Shadow Posted February 21, 2012 Share Posted February 21, 2012 Okay, this has confused me for a while now. Arctanh and arccotanh seem to be two different functions. Yet their derivatives are the same. What's more when I plot the two in WA, the plots overlap (at least the real parts). However, when I make WA solve the equation Re(arctanh(x)) = Re(arccotanh(x)), it gives me no real solutions, and a bunch of completely random complex solutions. So what's the deal? Link to comment Share on other sites More sharing options...

D H Posted February 21, 2012 Share Posted February 21, 2012 Of course the derivatives are the same. The two functions differ by a constant over any compact neighborhood in the complex plane where both functions are defined. Wolfram Alpha has a hard time seeing that the solution to Re(arctanh(x)-arccotanh(x)) = 0 is the complex numbers less the points at which arctanh(x) and arccotanh(x) are undefined. Link to comment Share on other sites More sharing options...

Shadow Posted February 22, 2012 Author Share Posted February 22, 2012 Thanks. I'm completely unfamiliar with hyperbolic functions, thus the confusion. Link to comment Share on other sites More sharing options...

DrRocket Posted February 27, 2012 Share Posted February 27, 2012 Of course the derivatives are the same. The two functions differ by a constant over any compact neighborhood in the complex plane where both functions are defined. Wolfram Alpha has a hard time seeing that the solution to Re(arctanh(x)-arccotanh(x)) = 0 is the complex numbers less the points at which arctanh(x) and arccotanh(x) are undefined. That is a nice simple concise explanation. If only it were so clear as you imply or not dependent on a number of choices. It has NOTHING to do with any compact sets, contrary to your "explanation". There is no "of course" about it. DH strikes again. Let us first consider the hyperbolic functions in the context of the question posed by the OP. That context is in terms of real-valued functions of a real variable. [math] cosh \ x = \frac {e^x+e^{-x}}{2}[/math] [math] sinh \ x = \frac {e^x-e^{-x}}{2}[/math] [math] tanh \ x = \frac {sinh \ x}{cosh \ x} = \frac {e^x-e^{-x}}{e^x+e^{-x}} = \frac {e^{2x}-1}{e^{2x}+1}[/math] [math]coth \ x = \frac {cosh \ x}{sinh \ x} = \frac {e^x+e^{-x}}{e^x-e^{-x}} = \frac {e^{2x}+1}{e^{2x} -1}[/math] Some observations: 1. All of the above functions are injective (one-to-one) 2. cosh is an even function which maps the real line to real numbers [math] \ge[/math] 1. 3. sinh is an odd function which maps the real line to the entire real line 4. tanh is an odd function which maps the real line to the real numbers of absolute value less than 1 5. coth is an odd function with a pole at 0 that maps the real line to the real numbers of absolute value greater than 1 Note that tanh and coth have disjoint images. Hence artanh and arcoth disjoint domains. It thus makes no sense to talk of them as differing by a constant, even though as we shall see, their derivatives have a common algebraic expression. Now consider the inverse functions arcosh, arsinh, artanh and arcoth. artanh is defined on [math]x \in \mathbb R : |x| < 1[/math] So suppose that [math] y=arcosh \ x [/math] then [math]x = cosh \ y[/math] and we can solve for [math] y [/math] as follows: [math] x = \frac {e^y+e^{-y}}{2}[/math] [math] x + \sqrt {x^2-1} = \frac {e^y+e^{-y}}{2} + \sqrt {\frac {e^{2x} +2 +e^{-2x}}{4} -1}[/math] [math] = \frac {e^y+e^{-y}}{2} + \sqrt {\frac {e^{2x} +- +e^{-2x}}{4} }[/math] [math] = \frac {e^y+e^{-y}}{2} + \frac {e^y-e^{-y}}{2}[/math] [math] = e^y[/math] Hence [math] y = arcosh \ x = ln(y+ \sqrt{x^2-1}[/math] for [math] x \ge 1[/math] Similarly [math]arsinh \ x = ln(y+ \sqrt{x^2+1}[/math] for [math] x \in \mathbb R[/math] With a bit more algebra you find that [math] artanh \ x = \frac {1}{2} \ ln(\frac {1+x}{1-x}) \ \ |x| <1 [/math] And [math] arcoth \ x = \frac {1}{2} \ ln(\frac {x+1}{x-1}) \ \ |x| > 1 [/math] Differentiating these expressions you find that [math] \frac {d}{dx} artanh \ x = \frac {1}{1-x^2} \\ |x| <1[/math] [math] \frac {d}{dx} arcoth \ x = \frac {1}{1-x^2} \\ |x| > 1[/math] So, while the expression may appear to be the same they have no common domain and it makes no sense to say that they "differ by a constant." This is the answer to the question as posted by the OP. For some reason the theory of functions of a complex variable has been injected into the discussion, so it seems necessary to explain the situation with respect to the extension of the real-valued functions of a real variable to the case of mermorphic funcctions of a complex variable. Now let us consider the extension of the hyperbolic functions to functions of a complex variable. It is obvious that there are meromorphic extensions simply because the exponential function, by virtue of the usual power series definition, extends from the real line to a holomorphic function on the entire complex plane. Since the exponential function is periodic with period [math] 2 \pi i[/math] we see immediately that sinh, cosh, tanh, coth are not injective when viewed as complex-valued functions of a complex variable. In particular they do not have an inverse unless one restricts the domain so that the restrictions are injective. As a practical matter this means restricting the domain to strips of width [math] 2 \pi [/math] parallel to the real axis in the complex plane (you could conceivably take smaller strips from several of these strips, but no one in their right mind would want to do that). One can also view the functions as being defined on a tube. A similar problem arises with the ordinary trigonometric functions of a real variable, so this is not an overwhelming problem, but it does call for care. Let us make the usual choice and take as our domain [math] D = \{ z \in \mathbb C : - \pi < Im(z) \le \pi \}[/math] With this restriction of the domain we apply the same definitions for the various hyperbolic functions, simply replacing "x" with the complex variable "z". The symmetry observed continues to hold but the notions of "even" and "odd" lose their utility. What we do see is that 1. All of the above functions are injective (one-to-one) 2. cosh maps D to the complex plain and has a single zero at [math] z= \frac {i \pi}{2}[/math] 3. sinh maps D to the complex plain and has a single zero at [math] z =0[/math] 4. tanh maps D to the complex plain and has a pole at [math] z =\frac { \pm i \pi}{2 }[/math] and a zero at [math] z =0[/math] 5. coth maps D to the complex plain and has a zero at [math] z =\frac {\pm i \pi}{2} [/math] and a pole at [math] z =0[/math] Now we work in analogy with the case of the real-valued functions of a real variable to solve for expressions giving us the inverse functions. We will restrict our domain to D less the set of points at which the specific functions that we consider have no poles. We look at the artanh and arcoth functions that precipitated the question and proceed formally (meaning that some steps in the calculation will be justified only later): Let [math]y = tanh \ z = \frac {e^z - e^{-z}}{e^z+e^{-z}} [/math] [math]\left( \frac {1+y}{1-y}\right )^{\frac{1}{2}}[/math][math] =\left( \dfrac {1 + \frac {e^z - e^{-z}}{e^z+e^{-z}}}{1-\frac {e^z - e^{-z}}{e^x+e^{-z}}} \right )^{\frac{1}{2}}[/math][math] = (e^{2z} )^{\frac {1}{2}} = e^z[/math] From which we will eventually conclude that [math]y = artanh \ z = \frac {1}{2} log\left( \frac {1+y}{1-y}\right)[/math] Similarly [math]y = arcoth \ z = \frac {1}{2} log\left( \frac {y+1}{y-1} \right)[/math] [math]= \frac {1}{2} log\left( \frac {-(1+y)}{1-y} \right)[/math] Now the claim is that artanh and arcoth differ by a constant additive factor, so one might want to say that [math] = log\left( \frac {-(1+y)}{1-y}\right) = log \left( \frac {(1+y)}{1-y}\right) \ + \ log(-1)[/math] Which, of course would require putting some meaning on the formal manipulations and on [math] \ log(-1)[/math] Remember that we have restricted the domain of our hyperbolic functions to strips in which the exponential function is bijective onto the entire complex plane. That is equivalent to what in the theory of functions of a complex variable is called "choosing a branch of the logarithm". The logarithm can be defined as an analytic function, but only by deleting from the complex plane a half-infinite ray extending from the origin. In essence this is because the "argument" of a complex number is only defined modulo [math] 2 \pi [/math] Our choice corresponds to deleting the negative real axis. So we cannot extend the logarithm analytically to make sense of [math] log (-1) [/math] Nevertheless with our choice for the domain there is a unique value of [math]z[/math] for which [math] e^z=-1[/math] and that value is [math] z= i \pi[/math]. With the (arbitary) choices made one can now show that with the (decidedly non-compact) domain that we have chosen for which tanh and coth are bijective that on the complex plane. artanh and arcoth then are defined on the complex plane, and with our choice of the domain of tanh and coth with demands that the argument lie between [math]-\pi[/math] not inclusive and [math]\pi[/math] inclusive, differ by only the additive constant [math] i\pi [/math] . This is the result of these various choices, and is likely not particularly obvious to the casual observer. Link to comment Share on other sites More sharing options...

D H Posted February 27, 2012 Share Posted February 27, 2012 That is a nice simple concise explanation. If only it were so clear as you imply or not dependent on a number of choices. It has NOTHING to do with any compact sets, contrary to your "explanation". There is no "of course" about it. DH strikes again. The only strike here is by you, and here it is: Let us first consider the hyperbolic functions in the context of the question posed by the OP. That context is in terms of real-valued functions of a real variable. The context is in clearly terms of a complex valued functions. (Otherwise, why would the OP be taking the real part?) The context is also in terms of what Wolfram Alpha computes. Read the original post. Again. And then again. You need to lose your attitude. Link to comment Share on other sites More sharing options...

DrRocket Posted February 27, 2012 Share Posted February 27, 2012 The only strike here is by you, and here it is: The context is in clearly terms of a complex valued functions. (Otherwise, why would the OP be taking the real part?) The context is also in terms of what Wolfram Alpha computes. Read the original post. Again. And then again. You need to lose your attitude. And the reason for your unhelpful answer is that "Wolfram Alpha made me do it." ? My "attitude" is that people deserve to have their question answered accurately and in the most straightforward manner available, even when that requires one to read a bit more into the question and the background of the poster than is openly presented. That is fairly common with mathematics questions -- the first step is to determine what the question really is. It also means that inaccurate or misleading responses by so-called "experts" (even if they come via Wolfram Alpha) are not on my list of the particularly useful. I am not likely to lose that attitude. In this case by first lookiing at the case of functions of a real variable, the essence of the issue comes out cleanly. The case of functions of a complex variable can be, and was, addressed, but requires much greater care and is more clearly presented once the theory of the real-valued functions is understood. -1 Link to comment Share on other sites More sharing options...

D H Posted February 27, 2012 Share Posted February 27, 2012 And the reason for your unhelpful answer is that "Wolfram Alpha made me do it." ? FFS, read the original post! I'll embiggen the key phrases. Okay, this has confused me for a while now. Arctanh and arccotanh seem to be two different functions. Yet their derivatives are the same. What's more when I plot the two in WA, the plots overlap (at least the real parts). However, when I make WA solve the equation Re(arctanh(x)) = Re(arccotanh(x)), it gives me no real solutions, and a bunch of completely random complex solutions. So what's the deal? Link to comment Share on other sites More sharing options...

Shadow Posted March 1, 2012 Author Share Posted March 1, 2012 I should have been more clear, I realize that my mentioning real parts might have been confusing. Nevertheless, I really was interested in the real valued function. Complex functions are still beyond my understanding, from a formal point of view. I only understand what I've picked up around here and elsewhere. Thank you DrRocket; it never occurred to me, and should have, to look at the domains of the functions and your explanation clears things up nicely (and also offers a peek at complex functions, for which thanks are also in order). However, although I'm as far from wanting to participate in this argument as I could be, I would like to point out that DH has been of great help to me in different issues, and was clearly not deliberately misinforming me. It seems unfair to insult him, if that was your intention. 5 Link to comment Share on other sites More sharing options...

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