Mathematical Food for Thought


			Mathematical Food for Thought

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Serves a Daily Special and an All-You-Can-Eat Course in Problem Solving. Courtesy of me, Jeffrey Wang.

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2007 MAO.	March 31st, 2007
Good job to all Bellevue people at MAO State! We placed 3rd in Sweepstakes! That’s the highest we’ve ever gotten in my four years . Posted in Announcements \|\| 1 Comment »
Condensation Sensation. Topic: Calculus/S&S.	March 27th, 2007
Theorem: (Cauchy Condensation Test) If $\displaystyle \{a_n\}_{n = 0}^{\infty}$ is a monotonically decreasing sequence of positive reals and is a positive integer, then $\displaystyle \sum_{n=0}^{\infty} a_n$ converges if and only if $\displaystyle \sum_{n=0}^{\infty} p^n a_{p^n}$ converges. ——————– Problem: Determine the convergence of $\displaystyle \sum_{n=2}^{\infty} \frac{1}{(\ln{n})^k}$ , where is a positive real. Solution: Well, let’s apply the Cauchy condensation test. Then we know that $\displaystyle \sum_{n=2}^{\infty} \frac{1}{(\ln{n})^k}$ converges if and only if $\displaystyle \sum_{n=2}^{\infty} \frac{p^n}{(\ln{p^n})^k} = \sum_{n=2}^{\infty} \frac{p^n}{(n \ln{p})^k} = \frac{1}{(\ln{p})^k} \sum_{n=2}^{\infty} \frac{p^n}{n^k}$ does. But this clearly diverges due to the fact that the numerator is exponential and the denominator is a power function. QED. ——————– Comment: This is a pretty powerful test for convergence, at least in the situations in which it can be applied. The non-calculus proof for the divergence of the harmonic series is very similar to the Cauchy condensation test; in fact, the condensation test would state that $\displaystyle \sum_{n=1}^{\infty} \frac{1}{n}$ converges iff $\displaystyle \sum_{n=1}^{\infty} \frac{p^n}{p^n}$ converges, which clearly shows that the harmonic series diverges. ——————– Practice Problem: Determine the convergence of $\displaystyle \sum_{n=2}^{\infty} \frac{1}{(\ln{n})^{\ln{n}}}$ . Posted in Calculus, Sequences & Series \|\| 5 Comments »
Hey Now. Topic: Inequalities. Level: AIME.	March 26th, 2007
Problem: Let be positive reals such that . Prove that $\sqrt{a}+\sqrt{b}+\sqrt{c} \ge ab+bc+ca$ . Solution: We play around and guess that $\sqrt{a} \ge \frac{ab+ac}{2}$ because that would be convenient. Indeed, replacing with , this is equivalent to $\sqrt{a} \ge \frac{a(3-a)}{2} \Leftrightarrow 4a \ge a^2(3-a)^2 \Leftrightarrow (a-1)^2(4-a) \ge 0$ , which is clearly true for . So we get the three inequalities $\sqrt{a} \ge \frac{ab+ac}{2}$ , $\sqrt{b} \ge \frac{bc+ba}{2}$ , $\sqrt{c} \ge \frac{ca+cb}{2}$ . Adding them up, we have the desired $\sqrt{a}+\sqrt{b}+\sqrt{c} \ge ab+bc+ca$ . QED. ——————– Comment: I found this solution to be pretty clever, as the initial “guess” is not trivially true. But the whole thing works out quite nicely with symmetry so it’s all good. Inequality problems usually require several random ideas and inspiration before finding the crux step. ——————– Practice Problem: Let be positive reals such that . Prove that $(a+b)(b+c)(c+a) \ge 2(1+a+b+c)$ . Posted in AIME, Inequalities \|\| 2 Comments »
This Integral Not-Diverges. Topic: Calculus/S&S.	March 20th, 2007
Problem: Show that the integral $\displaystyle \int_0^{\infty} \frac{\sin{x}}{x}dx$ converges. Solution: Consider the intervals $[2k \pi, (2k+2) \pi]$ for $k = 0, 1, 2, \ldots$ . We can rewrite the given integral as $\displaystyle \int_0^{\infty} \frac{\sin{x}}{x}dx = \sum_{k=1}^{\infty} \int_{2k \pi}^{(2k+2) \pi} \frac{\sin{x}}{x} dx+C$ , where is some unimportant constant. So how can we go about bounding the integral $\displaystyle \int_{2k \pi}^{(2k+2) \pi} \frac{\sin{x}}{x} dx$ ? Well, first note that $\sin{x} = – \sin{(x+\pi)}$ so we can say $\displaystyle \int_{2k \pi}^{(2k+2) \pi} \frac{\sin{x}}{x} dx = \int_{2k \pi}^{(2k+1)\pi} \left( \frac{\sin{x}}{x}+\frac{\sin{(x+\pi)}}{x+\pi} \right) dx = \int_{2k \pi}^{(2k+1)\pi} \left( \frac{\sin{x}}{x}-\frac{\sin{x}}{x+\pi}\right) dx$ . Then, putting the last expression under a common denominator, we get $\displaystyle \int_{2k \pi}^{(2k+1)\pi} \left( \frac{\sin{x}}{x}-\frac{\sin{x}}{x+\pi}\right) dx = \int_{2k \pi}^{(2k+1)\pi} \frac{\pi \sin{x}}{x(x+\pi)} dx$ , which we can easily bound with $\sin{x} \le 1$ and $x \ge 2k \pi$ . This gives us $\displaystyle \int_{2k \pi}^{(2k+1)\pi} \frac{\pi \sin{x}}{x(x+\pi)} dx < [(2k+1)\pi-2k \pi] \cdot \frac{\pi}{(2k \pi)(2k \pi + \pi)} = \frac{1}{2k(2k+1)}$ . Hence we know that $\displaystyle \int_0^{\infty} \frac{\sin{x}}{x}dx = \sum_{k=1}^{\infty} \int_{2k \pi}^{(2k+2) \pi} \frac{\sin{x}}{x} dx+C < \sum_{k=1}^{\infty} \frac{1}{2k(2k+1)}+C$ and this converges by a -series test. QED. ——————– Comment: A pretty neat problem, though it is a standard convergence/divergence exercise. I’m sure there are many ways of doing this, but it’s always nice to come up with a cool way of showing that a series converges or diverges. It’s also interesting to note that the practice problem integral, which is only slightly different from this one, diverges. ——————– Practice Problem: Show that the integral $\displaystyle \int_0^{\infty} \frac{\|\sin{x}\|}{x}dx$ diverges. Posted in Calculus, Sequences & Series \|\| 6 Comments »
LaTeX In Comments.	March 18th, 2007
The function to use LaTeX in comments has been added. Simply enclose your LaTeX code in `[ tex ]` and `[ /tex ]` tags (without spaces). Posted in Announcements \|\| 1 Comment »

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