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Oct
14
revised I don't understand why the solution to this probability question is set up in this manner
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Oct
13
reviewed Close A survey shows that $63$% of Americans like cheese and $76$% like apples. Which answer is true?
Oct
13
revised I don't understand why the solution to this probability question is set up in this manner
added 749 characters in body
Oct
12
answered I don't understand why the solution to this probability question is set up in this manner
Oct
11
comment How can I justify this set manipulation to show a result in probability?
Intersection is distributive over set difference: $$\begin{aligned}(A\cap B)\setminus(A\cap C)&=(A\cap B)\cap(A\cap C)'\\ &=(A\cap B)\cap(A'\cup C')\\ &=((A\cap B)\cap A')\cup((A\cap B)\cap C')\\ &=\emptyset\cup(A\cap(B\cap C'))\\ &=A\cap(B\setminus C).\end{aligned}$$
Oct
11
comment Generating Pythagorean Triples from Others via Dissections
Shouldn't the transformation in $(m,n)$ be $$ \begin{aligned} m &\mapsto 2m + n \\ n &\mapsto m ? \end{aligned} $$ For example, $(m,n)=(2,1)$ corresponds to the $(3,4,5)$ triple. Its descendent is the triple $(21,20,29),$ which corresponds to $(m,n)=(5,2).$ This is still an element of $\mathrm{SL}(2,\mathbb{Z}).$
Oct
9
comment Do mathematicians, in the end, always agree?
"For some reason the fact that someone having these opinions teaches and researches math disturbs me." If it disturbs you, then don't learn his theorems.
Oct
8
comment 7 white balls and 3 black balls in a box. You take out in sequence 3 balls at random without replacement. Probability that all 3 balls are black?
Suppose that you had a box with 10 black balls. If you choose 3 balls, it is clear that the probability they are all black is 1. But your method gives $10/10+9/9+8/8=3.$
Oct
6
revised Number of ways to choose a sequence of three letters from the letters of MISSISSIPPI
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Oct
6
revised Number of ways to choose a sequence of three letters from the letters of MISSISSIPPI
added 8 characters in body
Oct
6
answered Number of ways to choose a sequence of three letters from the letters of MISSISSIPPI
Oct
2
comment The relationship between ε and δ in the (ε, δ)-definition of limitor
... $\epsilon$ is, there will always be such a point beyond which $f(x)$ is within $\epsilon$ of $L.$ That's what the "there exists a $\delta$" business is all about.
Oct
2
comment The relationship between ε and δ in the (ε, δ)-definition of limitor
In $\epsilon$-$\delta$ arguments, there is no reason to have $\delta$ be as large as it can be. People often let $\delta$ be smaller than it could be for the sake of a simpler mathematical argument. Furthermore, if the $\epsilon$-$\delta$ definition of limit holds, then it is correct to say that that $f(x)$ gets arbitrarily close to $L$ as $x$ gets closer and closer to $a.$ That's because as $x$ gets closer and closer to $a,$ it will eventually be less than a distance of $\delta$ from $a.$ Beyond that point, $f(x)$ is guaranteed to be within $\epsilon$ of $L.$ No matter how small...
Oct
2
comment The relationship between ε and δ in the (ε, δ)-definition of limitor
In the comments to Umberto P.'s answer, I give an example where $\Delta(\epsilon)$ remains constant for all $\epsilon,$ not just for all $\epsilon$ in some range of values. Certainly there are many such examples. However, I think a better answer to your original question is that $\Delta(\epsilon)$ generally decreases with $\epsilon,$ but that there are examples where it remains constant rather than decreasing.
Oct
2
comment The relationship between ε and δ in the (ε, δ)-definition of limitor
To add to my previous comment: if $\lvert f(x)-L\rvert<\epsilon$ for all $x,$ that's the same thing as $\Delta(\epsilon)=\infty.$ I wanted to show an example where the condition $\Delta(\epsilon)=\infty$ held over a range of $\epsilon$ values (to show that $\Delta(\epsilon)$ doesn't always decrease). Equivalently, I wanted to exhibit a limit for which $\lvert f(x)-L\rvert<\epsilon$ held for all $\epsilon$ in a range of values with no restriction whatsoever on $x.$
Sep
30
awarded  Explainer
Sep
29
comment The relationship between ε and δ in the (ε, δ)-definition of limitor
Let $\Delta(\epsilon)=c$ for some positive constant $c.$ This implies that $\lvert f(x)-L\rvert<\epsilon$ for all $x\in (a-c,a+c)\setminus\{a\}.$ We claim that $f(x)=L$ for all $x\in(a-x,a+c)\setminus\{a\}.$ For suppose not. Then there is some $x_1\in(a-c,a+c)\setminus\{0\}$ such that $f(x_1)=M\ne L.$ Let $\epsilon=\frac{1}{2}\lvert M-L\rvert.$ Then $\lvert f(x_1)-L\rvert=\lvert M-L\rvert=2\epsilon>\epsilon,$ contradicting the assumption that $\lvert f(x)-L\rvert<\epsilon$ whenever $x\in(a-c,a+c)\setminus\{0\}.$ So $f$ is constant in the interval $(a-c,a+c),$ except possibly at $a.$
Sep
29
comment The relationship between ε and δ in the (ε, δ)-definition of limitor
@user1485853: OK, here's an example of a non-constant function for which $\Delta(\epsilon)$ takes a finite constant value. Let $f$ be defined by $$f(x)=\begin{cases}17 & x=5\\ 8 & x>5\text{ or }3<x<5\\ -39 & x=3\\ \frac{1}{x-3} & x<3.\end{cases}$$ Now $\lim_{x\to 5} f(x)=8,$ and if the $\epsilon$-$\delta$ definition is applied, we find that $\Delta(\epsilon)=2$ for all $\epsilon.$ Notice that the value of $f$ is constant in the set $(5-2,5+2)\setminus\{5\},$ which is necessary in order to have $\Delta(\epsilon)=2,$ but that it does not remain constant outside of this set.
Sep
29
comment A question regarding continuity and limits.
@ColeJohnson: The question is fine. It explicitly states that the formula only applies when $x$ is different from $4.$
Sep
27
comment The relationship between ε and δ in the (ε, δ)-definition of limitor
Are you asking about the case where $\Delta(\epsilon_1)=\Delta(\epsilon_2)$ for all $\epsilon_1,\epsilon_2>0,$ or the case where $\Delta(\epsilon_1)=\Delta(\epsilon_2)$ for some $\epsilon_1,\epsilon_2>0?$ Your question didn't specify the former, so I came up with an example for the latter.