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I have found a really interesting solution for interpolating cosh. First, the solver chooses the number of nodes, then, calculates Chebyshev polynomial roots on the desired interval, then finds the approximation using Lagrange construction of the polynomial. And then follows this method.

Let $\phi_1$ be the Lagrange polynomial the solver found before.

The solver began to calculate other $\phi$'s according to this recursive formula:

$$\phi_{i+1}(t) = 2\cdot\phi_i\bigg(\frac{t}{2}\bigg)^2-1$$

After he reached $\phi_7$, he stopped, I think because he reached the desired precision. And then he wrote the resulting interpolation as follows:

$$ \phi(t) = \begin{cases} \phi_7(-t), \space t<0 \\ \phi_7(t), \space t ≥0 \end{cases} $$

Of course this could be done because cosh is y-axis symmetrical and continuos in 0.

So my question is : what is the name of this method, and is it possible to apply it to other symmetrical functions except cosh.

petajamaja
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1 Answers1

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I would not call it interpolation; it is a kind of range reduction. Essentially you are using the half-angle formula (see e.g. http://dlmf.nist.gov/4.35.E21) $$ \cosh \frac{t}{2} = \left(\frac{\cosh t + 1}{2}\right)^{\frac{1}{2}} $$ to reduce the argument to a smaller range where you use an approximation for $\cosh$. You do not need interpolation, you can use e.g. the first terms of the Taylor expansion $$ \cosh t \approx 1 + \frac{1}{2}t^2 + \frac{1}{24}t^4 $$ With this method and 7 reduction steps you get $\cosh 1 = 1.543080634809$ with a relative error of about $-3.9\times 10^{-12}.\; $ With more steps or more Taylor terms it would be smaller ($1.4\times 10^{-16}$ with 4 terms), but the error increases for larger argument.

gammatester
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