the geometrical interpretation, specifically the area under a curve. Another perspective on the integral is ‘cumulative sum’ .

We will consider a curve under which are boundaries a and b,

and we will be concerned with the area under the curve and within those boundaries.

We also need other information, namely the bottom line, the x-axis, and the curve, defined in terms of the function f(x).

For out notation for this, we have an integral.

But it has limits on it. It starts at a and ends at b.

Then we write in the function f(x)dx.

This is the definite integral. Geometrically it is interpreted as the area under a curve. To compute that area we will follow three steps:

1. Divide region into “rectangles”
2. Add up area of rectangles
3. Take limit as rectangles become thin: “infinitesimally thin, very thin” [00.05.30]

Pictorially, we have a and b, and we have our curve.

We then chop the x-axis into little increments.

Then we chop it up into rectangles with some staircase pattern. “In some cases the rectangles overshoot; in some cases they're underneath. So the new area that I'm adding up is off. It's not quite the same as the area under the curve.”

So our region is here.

But it is missing this part here.

And it includes these extraneous pieces here.

But as the rectangles get thinner and thinner, the small off-parts shrink until they are negligible.

Our first example:

We will make a = 0, but to see the pattern, we will make b be arbitrary.

So we will draw it. Here is the parabola.

The piece under the curve stops at b.

We now divide the area into n pieces. We get to choose whether we want them directed to the right or left, and thus whether we overshoot or undershoot the curve. In this case we choose to direct them to the right.

Now we need to write formulas for these areas. Rectangles are base times height And our intervals are equal. The base length would then be b/n.

So now we make a table to look for patterns

The first place after zero is b/n. It is an x value.

The height is f(x), which is x². So in this case it would be (b/n)².

The next rectangle is 2b/n, and thus its height is (2b/n)².

And thus for the third: 3b/n, and (3b/n)².

Now we need to add up the areas. The area of each is the base times the height of each one, or each row of our column. And we add each together. The final term is nb/n, because there are a total of n rectangles.

The pattern is that we have the same base plus the first height, then plus the second height, with each new height having 2b, 3b, 4b, [+1b] up to nb in the numerator. So the rectangles get taller and taller with the last being the biggest.

Our formula begins complicated. But the limiting value is simple. What we notice is that each term in this series of additions has (b/n)(b/n)² or (b/n)³. So we will factor that out from each of the terms to get:

Now we will take the limit as n goes to infinity. The quantity that is hard to understanding is this massive quantity here.

We will now make a modest, miniscule change. We will write 1 as 1².

Now we will use a trick. It is not entirely recommended, but we will talk about that later. To understand how big this quantity is, we will use a geometric trick. We will draw a picture of the quantity by building a pyramid. The base will be n by n blocks. [This will at first be a bit confusing, because we starting by wanting to know the area under a curve, now we are trying to find the volume of a solid figure. We should first not confuse the slabs in the pyramid that we will build as being equivalent to the rectangles under the curve. Let’s think separately from the curve for a moment. Lets say we have a pyramid whose base is a square with length 5. We are also assuming the base length is the same as the height, for reasons we will see in a second. We can then find the area of this pyramid with the equation 1/3(base)(height) = 1/3(5²)(5) = 1/3(25)(5) = 1/3(125) = 125/3 ≈ 41.67.]

Then the next layer up will be n-1 by n-1.

“So the total number of blocks on the bottom is n squared. That's this rightmost term here.”

We did not write it in before, but let’s now write the next-most last term, n-1². It is the second layer of the pyramid.

Our first drawing was from the top view, but we can also picture it from the side view, with the bottom slab being n units long.

We keep piling up the slabs until we get to the top layer,

which is one giant block of stone, here as 1².

In our staircase pattern, we continue up to this top block.

[[Let’s for a moment note an ancient puzzle, Democritus’ staircase cone. We begin with a cone, and we cut it into slabs that get smaller and smaller until we reach the smallest. If we combine these slabs again, we would expect to obtain a cone. But the problem is determining their sizes. If they were all the same size, we would rebuild it as a cylinder. The sides would be smooth and even, but the shape would be wrong. If they were different size, we would get a staircase shape. The sides would be rough, but the shape would be right. If we take an infinitesimal view, the smallest parts are actually very tiny changes. Combining them will produce a cone with the right shape and with smooth sides.

(Moving diagram by Corry Shores, made with Open Office Draw and Unfreez)

]]

We will now use a trick to estimate the size of this figure; “it's sufficient in the limit as n goes to infinity”. [17.02] We can imagine the solid form underneath the staircase. It is an ordinary pyramid inside, and we know the formula for its volume since we know the volume for cones.

The formula for the volume within that space of the normal pyramid is 1/3(base)(height)

The base is a square of length n, meaning that the base is n². [In our original sequence there are n terms, each of which gets a slab. Apparently we are standardizing n as a unit of length and not just a tally of terms, (later he explains that the height of each slab is 1) and so] the height is n, giving us:

[Now notice that the pyramid lines go under the staircase edges, meaning that the volume of the pyramid will be less than the sum of the staircase.

] We found that whole sum is bigger than 1/3(n³).

The pyramid’s sides have slope 2, meaning you go 1/2 half over each time you go up 1.

We can also trap the staircase on the other side too, by drawing a parallel line  going out 1/2 more on each side, making the base (n+1)(n+1). And it will go up 1 higher.

We then use the same formulation for finding the area of the larger pyramid, and we can place all three into relation:

So we trap that volume between these two quantities:

Now we are ready to take the limit. First recall that we were summing the series of rectangles.

We then factored out (b/n)³.

[We will now distribute (b/n)³ to each of the terms, giving them a common denominator n and common factor b3, which we factor out in the numerator.]

{So the idea seems to be here that previously we excluded (b/n)³ from our pyramid formulation. Now we are adding it back, which will change our inequality. Previously it was:

So now we will divide this inequality by n³ (later we will also add the b³). When we do so, the largest term on the right will be reduced. It begins (n + 1)³ / n³ . Perhaps  then converts to [(n + 1)/n][(n + 1)/n] [(n + 1)/n], with each term also being [(n/n)+(1/n), reducing altogether to [1+(1/n)]³ .} So we now divide the inequality by n³, obtaining:

Now, look at the left-most part of the inequality. As n goes to infinity, we have 1/(infinity) or 0, meaning the left-most part of the inequality is now 1/3.

[Again recall our notation for the integral, here meaning its limits are a and b.

And we said that it is finding the sum for the function f(x)dx between a and b:

It seems now we might read this as the sum of all differentials from a and b, that is, the sum of the areas of all the very tiny rectangles whose x values have diminished arbitrarily or infinitesimally small. Our function was f(x)=x². We divided the curve from a to b into n rectangles with base b/n. Then we used the (base)(height) formula, and placed the sequence into a series of additions. The height, in accordance with the function, is the x value squared, and since the x value is determined by how many bases we have moved from a, the second one for example would be 2(b/n) or 2b/n, and thus its height is (2b/n)², and so on for rest of the series to the nth rectangle.

We then factored out (b/n)³.

We then wanted to find a value which was approximate to the right side of the formulation. We did that by reconceving the sequence of values as being successive slabs in a square pyramid. Under this interpretation, we found values for the volumes of slightly smaller and slightly larger pyramids.

It is not entirely clear how everything is rectified between our original area under the curve and our current formulations and approximations of a pyramid based on the curve’s rectangles’ values. However, we still now bring back the other parts of the original formulation which were factored out, namely, the (b/n)³ . But we began first with just the 1/n³. This gave us now the inequality:

We then made n go to infinity, meaning that in this pyramid conception we are increasing the slabs to infinity, but in the curve-rectangle conception it means we are increasing the number of rectangles to infinity. This means there are an infinity of very tiny rectangles with miniscule areas. But since it is very close to infinity, the right-most part of our inequality tends toward 1/3. So now we add the b3 factor that we had been excluding, and we get 1/3(b³). Hence} after adding b³, which was so far excluded during our other numerical operations, we get:

We will now look at summation notation, to help compress all the complicated elements of our computations.

{Before we continue with Jerison’s account, let’s first briefly review summation notation from Edwards & Penney.

Summation notation provides a concise representation for the sums of numerical sequences, and it takes the form:

The Σ is the Greek letter sigma, which means we are summing a sequence of terms. This whole sequence is signified by the a_i. The i (called the summation variable, summation index, or running index) is the variable part of the terms, and it is substituted firstly with a₁ (which is what the i = 1 means), and it is subsequently substituted with the successive integers. The sequence ends when the i value reaches the n value.

For example:

Here we see that the summation variable is substituted firstly with a₁, and each substitution is squared. The substitutions continue until reaching 10. Then all the terms are added, to produce 385. The variable-letters are arbitrary, so we may note it different ways:

The notation might be labeled thus:

}

Jerison will first give an example:

“So, the general notation is the sum of a_i, i = 1 to n, is = a₁ + a₂ + ... plus a_n. So this is the abbreviation.”

“And this is a capital Sigma.”

[Now recall the value that was difficult to determine and required the complicated procedure.

We will now give the summation notation for it.]

Our previous formulation is equivalent to:

[Shown more specifically:

]

“And so, this quantity here, for instance, is (1 / n³) times the sum i², i = 1 to n. So that's what this thing is equal to.” [We have 1/n³ at the beginning, because the summation is for the terms of the numerator, and that sum in total is placed over the denominator n³. In other words, the series begins with 1. All the terms are squared, and the series ends with n, the nth or final term. All of this is over n³]. What we showed in our calculations was that this tends toward 1/3 as n goes to infinity.

Now recall again that previously we had a very long sum:

We can render this into summation notation thus:

“one way of writing it is, it's the sum from i = 1 to n […]”

“[…] of, now I have to write down the formula for the general term. Which is (b/n)(ib/n)².”

[Again, we originally had:

Here b/n was the same factor in each case. But the coefficient in the numerator of the right-side part of each term in the series increased from 1 to n by increases of 1 each time. The summation notation is saying that this variable begins with 1 in the first term of the series and increases incrementally to n in the final term of the series.]

And we can still factor out the (b/n)³ like before.

[Note first this expression of the distributive property in Edwards and Penney:

]

On account of the distributive law, we factor it out like this:

We will do another example. The function is f(x) = x. If we draw it, it is a line with slope 1.

[The x value is b. Since the slope increases evenly, the height at that point will be b too, and the area will be of a triangle with base b and height b.]

[As the area of a triangle  1/2(b)(h),] the area of this triangle is 1/2(b)(b).

And thus 1/2(b)²

We do not need to do any elaborate summing here, because we already know how to find this area. [Since there are no irregularities of a curve falling outside the regular polygonal area, we do not need to make infinitesimal approximations].

Example 3: f(x) = 1

If we stop it at b, then we are just interested in this area.

0 to b,

We know it is height 1. The area for a rectangle is base times height, so the area here is:

And thus it simply equals b.

We will now look at the pattern of the function and the area under the curve.

In our prior examples, we found that:

To help see the pattern, let’s convert the left side terms to powers of x.

And let’s render the third row, right side, to match the format of those above it:

So let’s now guess for the next higher term [which according to the pattern will of course be b⁴/4]

Jerison then notes how Archimedes was attempting something similar for finding the area of a parabola, but his method was so complicated that it confused mathematicians for a long time and blinded them to this simple pattern. [See the Boyer discussion of Archimedes method and see Archimedes’ ‘Quadrature of the Parabola’]

“That's a reasonable guess, I would say. Now, the strange thing is that in history, Archimedes figured out the area under a parabola. So that was a long time ago. It was after the pyramids. And he used, actually, a much more complicated method than I just described here. And his method, which is just fantastically amazing, was so brilliant that it may have set back mathematics by 2000, years. Because people were so, it was so difficult that people couldn't see this pattern. And couldn't see that, actually, these kinds of calculations are easy. So they couldn't get to the cubic. And even when they got to the cubic, they were struggling with everything else. And it wasn't until calculus fit everything together that people were able to make serious progress on calculating these areas. Even though he was the expert on calculating areas and volumes, for his time. So this is really a great thing that we now can have easy methods of doing it.” [33.10 – 34.15]

Now, we will not need to labor to make pyramids as we did before. But before we get to those simpler methods next class, we will need a little more practice with the notation for definite integrals. We will look at Riemann sums, which provide the general procedure for definite integrals. So we consider this function with limits a and b.

We then break it into increments called Δx.

How many pieces are there? If there are n pieces, then the general formula will be Δx = 1/n times  the total length. So it would be b – a / n.

We will only allow ourselves one bit of flexibility: we will pick any height of f in each interval. We pick any value in between the intervals, call them c_i, and their level is f(ci). That is the rectangle that we choose.

So we pick f(c_i),

and we are going to add them all up.

It is the sum of the rectangles, because f(c_i) is the height and Δx is the base.

This is reminiscent of Leibniz notation for the integral. As Δx is replaced by dx at the limit:

The rectangles get thin as Δx goes to 0.

This is called a Riemann sum. In fact, our first example was a Riemann sum.

Example 4, a non-area example.

This example will show that integrals can be interpreted as cumulative sums.

We will consider a variable t which is time, in years. And we will consider a function f(t), which is dollars per year. And the unit dollars per year is a borrowing rate.

This example will show that there is a good reason for this dx that we append onto the definite integrals.

“It allows us to change variables, it allows this to be consistent with units. And allows us to develop meaningful formulas, which are consistent across the board.” [41.30]

So, we are borrowing money every day. Thus Δt is 1/365 [because change in time is one day.] “ That's almost 1 / infinity, from the point of view of various purposes.” [42.13] These are the time increments during which we borrow. But the rate [of how much you borrow each day] varies, because sometimes you need more money and other times less. So how much did we borrow? So consider day 45, which corresponds to t = 45/365. Our borrowing rate is f(45/365) [the amount of dollars for that fraction of the year. If it were the last day, it would be full amount.] Then we multiply that by change in time.

And Δt here is the change of one day, so

This final amount [right side of equation] is the final amount that we borrow in dollars. [I am not sure why we have day 45 if we are finding the final amount for all days. Perhaps either in this case he means total up to that day, or perhaps we only need to arbitrarily select one in order to figure out the whole.] We can render this into summation notation thus:

This formulation above gives us total amount borrowed. This will be very similar to the integral from 0 to 1 of f(t)dt.

It is equally important to model the amount that we owe the bank at the end of the year. The interest compounds continuously. We start with P as the principle. After time T, we owe Pe^rT, where r is interest rate, at 0.05/yr. [I do not know what the e is.]

To know how much we owe at the year:

We borrowed these amounts here [each day we borrowed a little, adding to the principle]

We owe more. We owe e to the power r [the interest rate] times the amount of time left in the year, which is 1 – i/365. (Or 365 minus i days left).

And since our formulation is e to the rT, we need to substitute it in.

[We need to add this up for each successive day] so we sum it:

This sum is essential the integral from 0 to 1.

[It seems in the following that we substitue t for i/365, and Δt becomes dt.] “The delta t comes out. And you have here e^{r(1 - t)} , so the t is replacing this i/365, f(t)dt”

“That is just an example. See you next time.”

David Jerison. ‘The Definite Integral’ Class 18 of 18.01 Single Variable Calculus. Fall 2006.

http://ocw.mit.edu/courses/mathematics/18-01-single-variable-calculus-fall-2006/

Videos: http://ocw.mit.edu/courses/mathematics/18-01-single-variable-calculus-fall-2006/video-lectures/

Lecture Notes:

http://ocw.mit.edu/courses/mathematics/18-01-single-variable-calculus-fall-2006/lecture-notes/
MIT OpenCourseWare.

Creative Commons Licence

Some summation notation from:

Edwards & Penney: Calculus. New Jersey: Prentice Hall, 2002, p.290b-291d.