I am trying to figure out what the problem with the following expression in C++ is:


My first intention was that there might occure a Cancellation due to the cosh(x) part, because it is definde as $\frac{e^x+e^{-x}}{2}$ and the computation of $e^x$ with double x results in Cancellation.

Am I on the right track? Or is there something different that causes Cancellation?

  • $\begingroup$ What is the problem? Does the compiler complain or is the resultant valurnorna erong? $\endgroup$
    – ghellquist
    Commented Sep 22, 2020 at 15:23
  • $\begingroup$ @ghellquist It has something to do with cancellation. Somehow this code can lead to bad results if executet with double values. $\endgroup$
    – Pepsilon7
    Commented Sep 22, 2020 at 15:33
  • 1
    $\begingroup$ @Pepsilon7 Can you give an example with a particular input value, the expected output, and the actual incorrect output? $\endgroup$ Commented Sep 22, 2020 at 16:30
  • $\begingroup$ std::cosh(x) is not computed as $\frac{e^x+e^{-x}}{2}$. For example, in the range $0\leq x\leq \ln(2)/2$ it is computed as $1+\frac{(e^x-1)^2}{2e^x}$, where the $e^x-1$ is not computed using std::exp and subtracting $1$, but using std::expm1. In the range $\ln(2)/2\leq x\leq 22$ is computed as $\frac{e^x+1/e^x}{2}$. From $22\leq x\leq \ln(\operatorname{maxdouble})$ it is computed as $e^x/2$. $\endgroup$
    – plop
    Commented Sep 22, 2020 at 16:40
  • 2
    $\begingroup$ @AaronRotenberg The problematic cases are when $\cosh(x)$ is large and when it is close to $1$. For example, $\log(\cosh(123456))$, when computed as std::log(std::cosh(123456)) overflows, while it is about $123455.3$, which is nowhere near to overflow. Similarly, $\log(\cosh(0.0000000000001234))$, when computed as std::log(std::cosh(0.0000000000001234)) it gives 0.0, while it is about $7.61378\times10^-{27}$, which is nowhere near to underflow. Those examples, when computed as in the comments above, give results close to the actual values. $\endgroup$
    – plop
    Commented Sep 22, 2020 at 18:34

1 Answer 1


Let me summarize what I wrote in the comments. It is not a complete answer, since the intervals on which to apply each formula still need to be investigated.

It is enough to assume $x\geq0$, since $\cosh(x)$ is even.

The type of cancellation that occurs when evaluating, in finite precision floating point (FP2), the expression



  1. When $x$ is large, in which case $\cosh(x)$ makes it even larger but $\log$ would bring the value back down. FP2 is more sparse for larger values. So, one should prevent $\cosh(x)$ making the value large.
  2. When $x$ is small the $\cosh(x)$ is close to $1$. This is cool on its own. FP2 is densest near $1$, but then $\log$ becomes close to $0$. In this case it is better to approximate the function $\log(1+x)$ and the function $\cosh(x)-1$ and compose those.

So, for $x$ large one can approximate $\cosh(x)$ by $\frac{e^x}{2}$. Composing with $\log(x)$ one gets $x-\log(2)$.

For $x$ small we can write $\cosh(x)=1+\frac{(e^x-1)^2}{2e^x}$ and compute $\log(\cosh(x))$ by composing $\log(1+x)$ and $\frac{(e^x-1)^2}{2e^x}$. The latter would be computed by approximating $e^x-1$ directly and not by evaluating $e^x$ and subtracting $1$. A C++ implementation could compute $\log(1+x)$ using std::log1p(x) and $e^x-1$ using std::expm1(x).

Finally, one needs to investigate what would be good value $x_1,x_2$ such that one would use

  1. the last computation on the interval $[0,x_1]$,
  2. the computation std::log(std::cosh(x)) on $(x_1,x_2]$
  3. and x-std::log(2) on the $[x_2,\infty]$.

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