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# s41 - 2 then 2 n is positive so T = 1 x 2 n 2 Ö(1 x 2 n 2...

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Solution to Problem 41 This week’s problem went unsolved by the Bradley students. Ray Kremer submitted a partial solution. A correct solution was submitted by Philippe Fondanaiche who also correctly solved the more challenging problems. Sorry, Philippe, the dollar must go to a Bradley student! Refer to the diagram in the statement of the problem. Let P = ( x , y ), O denote the origin, and R = ( x ,0). The triangles TOQ and PRQ are similar, giving us s T = s - x y or T = sy s - x = sy ( s + x ) s 2 - x 2 . From the right triangle POR we get x 2 + y 2 = s 2 which after substituting and simplifying gives T = x n (Ö( x 2 + x 2 n )) (x + Ö( x 2 + x 2 n )) x 2 n = 1 + x 2 n - 2 + Ö(1 + x 2 n - 2 ) x n - 2 Now look at cases. If n <

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Unformatted text preview: 2, then 2 - n is positive, so T = ( 1 + x 2 n- 2 + Ö(1 + x 2 n- 2 ) ) x 2 - n with the first term going to 2 and the second term going to 0, as x ® 0. So T ® in this case. If n > 2, the term 1 + Ö(1 + x 2 n- 2 ) x 2 n- 2 ® ¥ since the numerator approaches 2, as x goes to 0, while the denominator gets arbitrarily small. On the other hand x 2 n- 2 x n- 2 = x n ® 0 as x ® 0. So in this case T ® ¥ . Finally, if n = 2, then T = 1 + x 2 + Ö(1 + x 2 ) ® 2. Finally we have the answer. T ® 0, if n < 2 2, if n = 2 ¥ if n > 2 This proves, I guess, that 2 is between 0 and ¥. this page....
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s41 - 2 then 2 n is positive so T = 1 x 2 n 2 Ö(1 x 2 n 2...

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