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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Addition At Its Finest. Topic: Calculus/S&S.	June 29th, 2007
Problem: Evaluate $\displaystyle \sum_{n=1}^{\infty} \frac{x^n}{n(n+1)}$ where is a real number with . Solution: Looking at that all too common denominator, we do a partial fraction decomposition in hopes of telescoping series. The summation becomes $\displaystyle \sum_{n=1}^{\infty} \left(\frac{x^n}{n}-\frac{x^n}{n+1}\right)$ . Common Taylor series knowledge tells us that $\displaystyle \ln{(1-x)} = -\left(x+\frac{x^2}{2}+\frac{x^3}{3}+\cdots\right) = -\sum_{n=1}^{\infty} \frac{x^n}{n}$ , which convenient fits the first part of the summation. As for the second part, we get $\displaystyle \sum_{n=1}^{\infty} \frac{x^n}{n+1} = \frac{1}{x} \sum_{n=1}^{\infty} \frac{x^{n+1}}{n+1} = \frac{-\ln{(1-x)}-x}{x}$ from the same Taylor series. Combining the results, our answer is then $\displaystyle \sum_{n=1}^{\infty} \frac{x^n}{n(n+1)} = 1-\ln{(1-x)}+\frac{\ln{(1-x)}}{x}$ . QED. ——————– Comment: Even though the trick at the beginning didn’t actually get much to telescope, the idea certainly made it easier to recognize the Taylor series. Algebraic manipulations are nifty to carry around and can be applied in problems wherever you go. ——————– Practice Problem: Show that $\displaystyle \int_0^{\frac{\pi}{2}} \ln{(\tan{x})} = 0$ . Posted in Calculus, Sequences & Series \|\| 4 Comments »
One By One, We’re Making It Fun. Topic: Calculus/S&S.	May 28th, 2007
Theorem: (Stolz-Cesaro) Let $\{a_n\}$ and $\{b_n\}$ be sequences of real numbers such that $\{b_n\}$ is positive, strictly increasing, and unbounded. If the limit $\displaystyle \lim_{n \rightarrow \infty} \frac{a_{n+1}-a_n}{b_{n+1}-b_n} = L$ exists, then the following limit also exists and we have $\displaystyle \lim_{n \rightarrow \infty} \frac{a_n}{b_n} = L$ . ——————– Theorem: (Summation by Parts) If $\{f_k\}$ and $\{g_k\}$ are two sequences, then $\displaystyle \sum_{k=m}^n f_k(g_{k+1}-g_k) = [f_{n+1}g_{n+1}-f_mg_m]-\sum_{k=m}^n g_{k+1}(f_{k+1}-f_k)$ . ——————– Problem: Let $\{a_n\}$ be a sequence of real numbers such that $\displaystyle \sum_{k=0}^{\infty} a_k$ converges. Show that $\displaystyle \lim_{n \rightarrow \infty} \frac{1}{n+1} \sum_{k=0}^n k \cdot a_k = 0$ . Solution: Define the sequence $\{b_n\}$ by $\displaystyle b_n = \sum_{k=0}^n a_k$ and let $\displaystyle \lim_{n \rightarrow \infty} b_n = L$ . Then, by summation by parts with $\{f_n\} = \{n\}$ and $\{g_n\} = \{b_n\}$ , we have $\displaystyle \sum_{k=0}^n k \cdot a_k = \sum_{k=0}^n k \cdot (b_{k+1}-b_k) = (n+1)b_{n+1}-\sum_{k=0}^n b_{k+1}$ . The summation we wish to take the limit of is then $\displaystyle \frac{1}{n+1} \sum_{k=0}^n k \cdot a_k = b_{n+1}-\frac{1}{n+1} \sum_{k=0}^n b_{k+1}$ . But since, by Stolz-Cesaro, $\displaystyle \lim_{n \rightarrow \infty} \frac{1}{n+1} \sum_{k=0}^n b_{k+1} = \lim_{n \rightarrow \infty} b_{n+1} = L$ , we obtain $\displaystyle \lim_{n \rightarrow \infty} \frac{1}{n+1} \sum_{k=0}^n k \cdot a_k = \lim_{n \rightarrow \infty} \left(b_n-\frac{1}{n+1} \sum_{k=0}^n b_{k+1}\right) = L-L = 0$ . QED. ——————– Comment: Summation by parts is a very useful technique to change sums around so that they are easier to evaluate. If you hadn’t noticed, it is the discrete analogue of integration by parts and is in fact very similar. Stolz-Cesaro is powerful as well and seems like a discrete analogue to L’Hopital (but I’m not sure about this one). Applying well-known calculus ideas to discrete things can often turn into neat results. ——————– Practice Problem: If $\{a_n\}$ is a decreasing sequence such that $\displaystyle \lim_{n \rightarrow \infty} a_n = 0$ , show that $\displaystyle \sum_{k=1}^{\infty} a_k \cdot \sin{(kx)}$ converges for all . Posted in Calculus, Sequences & Series \|\| No Comments »
Ready For AP Calculus? Topic: Calculus/S&S. Level: AIME/Olympiad.	May 5th, 2007
Problem: Evaluate $\displaystyle \sum_{n=1}^{\infty} \frac{1}{n^2 \cdot 2^n}$ . Solution: Let $\displaystyle f(x) = \sum_{n=1}^{\infty} \frac{x^n}{n^2}$ . Then $\displaystyle f^{\prime}(x) = \sum_{n=1}^{\infty} \frac{x^{n-1}}{n} = -\frac{\ln{(1-x)}}{x}$ , a well-known Taylor series. So we want to integrate this: $\displaystyle \int \frac{\ln{(1-x)}}{x}dx = \ln{x} \ln{(1-x)}-\int \frac{\ln{x}}{x-1}dx$ by parts using $u = \ln{(1-x)}$ and . Substituting in the last integral, we have $\displaystyle \int \frac{\ln{(1-x)}}{x}dx = \ln{x} \ln{(1-x)}-\int \frac{\ln{(1-t)}}{t}dt$ . So $-f(x) = \ln{x} \ln{(1-x)}+f(t)+C = \ln{x} \ln{(1-x)}+f(1-x)+C$ . Thus $f(x)+f(1-x) = C-\ln{x} \ln{(1-x)}$ for some constant . Using our knowledge that , $\displaystyle f(1) = \frac{\pi^2}{6}$ , and $\displaystyle \lim_{x \rightarrow 1} \ln{x} \ln{(1-x)}= 0$ by L’Hopital twice, we see that $f(0)+f(1) = C \Rightarrow C = \frac{\pi^2}{6}$ Then setting $x = \frac{1}{2}$ we obtain $\displaystyle 2f\left(\frac{1}{2}\right) = \frac{\pi^2}{6}-\ln{\frac{1}{2}} \ln{\frac{1}{2}}$ and $\displaystyle \sum_{n=1}^{\infty} \frac{1}{n^2 \cdot 2^n} = f\left(\frac{1}{2}\right) = \frac{\pi^2}{12}-\frac{\ln^2{(2)}}{2}$ . QED. ——————– Comment: This was a pretty tough problem that required you to compound a lot of calculus knowledge all into a single problem – series, integration by parts, limits. Recognizing all the steps was the first part; following through with the right computations was another. Still, there weren’t really any super clever tricks, mostly just standard substitutions and approaches applied in a somewhat non-standard way. Makes for a very nice problem. ——————– Practice Problem #1: Show that $\displaystyle \lim_{x \rightarrow 1} \ln{x} \ln{(1-x)} = 0$ . Practice Problem #2: Evaluate $\displaystyle \sum_{n=1}^{\infty} \frac{x^n}{n}$ . Can you also find $\displaystyle \sum_{n=1}^{\infty} \frac{x^n}{n^3}$ ? Posted in AIME, Calculus, Olympiad, Sequences & Series \|\| No Comments »
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 »
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 »

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