MATH3283W LECTURE NOTES: WEEK 6 = 5 13, = 2 5, 1 13

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1 MATH383W LECTURE NOTES: WEEK 6 //00 Recursive sequences (cont.) Examples: () a =, a n+ = 3 a n. The first few terms are,,, 5 = 5, 3 5 = 5 3, Since 5 <, we suspect that a 3 5 n is a decreasing sequence. Let s prove it by induction: a < a is true. Suppose a n+ < a n, then we want to show that a n+ = 3 a n+ < a n+. First, we need to show that {a n } is bounded. Claim: a n (by induction). It s true for a =. If a n, then a n+ = 3 a n. So a 3 n is true by induction. Now 3 a n+ > 3 a n > 0. So a n+ = 3 a n+ < 3 a n = a n+. We also need to claim: 0 < a n, n. True for n =. Assume it is true for n, then 3 a n > 3 = and 3 a n > 0. By bounded convergence theorem, a n L for some L. Using recursive sequences, L = lim a n+ = = n 3 lim n a n 3 L L(3 L) = 3L L = L 3L + = 0 L = 3 ± 9 4 So L = (Q: why do we know L 3+ 5?)

2 MATH383W LECTURE NOTES: WEEK 6 Newton s Method formula: check calculus textbook. Examples Method of finding square roots. Consider f(x) = x. f(x) = 0 when x = ±. Using Newton method: guess x = 3, then x n+ = x n f(x n) f (x n ) = x n x n x n = x n x n + x n = x n + x n = (x n+ x n ) ( ) Rule: divide by guess and average with guess (divide and average method). Does it work? First, we show: n, x n >. x = 3 > true. Assume x n >, we want to show x n+ = (x n + x n ) >, i.e. x n+ x n >, same as x n x n + > 0. But x n x n + = (x n ) > 0 is always true. Second, we have to show that {x n } is decreasing, or x n+ = x n + x n x n i.e. x n + x n or x n But this is true by (). To compute L, we use (*) and take limits: L = L + L, L = L +, L = ± Since n, x n >, we know L =. How well does it work? =.4435 x = 3 =.5, x = ( 3 + 4) = 7 = ) = 577 = accurate to 5 decimal places! x 3 = ( Newton method and inversion. To find (if a > 0), we need to solve a Use Newton method: x n+ = x n f(x n) f (x n ) = x n f(x) = x a = 0 x n a x n = x n + x n ax n Note: replace division (inversion) by multiplication. a = 3, 3 = = x n ax n

3 MATH383W LECTURE NOTES: WEEK 6 3 x =.08, x = x ( 3x ) = = , x 3 = x ( 3x ) = = place accuracy Continued fraction expansion Consider recursive sequence a =, a n+ = + +a n. even terms odd terms a = + + =, a 3 = a 4 = = 5, a 5 = = 5 = 9 It turns out that a = x and a 4 = x from Newtons method to compute starting with x = 3. It can be shown that a n is decreasing, bounded below and a n. Also a n+ is increasing and a n+. This means = For example, a =, a = +, a + 3 = /4/00 Example: Find Cubic roots To find 3, solve f(x) = x 3 = 0. Newton method: x n+ = x n f(x n) f (x n ) = x n x3 n 3x n set x =, x = 3 ( + 4 ) = = 3 < x x n+ = 3 xn + 3x n Show by induction: x 3 = 8 > x 3 n >, x n, n 3 < x n if 3x n < 3 x n i.e. x 3 n = 3 (x3 n + ) = x n 3 (x n + ) x n So lim x n exists, L = 3 ( L3 + L ), 3L 3 = L 3 +, L 3 =, L = 3 values: x =, x = 3, x 3 = 3 ( ) = = 35 x 4 = 3 ( ) = = =.599, error.00 7 =.9696 Exercise: Ch3 3.7, 4.4, 4.5(a)-(d), 5.5, 5.6(a)-(e), 6.6, 6.7, 6.8, 6.

4 a m a n = (a m L) + (L a n ) a m L + L a n < ε + ε = ε 4 MATH383W LECTURE NOTES: WEEK 6 Cauchy sequences A sequence is called Cauchy if terms get closer and closer to another as n gets larger and larger. Formally, < a n > is Cauchy if ε > 0, n 0 N, (m, n n 0 a m a n < ε) Theorem 0.. If < a n > is convergent, then < a n > is Cauchy. Proof. Let ε > 0 and L = lim n a n. Since < a n > is convergent, If m, n n 0, then n 0 N, (n n 0 a n L < ε ) We now investigate whether Cauchy sequences converge. Theorem 0.. Let < a n > Cauchy. Then < a n > is bounded. Proof. Let ε =. < a n > Cauchy implies n 0 N, s.t. a m a n < if m, n n 0. In particular, for any m n 0, a m a n0 < and a m = a m a n0 + a n0 a m a n0 + a n0 < + a n0 A bound for a, a,, a n0 is So n N a n < M + a n0 + M = max{ a, a,, a n0 } Corollary 0.3. Let < a n > be Cauchy, then < a n > is monotone < a n > converges. Proof. Since < a n > is bounded, if < a n > is monotone, then < a n > converges by the bounded convergent theorem. A very hard result is: If < a n > is a sequence, then < a n > contains another sequence < b m > (a subsequence < a nm >, like odd or even terms), which is monotone. Now let < a n > Cauchy, and < b m > is a monotone subsequence of < a n >. < a n > is bounded < b m > is bounded So < b m > converges to some limit L. Since < a n > is Cauchy, the terms b m attract the terms a n and a n L.

5 MATH383W LECTURE NOTES: WEEK 6 5 Find monotone subsequences of Cauchy sequences: a m is called a peak of point if it satisfies (a m > a n if n > m). Thus a m is a strict upper bound for {a n n > m}. Let < a n > be a sequence. Then there are two cases: () Suppose < a n > has infinitely many peak points a n < a n < < a nk <. Then < a nk > is a (strictly) decreasing subsequence. () Suppose < a n > has only finitely many peak points a n, a n,, a nk (It may have none: a n = n, n ), then a n > a n > > a nk. Let m = n k +. Since a m is not a peak point, m > m s.t. a m a m. Since a m is not a peak point, m 3 > m s.t. a m3 a m. continue: if a mj a mj, then a mj is not a peak point. m j+ > m j with a mj+ a mj Thus, we can construct a subsequence < a mj >. Moreover, < a mj > is increasing. /6/00 Summation Notation Consider a sequence a, a, < a n,. We will look at ways to add these terms. First need nice ways to add them. Suppose m < n. Then sigma notation for adding is a i = a m + a m+ + + a n i=m i is a dummy variable. Thus a j = j=m a k = k=m a i = a m + a m+ + + a n i=m Examples. n k=m a k = n m k=0 a k+m. m j=m a j + 3m i=m+ a i = 3m k=m a k 3. Write n 4 in sigma notation. Examples n j 4 = (j + ) 4 = j= j=0 j=n+ (j n) 4

6 6 MATH383W LECTURE NOTES: WEEK 6 () Let s n = n 3 [( i n n ) + ]. Determine if lim n s n exists. s n = 3 n [ ( i n ) + = 3 n ] i n + 3 n n = 3 n ( i ) = 3 n(n + )(n + ) + 3 n 3 6 = 6n3 + 9n + 3n = 4 6n 3 () What is the value of n j=0 ( )j? ( ) j = ( ) 0 + ( ) + + ( ) n j=0 (3) Expand n (5i 5 i ) = ( ) n { if n is even = 0 if n is odd (5 i 5 i ) = 5 i 5 i = n 5 i 5 i = 5 n 5 0 = 5 n i=0 Theorem 0.4 (Generalized triangle inequality). Let n N and a, a,, a n real. Then a i a i Proof. Induction. True for n = : a i = a = a i

7 Assume true for n. Then n+ a i = a i + a n+ MATH383W LECTURE NOTES: WEEK 6 7 a i + a n+ n+ a i + a n+ = a i (induction) So it is true for all n by induction. Examples () Assume x. Show: for n 0, x k = xn+ x n+ x k = k= k= n = 0, = x true. x Assume true for n. Then k= x k +x n+ = xn+ + x n+ x n+ x () Simplify the following: (a) n ( n j= (i + j)) and n j= ( n (i + j)) ( (i + j)) = j= = n (ni + i + n = n j) = j= j j= n(n + ) ni + n = n (n + )(= (usual triangle inequality) = xn+ x = x(n+)+ x j j= ( (i + j)).) (b) i ( ) n+ 5 n n= (n+) 4 n+ n + appear several times. so we may want to write in terms of n + : 5 n = 5 n+, 4 n+ = 4 (n+)+, so the series looks like i n= ( ) n+ (n + ) 5(n+) 4 (n+)+ j=

8 8 MATH383W LECTURE NOTES: WEEK 6 j= i n= Now replace n + = j. Then j goes from to i + and ( ) n+ (n + ) 5(n+) 4 = i+ (n+)+ j= j=3 j= ( ) j j 4 = i+ ( 5 j+ 4 )j 00j 5j (c) n n j=i (i + j) Observe: (+j)+ (+j)+ (3+j)+ = n+ j+(n )+ j+3(n )+ j+ j= j= j= j=3