January 2003:
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01/06/03) Devise a formula for the nth term of the following sequence, and prove whether it converges or diverges.
{e , e1/2 , - e1/4 , - e1/8 , e1/16 , e1/32 , - e1/64 , ... }
It is rather obvious that the powers are geometrically decreasing, with a ratio of 1/2. Yet, the sign pattern is a little tricky. One way to handle the signs is to use a quadratic equation of (n+3)(n+2)/2 as an exponent of -1 for different nth terms. Thus the generalized formula would be:
tn = (-1)(n+3)(n+2)/2 * e1/2n-1 ,
where n takes integer values from 1 to infinity.
The limit test can be applied to test for convergence. Using limits, one can see that the positive values approach 1, while negative ones approach -1. Although convergence exists for individual positive and negative terms, the sequence diverges, due to oscillation of limit points.
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01/13/03) As opposed to the problem posted on 5/6/02 where the largest cylinder inscribed in a sphere was asked, find the radius and height of a cylinder with the largest lateral surface area inscribed in a given sphere.
Calculus is a great tool for solving geometrical problems. We know that the lateral surface area of a cylinder is 2pi*r*h where r and h are its radius and height respectively. There is also a relationship among r, h and R the radius of sphere: Pythagorean theorem.
(h/2)2 + r2 = R2
Solving and substituting this for r, the lateral area would be pi*h*sqrt(4R2-h2)/2. Differentiating this function and setting it equal to zero, the equation yields to h = R*sqrt(2). Hence, the radius r would be R*sqrt(2)/2.
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01/20/03) Find the area of the region bounded by the graphs of y = 0, x = 1, x = 2 and y = 1/sqrt(x2-1).
As we can tell, this is the integral of 1/sqrt(x2-1) dx from x = 1 to x = 2. We can use trigonometric substitution to find the integral.
Let x = sec u thus, dx = sec u tan u du
So the integral is equivalent to the integral of 1/(sec2-1) * sec u tan u du, which is trigonometrically equal to the integral of sec u du. The integral of sec u du is equal to ln| cos u / (sin u - 1) |. Substituting sec u for x, we get ln | cos(sec u) / (sin(sec u) -1) |.
Using a right triangle and applying the trig functions with some algebra, we obtain ln |1/(sqrt(x2 - 1) - x)|. Now we can evaluate this from x = 1 to x = 2. Hence the answer is ln(2 + sqrt(3)).
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01/27/03) Using calculus, prove that the number of diagonals of a polygon far outpaces the number of its sides.
The tool to compare the relative speeds of two functions in calculus is limits. Obviously, we have to know the functions of a polygon's diagonal number and number of sides, then, to do this.
Assuming that a polygon has n sides, its diagonals would be formed by the combinations of n dots in groups of 2, except for the sides itself. Mathematically this is equal to nC2 - n or n(n-1)/2 - n which is further simplified to (n2 - 3n)/2.
Comparing the paces of n and (n2 - 3n)/2 requires us to take the limit of their ratio as n approaches infinity. The limit of n over (n2 - 3n)/2 as n goes to infinity is zero, using the highest-power rule. Therefore, one can see that the latter far outpaces the former.
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