# help in interpreting x(t)

I dont know if the wording is confusing but I know x(.) is just $\mathbb{R} \rightarrow \mathbb{R}$. Can I say that as my precise definition.

and then say its just a mapping of real numbers to real numbers. I dont know if its right please help out

“Every magnitude which grows continually but not beyond all limits must certainly approach a limiting value.”

We can interpret “magnitude” as a real number x that changes in time, thus, as a real valued function x(t), t ∈ R. The above sentence is a theorem about the behaviour of such a function as t approaches $\infty$.

(a) The assumptions of the theorem are that the function x(t) is monotone (“grows”) and is bounded (“not beyond all limits”). Give a precise definition of each of these properties of the function x($\cdot$).

(b) State the theorem in your own words.

and also if someone could tell me what this line means

"The side and the diagonal of a square are incommensurable."

like how would i rephrase that if i want to explain this to someone?

-
a) $f$ is monotone increasing iff $f(x)<f(y) \quad \forall x<y$. Furthermore $f$ is bounded by $c<\infty$ from above , iff $f(x)\leq c \quad \forall x$. –  sonystarmap Feb 26 '13 at 18:18
Of course you have to adapt it, as $x$ is your function and $t$ the argument, thus instead of $f(x)$ you have to rewrite it as $x(t)$. –  sonystarmap Feb 26 '13 at 18:19
i dont get it. like i know it is giving you a function where the domain pertains to all real numbers, or in other words in x(t), t cannot be complex. but i dont know what it meant by put it as a precise definiton –  MathGeek Feb 26 '13 at 18:21

If a function $f:\mathbb{R}\to\mathbb{R}$ is monotonously increasing then it's first derivative (i.e. it's rate of change) is greater than or equal to $0$ for it's entire domain, i.e.:

$$f(t)\text{ is monotonously increasing} \iff f'(t)\geq0, \forall t\in\mathbb{R}$$

Similarly, if a function $g:\mathbb{R}\to\mathbb{R}$ is monotonously decreasing, then it's first derivative is less than or equal to $0$ for it's entire domain, i.e.:

$$g(t) \text{ is monotonously decreasing} \iff g'(t)\leq0, \forall t \in\mathbb{R}$$

For a function $h:\mathbb{R}\to\mathbb{R}$ to be bounded from above, it must be the case that for it's entire domain, i.e. $\forall t \in \mathbb{R}$, there exists some constant $x\in\mathbb{R}$ such that the function does not exceed $x$. That is:

$$h(t)\text{ is bounded above by }x\in\mathbb{R}\iff \exists x\in\mathbb{R}:\forall t \in \mathbb{R} \space h(t)\leq x$$

-
thank you @shaktal, that made sense. –  MathGeek Feb 26 '13 at 18:26
I was wondering would you know what this line means. its bothering me "The side and the diagonal of a square are incommensurable." –  MathGeek Feb 26 '13 at 18:26
• a) $x(\cdot)$ is monotone increasing iff $x(t_l)<x(t_r) \quad \forall t_l<t_r\quad t_l,t_r \in \mathbb{R}$. Furthermore $x$ is bounded by $c<\infty$ from above , iff $x(t)\leq c \quad \forall t\in \mathbb{R}$.
• b) If $x(\cdot)$ is montone increasing and bounded from above, then $\exists x_{\infty}\in \mathbb{R}: \lim_{t\rightarrow\infty}x(t) = x_{\infty}$

Note, that b) also reads: Every montone increasing (decreasing) sequence which is bounded from above (below) does converge.

Edit: From this source I quote

incommensurable

b. Having an irrational ratio.

This the case, because if a square as an side length of $a$, its diagonal has length $\sqrt{2}a$, thus the ratio is $\sqrt{2}$ which is irrational.

-