Math 131A: Homework 6There is no quiz for this homework but it will be examined on the final.1. SupposeP∞n=0anxnis a power series and that (|an|1n)∞n=1converges to a nonzero real number.Let β = lim |an|1nand R =1β. Prove that the power series converges for |x| < R, and divergesfor |x| > R.Help! Fix an x ∈ R and apply the root test to the seriesP∞n=0anxn.2. Let f : R \ {1} −→ R be defined by f(x) =11−x.(a) i. Calculate the Taylor series for f (x) at 0.ii. What is the radius of convergence R0of this Taylor series?iii. Does the Taylor series converge to f on (−R0, R0)?(b) i. Use part (a), the substitution x = 2y − 1, and the identity11 − y= 2 ·11 − (2y − 1)to calculate a power series for f centered about12.ii. What is the radius of convergence of R1/2of this power series?iii. Does the power series converge to f on (12− R1/2,12+ R1/2)?iv. You just calculated a Taylor series without calculating any derivatives!Use part i. to read off the values of f(n)(12) for n ∈ N ∪ {0}.3. Prove that there is an infinitely differentiable function g : R −→ R such that for all n ∈ N∪{0},g(n)(0) = n2.Help! Try to define g using a power series centered at 0.TURN OVER!14. (a) Use Taylor’s theorem for f (x) = sin x, n = 3, c = 0 and x = 1 to deduce that sin 1 > 0.(b) Use Taylor’s theorem for f (x) = sin x, n = 9, c = 0 and x = 4 to deduce that sin 4 < 0.(c) Use the intermediate value theorem to deduce that there is an x0∈ (1, 4) with sin x0= 0.You have just proved that π exists.5. Consider the inequality2(exp(x) − (1 + x))x2< exp(x) for x > 0.(a) Prove it using the definition of exp(x).(b) Prove it using Taylor’s theorem, with n = 2, for exp about 0, and the fact that exp isstrictly increasing.6. (a) Let f (x) =P∞n=1(−1)n+1x2n−2(2n+1)!. What is f (0)? What is x − x3f(x)?(b) Let g(x) =P∞n=1(−1)n+1x2n−2(2n)!. What is g(0)? What is 1 − x2g(x)?(c) Calculate limx−→0(x−sin x)701−cos(x105). Help! Ponder(x3f(x))70(x105)2g(x105)?7. Prove that there is an infinitely differentiable function h : R −→ R such that for all n ∈ N∪{0},h(n)(0) = n2, and such that h 6= g, where g is the function you constructed in question 3.Help! What should be true for h − g? Recall the “rubbish” Taylor series.THE LAST TWO ARE FOR THE KEEN!8. Calculate f0(0), f00(0) and f000(0) wheref(x) = (sin x) · (cos x)1808· exp(x601) · (1 + 3x3− 5x4+ 2754x232).Help! If you can figure out a1, a2, a3in a power series expansionP∞n=0anxnfor f (x) thenyou are basically done: f0(0), f00(0) and f000(0) are given by a1, 2a2and 6a3. How can youdo this? Multiply like polynomials; there are a lot of terms you can ignore.9. Prove that there is an infinitely differentiable function f : R −→ R such that f(0) = f0(0) = 0,f00(0) =196and x2f00(x) + xf0(x) + (x2− 4)f(x) = 0 for all x ∈ R.Help! Try to write f as a power series centered at 0.The function you have constructed is called the second Bessel function.2Math 131A: Homework 6 Solutions1. Fix x ∈ R. Then α = limn−→∞anxn1n= |x| · limn−→∞|an|1n= |x| · β =|x|R. Thus, the roottest tells us thatP∞n=0anxnconverges when|x|R< 1 and diverges when|x|R> 1.2. (a) i.P∞n=0xn.ii. 1.iii. Yes, because we know about geometric series.(b) i.11 − y= 2 ·11 − (2y − 1)= 2∞Xn=0(2y − 1)n=∞Xn=02n+1y −12n.ii.12.iii. Yes.iv. f(n)(12) = n! · 2n+1.3. Let g(x) =P∞n=0n2n!xn. This has infinite radius of convergence since β = 0. It is infinitelydifferentiable with the correct derivatives at 0 (by the off-syllabus theorem).4. (a) Taylor’s theorem tells us that there exists a y ∈ (0, 1) withsin 1 = 1 −cos(y)3!.Since − cos(y) ≥ −1, sin 1 ≥56> 0.(b) Taylor’s theorem tells us that there exists a y ∈ (0, 4) withsin 4 = 4 −433!+455!−477!+49cos(y)9!Since cos(y) ≤ 1,sin 4 ≤ 4 −433!+455!−477!+499!< 0.(c) By the intermediate value theorem we deduce that there is an x0∈ (1, 4) with sin x0= 0.5. (a) We note that2(exp(x) − (1 + x))x2=2x2∞Xn=2xnn!=∞Xn=22xn−2n!=∞Xn=02xn(n + 2)!.The result follows from the fact that, for all n ∈ N ∪ {0},2(n + 2)!≤1n!and the fact that this inequality is strict when n ≥ 1.3(b) Let x ∈ (0, ∞). Taylor’s theorem says that there is a y ∈ (0, x) such thatexp(x) = 1 + x +exp(y)2x2.Thus,2(exp(x)−(1+x))x2= exp(y) < exp(x), since exp is strictly increasing.6. (a) f (0) =16. x − x3f(x) = sin x.(b) g(0) =12. 1 − x2g(x) = cos x.(c) The off syllabus theorem says that f and g are continuous, so h(x) =f(x)70g(x105)is continuous.Thus,limx−→0(x − sin x)701 − cos(x105)= limx−→0(x3f(x))70(x105)2g(x105)= limx−→0f(x)70g(x105)=f(0)70g(0)=2670.7. Let h = f + g, where f is the function with the “rubbish” Taylor series, and g is the functionof question 3. For all n ∈ N ∪ {0}, we haveh(n)(0) = f(n)(0) + g(n)(0) = 0 + n2= n2.Since f is not the zero function, h 6= g.8. We consider the Taylor series of f(x) but ignore terms xnwith n ≥ 4:x1 −x23!1 −x22!1808. . . x1 −x23!1 −1808x22!We have f (x) = x − (13!+18082!)x3+P∞n=4anxnso that f0(0) = 1, f00(0) = 0 andf000(0) = −3! ·13!+18082!= −(1 + 3 · 1808) = −5425.9. Let f : R −→ R be defined byf(x) =∞Xn=0(−1)nn!(n + 2)!x22n+2.4A note on question 8Before being given a theorem that lets you know everything is okay, you should be concerned aboutmultiplying power series like polynomials. Here’s the theorem.Theorem. Suppose that f (x) =P∞n=0anxnand g(x) =P∞n=0bnxnhave a positive radii of con-vergences, Rf, Rg, respectively. For n ∈ N letcn=nXi=0aibn−iand set R = min{Rf, Rg}.The power series h(x) =P∞n=0cnxnhas radius of convergence bigger than or equal to R andfor x ∈ (−R, R) we haveh(x) = f(x)g(x).A very careful answer to 8 (definitely overkill) would make use of the following corollary.Corollary. Suppose f (x) =P∞n=0anxnand g(x) =P∞n=0bnxnhave infinite radii of convergence,that a0= b0= 1. Thenh(x) = f(x)g(x)has a power series expansionP∞n=0cnxnwith c0= 1 and infinite radius of convergence.You could then argue the existence of power series (the first four can be written down explicitly)g(x) =∞Xn=0pnxn, h(x) =∞Xn=0qnxn, j(x) =∞Xn=0rnxn, k(x) =∞Xn=0snxn, m(x) =∞Xn=0tnxnwith p0= q0= r0= s0= t0= 1, such thatsin x = x1 −x26· g(x)cos x = 1 −x22· h(x)exp(x601) = 1 + x601· j(x)1 + 3x3− 5x4+ 2754x232= 1 + 3x3· k(x)andf(x) = x1 −x26· g(x)·1 −x22·