Calculus Deconstructed: A Second Course in First-Year by Zbigniew H. Nitecki

By Zbigniew H. Nitecki

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We also know that the first term is less than 1. since x99 ≤ 1, we know that x100 ≤ 1—so there! This reasoning is formalized in proof by induction, a technique for proving statements about all natural numbers. It consists of two stages. First, we establish that our statement is true in the initial case: in the argument above, we showed (essentially by inspection) that x0 ≤ 1. Then we prove an abstract assertion: that if our statement holds for some particular (but unspecified) number, then it also holds for the next number.

1(1) give us |(xk + yk ) − (x + y)| = |(xk − x) + (yk − y)| ≤ |xk − x| + |yk − y| . One way to make sure that our desired estimate holds is by making each of the terms in this last expression less than ε/2. But we can do this: since xk → x, we can find N1 so that k ≥ N1 guarantees |xk − x| < ε 2 and since yk → y, we can also find N2 so that k ≥ N2 guarantees |yk − y| < ε . 2 Now, let N ≥ max{N1 , N2 }. Then k ≥ N guarantees |(xk + yk ) − (x + y)| ≤ |xk − x| + |yk − y| < ε ε + =ε 2 2 as required.

The sequence of odd integers could be written {2j + 1j }∞ j=0 or {2j − 1j }∞ j=1 . ) These sequences were specified in closed form, telling us how to determine the value of each element directly from its index. Some sequences are more naturally specified by a recursive formula, in which an element is calculated from its immediate predecessor(s) in the sequence. 05)bm−1 . Note that this doesn’t help us calculate the starting balance (since it has no predecessor); we need to determine the value of bm when m is at its initial value “by hand”.

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