Group Abstract Group Abstract

Message Boards Message Boards

Mathematica beyond mathematics

POSTED BY: JOSÉ GUILLERMO SÁNCHEZ LEÓN
12 Replies
Sort By:
POSTED BY: JOSÉ GUILLERMO SÁNCHEZ LEÓN
Updating Name
Updating Name
Posted 7 years ago
POSTED BY: Updating Name

This option is explained later in the book, for this reason I avoid to use a function until the concept it has not been explained.

In any case supplementary material and corrections (very few it has been necesary) are available at
http://Mathematicabeyondmathematics.com

Thanks

POSTED BY: JOSÉ GUILLERMO SÁNCHEZ LEÓN
Rohit Namjoshi
Rohit Namjoshi
Posted 7 years ago

Or apply the rule to the right level.

Replace[list, {a_, b_, c_} -> a/(b + c), 2]
POSTED BY: Rohit Namjoshi

Thank you Paul

I have clarified this problem in the book web site: http://Mathematicabeyondmathematics.com

The rule works fine except if the list has exactly the same size that the number of sublist, three in this case. You can test with a list with more than 3 sublist.*) type list={{a1, b1, c1}, {a2, b2, c2}, {a3, b3, c3}, {a4, b4, c4}};

To avoid this type of problems I suggest in the book use pure functions: e.g. (#1/(#2 + #3)) & @@@ list

But your proposal it is a good option
POSTED BY: JOSÉ GUILLERMO SÁNCHEZ LEÓN
Paul Horth
Paul Horth
Posted 7 years ago

Hi José,

Just to say, I'm thoroughly enjoying your book.

One thing confused me though: on pages 78 and 79, you define a list

list = {{a1, b1, c1}, {a2, b2, c2}, {a3, b3, c3}}

and then use replacement list /. {a, b, c_} -> a/(b + c) But this produces

{a1/(a2 + a3), b1/(b2 + b3), c1/(c2 + c3)}

Not

{a1/(b1 + c1), a2/(b2 + c2), a3/(b3 + c3)}

as suggested should be the answer.

From my limited knowledge, I think it is because it sees a, b and c being the three lists inside rather than the elements of each list.

Am I on the right track?

Also, from a bit of research, I think this would cure the problem:

list /. {a_Symbol, b_Symbol, c_Symbol} -> a/(b + c)

Thank you for your time.

Paul
POSTED BY: Paul Horth

Dear friend I have run all notebooks included in my book Mathematica Beyond Mathematics in Mathematica 11.3.0. I have updated the Mathematica beyond mathematics (it is necessary to modify only two functions using 11.3). Supplementary materials: http://diarium.usal.es/guillermo/mathematica/

In this occasion I have found very few differences with the previous version. I would like to share these differences not only for my readers but also for all Mathematica users.

a) In Mathematica 11.3 in some occasions it could be necessary to include the WorkingPrecision NIntegrate[…., WorkingPrecision -> 10]

NIntegrate[PDF[NormalDistribution[0.49, 0.02], p]*PDF[BinomialDistribution[3030, p], 1515], {p, 0, 1}, WorkingPrecision -> 10] In Math 11.2 you can avoid to including WorkingPrecision

b) According to Mathematica support:"hanges in Import and Export of XLSX files have been implemented to highly improve performance. ...Our development team has decided to include the legacy functionality of previous versions as well,You can use the legacy importer with SheetsLegacy."

Example (included in my book): Import["bp-statistical-review-of-world-energy-2016-workbook.xlsx", {"SheetsLegacy", "Primary Energy Consumption "}];

c) I have found a problem in Dataset that I hoppe be fixed in a next version. Dataset[EntityValue[EntityClass["Particle", "Lepton"], {"Symbol", "Charge", "Mass"}, "EntityPropertyAssociation"]] (* Note, RowBox and FractionBox in the output if you use Math 11.3.0. *)

J. Guillermo Sanchez
POSTED BY: JOSÉ GUILLERMO SÁNCHEZ LEÓN

In fairness, it should be noted that in his presentations, Stephen Wolfram often emphasizes using Mathematica in various areas outside of mathematics proper; indeed, he sometimes almost grudgingly shows new mathematical features. See, e.g.: https://www.twitch.tv/stephen_wolfram
POSTED BY: Murray Eisenberg
POSTED BY: JOSÉ GUILLERMO SÁNCHEZ LEÓN

Math 11.0.1 (tested with 11.1)

Guillermo
POSTED BY: JOSÉ GUILLERMO SÁNCHEZ LEÓN
Eric Laarz
Eric Laarz
Posted 9 years ago

Which recent versions of Mathematica are covered by the book?

POSTED BY: Eric Laarz

Looks like a great book.

About (iii), I see from the table of contents that this is a big book. To learn Wolfram Language, you are probably not suggesting to read the whole book, but it is enough to do chapters 1-3 (and maybe skip part of 3). One doesn't need to learn 6000 functions. I am wondering how many functions do you think someone needs to learn?

Out of curiosity, are you planning to make a Spanish version?
POSTED BY: Todd Rowland
Reply to this discussion
Community posts can be styled and formatted using the Markdown syntax.
Tag limit exceeded
Note: Only the first five people you tag will receive an email notification; the other tagged names will appear as links to their profiles.
Reply Preview
Attachments
Remove
or Discard