Mathematics for Enzyme Reaction Kinetics and Reactor Performance. F. Xavier Malcata

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when x < −1, but for x > 1 – so only the latter can be taken as domain for cosh⁻¹ x, so that logarithm thereof can be defined. However, , so two possible values actually arise, i.e. cosh⁻¹ x is a double‐valued function; only the positive value of the logarithm is usually considered, thus implying use of the positive sign preceding the square root in its argument.

      Following a similar rationale, one may calculate the inverse hyperbolic tangent – by, once again, setting

(2.580)

      at startup, in parallel to Eq. (2.566) – thus implying that

(2.581)

      as per composition of functions; consequently,

      (2.582)

      owing to Eqs. (2.482) and (2.581), which becomes

(2.583)

after elimination of denominator. Upon insertion of Eq. (2.581), one obtains

      (2.584)

      from Eq. (2.511), with x replaced by y – which readily becomes

(2.585)

      only the plus sign preceding the square root was taken here, because sech y only takes positive values (see Fig. 2.14 c). One may now revisit Eq. (2.479) as

(2.586)

at the expense of Eq. (2.583), along with factoring out of cosh y; in view of Eq. (2.487), one may redo Eq. (2.586) to

      (2.587)

where insertion of Eq. (2.585) permits further transformation to

(2.588)

Since a difference of squares is expressible as the product of two conjugate binomials, Eq. (2.588) unfolds

      (2.589)

      or else

(2.590)

after cancelation of between numerator and denominator; following application of logarithms to both sides, Eq. (2.590) turns to

      (2.591)

where Eqs. (2.25) and (2.580) support final transformation to

(2.592)

– defined for ∣x ∣ < 1 only, so as to guarantee a positive argument for the logarithm. Equation (2.592) is illustrated in Fig. 2.15 b; note the monotonically increasing pattern of tanh⁻¹ x, spanning]−1,1[ as domain; at either x = −1 or x = 1, a vertical asymptote arises – according to

      (2.593)

      that drives the curve toward −∞ at x = −1, coupled with

      (2.594)

      that drives the curve toward ∞ at x = 1.

      The inverse hyperbolic cotangent may be obtained after applying the hyperbolic tangent operator to both sides of Eq. (2.592), namely,

(2.595)

once reciprocals are taken of both sides, Eq. (2.595) becomes

(2.596)

– also with the aid of Eq. (2.483). Division of both numerator and denominator of the argument of the logarithm function by x converts Eq.

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