The following is my understanding of what happens: if I take a "two dimensional problem" e.g. I have $X$ as inputs and Y as the outcome and I add a feature $x^2$. This gives a problem an additional dimension and the linear fit on the $x$ and $y$ values define a line as well as the linear fit on $x^2$ and $y$ values and the two lines define a plane which is the best fit. Is this correct? How does this translate back to the 2 dimensional space? Does this somehow show up in two dimensions as curvy? How?

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    $\begingroup$ $x^2$ is not an additional dimension because it's determined by $x$. the dimensions must be independent to some degree at least $\endgroup$
    – Aksakal
    May 31, 2019 at 19:57
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    $\begingroup$ @Aksakal In the sense of dimensions of the column space of the model matrix, though, $x^2$ usually does introduce an additional dimension. That seems like a natural and useful way to understand this question. $\endgroup$
    – whuber
    May 31, 2019 at 20:15
  • $\begingroup$ If we're thinking in terms of the design matrix $X$ that has observations as rows and variables as columns, then $x^2$ has its own column and in this regard adds a dimension. for instance, a covariance matrix $p\times p$ will one more dimension. moreover, in many cases the matix will even retain its rank $p$ despite $x^2$ being dependent on $x$, because it is not linearly dependent. that's why polynomial regression often works. however, it may sometimes fail due to collinearity or condition. $\endgroup$
    – Aksakal
    May 31, 2019 at 20:21
  • $\begingroup$ I would suggest using orthogonal polynomials though. they're free of dependency issues of simple polynomials $\endgroup$
    – Aksakal
    May 31, 2019 at 20:22
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    $\begingroup$ Using orthogonal polynomials instead of simpler ones doesn't change the result - that is, the estimated fit is the same -, although orthogonal polynomials have some practical advantages. That's not different from most multivariate regression problems, where predictors are correlated. $\endgroup$
    – Pere
    Jun 1, 2019 at 8:19

2 Answers 2


This is a piece of a plane in 3D.

Figure 1

Here is the same plane with coordinates shown and a set of points selected along its $x$ axis.

Figure 2

The third coordinate is used to plot the squares of these $x$ values, producing points along a parabola at the base of the coordinate box.

Figure 3

A vertical "curtain" through the parabola intersects the plane at all the points directly above the parabola. This intersection is a curve.

Figure 4

A polynomial model supposes the response $y$ (graphed in the vertical direction) differs from the height of this plane by random amounts. The values of $y$ corresponding to these $x$ coordinates are shown as red dots.

Figure 5

Consequently, the $(x,y)$ points lie along a curve--this projection--rather than a line, even though the model of the response is based on the plane originally shown.

Figure 6


When the explanatory variables clearly lie on a curve, the responses will appear to lie on a curve, too.

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    $\begingroup$ Thank you so much, this was so helpful. $\endgroup$
    – user412953
    Jun 2, 2019 at 12:08

If you have a single independent variable x and a single dependent variable y, then "y = f(x)" is usually considered as two dimensional even if the relationship between these two variables is complicated. As a hypothetical example, if an experimental model is "pressure = a * temperature + b * log(temperature) - c * sine(temperature)" there are only two variables, temperature and pressure. For this reason, such a relationship could be plotted as a curved line on a flat plane.

If the model had two independent variables, such as "pressure = a * log(temperature) - b * exp(altitude)", this has the form of "z = f(x,y)" and could be plotted as a 3D surface.


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