By Jonathan M. Borwein

Thirty years in the past mathematical, rather than utilized numerical, computation used to be tough to accomplish and so really little used. 3 threads replaced that: the emergence of the non-public laptop; the invention of fiber-optics and the resultant improvement of the fashionable net; and the development of the 3 “M’s” Maple, Mathematica and Matlab.

We intend to cajole that Mathematica and different comparable instruments are worthy understanding, assuming basically that one needs to be a mathematician, a arithmetic educator, a working laptop or computer scientist, an engineer or scientist, or an individual else who wishes/needs to take advantage of arithmetic greater. We additionally wish to provide an explanation for easy methods to turn into an "experimental mathematician" whereas studying to be higher at proving issues. to complete this our fabric is split into 3 major chapters through a postscript. those hide effortless quantity concept, calculus of 1 and several other variables, introductory linear algebra, and visualization and interactive geometric computation.

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**Extra resources for An Introduction to Modern Mathematical Computing: With Mathematica®**

**Sample text**

Now, returning to the question of convergence of 1/k 2 , we have yet to verify analytically whether the series converges or diverges. At this stage it is often a good idea to postpone our desire for an analytic answer in favor of an more computational (or experimental) approach. With this in mind, recall that an inﬁnite sum is a limit of partial sums. That is, ∞ N a(k) = lim k=1 N →∞ a(k) k=1 As our next attempt we attempt to see how the partial sums behave. To do this we begin by creating a function to compute the N th partial sum of the 1/k 2 series.

This is something we already know how to express in Mathematica. In order to see if, say, 3 was a divisor of our 6 then we could issue the command In[169]:= If[6 ~Mod~ 3 == 0, 6 / 3] Out[169]= 2 The above code should be read as “If 6 is equal to 0 modulo 3 then calculate 6/3,” and because 6 is most certainly equivalent to 0 modulo 3, Mathematica has correctly gone on to calculate 6/3 = 2. Note that if, instead of 6, we had used some other number such that was not equivalent to 0 (modulo 3) then Mathematica would have performed no calculation at all.

We use the Range function to ﬁrst compute the list of integers from 1 to N , and then use Cases to choose from that list the elements which match our criterion. In[193]:= With[{N = 10 000}, Cases[Range[N], n_Integer /; (Divisors[n ] // Total) == 2 * n ] ] Out[193]= {6, 28, 496, 8128} Finally, we apply these same concepts directly to a new problem. Suppose we have a natural number, n, that is not perfect. Let m be the sum of the proper divisors of n. It is possible (but not necessarily likely) that n also happens to be equal to the sum of the proper divisors of m.