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Nov
16
comment Finding $\int_0^{\pi/2} \sin x\,dx$
$\lim_{x^\circ \rightarrow 0} \frac{\sin(x^\circ )}{x^\circ}$ is most certainly $1$, by any standard. Whatever your interpretation of $x^\circ$ is, it must be as a real variable, and $\lim_{y\rightarrow 0} \frac{\sin y}{y}=1$ for any real variable $y$. Typically (e.g. by Mathematica), $x^\circ$ is interpreted via $x^\circ=x\times\frac{\pi}{180}$. Perhaps you meant to write $\lim_{x\rightarrow 0} \frac{\sin (x^\circ )}{x} = \frac{\pi}{180}$ or $\lim_{x^\circ\rightarrow 0} \frac{\sin (x^\circ )}{x} = \frac{\pi}{180}$? Either would be correct. What you have is not.
Nov
16
comment Intuitive understanding of the derivatives of $\sin x$ and $\cos x$
I disagree with the cynical analysis of people's calculus education. I think the reason many people answer this question poorly is because maths students are simply not familiar with degrees, as they are abandoned immediately upon entering a calculus course. And rightly so. Reasoning that an inability to answer questions about degrees is evidence of a poor calculus education is like asking European children to convert between cups and gallons, or about the Fahrenheit scale, and then concluding when they give you puzzled looks that it is "a sign of how badly the French learn arithmetic".
Nov
16
revised Find $\int_0^2 x^2f''(2x)dx $ given $f(2)$, $f'(2)$, and $\int_0^2 f(x)dx$
typo
Nov
16
revised Find $\int_0^2 x^2f''(2x)dx $ given $f(2)$, $f'(2)$, and $\int_0^2 f(x)dx$
typo
Nov
16
answered Find $\int_0^2 x^2f''(2x)dx $ given $f(2)$, $f'(2)$, and $\int_0^2 f(x)dx$
Nov
16
comment Uncountable reals in the theory
If you want the reals to really be the reals, so in particular have theory RCF, then no. RCF is not expressive enough. If you are just asking for the sake of asking, fix a countable model of ZF and a bijection between its universe and $\mathbb{R}$. Then look at the induced structure.
Nov
16
revised Any 3 letters shuffled in a word- What is the probability that the word remains the same?
added 116 characters in body
Nov
16
revised Any 3 letters shuffled in a word- What is the probability that the word remains the same?
added 58 characters in body
Nov
16
comment Any 3 letters shuffled in a word- What is the probability that the word remains the same?
There are 3*2*5 ways to select exactly 2 A's. But I get your point now. Fixed, thank you!
Nov
16
comment Any 3 letters shuffled in a word- What is the probability that the word remains the same?
15/56 is not even an integer ...
Nov
16
comment Probability that $0$ won't be chosen out of $k$ picks from the set $\{0, 1, \ldots, 9\}$
@perfect6 If you insist on doing it without order, that's fine. The quotient probability measure will give the same answer. It just won't have a uniform law anymore. For example, for $k=3$ the unordered $k$-tuple {9,9,9} is obviously much less likely than the unordered $k$-tuple {1,2,3} (there's only one way to pick {9, 9, 9}, but six ways to pick {1, 2, 3}). The problem tells you that the P-measure is the $k$-fold product of the uniform P-measure on 0 through 9, so dinoboy used that directly. You are welcome to pass through the quotient if you like to make your work harder for yourself.
Nov
16
comment Any 3 letters shuffled in a word- What is the probability that the word remains the same?
There are six ways to place three pegs in three holes. If one of the pegs must go in one of the holes, but the other two pegs can be freely transposed, there are two ways that work. So the conditional probability of success given Case 2 is 2/6, and similarly for the conditional probability of success given Case 3.
Nov
16
answered Any 3 letters shuffled in a word- What is the probability that the word remains the same?
Nov
15
comment Why isn't $\lim \limits_{x\to\infty}(1+\frac{1}{x})^{x}= 1$?
There's no reason to think $(1+1/n)^n$ should be close to $1$ just because $n$ gets large and $1/n$ gets small. $0.00001$ is a very small number, and $1,000,000$ is a very big number, yet $1.00001^{1,000,000}=22025.3645063913326\ldots$ which is not close at all to $1$.
Oct
12
comment Why König's lemma isn't “obvious”?
I don't think it's obvious at all. Intuitively, one can imagine building a tree, continually adding on longer and longer branches, but being careful never to extend any one branch too far. Indeed, this intuition is precisely how one constructs an unbounded, computable, finitely branching tree with no computable infinite branch. Or an infinitely branching counterexample. The assumptions are in some sense tailored to the proof, so I would say precisely the opposite: the proof is obvious (to someone who understands induction), but the lemma requires careful formulation.
Sep
30
awarded  Explainer
Sep
27
comment why have we chosen our number system to be decimal (base 10)
@J.M. The word digit is used to refer to the numerals below ten because you can count them on your fingers, yes, but using that as evidence for why we selected ten as a base in the first place is bogus. Using the multiple meaning of "digitus" as evidence that we originally selected base ten because of our fingers is like answering "Why do we use strings of numerals to assign unique identifiers to individual phone lines?" by pointing out that "digits" is common slang for "phone number". Certainly the slang came after the choice, not the choice to justify the slang.
Sep
27
comment why have we chosen our number system to be decimal (base 10)
@MichaelEdenfield The positional system was new, yes, and made arithmetic much easier. But the choice of ten as a base goes back to Mesopotamian civilizations, and this particular choice doesn't make arithmetic easier at all.
Sep
26
comment Weak convergence vs Convergence in Measure
@smiley06 $X_n$ have associated push-forward measures $\mu_{X_n}$ defined by $\mu_{X_n}(A) = P(X_n \in A)$.
Sep
20
comment Why some people don't like proofs by contradiction
In many cases, the direct proof actually establishes a stronger result and then applies modus ponens. For example, a useful way to prove that some particular set $A$ does not belong to a class $\mathcal{S}$ is by proving that every member of $\mathcal{S}$ contains an element not in $A$. If I did the argument by contradiction, I would assume $A\in \mathcal{S}$ and then deduce that $A$ contains an element not in $A$, a contradiction. The stronger statement--that every set in $\mathcal{S}$ contains elements outside of $A$--is obscured.