1873_solutions

Any positive integer n if the number x belongs to one

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Unformatted text preview: square 0, 1 0, 1 0, 1 the such that for every x 0, 1 the vertical x-section S x of S has measure zero and for every y horizontal y-section S y contains almost every number in 0, 1 . This example is quite easy to produce if you have studied enough set theory and it can be found here. Unless you have read the set theory, must assume the existence of the Sierpinski example for the purposes of this exercise. Using Sierpinski’s example, prove that the statement of Fichtenholz’s theorem becomes false if we widen it to include functions that are almost Riemann integrable. Exercises 5 and 6 are offered as a resource for a special project. 18 Calculus of Several Variables Exercises on Velocity, Speed and Curve Length 1. Point at the equation  t  t cos t, t sin t, t 2  and assign it as a definition in Scientific Notebook. Then point at the integral Þ  t dt and click on 0 Evaluate and then point at the result and then click on Simplify to show that the length of the curve is 401 5 2  1  1 5 ln 10 1 2 5 5 2  1 2. Prove that every piecewise smooth curve in R k is rectifiable. This assertion follows at once from the relationship between speed and curve length and the fact that a function that is piecewise continuous on an interval must be itegrable. 3. Given that if 0  t 1 t t, t sin t   if t  0 0, 0 , prove that  is not rectifiable. 1. Solution: We observe first that for each t t   t  and therefore, whenever 0  t  1 4 t 0,  we have 2 1  1 t sin 1 cos 1 t t t2 we have 1 t sin 1 cos 1 1 t t t t 1 t sin 1 t t cos 1 t cos 1 t 1 4 Now given any positive integer n, if  4 2n 2n   4 1 t then cos 1 t  cos 1 t In other words, if n is a positive integer and if 4 8n  1  then 4 t cos 1 t 1 t t 1. 2 8n 1 4 1 1 t 1 2 1 4 and so Þ 4 8n 1  4 8n1  Þ  t dt 4 8n 1  4 8n1  1 t 1 4 1 2 dt  1 2 1 4 log  1 2 1 4 log 8n  1 8n 1 4 8n log 1 1 2  Now since lim nÝ log 1  2 8n 1 1 4 1 n it follows from the comparison test that the series log 1  2 8n 1 is divergent and we deduce that the series Þ 402 4 8n 1  4 8n1   t dt 4 8n  1  1 4 log 1  2 8n 1 is divergent. Therefore  Þ  t dt  Ý. 0  lim  4. Prove that if f is an increasing continuous function from an interval a, b into R and if  t  t, f t for all t a, b then  is rectifiable. 1. Solution: Suppose that t 0 , t 1 , t 2 , , t n is a partition of the interval a, b . We have n n  tj  tj  1 j1 tj f tj tj 1, f tj 1 j1 n |t j t j 1 |  |f t j f tj 1 | f tj 1 j1 n  tj tj 1  f tj j1 b af b fa. 5. Suppose that  is a rectifiable curve with domain a, b in a metric space X and that, for each t a, b we have defined  t to be the length of the restriction of the curve  to the interval a, t . Prove that the function  must be continuous on the interval a, b . 1. Solution: We shall explain the continuity of  at each number t in the open interval a, b and leave as an exercise the task of proving the one sided continuity of  at the numbers a and b. Suppose that a ...
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