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June 3

Tangent/secant functions

Who discovered the tangent and secant functions? According to the article Trigonometric functions#History, it is Muhammad ibn Mūsā al-Khwārizmī and Abū al-Wafā' Būzjānī, respectively; according to the article Thomas Fincke, it is Thomas Fincke. Which is correct here? I try to ask here to clear up the discrepancies and to be able to correct the articles in question. Iceblock (talk) 04:53, 3 June 2010 (UTC)[reply]

Much mathematics was known to eastern scholars long before it was formally introduced to western civilisation. I suspect that this is true of these functions, though I don't know whether Thomas Fincke independently discovered the functions or just published the knowledge. Perhaps someone has access to his publication? I think he probably just introduced the words (and abbreviations). They were formerly known as "umbra versa" & "umbra recta", translations into Latin of the Arabic "tables of shadows" which were, in turn, developments of work by Indian mathematicians. Our article History of trigonometry has more detail of the development. Dbfirs 06:24, 3 June 2010 (UTC)[reply]

It's not a huge mystery, the Thomas Fincke article says that he 'introduced' secant and tangent but what that means is he introduced the terms, not that he invented or discovered the functions. The article should be rephrased to avoid confusion but it's not a discrepancy.--RDBury (talk) 06:49, 3 June 2010 (UTC)[reply]

Yes, just the terms. Regiomontanus had prevoiusly published a table of tangents in the west. I'll make the adjustment to the article if no-one else has done so. Dbfirs 06:57, 3 June 2010 (UTC)[reply]

Thank you very much, both of you! Iceblock (talk) 07:26, 3 June 2010 (UTC)[reply]

What is the name?

Is there a name for $\scriptstyle {\frac {a-b}{a+b}}$ when $\scriptstyle a>b>0$ ? Bo Jacoby (talk) 16:09, 3 June 2010 (UTC).[reply]

It does come up often but I haven't seen a name. Half the relative difference would be one name - but there's all sorts of definitions of relative difference. You might also like to look at the law of tangents Dmcq (talk) 19:56, 3 June 2010 (UTC)[reply]

Is there a name for the ratio $\scriptstyle {\frac {\sigma }{\mu }}$ between the standard deviation and the arithmetic mean? That would also solve the first problem. Bo Jacoby (talk) 09:28, 4 June 2010 (UTC).[reply]

Coefficient of variation. I don't see how that solves the first problem though. --Qwfp (talk) 10:26, 4 June 2010 (UTC)[reply]

Thank you very much! Just what I needed. If a population contains only two numbers, $\scriptstyle a>b$ , then the mean value is $\scriptstyle \mu ={\frac {a+b}{2}}$ and the standard deviation is $\scriptstyle \sigma ={\sqrt {\frac {(a-\mu )^{2}+(b-\mu )^{2}}{2}}}={\sqrt {\frac {\left(a-{\frac {a+b}{2}}\right)^{2}+\left(b-{\frac {a+b}{2}}\right)^{2}}{2}}}={\sqrt {\frac {\left({\frac {a-b}{2}}\right)^{2}+\left({\frac {-a+b}{2}}\right)^{2}}{2}}}={\sqrt {\frac {2\left({\frac {a-b}{2}}\right)^{2}}{2}}}={\sqrt {\left({\frac {a-b}{2}}\right)^{2}}}={\frac {a-b}{2}}$ , so the coefficient of variation is $\scriptstyle {\frac {\sigma }{\mu }}={\frac {\frac {a-b}{2}}{\frac {a+b}{2}}}={\frac {a-b}{a+b}}.$ Bo Jacoby (talk) 08:25, 5 June 2010 (UTC).[reply]

That's quite a long derivation for σ. It's obvious that the absolute deviation from the mean is

{\tfrac {a-b}{2}}

for both points, so σ, the rms of the deviation, is also

{\tfrac {a-b}{2}}

. -- Meni Rosenfeld (talk) 20:03, 5 June 2010 (UTC)[reply]

True. But when Qwpf says: I don't see how that solves the first problem it is perhaps not helpful to answer that it's obvious. Obviousness is subjective. Note also that the absolute deviation does not enter into the derivation. Only the square of the signed deviation enters. Bo Jacoby (talk) 23:54, 5 June 2010 (UTC).[reply]

Integrating

I'm trying to integrate 1/(x-x^2), but can't remember which method to use. I considered u-substitution and integ. by parts, but they don't work. Please advise. —Preceding unsigned comment added by 142.58.43.148 (talk) 21:33, 3 June 2010 (UTC)[reply]

Have a look at partial fractions and partial fractions in integration Dmcq (talk) 21:56, 3 June 2010 (UTC)[reply]

\int {\frac {1}{x-x^{2}}}dx=\ln \left|x-x^{2}\right|+C

. 76.199.147.211 (talk) 22:01, 3 June 2010 (UTC)[reply]

'fraid not. 1/(x-x^2) turned into partial fractions is

{\frac {1}{x-x^{2}}}={\frac {1}{x}}+{\frac {1}{1-x}}

You get a minus on the second term when you integrate it. Dmcq (talk) 22:12, 3 June 2010 (UTC)[reply]

...and therefore you get a logarithm of a quotient:

\log |x|-\log |1-x|=\log \left|{\frac {x}{1-x}}\right|.

It's easy to imagine "76.199.147.211" seeing 1/x + 1/(1 − x) and ultimately getting the same wrong answer that he or she posted originally. Michael Hardy (talk) 17:16, 7 June 2010 (UTC)[reply]

It can also be done by u-substitution, try

u={\frac {1-x}{x}}.

Martlet1215 (talk) 16:10, 4 June 2010 (UTC)[reply]

You'd have to already know the answer before you'd think of using that transformation! I think the usual term for that is 'magic'. Dmcq (talk) 19:32, 4 June 2010 (UTC)[reply]

Speak for yourself. Writing it in terms of

u

was a bit pedagogic but I spotted that I could factor out

1/x^{2}

and integrate directly. Martlet1215 (talk) 23:37, 8 June 2010 (UTC)[reply]

That's my point, it isn't how you worked it out. For pedagogy you should show a way a person could work it out for themselves. Dmcq (talk) 09:41, 9 June 2010 (UTC)[reply]

Ok, you're entitled to your opinion. I only posted it because the OP stated that u-substitution doesn't work. Personally, I see my method as substitution where I've skipped steps because I'm familiar with the technique. Martlet1215 (talk) 18:32, 9 June 2010 (UTC)[reply]