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Application Research of Mathematica Software in Calculus Teaching

Data publikacji: 15 Jul 2022
Tom & Zeszyt: AHEAD OF PRINT
Zakres stron: -
Otrzymano: 05 Apr 2022
Przyjęty: 18 Jun 2022
Informacje o czasopiśmie
License
Format
Czasopismo
eISSN
2444-8656
Pierwsze wydanie
01 Jan 2016
Częstotliwość wydawania
2 razy w roku
Języki
Angielski
Introduction

In recent years, mathematics has become an essential introductory theoretical course in universities. It can help students improve their math skills. The mathematics courses offered by liberal arts majors in colleges and universities are simplified versions of science and engineering mathematics courses. The Ministry of Education hopes to introduce more basic mathematics knowledge [1]. At the same time, the compression of class hours makes content simplification inevitable. Daily teaching focuses on teaching conclusions and operational forms and weakening the derivation of logical reasoning and proof processes. But this requires students to have an excellent theoretical foundation. But humanities students generally have weak theoretical foundations, and many people are afraid of mathematics, which requires logical thinking and theoretical foundations. Many students give up mathematics because of this. The complex numerical calculations and formula theorems make them shunned, and some students are afraid of mathematics.

Using computer software in the teaching process, the calculation software of different symbols can fully play the role of auxiliary teaching, which teachers and students deeply love. But it is worth mentioning that, at this stage, the expansion of computing software in education is mainly aimed at science and engineering students. At present, there are not many examples used in teaching liberal arts mathematics. Compared with science and engineering mathematics teaching, using symbolic computing software in liberal arts mathematics teaching can produce better teaching effects. Extended computing software can make the effect of software use more intuitive [2]. Make abstract problems concrete. This helps to understand conceptual issues. Using computers and software to model and experiment lets students feel that mathematics is no longer just an abstract and obscure theory. With the help of symbolic computing software, precise and scientific number operations and derivation can be carried out in a short time, and no more time-consuming “details.” In addition, the use of relevant mathematical software can stimulate students’ enthusiasm for learning in the teaching process, especially in a relatively challenging course such as calculus, which is of great significance. Mathematica can play the role of auxiliary teaching.

Introduction to Mathematica Symbolic Computing Software

Mathematica mathematical calculation software launched by Wolfram Research has potent functions; specifically, in addition to graphics and calculus, it also includes numerical calculation. Mathematica's computing function is powerful, and it can effectively identify and calculate the algorithm research of different calculus. It has been widely used and studied in university mathematics teaching. In the stage of university study, most mathematical problems can be solved by using Mathematica. Mathematica has a drawing function [3]. The production of unary and binary operations can be expressed in different ways. In the actual use process, you can choose the appropriate precision and range according to your needs. Many mathematical problems can be made easier to understand with the help of these numbers. Mathematica as an emerging interactive computing system. The application interface is simple and orderly, with a distinct personality, which allows new users to operate flexibly and get started in time. The command icons are all simple English words, allowing liberal arts students to learn how to manage the software.

Applications of Mathematica in Calculus

For the current liberal arts-related majors, the knowledge points involved in calculus are differentiation, integration, limit theory, etc. [4]. Next, the application of Mathematica in the teaching of these knowledge points is introduced.

Limits and Continuity of Functions

Limits are the foundation of calculus and the difficulty of teaching. In particular, the definitions of limits ɛN and ɛδ are abstract. Calculus is developed based on limits and is a problematic point in teaching. Among them, the concept of limit and limit is challenging and obscure. It is difficult for students to understand and differentiate in learning [5]. Using Mathematica to make images and sequences in the teaching process can make students understand intuitively.. For example, we can use the command

data = Table[{n,1 + (−1) ^ n / n},{n,1,100}];

aa = ListPlot [data, PlotStylePo int Size [0.0125], DisplayFunctionIdentity];

cc = Show [GraphicsArray [{aa}, FrameTrue]]

data = Table [n. 1/(2$n)}, {n, 1, 100}];

aa = ListPlot [data, PlotStylePo int Size [0.0125], DisplayFunctionIdentity];

cc = Show [GraphicsArray [{aa}, FrameTrue]]

data = Table [{n, (1−1/(2$n)}, {n, 1, 100}];

aa = ListPlot [data, PlotStylePo int Size [0.0125], DisplayFunctionIdentity];

cc = Show [GraphicsArray [{aa}, FrameTrue]]

At this point, we draw the image of the sequence {1+(1)nn},{12n},{1122} \{ 1 + {{{{\left( { - 1} \right)}^n}} \over n}\} ,\{ {1 \over {{2^n}}}\} ,\{ 1 - {1 \over {{2^2}}}\} (Figure 1, Figure 2, Figure 3). This paper also defines the concept of limit, which is convenient for subsequent students to learn and comprehend[6]. For calculating the limit, we can directly use the function Limit. For example, the limit of limn+(1+(1)nn) \mathop {\lim }\limits_{n \to + \infty } (1 + {{{{\left( { - 1} \right)}^n}} \over n}) can be found now:

In [1]: = Limit [1+(−1)^n/n, n → ∞]

Out [1] = 1

Figure 1

Image of {1+(1)nn} \{ 1 + {{{{\left( { - 1} \right)}^n}} \over n}\}

Figure 2

Image of {12n} \{ {1 \over {{2^n}}}\}

Figure 3

Image of {1122} \{ 1 - {1 \over {{2^2}}}\}

In this process, This article effectively solve the function of changing the point through the process of calculating the absolute value and then use it to analyze the limit value of this point. The relevant conclusions can be drawn through the following calculation, using the formula [7], For example, we can use Limit [(1 − x)^x, x → ∞] to directly find the limit of limx(11x)x \mathop {\lim }\limits_{x \to \infty } {(1 - {1 \over x})^x} is 1e {1 \over e} . This calculation is more complicated. At this time, the limits we need to calculate with Robida's rule can also be directly calculated by Limit. For example to find the limit of limxπ2lnsinx(π2x)2 \mathop {\lim }\limits_{x \to {\pi \over 2}} {{\ln \sin x} \over {{{\left( {\pi - 2x} \right)}^2}}} :

In [2]: = Limit [Log [Sin [x]] / ((Pi − 2x) ^2), xPi / 2]

Out [2] = 18 - {1 \over 8}

A function is continuous at a point if the limit is the process value. Therefore, this paper can directly use the limit to determine whether the value of the limit and the function are equal, thereby determining the continuity of the operation [8]. So the discontinuity boundary can still be used to determine what type of discontinuity it is.

Example 1: Solving for the continuity of the function f(x) f(x)={x1x<00,x=0x+1x>0 f\left( x \right) = \left\{ {\matrix{ {x - 1} & {x < 0}\hfill \cr {0,} & {x = 0}\hfill \cr {x + 1} & {x > 0}\hfill \cr } } \right. . We use Limit to compute the left and proper limits of ( ) f (x):

In [3]: Limit [x − 1, x → 0, Direction → 1] // left limit

Out [3] = −1

In [4]: Limit [x + 1, x → 0 Direction → −1] // right limit

Out [4] = 1

Then the left and right limits are not equal, so it is a jump discontinuity point.

Example 2: Solving for the continuity of a function y = tan x at point π2 {\pi \over 2} . We use Limit to calculate the limit of tan x at point π2 {\pi \over 2} .

In [5]: Limit [Tan [x], xPi / 2]

In [5] = − ∞

The limit of tan x at point π2 {\pi \over 2} does not exist and is therefore discontinuous at the moment π2 {\pi \over 2} .

A discontinuity is an infinite discontinuity.

Similarly, other discontinuity types can also be judged.

Derivatives and Differentiation

In the research background of the new era, the derivative has always occupied a place in the field of calculus research. The entry-level low-level product operations are simple but slightly complex, or high-level results, even binary results, require a lot of effort. The binary function is more complex, and the rendering of three-dimensional images is also highly complicated [9]. The ability of spatial imagination will be a big help at this time. We can use the software's plotting function Plot3D to ultimately draw the three-dimensional image of the binary function, which is convenient for the subsequent understanding of the calculus content.

As shown in Figures 4 and 5 below, this is a graph of two binary functions drawn by the software f (x, y) = x2y2, x2 + y2 + z2 = 1. Researchers use the Mathematica command D to calculate the derivative in this graph. Through accurate calculation, it can be concluded that the solution process of this function can be obtained. The paper wants to end that the continuous order algorithm of the calculation function has an apparent linear product relationship [10]. After that, this paper can derive the corresponding results by reasoning, calculating, and drawing conclusions.

Figure 4

Image of f (x, y) = x2y2

Figure 5

x2 + y2 + z2 = 1 Image

Example 3: Find the second derivative of the function f(x,y)=lnx2+y2 f\left( {x,y} \right) = \ln \sqrt {{x^2} + {y^2}} concerning x and the mixed derivative of the first order y with respect to x.

In [6]: f = Log [Sqrt [x ^ 2 + y ^ 2]];

D [f, {x, 2}]

Out [6] = 2x2(x2+y2)2+1x2+y2 - {{2{x^2}} \over {{{\left( {{x^2} + {y^2}} \right)}^2}}} + {1 \over {{x^2} + {y^2}}}

In [7]: f = Log [Sqrt [x ^ 2 + y ^ 2]];

D [f, {x, 1}, {y, 3}]

Out [7] = 48xy3(x2+y2)4+24xy(x2+y2)3 - {{48x{y^3}} \over {{{\left( {{x^2} + {y^2}} \right)}^4}}} + {{24xy} \over {{{\left( {{x^2} + {y^2}} \right)}^3}}}

Commands to compute derivatives can be used with controls to solve or root equations to compute extrema and conditional extrema of functions in one and two variables.

Indefinite and definite integrals

This paper finds that the most challenging part is the formulas of indefinite integrals in this calculation process. These formulas are generally complicated to understand in the calculation process. The solution is complex and requires specific skills to obtain the corresponding results. [11]. The relevant conclusions can be drawn by using Mathematica software for simulation operation and the image to determine the critical area [12].

Example 4: Find tan2 xdx.

In [8]:= Integrate [Tan [x] ^ 2, x]

Out [8] = x + Tan [x]

Example 5: Calculate Dxydxdy \int\!\!\!\int\limits_D {xydxdy} , where D is the area determined by y = x2, y = x.

Then it can be seen from the above figure 6 that x ∈ [0, 1], y ∈ [x2, x], then

In [8]:= Integrate [x y, (x, 0, 1}, {y, x ^ 2, x}]

Out [2] = 124 {1 \over {24}}

Figure 6

x ∈ [0, 1], y ∈ [x2, x]

Using the plot function, we can further determine more complex integration regions.

Conclusion

The learning of calculus has always been a difficult point in the higher education teaching system, and it is easy to make people feel intimidated. The combination of symbolic calculation software and calculus brings intuitive effects to students. In addition, the application of such software can help students understand knowledge in simple terms, make students have a strong interest in learning, and help them improve their academic performance. In this way, students can better solve various problems encountered. But at the same time, we should also pay attention to the two sides of everything, avoid students’ bad learning attitude caused by over-reliance on software, and help students understand this kind of computer software correctly.

Figure 1

Image of 



{1+(−1)nn}
\{ 1 + {{{{\left( { - 1} \right)}^n}} \over n}\}
Image of {1+(−1)nn} \{ 1 + {{{{\left( { - 1} \right)}^n}} \over n}\}

Figure 2

Image of 



{12n}
\{ {1 \over {{2^n}}}\}
Image of {12n} \{ {1 \over {{2^n}}}\}

Figure 3

Image of 



{1−122}
\{ 1 - {1 \over {{2^2}}}\}
Image of {1−122} \{ 1 - {1 \over {{2^2}}}\}

Figure 4

Image of f (x, y) = x2 − y2
Image of f (x, y) = x2 − y2

Figure 5

x2 + y2 + z2 = 1 Image
x2 + y2 + z2 = 1 Image

Figure 6

x ∈ [0, 1], y ∈ [x2, x]
x ∈ [0, 1], y ∈ [x2, x]

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