C语言实现数值分析入门

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"这是一本关于使用C语言进行数值分析的英文电子书，由Attila Máté编写，包含了原理解析、代码示例，适合学习数值分析和C语言编程的人群。书中涵盖浮点数、误差分析、泰勒公式余项、插值法（拉格朗日、牛顿、赫尔米特）、有限差分、非线性方程求解、牛顿法、固定点迭代、加速收敛技术、数值微分以及数值积分等多个主题。" 在这本《Introduction to Numerical Analysis with C Programs》中，作者深入浅出地介绍了数值分析的基础概念和方法，并结合C语言编程实践，使得读者能够理解并应用这些理论。以下是部分关键知识点的详细说明： 1. 浮点数：讲解了计算机如何存储和处理浮点数，包括浮点数的表示形式、精度限制及其对计算的影响。 2. 绝对和相对误差：讨论了衡量数值计算结果准确性的两种主要方式，绝对误差是实际值与近似值之间的差值，而相对误差则考虑了误差相对于真实值的比例。 3. 舍入和截断误差：分析了在数值计算过程中，由于四舍五入或截断操作引入的误差，以及它们如何影响结果的精度。 4. 泰勒公式余项：介绍了泰勒级数，特别是泰勒公式的误差部分，它可以帮助我们理解近似值的精确度。 5-7. 插值法：详细阐述了拉格朗日插值、牛顿插值和赫尔米特插值，这些方法用于通过有限数据点构建函数近似。 8. 牛顿插值多项式误差：讨论了牛顿插值法的误差性质，包括其产生的原因和影响。 9-11. 有限差分、等距点上的牛顿插值与赫尔米特插值的拉格朗日形式：这些章节涉及使用有限差分来近似导数，以及在等距离点上进行插值的方法。 12-13. 非线性方程求解：介绍了解决非线性方程的基本方法，如牛顿法和用于编程实现的算法。 14-15. 牛顿法和固定点迭代：这两种方法常用于求解方程，牛顿法以其快速收敛性著称，而固定点迭代则适用于某些特定问题。 16. Aitken的加速：这是一种加速迭代过程收敛的技术，可以更快地找到方程的根。 17. 牛顿法的收敛速度：分析了牛顿法的收敛特性，讨论了影响收敛速度的因素。 18-19. 数值微分：介绍了如何利用数值方法求解函数的导数，分为表格数据的数值微分和连续函数的数值微分。 20-22. 数值积分：涵盖了简单的数值积分公式、复合积分和自适应积分，这些都是解决实际问题中的重要工具。这本书不仅提供理论知识，还包含了许多实际的C语言程序示例，帮助读者将所学应用于实践，提升数值计算技能。对于希望在C语言环境中进行数值计算的学习者来说，这是一份宝贵的资料。

10 Introduction to Numerical Analysis

3. ROUNDOFF AND TRUNCATION ERRORS

The main types of errors are roundoﬀ error, and truncation error. The forme r refers to the fact

that real numbers, which are u su ally inﬁnite decimals , are re pr e s ented by a ﬁnite number of digits

in the computer. The latter refers to the error committed in a calculati on when an inﬁnite process

is replaced by a ﬁnite process of calculation. A simple example for this latter is when the sum of an

inﬁnite series is evaluated by adding ﬁnitely many terms in the series. More generally, continuous

processes in the computer are represented by ﬁnite, discrete processes, resu lt in g in truncation error.

A third type of error, error resulting from a mistake or blunder is an altogether diﬀerent matter;

such errors ought to be avoided in so far as possible, whereas roundoﬀ and trun cation errors cannot

be avoided.

4. THE REMAINDER TERM IN TAYLOR’S FORMULA

In Taylor’s formula, a function f(x) is approximated by a polynomial

k=0

(k)

(a)

(x − a)

The goodness of this approximation can be measured by the remainder term R

(x, a), deﬁned as

(x, a)

def

= f(x) −

k=0

(k)

(a)

(x − a)

To estimate R

(x, a), we need the following lemma.

Lemma. Let n ≥ 1 be an integer. Let U be an open interval in R and let f : U → R be a function

that is n + 1 times diﬀerentiable. Given any b ∈ U , we have

(1)

(b, x) = −

(n+1)

(x)(b − x)

for every x ∈ U.

Proof. We have

(b, x) = f(b) −

k=0

(k)

(x)

(b − x)

= f( b) −f (x) −

k=1

(k)

(x)

(b − x)

We separated out t he term for k = 0 since we are going to use the product rule for diﬀerentiation,

and the term for k = 0 involves no product. We have

(b, x) =

f(b) −

f(x)

−

k=1



(k)

(x)

(b − x)

+ f

(k)

(x)

(b − x)



= −f

′

(x) −

k=1



(k+1)

(x)

(b − x)

+ f

(k)

(x)

−k(b − x)

k−1



= −f

′

(x) −

k=1



(k+1)

(x)

(b − x)

− f

(k)

(x)

(b − x)

k−1

(k −1)!



12 Introduction to Numerical Analysis

Note. The above argument can be carried out in somewhat more generality. Let g(x) be a

function t hat is diﬀerentiable on (a, b), g

′

(x) 6= 0 for x ∈ (a, b), and g(b) = 0. Note that, by the

Mean-Value Theorem,

−

g(a)

b − a

g(b) − g(a)

b − a

= g

′

(η)

for some η ∈ (a, b). Since we have g

′

(η) 6= 0 by our assumptions, it follows that g(a) 6= 0. Instead of

(3), determine K now such that

(4) R

(b, a) = Kg(a) .

As g(a) 6= 0, it is possible to ﬁnd such a K. Write

φ(x) = R

(b, x) −Kg( x).

We have φ(a) = 0 by the choice of K. We have φ(b) = 0 since R

(b, b) = 0 and g(b) = 0 (the latter

by our assumptions). As φ is diﬀerentiable on (a, b), there is a ξ ∈ (a, b) such that φ

′

(ξ) = 0. Thus,

by (1) we have

0 = φ

′

(ξ) = −

(n+1)

(ξ)(b −ξ)

− Kg

′

(ξ).

As g

′

(ξ) 6= 0 by our assumptions, we can determine K from this equation. Sub s tit ut in g the value of

K s o obtained into (4), we can see that

(b, a) = −

(n+1)

(ξ)(b −ξ)

g(a)

′

(ξ)

Note that in the argument we again ass ume d that a < b, but this was unessential. Further, note

that the function g can depend on a and b. We can restate the result just obtained in the following

Corollary 2. Let n ≥ 1 be an integer. Let U be an o pen interval in R and let f : U → R be a

function that is n + 1 times diﬀerentiable. For any any a, b ∈ U with a < b, let g

a,b

(x) be a function

such that g

a,b

is continuous on [a, b] and diﬀerentiable on (a, b). Assume, further, that g

a,b

(b) = 0

and g

′

a,b

(x) 6= 0 for x ∈ (a, b). Then there is a ξ ∈ (a, b) such that

(b, a) = −

(n+1)

(ξ)(b −ξ)

a,b

(a)

′

a,b

(ξ)

If b < a, then the same result holds, except that one needs to write (b, a) instead of (a, b) and

[b, a] instead of [a, b] for the intervals mentioned (but the roles of a and b should not otherwise be

interchanged).

This result is given in [B]; see also [M]. The Wikipedia entry [Wiki] also discusses the result

(especially under the subheading Mea n value theorem), without giving attributions. Taking g

a,b

(x) =

(b − x)

for an integer r with 0 < r ≤ n + 1, we obtain that

(b, a) =

(n+1)

(ξ)

rn!

(b − ξ)

n−r+1

(b − a)

This is called the Roche–Schl¨omilch Remainder Term of the Taylor Se r ie s . Here ξ is some numbe r

in the interval (a, b) or (b, a); it is important to realize that the value of ξ depends on r. Taking

r = n + 1 here, we get formula (2); this is called Lagrange’s R emaind er Term of the Taylor S eries.

Taking r = 1, we obtain

(b, a) =

(n+1)

(ξ)

(b − ξ)

(b − a);

5. Lagrange interpolation 13

this is calle d Cauchy’s Remainder Term of the Taylor Series.

The diﬀerent forms of the remainder term of Taylor’s series are useful in diﬀerent situations. For

example, Lagrange’s Remainder Term is convenient for establishing the convergence of the Taylor

series of e

, sin x, and cos x on the whole real line, b ut it is not suited to establish the convergence

of the Taylor series of (1 + x)

, where α is an arbitrary real number. The Taylor series of t hi s last

function is convergent on the interval (−1, 1), and on this interval it does converge to the function

(1 + x)

(this series is called the Binomial Series). This can be established by using Cauchy’s

Remainder Term.

5. LAGRANGE INTERPOLATION

Let x

, x

, . . . ,x

be distinct real numbers, and let y

, y

, . . . ,y

be also be real numbers, these

not necessarily distinct. The task of polynomial interpolation is to ﬁnd a polynomial P (x) such that

P (x

) = y

for i with 1 ≤ i ≤ n. It is not too hard to prove that there is a unique polynomial of

degree n−1 satisfying these conditions, while a polynomial of lower degree usually cannot be found.

The polynomial interpolation problem was ﬁrst solved by Newton. A diﬀerent, elegant solution was

later found by Lagrange. Here we consider the latter solution.

Lagrange c onsi de r s the polynomials

(x) =

j=1

j6=i

x − x

− x

It is clear that l

(x) is a poly nomial of degr ee n, since the numb e r s in the denominator do not depend

on x. Further, for any integer j with 1 ≤ j ≤ n we have

) =



1 if j = i,

0 if j 6= i.

Indeed, if x = x

then each of the fractions in the product expr e ss in g l

(x) is 1, and if x = x

then

one of the fractions in this product has a zero numerator. For this reason, the polynomial P (x)

deﬁned as

P (x) =

i=1

(x)

satisﬁes the requirements; that is P (x

) = y

for i with 1 ≤ i ≤ n.

Example. Find a polynomial P (x) of degree 3 such that P (−1) = 7, P (2) = 3, P (4) = −2, and

P (6) = 8.

Solution. Writing x

= −1, x

= 2, x

= 4, and x

= 6, we have

(x) =

(x − 2)(x − 4)(x − 6)

(−1 − 2)(−1 − 4)(−1 − 6)

= −

105

(x − 2)(x − 4)(x − 6),

(x) =

(x + 1)(x − 4)(x − 6)

(2 + 1)(2 − 4)(2 − 6)

(x + 1)(x − 4)(x − 6),

(x) =

(x + 1)(x − 2)(x − 6)

(4 + 1)(4 − 2)(4 − 6)

= −

(x + 1)(x − 2)(x − 6),

These polynomials are sometimes called Lagrange fundamental polynomials.

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