In reading an article in the New Scientist yesterday, I came across a description of the 'delta function'. It seems as if it has this property:{b,a} (f(x) )dx =f(t) What on Earth would you want that result for? How does the delta function do it, anyway? [Simon on Compuserve, around 1993] --------------------------------------- I hope that after my answer you shall receive also the opinion of other mathematicians, of some physicist or engineer. There is nothing mystical in the delta function, but this mathematical concept belongs to a very huge area of mathematical disciplines, some of which are really hard. The easiest way, I think, to understand this sort of things, is to concentrate on two aspects: 1. Summation rules, 2. Transformation of problems. Consider some set V of functions (which may be defined on the real axis, or also only on the natural numbers 0,1,2,..., so that, in this second case, our functions are simply sequences, or even only on the numbers 1,2,...,N, so the functions are vectors in an N-dimensional space), whose values may be real or complex numbers. In each of these cases you will assume that also the sum of two functions belonging to V belongs to V, and any scalar multiple of such a function. One says then that V is a vector space of functions. For simplicity, assume the last case considered (functions defined on a finite set), so that a function f is given by its values f(1),...,f(N). You may then consider the sum Sf of these values, hence Sf = f(1)+...+f(N). It is clear, that S(f+g)=Sf+Sg and that S(af)=aS(f), if a is any number. One says that the summation operator S is a linear operator. In many applications it happens now that instead of this simple sum you consider a weighted sum, i.e., for fixed w(1),...,w(N), you consider Pf = w(1)f(1)+...+w(N)f(N), and it is clear that also this operator P is linear. In probability theory this is a mean or average. So one is lead to define an average, or mean operator or a summation rule as a linear operator T, defined on the vector space V, with real or complex values (the same as for the single functions). In the case that the functions are defined for infinitely many arguments, e.g., if they are functions of a real variable, one may (and this becomes then the really difficult part) require some continuity condition, i.e., a nice behaviour with respect to limits. Now go back to our simple example, and take a fixed argument, say i, with weight 1, and put all other weights equal to 0. Then Pf = w(i)f(i) = f(i), that is, the operator P picks out the value of f at i. And this is, in this case, the delta function centered at i (you have, also in the general case, a delta function for each argument of the function). Therefore such a delta function is really a very simple object, but the important thing is that it is still a summation rule as the other ones and can be treated together with them: it is the simplest sort of them, similarly as the number 1 in multiplication. Fix now some function w and associate to each function f the integral (in some sense) Pf of the product wf. Then (not bothering about the nature of that integral) you shall still have a summation rule P. Now also in this case you may consider the weighted sum which associates to each f its value f(0). And this is here the delta function! But now we have a difficulty, because (in this infinite case) you cannot more consider delta as a function w! Only as a summation rule! Probably your phantasy is already satisfied and can fill in details alone. Try now to find the coefficients c(n) of the product of two polynomials a(0)+a(1)x+..., b(0)+b(1)x+..., and you shall find that c(n) is the sum of all terms a(k)b(n-k), where one has to put equal to 0 all not really existing terms. For functions f,g defined on the real numbers there exists a similar operation: you define (g*f)(s) as the integral from - to + infinite with integrand g(t)f(s-t). g*f, considered as a function of s, is called the convolution of g and f. What is then delta*f? Again, instead of "integral" you have to use the word "summation rule": (delta*f)(s) is then the value at t=0 of the function t-->f(s-t), that is (delta*f)(s)=f(s), hence delta*f=f. So delta is the 1-element for convolution. The entire discipline we entered belongs to transform theory. For example you may associate to a function f its Fourier transform Ff, which has the fundamental properties that F(f*g)=Ff.Fg and that it transforms derivation into multiplication by the variable (approximately). Often (this is a very subtle point) f and Ff determine each other, and one of them is more easy to study. So you transform a problem, which for f seems very hard, into a problem, which for Ff is easy, solve that problem, and then go back to f. Well, the philosophy is not difficult, but already in this simplicistic distillation rather long, and the mathematics involved is almost all of mathematics, and the applications involved are almost all of applied mathematics! Mathematicians call distributions, what I called ummation rules. Josef Eschgfäller --------------------------------------- I'm sorry, my message was so long that I forgot the most evident reason for the delta function being necessary. Philosophically speaking, harmonic analysis is the mathematical theory of transforming problems, usually it means that one tries to represent an arbitrary function as a sum or integral of trigonometric functions. Things become most manageable if one works with the complex exponential, for which e^ix = cos(x)+isin(x), the two main reasons for introducing complex numbers being here that e^i(x+y)=(e^ix)(e^iy) (from which you may recover the addition formulas for sin and cos) and that multiplication by i is rotation by a right angle. Well, let us therefore try to represent a function f as a weighted sum ("integral") of functions of the form x-->e^ivx, using many v's, and call the weight of the v-th function (Ff)(v). Then you obtain a "function" Ff, which is the Fourier transform of f, and in many cases you shall have f(x) = (integral) ((Ff)(v) e^ivx dv) (maybe with some constant factor). But what happens, if the function f is already of the form x-->e^ivx? It is clear that in this case you have the situation, where our weight is degenerated, i.e. 1 in v and 0 elsewhere. And as I mentioned in the other message, in this case Ff is not more a function, but of course a (very simple) summation rule, the delta function centered at v! And this comes out: The Fourier transform of the constant function 1 (which corresponds to v=0) is the delta function centered at 0 ("the" delta function), the Fourier transform of the function x-->e^ivx is the delta function centered at v. Usually one has to include some norming factor, for example 2PI, forget about it here. Let's try to write it down: (integral) (delta(w) e^iwx dw) = (e^iwx) (evaluated at w=0) = 1 (indep. of x), and, more generally, (integral) (delta)(w-v) e^iwx dw) = (e^iwx) (evaluated at w=v) = e^ivx, and remember that the notation (delta)(w-v) is only symbolic, it does not mean that delta is a function. Delta is a summation rule. The second equation means simply that the function x-->e^ivx may be obtained as a weighted sum, whose only summand (with weight 1) is this function itself. The power and the difficulty of the theory mathematicians constructed is that all things can be combined together, that one can even differentiate such "generalized functions" and that many other goodies (and headaches) are found. Josef Eschgfäller