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454
Homework Math #12 Solutions
Spring, 2009
Exercise 6.5.1. Consider the function g defined by the power series x 2 x3 x 4 x5 g ( x) = x - + - + - 2 3 4 5 (a) Is g defined on (-1,1)? Is it continuous on this set? Is g defined on (-1,1]? Is it continuous on this set? What happens on [-1,1]? Can the power series for g(x) possibly converge for any other points |x| > 1? Explain. Solution: Maybe the easiest test to run is the ratio test. If we let an = xn/n! and examine the limit of |an+1/an|, we get |x|. So there is absolute convergence on (-1,1). At x=1 we have the alternating harmonic series, known to converge, while at x = -1 we have the harmonic series, so divergence. Thus the series for g(x) converges on (-1,1] and diverges elsewhere. (b) For what values of x is g'(x) defined? Find a formula for g'. Solution: By Thm 6.5.7, on (-1,1) we have g ' ( x) = 1 - x + x 2 - x 3 + x 4 - via termwise differentiation. Exercise 6.5.2. Find suitable coefficients (an) so that the resulting power series a n x (a) converges absolutely for all x [-1,1] and diverges off of this set. Solution: an = 1/np, where p>1 (b) converges conditionally at x = -1 and diverges at x = 1. Solution: an = 1/n. (c) converges conditionally at both x = -1 and x = 1. (-1) n 2 n Solution: I am stumped, at least for now. The series x does converge conditionally n as required, though it does not have the form requested. I'll look some more. (d) Is it possible to find an example of a power series that converges conditionally at x = -1 and converges absolutely at x = 1? n n Solution: No, if it converges absolutely at x=1, then | a n 1 | , converges, so a n | (-1) | =
n
| a 1
n
n
| converges, that is, the series converges absolutely (not conditionally) at -1.
Exercise 6.5.3. Explain why a power series can converge conditionally for at most two points. Solution: Conditional convergence occurs only at the "endpoints" of an interval of convergence there are at most two of them. Exercise 6.5.5. Use the Weierstrass M-Test to prove Theorem 6.5.2 (if a power series converges absolutely at a point xo, then it converges uniformly on [-c,c], where c = | xo|.) Solution: For Weierstrass we need a sequence of constants, commonly denoted by Mn which "dominate" the power series and for which M n converges. Toward this end we let Mn = |anxon|. Clearly (?), for any x in the interval [-c,c], we have | a n x n || a n xo |= M n . By hypothesis,
n
M
n
converges, and so we have the result.
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Math 454
Homework #12 Solutions
Spring, 2009
Exercise 6.5.7. (a) The Ratio Test states that if (bn) is a sequence of nonzero terms satisfying lim|bn+1/bn| = r < 1, then the series bn converges. Use this to argue that if s satisfies 0 < s < 1, then n sn-1 is bounded for all n 1. Proof: We apply (as suggested) the ratio test to the sequence an = n sn-1, and get a ns n + s n 1 lim n +1 = lim = lim s + s = s < 1 . By the ratio test, the series a n converges. n -1 n a n n n ns n Therefore the sequence (an) converges to zero, so is bounded, which was to be shown. (b) Given an arbitrary x (- R, R ) , pick t to satisfy |x| < t < R. Use the observation 1 x n -1 | na n x n -1 |= n n -1 | a n t n | to construct a proof of Theorem 6.5.6. (The t t differentiated series of a convergent series on (-R,R) also converges.) Proof: (Once we make the requested observation, which is clear). Since |x/t| < 1, part (a) above n -1 n -1 x x insures that n is bounded for all natural numbers n, say n B . Then, t t
n -1 1 x | a n t n | B | a n t n | . The series on the right converges | na n x n -1 | = n t t n =1 n =1 n =1 t since t (- R, R ) . We have that the differentiated series converges absolutely and therefore it converges, which was to be shown.
Exercise 6.6.1. Assuming that arctan(x) is continuous, explain why the value of the series at x = 1 must necessarily be arctan(1). What interesting identity do we get in this case? Solution: By Abel's Theorem we have uniform convergence on [0,1], so the series represents a 1 1 1 continuous function. We set x=1 and get = 1 - + - + - , widely known as 4 3 5 7 "Leibnitz's formula". 1 = 1 + t + t 2 + t 3 + of this section, find a 1- t power series representation for ln(1+x). For what values of x is this expression valid? Solution: Replace the `t' by `-t', then integrate term-by-term to get x 1 x 2 x3 x4 x5 ln(1 + x) = dt = x - + - + - + Convergence on (-1,1]. 0 1+ t 2 3 4 5 Exercise 6.6.2. Starting with the identity (1) Exercise 6.6.3. Exercise 6.6.4.
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Math 454
Homework #12 Solutions
Spring, 2009
Exercise 6.6.5. Prove that the partial sums SN(x) converge uniformly to sin(x) on [-2,2]. Generalize this proof to show that the convergence is uniform on any interval of the form [-R,R]. Proof: Notice first that all derivatives of sin x are bounded by 1. Now we reference Lagrange's f N +1 (c) N +1 1 2 N +1 x | x N +1 | remainder theorem: | E N ( x) |= . Since the ( N + 1)! ( N + 1)! ( N + 1)! factorials the in denominator "grow faster" than the exponential expression in the numerator, we see that the magnitude of EN(x) can be made uniformly smaller than on [-2,2]. This is what we were to prove. Exercise 6.6.6. (a) Generate the Taylor coefficients for the exponential function f(x) = ex, and then prove that the corresponding Taylor series converges uniformly to ex on any interval of the form [-R,R]. Proof: Notice first that all derivatives of ex are again ex, so the Taylor coefficients are all given xn x 0 by an = e /n! = 1/n!. This gives the series e = . On the interval [-R,R], we n = 0 n! ec eR N +1 x R N +1 . This error bound examine the remainder sum, | E N ( x) |= ( N + 1)! ( N + 1)! once again (as in Exercise 6.6.5) converges to zero, so the remaider term EN(x) converges uniformly to zero, and the uniform convergence of the Taylor series on [-R,R] is assured. (b) Verify the formula f'(x) = ex. Solution: (Merely differentiate the original series term-byterm, and it falls out.) (c) Use a substitution to generate the series for e-x, and then calculate ex . e-x by multiplying together the two series and collecting common powers of x. Solution: (This is pretty much manipulation of very long polynomials. We write the series for the two exponential functions, then begin to multiply. For example, the coefficient of x7 is 1 1 1 1 1 1 1 1 - + - + - + - , which we notice lacks a factor of 7! from 7! 6!1! 5!2! 4!3! 3!4! 2!5! 1!6! 7! 7 being the binomial coefficients . Indeed a closer examination shows that the k 1 n coefficient for xn is given as (1 + (-1)) = 0n for n 1. Thus the coefficients are zero n! except for the constant term 1. So we get the well-known fact that ex . e-x = 1. Exercise 7.2.1. Let f be a bounded function on [a,b], and let P be an arbitrary partition of [a,b]. First, explain why U(f) L(f,P). Proof: Fixing partition P, we know that for any partition P*, L(f,P) U(f,P*) lemma 7.2.4. Since L(f,P) is a lower bound for all upper sums it is no larger than the infimum of them; i.e., L(f,P) U(f).
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Math 454
Homework #12 Solutions
Spring, 2009
Now, prove Lemma 7.2.6 (For any bounded function f on [a,b], it is always the case that U(f) L(f). Proof: Since the partition P above is arbitrary, it follows that the supremum of all such lower sums is also less than U(f). Exercise 7.2.2. Consider f(x) = 2x + 1 over the interval [1,3]. Let P be the partition consisting of the points {1,3/2,2,3}. (a) Compute L(f,P) = 17/2, U(f,P) = 23/2, and U(f,P) L(f,P) = 3. (b) What happens to the value of U(f,P) L(f,P)= 2 when we add the point 5/2 to the partition? (c) Find a Partition P' of [1,3] for which U(f,P) L(f,P)<2. Solution: Refine with any other intermediate point Exercise 7.2.3. Show directly (without appealing to Theorem 7.2) that the constant function f(x) = k is integrable over the over any closed interval [a,b]. What is Solution: For any partition P, the lower and upper sums are k(b-a). That common value ensures that f is integrable and the integral has value k(b-a). Exercise 7.2.5. Given a sequence (fn) of integrable functions that converge uniformly on [a,b] to a function f. Prove that f is integrable on [a,b]. Proof: The approach is to show that for every > 0 there is a partition P such that U(f,P) - L(f,P) < . And the angle is to do some add-subtract and use the triangle inequality: U(f,P) - L(f,P) = [U(f,P) - U(fN,P)] + [U(fN,P) - L(fN,P)] + [L(fN,P) - L(f,P)] |U(f,P) - U(fN,P)| + [U(fN,P) - L(fN,P)] + |L(fN,P) - L(f,P)| Let > 0 be given. By uniform convergence we can choose N so that | f N ( x) - f ( x ) |
b
a
f ?
3(b - a) for all x in [a,b]. (This will be helping to insure that the first and third differences above will be smaller than /3). Now for this N, since fN is integrable, there is a partition P for which U ( f N , P ) - L( f N , P ) < .(This will handle the second term). In the end, the 3 difference U(f,P) L(f,P) < /3 + /3 + /3.
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Math 454
Homework #12 Solutions
Spring, 2009
Exercise 7.2.6. Let f:[a,b] R be increasing on the set [a,b] (i.e., f(x) f(y) whenever x < y). Show that f is integrable on [a,b]. Proof: (We appeal to the characterization of integrability in Theorem 7.2.8 for ever >0 there is a partition P for which the upper sum and lower sum differ by less than .) The key to this argument is that on any subinterval the infimum mk and supremum Mk occur at the left and right endpoints, respectively. Let > 0 be given and choose a partition P that are equally spaced so that x = xk xk-1< /(f(b)-f(a)) for all k (to do this we choose the number n of equally spaced points appropriately). Then, U ( f , P ) - L( f , P) = (M1 m1) x + (M2 m2) x + (M3 m3) x + ... + (Mn mn) x = {[f(x1) f(x0)] + [f(x2) f(x1)] + [f(x3) f(x2)] + ... [f(xn) f(xn-1)] } x. We notice the "telescoping sum" and observe that x0 = a while xn = b. We have U ( f , P ) - L( f , P) = [f(b) f(a)] x < [f(b) f(a)] { /(f(b)-f(a))} = . The result follows by Theorem 7.2.8.
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