Advanced Calculus: Second Edition [NOOK Book]


This classic text by a distinguished mathematician and former Professor of Mathematics at Harvard University, leads students familiar with elementary calculus into confronting and solving more theoretical problems of advanced calculus. In his preface to the first edition, Professor Widder also recommends various ways the book may be used as a text in both applied mathematics and engineering.
Believing that clarity of exposition depends largely on precision of statement, the ...

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Advanced Calculus: Second Edition

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This classic text by a distinguished mathematician and former Professor of Mathematics at Harvard University, leads students familiar with elementary calculus into confronting and solving more theoretical problems of advanced calculus. In his preface to the first edition, Professor Widder also recommends various ways the book may be used as a text in both applied mathematics and engineering.
Believing that clarity of exposition depends largely on precision of statement, the author has taken pains to state exactly what is to be proved in every case. Each section consists of definitions, theorems, proofs, examples and exercises. An effort has been made to make the statement of each theorem so concise that the student can see at a glance the essential hypotheses and conclusions.
For this second edition, the author has improved the treatment of Stieltjes integrals to make it more useful to the reader less than familiar with the basic facts about the
Riemann integral. In addition the material on series has been augmented by the inclusion of the method of partial summation of the Schwarz-Holder inequalities, and of additional results about power series. Carefully selected exercises, graded in difficulty, are found in abundance throughout the book; answers to many of them are contained in a final section.

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Product Details

  • ISBN-13: 9780486134666
  • Publisher: Dover Publications
  • Publication date: 4/25/2012
  • Series: Dover Books on Mathematics
  • Sold by: Barnes & Noble
  • Format: eBook
  • Pages: 544
  • Sales rank: 862,824
  • File size: 46 MB
  • Note: This product may take a few minutes to download.

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By David V. Widder

Dover Publications, Inc.

Copyright © 1989 David V. Widder
All rights reserved.
ISBN: 978-0-486-13466-6


Partial Differentiation

§1. Introduction

We shall be dealing in this chapter with real functions of several real variables, such as u = f(x, y), u = f(x, y, z), etc. In these examples the variables x, y, z, ... are called the independent variables or arguments of the function, u is the dependent variable or the value of the function. Unless otherwise stated, functions will be assumed single-valued; that is, the value is uniquely determined by the arguments. Multiple-valued functions may be studied as combinations of single-valued ones. For example, the equation

(1) u2 + x2 + y2 = a2

defines two single-valued functions,

(2) u = +√a2 - x2 - y2

(3) u = -√a2 - x2 - y2 x2 + y2 ≤ a2

A function of two variables clearly represents a surface in the space of the rectangular coordinates x, y, u. In the study of functions of more than two variables, geometrical language is often retained for purposes of analogy, even though geometric intuition then fails.


A partial derivative of a function of several variables is the ordinary derivative with respect to one of the variables when all the rest are held constant. Various notations are used. The partial derivatives of u = f(x, y, z) are

[partial derivative]u/[partial derivative]x = f1(x, y, z) = [partial derivative]f/[partial derivative]x = [partial derivative]/ [partial derivative]x f(x, y, z)

[partial derivative]u/[partial derivative]y = f2(x, y, z)

[partial derivative]u/[partial derivative]z = f3(x, y, z)

An important advantage of the subscript notation is that it indicates an operation on the function that is independent of the particular letters employed for the arguments. Thus, if f(x, y, z) = xzy, we have

f2(x, y, z) = xzy log z

f2(r, s, t) = rts log t

It shares this advantage with the familiar f'(x) for the derivative of a function of one variable. The notations for the value of a derivative at a point are illustrated by


For example,





The example of §1 serves to illustrate how a function may be defined implicitly. Thus, equation (1) defines the two functions (2) and (3), which are said to be defined implicitly by (1) or explicitly by (2) and (3). In other cases, a function may be defined implicitly even though it is impossible to give it explicit form. For example, the equation

(4) u + log u = xy

defines one single-valued function u of x and y. Given any real values of the arguments, the equation could be solved by approximation methods for u. Yet u cannot be given in terms of x and y by use of a finite number of the elementary functions.

The partial derivatives of a function defined implicitly may be obtained without using an explicit expression for the function. One has only to differentiate both sides of the defining equation with respect to the independent variable in question, remembering that the dependent variable is really a function of the independent ones. For example, differentiating equation (1) gives

2x + 2u[partial derivative]u/[partial derivative]x = 0, [partial derivative]u/[partial derivative]x = -x/u

2y + 2u[partial derivative]u/[partial derivative]y = 0, [partial derivative]u/[partial derivative]y = -y/u

These results can be checked directly by use of equation (2) or of equation (3). From equation (4) we would have

[partial derivative]u/[partial derivative]x = uy/u + 1, [partial derivative]u/[partial derivative]y = ux/u + 1

The method applies equally well if several functions are defined by simultaneous equations.


One could also solve for [partial derivative]v/[partial derivative]x. To obtain the derivatives with respect to y, one has only to differentiate the defining equations with respect to that variable.


Partial derivatives of higher order are obtained by successive application of the operation of differentiation defined above. The notations employed will be sufficiently illustrated by the following examples. If u = f(x, y, z),


A function of two variables has two derivatives of order one, four of order two, and 2n of order n. A function of m independent variables will have mn derivatives of order n. Later we shall see that many of the derivatives of a given order will be equal under very general conditions. In fact, the number of distinct derivatives of order n is then the same as the number of terms in a homogeneous polynomial in m variables of degree n:





1. Find. [partial derivative]/ [partial derivative]x sin xy/cos(x + y).

2. Find [partial derivative]2/[partial derivative]r2 log (r2 + s).

3. Find f1(x, y) and f2(1, 2) if f(x, y) = tan2 (x2 - y2).

4. If z2 = x2 - 2xy - 1, find [partial derivative]z/[partial derivative]x when x = 1, y = -2, z = -2 first by the explicit and then by the implicit method.

5. If

u - v + 2w = x + 2z

2u + v - 2w = 2x - 2z

u - v + w = z - y

find [partial derivative]u/[partial derivative]y, [partial derivative]v/[partial derivative]y, [partial derivative]w/[partial derivative]y.

6. If f(x, y) = x tan-1 (x2 + y), find f1(1, 0), f2(x, y).

7. If f(x, y, z) = x log y2 + yez, find f1(1, -1, 0), f2(x, xy, y + z).

8. If [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII], find [partial derivative]u/[partial derivative]x, [partial derivative]u/[partial derivative]y, [partial derivative]u/[partial derivative]z.

9. If u = xu + uy, find [partial derivative]u/[partial derivative]x, [partial derivative]u/[partial derivative]y.

10. If

u2 + x2 + y2 = 3

u - v3 + 3x = 4

find [partial derivative]u/[partial derivative]x, [partial derivative]u/[partial derivative]y, [partial derivative]v/[partial derivative]x, [partial derivative]v/[partial derivative]y.

11. If u = xy, show that

[partial derivative]3u/[partial derivative]x2[partial derivative]y = [partial derivative]3u/[partial derivative]x [partial derivative]y [partial derivative]x.

12. Prove the statement in the text about the number of terms in a homogeneous polynomial.

13. In Example C find [partial derivative]2u/ [partial derivative]x2.

14. If

2u + 3v = sin x

u + 2v = x cos y,

find u(x, y) explicitly and then u1(π/2, π). Find the same derivative by the implicit function method.

15. If

u2 + v2 = x2

2uv = 2xy + y2

find u1(x, y) when x = 1, y = -2, u = 1, v = 0, by the two methods of the previous problem.

§2. Functions of One Variable

We recall here certain notions about functions of one variable, which the student is assumed to have met before, perhaps in a less precise form. We shall also introduce certain abbreviating notations that will facilitate the statement of theorems.


A function f(x) approaches a limit A as x approaches a if, and only if, for each positive [member of] number e there is another, δ, such that whenever 0 < | x - a | < δ we have | f(x) - A | < [member of]. That is, when x is near a (within a distance δ from it), f(x) is near A (within a distance [member of] from it). In symbols we write



For, in this example, we may choose δ equal to the given [member of]. We have

|f(x) - A| = |√x - 1| = |x - 1|/√x + 1 0 < x< 2

If 0 < | x - 1 | < δ = [member of], we obtain

|√x - 1| < ε/√x + 1 < ε

Example B.f(x) = sin (1/x) x ≠ 0

Here f(x) has no limit as x approaches zero. Since f(x) takes on the values -1 and +1 infinitely often in every neighborhood of the origin, it is certainly not within a distance less than 1 from any number throughout any neighborhood of the origin.

If, in the definition of limit, the first inequalities are replaced by 0 < x - a< δ (0 < a - x< δ), we say that lim f(x) = A as x approaches a from above (below) and write



It is now easy to formulate what is meant by a continuous function. Let vis first introduce the following symbols:

[member of]—"belongs to" or "is a member of."


[??]—"implies and is implied by" or "if, and only if."

C—"the class of continuous functions."



This may be read, "f(x) belongs to the class of functions continuous at x = a" (or "f(x) is continuous at x = a") "if, and only if, the limit of f(x) is f(a) as x approaches a."

In Example A, the function √x is continuous at x = 1, since also √1. Observe that the last equality in Definition 1 is equivalent to


For a function to be continuous at x = a, it certainly must be defined there. Thus, f(x) = (sin x)/x is not continuous at x = 0 in the first instance, since division by zero is undefined. However, if f(0) is defined as 1, f(x) becomes continuous at x = 0. In Example B, f(x) is discontinuous at x = 0 on two counts: f(0) is undefined, and the limit involved does not exist. No choice of definition for f(0) could make f(x) continuous at x = 0.


Excerpted from ADVANCED CALCULUS by David V. Widder. Copyright © 1989 David V. Widder. Excerpted by permission of Dover Publications, Inc..
All rights reserved. No part of this excerpt may be reproduced or reprinted without permission in writing from the publisher.
Excerpts are provided by Dial-A-Book Inc. solely for the personal use of visitors to this web site.

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Table of Contents


1 Partial Differentiation,
2 Vectors,
3 Differential Geometry,
4 Applications of Partial Differentiation,
5 Stieltjes Integral,
6 Multiple Integrals,
7 Line and Surface Integrals,
8 Limits and Indeterminate Forms,
9 Infinite Series,
10 Convergence of Improper Integrals,
11 The Gamma Function. Evaluation of Definite Integrals,
12 Fourier Series,
13 The Laplace Transform,
14 Applications of the Laplace Transform,
Selected Answers,
Index of Symbols,

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    Posted September 26, 2009

    As expected

    The book is comprehensive and well-written, but it sometimes hard to get through. I would buy it again.

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