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My name is Chris, and I have general interests in some things.


Jan
14
comment Does maximum or supremum of an infinite set exit?
Glad you solved the problem. Don't forget to upvote and accept the answer you think is the most helpful.
Jan
14
comment Does maximum or supremum of an infinite set exit?
@Amanda for another example, take the set of all prime integers. There are countably infinitely many, and each prime integer is "finite", yet there is no maximum or supremum of the set.
Jan
14
comment Does maximum or supremum of an infinite set exit?
Yes. By the pattern, $1, 2, 3, 4, 5, \dots$, each is a natural number. Natural numbers are known to be "finite," that is, informally, "less than infinity."
Jan
14
revised Does maximum or supremum of an infinite set exit?
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Jan
14
comment Does maximum or supremum of an infinite set exit?
Agreed, but I think from anyone asking this type of question, whether the issue might have some involvement is unclear.
Jan
14
comment Does maximum or supremum of an infinite set exit?
Thanks for clarifying. You should specify this in your question so that users won't give examples such as $(a,b)$ or $[a,b]$, as each of those contains an uncountably infinite number of points.
Jan
14
comment Does maximum or supremum of an infinite set exit?
Because the OP describes $S$ as $\left\{k_1, k_2, \dots, k_n\right\}$, it seems that $S$ must contain a countably infinite number of points, thus intervals such as $[a,b]$ or $(a,b)$ are not good examples.
Jan
14
comment Does maximum or supremum of an infinite set exit?
By the way you describe $S$, it looks like you want it to have a countably infinite number of elements. Thus $S$ may not take the form of any interval, open or closed or otherwise, on the Real line. Is this correct?
Jan
14
answered Does maximum or supremum of an infinite set exit?
Jan
12
revised How to understand point functions
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Jan
12
answered How to understand point functions
Jan
2
awarded  Popular Question
Dec
9
revised Prove if $\left\{x_n\right\}$ converges to $2$, then $\left\{\frac{1}{x_n}\right\}$ converges to $\frac{1}{2}$
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Dec
9
answered Prove if $\left\{x_n\right\}$ converges to $2$, then $\left\{\frac{1}{x_n}\right\}$ converges to $\frac{1}{2}$
Dec
7
comment Does this piecewise function contradict the fact that all differentiable functions are continuous?
Great use of algebra to demonstrate the fact that the limit of the difference quotient does not exist. What might add to one's conceptual understanding is a geometric explanation: the slope of the line between $\bigl(1, f(1)\bigr)$ and $\bigl(1+h, f(1+h)\bigr)$ increases without bound (the line gets steeper) as $h\to 0$. See this graph for a demo.
Dec
7
comment Does this piecewise function contradict the fact that all differentiable functions are continuous?
It would be nice if this answer demonstrated why the function is not differentiable at 1. Also, the fact that it is discontinuous there is unnecessary. The OP knows it is discontinuous, which is why they asked the question in the first place: it was an attempt at a counterexample.
Dec
7
comment The best advanced maths library in java
while this is a good question, I don't believe it fits the model of the StackExchange network, as it is likely to invoke opinionated answers. try asking it in a discussion forum, or better yet, the chatrooms here at SE.
Dec
7
comment Does this piecewise function contradict the fact that all differentiable functions are continuous?
Remember, when we say "a function $f$ is differentiable" what we mean to say is that for each $c\in\mathbb{R}$, the limit $$\lim_{x\to c}\frac{f(x)-f(c)}{x-c}$$ exists and is a real number for all $x\in\mathbb{R}$. What is confusing here is that this fact is not necessarily implied by the fact that the derivative function $f^{\prime}$ is continuous for all $x\in\mathbb{R}$. In your example, you are correct: $f^{\prime}(x) = 2x$ for all $x$ but that does not mean that the derivative limit (above) exists for all $x$.
Nov
25
comment Are the definitions of dot product and cross product the wrong way round?
@Meelo, however, for one-dimensional vectors, $(a)\cdot(b) = ab$, which is analogous to "normal multiplication" (multiplication of real numbers). Also keep in mind that cross products are only defined for three-dimensional vectors, whereas dot products are defined for any n-dimensional vectors.
Nov
21
revised Linear transformation of normal distribution
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