CP 8.3, 8.4 - Inverse hyperbolic functions as logarithms

CP 8.3, 8.4 - Inverse hyperbolic functions as logarithms

Define the inverse hyperbolic functions with their correct domains and ranges (CP 8.3), then derive and use their logarithmic forms (CP 8.4). The logarithms make exact values accessible, while branch restrictions explain which solutions are allowed.

Choosing a one-to-one branch

Why is arcosh⁡(cosh⁡(−2))=2\operatorname{arcosh}(\cosh(-2))=2, rather than −2-2? An inverse can reverse a function only when each output comes from one permitted input. The full cosh⁡\cosh graph has two opposite inputs for every output above 11, so we must choose a branch.

The functions sinh⁡\sinh and tanh⁡\tanh are strictly increasing on the real line, so they are already one-to-one. For cosh⁡\cosh, choose the increasing branch x≥0x\ge0. Its outputs cover [1,∞)[1,\infty), and its inverse therefore takes inputs in that interval and returns non-negative values.

You already know

From A-level Mathematics, an inverse exchanges a function's domain and range and reflects its graph in y=xy=x; it is not a reciprocal. The hyperbolic ranges are R\mathbb R for sinh⁡\sinh, [1,∞)[1,\infty) for cosh⁡\cosh, and (−1,1)(-1,1) for tanh⁡\tanh.

Use Pearson's inverse names arsinh⁡\operatorname{arsinh}, arcosh⁡\operatorname{arcosh} and artanh⁡\operatorname{artanh}. Some calculators display an inverse using a superscript, but the function is distinct from 1/sinh⁡x1/\sinh x, 1/cosh⁡x1/\cosh x or 1/tanh⁡x1/\tanh x.

Inverse hyperbolic functions

The inverse relations are

RelationDomain of inverseRange of inversey=arsinh⁡x  ⟺  x=sinh⁡yRRy=arcosh⁡x  ⟺  x=cosh⁡y, y≥0[1,∞)[0,∞)y=artanh⁡x  ⟺  x=tanh⁡y(−1,1)R\begin{array}{c|c|c} \text{Relation}&\text{Domain of inverse}&\text{Range of inverse}\\ \hline y=\operatorname{arsinh}x\iff x=\sinh y&\mathbb R&\mathbb R\\ y=\operatorname{arcosh}x\iff x=\cosh y,\ y\ge0&[1,\infty)&[0,\infty)\\ y=\operatorname{artanh}x\iff x=\tanh y&(-1,1)&\mathbb R \end{array}

For example, arcosh⁡1=0\operatorname{arcosh}1=0, while arcosh⁡(1/2)\operatorname{arcosh}(1/2) has no real value. The inverse arcosh⁡\operatorname{arcosh} returns a single non-negative value, not a pair.

Their graphs reflect the appropriate original graphs, including the restriction on cosh⁡\cosh. The two horizontal asymptotes of tanh⁡\tanh become vertical asymptotes of artanh⁡\operatorname{artanh}, at x=±1x=\pm1.

[DIAGRAM: asset_name: The inverse hyperbolic graphs and their domains; asset_slug: 9fm0-inverse-hyperbolic-graphs; caption: Inverse graphs exchange the permitted inputs and outputs of the original functions; description: Three separate panels show arsinh x increasing through the origin with domain all real numbers, arcosh x starting at the included point (1,0) and increasing in the first quadrant, and artanh x increasing through the origin with dashed vertical asymptotes x equals minus 1 and x equals 1. Domain and range statements appear below each panel.]
Inverse graphs exchange the permitted inputs and outputs of the original functions

The permitted branches give arsinh⁡(sinh⁡x)=x\operatorname{arsinh}(\sinh x)=x and artanh⁡(tanh⁡x)=x\operatorname{artanh}(\tanh x)=x for all real xx. For hyperbolic cosine, however, arcosh⁡(cosh⁡x)=∣x∣\operatorname{arcosh}(\cosh x)=|x|. These domain, range and branch conventions are knowledge to learn; the booklet supplies the logarithmic forms and their input restrictions.

Key idea

A real equation cosh⁡y=x>1\cosh y=x>1 has two opposite solutions, but arcosh⁡x\operatorname{arcosh}x names only the non-negative one.

Check the branch before using a formula.

Deriving the logarithm for arsinh

The equation x=sinh⁡yx=\sinh y becomes a quadratic in the positive quantity ey\mathrm e^y, and exactly one quadratic root is admissible.

Let t=ey>0t=\mathrm e^y>0. Then

x=t−t−12exponential definitiont2−2xt−1=0multiply by tt=x±x2+1quadratic formula\begin{aligned} x&=\frac{t-t^{-1}}2 && \text{exponential definition}\\ t^2-2xt-1&=0 && \text{multiply by }t\\ t&=x\pm\sqrt{x^2+1} && \text{quadratic formula} \end{aligned}

Because x2+1>∣x∣\sqrt{x^2+1}>|x|, the plus choice is positive and the minus choice is negative for every real xx. Rejecting the negative value is essential: it cannot equal a real exponential. Taking the logarithm of the positive value gives the inverse.

Logarithmic form of arsinh

For every real xx,

arsinh⁡x=ln⁡(x+x2+1)\operatorname{arsinh}x=\ln\left(x+\sqrt{x^2+1}\right)

The logarithm's argument is positive even when x<0x<0, because x2+1>∣x∣\sqrt{x^2+1}>|x|.

In the formula booklet.

There is an elegant symmetry hidden inside this apparently asymmetric formula:

(x2+1+x)(x2+1−x)=1\left(\sqrt{x^2+1}+x\right)\left(\sqrt{x^2+1}-x\right)=1

The two positive quantities are reciprocals, so their logarithms are opposites. This proves arsinh⁡(−x)=−arsinh⁡x\operatorname{arsinh}(-x)=-\operatorname{arsinh}x directly from the logarithmic form.

Worked example 1: an exact positive inverse

Question. Express arsinh⁡(3/4)\operatorname{arsinh}(3/4) as a simplified natural logarithm.

Plan. Substitute into the logarithmic form and simplify the square root before adding.

arsinh⁡34=ln⁡(34+916+1)logarithmic form=ln⁡(34+54)positive square root=ln⁡2combine exact fractions\begin{aligned} \operatorname{arsinh}\frac34 &=\ln\left(\frac34+\sqrt{\frac9{16}+1}\right) && \text{logarithmic form}\\ &=\ln\left(\frac34+\frac54\right) && \text{positive square root}\\ &=\ln2 && \text{combine exact fractions} \end{aligned}

Why this step? The radical is the positive square root 5/45/4; the inverse's sign comes from the logarithm, not from attaching a second square-root sign.

Answer. arsinh⁡(3/4)=ln⁡2\operatorname{arsinh}(3/4)=\ln2.

Check. sinh⁡(ln⁡2)=(2−1/2)/2=3/4\sinh(\ln2)=(2-1/2)/2=3/4.

Change only the input's sign. The logarithm remains real; its argument becomes the reciprocal of the previous one.

Worked example 2: an exact negative inverse

Question. Express arsinh⁡(−3/4)\operatorname{arsinh}(-3/4) as a simplified natural logarithm.

Plan. Keep the same square root and subtract the input's magnitude.

arsinh⁡(−34)=ln⁡(−34+916+1)retain signed input=ln⁡12positive logarithm argument=−ln⁡2reciprocal logarithm\begin{aligned} \operatorname{arsinh}\left(-\frac34\right) &=\ln\left(-\frac34+\sqrt{\frac9{16}+1}\right) && \text{retain signed input}\\ &=\ln\frac12 && \text{positive logarithm argument}\\ &=-\ln2 && \text{reciprocal logarithm} \end{aligned}

Why this step? A negative inverse value is allowed: it comes from taking the logarithm of a positive number below 11.

Answer. arsinh⁡(−3/4)=−ln⁡2\operatorname{arsinh}(-3/4)=-\ln2.

Check. Odd symmetry gives sinh⁡(−ln⁡2)=−3/4\sinh(-\ln2)=-3/4, as required.

A derivation must explain the rejected exponential root, rather than simply erase it between two lines.

Watch out

Wrong: arsinh⁡x=ln⁡(x±x2+1)\operatorname{arsinh}x=\ln(x\pm\sqrt{x^2+1}). The minus choice is always negative, so its real logarithm does not exist. Right: use x+x2+1x+\sqrt{x^2+1} alone.

Try another exact square root.

Deriving arcosh and choosing its branch

For x=cosh⁡yx=\cosh y, both exponential roots can be positive, so positivity alone does not select the inverse: the condition y≥0y\ge0 makes the decision.

Continue this lesson.

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