Talk:Hyperbolic functions/Archive 1

Archive 1

Suggestion

Why are you using sinh^2 x for (sinh x)^2?? Go to the inverse function article and it says the definition for f^2 x when not in trigonometry. Then, it has the exception and it says trigonometric functions, not hyperbolic functions. 66.245.25.240 15:42, 12 April 2004 (UTC)

By convention this form is used with hyperbolic functions as well. -- Decumanus | Talk 15:43, 12 April 2004 (UTC)

The parenthetical comments about pronounciation could be clearer... what type of ch sound in sech, for example. — Preceding unsigned comment added by 128.163.161.40 (talk) 06:22, 12 October 2004 (UTC)

I have this problem

As can be seen from plots the -cosh(x) function is concave in x.

So, if we have w positive then exp{-wcosh(x)} should alos be convex in x, right? Apparantly not always ... according to this article I have this is only valid iff w{sinh(x)}^2<cosh(x), does anybody see why?? — Preceding unsigned comment added by CCMartijn (talk • contribs) 09:53, 3 November 2004 (UTC)

No Period?

"However, the hyperbolic functions are not periodic." I'm not really an expert in this, but someone told me that the hyperbolic functions do have a period, but it is imaginary. Is this true? --Dragglebaggle 03:09, 20 September 2005 (UTC)

Yes, 2πi. --Macrakis 15:52, 20 September 2005 (UTC)

ArcTanh(adj,opp)

I needed a hyperbolic analogue of ArcTan[adjacent,opposite], and "Klueless" provided a relationship that I coded as arcTanh[a_,0]:=If[a==0,∞,0,0]; arcTanh[a_,o_]:=If[Chop[a^2-o^2]==0,∞,Log[(a+o)/SQRT[a^2-o^2]] in Mathematica (ref). (The formula is not in the Mathematica functions library) This allowed me to show that, in the algebra with the Klein 4-group as multiplication table, there is a hyperbolic dual of the Argand diagram, with {a,o} <=> {u=Sqrt[a^2-o^2],theta=arcTanh[a,o]} in which "ulnae" (u) multiply and hyperbolic angles (theta) add on multiplication. The hyperbolic plane is covered by two pairs of hyperbolae. Would a referenced write-up of this be acceptable? It seems to be a significant generalization of hyperbolic functions.

(ref) http://library.wolfram.com/infocenter/MathSource/4894

Roger Beresford. 195.92.168.164 08:22, 7 October 2005 (UTC)

Good stuff, but my impression is that Wikipedia should concentrate on widely accepted definitions, not novel insights. By the way, your formulae might be clearer if you didn't use "o" as a variable name and if you didn't you Mathematica-specific conventions -- I assume that If[a==0,∞,0,0] means "if a==0 then ∞ else 0", but I don't understand why there are two 0's there. --Macrakis 15:10, 8 October 2005 (UTC)

ArcTanh[y,x] is not original research, but plagiarism! (I copied from only one source.) ArcTan[y,x] is accepted as it is needed to cope with different angles in opposing quadrants. ArcTanh[y,x] apparently is not, although ArcTanh[-y,-x] is a different angle. Your veto can stand, though Wiki is poorer for it. 195.92.168.168 09:50, 12 October 2005 (UTC)

Roger, I have no veto, only my judgement based on the arguments above. ArcTanh2 simply doesn't seem widely enough known or used to merit a mention in this article. --Macrakis 11:37, 12 October 2005 (UTC)

ERROR in formula

There is a problem with the expansion series for arccosh. There is a switch of sign in the sum (right), but not in the examples of the first terms (left), and the values do not match. I don't know which is correct. — Preceding unsigned comment added by 132.165.76.2 (talk) 07:16, 14 December 2005 (UTC)

Origin of pronunciations

Does anyone know how the pronunciations for sinh, cosh, etc came about? Do they differ between British and American usages? 128.12.20.195 (talk) 05:47, 22 January 2006 (UTC)

Arc{hyperbolic function} is a misnomer

Arc{hyperbolic function} is a misnomer

The article states correctly that the parameter t represents an area, rather than a (circular) angle. Note also that t does not represent an arc length. As such, it is actually a misnomer to refer to the inverse hyperbolic functions using the prefix "arc".

OTOH, for the inverses of the trigonometric functions, the prefix "arc" is not a misnomer, since their values may indeed be thought of as representing arc lengths (or circular angles).

Arc{hyperbolic function} probably originated due to a false analogy with trigonometric functions. In any event, it is sometimes used nowadays. But IMO its usage should be deprecated in favor of one of several better alternatives. I prefer the simplest alternative: a{hyperbolic function}, in which "a" may correctly be taken to represent "area".

What do other people think? Should we change the article's currect arc{hyperbolic function} notation to my preference, or some other alternative, or leave the current notation as is?

--David W. Cantrell 07:23, 31 December 2004 (UTC)

Hi, I think it the prefix for inverse hyperbolic functions used to be Sect, as in "sector". Why exactly I'm not sure, but that's what I grew up with. Orzetto 09:14, 16 March 2005 (UTC)

Probably because sect was easily confused with sec as in secant. Especially if the angle was t, e.g. sectsint would be very ambiguous. 218.102.221.84 07:04, 30 December 2005 (UTC)

The prefix for inverse hyperbolic functions is ar. The original latin names are area sinus hyperbolicus, area cosinus hyperbolicus and the respective functions are named arsinh, arcosh etc. In the US, this knowledge seems to have been lost, so I'm not sure what we should opt for. --Tob 08:39, 7 December 2005 (UTC)

The prefix should be ar not arc for hyperbolic functions. My opinion is that arcsinh ect. is wrong and misleading. SKvalen 18:41, 11 December 2005 (UTC)

It is wrong, but in the US they all use arc. 218.102.221.84 07:04, 30 December 2005 (UTC)

asinh etc. is computer science use and there seems to be a consensus that arc is wrong. Changing to ar. --Tob 14:00, 4 January 2006 (UTC)

An old (American) calculus textbook (Spivak's) has the following to say: "The functions sinh and tanh are one-one; their inverses, denoted by arg sinh and arg tanh (the "argument" of the hyperbolic sine and tangent) ... If cosh is restricted to [0,∞), it has an inverse, denoted by arg cosh...". This seems to confuse the matter further, and a brief Google search suggests that hardly anyone uses this, at least, not on the Internet. 82.12.108.243 15:05, 13 February 2007 (UTC)

Error in the LaTeX and normal images

I can't see them, look at the error: Failed to parse (Can't write to or create math output directory): \sinh(x) = \frac{e^x - e^{-x}}{2} = -i \sin(i x) Why is this? — Preceding unsigned comment added by 161.111.190.140 (talk) 17:32, 19 April 2006 (UTC)

I found something similar, not with this equation, but with \sech . I added braces around ix to get \sec {ix} and that seemed to work. What is strange is that all of the other functions with the same format work well for me. Garethvaughan (talk) 22:50, 18 July 2007 (UTC)

Error in plot

Mmm, I think that there is an error in the image of the iperbola: sinh(a) should refer to the ordinate of the intercepta!
In fact $\cosh ^{2}(a)-\sinh ^{2}(a)=1$, and $\cosh(a)\neq 0$ always. I'm not changing it because I'm new at editing wikipedia, but if this message will not be answered in one week, I'll be back and mod it.

--Astabada 22:15, 28 October 2007 (UTC)

.......and that is exactly what the image shows. Hyp.sine is the length of the vertical red line, that is the ordinate. Hyp.cosine is the length of the horizontal red line, and it is always different from zero — hyperbola does not touch OY axis in any point. In fact it does not approach closer than to x=1, so cosh(a)≥1 always. --CiaPan 07:34, 29 October 2007 (UTC)
BTW I slightly corrected your LaTeX notation. CiaPan

.......yes, I'm sorry, it can be viewed in both ways... I would have written $\sinh(a)$ as near as possible to the Y axis, but I recognize it may depend on habits or conventions used in countries different from mine...

--Astabada 11:41, 29 October 2007 (UTC)

Right, that's a matter of habits. You can write the symbol at the axis if you want to emphasize the co-ordinate, or at the construction line if you want to emphasize the specific length. Tha latter is useful if some lengths are not perpendicular to any axis or the line does not start from an axis (see Image:Circle-trig6.svg for example).

In the case we discuss here both methods would be equivalent. --CiaPan 10:41, 30 October 2007 (UTC)

calculus

I use maple and am not sure how to use latex. The article should give the differentials of sinh and cosh, writen in maple as: diff(sinh(x),x)=cosh(x); diff(cosh(x),x)=sinh(x); Sorry I cant edit it myself

<math> d \cosh(x) = \sinh(x), \; \; d \sinh(x) = \cosh(x) </math> as per Latex. Maple will convert output into latex code for you; see the help pages.---CH 16:14, 10 June 2006 (UTC)

You possibly meant LaTeX, CH? ;) --CiaPan 09:37, 11 November 2007 (UTC)

???

well, ppl look for this page are generally for a glance of the meaning of cosh. wth u described so many complicated information (not professional tho) but not instead just fukcing tell me how to type cosh in a standard calculator?? —Preceding unsigned comment added by 203.122.102.163 (talk) 02:06, August 29, 2007 (UTC)

its there. twice. in big print.

Of course, if your standard calculator is a TI-83 graphing calculator, you can type in anything by hitting the CATALOG button and finding it in the list, hyperbolic functions included. I think most scientific calculators don't have it built in, but if they can handle complex numbers (look for an "i" key), then I suppose you could calculate cosh, sinh, and tanh from the exponential-form definition. It would be labour-intensive though.--24.83.219.223 (talk) 08:05, 3 October 2008 (UTC)

cosh in nature

Someone should write a section about cosh as a solution to a differential equation which naturally appears in a common physics problem. If you take both ends of a rope and keep them at a distance to each other, the rope will hang in a bow like shape which happens to be the curve $y=cosh(a\cdot x)/a$, where 1/a is the bend radius at the bottom of the curve. The differential equation which appears is $y''=a{\sqrt {1+(y')^{2}}}$, and when a = 1, y(0) = 1, y'(0) = 0, you can easily se that y = cosh(x) solves the system, by using substitution. This is a top-down solution, I don't know how to make it the other way though. --Kri (talk) 12:09, 3 November 2008 (UTC)

Perhaps you didn't notice that in the second paragraph of the article, it says:

Hyperbolic functions are also useful because they occur in the solutions of some important linear differential equations, notably that defining the shape of a hanging cable, the catenary,...

--macrakis (talk) 14:38, 3 November 2008 (UTC)

What does this mean?

This phase was added a short time ago: "One important point that must be made regarding Hyperbolic functions is that the derivatives are never the sum equals of the square of the function divided by two." What on earth does this mean??? cojoco (talk) 23:40, 3 November 2008 (UTC)

New error in plot

The plot currently shows alpha as an angle between -pi/4 and pi/4. I understand, from reading other comments on this talk page, that the argument should be the shaded area. The animated plot contains the same error. I'm new to editing Wikipedia, and my understanding of the correct interpretation is based only on what I read here, so I won't correct it, but clearly cosh and sinh are defined for all real numbers, but the graphic doesn't illustrate that. Dobrojoe (talk) 21:47, 5 December 2009 (UTC) I just looked again, and don't see a problem with the animated plot after all. I don't know what I thought I was seeing. Dobrojoe (talk) 07:59, 22 December 2009 (UTC)

Names of inverses

It appears (from Google, etc.) that asinh etc. are the most common form of the inverse functions, not arsinh or arcsinh. WP doesn't try to prescribe best usage, but record actual usage. Thanks to SKvalen for pointing out that arcsinh is not the most common form. --Macrakis 02:54, 22 December 2005 (UTC)

If you take Google as a reference, please note that the reason for having the most hits for asinh etc. is that this is the way these functions are called in computer libraries. I've never seen a mathematician use asinh etc. --Tob 14:02, 4 January 2006 (UTC)

Tob is right. I was taught it was "arsinh". Deskana (talk) 19:26, 8 February 2006 (UTC)

who cares? 128.197.127.74

I've never seen the variants with only ar instead of arc before. Do we have a reference for this usage? Do we have any reference as to what usage should be preferred/is more prevalent? --MarSch 12:09, 3 May 2006 (UTC)

Scanning back up through talk, I see that or the hyperbolic variants ar shoud be preferred over arc. It would be good to have a source for this. --MarSch 12:21, 3 May 2006 (UTC)

Arc prefix is definitely wrong! It applies to trigonometric functions only.

Trigonometric functions are (or: can be) defined in terms of the unit circle's arc length as a parameter, so inverse trigonometric functions give the arc length as their value (output); that's why they are arc–functions, and their names have the arc prefix.

On the other hand, hyperbolic functions are (or: can be) defined in terms of some hyperbolic figure area as a parameter (see Image:Hyperbolic functions.svg), so inverse hyperbolic functions give an area as their value (output); that's why they are area–functions, and their names have ar prefix, distinct from trigonometric arc-functions. ---CiaPan 12:36, 22 October 2007 (UTC)

Definatly true. So is there any reason to write the integrals in "sin^-1"-Form? IMHO it'd be preferable to write arsinh etc. to avoid misunderstandings (1/sinh). —Preceding unsigned comment added by 129.217.132.38 (talk) 10:35, 16 February 2010 (UTC)

Tanh x

Under hyperbolic functions tanhx is said to equal (e^x-e^-x)/ (e^x+e^-x)...correct

The writer then goes on to say, thus

tanhx = (e^2x -1)/(e^2x +1)

I dont think thats right

(----Mike B) —Preceding unsigned comment added by 41.241.108.243 (talk) 16:43, 12 March 2010 (UTC)

Multiply top and bottom by e^x. -- Dr Greg  talk  18:34, 12 March 2010 (UTC)

Plot Colors

The red and green in the plots are really bad for people who are color blind. Any chance somebody could redo them with more accessible colors?

Sure. What colors are better? I'm not colorblind, so I can't tell. --M1ss1ontomars2k4 ^{(T | C | @)} 01:11, 1 December 2006 (UTC)

As a rule of thumb, dark and deeply-saturated pure greens are a problem for most red-green people. Moving the hue more towards blue-green and/or desaturating the green are usually pretty effective. That might not cut it for blue-yellow people though. Generally, if your colors all have different brightness and saturation levels, you're probably OK. (Wikipedia should really have a set of recommended colors for diagrams and such.) Shaunm 20:37, 20 February 2007 (UTC)

Upon observation, the graphs not only have color varience, but the lines are dotted in different patterns. This removes the alleged problem. Inthend9 (talk) 17:57, 31 March 2010 (UTC)

The Maloney Formula

There's a reference to "The MaloneyFormula" sinh(x)-r*cosh(x)=x. But I couldn't find anything like it anywhere else? Any ideas? —Preceding unsigned comment added by 78.56.23.241 (talk) 21:57, 9 April 2010 (UTC)

Looks more like the baloney formula. Fredrik Johansson 15:12, 13 April 2010 (UTC)

Yeah, it's obviously false, for example for r=0, removed. It was added by 128.46.111.18, and constitutes his only contribution. Sergiacid (talk) 08:15, 26 April 2010 (UTC)

Unaccesable?

I thought wikipedia was supposed to be an easy to use encyclopedia, these won't be easy for anyone not studying maths to understand

I have added some additional explanatory text to the introduction of the article. I hope this helps, though there is no doubt room for more. Beyond that, it is true that the content gets moderately technical, but then this is a technical subject. --Macrakis 20:49, 19 October 2005 (UTC)

Thanks that has helped somewhat --Albert Einsteins pipe

As I understand it, math is a tiered study. You can't understand multiplication without an understanding of addition. This description of hyperbolic functions is not unaccessible provided that you already understand the math leading up to it... and if you just want to understand it intuitively without being able to use it, I think it still does a fairly good job. But remember, Wikipedia gives anything and everything about a topic... it never says you have to understand everything on the page at once. If you wanted to learn addition, you could look it up, but at the age of 5 or 6, you wouldn't be able to understand that it's commutative and associative (let alone the comments about Dedekind cuts used to add irrational numbers.) Sobeita (talk) 02:47, 16 December 2010 (UTC)

log on inverse functions.

log implies that this is to the base 10, when it is actually to the base e so i suggest that this is changed to ln.

In pure mathematics, log implies base e. But since this topic is relevant for non-mathematicians as well, you have a point. Fredrik Johansson 15:21, 21 August 2006 (UTC)

Well, not really. log as it's written like this implies it has some arbitrary (but valid) base. log standing for log₁₀ is a convention while log standing for log_e is another convention; unless it's written out clearly (eg at the beginning of an article) which implicit base is used.

It's not a matter when the base isn't important like the expression
log (ab) = log a + log b
as this is always true whatever base you choose, be it log₂, log₃, log₄ or even log_π! However, it's not the case for our log functions in the article. So either log_e or ln should be used.

Oh yes, ln standing for log_e is also another convention, but with the difference that this convention is universally accepted and without confusion.

石庭豐 (talk) 17:46, 30 December 2010 (UTC)

Some suggestions

What about changing the definition of arccsch(x) to ln(1/x + sqrt(1+x^2)/abs(x))?

Also, why does the definition of arcsech(x) contains the plusminus sign? As far as I know, the definition of arcsech(x) is "the inverse of (sech(x) restricted to [0,+inf))" (Howard Anton et.al., Calculus 7th ed, Chapter 7). In that case we should replace plusminus with plus. — Preceding unsigned comment added by Agro r (talk • contribs) 13:05, 11 February 2005 (UTC)

Definitions can be different. In order to find an inverse function of a many-to-one function, you need to restrict the interval in which the values of the input are defined in order to turn it into a one-to-one function (since a function has to be one-to-many or one-to-one by definition). You can use different values of the interval in order to turn it into a one-to-one function. For example, with cos, you could use [0,pi) or [-pi/2,pi/2). They're both right, I assume. Deskana (talk) 19:21, 8 February 2006 (UTC)

Agro_r: The plus-minus sign is essential to show that there're two possible output values to the inverse of the csch. It's only when we want to simplify thing (eg using simplistic approach) that the inverse is limited to have only one output for every input and it's at this moment that we define the function. But I see that the expression to arcsch has already been changed. What a pity!

Deskana: No, it's false to say that using [0,pi) or [-pi/2,pi/2) for cosine are both right. They don't have the same target domains. I'll let you see why.

石庭豐 (talk) 13:00, 9 January 2011 (UTC)

Readability through visualization

Regarding the excerpt, "For contrast, in the terminology of topological groups, B forms a compact group while A is non-compact since it is unbounded." I happened to notice that the unbounded set had unbounded functions giving its coordinates. Am I right in thinking one causes the other? I compared the hyperbolic and trigonometric functions and noticed the functions' (un)boundedness, having not known already. Especially by someone who knows what connections to make, matching images to text could be a useful guide for elucidation. ᛭ LokiClock (talk) 12:16, 11 July 2011 (UTC)

Survey: New Images and Sech(x)

Hi all,

I think we ought to have a graph of sin/cos/sinh/cosh in complex 4-space (x_Re, x_Im, y_Re, y_Im) that shows the relationship between the four. Do you think this will make a good picture for this article? Please vote below!

Doesn't the Hyperbolic Secant curve look like the Gaussian distribution (bell) curve?

Thanks,

The Doctahedron, 04:30, 18 December 2011 (UTC)

Is there someplace we can already see this graph? I've noticed the similarity before. Many shapes look similar, but when attached to physics are apparently quite different. The Cauchy distribution also looks like the Gaussian distribution, and parabolas look like catenaries, but somehow I'd never mistake a slack chain for a bouncing ball. It's how the image implies forces producing the curve, the mechanical distinction. For that same reason we better sort Cauchy vs. Gaussian by their sources. Plus, sech has a sensual slouch, where the Gaussian is quite binary in the way it inflects at the base. ᛭ LokiClock (talk) 23:20, 19 December 2011 (UTC)

In the Mathematical Gazette, problem 83B starts with the premise (based on March 1997 note 81.10 by Tony Robin) that $f(x)={\frac {1}{2}}\lambda {\text{sech}}^{2}(\lambda x)$ gives an approximation to the standard normal density function (with mean 0 and variance 1). The problem is to show that the variance of the approximation is the same as the actual variance (=1) if $\lambda =\pi /{\sqrt {12}}$. Math. Gaz. 83, November 1999, p. 515 gives a proof. I doubt that this belongs in this article, though. Duoduoduo (talk) 17:00, 13 January 2012 (UTC)

I don't know about that. Address the perception, but come back with information about the similarity instead of the difference. That, I think, is truly encyclopedic. ᛭ LokiClock (talk) 15:41, 29 January 2012 (UTC)

question about the diagram of circular and hyperbolic angle

This is a discussion on Talk:Hyperbolic_angle.

There might be a mistake in the diagram of circular and hyperbolic angle: File:Circular and hyperbolic angle.svg.

This diagram show the use of a "hyperbolic triangle" to define the hyperbolic functions

There are two angles in the diagram. One is the hyperbolic angle related to the hyperbola xy=1, and let's call this hyperbolic angle u_hyp. The other is the circular angle related to the circle xx+yy=2, and let's call this circular angle u_cir. Right now in this diagram, these two angles are shown to be indentical, i.e. u_hyp=u_cir=u. I think such indentity should not exit in general, i.e. u_hyp is usually not equal to u_cir. By equations, u_hyp=ArcTanh[Tan[u_cir]]>=u_cir, and u_cir=ArcTan[Tanh[u_hyp]]<=u_hyp, where the equals sign holds only when u_hyp=u_cir=0. In other words, u_hyp is equal to the area of the yellow and red regions, while u_cir is equal to the area of yellow region only. Armeria wiki (talk) 05:30, 10 June 2013 (UTC)

Two lines that meet form an angle, but the measure of that angle depends on whether it is viewed as a circular angle or a hyperbolic angle. These measures are very different functions. As you note u_cir and u_hyp are different measures of the "same angle" formed by two lines that meet. The measure of this angle is the yellow area when the angle is circular, and the measure is the red area when it is hyperbolic.Rgdboer (talk) 21:28, 10 June 2013 (UTC)

The measure of angle is the red area plus the yellow area when the angle is hyperbolic.Armeria wiki (talk) 23:50, 10 June 2013 (UTC)

The problem is the 'hyperbolic angle' is NOT AN ANGLE, but rather AN AREA OF A HYPERBOLIC SECTOR. When you say 'hyperbolic angle' you should understand 'hyperbolic argument', and drop the geometric meaning of the word 'angle'. The hyperbolic functions have similar names to trigonometric functions and they fit similar relationships, but they are themselves NOT SIMILAR to trigonometric functions, and their argument is no way similar to angle. Period. --CiaPan (talk) 19:34, 11 May 2014 (UTC)

Yes, the use of "angle" in this context is confusing since "hyperbolic angle" does not have the uniform density seen with circular angle. But usage is our guide in WP so we must describe it. The usage as argument of a function for these hyperbolic functions supports the "angle" usage since they parallel the circular functions. Anything we can do to imply "hyperbolic measure" (as Klein says), and to provide clarity to readers on this important juncture in mathematics, is encouraged. Currently the article says:

The hyperbolic functions take real values for a real argument called a hyperbolic angle. The size of a hyperbolic angle is the area of its hyperbolic sector. The hyperbolic functions may be defined in terms of the legs of a right triangle covering this sector.

While the above discussion with Ameria was concluded last year, suggestions are welcome,Rgdboer (talk) 20:54, 11 May 2014 (UTC)

Circular angles stay the same under rotation. Hyperbolic angles are preserved by squeeze mapping. One of the ways that the two types of "angle" are gathered together is as invariant measures. Comparison of circle and hyperbola is necessary to give hyperbolic functions a basis for comprehension.Rgdboer (talk) 00:58, 17 May 2014 (UTC)

Maclaurin not Taylor ... Laurent?

The series presented here are Maclaurin series, not Taylor series. We should either add in an "a" term - the taylor series about "a", or call them Maclaurin series. Gareth 139.80.48.19 (talk) 21:00, 14 November 2007 (UTC)

Maclaurin Series are a special case of Taylor series. It is not incorrect to call them Taylor series. —Preceding unsigned comment added by 67.244.58.252 (talk) 15:42, 5 February 2009 (UTC)

My understanding is that the expansions given for coth and csch in the "Taylor series expansions" are actually Laurent series, and as such accommodate singularities in the complex plane. These formulas could be written with a "z" variable (for complex) rather than just an "x" variable (usually denoting real). Just trying to get things right, Grandma (talk) 15:15, 21 November 2014 (UTC)

Fanciful pronuciations not referenced

The following pronunciations given in the article are ridiculous baby-talk:

(/ˈsɪntʃ/ or /ˈʃaɪn/) (/ˈtæntʃ/ or /ˈθæn/) (/ˈʃɛk/ or /ˈsɛtʃ/) (/ˈkoʊθ/ or /ˈkɒθ/) (/ˈkɒʃ/) (/ˈkoʊʃɛk/ or /ˈkoʊsɛtʃ/)

I can imagine a professor babbling while writing on a blackboard (i.e. paying attention to what he is writing, rather than what he is saying,) but there is nothing meaningful or encyclopedic about such pronunciations. If it is necessary in the article, by all means give the correct pronunciation for the full words, but this is absurd for what is supposed to be an encyclopedia article, and completely lacking reliable references as well. 75.164.231.22 (talk) 23:48, 2 December 2014 (UTC)

Undoing reversion by "I'm your Grandma." This is not a joke. The joke is that somebody thinks these pronunciations are legitimate. Say "hyperbolic sine" instead of "cinch" or "shine" if you want your speech to be understood and you don't want people to groan when they hear you. Furthermore, find a reliable reference to these "pronunciations" if you want them to stay in the article. (Same person, different IP) 71.222.77.91 (talk) 06:28, 5 December 2014 (UTC)

Anonymous IP person, I apologize, and I now see what you are getting at. Should we have pronunciation guides for these words? When I was a kid (long ago) we learned that certain symbols stood for certain sounds. The symbols that Wiki uses, however, are not familiar to me. Still, I occasionally see the pronunciation symbols in some articles, never with a citation for reference, just used. I also note that the use of the pronunciation symbols (whatever they are called) is not common in mathematics articles (but see Fourier analysis where the pronunciation symbols for the name "Fourier" are shown). I'm not sure that any of this is too big of a deal, but other editors might have an opinion. Still ticking, Grandma (talk) 14:59, 5 December 2014 (UTC)

These pronunciations are standard, probably used by all professional mathematicians. I've added reliable references. -- Dr Greg  talk  19:17, 5 December 2014 (UTC)

Thank you, Grandma and Dr. Greg. I'm satisfied that there is now a reference. The abbreviations themselves (sinh, cosh, tanh, coth, sech, csch) are very standard, but I wasn't aware that their pronunciations were so standardized. 71.222.77.91 (talk) 20:38, 5 December 2014 (UTC)

Why is it defined so?

For

$\sinh(x)={\frac {e^{x}-e^{-x}}{2}}=-i\sin(ix)$

Could anyone tell me why? --HydrogenSu 07:48, 21 December 2005 (UTC)

Hyperbolic functions that are defined in terms of $e^{x}$ and $e^{-x}$ that bear similarities to the trigonometric (circular) functions. When plotted parametrically, trigonometric functions can be used to create circles. When plotted parametrically, hyperbolic functions can be used to create hyperbolas. That's the best explanation I can offer, as I've only recently been introduced to them. Deskana (talk) 19:25, 8 February 2006 (UTC)

well,

$\sin(x):={\frac {e^{ix}-e^{-ix}}{2i}}$

sanity check: $e^{i\pi /2}=i=>\sin(\pi /2)=1$ okay :) Now the hyperbolic variants are the same except without i. --MarSch 12:16, 3 May 2006 (UTC)

I think the "best" explanation is that the "usual" trig functions are all areas and lengths on a unit circle. The hyperbolic functions are all areas and lengths on a unit hyperbola. The fact that these have staying power is due to them being useful in some areas of non-mathematical study (see the above "Why?" question). Svyatoslav (talk) 02:56, 18 September 2016 (UTC)