Functions based on sin ( π. and cos

Transcription

1 Functions based on sin and cos. Introduction In Complex Analysis if a function is differentiable it has derivatives of all orders. In Real Analysis the situation is very different. Using sin (π/ and cos (π/ it is possible, for any m 1, to construct a function, f, which has m derivatives on R but for which f (m+1 ( does not exist for all x R. Though f (m ( will not be differentiable for all x R we can construct it so that it is continuous on all of R, or even not continuous on all of R. The functions. Define, for n Z, { x n sin if x 0 f n ( = 0 if x = 0. and g n ( = { x n cos if x 0 0 if x = 0. There functions have derivatives of all orders on R \ {0}. The interest comes when x = 0. Continuity on R. If n 1 then lim x 0 f n ( and lim x 0 f n ( do not exist. In fact, write n = m with m 1, then with k Z, ( 1 f n = (k m sin (kπ = 0, k ( ( m (( f n = sin k + 1 ( m π = ( 1 g n k, since m 1, (1 = (k m cos (kπ = (k m ( f n k, since m 1, ( ( m (( = cos k + 1 π = 0. 1

2 Lemma If n 1 then lim x 0 f n ( does not exist. Proof Assume that lim x 0 f n ( exists, called l say. Choose ε = 1 in the definition of limit to find δ > 0 such that 0 < x 0 < δ f n ( l < 1. But then, for these x we have f n ( = f n ( l + l f n ( l + l 1 + l. (3 Choose k Z sufficiently large such that / (4k + < δ and ( / > 1 + l. Then we get a contradiction since, by (3, 1/k < δ implies f n (1/k 1 + l which by (1 implies ( / 1 + l. So we have both ( / > 1 + l and ( / 1 + l, a contradiction. Hence the last assumption is false and thus lim x 0 f 1 ( does not exist. Similarly you can show that if n 1 then lim x 0 g n ( does not exist. If n = 0 then for x 0, f 0 ( = sin (π/ which has no limit as x 0, as seen in the notes. Similarly g 0 ( = cos (π/ has no limit as x 0. If n 1 then f n ( and g n ( are continuous only on R \ {0}. f 0 ( and g 0 ( are continuous only on R \ {0}. If n 1 then we can combine 1 sin 1 and x n x n x n to get x n x n sin x n. Then as x 0 the Sandwich Rule shows that lim x 0 f n ( = 0. Similarly for g n, hence For n 1, the f n ( and g n ( are continuous on R.

3 Differentiability. First derivatives. If n 0 the functions f n and g n are not continuous at x = 0 and are thus differentiable only on R \ {0}. If n 1 then, for x = 0, we look at the definition of derivative as a limit f n ( f n (0 lim x 0 x 0 = lim x n 1 sin = lim f n 1 (. (4 x 0 x x 0 As we ve seen above, this limit exists only when n 1 1, i.e. n, when the value is 0. Similarly for g n, hence If n = 1 then f 1 (0 and g 1 (0 are not defined and so f 1 and g 1 are differentiable only on R \ {0}. If n, then (4 and the similar result for g n give f n (0 = 0 and g n (0 = 0. Thus f n and g n are differentiable at x = 0 and thus on R. Continuity of the First derivatives. If n 1 then f n (0 and g n (0 do not exist so f n and g n are continuous only on R \ {0}. If x 0 then, from the definition we have, for all n 1, f n ( = nx n 1 sin πx n cos. (5 which is continuous on R \ {0}. Similarly for g n. With n = 1, (5becomes f 1 ( = sin πx 1 cos Lemma lim x 0 f 1 ( and lim x 0 g 1 ( do not exist. Proof We only show this for f 1 and leave g 1 to the Student. Note that for any k Z we have ( 1 f 1 = sin (kπ kπ cos (kπ = kπ. (6 k 3

4 Assume that lim x 0 f 1 ( exists, called l say. Choose ε = 1 in the definition of limit to find δ > 0 such that 0 < x 0 < δ f ( l < 1. But then, for these x we have f ( = f ( l + l f ( l + l 1 + l. (7 Choose k Z sufficiently large such that 1/k < δ and kπ > 1+ l. Then we get a contradiction since, by (7, 1/k < δ implies f (1/k 1 + l which by (6 implies kπ 1 + l. So we have both kπ > 1 + l and kπ 1 + l, a contradiction. Hence the last assumption is false and thus lim x 0 f 1 ( does not exist. If n = then (5 becomes f ( = x sin π cos. Lemma lim x 0 f ( and lim x 0 g ( do not exist. Proof We only show this for f and leave g to the Student. Note that for any k Z we have ( 1 f f k ( 1 k + 1 = sin (kπ π cos (kπ = π, (8 k = sin ((k + 1 π π cos ((k + 1 π = π. (9 k + 1 Assume that lim x 0 f ( exists, called l say. Choose ε = π/ in the definition of limit to find δ > 0 such that 0 < x 0 < δ f ( l < π/. (10 Choose k Z sufficiently large such that 1/k < δ for then (10 implies f (1/k l < π/, i.e. π l < π/ by (8. Next choose k Z sufficiently large such that 1/ (k + 1 < δ for then (10 implies f (1/ (k + 1 l < π/, i.e. by (8. π l < π/ 4

5 So l is both close to π and π! We see this contradiction more clearly by using the triangle inequality to say π = π = π l + π + l π l + π + l < π/ + π/ = π. This contradiction means that the last assumption is false and thus lim x 0 f ( does not exist. If n 3 then the Sandwich Rule gives both of lim x 0 nxn 1 sin = 0 and lim πx n cos x x 0 x Thus lim x 0 f n ( = 0 = f n (0. All the above also hold for g n, hence If n = then f and g are continuous only on R \ {0}. If n 3, then f n and g n are continuous on R. We can sum up the above as f n and g n are continuous on R iff n 1, f n and g n are differentiable on R iff n, f n and g n are continuous on R iff n 3. = 0. Higher Derivatives. Note that if x 0 then (5 can be written as Similarly Differentiating (11 then gives f ( n ( = nf n 1 ( πg n ( f n ( = nf n 1 ( πg n (. (11 g n ( = ng n 1 ( + πf n (. (1 = n ((n 1 f n ( πg n 3 ( π ((n g n 3 ( + πf n 4 ( on substituting (11 and (1, = n (n 1 f n ( π (n g n 3 ( + π f n 4 (, (13 on collecting terms. 5

6 Continue by differentiating (13 to get f (3 n ( = n (n 1 f n ( π (n g n 3 ( + π f n 4 ( = n (n 1 ((n f n 3 ( πg n 4 ( π (n ((n 3 g n 4 ( + πf n 5 ( +π ((n 4 f n 5 ( πg n 6 ( = n (n 1 (n f n 3 ( π ( 3n 9n + 6 g n 4 ( π (n + f n 5 ( π 3 g n 6 (. Though the details look complicated, these few results indicate that on continued differentiating, f n (j ( will be a sum of f n j, g n j 1, f n j, g n j 3, etc. down to either f n j if j even or g n j if j odd. In particular, the properties of f n (j will come from the properties of the f n k or g n k, j k j. Thus f n (j ( will be continuous on R iff all the f n k and g n k, j k j are continuous, which by above is equivalent to n j 1, i.e. j n 1. Similarly f n (j ( will be differentiable on R iff all the f n k and g n k, j k j are differentiable, which by above is equivalent to n j, i.e. j n. Hence f (j n ( is continuous on R iff j n 1, f (j n ( is differentiable on R iff j n. Example If n = k + 1, then f n (k ( is continuous on R but is differentiable only on R \ {0}. If n = l then f (l 1 n only on R \ {0}. ( is differentiable on R, but f n (l ( is differentiable Example If you required a function that is differentiable on R 100 times but not 101 times, and with a 100th derivative that is continuous on all of R choose f 01.. If you required a function that is differentiable on R 100 times but not 101 times, and with a 100th derivative that is not continuous on all of R choose f 0.. Finally. In the examples if something goes wrong, i.e. non-continuous or non-differentiable, it happens at x = 0. By looking at f n (x a the problems will occur at x = a. By summing over a, we get functions with problems that occur in subsets of R. Further, using the ideas that Wierestrass uses to give continuous functions nowhere differentiable, we can sum over countable 6

7 sets of a. This leads to some very interesting pathological functions, i.e. f differentiable m times but for which f (m is nowhere differentiable. Finally, Finally Examples are sometimes given using f n,m ( = x n sin and g x m n,m ( = x n cos. x m I leave it to the Student to check the properties of these functions. 7