Non-trigonometric parametric equation of a circle

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Discussion Overview

The discussion revolves around the possibility of expressing the equation of a circle, specifically 1=√(x²+y²), as a parametric equation without utilizing sine and cosine functions. The scope includes mathematical reasoning and exploration of alternative parametrizations.

Discussion Character

  • Exploratory, Mathematical reasoning, Debate/contested

Main Points Raised

  • One participant questions whether it is possible to write the circle's equation as a parametric equation without sine and cosine.
  • Another participant suggests that while using the Maclaurin series for sine and cosine could work, it would require an infinite sum to be accurate.
  • A different parametrization is proposed, with the equations x={1-t² / 1+t²} and y={2t / 1+t²}, although it is noted that this represents a semicircle for -1
  • Further clarification is provided that the semicircle can be extended with t ∈ ℝ or t ∈ ℝ ∪ {∞} for a more complete representation.

Areas of Agreement / Disagreement

Participants express differing views on the feasibility of the task, with some proposing alternative parametrizations while others maintain that the original request may not be achievable without sine and cosine.

Contextual Notes

The discussion highlights limitations regarding the completeness of the proposed parametrizations and the conditions under which they apply, particularly concerning the range of t.

johann1301
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I wish to write this equation:

1=√(x2+y2)

as a parametric equation but WITHOUT the use of sine and cosine.

Is this possible?
 
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I don't think so. You could use the maclauren series for sine and cosine, but you'd need an infinite sum for it to be completely correct.

Edit: A generic circle can be drawn in the complex plane without sine/cosine, but it wouldn't be whatsoever equivalent to the formula you gave.
 
There is another parametrization:
$$x={1-t^2 \over 1+t^2}$$
$$y={2t \over 1+t^2}$$
 
I like Serena said:
There is another parametrization:
$$x={1-t^2 \over 1+t^2}$$
$$y={2t \over 1+t^2}$$

Almost - this is a semicircle. -1<t<1.
 
mathman said:
Almost - this is a semicircle. -1<t<1.

It's a little more with ##t \in \mathbb R##, or even better with ##t \in \mathbb R \cup \{\infty\}##.
 

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