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Sequence for arccos (inverse cosine) close to 1


Suppose you might want to estimate the inverse cosine of an argument close to 1. There’s a sequence for that:

You could find this sequence, for instance, right here.

This is useful, for instance, when working with the analog of the Pythagorean theorem on a sphere.

You possibly can simply use the sequence and be in your manner. However there’s much more occurring than instantly meets the attention.

Why this shouldn’t be attainable

We’re a sequence for inverse cosine centered at 1, and but inverse cosine is multivalued in a neighborhood of 1: for an argument barely lower than 1, there are two attainable angles which have such a cosine. That issues as a result of in an effort to have an influence sequence at a degree, a perform must be properly behaved in a disk round that time within the advanced aircraft. Not solely is our perform not properly behaved, it’s not even properly outlined till we contemplate department cuts. Additionally, the perform arccos(z) isn’t differentiable at 1; the by-product has a singularity at 1.

In a nutshell, we’re attempting to broaden a perform in a sequence at a degree the place the perform is badly behaved. That sounds inconceivable, or at the very least ailing suggested.

Why this can be attainable

The sequence above shouldn’t be a sequence for arccos per se. If we divide either side of the sequence by √(2-2z) we see that we even have a sequence for

frac{arccos(z)}{sqrt{2-2z}}

Though arccos is badly behaved at 1, so is √(2-2z), and their ratio is properly behaved at 1. The truth is, it’s an analytic perform and so it has an influence sequence.

Inverse cosine has a sequence enlargement but it surely doesn’t have a energy sequence. The sequence on the high shouldn’t be an influence sequence as a result of it doesn’t consist solely of powers of z. There’s a sq. root perform on the fitting facet as properly, and this perform is essential to creating issues work.

Motivation

Why would anybody assume to discover a sequence this fashion, i.e. why divide by √(2-2z)?

For small values of z,

cos(z) approx 1 - frac{z^2}{2}

and so

arccos(z) approx sqrt{2 - 2z}

One may hope, accurately it seems, that by dividing arccos(z) by approximation may yield a perform good sufficient to broaden in an influence sequence.

Why it’s attainable

Let’s plot

frac{arccos(z)}{sqrt{2-2z}}

to see whether or not it appears like an inexpensive perform. There’s often no center floor in advanced variables: capabilities are both analytic or badly behaved. Each the numerator and denominator have department cuts, however hopefully the cuts coincide and the ratio could be prolonged easily throughout the cuts.

The plot beneath suggests that’s the case.

This plot was produced with

    f[z_] := ArcCos[z] / Sqrt[2 - 2 z]
    ComplexPlot3D[f[z], {z, 0 - I, 2 + I}]

The white streak throughout the plot shouldn’t be an unintended artifact of plotting however illustrates one thing vital.

It isn’t attainable to increase arccos(z) to a perform that’s analytic for all z. It’s a must to exclude some values of z from the area, i.e. it’s important to make department cuts, and Mathematica makes these cuts alongside the actual axis for z ≤ -1 and z ≥ 1.

The sq. root perform additionally requires a department lower, and Mathematica chooses that department lower to be alongside the unfavourable actual axis, with means we’ve to exclude z ≥ 1. So the department cuts of our numerator and denominator do coincide. (Inverse cosine has a further department lower, but it surely’s not close to 1 so it doesn’t matter for our functions.)

In abstract, the sequence on the high of the put up expands arccos at a degree the place the perform is badly behaved, by dividing it by one other perform that’s badly outlined in the identical manner, making a perform that’s properly behaved.

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