# How do I find the function that is perfectly between y = x^2 and y = x?

At first, I thought it was as simple as taking both functions adding them together then dividing by two, but this is not the case for what I am looking for. Here is a plot of the following:

y = x
y = x^2
y = 1/2*(x^2+x)
points exactly in between y = x and y = x^2

As you can see the red line representing y = 1/2*(x^2+x) does not land on the green points which are exactly in between the two functions y = x and y = x^2. What I am trying to learn how to do is figure out how to find the function which represents the exact middle between the two equations y = x and y = x^2.

I have already tried using an excel sheet to fit a line to all the green points I have calculated and still can't come up with a good line that fits.

I have looked into calculating the midpoint between two given points and that only helped calculate the points between the two equations, and didn't help towards obtaining a function that perfectly represents the line between y = x and y = x^2.

Thanks for any help or suggestions towards the right domain of math reserved for solving cases like this one.

cheers!!

• How do you define “exactly in between?” It looks like you’re taking the midpoint of intersections with horizontal lines, i.e., the average of the $x$-values that produce the same value of $y$.
– amd
Jun 29 '19 at 19:30

[A simple way to derive the function in the accepted answer.] It looks like you’re defining “exactly in between” as being halfway between the two graphs along a a horizontal line. That is, the $$x$$-coordinate of the in-between point is the average of the values of $$x$$ that produce the same value of $$y$$ for the two functions. Inverting the two functions, we then have for this midpoint $$x = \frac12(y+\sqrt y).$$ Solving for $$x$$ and taking the positive branch gives $$y = \frac12(1+4x+\sqrt{1+8x}).$$

• Thank you, this makes perfect sense to me Jun 30 '19 at 3:21

The green dots in your plot have the coordinates

a = {{0, 0}, {1, 1}, {3, 4}, {6, 9}, {10, 16}, {15, 25}, {21, 36}, {28, 49}, {36, 64}, {45, 81}}

which can be calculated with the formula

f[x_] = (1 + 4 x - Sqrt[1 + 8 x])/2

Test:

f[10]
(*    16    *)
f[45]
(*    81    *)

How to find this formula: in Mathematica,

FindSequenceFunction[a, x]
(*    (-1 + 1/2 (1 + Sqrt[1 + 8 x]))^2    *)
• How did you come up with the equation f[x_] = (1 + 4 x - Sqrt[1 + 8 x])/2 ?? Thanks btw! Jun 29 '19 at 14:38
• @Roman, I do not see anything inpolite in that. What is inpolite is to teach ethics people who did not ask for that. I didn't touch your parts and made amendments while it was on MMA. Additions did not deserve the new answer. Pardon if it offended you. Jun 29 '19 at 15:28
• @vternal3 Also Solve[{y == t^2, x == Sum[i, {i, t}]}, {y}, {t}]. Jun 29 '19 at 15:52

The point between two points is simply $$\frac{1}{2} \left(p_1+p_2\right)=p_m$$

For example:

$$\frac{1}{2} (\{1,2\}+\{5,6\})=\{3,4\}$$

So an easy solution is to write a mean of the functions your have and we should get something in the middle. You say a line...but I assume you mean a curve?

f[x_] := Mean[{x^2, x}]
Plot[{x, x^2, f[x]}, {x, -10, 10}, ImageSize -> Large, PlotLegends -> "Expressions"]

There are no perfectly straight lines you can build between x and x^2, x^2 grows quadratically and x linearly, to stay perfectly on the mean of the two, you will end up building a curve and not a line.

• $x^2$ grows quadratically, not exponentially. Jun 29 '19 at 10:22
• yes, thats what I meant! will correct :) Jun 29 '19 at 10:23
• right I do indeed mean a curve not a line... or maybe a curved line hehe.
– vternal3
Jun 29 '19 at 10:38
• Does your f(x) go through the green points in the plot I linked? Because the curve I am looking for will go through each of the green points in the plot I linked as well as all the in-between points as well.
– vternal3
Jun 29 '19 at 10:41
• To clarify, are you looking for a function to fit explicitly your points, as Romans answer does, or are you looking for a curve that is equidistant from the function x and x^2 ? As these are very different solutions. Jun 29 '19 at 12:46