Multivariable Advanced Calculus

410 LINE INTEGRALS6. In the situation of the above problem, show that if f (x) = ∇F (x) where F is apotential, then if the motion is governed by the Newton’s law it follows that forγ (t) the motion,is constant.−F (γ (t)) + 1 2 m |γ′ (t)| 27. Generalize Stoke’s theorem, Theorem 14.4.4 to the case where R is only assumedC 1 .8. Given an example of a simple closed rectifiable curve Γ and a horizontal line whichintersects this curve in infinitely many points.9. Let Γ be a simple closed rectifiable curve and let U i be its inside. Show you canremove any finite number of circular disks from U i and what remains will stillbe a region for which Green’s theorem holds. Hint: You might get some ideasfrom looking at the proof of Lemma 14.3.6. This is much harder than it looksbecause you only know Γ is a simple closed rectifiable curve. Begin by punchingone circular hole and go from there.10. Let γ : [a, b] → R be of bounded variation. Show there exist increasing functionsf (t) and g (t) such thatγ (t) = f (t) − g (t) .Hint: You might let f (t) = V (γ; [a, t]) . Show this is increasing and then considerg (t) = f (t) − γ (t) .11. Using Problem 10 describe another way to obtain the integral ∫ fdγ for f a realγvalued function and γ a real valued curve of bounded variation as just describedusing the theory of Lebesgue integration. What exactly is this integral in thissimple case? Next extend to the case where γ has values in R n and f : γ ∗ → R n .What are some advantages of using this other approach?12. Suppose f is continuous but not analytic and a function of z ∈ U ⊆ C. Show f hasno primitive. When functions of real variables are considered, there are functionspaces C m (U) which specify how many continuous derivatives the function has.Why are such function spaces irrelevant when considering functions of a complexvariable?13. Analytic functions are all just long polynomials. Prove this disappointing result.More precisely prove the following. If f : U → C is analytic where U is an openset and if B (z 0 , r) ⊆ U, then∞∑f (z) = a n (z − z 0 ) n (14.46)for all |z − z 0 | < r. Furthermore,n=0a n = f (n) (z 0 ). (14.47)n!Hint: You use the Cauchy integral formula. For z ∈ B (z 0 , r) and C the positivelyoriented boundary,∫1 f (w)f (z) =2πi C w − z = 1 ∫f (w) 12πi C w − z 0 1 − z−z dw0w−z 0∫1∞∑ f (w)=2πi (w − z 0 ) n+1 (z − z 0) n dwC n=0