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Take a concave function e.g. the quadratic below (picture attached).

Why does the chord between two values $f(x₁)$ & $f(x₂)$ give us the linear relationship between the convex combination of the two points $x(λ) = (1-λ)x₁ + (λ)x₂$, and the convex combination of the functions value at those two points $(1-λ)f(x₁) + (λ)f(x₂)$?

Example

If we take the concave function $f(x) = -(x - 4)² + 16$, and we take $x₁ = 2, x₂ = 4$

Connecting $f(x₁) = 12$ & $f(x₂) = 16$ with a chord we get the line with equation $g(x) = 2x + 8$

For values between $x₁$ & $x₂$ this ‘$x$’ can be described as $x(λ) = (1-λ)x₁ + (λ)x₂$

I.e. for $x₁≤ x ≤ x₂, g(x) = 2x(λ) + 8$

Which we note can even be written: $g(x) = x₁x(λ)+ x₁ x₂$

  • Q1) Why is this linear equation $g(x)$ equal to the convex combination of the value of the quadratic (between $x₁$ & $x₂$). Essentially, why does this chord magically give us the values $(1-λ)f(x₁) + (λ)f(x₂)$?

I can't wrap my head round the fact that a a convex combination of points, is mapped to the same convex combination of the quadratic of those points, by a linear function.

Edit: I have wrapped my head around the core part of the question but am still trying to understand these questions below. Have posted an answer this progress so far. Thoughts appreciated!

  • Q2) Why is the $y$ intercept for $g(x) = x₁ x₂$

  • Q3) Why is the slope equal to $x₁$

  • Q4) I feel like the formulation: $f(x₁) + λ(f(x₂) - f(x₁))$ gets me closer but I’m not there yet.

I have verified this for values of $x₁ = 1$ and 2, with a range of values for $x₂$ (for example the picture includes $x₁ = 1, x₂ = 6$

Cheers!

enter image description here

CormJack
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    If you have any two points $p=(a,b)$ and $q=(c,d)$ then the line segment between them is parameterized by $\lambda p+(1-\lambda)q = (\lambda a+(1-\lambda)c, \lambda b+(1-\lambda)d)$, $0\leq \lambda \leq 1$. For instance for $\lambda=0,1$ you have the endpoints, for $\lambda=0.5$ the midpoint etc. Maybe that is what you need to convince yourself of first. Then you apply this observation to the points $p=(x_1,f(x_1))$, $q=(x_2,f(x_2))$. – Michal Adamaszek Jan 06 '23 at 14:10
  • Okay will do! Thanks for commenting @MichalAdamaszek will feedback! – CormJack Jan 06 '23 at 14:11
  • @MichalAdamaszek Wow I just walked through it. Mind blowing! Thanks for such a clear and strait forward path to the answer. It's a shame, it isn't an answer I would have voted it. I'm going to post it myself as an answer, please do let me know if it's correct thanks! – CormJack Jan 07 '23 at 10:43

1 Answers1

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Okay this was bugging me for ages, so maybe the answer will help others. Special thanks to @Michal Adamaszek for the superb comment.

  1. Take any two points$p = (x_1, y_1) \in \mathbb{R^2}$ and $q = (x_2, y_2) \in \mathbb{R^2}$

  2. We can describe the line segment between these two points as: $(\lambda p+(1-\lambda)q = (\lambda x_1+(1-\lambda)x_2,\; \lambda y_1 +(1-\lambda)y_2$), where the first entry describes our point on the x axis, between $x_1$ & $x_2$, and the second entry is the corresponding point for the y axis.

  3. We then chose our particular points of interest: $p=(x_1,f(x_1))$ & $q=(x_2,f(x_2))$ which are the two points on the function between which we draw the chord.

  4. Parametrising these in the same way, we get: $p+(1-\lambda)q = (\lambda x_1+(1-\lambda)x_2,\; \lambda f(x_1) +(1-\lambda)f(x_2)$). Where the second entry is the desired point on the y axis between $f(x_1)$ & $f(x_2)$.

This answers my main question. I'm still working on:

  • Q2) why the intercept equals $x_1x_2$.
  • Q3) why the slope equals $x_1$.

Will update, but contributions are welcome!

Edit: To understand why the slope equals $x_1$ (I can't tell if this is fluke or not) I set the slope $\frac{f(x_2) - f(x_1)}{x_2 - x_1} = x_1$. Inputing the actual function for $f$ I end up getting $x_2 = 8$?

CormJack
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