New guy and programming problem

[Thomas Klemm]

(09-01-2018 03:18 PM)sasa Wrote:  Even result is good enough, there is still too much trouble to get accurate result to the last significant digit with arguments near 90 degree.

For sin(0.00025°) WolframAlpha gives:

4.3633231299719786898764073194250596667396107572873548... × 10^-6

The HP-41C returns:

4.363323130e-06

This result is correctly rounded to the last digit.

Then we square the rounded number and get:

1.90385887367929969 × 10^-11

The HP-41C returns:

1.903858874e-11

Again the result is rounded correctly.

Last thing to do is to multiply the number by 2:

3.807717748 × 10^-11

Which is exactly what the HP-41C returns:

3.807717748e-11


As we can see the HP-41C rounds these intermediate results correctly.

The problem with the cancelation when calculating \(1-sin(x)\) is not related to the accuracy of calculating \(sin(x)\) for \(x\) close to \(90°\).

---

Similar to Albert's trick we can multiply \(\sqrt{1+x^2}-1\) by \(\frac{\sqrt{1+x^2}+1}{\sqrt{1+x^2}+1}=1\) and get:

\(\begin{align*}
\sqrt{1+x^2}-1 &= \sqrt{1+x^2}-1 \cdot \frac{\sqrt{1+x^2}+1}{\sqrt{1+x^2}+1} \\
&= \frac{(\sqrt{1+x^2}-1)(\sqrt{1+x^2}+1)}{\sqrt{1+x^2}+1} \\
&= \frac{1+x^2-1}{\sqrt{1+x^2}+1} \\
&= \frac{x^2}{\sqrt{1+x^2}+1}
\end{align*}\)

And again cancelation for small \(x\) is avoided.