3
$\begingroup$

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!!

$\endgroup$

migrated from mathematica.stackexchange.com Jun 29 at 13:11

This question came from our site for users of Wolfram Mathematica.

  • $\begingroup$ 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 $y$. $\endgroup$ – amd Jun 29 at 19:30
3
$\begingroup$

[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 $$y = \frac12(1+4x+\sqrt{1+8x}).$$

$\endgroup$
  • $\begingroup$ Thank you, this makes perfect sense to me $\endgroup$ – vternal3 Jun 30 at 3:21
4
$\begingroup$

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    *)
$\endgroup$
  • $\begingroup$ How did you come up with the equation f[x_] = (1 + 4 x - Sqrt[1 + 8 x])/2 ?? Thanks btw! $\endgroup$ – vternal3 Jun 29 at 14:38
  • 1
    $\begingroup$ @garej please add your own answer instead of editing mine, it's not polite. $\endgroup$ – Roman Jun 29 at 15:16
  • $\begingroup$ @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. $\endgroup$ – garej Jun 29 at 15:28
  • 1
    $\begingroup$ @vternal3 Also Solve[{y == t^2, x == Sum[i, {i, t}]}, {y}, {t}]. $\endgroup$ – Michael E2 Jun 29 at 15:52
0
$\begingroup$

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"]

plot

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.

$\endgroup$
  • $\begingroup$ $x^2$ grows quadratically, not exponentially. $\endgroup$ – Roman Jun 29 at 10:22
  • $\begingroup$ yes, thats what I meant! will correct :) $\endgroup$ – morbo Jun 29 at 10:23
  • $\begingroup$ right I do indeed mean a curve not a line... or maybe a curved line hehe. $\endgroup$ – vternal3 Jun 29 at 10:38
  • $\begingroup$ 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. $\endgroup$ – vternal3 Jun 29 at 10:41
  • 1
    $\begingroup$ 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. $\endgroup$ – morbo Jun 29 at 12:46

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.