# Looking for a function that approximates a parabola

I have a shape that is defined by a parabola in a certain range, and a horizontal line outside of that range (see red in figure).

I am looking for a single differentiable, without absolute values, non-piecewise, and continuous function that can approximate that shape. I tried a Gaussian-like function (blue), which works well around the maximum, but is too large at the edges. Is there a way to make the blue function more like the red function? Or, is there another function that can do this?

• What does it mean for a function to be piecewise?
– user736690
Nov 26, 2020 at 17:51
• @CarlaCvekla This is a rhetorical question, right? There is no such thing as a piecewise function. I've seen more questions where someone asks for a function, but what they really mean is a function that can be constructed from a finite combination of standard operations like addition, multiplication, powers and trigonometric functions. In this case OP does not allow piecewise definition to be one of those operations. Gimelist, you should know that your question can strictly not be answered without specifying a list of allowed operations.
– Paul
Nov 26, 2020 at 20:12
• @Gimelist I think the confusion is because "piecewise" is a way of describing how you write a function, not a function itself. For example, the absolute value function can be written as $\begin{cases}x,&x\ge 0\\-x,&x<0\end{cases}$ (piecewise) or as $\mathrm{abs}(x)$ or $\sqrt{x^2}$ (both non-piecewise). If you mean you want to avoid functions whose graphs consist of different lines/curves that are not connected to each other, the word you're looking for is "discontinuous function" - but I think it's clear to everyone that that's not the type of function you want. Nov 27, 2020 at 3:22
• I think that we can assume that "non-piecewise" in the question means "analytic." The great majority of "ordinary, everyday" functions are piecewise analytic, and a piecewise analytic function which is "non-piecewise" is an analytic function. Nov 27, 2020 at 17:46
• It's not an answer to this, but I thought I'd add it here as a fun fact - you can express the function (for the case of $1-x^2$ for $(-1,1)$ and zero otherwise) exactly as $(1-x^2)((\tan^{-1}((1+x^2)/(1-x^2))-\tan^{-1}(x^2))/\pi+3/4)$. Nov 29, 2020 at 1:21

I would suggest just approximating the $$\max(0,\cdot)$$ function, and then using that to implement $$\max(0,1-x^2)$$. This is a very well-studied problem, since $$\max(0,\cdot)$$ is the relu function which is currently ubiquitous in machine learning applications. One possibility is $$\max(0,y) \approx \mu(w)(y) = \frac{y}2 + \sqrt{\frac{y^2}4 + w}$$

One way of deriving this formula: it's the positive-range inverse of $$x\mapsto x-\tfrac{w}x$$.

Then, composed with the quadratic, this looks thus:

Notice how unlike Calvin Khor's suggestions, this avoids ever going negative, and is easier to adopt to other parabolas.

Nathaniel remarks that this approach does not preserve the height of the peak. I don't know if that matters at all – but if it does, the simplest fix is to just rescale the function with a constant factor. That requires however knowing the actual maximum of the parabola itself (in my case, 1).

To get an even better match to the original peak, you can define a version of $$\mu$$ whose first two Taylor coefficients around 1 are both 1 (i.e. like the identity), by rescaling both the result and the input (this exploits the chain rule): \begin{align} \mu_{1(1)}(w,y) :=& \frac{\mu(w,y)}{\mu(1)} \\ \mu_{2(1)}(w,y) :=& \mu_{1(1)}\left(w, 1 + \frac{y - 1}{\mu'_{1(1)}(w,1)}\right) \end{align}

And with that, this is your result:

The nice thing about this is that Taylor expansion around 0 will give you back the exact original (un-restricted) parabola.

import Graphics.Dynamic.Plot.R2
import Text.Printf

μ₁₁, μ₁₁', μ₂₁ :: Double -> Double -> Double
μ₁₁ w y = (y + sqrt (y^2 + 4*w))
/ (1 + sqrt(1 + 4*w))
μ₁₁' w y = (1 + y / sqrt (y^2 + 4*w))
/ (1 + sqrt(1 + 4*w))
μ₂₁ w y = μ₁₁ w $1 + (y-1) / μ₁₁' w 1 q :: Double -> Double q x = 1 - x^2 main :: IO () main = do plotWindow [ plotLatest [ plot [ legendName (printf "μ₂₍₁₎(w,q(x))") . continFnPlot $$μ₂₁ w . q , legendName (printf "w = %.2g" w) mempty ] | w <- (^1500).recip<$$>[1,1+3e-5..]] , legendName "max(0,q x)" . continFnPlot$ max 0 . q, xAxisLabel "x"
, yInterval (0,1.5)
, xInterval (-1.3,1.3) ]
return ()

• This is an excellent answer. Also has the benefit of not raising things to large powers in order to get good accuracy! Nov 26, 2020 at 23:52
• And a question - if the approximation gets better as $w$ approaches zero, why do we even need it? Is there any reason why excluding it would not work and just use $\frac{y}2 + \sqrt{\frac{y^2}4}$? Nov 27, 2020 at 2:27
• $\sqrt{y^2/4} = |y/2|$, which fails to be differentiable at $y=0$. Adding the positive $w$ under the square root sign shifts things so that this singularity is avoided. Nov 27, 2020 at 2:32
• Nov 28, 2020 at 0:18
• @XYZT I added the source code, in case you're interested. Nov 28, 2020 at 16:25

Here is a family of real analytic functions that does the job; making the parameter $$N=1,2,3,4,\dots$$ larger makes the approximation better. Lets just say the quadratic is $$q(x) = 1-x^2$$, so the function is $$f(x)=\max(0,q(x))$$. Its easy to do a more general case. Then set $$f_N(x) = \frac1{1+x^{2N}}q(x).$$ This works because for $$N$$ very big, when $$|x|<1$$, $$x^{2N}\approx 0$$, so $$\frac1{1+x^{2N}}\approx 1$$. But the moment $$|x|>1$$, $$x^{2N}$$ becomes huge, so then $$\frac1{1+x^{2N}} \approx 0$$. It's not exactly 0 outside $$|x|<1$$, but instead decays like $$O(|x|^{-2N+2})$$ (again, crank up $$N$$ to get a better approximation)

This is what it looks like for $$N=10$$:

The general case: the quadratic with height $$h$$ above $$y=c$$, base $$d$$ centered around $$x_0$$ is $$q_{h,c,d,x_0}(x)=h q\left(\frac{x-x_0}{d/2}\right)+c.$$ The quadratic truncated at $$y=c$$ is then $$f_{h,c,d,x_0}(x) = h f\left(\frac{x-x_0}{d/2}\right)+c,$$ and the corresponding approximation I am proposing is $$f_{N,h,c,d,x_0}(x) = h f_N\left(\frac{x-x_0}{d/2}\right)+c.$$

You can adjust these parameters in this interactive Desmos graph.

A further variation is using $$1/(1-x^{2N})$$ instead to define $$f_N$$; this makes $$f_N$$ everywhere positive (because it changes sign exactly where $$q$$ does, and $$f_N$$ is differentiable at the truncation points by l'Hopital), which could be useful. But the approximation is worse near the boundary.

Here's a graph of this last variant-

The first variant is exactly 1 at $$x=0$$, and exactly 0 at $$x=\pm 1$$. It has the same sign as the original polynomial $$1-x^2$$ everywhere.

The second is exact only at $$x=0$$, (but even if $$N$$ is small), and is everywhere positive. (In particular, its a slight mischaracterisation to say that "unlike Calvin Khor's suggestions, this avoids ever going negative" :) but I didn't include a graph of this variant earlier, and their solution is great too!)

There's no such thing as a "piecewise function". In the phrase "piecewise defined function", the word "piecewise" modifies the word "defined", not "function". The term "piecewise" refers to how the function is defined, not the function itself.

$$t(x) = \begin{cases} -x & x < 0 \\ x& x \geq 0 \end{cases}$$

and

$$t(x)=|x|$$

are the same function. In the first case, the definition is given piecewise, and in the second, it isn't. You can make any piecewise definition non-piecewise with enough absolute values. For instance, the following gives a graph along the lines of yours:

$$f(x) = \frac 12(200-||x-50|-|x-150||)(100-|x-50|)$$

• I added the condition that absolute values are not to be used. Nov 27, 2020 at 0:36
• While this is true, a definition of the absolute value is usually given piecewise so I decided against things like this; my choice of $1/(1+x^{2N})$ was meant to mimic a compactly supported bump function which would afaik need similar trickery to do ‘non piecewise’ Nov 27, 2020 at 11:00

As an expansion of Calvin Khor's answer, you need to generate a suitable approximation to the square pulse function, which can be done in a number of different ways.

Consider that the error function goes from -1 to 1, with a relatively rapid transition from one side to the other. Using this function, we can write $$f_a(x)=\frac{\text{erf}(ax)+1}2-\frac{\text{erf}(a(x-1))+1}2=\frac{\text{erf}(ax)-\text{erf}(a(x-1))}2$$ Here is what that function looks like for $$a=5$$ (red), $$a=50$$ (blue), and $$a=500$$ green):

We can translate this as needed, then multiply by our required polynomial, in order to get our fit. Fitting $$(1-x^2)$$ leads us to $$f(x)\approx (1-x^2)f_a(2x+1)$$ which, for the same values of $$a$$, gives

Any sequence that converges towards the square pulse will work. Instead of $$\text{erf}$$, you could use $$\tan^{-1}$$ (with appropriate rescaling), for example, or $$\tanh$$.

So, you have a smoothness constraint. But - don't take it farther than you need to. That is, there's a natural (?) tendency to assume that if you're asked for a smooth approximation than it has to be infinitely-differentiable. So, you went for a Gaussian. I'm not saying that's a bad choice, but - you should certainly consider a solution which "just" meets the constraint and not much beyond that.

What immediately comes to my mind is "nearly-cheating": taking the constraint-breaking piecewise function (say its pieces are $$p_1$$ and $$p_2$$) and smoothing its pieces out, like so: $$f(x) = (1-t(x)) \cdot p_1(x) + t(x) \cdot p_2(x)$$ with some function $$t : \mathbb{R} \to [0,1]$$. that is, $$p_1$$ will gradually transition into $$p_2$$ instead of just switching to it.

In your case, a super-simple $$t(x)$$ would be the following piecewise linear affair: $$t(x) = \begin{cases} 0 & x < -100 \\ \frac{x+100}{100} &- 100 \leq x < 0 \\ \frac{100-x}{100} &0 \leq x < 100 \\ 0 & x \geq 100 \end{cases}$$ (seeing how your parabola segment crosses at -50 and 50.)

This is of course not the greatest choice of $$t$$. You can either narrow the linear pieces going from 0 to 1 and back; or you could add more pieces; or you could make it a spline of higher degree. Narrowing the transition segment is fine - since the smoothed function will still meet your constraints (you were not limited on the absolute value of the derivative). You will then have a fifth, central linear piece with value 1.

If you wanted to play nicer, you might want to increase the degree of the piecewise linear functions, to get a higher-degree spline. When you think in that direction, you might notice that what you have already is a degree-1 Bernstein polynomial - a basis for the space of polynomials from [0,1] to $$\mathbb{R}$$. So, IIANM your smoothing coefficients which generalizes into making Bezier splines of your chosen degree of smoothness.

Obligatory "not an answer but I wanted to share the image".

How about a transformation of the $$\text{sinc}$$ function?

• Looks close, but it has to be zero or nearly so outside the range (ie below 2 and above 4 in your case) Nov 26, 2020 at 4:04

I am looking for a single differentiable, without absolute values, non-piecewise, and continuous function that can approximate that shape

An explicit function for what you want using the Heaviside Step Function :

$$y(x) = y_0\left(1-\frac{x^2}{a^2}\right)\theta(a^2-x^2)$$

giving :

\begin{align} y(0) & =y_0 \\ y(\pm a) & =0 \end{align}

This is differentiable and is exact, not just approximate, as the Heaviside function differentiates to the Dirac Delta function. This may be all you need without recourse to approximations depending on your precise application.

My smoothness constraint should be taken seriously, as it is a requirement for the function downstream in the thing I need it for.

Any smooth approximation for the Heaviside function should be usable as a replacement. Wikipedia quotes the logistic function as an example :

$$\theta(t)\approx \frac 1 {1+e^{-2kt}}$$

Precisely what function you use as a smooth approximation is going to depend on your precise needs. Mathworld gives several more limiting expressions you may be able to use.

• The Heaviside function is not considered differentiable, as it isn't even continuous. To be differentiable, the limit that defines the derivative must exist at every point. If the limit goes to infinity, it does not converge, and thus the limit does not exist. Nov 30, 2020 at 3:04
• @GlenO Both Mathworld and Wikipedia (and elsewhere) state the Heaviside function is differentiable and equate that to the Dirac delta function, so your issue is not with me. :-) However the point I am making is that a "natural" way to construct a continuous approximation the OP can use for their purposes is to express their exact function using the Heaviside function and replace the Heaviside function with some continuous approximation tuned to their own needs. Nov 30, 2020 at 7:55
• "Differentiable" has a mathematical meaning, and it's not the one you're thinking of. Mathworld and Wikipedia both say that the derivative of the Heaviside function is the Dirac delta function... but that's not the same as the Heaviside function being differentiable. It is not differentiable at zero. Please see "Differentiable function": en.wikipedia.org/wiki/Differentiable_function Nov 30, 2020 at 23:52
• @GlenO I understand the mathematical rigor you're discussing, but this question does not need it. What happens at the boundary in the Heavaside function and Dirac function makes no practical difference as the OP is concerned with approximation not rigor. In many practical cases using the Heaviside and Dirac functions in this way (related by differentiation) "just works". I know many on Mathematics SE seek rigor above all else, but many will be using it as a tool for practical results. The later I feel is the correct context here. YMMV and I naturally respect that viewpoint as well. Dec 1, 2020 at 0:20
• All due respect, but I'm trying to get you to correct a statement in your answer that is false. The context is irrelevant, and it's blatantly obvious that Gimelist is using the real meaning of "differentiable", given that $\text{max}(0,1-x^2)$ is a well-defined function with a derivative by your reasoning. Given you still insist on leaving a false statement in your answer, I feel I must award your answer a "-1". Dec 1, 2020 at 0:30