New guy and programming problem
09-01-2018, 04:28 PM (This post was last modified: 09-01-2018 04:35 PM by Thomas Klemm.)
Post: #30
 Thomas Klemm Senior Member Posts: 2,078 Joined: Dec 2013
RE: New guy and programming problem
(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.
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