From: kovarik@mcmail.cis.McMaster.CA (Zdislav V. Kovarik) Subject: Re: ordinary and partial diff eq. and the meaning of dt. Date: 5 Aug 1999 04:06:53 -0400 Newsgroups: sci.math In article <37A76BCA.13A80BBE@i.am>, wis wrote: >First, excuse my poor english and my poor ascii art. >Then remove .poutou in my email to reply. > This may be of interest to others, so no e-mail. [...] >What is dy/dt exactly, for me it is JUST a notation for > y(t+h)-y(t) > lim ----------. > h->0 h > > 1. so why can we multiply both part by dt ? > 2. why can we put both integral sign ? The definition of the derivative of a real-valued function of a single real variable is correct (an assumption about t being an interior point of the domain of f needs to be included). Bad news: As stated, the definition is useless or exceedingly clumsy to extend to many variables. Why? You can't divide by vectors in any way useful for calculus. So, a reform was needed; done by Frechet or someone before him, but the news did not reach the chambers of most educators, or most textbook authors. What should the derivative be, according to Frechet? A linear transformation (call it L for the time being and remember that it depends on x) acting on the increment h in the "short Taylor's expansion with remainder" f(x+h) = f(x) + L(h) + R(x;h) where R(x;h)/norm(h) goes to 0 as norm(h) goes to 0. (Norm: usually the Euclidean length.) Example: let f(x) = 1/x where x is not 0. Then, subtracting one term at a time and simplifying, we get (check it out!) 1/(x+h) = 1/x - h/x^2 + h^2/(x^2 * (x+h)) Here L(h) is defined by L(h) = h/x^2, and it is linear in h indeed, while the term h^2/(x^2 * (x+h)) can be declared the remainder term (divide by abs(h), and it will still go to 0 as h goes to 0). Who has difficulties with general x, might start by replace x with 2 and write: 1/(2+h) = 1/2 - h/4 + h^2/(4*(4+h)) so the linear transformation is L(h) = (-1/4) * h. *** Here is the point that clouds the understanding of the derivative in *** one variable: *** A linear transformation in one variable is simply multiplication by a *** constant. This constant is traditionally (and simplistically) *** presented as the derivative. Instead of the concept of a *** proportionality relation, one is shortchanged by giving just that *** constant of proportionality. And back to dx and dy: In the context of the relation y = f(x) , one writes the "derivative relation" using the linear transformation L, as dy = L(dx) = f'(x) * fx A fine point of domain of function: The variable h in Taylor's expansion is often restricted, so that (x+h) is still in the domain of f. No such restriction is needed for the variable dx: it can run through all real numbers. The punchline: If a (user-supplied) value of dx is not 0 then we can divide by dx, and reconcile the L-notation with the prime notation: dy / dx = f'(x) keeping in mind that a specific x is considered, but otherwise with no further fuss. Division as we know it. Let's go for 2 variables, and let f(x, y) = x * y^2 . Then the polynomial routine gives (the increment of the vector (x,y) is called by me (h,k): f(x+h, y+k) = f(x,y) + y^2 * h + 2*x*y * k + h*k*(2*y + k) Immediately, the second line expression + y^2 * h + 2*x*y * k is a linear transformation of a vector (h,k) into a number, and the third line term can be shown to go to 0 even after being divided by norm((h,k)) = sqrt(h^2 + k^2) . So, L defined by L(h,k) = (y^2) * h + (2*x*y) * k is the derivative. (As a piece de resistance, I like to show the students how concise the derivative of the matrix inversion function (matrix-valued function of a matrix variable!) looks when you present it as a linear transformation: Let Inv be defined as Inv(X) = X^(-1) for all invertible matrices. Then d Inv(X) (H) = - Inv(X) * H * Inv(X) and writing it down in terms of cofactors is pure torture.) With the extension as before, replace h by dx and k by dy, and from the relation z = f(x,y) we obtain dz = L(dx, dy) = (y^2) * dx + (2*x*y) * dy One should recognize the coefficients of the linear transformation as the "partial derivatives", and prove familiar theorems about them. So, in case of differentiable f, df(x,y) = (partial d)f/(partial d)x * dx + (partial d)f/(partial d)y * dy The dx and dy are, respectively, the increments of x and y, extended for the purposes of linear transformation to all pairs of real numbers. > for me > / > | f dt > /is only a NOTATION for the primitive or the RIEMANN integral if >borned, the dt has no sense. [ 3D example skipped because 2 variables already illustrate the point ] The integral question is a case of fusing two concepts into one (Riemann's integral and "Newton's integral") because they seem to lead to the same formula manipulation to obtain results: On a straight line segment, the function of two variables x and dx, linear in dx, defined by f(x) * dx is the same as what is called in differential geometry a "differential form". It is this differential form that is being integrated by "boundary evaluation of the primitive form": finding F such that (the notation is deliberately redundant) F'(x) * dx = f(x) * dx and then int[a to b] f(x) * dx = F(b) - F(a) Observe: No mention of Riemann approach, and the problem of the existence of F is postponed. We can still prove and use the sum formula, substitution rule, and integration by parts, before we learn about Riemann. It is a fundamental achievement of early Calculus developers, showing that this (Newton's, if you need a name) integral of f(x)*dx comes out the same as Riemann's integral of f (and the question of existence is resolved). Remark: To emphasize the distinction, some textbooks write (replace int by that tall S and edit graphically) int[a to b] f , or even int_J f where J = [a, b] for Riemann's integral, to indicate by the absence of dx (or d(whatever)) that Riemann's or related ideas are followed here. (In advanced integration theory, dx is short for d(mu)_1 where (mu)_1 is one-dimensional Lebesgue measure (restricted if required). Tradition, rather than notational consistency, prevailed.) Now the manipulations which treat dy/dx as a true quotient become legitimate (as long as dx is not 0). Hope some of it helped, ZVK(Slavek).