Category: Blarghing

Let $a, b, c$ be real numbers. Prove that three roots of the equation
\begin{align*}
\frac{b+c}{x-a} + \frac{c+a}{x-b} + \frac{a+b}{x-c} = 3
\end{align*}
are real.

Solution: The solution which the book lists is actually much more elegant than what I will present here. We let $s = a+b+c$ to be the sum. Inserting this into the equation, we see that
\begin{align*}
\frac{s-a}{x-a} + \frac{s-b}{x-b} + \frac{s-c}{x-c} = 3
\end{align*}
Now, multiplying both sides by $(x-a)(x-b)(x-c)$ and rearranging results in
\begin{align*}
(s-a)(x-b)(x-c) + (s-b)(x-a)(x-c) + \\
(s-c)(x-a)(x-b) – 3(x-a)(x-b)(x-c) = 0.
\end{align*}
By inspection, it’s pretty clear that $x = s$ is a root, hence if we find one additional real root we are done as imaginary roots must exist in pairs.

Here is where I deviated against the solutions. I plugged in $x = a, b, c$ respectively and considered all the cases separated by magnitude and showed that there must exist a pair to the left/right of $s$ such that it is positive and negative, hence there exists another real root.

We started a new problem recently, but still within the realms of preconditioning. The exact question and algorithms will remain under wraps right now, but the preliminary results I got were incredibly curious; here’s a plot of condition number below:

Note that a constant condition number as we increase order is extremely good, but my adviser and I were not rejoicing.  It’s just that the result was too good! We were not expecting the condition number to not increase, as that’s far better than what current literature (and our intuition) suggests.

After a week of fumbling around, I’ve extended the computation and produced the following plot:

Not that the above plot is the same algorithm, just with extended order (or p). And ahhh, there it is,  the growth that we were looking for.

Even though it’s a worse result that what we initially saw, we can actually prove this version!

This is a problem from Putnam and Beyond:

Given that $9a^2 + 8ab + 7b^2 \le 6$, show that $7a + 5b + 12ab \le 9$.

My solution is extremely janky; nevertheless, it is still a solution that is slightly different from the book’s.

Our hypothesis implies that $-12ab \ge -9 + \frac{27}{2}a^2 + \frac{21}{2}b^2$ by manipulation. Note that our conclusion is equivalent to $9 – 7a – 5b – 12ab \ge 0$. Hence, we have

\begin{align*}
9 – 7a – 5b – 12ab &\ge \frac{27}{2}a^2 + \frac{21}{2}b^2- 7a – 5b \stackrel{?}{\ge} 0.
\end{align*}

Now consider the two quadratics $\frac{27}{2}a^2 – 7a$ and $\frac{21}{2}b^2 – 5b$. They are negative between $0$ and their other root, which is $a = \frac{14}{27}$ and $b = \frac{10}{21}$. Since the objective function is strictly increasing for in the positive numbers, checking the last root reveals that it is indeed satisfied as $7a + 5b + 12 ab = \frac{1696}{189} < 9$. Again, janky but doable in a reasonable time. (Of course, there's more proper methods with Lagrange multipliers and such...)

Here’s an excellent question from AoPS:

In the figure to the left, circle $B$ is tangent to circle $A$ at $X$, circle $C$ is tangent to circle $A$ at $Y$, and circles $B$ and $C$ are tangent to each other. If $AB = 6, AC = 5,$ and $BC = 9$, what is $AX$?

The solution really requires no geometry besides the very simple definitions of a circle and a little bit of algebra. Let $r_B, r_C$ be the radius of circles $B, C$ respectively, then we have $r_B + r_C = BC = 9$. Second, $AX = AY = AC + CY = 5 + r_C = 6 + r_B$. Solving these two systems of equation gives 4 and 5. Hence the radius is 10.

I grew up with basic cable, and strict parents. Mr. Rogers’ show (or PBS in general) was never the favorite if I had the luxury of choice; Saturday morning cartoons takes that award. But as an adult, I wished that I watched his show a bit more as a child.

The more I learned about him, the more I think he is the moder day equivalent of Jesus. From his heartfelt reaction to meeting a former guest on the show to washing Officer’s feet, I really could have learned a lot from him. When Cassady and I spent Thursday night at the dollar theater seeing the 2018 film, I was a bit wary at first.

How do you portray someone with no scandals and blemishes? I didn’t want a film which was obsequious, but  rather show the more human-side of him.

I think the film did a good job.

It’s an extremely touching film which showed a lot of the behind-the-scenes of the iconic TV show. More importantly, they showed the humorous side of Fred, along with the pragmatic side. The writers showcased not only his willingness to rise against racism, but his reluctance of gay rights in an era where that was a dark mark.

I really do wonder what he thinks about the current landscape. Would he think he failed? Or just get back to work educating children?

My god.; this weekend was utterly wasted for the most part. I spent a good 8 hours in a training session on Saturday, then afterwards started a game of Civ5 as Poland on emperor difficulty.

My position started out pretty precarious, with two warmongering civilizations Songhai AND Mongols neighboring me. But luckily, it seems that they started going to war against each other with the Mongols capturing a city-state to the east of me, and the Songhai’s capital actually.

I got my first three cities running, then just started rolling over the Mongols once I researched and built crossbowmen. Not sure where the Khan’s troops were, but a combination of longswords men and crossbowmen really just steamed rolled the Mongols.

By that time, I obtained the technology for producing winged Hussars and cannons. This duo was literally impossible to stop by the AIs. I conquered the capital of Byzantine, then rolled over to the Shoshones. And then somehow everyone in the world decided to declare war on me.

And then I conquered all of them by the 1950s in-game-time.

I guess the most fun part of civ 5 is reaching late game; by then, the juggernaut becomes impossible to stop. The AI is so bad at naval combat that my battleships were shelling every city with no counterplay from them.

It’s been a rough few days for my body. My left arm has some ulnar nerve impingement issues, and have been extra sensitive.

This past Sunday I fell from my bike and scraped my right knee, and injured my right arm’s wrist.

I have one remaining non-impaired limb left.

(Work has been going slow lately… hence all these math posts.)

Let $n$ be a positive integer. Starting with $1, \frac{1}{2}, \cdots, \frac{1}{n}$, form a new sequence of $n-1$ entries $3/4, 5/12, \cdots, (2n-1)/(2n(n-1))$ by taking the averages of two consecutive entries in the first sequence. Repeat the averaging process on the second sequence to obtain a third sequence of $n-1$ entries, and continue until the final sequence of only a single number $x_n$. Show that $x_n < 2/n$. Solution: Given a sequence $a_1, a_2, \ldots, a_n$, the averaging process will give us $(a_1 + a_2)/2, \ldots, (a_{n-1} + a_n)/2$. If we proceed with the averaging procedure, it is not hard to see that the binomial coefficients show up. In fact, we can see that \begin{align*} x_n &:= \frac{\sum_{k=1}^n \binom{n-1}{k-1} \frac{1}{k}}{2^{n-1}} \\ &= \frac{\sum_{k=1}^n \frac{(n-1)!}{(k-1)!(n-k)!} \frac{1}{k}}{2^{n-1}} \\ &= \frac{\sum_{k=1}^n \frac{1}{n}\binom{n}{k}}{2^{n-1}} = \frac{2^n - 1}{n 2^{n-1}} < 2/n. \end{align*} (Definitely one of the easier problems.)