By Kenneth S. Miller.
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Additional info for An introduction to the calculus of finite differences and difference equations
Proof of (3): Again, given ε > 0 we need to find N so that k ≥ N guarantees |xk yk − xy| < ε. 4. FINDING LIMITS 49 This doesn’t break up as nicely as the expression in the first part, but a trick helps us considerably: we subtract and add the quantity xyk in the middle: |xk yk − xy| = |xk yk − xyk + xyk − xy| ≤ |xk yk − xyk | + |xyk − xy| . Now we can factor each term to get |xk yk − xy| ≤ |xk − x| |yk | + |x| |yk − y| . In the spirit of the proof of the first part, we wish to make each term less than ε/2.
Which alternates the two values +1 and −1. This is bounded, since the absolute value of all elements is 1. Nonetheless the sequence diverges. To see this, consider the possibilities. , when k is odd) satisfies (−1)k − 1 = 2 (k odd) so it cannot approximate +1 with accuracy, say, ε = 1. However, if y = 1, we see that every even index term has (−1)k − y = |y − 1| > 0 (k even) and so we can’t approximate y with accuracy, say, half the distance from y to 1: ε = |y − 1| /2. This particular sequence repeats a regular pattern: we call it a periodic sequence.
Bn := 17. EN := n 1 k=1 k , n = 1, ... N 1 n=0 n! , k = 0, ... 10. αn := sin n, n = 0, ... 11. βn := sin πn , n = 1, ... 15. Cn := k2 k+1 , 1 k sin k, k = 1, ... n i i=0 i+1 , 16. DN := N 2i+1 i=0 3i , 18. Fk := kj=1 k = 1, ... N = 0, ... 19. Gn := nk=0 (−2)−k , n = 0, ... 20. XK := n = 0, ... j j+1 N = 0, ... − K k2 k=0 k+1 , j+1 j , k = 0, ... 21. 05)xm−1 , m = 1, ... 22. x0 = 1, xn := xn−1 xn−1 +1 , n = 1, ... 23. x0 = 2, xn := xn−1 − 1 xn−1 , n = 1, ... 24. x0 = 1, xn := xn−1 + 1 xn−1 , n = 1, ...