\[\frac{\varepsilon \cdot \left(e^{\left(a + b\right) \cdot \varepsilon} - 1\right)}{\left(e^{a \cdot \varepsilon} - 1\right) \cdot \left(e^{b \cdot \varepsilon} - 1\right)}\]
Test:
NMSE problem 3.4.2
Bits:
128 bits
Bits error versus a
Bits error versus b
Bits error versus eps
Time: 35.8 s
Input Error: 27.9
Output Error: 8.0
Log:
Profile: 🕒
\(\begin{cases} {\left(\sqrt[3]{\frac{(e^{\varepsilon \cdot \left(b + a\right)} - 1)^*}{(e^{a \cdot \varepsilon} - 1)^*}}\right)}^3 \cdot \frac{\varepsilon}{(e^{b \cdot \varepsilon} - 1)^*} & \text{when } a \le -1639.1333f0 \\ \frac{(e^{\varepsilon \cdot \left(a + b\right)} - 1)^*}{(e^{b \cdot \varepsilon} - 1)^* \cdot a} - \frac{1}{2} \cdot \frac{(e^{\varepsilon \cdot \left(a + b\right)} - 1)^*}{\frac{(e^{b \cdot \varepsilon} - 1)^*}{\varepsilon}} & \text{when } a \le 3.581931f-06 \\ {\left(\sqrt[3]{\frac{(e^{\varepsilon \cdot \left(b + a\right)} - 1)^*}{(e^{a \cdot \varepsilon} - 1)^*}}\right)}^3 \cdot \frac{\varepsilon}{(e^{b \cdot \varepsilon} - 1)^*} & \text{otherwise} \end{cases}\)

    if a < -1639.1333f0 or 3.581931f-06 < a

    1. Started with
      \[\frac{\varepsilon \cdot \left(e^{\left(a + b\right) \cdot \varepsilon} - 1\right)}{\left(e^{a \cdot \varepsilon} - 1\right) \cdot \left(e^{b \cdot \varepsilon} - 1\right)}\]
      25.6
    2. Applied simplify to get
      \[\color{red}{\frac{\varepsilon \cdot \left(e^{\left(a + b\right) \cdot \varepsilon} - 1\right)}{\left(e^{a \cdot \varepsilon} - 1\right) \cdot \left(e^{b \cdot \varepsilon} - 1\right)}} \leadsto \color{blue}{\frac{(e^{\varepsilon \cdot \left(b + a\right)} - 1)^*}{(e^{a \cdot \varepsilon} - 1)^*} \cdot \frac{\varepsilon}{(e^{b \cdot \varepsilon} - 1)^*}}\]
      8.7
    3. Using strategy rm
      8.7
    4. Applied add-cube-cbrt to get
      \[\color{red}{\frac{(e^{\varepsilon \cdot \left(b + a\right)} - 1)^*}{(e^{a \cdot \varepsilon} - 1)^*}} \cdot \frac{\varepsilon}{(e^{b \cdot \varepsilon} - 1)^*} \leadsto \color{blue}{{\left(\sqrt[3]{\frac{(e^{\varepsilon \cdot \left(b + a\right)} - 1)^*}{(e^{a \cdot \varepsilon} - 1)^*}}\right)}^3} \cdot \frac{\varepsilon}{(e^{b \cdot \varepsilon} - 1)^*}\]
      8.8

    if -1639.1333f0 < a < 3.581931f-06

    1. Started with
      \[\frac{\varepsilon \cdot \left(e^{\left(a + b\right) \cdot \varepsilon} - 1\right)}{\left(e^{a \cdot \varepsilon} - 1\right) \cdot \left(e^{b \cdot \varepsilon} - 1\right)}\]
      30.0
    2. Applied simplify to get
      \[\color{red}{\frac{\varepsilon \cdot \left(e^{\left(a + b\right) \cdot \varepsilon} - 1\right)}{\left(e^{a \cdot \varepsilon} - 1\right) \cdot \left(e^{b \cdot \varepsilon} - 1\right)}} \leadsto \color{blue}{\frac{(e^{\varepsilon \cdot \left(b + a\right)} - 1)^*}{(e^{a \cdot \varepsilon} - 1)^*} \cdot \frac{\varepsilon}{(e^{b \cdot \varepsilon} - 1)^*}}\]
      15.2
    3. Using strategy rm
      15.2
    4. Applied expm1-udef to get
      \[\frac{(e^{\varepsilon \cdot \left(b + a\right)} - 1)^*}{\color{red}{(e^{a \cdot \varepsilon} - 1)^*}} \cdot \frac{\varepsilon}{(e^{b \cdot \varepsilon} - 1)^*} \leadsto \frac{(e^{\varepsilon \cdot \left(b + a\right)} - 1)^*}{\color{blue}{e^{a \cdot \varepsilon} - 1}} \cdot \frac{\varepsilon}{(e^{b \cdot \varepsilon} - 1)^*}\]
      29.8
    5. Applied taylor to get
      \[\frac{(e^{\varepsilon \cdot \left(b + a\right)} - 1)^*}{e^{a \cdot \varepsilon} - 1} \cdot \frac{\varepsilon}{(e^{b \cdot \varepsilon} - 1)^*} \leadsto \frac{(e^{\varepsilon \cdot \left(b + a\right)} - 1)^*}{a \cdot (e^{\varepsilon \cdot b} - 1)^*} - \frac{1}{2} \cdot \frac{(e^{\varepsilon \cdot \left(b + a\right)} - 1)^* \cdot \varepsilon}{(e^{\varepsilon \cdot b} - 1)^*}\]
      7.2
    6. Taylor expanded around 0 to get
      \[\color{red}{\frac{(e^{\varepsilon \cdot \left(b + a\right)} - 1)^*}{a \cdot (e^{\varepsilon \cdot b} - 1)^*} - \frac{1}{2} \cdot \frac{(e^{\varepsilon \cdot \left(b + a\right)} - 1)^* \cdot \varepsilon}{(e^{\varepsilon \cdot b} - 1)^*}} \leadsto \color{blue}{\frac{(e^{\varepsilon \cdot \left(b + a\right)} - 1)^*}{a \cdot (e^{\varepsilon \cdot b} - 1)^*} - \frac{1}{2} \cdot \frac{(e^{\varepsilon \cdot \left(b + a\right)} - 1)^* \cdot \varepsilon}{(e^{\varepsilon \cdot b} - 1)^*}}\]
      7.2
    7. Applied simplify to get
      \[\frac{(e^{\varepsilon \cdot \left(b + a\right)} - 1)^*}{a \cdot (e^{\varepsilon \cdot b} - 1)^*} - \frac{1}{2} \cdot \frac{(e^{\varepsilon \cdot \left(b + a\right)} - 1)^* \cdot \varepsilon}{(e^{\varepsilon \cdot b} - 1)^*} \leadsto \frac{(e^{\varepsilon \cdot \left(a + b\right)} - 1)^*}{(e^{b \cdot \varepsilon} - 1)^* \cdot a} - \frac{1}{2} \cdot \frac{(e^{\varepsilon \cdot \left(a + b\right)} - 1)^*}{\frac{(e^{b \cdot \varepsilon} - 1)^*}{\varepsilon}}\]
      7.2

    8. Applied final simplification

  1. Removed slow pow expressions

Original test:


(lambda ((a default) (b default) (eps default))
  #:name "NMSE problem 3.4.2"
  (/ (* eps (- (exp (* (+ a b) eps)) 1)) (* (- (exp (* a eps)) 1) (- (exp (* b eps)) 1)))
  #:target
  (/ (+ a b) (* a b)))