Lecture 25: Power Series and the Weierstrass Approximation Theorem

18.100A: Complete Lecture Notes Lecture 25: Power Series and the Weierstrass Approximation Theorem Last time, we asked three questions about interchanging limits: Question 1. Hence, we ask three questions about interchanging limits: 1. If fn : S → R, fn continuous and fn → f pointwise or uniform, then is f continuous? 2. If fn : [a, b] → R, fn differentiable, and fn → f with fn0 → g, then is f differentiable and does g = f 0 ? 3. If fn : [a, b] → R, with fn and f continuous such that fn → f , then does Z b Z fn = a b f? a The answer to the above questions are all no, if the convergence is pointwise as seen by the following counterexamples:  0 x ∈ [0, 1) 1. Let fn (x) = xn on [0, 1] is continuous ∀n. As we noted earlier, fn (x) → f (x) = . Notice that 1 x = 1 f is not continuous. 2. Let fn (x) = xn+1 n+1 on [0, 1]. Then, fn → 0 pointwise on [0, 1]. However,  0 x ∈ [0, 1) fn0 (x) → g(x) = . 1 x = 1 Thus, g(x) 6= (0)0 = 0 at x = 1. 3. Consider the functions   1   4n2 x x ∈ 0, 2n    1 1 fn (x) = 4n − 4n2 x x ∈ 2n ,n      0 x ∈ n1 , 1 as described in the previous lecture. Then, fn (x) → 0 pointwise on [0,1] as we showed last time. However, Z 0 1 1 1 fn = (base)(height) = · 2n = 1 6→ 0 = 2 2n Z 1 0. 0 We now prove that the answer to the three questions above is yes if convergence is uniform. Theorem 2 If fn : S → R is continuous for all n, f : S → R, and fn → f uniformly, then f is continuous. Proof : Let c ∈ S and let  > 0. Since fn → f uniformly, ∃M ∈ N such that ∀n ≥ M , ∀y ∈ S, |fn (y) − f (y)| < 1  . 3 Since fM : S → R is continuous, ∃δ0 > 0 such that ∀|x − c| < δ0 , |fM (x) − fM (c)| <  . 3 Choose δ = δ0 . If |x − c| < δ, then |f (x) − f (c)| ≤ |f (x) + fM (x)| + |fM (c) − f (c)| + |fM (x) − fM (c)|    < + + = . 3 3 3 Theorem 3 If fn : [a, b] → R is continuous for all n, f : [a, b] → R and fn → f uniformly, then Z b Z b fn → a f. a Proof : Let  > 0. Since fn → f uniform, ∃M0 ∈ N such that ∀n ≥ M0 , ∀x ∈ [a, b],  . b−a |fn (x) − f (x)| < Then, for all n ≥ M = M0 , we have Z b b Z fn − a b Z f ≤ a b  = . b−a Z b Z |fn − f | < a a Remark 4. Notationally, this states that Z lim n→∞ b b Z fn = a lim fn = a n→∞ f. a Theorem 5 If fn : [a, b] → R is continuously differentiable, f : [a, b] → R, g : [a, b] → R, and fn → f pointwise, fn0 → g uniformly, then f is continuously differentiable and g = f 0 . Proof : By the FTC, ∀n∀x ∈ [a, b], Z fn (x) − fn (a) = x fn0 . a Thus, by the previous two theorems, f (x) − f (a) = lim (fn (x) − fn (a)) n→∞ Z x = lim fn0 n→∞ a Z x = g. a 2 Rx Rx Therefore, f (x) = f (a) + a g. Thus, by the FTC, f is differentiable and f 0 = ( a g)0 = g. We now return back to our study of power series, answering some questions we asked at the beginning of Lecture 23. Theorem 6 P∞ P∞ Let j=0 aj (x − x0 )j be a power series of radius of convergence p ∈ (0, ∞]. Then, ∀r ∈ (0, p), j=0 aj (x − x0 )j converges uniformly on [x0 − r, x0 + r]. Proof : Let r ∈ [0, p). Then, ∀j ∈ N ∪ {0}, ∀x ∈ [x0 − r, x0 + r], |aj (x − x0 )j | ≤ |aj |rj =: Mj . Now, 1/j lim j → ∞Mj = lim |aj |1/j r = j→∞  r p p<∞ 0 p=∞ since p−1 = limj→∞ |aj |1/j . Since r < p, we have 1/j lim Mj j→∞ By the Weierstrass M-test, it follows that ∞ X < 1 =⇒ Mj converges. j=0 P∞ j=0 aj (x − x0 )j converges uniformly on [x0 − r, x0 + r]. Theorem 7 P∞ Let j=0 aj (x − x0 )j be a power series with radius of convergence p ∈ (0, ∞]. Then, P∞ 1. ∀c ∈ (x0 − p, x0 + p), j=0 aj (x − x0 )j is differentiable at c and ∞ ∞ X d X jaj (x − x0 )j−1 . aj (x − x0 )j = dx j=0 j=0 2. ∀a, b such that x0 − p < a < b < x0 + p, Z ∞ bX j aJ (x − x0 ) dx = a j=0 ∞ X  aj j=0 (b − x0 )j+1 (a − x0 )j+1 − j+1 j+1  . Remark 8. Since lim ((j + 1)|aj+1 |)1/j = lim j→∞ 1. implies P  j→∞ ((j + 1)|aj+1 |1/(j+1) (j+1)/j = lim |ak |1/k = p, k→∞ aj (x − x0 )j is infinitely differentiable and  k!ak =  dk X j aj (x − x0 ) dxk x=x0 . Weierstrass Approximation Theorem Remark 9. This theorem essentially states: "Every continuous function on [a, b] is almost a polynomial." 3 Theorem 10 (Weierstrass Approximation Theorem) If f ∈ C([a, b]), there exists a sequence of polynomials {Pn } such that Pn → f uniformly on [a, b]. The idea of the proof is to choose a suitable sequence of polynomials {Qn }n such that Qn behaves like a ‘Dirac R1 delta function’ as nto∞. Then, the sequence of polynomials Pn (x) = 0 Qn (x − t)f (t) dt converges to f (x) as n → ∞. We will prove this momentarily, but first we need to do the ground work. Notice that we only need to consider a = 0 and b = 1, with f (0) = f (1) = 0. If we prove this case, then for a general f˜ ∈ C([0, 1]), ∃ a sequence of polynomials Pn (x) → f˜(x) − f˜(0) − x(f˜(1) − f˜(0)) uniformly. Hence, P̃n (x) = Pn (x) + f˜(0) + x(f˜(1) − f˜(0)) → f˜(x) uniformly. Theorem 11 R1 Let cn := ( −1 (1 − x2 )n dx)−1 > 0, and let Qn (x) = cn (1 − x2 )n . Then, 1. ∀n, R1 −1 Qn = 1. 2. ∀n, Qn (x) ≥ 0 on [−1, 1], and 3. ∀δ ∈ (0, 1), Qn → 0 uniformly on δ ≤ |x| ≤ 1. Before we prove this, here is a picture of Qn : Proof : 2. Immediately clear. R1 R1 1. −1 Qn = cn −1 (1 − x2 )n dx = 1 by definition of cn . 3. We first estimate cn . We have for all n ∈ N and ∀x ∈ [−1, 1], (1 − x2 )n ≥ 1 − nx2 . We proved this way earlier in the course by induction, but it also follows from the calculus we have proven as g(x) = (1 − x2 )n − (1 − nx2 ) 4 satisfies g(0) = 0, and g 0 (x) = n · 2x(1 − (1 − x2 )n−1 ) ≥ 0 in [0,1]. Thus, g(x) ≥ 0 by the MVT. Then, 1 = cn Z 1 (1 − x2 )n dx −1 1 Z (1 − x2 )n dx =2 0 Z √ 1/ n (1 − x2 )n dx >2 0 Z √ 1/ n (1 − nx2 ) dx 0  1 n = 2 √ − · n−3/2 3 n √ 4√ n > n. = 3 ≥2 Therefore, cn < √ n. Let δ > 0. We note limn→∞ √ n(1 − δ 2 )n = 0. Then, √ lim ( n(1 − δ 2 )n )1/n = lim (n1/n )1/2 (1 − δ 2 ) n→∞ n→∞ = 1 − δ 2 < 1. Therefore, lim n→∞ √ n(1 − δ 2 )n = 0. Let  > 0, and choose M ∈ N such that for all n ≥ M , √ n(1 − δ 2 )n < . Then, ∀n ≥ M and ∀δ ≤ |x| ≤ 1, |cn (1 − x2 )n | < √ √ n(1 − x2 )n ≤ n(1 − δ 2 )n < . We now prove the Weierstrass Approximation Theorem. Proof : Suppose f ∈ C([0, 1]), f (0) = f (1) = 0. We extend f to an element of C(R) by setting f (x) = 0 for all x∈ / [0, 1]. We furthermore define Z 1 f (t)Qn (t − x) dt Pn (x) = 0 Z = 1 f (t)cn (1 − (t − x)2 )n dt. 0 Note that Pn (x) is in fact a polynomial. 5 Furthermore, observe that for x ∈ [0, 1], Z 1 f (t)Qn (t − x) dt Pn (x) = 0 Z 1−x = f (x + t)Qn (t) dt −x Z 1 = f (x + t)Qn (t) dt. −1 The second equality is true by a change of variable, and the last equality is true as f (x + t) = 0 for t ∈ / [−x, 1 − x]. We now prove Pn → f uniformly on [0, 1]. Let  > 0. Since f is uniformly continuous on [0, 1], ∃δ > 0 such that ∀|x − y| ≤ δ, |f (x) − f (y)| < 2 . Let C = sup{f (x) | x ∈ [0, 1]}, which exists by the Min/Max theorem i.e. the EVT. Choose M ∈ N such that ∀n ≥ M , √  n(1 − δ 2 )n < . 8C Thus, ∀n ≥ M, ∀x ∈ [0, 1], by the previous theorem, Z 1 (f (x − t) − f (t))Qn (t) dt |Pn (x) − f (x)| = Z −1 1 ≤ |f (x − t) − f (x)|Qn (t) dt Z−1 ≤ |t|≤δ Z |f (x − t) − f (x)|Qn (t) dt + Z √ Qn (t) dt + n(1 − δ 2 )n Z  2 |t|≤δ √  < + 4C n(1 − δ 2 )n 2   < + = . 2 2 ≤ |f (x − t) − f (x)|Qn (t) dt δ≤|t|≤1 2C kδ|≤|t|≤1 In the last minute of the course, Casey Rodriguez stated: "This was quite an experience; teaching to an empty room. I hope you did get something out of this class. Unfortunately I wasn’t able to meet a lot of you, and that’s one of the best parts of teaching...." 6