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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Symmetry For The Win. Topic: Calculus.	October 30th, 2006
Problem: Evaluate the improper integral $I = \displaystyle \int_0^{\pi} \ln{(\sin{x})} dx$ . Solution: As the topic suggests, we will look for a symmetry to simplify the problem. Notice the identity $\displaystyle \int_0^a f(x) dx = \int_0^a f(a-x) dx$ (since we’re just integrating across the interval from different “directions.” Using this, we have $\displaystyle I = \int_{0}^{\frac{\pi}{2}}\ln{(\sin{x})}dx = \int_{0}^{\frac{\pi}{2}}\ln{\left(\sin{\left(\frac{\pi}{2}-x\right)}\right)}dx = \int_{0}^{\frac{\pi}{2}}\ln{(\cos{x})}dx$ . Adding the old integral to the new one, we have $\displaystyle 2I = \int_{0}^{\frac{\pi}{2}}\ln{(\sin{x})}dx+\int_{0}^{\frac{\pi}{2}}\ln{(\cos{x})}dx = \int_{0}^{\frac{\pi}{2}}\ln{\left(\frac{1}{2}\sin{2x}\right)}dx$ from the property of logs and the double-angle identity (here it is again!). But in fact this expression is simply $\displaystyle 2I =-\frac{\pi}{2}\ln{2}+\frac{1}{2}\int_{0}^{\pi}\ln{(\sin{u})}du$ () by the substitution . Taking into the account of the symmetry of $\sin{x}$ from to $\pi$ , we get $\sin{x} = \sin{(\pi-x)}$ so $\displaystyle \frac{1}{2}\int_{0}^{\pi}\ln{(\sin{u})}du = \int_{0}^{\frac{\pi}{2}}\ln{(\sin{u})}du = I$ . Plugging back into () we obtain $\displaystyle 2I =-\frac{\pi}{2}\ln{2}+I$ so we get $I = -\frac{\pi}{2}\ln{2}$ . QED. ——————– Comment: The function is not nice to actually integrate; it involves the polylogarithm function if you try here. This is an important example of how symmetry can help a ton in integration because there are so many functions that cannot be integrated with elementary functions but can be evaluated over an interval through different techniques. ——————– Practice Problem: Evaluate $\displaystyle \int_0^{\frac{\pi}{2}} \sin^4{x} dx$ without actually finding the antiderivative. Posted in Calculus \|\| No Comments »
Out Of Place Trig. Topic: Geometry/Trigonometry. Level: AMC/AIME.	October 29th, 2006
Problem: (2006-2007 Warm-Up 4) Two sides of a triangle are and $\|\cos{\theta}\|$ and the angle between them is $\theta$ . If $0 < \theta < \pi$ , what is the maximum area of the triangle? Solution: Well, given two sides and the angle between them, there is only one formula for the area of a triangle that comes to mind. $[ABC] = \frac{1}{2}ab \sin{(\angle C)}$ . So plugging this in, we get the area to be $\frac{1}{2} (4) \|\cos{\theta}\| \sin{\theta} = 2 \sin{\theta} \|\cos{\theta}\|$ . By the double-angle identity, this is equal to $\|\sin{(2\theta)}\|$ for $0 < \theta < \pi$ . But the maximum of this is clearly and it is obtained when $\theta = \frac{\pi}{4}, \frac{3 \pi}{4}$ . QED. ——————– Comment: This problem shouldn’t have been too hard; finding that specific formula for the area of the triangle could have been derived by drawing an altitude. And the double-angle identity should always be known. Lastly, maximizing the sine function is a given. ——————– Practice Problem: (2006 Skyview) If $\sin{x} = \frac{3}{5}$ and $\cos{x} = \frac{4}{5}$ , what is $\sin{(3x)}$ ? Posted in AIME, AMC, Geometry, Trigonometry \|\| 1 Comment »
Bellevue Wins First At Skyview!	October 28th, 2006
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Square Sum Stuff. Topic: Polynomials/S&S. Level: AMC/AIME.	October 27th, 2006
Problem: Evaluate the summation $\displaystyle \sum_{k=1}^{\infty} \frac{k^2}{2^k} = \frac{1^2}{2^1}+\frac{2^2}{2^2}+\frac{3^2}{2^3}+\frac{4^2}{2^4}+\cdots$ . Solution: Here’s a technique that will help you evaluate infinite series that are of the form polynomial over exponential. It’s based on the idea of finite differences: If is a polynomial with integer coefficients of degree then is a polynomial of degree (not hard to show; just think about it). So let $S = \frac{1^2}{2^1}+\frac{2^2}{2^2}+\frac{3^2}{2^3}+\frac{4^2}{2^4}+\cdots$ . Then consider by simply multiplying each term by : $2S = 1^2+\frac{2^2}{2^1}+\frac{3^2}{2^2}+\frac{4^2}{2^3}+\cdots$ . And now find the difference by subtracting the terms with equal denominators. We get $S = 2S-S = 1+\frac{2^2-1^2}{2^1}+\frac{3^2-2^2}{2^2}+\frac{4^2-3^2}{2^3}+\cdots$ $S = 1+\frac{3}{2^1}+\frac{5}{2^2}+\frac{7}{2^3}+\cdots$ . Notice that the numerator is now a polynomial of degree instead of . Repeating this, we have $2S = 2+3+\frac{5}{2^1}+\frac{7}{2^2}+\cdots$ and $S = 2S-S = 2 + 2 + \frac{5-3}{2^1}+\frac{7-5}{2^2}+\cdots$ $S = 2 + (2 + 1+\frac{1}{2}+\cdots)$ . Notice that the latter part is just a geometric series which sums to $2 + 1 + \frac{1}{2} + \cdots = 4$ so . QED. ——————– Comment: The method of finite differences is extremely useful and is basically a simplified version of calculus – $P(x+1)-P(x) \approx P^{\prime}(x)$ in a very approximating sense. It’s a good thing to know, though, because then you have a better understanding of how polynomials work. ——————– Practice Problem: Let be a polynomial with integer coefficients. Using the method of finite differences, predict the degree of . $P(1) = 1 \mbox{ } P(2) = 9 \mbox{ } P(3) = 20 \mbox{ } P(4) = 36 \mbox{ } P(5) = 59 \mbox{ } P(6) = 91$ . Posted in AIME, AMC, Polynomials, Sequences & Series \|\| 2 Comments »
Serious Substitution. Topic: Calculus.	October 25th, 2006
Problem: Use the substitution $z = \sqrt{x}$ to solve the differential equation $4x \frac{d^2y}{dx^2}+2 \frac{dy}{dx}+y = 0$ . Solution: Well, let’s do what it says. From the chain rule, we have $\displaystyle \frac{dy}{dz} = \frac{dy}{dx} \cdot \frac{dx}{dz} = 2z \frac{dy}{dx}$ . Then we also have by the product rule and chain rule again $\displaystyle \frac{d^2y}{dz^2} = 2 \frac{dy}{dx}+2z \frac{d^2y}{dx^2} \cdot \frac{dx}{dz} = \frac{1}{z} \cdot \frac{dy}{dz}+4z^2 \frac{d^2y}{dx^2}$ . So we can make the substitutions $\displaystyle 4x \frac{d^2y}{dx^2} = 4z^2 \frac{d^2y}{dx^2} = \frac{d^2y}{dz^2}-\frac{1}{z} \cdot \frac{dy}{dz}$ and $\displaystyle \frac{dy}{dx} = \frac{1}{2z} \cdot \frac{dy}{dz}$ to obtain the differential equation $\frac{d^2y}{dz^2}-\frac{1}{z} \cdot \frac{dy}{dz} + 2 \cdot \frac{1}{2z} \cdot \frac{dy}{dz}+y = \frac{d^2y}{dz^2}+y = 0$ . But we know the solution to this is $y = C_1 \cos{z}+C_2 \sin{z}$ so our final solution is $y = C_1 \cos{\sqrt{x}}+C_2 \sin{\sqrt{x}}$ . QED. ——————– Comment: This substitution is, of course, not really natural but was actually found after solving the ODE in another way. Fortunately, it simplifies the problem rather greatly and seems to be a useful technique to look out for. ——————– Practice Problem: Find another way to solve the differential equation. Posted in Calculus \|\| 3 Comments »

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