given something like $\int_0^2 \int_{\frac{y}{2}}^{1} {e^{x}}^{2} dxdy $ in order to sketch this, I've drew in the functions & values for the bounds. Without evaluating what else do I do again?
yeah, and she just got out of bed at almost 10 pm to rant to me about something.
obliv, i am not sure what the instructions were, but as far as 'sketch this' i think sketching the region of integration is appropriate. if you actually evaluate that thing, it would be a number, which isn't very fun to sketch.
[draws number line with point on it labeled 'the value of the integral']
uh, i would be mildly surprised if they wanted you to do that. it's more common to ask just for a sketch of the region of integration (which would be the same no matter what the integrand is)
if they really want a picture of the graph of z = e^(x^2) over the region of integration, presumably accuracy is not that big of a deal and basically any increasing concave function of x would give the general spirit.
well, if just passing is enough to get you where you want to go, i hope you at least pass.
i only taught multivar a few times, but even solid A students would goof up quite a bit in exam settings. there are too many things that can potentially go wrong.
it's really hard to write problems that don't invite mishaps that complicate everything, without giving just incredibly easy questions that chatGPT could do.
Let Q be the solid bounded by $x^2+z^2=1, y=x, x=0$ in the first octant. Set up the integral of its volume and sketch it. So this problem gave me a headache
I ended up using up too much time so I just set up two integrals for it which is probably correct but not the way he intended.
I did two cylinders $\frac{1}{2}\int_0^1\int_{0}^{\frac{\pi}{2}}\int_{0}^{1}rdrd\theta dz + \int_{0}^{\infty}\int_{0}^{\frac{\pi}{2}}\int_{0}^{1}rdrd\theta dz$ but looking back on it it should have been 1/8 and 1/4 in front of the integrals
Put $z$ on the inside. You see the projection on the xy-plane, so $dydx$ should be outside.
That picture doesn’t help me. I draw $yz$ in the usual $xy$ positions, with $x$ coning out of the paper/board. I did several problems like this in my lectures.
1st and 2nd year, but a very advanced course. However, the standard computational stuff (plus a lot of linear algebra) is in there. I’m suggesting you watch the computational examples, not the theory.
Is it true that for a continuous function that x < y implies f(x) < f(y)? We can also assume that the function is bounded. I was reading the definition of a lower and uppoer riemann integral which was given using supremum and infimum of a continuous bounded function from a closed subset of R to R. I haven't yet studied continuous functions much. So, was just wondering if that is the case.
again, think graphically, vertically shifting a graph up or down is not going to change continuity or whether the property you want to hold actually does hold, but it can be used to change whether a bounded function is or isn't nonnegative
There is this lemma that if 0≤a<b is true for all b>0, then a must be 0. I know one very short proof, but here is another attempt i saw online: b-a>0 which means a-b<0. Find two numbers n€N and c€W such that nc=a, we get nc-b<0 which is not true from archimedian principle. Is this proof correct to conclude nc must be zero?
Let x be a non-negative real number such that, given any real number r, we have that x is either at 0 or between 0 and x.
Suppose x is greater than 0. Clearly, there is a real number r_1 between 0 and x. But this contradicts our definition that, if x is not at 0, then for any real number, x is between 0 and that real number. Therefore, x cannot be greater than 0, and since it was either greater than 0 or at 0, x is at 0. Since both x is r_1 were arbitrary non-negative real numbers, this means that this holds if x is any real number defined as above.
@nickbros123 it helps to clearly distinguish between a statement that contains a quantified variable, and the instantiated variable within the proof
the proof you mentioned hinges on saying that a - b < 0 implies that nc - b < 0 is a contradiction, but that makes no sense if nc = a, making those two statements are exactly the same
My teacher just exposed me to archimedean principle, and this lemma. Where do I find problems pertaining to usage of these, Spivak doesn't talk about this, as far as I have come.
I feel like I understand more of the theorem if i do problems using it
my gut feeling is that all of those fundamental inequalities can probably be proved in terms of one another (assuming that you have the one you're assuming in an appropriately general form, e.g. for arbitrary tuples of numbers)
This is Exercise 1.8.4(1) of Springer's, "Linear Algebraic Groups (Second Edition)". It is not a duplicate of
The dimension of $\mathbb P^n$ is $n$
because I'm after a particular perspective; namely, the approach Springer takes (using transcendence degree and not Krull dimension).
The Question:
...
@Tink My favourite quote from that Twitter thread is "Because if ChatGPT is, as it seems to be, a consummate bullshitter, it's also—definitionally—a bullshitter who doesn't know when its bullshitting. And we all know that that's the most dangerous kind." — PM 2RingDec 5 at 17:27
PM i can see the risks in other fields. in early math it might be a useful pedagogical tool. 'spot the error in the proof' type exercises are sometimes too easy, because humans will signal gaps in their understanding with obfuscating language or qualifiers. chatGPT has no qualms. see e.g. chat.stackexchange.com/transcript/message/62510796#62510796
The problem witj ChatGPT on SO (& a few other sites) is people were posting large numbers of answers in small timespans, without bothering to verify that the answers were correct, or even self-consistent. And because it writes with good grammar & sentence structure it bypasses some of our usual crap detection methods.
I doubt we'd see people posting raw ChatGPT answers on Math.SE, but I wouldn't be surprised to see it on Physics.SE. In fact, I suspect I saw it in action there a couple of weeks ago.
I must show that $\lim_{x \to 1^{-1}} \sum_{n=1}^{+\infty} \frac{x^n}{\sqrt{n}}=+\infty$. I tried this: since $\sum_{n=1}^{+\infty}\frac{|x|^n}{\sqrt{n}} \le \sum_{n=1}^{+\infty}|x|^n$ and the latter series is uniformly convergent in any compact $[0,A] \subset [0,1)$, the series $\sum_{n=1}^{+\infty} \frac{x^n}{\sqrt{n}}$ converges uniformly in any compact set $[0,A]$ and so I can exchange the limit and the series when $x \to 1^-$.
Hence $\lim_{x \to 1^-} \sum_{n=1}^{+\infty} \frac{x^n}{\sqrt{n}}=\sum_{n=1}^{+\infty} \lim_{x \to 1^-} \frac{x^n}{\sqrt{n}}=\sum_{n=1}^{+\infty} \frac{1}{\sqrt{n}}=+\infty$. Does this work?
PM: there have been some indications of chatbot posts on MSE already. nothing definitive, as far as i know. and nothing resembling the SO scale.
zawarudo: you can't let x go to 1 without leaving every interval of the form [0,A] with A < 1. this might affect your uniform convergence argument, unless by referring to 'exchange the limit and the series' you are referring to something subtler than what i'm thinking of.
Well, it's good with grammatical structure, so it's not too surprising that it can do decent Latex. Similarly, it usually writes code that doesn't throw syntax errors. And sometimes the code even does the job it's supposed to do. Or it does something quite different. ;)
Howdy, @Xander. I think you scared off that guy with the parabola/envelope. The second post seems also to have vanished — at least, I couldn't find it easily.
This is Exercise 1.8.4(1) of Springer's, "Linear Algebraic Groups (Second Edition)". It is not a duplicate of
The dimension of $\mathbb P^n$ is $n$
because I'm after a particular perspective; namely, the approach Springer takes (using transcendence degree and not Krull dimension).
The Question:
...
@leslietownes thanks for the answer, so the problem is that even if $0<A<1$ is arbitrary it is fixed before taking the limit and so when $x \to 1^-$ it will be $A<x<1$ eventually and so I lose the uniform convergence?
@Shaun I might have said a few things about transcendence degree that seem to be useful, or were useful in similar problems. Likewise for P^n (i.e. stating whether you know and understand the standard affine cover would be helpful). It's tough (for me) to answer a question in which the asker does not state what they know already.
In principle, I totally agree, @Karl. This is why I insist that OPs show me some effort and indicate what's in their toolbox. This is an issue in most areas of mathematics, but very much so in differential and algebraic geometry.
@ZaWarudo Have you tried copying the argument in the proof of Abel's Theorem (normally done with a convergence hypothesis)? The key notion is summation by parts.
@KarlKroningfeld Thank you for the feedback. I tried to supply all I knew but, down to ignorance (but not lack of effort), I just couldn't include enough, evidently.
@TedShifrin thank you Ted, you're like a fortune teller because I remembered some result from series of function that helped in a situation like this but I couldn't remember the result, so thanks for reading my mind) I will try. So your "yes" was referring to my answer to leslie?
@ZaWarudo if you want to go overkill, note that you can apply Beppo Levi wrt the counting measure, because if we define $f_n(x)=x^n/\sqrt{n}$, then $f_n$ is monotonically decreasing on $(0,1]$
How does this solve the following conundrum: Let f be a (sequi) form on finite dimensional vector space V over F. Matrix of f is $(a_{ij}$ w.r.t. some ordered basis $B=\{b_1,...,b_n\}$, where $a_{ij}= f(b_j,b_i)$.
But if f were a bilinear form (say for example F is R), then $a_{ij}:=f(\color{red}{b_i, b_j})$
koro there is no consensus on which side of the sesquilinear form is going to be conjugate linear. they could want the "b_i" part to be the linear part and have the conjugate linearity in the first entry.
i guess i'd want to see a situation where it comes up in a way that feels like it makes a substantive difference. if this is just a definition looking weird, it just looks weird.
sesquilinearity is just like bilinearity, except when it's not.
I'm referring to chapter 9 (where $a_{ij}= f(b_j, b_i)$) and chapter 10 (here arguments of f are reversed to get ij-th entry of its matrix) of Hoffman and Kunze's.
well, the difference will never be substantive, it will just be annoying to remember that you recover a bilinear form via $f(v,w)=v^T A w$ and a sesquilinear form via $f(v,w)=\overline{w}^TAv$, that is all
I mean in our linear algebra class, an orthonormal basis was not actually a basis! because an orthogonal basis was defined to be ordered and of course an ordered basis is not a basis, it just gives rise to a basis because an $n$-tuple is not necessarily an $n$-element set. Does anyone care? no
@AkivaWeinberger Last week, you were distributing points uniformly on a sphere. One way to do that is with a Fibonacci spiral. Here's a demo. The epsilon=0.36 (allegedly) makes the mean closest neighbour distance the most uniform.
This is probably a trivial question, but my search has been fruitless so far, including this wikipedia article, and the "Compendium of Distributions" document.
If $X$ has a uniform distribution, does it mean that $e^X$ follows an exponential distribution?
Similarly, if $Y$ follows an exponenti...
either with the random.exponential () or with inverse transformation method, using the inverse of the CDF of the pdf of the distribution
but both give me nr in the range 0 and 1
I will briefly write what it is written
The lifetimes of particles are measured, which are stopped in an absorber after passing through a detector. The decays are again registered by the same detector via the decay products. The theoretical lifetime of the particles is 𝜏= 2µs, which roughly corresponds to the lifetime of the muon. Due to the overlapping of the detector signals, lifetimes smaller than 𝑡min= 1µs cannot be reliably measured.
The measurement electronics are only active up to the point in time 𝑡max= 10 µs. The number of decays registered in this way is only very small with 𝑁=50 registered events, so that in an unbinned maximum likelihood fit (i.e. all data points are considered in the likelihood function, not just the entries in bins of a histogram) an exponential function should be adjusted to the lifetimes measured in the interval [𝑡min,𝑡max].
Write a function that generates 50 exponentially distributed random numbers in the sensitive detector interval [𝑡min,𝑡max].
@Xander The one at 2:50 CST (my time in central standard time) in the mods' chat. It's okay. And @Ted I'm more than pung out.
@XanderHenderson It's fine. feel free to delete my comments, but please know that many of keep noting such protection of some users, often at the expense of others. We're not all "stupid".
@Xander See also my comment in the cafe. You, among others, have wanted me to depart/give up. You'll have your wish shortly.
This is probably a trivial question, but my search has been fruitless so far, including this wikipedia article, and the "Compendium of Distributions" document.
If $X$ has a uniform distribution, does it mean that $e^X$ follows an exponential distribution?
Similarly, if $Y$ follows an exponenti...
