(1) ^ ^ ^ ^ ^ ^ ^ ^ ^

1 - tanh^2(x) = 1 - sinh^2(x)/cosh^2(x)

. . . . . . . . . . . =( cosh^2(x) - sinh^2(x) ) / cosh^2(x)

. . . . . . . . . . . = 1/ cosh^2(x)

(2) ^ ^ ^ ^ ^ ^ ^ ^ ^

cosh(x) = (1+exp(2x)) exp(-x) /2

(3) ^ ^ ^ ^ ^ ^ ^ ^ ^

therefore, for ALL x (including 0)we have

ln(1-tanh^2(x)) = - 2 ln(cosh(x)) . . . . . . . by (1)

. . . . . . . . . . .= 2 ln 2 + 2 x - 2 ln(1+exp(2x)) . . . . . by(2)

^ ^ ^ ^ ^ ^ ^ ^ ^

The 'blue' answer is ok but not complete. Indeed, for all x (including 0)

cosh(-x) = (exp(-x) + exp(x)) / 2 = exp(x) (1+ exp(-2x))/2

However, the blue answer omits the case of x=0. Therefore it cannot be called perfectly correct :-)

Joachim Domsta

Dear Professor Domsta:

I am very grateful for your reply! You gave two calculation results, and the derivation process is very concise. I learned a lot from it.

However, I still have a question to ask you. In the “correct answer”, when x>0, why is ln(1-tanh^2(x)) equal to: ln4 -2x-2 ln(1+exp(-2x))? ( In our calculations，ln(1-tanh^2(x))=2 ln 2 + 2 x-2 ln(1+exp(2x)) for ALL x including 0.)

Thank you for your time.

@Wei Xin 

Dear Wei Xin

Both formulas are correct for all x since they are equivalent. In particular YOUR formula is correct for all x.

Some details: If you

1) replace in your derivation the fraction in parantheses

( (1 + exp(2x))/ exp(x) )

by the equivalent product

( (1 + exp(-2x)) exp(x) )

2) and then continue with using properties of ln

you will arrive at the second formula also valid for all x.

Simply, these formulas are exchangeable.

Regards, Joachim Domsta

Dear Joachim Domsta, the expression ln(1-tanh2(x)) is clearly nonpositive as a result of 1-tanh2(x)0. Similarly, the formula lnexp(-x)=-x holds for x0. Finally, we may note that for x=0, the two expressions also hold. Therefore, ln(1-tanh2(x))= ln4-2ln(1+exp(2x))+2x if x=0.

Dear @Spiros Konstantogiannis

If you see any false equality in my or @Wei Xin's derivation, please copy it and find at least one value of x which makes the equality false. It would be useful for all participants :-)

In your outline the following is an example of incomplete statements:

ln exp(x)=x for x>0.

You can easily check that this holds for ALL real values of x. For example, if x= -1, then exp(x) = 1/e > 0 and therefore the ln of exp(x) exists and equals -1, which is the value of the RHS.

Regards, Joachim Domsta

ln4-2ln(1+exp(2x))+2x = 0 for x= 0

(ln4-2ln(1+exp(2x))+2x)' = 0 for x=0

(ln4-2ln(1+exp(2x))+2x)'' =

. . . . . . =- (2ln(1+exp(2x)))''

. . . . . . = - ( 4 exp(2x) / (1+exp(2x)) )'

. . . . . . .= - 8 exp(2x) / (1 + exp(2x))^2

. . . . . . . = - 2 for x=0.

By Taylor theorem, in some vicinity of x=0 the expression is negative.

Joachim Domsta

Dear Professor Domsta:

I am very grateful for your guidance. Therefore, the two formulas in blue font are equivalent for all x, there is no need to divide cases.

Thank you for your time.

Dear Wei Xin, you are welcome.

Joachim Domsta

Dear Joachim Domsta , you are right. I got confused by the linear approximation exp(x)=1+x I used. If more terms are added, it is seen that the expression ln4-2ln(1+exp(2x))+2x is indeed negative near 0. Actually, the function f:R->R defined by f(x)=ln4-2ln(1+exp(2x))+2x is even, as it is straightforwardly seen. Its graph can be seen in the following link:

https://www.wolframalpha.com/input?i=plot+ln4-2ln%281%2Bexp%282x%29%29%2B2x

By the way, I have never heard about "incomplete" statements; I know about true or false statements only. Probably, you mean that the truth set of the respective predicate is a superset of the given set.

Yes.

Hi all.

The following addition is about my statements that

lnexp(x)=x for every x>0 and lnexp(-x)=-x for every xB and a function g:C->D, if the image f(A) of f, which is a subset of B, is also a subset of C, then f(x) belongs to C for every x belongs to A, and the composition g*f is a function from A to D.

(i) Let f:R->R defined by f(x)=exp(x) and let g:(0,infinity)->R defined by g(x)=lnx. Since exp(x)>0 for every x belongs to R, it follows that f(R) is a subset of (0,infinity) (in particular, f(R)=(0,infinity)) and g*f is a function from R to R. As a result, (g*f)(x)=g(f(x))=lnexp(x)=x for every x belongs to R.

(ii) Let f:R->R defined by f(x)=exp(-x) and let g:(0,infinity)->R defined by g(x)=lnx. Since exp(-x)>0 for every x belongs to R, it follows that f(R) is a subset of (0,infinity) (in particular, f(R)=(0,infinity)) and g*f is a function from R to R. As a result, (g*f)(x)=g(f(x))=lnexp(-x)=-x for every x belongs to R.

(iii) Let f: (0,infinity)->R defined by f(x)=lnx and let g:R->R defined by g(x)=exp(x). By the previous reasoning, g*f is a function from (0,infinity)->R and (g*f)(x)=exp ln(x)=x for x>0.

(iv) Let f: (-infinity,0)->R defined by f(x)=ln(-x) and let g:R->R defined by g(x)=exp(x). By the previous reasoning, g*f is a function from (-infinity,0)->R and (g*f)(x)=exp ln(-x)=-x for x0, and

expln(-x)=-x for every x

It's pretty much the same thing (see enclosed file).

But two comments need to be made:

1. Note that ln(1-th(x)^2)=ln(1/ch(x)^2)=-2ln(ch(x))

(the rest of the conversions, as it follow from your question, will not be difficult for you)

ch(x) - even function, ch(-x)=ch(x)

if x>0, then ln(1-th(x)^2)=-2ln(ch(x))

let's y=-x, then y

The two cases written in bule are same, in case of x>0, you can easily obtain the your

answer by extracting exp(-2x) in the ln term.

M. B. Laskin

Dear Dr. Laskin:

Thank you for your detailed analysis and clear and intuitive images!

The two calculation results in blue, which are both correct for all ranges of x, just two forms of mathematical expression.

Thank you for your time!

Zilong Zhong

Dear Zilong Zhong:

You're right. They're equivalent！

Spiros Konstantogiannis

Dear Konstantogiannis:

Thank you for adding some properties and proofs about exponential and logarithmic functions, which are the basis for solving the problem.

Thank you for your time!