Is monotonically increasing is same as non-decreasing? Thank you for answer beforehand.
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$\begingroup$ No. Monotonically increasing is the same as increasing tough. $\endgroup$– KasperFeb 13, 2013 at 20:22
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2$\begingroup$ To avoid danger of misunderstanding, I tend to use "strictly increasing". $\endgroup$– Robert IsraelFeb 13, 2013 at 20:28
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2$\begingroup$ It is slightly ambiguous, depending on the author. Some folks use increasing for non-decreasing. I prefer the use of strictly- and non- when the ambiguity matters. $\endgroup$– copper.hatFeb 13, 2013 at 20:29
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2 Answers
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As noted in the comments, for a sequence to be "monotonically increasing", what that requires is a bit ambiguous, depending on authors.
For example, Rudin, in his Principles of Mathematical Analysis (Definition 3.13), defines a sequence $\{s_n\}$ of real numbers to be:
(a) monotonically increasing if $s_n \leq s_{n+1}, (n = 1, 2, 3, \cdots)$ [i.e., non-decreasing].
(b) monotonically decreasing if $s_n \geq s_{n+1}, (n = 1, 2, 3, \cdots)$ [i.e., non-increasing].
To be explicit (and avoid ambiguity), one can specify that a monotonic sequence is "strictly increasing" or that it is non-decreasing, or non-increasing, or strictly decreasing.
Note that the Monotone Convergence Theorem applies regardless of whether the above interpretations: a non-decreasing (or strictly increasing) sequence converges if it is bounded above, and a non-increasing (or strictly decreasing) sequence converges if it is bounded below.
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No. $x(0)=0, x(1)=0, x(2)=1$ is non-decreasing, but not monotonically increasing. It is monotonically increasing if $x(1)=0.5$ instead, for example.
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2$\begingroup$ Depends. There are some authors who use "increasing" to describe a function that is either non-decreasing or increasing tough. $\endgroup$– LudolilaFeb 13, 2013 at 20:26