If I plot some points randomly on the graph and then join them, can that curve be represented as a polynomial? I know that a straight line can be expressed algebraically as a linear equation, and quadratic equation as a parabola, but can you express a smooth curve which is made by joining some random points on the graph algebraically ?
Can every curve be represented as an equation?
algebra-precalculusgeometry
Related Solutions
Kindly ask Wolfram Alpha to
Collect[Expand[Eliminate[{
x1*t^2 + x2*t*(1-t)*2 + x3*(1-t)^2 == x,
y1*t^2 + y2*t*(1-t)*2 + y3*(1-t)^2 == y
}, t]], {x, y}]
and it will tell you
which you can reformulate to something close to
\begin{align*}F(x, y) =&\phantom{+} x^2 (y_1^2-4 y_1 y_2+2 y_1 y_3+4 y_2^2-4 y_2 y_3+y_3^2) \\ &+xy (-2 x_1 y_1+4 x_1 y_2-2 x_1 y_3+4 x_2 y_1-8 x_2 y_2+4 x_2 y_3-2 x_3 y_1+4 x_3 y_2-2 x_3 y_3) \\ &+x (2 x_1 y_1 y_3-4 x_1 y_2^2+4 x_1 y_2 y_3-2 x_1 y_3^2+4 x_2 y_1 y_2-8 x_2 y_1 y_3+4 x_2 y_2 y_3-2 x_3 y_1^2+4 x_3 y_1 y_2+2 x_3 y_1 y_3-4 x_3 y_2^2) \\ &- y^2 (-x_1^2+4 x_1 x_2-2 x_1 x_3-4 x_2^2+4 x_2 x_3-x_3^2) \\ &- y (2 x_1^2 y_3-4 x_1 x_2 y_2-4 x_1 x_2 y_3-2 x_1 x_3 y_1+8 x_1 x_3 y_2-2 x_1 x_3 y_3+4 x_2^2 y_1+4 x_2^2 y_3-4 x_2 x_3 y_1-4 x_2 x_3 y_2+2 x_3^2 y_1) \\ &+(x_1^2 y_3^2-4 x_1 x_2 y_2 y_3-2 x_1 x_3 y_1 y_3+4 x_1 x_3 y_2^2+4 x_2^2 y_1 y_3-4 x_2 x_3 y_1 y_2+x_3^2 y_1^2) \end{align*}
So the parameters of your conic will be
\begin{align*} A =& y_1^{2} - 4 \, y_1 y_2 + 2 \, y_1 y_3 + 4 \, y_2^{2} - 4 \, y_2 y_3 + y_3^{2} \\ B =& x_1^{2} - 4 \, x_1 x_2 + 2 \, x_1 x_3 + 4 \, x_2^{2} - 4 \, x_2 x_3 + x_3^{2} \\ C =& -2 \, x_1 y_1 + 4 \, x_1 y_2 - 2 \, x_1 y_3 + 4 \, x_2 y_1 - 8 \, x_2 y_2 + 4 \, x_2 y_3 - 2 \, x_3 y_1 + 4 \, x_3 y_2 - 2 \, x_3 y_3 \\ D =& 2 \, x_1 y_1 y_3 - 4 \, x_1 y_2^{2} + 4 \, x_1 y_2 y_3 - 2 \, x_1 y_3^{2} + 4 \, x_2 y_1 y_2 - 8 \, x_2 y_1 y_3 \\ &+\, 4 \, x_2 y_2 y_3 - 2 \, x_3 y_1^{2} + 4 \, x_3 y_1 y_2 + 2 \, x_3 y_1 y_3 - 4 \, x_3 y_2^{2} \\ E =& -2 \, x_1^{2} y_3 + 4 \, x_1 x_2 y_2 + 4 \, x_1 x_2 y_3 + 2 \, x_1 x_3 y_1 - 8 \, x_1 x_3 y_2 + 2 \, x_1 x_3 y_3 \\ &-\, 4 \, x_2^{2} y_1 - 4 \, x_2^{2} y_3 + 4 \, x_2 x_3 y_1 + 4 \, x_2 x_3 y_2 - 2 \, x_3^{2} y_1 \\ F =& x_1^{2} y_3^{2} - 4 \, x_1 x_2 y_2 y_3 - 2 \, x_1 x_3 y_1 y_3 + 4 \, x_1 x_3 y_2^{2} + 4 \, x_2^{2} y_1 y_3 - 4 \, x_2 x_3 y_1 y_2 + x_3^{2} y_1^{2} \end{align*}
The above was computed using sage, where I could get the output into the form I needed more easily. This should agree with the first formula, but if it does not, this here is the more reliable one. You can check that $4AB-C^2=0$ which proves that this conic is indeed a parabola.
Not true. See: Wikipedia article: we normally define the length of a curve as a (potentially infinite) supremum of all the lengths (in the "standard" sense) of polygonal lines joining the beginning and the end of the curve and "inscribed" in the curve. We say that the curve is "rectifiable" if this supremum is finite.
With that definition, the fact that the straight line is the shortest follows pretty much immediately from the definition. Pretty much, the only slightly nontrivial bit is to prove that the length of the straight line (in this new sense) is the same as the length of the straight line (in the standard sense). This is true because, for every polygonal line "inscribed" in a straight line, the (standard) length of it is exactly the same as the (standard) length of the original straight line.
This consideration works in any $\mathbb R^n$. However, it does not scale well to other manifolds, because those may be locally homeomorphic with $\mathbb R^n$ but not locally isometric with $\mathbb R^n$. The common way of adding metric to manifolds (which is: adding a metric tensor) requires differentiability of the maps between charts in the atlas (i.e. the manifold must be differentiable), in order for the metric tensor to allow for a change in co-ordinates when switching from one chart to another. That approach scales a lot better, but requires differentiability, and in that approach, the length of the shortest path is calculated using calculus of variations. This is probably what you had in mind in your question.
Best Answer
There are functions like trigonometric and logarithmic functions which cannot be represented as polynomials, yet they can be graphed:
To illustrate it by an example, we may take the parabolic graph of $x^2$.