Question on 1.3-14 > on problem 1.3-14 i'm kind of confused on what its asking for when it > says the existence nor uniqueness is guaranteed in some neighborhood of > x=0. Do i make a domain with the center being 0,0? > Is this the same equation you gave in class when you did > the example y'=y^(1/3) and y(0)=0? ANSWER: Create a box B with center x=0, y=0 and verify that f(x,y)=y^(1/3) is continuous on the box. Then consider any box C with center x=0, y=0. Verify that the partial of f on y is discontinuous in C. These two checks verify that Peano's theorem applies but Picard-Lindelof fails to apply. Please refer to the Peano and Picard slides, instead of the textbook, which since the second edition has had some disastrous edits applied to Theorem 1, making problem 1.3-14 unintelligible. The original Theorem 1, to which the problem refers, contained both statements: Peano and Picard-Lindelof. Functions and Continuity Background FUNCTION. ======================= A function is a rule or equation that defines one value for each instance of its variables. For example, y=1+x defines one y for each value of the variable x. And z=x+y+1 defines one value z for each instance of variable values x,y. FUNCTION NOTATION. ======================= Function notation f(x) or f(x,y) or f(x,y,z) was introduced by Peter Gustav Lejeune Dirichlet (1805-1859). The history can be found at link http://www.amt.canberra.edu.au/dirichle.html Historical remarks about Dirichlet and be found at Wikipedia: http://en.wikipedia.org/wiki/Johann_Peter_Gustav_Lejeune_Dirichlet The variable names x,y,z are supposed to be dummy variables subject to substitution of actual values. For example, if the defining equation is f(x,y) = x + y, then f(0.1,0.7) means Substitute x=0.1 and y=0.7 into the equation x+y, simplify, and report the numerical answer. We would write f(0.1,0.7)= (0.1) + (0.7) = 0.8 and say f of 0.1, 0.7 equals 0.8 This same fundamental idea applies to modern computer languages like C. Here, a function is a subroutine which computes some answer, to be used elsewhere. An example: float f(x,y) float x,y; { return (x+y); } The variable names x,y are dummy variables representing actual floating point numbers. For instance, f(0.1,0.7) represents 0.8, the computer calculation exactly parallel to the mathematical discussion above. CONTINUITY. ======================= A working definition for most applied work is that a function is continuous provided we can legally find its derivative(s). Then all the elementary functions of calculus are continuous, because they appear in the derivative table. An example of a known continuous function that cannot be justified to be continuous based upon differentiation is |x|. We can't legally take its derivative at x=0. However, continuity of |x| can be justified by other means, for example, by viewing it as the pasting together of the differentiable functions x and -x. There are useful theorems about continuity that apply to decide if some complex expression is the equation of a continuous function. These theorems also decide for |x+1|, an expression that is subtly simple. Some notable theorems: The sum of continuous functions is continuous. The same is true for products and quotients. Compositions of continuous functions are continuous. To elaborate on these statement, an expression z = (x+2y)^3/(1+x^2y^2) is continuous, because it is the quotient of continuous functions (x+2y)^3 and 1+x^2y^2. The power (x+2y)^3 is the composition of u^3 and u=x+2y, both of which are continuous (because they are differentiable). The polynomial 1+x^2y^2 is differentiable (in variables x,y) and therefore continuous. Functions like z=1/(x+y) have a natural domain of definition, namely where the fraction is defined, in this case where x+y is nonzero. Because we can legally find its partial derivatives on this domain, we justify it is continuous there. A function like z=(y-1)^(1/3) makes sense everywhere and because it is a composition of u^(1/3) and u=y-1, then it is justified to be continuous everywhere by calculus theorems (see above). It cannot be justified to be continuous everywhere by taking partial derivatives, because the y-partial of z has y-1 in the denominator, causing it to be infinity at y=1. However, the partial derivative argument applies at all other x,y (y not equal to 1). A similar example is z=y^(1/2) in the box defined by 0