Activity for whybecauseâ€
Type | On... | Excerpt | Status | Date |
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Edit | Post #286132 | Initial revision | — | almost 3 years ago |
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A: How can I justify whether I'm able to apply the Gaussian elimination method to this system? I may be misunderstanding the question, but technically you can always use the Gaussian elimination method. But perhaps what you meant to ask is: Will the method give you a unique solution? Sometimes you get no solutions, sometimes you get infinitely many solutions, and sometimes you get precis... (more) |
— | almost 3 years ago |
Edit | Post #285986 |
Post edited: -1 |
— | almost 3 years ago |
Edit | Post #285986 |
Post edited: corrected -1 |
— | almost 3 years ago |
Comment | Post #285986 |
That is certainly another helpful perspective, sure. (more) |
— | almost 3 years ago |
Edit | Post #285987 | Initial revision | — | almost 3 years ago |
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A: How to write the big Xi notation in MathJax? If you google the Greek alphabet, the LaTeX rendered Xi is a slightly stylized version of the usual print Greek character. This doesn't seem bad to me. Note that in many languages, Greek included, you often produce a symbol in hand-writing somewhat differently than you produce it in typesetting (... (more) |
— | almost 3 years ago |
Edit | Post #285986 |
Post edited: elaboration of i |
— | almost 3 years ago |
Edit | Post #285986 | Initial revision | — | almost 3 years ago |
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A: How $ijk=\sqrt{1}$? The mistake here is taking $i=\sqrt{-1}$. This is not correct, even though $i^2=-1$. How is that possible?! Because $i$ in this page is NOT meant as the complex number, even though it is similar. $i$ is meant here as a purely symbolic object satisfying an algebraic property. Put another wa... (more) |
— | almost 3 years ago |
Edit | Post #285938 | Initial revision | — | almost 3 years ago |
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A: Is it impossible to prove Jensen's inequality by way of step functions? Right, due to the comment, we know the convex function is continuous and therefore the rest of the proof can be done using the MCT. In particular this is because every function is approached from below by a sequence of step functions. (more) |
— | almost 3 years ago |
Edit | Post #285935 | Initial revision | — | almost 3 years ago |
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A: "Pointwise equicontinuity implies uniform" implies compact Oh wait never mind. The natural number example works because every family of functions is equicontinuous, because delta=1 works for everything. (more) |
— | almost 3 years ago |
Edit | Post #285925 | Initial revision | — | almost 3 years ago |
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"Pointwise equicontinuity implies uniform" implies compact Suppose $(M,d)$ is a metric space such that every sequence $fn:M\to \Bbb R$ which is pointwise equicontinuous is also uniformly equicontinuous. Does this imply that $M$ is a compact metric space? My thoughts: If true, we could try to show compactness directly or show sequential compactness ... (more) |
— | almost 3 years ago |
Comment | Post #285755 |
It's only zero. If ax=b and a is not zero, then x=b/a. This effectively guarantees that division by a nonzero number preserves any solution, regardless of what x is (or what a and b are, again assuming a is nonzero).
This comes down to the definition of division. What does 15/5 = 3 mean? It ... (more) |
— | almost 3 years ago |
Comment | Post #285755 |
To my knowledge this rule does not have a name. Rather, to talk about it, we merely point out that division by zero could be a problem, so we proceed to separate the solution method into cases. Each case avoids the issue of dividing by zero.
Alternately you can characterize this by the zero-pr... (more) |
— | almost 3 years ago |
Comment | Post #285755 |
To answer this, it is important to remember what any of this even means. What do we mean by any equation, like x-2=3?
Actually, without any context, this doesn't mean anything at all. But usually when we write "x-2=3" this is like a short-hand for a question. Namely "What are all of the value... (more) |
— | almost 3 years ago |
Edit | Post #285819 | Initial revision | — | almost 3 years ago |
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Is it impossible to prove Jensen's inequality by way of step functions? Jensen's Inequality: Let $\varphi:\Bbb R\to \Bbb R$ be convex, and $f:[0,1]\to\Bbb R$ be integrable, and suppose $\varphi\circ f$ is integrable over [0,1]. Then $$ \varphi\left(\int{[0,1]} f\right)\le \int{[0,1]}\varphi\circ f $$ A proof from step functions: I have seen a proof of this inequal... (more) |
— | almost 3 years ago |
Edit | Post #285759 |
Post edited: |
— | almost 3 years ago |
Edit | Post #285759 | Initial revision | — | almost 3 years ago |
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A: $\{f_n\}\to f$ in $X$ implies $||f||\le\liminf||f_n||$ Oh I think I got it. $$ \lim |T(fn)| = \liminf |T(fn)| \le \liminf ||fn|| $$ (more) |
— | almost 3 years ago |
Edit | Post #285756 |
Post edited: Added duh |
— | almost 3 years ago |
Edit | Post #285756 | Initial revision | — | almost 3 years ago |
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$\{f_n\}\to f$ in $X$ implies $||f||\le\liminf||f_n||$ Let $X$ be a normed linear space and suppose that, for each $f\in X$ there exists a bounded linear functional $ T\in X^ $ such that $T(f)=||f||$ and $||T||=1$. Prove that if $\\{fn\\}\to f$ in $X$ then we have $||f||\le\liminf||fn||$. I think the only $T$ that makes sense to work with is that ... (more) |
— | almost 3 years ago |
Comment | Post #285755 |
If b=0 then you may get invalid solutions, yes.
So you should divide this solution method into two cases. Case 1 is that x=5. In that case you cannot divide by x-5, but it doesn't matter, because you already know the solution. Case 2 is that x is not 5. Then you can divide by x and then the ... (more) |
— | almost 3 years ago |
Edit | Post #285755 | Initial revision | — | almost 3 years ago |
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A: Missing a solution: Are A and B always equal if A-B=0 Dividing by 5-x is valid only if x is not 5. But x=5 is another solution to the original equation. (more) |
— | almost 3 years ago |
Edit | Post #285678 | Initial revision | — | almost 3 years ago |
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A: Why 1. multiply the number of independent options? 2. add the number of exclusive options? Look at the sub-picture with the beef burger, and all the nodes which extend from it. There are four options. Now look at the sub-picture with the chicken burger. Same number of options. In fact, for each one of those sub-pictures, there are four options, one for each of the different cho... (more) |
— | almost 3 years ago |
Comment | Post #285015 |
@#36356 I think the point you're making, which I would agree with, is that not all things *can* be made intuitive. That's right, and it's right to point that out to a student, in such cases. (more) |
— | almost 3 years ago |
Edit | Post #285525 |
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— | almost 3 years ago |
Edit | Post #285525 |
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— | almost 3 years ago |
Edit | Post #285525 | Initial revision | — | almost 3 years ago |
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A: How's it possible to arrange 0 objects? How can 0! = 1? This explanation may work for some but not for others--it is perhaps a matter of taste or intuition whether you feel that it is sensible. Is doing nothing a "way of arranging" zero objects? Eh, you can get very philosophical about this and probably not in a productive way. I find the following... (more) |
— | almost 3 years ago |
Comment | Post #285447 |
I suspect this question isn't stated fully enough, and also should probably be stated as a pure math problem rather than relying on any domain knowledge about ETF's. If you're minimizing overlap and want to buy as few as possible, and those are literally the only constraints ... then choose zero of ... (more) |
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Edit | Post #285436 |
Post edited: and the universe |
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Edit | Post #285436 |
Post edited: countable -> finite |
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Comment | Post #285436 |
Right, wasn't being careful, I'll edit. (more) |
— | almost 3 years ago |
Edit | Post #285436 |
Post edited: writing style |
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Edit | Post #285436 | Initial revision | — | almost 3 years ago |
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Are we in a "history-valley" for Topology? Here is my current rough timeline of Topology: Newton invents the calculus People like Riemann and Cauchy make it rigorous, and by this time, we have the $\varepsilon,\delta$ definition of continuity of a real valued function of a real variable. We realize that we want a definition of the con... (more) |
— | almost 3 years ago |
Edit | Post #285435 |
Post edited: |
— | almost 3 years ago |
Edit | Post #285435 | Initial revision | — | almost 3 years ago |
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A: Intuitively, why does $p$ vary inversely with $P(C_3 \mid D_2)$? But directly with $P(C_2 \mid D_3)$? $P(C3|D2)$ is the probability of the car being behind door 3 given that Monty opened door 2. $p$ is the probability that Monty opens door 2, under the assumption that he has a choice between it and door 3. That is to say $p$ is the probability that Monty opens door 2 under the conditions that: (i) ... (more) |
— | almost 3 years ago |
Edit | Post #285428 | Initial revision | — | almost 3 years ago |
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A: Book suggestion category proposal As long as we only answer it once per field, and are good about closing duplicate questions and redirecting them. Because this kind of question gets asked about once per second, somewhere on a math-related discussion forum. (more) |
— | almost 3 years ago |
Edit | Post #285427 | Initial revision | — | almost 3 years ago |