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October 29

About the Word 'Rigorize'

Hello! I saw the word 'rigorize' widely used in wiki, but when I refer to dictionaries, I can't find it anywhere. So I wonder whether this is an official word in English. — Preceding unsigned comment added by 113.68.24.244 (talk) 09:11, 29 October 2013 (UTC)[reply]

Wiktionary contains an entry on "rigorize" and defines it as "To make rigorous; for example, to add further detail or elaborate on a proof or demonstration". Gandalf61 (talk) 10:26, 29 October 2013 (UTC)[reply]

I asked several people, but they did not reach an agreement. I wonder if this word is often used by mathematicians. Does it appear in some publications? — Preceding unsigned comment added by 113.67.114.215 (talk) 12:23, 29 October 2013 (UTC)[reply]

Google only finds 4,000-and-something instances of "rigorize", so it is certainly not a common word. Choice of words is very subjective, but I would probably use a synonym such as "formalise" (or "formalize" in American English). Gandalf61 (talk) 12:52, 29 October 2013 (UTC)[reply]

There is no official body that recognizes words in English so it's a matter of what is commonly accepted as proper. Given that it's not found in most dictionaries I would hesitate to use it in a formal setting, but the meaning is self-evident so there's no reason not to use it otherwise. I'm not sure what you (the OP) mean by 'widely used in wiki', I got only 7 matches on WP including talk pages. --RDBury (talk) 17:07, 29 October 2013 (UTC)[reply]

I don't think rigorous means the same thing as formal. I would say "make more rigorous". StuRat (talk) 21:28, 30 October 2013 (UTC)[reply]

convergence order

if G(x)= x -[x-g(x)]/[1-g'(x)/r] has convergence order r+1 ,prove that g(x) has convergence order r . how we can prove this please give some idea.True path finder (talk) 17:51, 29 October 2013 (UTC)[reply]