Archimedes studied integrals for calculating areas in the 3rd century BC, and Newton and Leibniz did not come up with differential methods until the 17th century. The difference between the two is more than 1800 years.
01
Why start with points?Almost all college mathematics textbooks explain differentiation first and then introduce its inverse operation—indefinite integral. Also, the definite integral used to calculate the area is defined as the difference of indefinite integrals. While it certainly makes sense to teach mathematics logically in the above order, historically the order of development has been just the opposite. Archimedes studied integrals for calculating areas in the 3rd century BC, and Newton and Leibniz did not come up with differential methods until the 17th century. The difference between the two is more than 1800 years.
Integrals were discovered first in history, and there are certain reasons for this. Integrals are directly related to the calculation of specific quantities such as area and volume. In addition, before studying differentiation, one must first understand concepts such as infinitesimal and limit accurately. For example, the speed of an object needs to be defined by differentiation, but because there was no concept of determining the limit in ancient Greece, Zeno's "flying arrow does not move" paradox appeared.
I think that before learning complex differentials, it is best to correctly grasp integrals that are relatively easy to understand intuitively, and then think about their inverse differentials. So first explain the integral. Whether you're stumped in your college calculus class or you're planning to start learning calculus, try "starting with calculus first."
02
How exactly is the area calculated?Integration begins with calculating the area of the graph. The units of area include square meters, square kilometers, etc., that is, they all have the word "square". The area of a square with side length 1 meter is equal to 1 square meter. That is to say, the area is based on the square as the unit, and the area of the calculated figure is equivalent to several squares. If it is a rectangle, how to calculate the area? In elementary school, we learned that the area of a rectangle is the product of its length and width, but let’s pretend we haven’t learned this formula for now.
Assuming that the known rectangle is 1 meter wide and 2 meters long, then divide the rectangle into two parts vertically from the middle to get two squares with a side length of 1 meter, so the area of the rectangle is equal to 2 square meters. That is, the product of length and width equals the area of the rectangle.
Next, assume that n and m are both natural numbers. It is known that the width of the rectangle is n meters and the length is m meters. Then as long as the width is divided into n parts and the length is divided into m parts, we can get n × m sides with a length of 1 meter. square (Figure 7-2). The area of the rectangle is exactly equal to n × m times the area of the square, ie n × m square meters. The result is again equal to the product of length and width.
Even if the values of length and width are fractions, as long as you use approximate values, you can calculate the area of a rectangle equal to an integer, and its area is still equal to the product of length and width. In addition, as long as the limit is considered, even if the value of length and width is an irrational number similar to (square root 2), the area can be calculated by multiplying the length and width.
We also learned in elementary school that the area of a triangle is equal to "base multiplied by height divided by 2". As shown in Figure 7-3, if the area of the triangle is doubled, it is equal to the area of the rectangle.
Not only rectangles and triangles, as long as it is a figure surrounded by broken lines, no matter what shape it is, the ancient Greeks could find out the relationship between its area and the area of a triangle. As shown in Figure 7-4, even if the figure surrounded by polylines is an irregular figure, it can be represented by a set of triangles, so as long as the area of all triangles is calculated, the total area of the figure can be obtained by adding the calculation results .
03
How to calculate any shape is OK, Archimedes' clamping theoremNow we have known that as long as it is a figure surrounded by polylines, it can be divided into triangles and then the area can be calculated. Then, if it is a figure surrounded by smooth curves, such as a parabola and a circle, how to calculate its area? Regarding this problem, we can immediately think of approximating the curve as a broken line, as shown in Figure 7-5. Therefore, when calculating the area of the graph enclosed by the curve, it is only necessary to calculate the approximate area of the broken line graph.
While this is a good idea, since approximations always have errors, it is necessary to estimate the magnitude of the errors. Try to reduce the error to 0 if you can. At this time, let's first explain the "method" developed by Archimedes.
As shown in Figure 7-6, assuming that the graph A surrounded by curves is known, first draw graph B surrounded by polylines in graph A. Then draw a figure C around the outside of figure A. There is an inequality between the above three figures, that is, area (B) ≤ area (A) ≤ area (C). We cannot calculate the area of graph A correctly for the time being, but the area of graph A is larger than the area of graph B (it can be calculated because it is a broken line graph), and smaller than the area of graph C (it can be calculated because it is also a broken line graph). Therefore, we get the approximation error of the line graph.
But how can we reduce the error to 0? The method Archimedes thought of was not limited to a set of broken line graphs B and C, but increasing the number of vertices as shown in Figure 77, so that the broken line graph is constantly approaching the curved graph A. Therefore, these line graphs can be recorded as
In each group of figures, figure A contains figure Bn, and is contained by figure Cn at the same time. So,
The relationship between the three is the same as before, namely
The larger the value of the graph group (Bn, Cn), the more accurate the approximation. The more accurate the approximation, the closer the areas of the graphs Bn and Cn are to the area of the graph A. However, we don't know how big the area of graph A is. So how can we ensure that the area of the broken line graph will continue to approach the unknown area?
Archimedes believed that in the process of increasing n,
The value will keep decreasing. So when n is infinite, the difference of the above formula will be equal to 0. Since the area of graph A is between area (Bn) and area (Cn), the value when both reach the limit should be the area of graph A. Archimedes is said to have developed the above method by borrowing the theories of the 4th century BC mathematician Eudoxus. Because Archimedes successfully solved many problems in geometry with the above method, this method is called "Archimedes' pinching theorem".
Next we illustrate how to calculate the area of a circle with an example. As shown in Figure 77, (B1, C1) is a square, (B2, C2) is a regular octagon, it can be inferred that (B3, C3) is a regular hexagon, then (Bn, Cn) is a regular 
. Use (Bn, Cn) to approximate the area of a circle.
The graphs Bn and Cn can be divided into triangular sets by straight lines connecting the centers and vertices. We can find that every time the value of n increases by 1, the area difference between the two
will be reduced to less than half. The larger the value of n, the error will be halved in turn, and the error will be smaller and smaller. Therefore, when n is infinite, the values of area (Cn) and area (Bn) are exactly equal, and this value is equal to the area of the circle. This is how Archimedes calculated the area of a circle.
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What exactly do points count?Using Archimedes' clamping theorem, the area of more complex curves can be calculated. Expressed in Cartesian coordinates, as shown in Figure 78, a straight line can be expressed as y = ax + b, and a parabola can be expressed as
Assuming that a function f(x) is known, let's think about the curve y = f(x). As shown in Figure 7-9, assuming that the value of f(x) is always greater than 0 on the interval a ≤ x ≤ b, then let’s study the curve y = f(x) and y = 0, x = a, x = b Figure A enclosed by these three straight lines (the shaded part in the figure). If you know how to calculate the area of graph A, you can calculate the area of any graph enclosed by any curve by combining them.
The curve y = f(x) rises or falls along the y-axis. For ease of calculation, assume that y = f(x) is always increasing on the interval a ≤ x ≤ b. In other cases, split the interval a ≤ b into two parts, an interval with increasing f(x) and an interval with decreasing f(x). As long as the following methods are respectively substituted into the above two intervals.
In order to use Archimedes' clamping theorem to calculate the area of graph A, first divide the interval a ≤ x ≤ b into n parts, such as graphs Bn and Cn in Figure 710. Graph A contains graph Bn, and is contained in graph Cn at the same time. Graphics Bn and Cn are both rectangular sets, so the area can be calculated.
As shown in Figure 711, the difference between area (Cn) and area (Bn) is equal to
That is, the area of a rectangle with base ε = (b − a)/n and height = (f(b) − f(a)). The larger the value of n, the smaller the value of ε, so the areas of graph Bn and graph Cn are closer. When the value of ε reaches the limit and is equal to 0, the areas of the two figures are equal. The value when reaching the limit is the area of graph A.
The area of the graph A calculated according to the above method is called "the integral of the function f(x) on the interval a ≤ x ≤ b", which is written as
Leibniz, who created calculus at the same time as Newton, invented the notation
It is an elongation of the initial letter "S" of "sum". Also, the d of "dx" refers to the initial letter of "difference". When the figure is approximated as a set of rectangles, the base length of a rectangle is equal to x + ε and
The "difference" of x. The area of a rectangle with height f(x) and base e is equal to f(x)e, so the symbol dx can be used instead of e, ie f(x)dx. That is to say,
Contains Leibniz's ideas, that is, "Integral refers to arranging rectangles whose height is equal to f(x) and base is dx on the interval from x = a to b, and finding the sum of their areas".
The integral explained above follows the definition of the 19th century German mathematician Bernhard Riemann, so it is called "Riemann integral". In fact, integrals include many types, such as the "Lebesgue integral" proposed by the French mathematician Henri Lebesgue, the "Ito integral" proposed by the Japanese mathematician Kiyoshi Ito, etc. The Riemann integral is enough to deal with the function problems we learned in high school, but when we need to deal with randomly fluctuating values like stock prices, we need to use the Ito integral. Ito points are also used to determine the price of options, so Ito Kiyoshi is considered "the most famous Japanese on Wall Street".
05
Integrals and functionsLet's calculate various integrals according to Riemann's definition. First, what happens if you integrate the function y = x over the interval x = 0 to x = a? As shown in Figure 7-12, this is the area of a right triangle with base a and height a, so it should be equal to a2/2. Next, let's check it out. At this time, the graph Cn is a set of rectangles whose base length is ε = a/u and height is ε,2ε , ··· , then
Since the time of Pythagoras, people have known how to calculate the value of (1 + 2 + . . . + n). When calculating, first multiply it by 2, that is,
The first term in the first square bracket on the right side of the equation is n, and the first term in the second square bracket is 1, so the sum of the two terms is equal to (n + 1). The second item in the two square brackets is (n − 1) and 2 respectively, and the sum of the two items is also equal to (n + 1). There are a total of n combinations on the right side of the equation and the sum is equal to (n + 1), so the right side is equal to n × (n + 1). Because the left side is multiplied by 2, it is divided by 2 at the end, that is
Substituting into the above formula, that is
The larger the value of n, the more the value of 1/n in parentheses can be ignored. So when n is infinite, the area is equal to
The result is equal to the area calculated directly with the triangle. Expressed in integral notation, that is
find quadratic function
The area on the interval x = 0 to x = a also uses the same method, but the calculation process is a bit long to explain in detail. Archimedes discovered the integral formula of the quadratic function in the 3rd century BC, so it would be a pity not to explain it. So let me explain briefly. In the case of the quadratic function y = x2, the graph Cn is a set of rectangles with a base length ε = a/n and a height ε2, (2ε)2, ..., namely
The sum occurring here can be calculated as
So
So when n is infinite, the integral can be computed
For details, please refer to the supplementary knowledge in the appendix of this book. Using the same method, it is also possible to compute integrals of higher order functions. calculate
When integrating over the interval x = 0 to x = a, one must first compute
In 1636, Fermat, who became famous for his "Fermat's Last Theorem", mentioned in a letter to his friends that because
so can calculate
points. In fact, as long as we use the above inequality and Archimedes' clamping theorem, we can get
Finally, Euler proved Guan Xiaohe and Bernoulli's formula. Because Japan adopted a closed country policy at that time, Euler had no way of knowing Guan Xiaohe's performance. Because of this, he called the coefficients in the above formula Bernoulli numbers, and this name is still used today. Since Guan Xiaohe and Bernoulli independently discovered the above formula, it should be called Guan-Bernoulli number.
06
Feiya not moving?Integrals are concerned with calculations of area and volume, while differentials are concerned with calculations of velocity. Since speed is to be considered, Zeno is about to play again.
Let's consider the situation of the arrow that leaves the string. Time is a collection of moments, so the arrow after leaving the string is in a fixed position at any moment during its flight. Since it is a fixed position, it is no different from not moving, so "a flying arrow does not move" is a paradox. Of course, the above proposition sounds pretty silly. So what went wrong?
First, let's reflect on what speed is. Assuming that you can walk 3.6 kilometers in 1 hour, then the speed is 3.6 kilometers per hour. Velocity is the distance traveled divided by the time it takes, i.e.
If the time is shortened again, i.e.
The velocity becomes 1 m/s. As the time shortens, the distance traveled also decreases, and if the movement speed remains the same, the ratio of time to distance does not change. As long as the time is gradually reduced to 0, which is the limit, it should be able to define the speed at a certain instant.
Assume moving from left to right on a straight line, use the coordinate x of the straight line to judge the position. Denote the position at time t as x(t), and only move (x(t') −x(t)) in the interval from time t to t′. The average velocity while moving is equal to (x(t′) − x(t)) ÷ (t′− t). Therefore, if t' keeps approaching t, then the velocity at time t should be able to be calculated at the limit t' = t. However, since the results of x(t′) − x(t) and t′− t are both equal to 0 in the limit, if we calculate 0 ÷ 0, the calculation process becomes inexplicable. Care should be taken when calculating. Suppose x(t) = t, then
The reason why the error of 0 ÷ 0 is easy to appear in the limit is because both the numerator and the denominator exist (t′− t). However, as long as the (t′− t) of the two parties are offset in advance, there will be no problem even assuming t′= t.
Then consider the situation when x(t) = t2, that is
The numerator and denominator in the above formula cancel out (t′ − t). After cancellation, assuming t′ = t, then the velocity is equal to 2t. That is, velocity is proportional to t.
In order to calculate the speed at a certain moment, when thinking about the limit of t′→t in (x(t′) − x(t)) ÷ (t′− t), the problem of 0 ÷ 0 is easy to appear. However, the previous examples have shown that as long as the (t′− t) in the numerator and denominator are canceled first, and then the limit is obtained, there will be no problem. Using the differential definition, it can be expressed as
The symbol lim on the right side of the equation means "limit". Although Newton in England and Leibniz in Germany independently discovered differentiation, like integrals, the expression dx/dt of differentiation was also invented by Leibniz.
Returning to Zeno's paradox, the problem is that
and how to calculate the limit of t′ → t. If the limits of the numerator and denominator are calculated arbitrarily, it is easy to cause paradox. If the numerator x(t′) − x(t) is assumed to be equal to 0, then the above formula is 0÷ (t′− t), and then no matter how small the value of the denominator (t′− t) is, the formula is equal to 0. This is the fundamental meaning of "flying arrows don't move". That is to say, Zeno's paradox lies in the way of dealing with limits. Instead of thinking about the limit of the numerator and denominator separately, but consider the numerator and denominator of (x(t′) − x(t)) ÷ (t′− t) as a whole, and calculate the limit of t′→ t for it , "instantaneous speed" is meaningful. About 2,100 years passed between Zeno's time and when Newton and Leibniz really understood what it meant.
07
Differentiation is the inverse of integralDifferentiation is the inverse of integration, one of the most important discoveries of Newton and Leibniz. Assuming a known function f(x), the integral of f(x) on the interval from 0 to a is expressed as
It can be viewed as a function of a. Then differentiate a to get
Substitute into the original function f(x), ie x = a. This means that differentiation is the inverse operation of integration. Prove it next. If graph A can be divided into graph B and graph C, that is
Suppose graph A is the interval 0 ≤ x ≤ b under the curve y = f(x). Split the interval into two parts, 0 ≤ x ≤ a and a ≤ x ≤ b. Then the area is also divided into
Thus, moving the first term on the right-hand side of the equation to the left, we get
Substituting the above equation into the definition of differential, then
In order to calculate the differential, the value of a' keeps decreasing and approaches a, then in the short interval a<x<a', the value of f(x) is almost unchanged. Therefore points
can be approximately equal to the area of a rectangle whose base is (a′− a) and height is f(a), namely
Substituting the above formula into the above formula, we get
We can find that after differentiating the integral, we get back to the original function. Conversely, if the function is differentiated and then integrated, it will return to the original state.
Newton and Leibniz's "Fundamental Theorem of Calculus" states that differentiation and integration are inverse operations. In Japanese high school textbooks, differentiation is usually defined first, and then integral is its inverse operation. So in Japanese high school mathematics, the "Fundamental Theorem of Calculus" is not a theorem, but a definition of integral. In this chapter, we have defined the integral as "the area under the curve y = f(x)", so the "Fundamental Theorem of Calculus" is a theorem.
08
Differentiation and Integration of Exponential FunctionsDifferentiation is a more advanced mathematical concept than integration and requires more attention to the way limits are dealt with. Even so, Japanese high schools still teach differential first, and one of the reasons is that the calculation of differential is relatively simple. exponential function using the natural constant e
Next, we calculate the differential of the exponential function, first substitute it into the definition of the differential, and get
Exponential functions have the following properties:
Using the above formula, the numerator on the right-hand side of the equation is equal to
So the differentiation of the exponential function can be expressed as
Suppose x′− x = ε on the right side of the equation, then the limit of x′→ x is equal to ε → 0. so
The limiting value on the right-hand side of the above equation is
The proof process, while simple, still requires computation. Therefore, please refer to the supplementary knowledge in the appendix of this book for details. Using the above formula, we can find that the differential of the exponential function ex is itself.
Now that the differential can be calculated, the integral can be simply calculated according to the "Fundamental Theorem of Calculus". First, integrate both sides of the above differential formula separately, namely
Using the Fundamental Theorem on the left-hand side of the above equation, we get
therefore,
The integral of an exponential function can be computed directly using the theorem without resorting to differentiation. For details, please refer to the supplementary knowledge in the appendix of this book. After reading it, you will realize how complex the calculation of integrals is compared to differentials!
For the trigonometric functions sin x, cos x, and tan x, integrals can also be computed as the inverse of differentiation.
For functions learned in high school such as exponential functions and trigonometric functions, the calculation of differential is simpler than integral. However, with the exception of power functions, exponential functions, and trigonometric functions, there are very few functions that can be calculated precisely in calculus. Functions that appear in mathematical applications, although some of them can be approximated by one of the three functions mentioned above, most of them can only be calculated numerically by computer.
Although it is helpful to know how to calculate the calculus of exponential functions and trigonometric functions, as the knowledge that must be mastered, differentiation and integration are particularly important, so we must first correctly understand the meaning of differentiation and integration. Because of this, the content of this chapter begins with integrals.
recommended reading
Author: [J] Hiroshi Oguri
Translator: You Binbin
Director of the Institute of Theoretical Physics, California Institute of Technology, Professor Hiroshi Oguri, Research Director of the Kavli Mathematical Physics Joint Universe Research Institute, University of Tokyo, Japan
Break through the traditional teaching order and methods of mathematics education / Explain the core concepts and principles of mathematics with "language thinking" / Return to the "basic principles" to re-understand the essence of mathematics
