Discrete Weighted Minimal Riesz Energy Problems on Rectifiable Sets - Assignment Example

Since f is differentiable at these two values must be equal, so that f’ (a) = 0. On should  take care here with the logic of the proposition! It does not claim that each point at which f’ (a) = 0 will automatically be extreme: this would be contradicted, for example, by f (x) = x3, which has f’ (x) = 3x2, which is 0 at x = 0, although 0 is not an extreme point for f, since f(x) < 0 whenever x < 0, and f (x) > 0 whenever x > 0. Nonetheless, proposition 1 is very useful in ruling out points from being extreme: if f is differentiable at a and f’ (a) ≠ 0 then a cannot be an extreme point for f.

So the extreme points come from two classes: they will either be among the stationary points or they will be points at which f fails to be differentiable. Points that fall into one of these two classes will be called critical points. The following very simple example contains both classes: let f(x) = (|x|-1)2, which has the domain, and is differentiable except at x =0. For x ≠0, x|x| has derivative sgn(x) = so by the chain rule we obtain f’(x) = 2 (|x|-1), which is 0 only if x = 1 or -1.

Hence the three possible extreme points are -1, 0, and 1. But f (x) ≥ 0 for all x, and f (-1) = f(1) =0, so both of these are minimum points, while there is a strict local maximum at x =0 since f(x) < 1 = f (0) for x  N (0,2),. This maximum is not global, since f(3) > 1. In this case, one is able to deduce what happens ‘near’ the critical points simply by inspection. But we can use the sign off’ to gain information about the behavior off in a neighborhood of a: for example, if f’(a) > 0, then the continuity of the chord-slope function F at a tells us that the ratios must be positive for x  N(a,δ), provided δ > 0 is sufficiently small.

Hence we must have f(x) < f(a) for a – δ < x < a and f(x) > f (a) for a < x < a + δ. Thus f is strictly increasing throughout some neighborhood N(a, δ) if f’ (a) > 0. Our next task will be to extend this idea when f’ > 0 throughout any given set, i.e. we want to examine what ‘global’ deductions we can make from the properties of the derivative.

As with continuous functions, we need the set in question to be an interval for these results. Let sn =  for each n  . Then 0 ≤ sn ≤tn for each n, hence if (tn) is bounded above, so is (sn). Hence sn→supn tn. On the other hand, if  diverges, (sn) is unbounded above, hence so is (tn), and thus  diverges. (i) 0 <  ≤  for all n > 1. Now the terms on the right have sum  =  = 1, by further exercise 4 (i) on chapter 2. /this implies that  converges, by comparison, and that  ≤ 1, i.e. 1+  + + + … =  ≤ 2.

Note that in this example we compared series term by term form n=2, rather than from n = 1. It should be clear by now that the Comparison Test applies equally to series with an ≤ bn for n ≥ n0 for some n0 – though we need to remember that this will only yield  ≤  and that the first n0 terms have to be handled separately in that case. (ii)  has  ≤  = , so the series will converge by comparison with . On the other hand,  will diverge, since  ≥  for n > 10 we have n2 – 5n = n(n-5) > ½ n2 if n> 10, and the series dominates the harmonic series term by term. (iii) n satisfies  ≤  for all n ≥ 3, and therefore n ≤ (⅚)n for all such n.

Hence this series converges by comparison with the geometric series for x = ⅚, and since n =  = , it follows that n ≤ (1 + ½ ) + ( ½ + ½ )2 +  =  < 6. The next three tests are simple consequences of the Comparison Test, and are proved by describing conditions under which we can compare the series to a geometric one. We shall keep to series with positive terms for the time being. MAIN RESULTS A) The evaluation of  causes a new problem; the integrand f (t) = (t-1) -1/3 is undefined at t = 1.

We can hope to overcome this by first finding the sum of the intergrals  and  and then hoping for a limit as   0. In this case the ploy works, and the limit turns out to be (t) dt =  (9. However, this technique also fails frequently, since we have no guarantee that the limit will exist: for example, -2 dt diverges. Not that a careless ‘application’ of the Fundamental Theorem would give the value  (!). If f ≥ 0 is monotone decreasing, then the sequence (an) defined by an = (k) - (t) dt is non-negative, monotone increasing and convergent, and a = limn→ ∞ an satisfies 0 ≤ a ≤ f(1).

Therefore the improper integral  (t) dt converges if and only if the series (k) converges. We can also use the Intergal Test to distinguish between the series  and α for α > 1: we compare the behaviour of the first with that of the integral  , which diverges, since  =  = log (logb) – log(log2) On the other hand, the integral  converges, since  =  =  – (log 2) –α+1} →  as b → ∞. Hence  diverges, while  converges for every α > 1. Just as for series, we define the improper integral  to be absolutely convergent if  converges.

The integral  is conditionally convergent if it converges, but does not converge absolutely. We leave it for the exercise below to show that absolutely convergent integrals converge. Examples of such integrals include Iα =  dx for α > 1. This is most easily seen by comparing J α =  dx with the convergent integral  since |sinx|≤1. However, this does not extend to the case 01,  dx = dx ≥ dx ≥  Since |sin x|≥  and xα ≤ (k + )α on each subinterval.

But since the series  diverges when 0 < α ≤ 1, so does the integral, so Jα is divergent. To see that Iα converges for 0 < α ≤ 1, we need only integrate by parts to obtain:  dx =  -   dx and the last integral converges absolutely, by comparison with  for β = α + 1 > 1. Hence I α converges conditionally. B)Given a series, with fk   for all k0, define the function f: [0, ∞)  by setting f(t) = fk on the interval (k-l,k) for each k  1.