商务科学与技术数学：MATLAB与Excel计算详解

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"《数学应用于商业、科学与技术：MATLAB与Excel计算》第三版是一本专为不同背景的学习者和专业人士设计的教材。它旨在满足那些准备在社区学院或大学攻读商业或科学课程的高中毕业生的需求，包括正在寻求职业提升的在职人员，以及在商务管理、会计等领域继续教育的学生。此外，这本书也适合那些投身于科学、电子和计算机技术职业的人群。本书的核心内容涵盖了数学基础知识，特别是如何运用MATLAB®（一种强大的数值计算软件）和Excel®（办公软件中的数据分析工具）进行实际问题的解决。这使得它不仅是一本理论教材，更是实践技能的教科书。书中对概率论和统计学进行了全面而深入的讲解，并通过大量的例子来阐述这些概念，帮助读者理解和掌握它们的实际应用。随着每版的更新，作者Steven T. Karris确保内容紧跟行业需求，反映了最新技术和教学标准。从图表中可以看出，该书自出版以来销量稳步增长，表明其在市场上的受欢迎程度。版权方面，所有的复制和分发必须得到Orchard Publications的明确许可，以尊重知识产权。《数学应用于商业、科学与技术：MATLAB与Excel计算》第三版是一本实用性强、适用于各种背景的学习者的专业书籍，无论是初学者还是经验丰富的专业人士，都能从中受益匪浅，提升他们在各自领域内的数学素养和计算能力。通过这本书，读者将学会如何利用现代科技工具来解决实际问题，增强他们的职业竞争力。"

Mathematics for Business, Science, and Technology, Third Edition 1−3

Algebraic Equations

Example 1.2

Given the equality

prove that Property 1 holds if we

a. add to both sides

b. subtract from both sides

Proof:

a. Adding to both sides we get

and thus, the equality holds.

b. Subtracting from both sides we get

and thus, the equality holds.

Example 1.3

Solve the following equation, that is, find the value of .

Solution:

We need to find a value for so that the equation will still hold after the unknown has been

replaced by the value that we have found. For this type of equations, a good approach is to find a

number that when added to or subtracted from both sides of the equation, the left side will con-

tain only the unknown . For this example, this will be accomplished if we add to both sides of

the given equation. When we do this, we get

and after simplification, we obtain the value of the unknown x as

We can check this answer by substitution of into the given equation. Thus, or

. Therefore, our answer is correct.

75+ 12=

753++ 12 3+= 15 15=

752–+ 12 2–= 10 10=

x5– 15=

x5– 5+ 15 5+=

x20=

x205– 15=

15 15=

Mathematics for Business, Science, and Technology, Third Edition 1−7

Laws of Exponents

(1.13)

where

PV = cost of the annuity at present

Payment = amount paid at the end of the year

Interest = Annual interest rate which the deposited money earns

n = number of years the Payment will be received.

Example 1.8

Suppose that an insurance company offers to pay us at the end of each year for the next

years provided that we pay the insurance company now, and we are told that our

money will earn per year. Should we accept this offer?

Solution:

, , and . Let us calculate PV using (1.13).

(1.14)

We will use Excel to compute (1.14). We select any cell and we type in the following expression

where the asterisk (*) denotes multiplication, the slash ( / ) division, and the caret (^) exponenti-

ation.

=12000*(1−(1+0.08)^(−25))/0.08

Alternately, we can use MATLAB

with the statement

PV=12000*(1−(1+0.08)^(−25))/0.08

Both Excel and MATLAB return and this is the fair amount that if deposited now at

interest, will earn per year for years. This is less than which the insur-

ance company asks for, and therefore, we should reject the offer.

Excel has a “build-in” function, called PV that computes the present value directly, that is, with-

out the formula of (1.13). To invoke and use this function, we perform the sequential steps

>Financial>PV>Rate=0.08>Nper=25>Pmt=12000>OK. We observe that the numeri-

* An introduction to MATLAB is provided in Appendix B. MATLAB applications may be skipped without loss

of continuity. However, it is highly recommended since MATLAB now has become the standard for advanced

computation. The reader may begin learning it with the inexpensive MATLAB Student Version.

PV Payment

1 1 Interest+()

n–

–

Interest

------------------------------------------------

×=

$12,000

25 $130,000

Payment $12,000= Interest 0.08 (8%)= n 25 years=

PV 12000

110.08+()

25–

–

0.08

----------------------------------------

×=

$128,097

8% $12,000 25 $130,000

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