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Does ln(x - √(1 + x^2)) really have a real part?

sweatingbear

Member
May 3, 2013
91
Hello forum. In my studies of hyperbolic functions, I stumbled upon this guy

\(\displaystyle \ln \left( x - \sqrt{1 + x^2 } \right) \, .\)

Trying different values for \(\displaystyle x\) in my head, I confidently concluded that this function does not have any outputs in the real numbers. I attempted to plot it in Maple but Maple showed me a blank coordinate system (expectedly). However, this is what WolframAlpha outputs.

It looks like a flipped hyperbolic sine, but is WolframAlpha even displaying correctly to begin with?
 

Prove It

Well-known member
MHB Math Helper
Jan 26, 2012
1,403
You require that \(\displaystyle \displaystyle \begin{align*} x - \sqrt{ 1 + x^2} > 0 \end{align*}\), so

\(\displaystyle \displaystyle \begin{align*} x - \sqrt{ 1 +x^2} &> 0 \\ x &> \sqrt{1 + x^2} \\ x^2 &> 1 + x^2 \\ 0 &> 1 \end{align*}\)

This is an obvious contradiction, so that means you are correct, there is no real part to this...
 

chisigma

Well-known member
Feb 13, 2012
1,704
Hello forum. In my studies of hyperbolic functions, I stumbled upon this guy

\(\displaystyle \ln \left( x - \sqrt{1 + x^2 } \right) \, .\)

Trying different values for \(\displaystyle x\) in my head, I confidently concluded that this function does not have any outputs in the real numbers. I attempted to plot it in Maple but Maple showed me a blank coordinate system (expectedly). However, this is what WolframAlpha outputs.

It looks like a flipped hyperbolic sine, but is WolframAlpha even displaying correctly to begin with?
It is immediate to verify that...

$$\ln (x - \sqrt{1+x^{2}}) = \ln \{-1\ (\sqrt{1+x^{2}} - x)\} = \ln (\sqrt{1+x^{2}} - x) + \pi\ i\ (1)$$

... so that the plot of 'Monster Wolfram' is perfectly plausible...


Kind regards


$\chi$ $\sigma$
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,707
Hello forum. In my studies of hyperbolic functions, I stumbled upon this guy

\(\displaystyle \ln \left( x - \sqrt{1 + x^2 } \right) \, .\)

Trying different values for \(\displaystyle x\) in my head, I confidently concluded that this function does not have any outputs in the real numbers. I attempted to plot it in Maple but Maple showed me a blank coordinate system (expectedly). However, this is what WolframAlpha outputs.

It looks like a flipped hyperbolic sine, but is WolframAlpha even displaying correctly to begin with?
In studying hyperbolic functions, you may have come across the fact that inverse hyperbolic functions can be expressed as logarithms. In particular, $\sinh^{-1}x = \log\bigl(\sqrt{1+x^2}-x\bigr)$. Your logarithm has the negative of that one, so $$\log\bigl(x-\sqrt{1+x^2}\bigr) = \log\bigl((-1)\bigl(\sqrt{1+x^2}-x\bigr)\bigr) = \log(-1) + \log\bigl(\sqrt{1+x^2}-x\bigr) = i\pi + \sinh^{-1}x.$$ That agrees with the WolframAlpha output.
 

sweatingbear

Member
May 3, 2013
91
@chisigma, Opalg: Ok, but nevertheless, the function clearly cannot have a real part as WolframAlpha purports?
 

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
Let us say the following :

\(\displaystyle \ln(z_1 z_2) = \ln(z_1)+\ln(z_2) \)

Proof :

\(\displaystyle \ln(z_1 z_2) = \ln|z_1z_2|+ i \arg(z_1z_2) =\ln|z_1z_2|+ i arg(z_1)+i \text{arg}(z_2) \)

\(\displaystyle \ln|z_1|+i \text{arg}(z_1) + \ln|z_2|+ i \text{arg}(z_2) = \ln(z_1)+\ln(z_2) \, \square \)

[HR][/HR]

Generally this is not true in the case of principle logarithm

\(\displaystyle \text{Log}(z_1 z_2) \neq \text{Log}(z_1)+ \text{Log} (z_2) \)

\(\displaystyle \text{Log}(z_1 z_2) = \ln|z_1 z_2| + i \text{Arg}(z_1 z_2) \)

Now in order to be able to do that we must have

\(\displaystyle -\pi < \text{Arg}(z_1 z_2) \leq \pi \)

That means using polar coordinates

\(\displaystyle -\pi < \theta_1 \,+\theta_2 \leq \pi \)

\(\displaystyle \text{Log}(-1 \cdot -1) =0 \)

\(\displaystyle \text{Log}(-1 \cdot -1) \neq \text{Log}(-1) + \text{Log}(-1) = 2i\pi \)

This is completely justifiable since \(\displaystyle \text{Arg(-1)}+\text{Arg(-1)}= 2\pi \)

In contrast that example might work perfectly if we use another branch cut i.e. \(\displaystyle \text{Log}_{0}\)
 

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
But I still cannot get why W|A draws the imaginary part near 3 , it should be at an integer , does anybody know ?
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,707
@chisigma, Opalg: Ok, but nevertheless, the function clearly cannot have a real part as WolframAlpha purports?
But I still cannot get why W|A draws the imaginary part near 3 , it should be at an integer , does anybody know ?
WA interprets the logarithm as a complex-valued function and evaluates it as $ i\pi + \sinh^{-1}x$ (see my previous comment). It then renders the real part as $\sinh^{-1}x$ (the blue curve in the WA plot) and the imaginary part as the constant $\pi$ (the orange line in the plot).
 

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
WA interprets the logarithm as a complex-valued function and evaluates it as $ i\pi + \sinh^{-1}x$ (see my previous comment). It then renders the real part as $\sinh^{-1}x$ (the blue curve in the WA plot) and the imaginary part as the constant $\pi$ (the orange line in the plot).
That was stupid of me :eek: , I completely discarded \(\displaystyle \pi\) !

So if we take the principle logarithm \(\displaystyle \text{Log}(z) \)

\(\displaystyle \text{Log}\left(-1 \cdot (\sqrt{x^2+1} -x ) \right)\)

Now since \(\displaystyle -\pi < \text{Arg}\left(-1 \cdot (\sqrt{x^2+1} -x) \right) \leq \pi \)

We can rewrite it as

\(\displaystyle \text{Log}\left(-1 \cdot (\sqrt{x^2+1} -x ) \right) = \text{Log}\left(-1 \right)+\text{Log}\left( \sqrt{x^2+1} -x \right)\)

\(\displaystyle \text{Log}\left(-1 \right)= \ln|1|+ i\pi = i\pi \)

\(\displaystyle \text{Log}\left(\sqrt{x^2+1} -x\right) = \ln \left|\sqrt{x^2+1}-x \right|+i 0 = \ln \left( \sqrt{x^2+1}-x \right)\)

Hence we have :

\(\displaystyle \text{Log}\left(x-\sqrt{x^2+1}\right) = \ln \left( \sqrt{x^2+1}-x \right) + i\pi \)
 

sweatingbear

Member
May 3, 2013
91
@Opalg: Is not WolframAlpha plotting \(\displaystyle \text{arcsinh} (-x)\) as opposed to \(\displaystyle \text{arcsinh} (x)\)?
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,707
@Opalg: Is not WolframAlpha plotting \(\displaystyle \text{arcsinh} (-x)\) as opposed to \(\displaystyle \text{arcsinh} (x)\)?
Yes. Naturally, WA is right and I was wrong. The logarithm form of $\sinh^{-1}x$ is $\ln\bigl(\sqrt{1+x^2}+x\bigr)$, which is equal to $-\ln\bigl(\sqrt{1+x^2}-x\bigr)$. So I should have used $-\sinh^{-1}x$ (with a minus sign) in my previous comments.
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,780
Just for the record, from wiki:

The abbreviations arcsinh, arccosh, etc., are commonly used, even though they are misnomers, since the prefix arc is the abbreviation for arcus, while the prefix ar stands for area.[1][2][3]


So it should be $\text{arsinh }x$, which is how I've learned it.
 

sweatingbear

Member
May 3, 2013
91
Just for the record, from wiki:

The abbreviations arcsinh, arccosh, etc., are commonly used, even though they are misnomers, since the prefix arc is the abbreviation for arcus, while the prefix ar stands for area.[1][2][3]


So it should be $\text{arsinh }x$, which is how I've learned it.
Thanks, but does this mean that the outputs of the inverse hyperbolic sine can be interpreted as areas of some sort or what?
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,780

TheBigBadBen

Active member
May 12, 2013
84

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,780

sweatingbear

Member
May 3, 2013
91
Oh, ok! Sorry for my lack of Googling-skills... I will read up on the hyperbolic sine soon. Thanks!