Talk:Operational calculus

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Sources contribution

This article is very incomplete. You should have a lager article that shows a fuller scope of operational calculus

I have added a very rough sketch of what Heaviside's operators are, and mentioned briefly some of the rules to compute the action of operators on functions. I have also mentioned connections to Laplace transformation and fractional calculus and the applications of the method in electrical engineering. I have added a list of references, most of them available on the Internet that may help to flesh out this article.

86.206.180.78 19:56, 24 July 2007 (UTC)[reply]

Removing the example

I have removed the example of functional calculus, despite the fact that it was present from the very start of this article. The reason is that the example is, unfortunately, incomprehensible in its present form. Here is the example in question:

(start of example)

An example of this calculus is given below: Problem:

$L(2n)=L^{2}(n)+2\times (-1)^{n+1}$

Solution:

$\left(e^{D}+e^{-D}\right)F(n)=e^{nD}-(-1)^{n}e^{-nD}$

For function $F(a)$ we have:

$\left[F(a+1)+F(a-1)\right]F(n)=F(a+n)-(-1)^{n}F(a-n)$

or

$L(a)F(a)=F(a+n)-(-1)^{n}F(a-n)$

for $a=2n$ and $F(3n)=\left[L^{2}(n)+(-1)^{n+1}\right]F(n)$

$L(2n)=L^{2}(n)+2\times (-1)^{n+1}$

(end of example)

What is the statement of the problem? Is it to solve the functional equation for $L(x)$ ? If so, what is the solution of the problem? The solution should ideally be written as “ $L(x)=$ an explicit expression involving only known quantities”; no such statement appears in the write-up. What is the nature of the function $F(x)$ ? Is it an operator? What does it have to do with $L(x)$ ?

If anyone understands what this example is supposed to be saying, please rewrite it, with explanations. Reuqr (talk) 03:07, 9 April 2010 (UTC)[reply]

Wrong link

The link given here - O Heaviside Proc. Roy. Soc. (London) 52. 504-529 (1893), 54 105-143 (1894). [Original articles] is wrong. 86.132.223.249 (talk) 16:15, 16 March 2017 (UTC)[reply]