Hubbard Vector Calculus Solution Manual (optional) Lial, Greenwell, and Ritchey, Student Solutions Manual for Finite Hubbard and Hubbard, Vector Calculus, Linear Algebra and Differential Forms: A. 淘豆网网友bolee65近日为 您收集整理了关于Hubbard-Vector Calculus, Linear Algebra, and Differential Forms-A Unified Approach Solution manual的文档，希望. Vector calculus linear algebra and differential forms a unified approach Details Category: Mathematics Vector calculus linear algebra and differential forms a unified approach Material Type Book Language English Title Vector calculus linear algebra and differential forms a unified approach Author(S) John Hamal Hubbard (Author) Barbara. Ful lls the vector calculus requirement for the math major. The textbook for the course is Vector Calculus, Linear Algebra, and Di erential Forms, by Hubbard and Hubbard (4th edition). We will cover the contents of Chapters 1, 2, 4, and 6. Homework: Homework sets will be distributed each Friday and collected the following Wednesday. John Hamal Hubbard, Barbara Burke Hubbard. Upper Saddle River, NJ: Prentice Hall; Toronto: Pearson Education Canada, c2002. We are aware that students taking courses in other fields need to master the language of vector calculus, and we devote three sections of Chapter 6 to integrating the standard vector.

1. Vector Calculus Identities
2. Hubbard And Hubbard Vector Calculus

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Vector Calculus Identities

May 12, 2011Ronald Lett rated it it was amazing · review of another edition

Hubbard And Hubbard Vector Calculus

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I have a question about the solution to part (a) of exercise A12.5 of Hubbards' Vector Calculus, Linear Algebra, and Differential Forms.Here is the exercise:

Let $f $ be the function $f begin{pmatrix} x y end{pmatrix}=sgn(y) sqrt{ frac {-x+sqrt{x^2+y^2}}{2}} $ where $sgn(y)$ is the sign of y (i.e., $+1$ when $y>0$ , $0$ when $y=0$ ,$-1$ when $y<0$ )

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a. show that $f$ is continuously differentiable on the complement of the half line $y=0, xle 0$.

I've included part of the solution on the student solution manual to this exercise below:

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To show that $f$ is continuously differentiable on the locus where $y=0,x>0$:

In a neighborhood of a point $begin{pmatrix} x_0 y_0 end{pmatrix}$satisfying $x_0>0 $ , $y_0=0$ , we can write
$-x+sqrt{x^2+y^2}=-x+xsqrt{1+frac{y^2}{x^2}}= -x+x(1+frac{1}{2}frac{y^2}{x^2}+o(frac{y^2}{x^2}))$
$ =frac{y^2}{2x}+o(frac{y^2}{x})$,
and since $sgn(y)y^2$ is of class $C^1$, the function is of class $C^1$ on the half axis $y=0, x> 0$.

I don't understand the bold part of the solution. Could you explain why the fact that $sgn(y)y^2$ is of class $C^1$ means that the function $f$ is continuously differentiable on the half-axis $y=0, x>0$ ?

(The book uses the following definition for little $o$ notation: a function $f$ is in $o(h)$ if $lim limits_{x to 0}frac{f(x)}{h(x)}=0.$)

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2 Answers

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In order to better understand what's going on here introduce polar coordinates for the moment. Then$$x=rcosphi,qquadsqrt{x^2+y^2}=r$$and therefore$${1over2}bigl(sqrt{x^2+y^2}-xbigr)=r{1-cosphiover2}=rsin^2{phiover2} .$$It follows that$$sqrt{{1over2}bigl(sqrt{x^2+y^2}-xbigr)}=sqrt{r}left sin{phiover2}right .$$As $ {rm sgn}bigl(sin{phiover2}bigr)={rm sgn}(y)$ we therefore obtain$$f(x,y)=sqrt{r}sin{phiover2}={root 4of {x^2+y^2}} sin{{rm Arg}(x,y)over2} .$$As$$nabla{rm Arg}(x,y)=left({-yover x^2+y^2},{xover x^2+y^2}right)$$the function $f$ is even real analytic in the indicated domain.

Christian BlatterChristian Blatter$endgroup$$begingroup$

Definitely strange. In fact, $f_y(x_0,y_0)$ does not exists for $x_0 > 0$, $y_0 = 0$:

First, ignore the irrelevant 2 and consider $$f(x,y) = hbox{sgn}(y)sqrt{sqrt{x^2+y^2} - x}.$$$$lim_{yto 0}frac{sqrt{sqrt{x_0^2+y^2} - x_0} - 0}{y - 0} = lim_{yto 0}sqrt{frac1{2x_0} + frac{o(y^2/x_0)}{y^2}} = frac1{sqrt{2x_0}},$$so the lateral limits of$$lim_{yto 0}frac{hbox{sgn}(y)sqrt{sqrt{x_0^2+y^2} - x_0} - 0}{y - 0}$$are different.Using$$sqrt{x_0^2+y^2} - x_0 =frac{(sqrt{x_0^2+y^2} - x_0)(sqrt{x_0^2+y^2} + x_0)}{sqrt{x_0^2+y^2} + x_0} =frac{y^2}{sqrt{x_0^2+y^2} + x_0}$$gives the same result.

Martín-Blas Pérez PinillaMartín-Blas Pérez Pinilla$endgroup$

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