Math Challenge - October 2021

[fresh_42]

Why does the sequence converge at all? I believe it must converge because of the series is strictly decreasing and it contains only positive values.

When the series is strictly decreasing, the value of successive elements must either keep decreasing without a bound (and so tend to $-\infty$) or must decrease indefinitely with some finite bound. Since the series contains only positive values, it cannot touch or drop below zero, meaning 0 acts as a strict lower bound to the limit toward which such a series can tend towards.

Convergence needs a certain property of the real numbers that should be mentioned. I don't want to hear the technical term but the principle. That's why I wrote that a heuristic is sufficient.

 

[Not anonymous]

Extra: (open)

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Consider the two elliptic curves and observe that one has two connection components and the other one has only one. Determine the constant $c \in [1,3]$ in $y^2 = x^3 - 2x + c$ where this behavior exactly changes. What is the left-most point of this curve?

Spoiler: Question 14 (Extra)

In $y^2 = x^3 -2x + 1$, there are two components (disconnected from one another) because the curve is defined only when $x^3 -2x + 1 \geq 0$ (due to $y$ being the square root of that polynomial in $x$) and there is a range of $x$-values (incidentally including $x=0$) such that $x^3 -2x + 1$ is negative over this range (this is the "gap" between the x-ranges of the two components) but is non-negative for all x-values greater than this range and also for a some range to the left of this gap.

Now, to find the value of constant $c \in [1,3]$ where curves defined by $y^2 = x^3 - 2x + c$ change from containing one component to two components, we use the fact that the .

Let us find the minima of the polynomial $f(x) = x^3 -2x + 1$. At the minima, $f'(x) = 0$ and $f''(x) > 0$.
$f'(x) = 3x^2 - 2 = 0 \Rightarrow x = \pm \sqrt{\dfrac{2}{3}}$
$f''(x) = 6x \Rightarrow f''(x) > 0 \iff x > 0$

Therefore $x=\sqrt{\dfrac{2}{3}}$ must correspond to a local minimum point while $x=-\sqrt{\dfrac{2}{3}}$ must correspond to a maximal point.

In the case where the curve $y^2 = f(x)$ consists of 2 components, the minimal value $f(\sqrt{\dfrac{2}{3}})$ must be negative, because otherwise, the $f(x) < 0$ only for $x < x_1$ for some $x_1 < - \sqrt{\dfrac{2}{3}}$ and so the curve be defined for all $x \geq x_1$ and won't have 2 separate components.

$f(\sqrt{\dfrac{2}{3}}) < 0 \Rightarrow \dfrac{2}{3}\sqrt{\dfrac{2}{3}} - 2\sqrt{\dfrac{2}{3}} + c < 0 \Rightarrow c < \dfrac{4}{3}\sqrt{\dfrac{2}{3}}$.
Thus, $c_0 = \dfrac{4}{3}\sqrt{\dfrac{2}{3}} \approx 1.088662108$ is the value of the constant where the change in the nature of the curve (change between consisting of two components vs. one component alone) happens.

(2)​

At the left-most point of the curve $y^2 = x^3 -2x+c_0$, we must have $y^2 = y = 0$. We solve for $x$ accordingly.

$x^3 - 2x + c_0 = 0 \Rightarrow x(x^2-2) = -c_0 = -\dfrac{4}{3}\sqrt{\dfrac{2}{3}}$.

(2)​

$x=\sqrt{\dfrac{2}{3}}$ is obviously a solution for this equation but it is not the leftmost point. With some trial and error, or informed guesswork, followed by validation, I found that $x=-2\sqrt{\dfrac{2}{3}}$ is the other solution for equation (1). Therefore, $(x=-2\sqrt{\dfrac{2}{3}}, y=0)$ is the left-most point of the curve $y^2 = x^3 - 2x + \dfrac{4}{3}\sqrt{\dfrac{2}{3}}$