agreed. and you got it. i sometimes write tae instead of tai but it's tai.
around the office some of the younger people would suggest a lunch trip there by asking "hey.. DTF?" the millennials find this funny. i don't know why.
The bottom of the ladder is falling down at a speed of 3m/s. The length of the ladder is 17m. What is the speed of the top of the ladder falling down when the bottom is 8m away from the wall?
i interpreted dx/dt as 3. i probably messed up the numbers.
maybe the ladder is in freefall and both the bottom and the top are falling down. we would need to do a surface integral to figure out the net rotational force on the ladder.
@Thorgott I'm afraid I didn't see why it is separable (am really out of it), but it was quite easy using some equivalent characterisations of purely inseparable field extensions, so I'm content with that
Hello, I have a question. Apollonius showed that a circle can be defined as the set of points in a plane that have a specified ratio of distances to two fixed points. https://en.wikipedia.org/wiki/Circles_of_Apollonius Is there a 3d equivalent for the sphere? A set of points in 3d that have a specified ratio of distances to three fixed points? I tried googling but couldn't find much.
In the context of elliptic curves, what would be a local field complete w.r.t a discrete valuation? I can't seem to find a clear definition of a local field. My guess now is: we have a discrete valuation on our field which induces a metric (and therefore a topology), and w.r.t. to this metric the space is complete. If this is correct: is there a reason why they add that the field has to be local (i.e., would there be examples of non-local fields (whatever that me may) that arise this way?)
Suppose $Z\sim N(\mu,1)$ and $V$ is independent of $Z$ with distribution $\chi^2_m$. Then $T=\frac{Z}{(V/m)^{1/2}}$ is said to have a noncentral $t$ distribution with noncentrality $\mu$ and $m$ degrees of freedom. I want to show that $$P(T\leq t)=2m\int_0^\infty \Phi(tw-\mu)f_m(mw^2)wdw$$
where ...
I think what I said is correct (considering some more notation in the book), and I'll assume that the assumption that the space is locally compact is necessary
Prove that $a^2x^2+ (b^2 + a^2 - c^2)x + b^2 = 0$ does not have real roots if $a+b>c$ and $|a-b|<c$. $a,b,c \in \mathbb{R}$.
Solution I found online:
The discriminant is given as:
$$D = (b^2 + a^2 - c^2)x - 4a^2b^2$$
$$(a^2 + b^2)^2 - 2c^2(a^2 + b^2) + c^4 - 4a^2b^2$$
Let $c^2= t$.
You get : ...
Suppose $Z\sim N(\mu,1)$ and $V$ is independent of $Z$ with distribution $\chi^2_m$. Then $T=\frac{Z}{(V/m)^{1/2}}$ is said to have a noncentral $t$ distribution with noncentrality $\mu$ and $m$ degrees of freedom. We deduced that
$$P(T\leq t)=2m\int_0^\infty \Phi(tw-\mu)f_m(mw^2)wdw$$
where $fm(...
On Page 31 here, while proving the one-dimensional Brunn-Minkowski inequality, they assume that A and B are compact. I read that this is due to the inner regularity of the Lebesgue measure on R^n. Could someone explain please?
there's no such y. if you write sinh in terms of exponentials its clear that sinh y - e^y is always negative.
i mean i guess you're finding slopes of level curves of something, but that equation doesn't define y as a function of x.
there were a number of problems like that in the 'calculus for the business and social sciences' textbook i had to teach out of. not with sinh but same technical problem. the author probably didn't know they were joking.
earth: i dunno, historical accident? many equations just more naturally present themselves in differential form? once you get beyond the 'standard' stuff that everybody who is math-adjacent is forced to take, there is far less of a division between the two.
for many simple families of ODE, the mathematically simplest way of proving existence is by rewriting as an integral equation. but this is simple in terms of theorems and proofs, maybe not in terms of expressions you write down.
the fact that if a function has a nice expression, its derivative likely does too, although its integral might not, may reinforce the bias, at least in introductory classes.
there's this idea that if you don't have an elementary formula for something, it's "worse," creating an impression that differentiation is 'easier' than integration because it is less likely to mess up clean formulas.
by most technical standards, in terms of proving theorems, differentiation is "worse"
d/dx e^(kx) = k e^(kx) is a nice formula. it also shows you that if you put any norm on any function space including all functions e^(kx), d/dx is going to be an unbounded operator, which kinda sucks
also d/dx tends to kick out stuff that isn't necessarily in the domain of d/dx anymore, which also sucks
integration would never$^{\star}$ do this to you
$^{\star}$ Narrator: integration will sometimes do this to you
it bothered me in my early engineering days to look at a forced ode where the input had noise, because surely this is not differentiable. those were nice simple days.
most of the differential equations ive looked at were motivated or came directly from physics where they were reductions of some system based on $F=ma$, but there is something that feels more physical about integrals imo.
@epsilon-emperor You want $\mu(A+B)^{1/n}\geq\mu(A)^{1/n}+\mu(B)^{1/n}$, fix a sequence of compact sets $K_m\subseteq A$ with $\mu(K_m)\to\mu(A)$ and $C_m\subseteq B$ with $\mu(C_m)\to\mu(B)$. Then you prove Brunn-Minkowski for those, getting, for all $m$, $\mu(K_m+C_m)^{1/n}\geq\mu(K_m)+\mu(C_m)$ and you take the limit as $m\to\infty$
@AlessandroCodenotti I forgot a few exponents there, anzway the point is that you get to $\mu(A+B)^{1/n}\geq\mu(K_m+C_m)^{1/n}\geq\mu(K_m)^{1/n}+\mu(C_m)^{1/n}$ and take $m\to\infty$ on the rhs
I am building a statistical model. From the theortical properties of the entity being modelled, two features which when interchanged must lead to an exactly negated prediction. The model however need not meet this requirement but ideally should.
I have decided to call this "anti-symmetry consistency" or "symmetric consistency". I want to know if this is an abuse of existing mathematical terms.
If the input is (x, y) and the prediction is z, the prediction must "ideally" be -z for the input (y, x). I am not sure if there is an existing term for this and whether the names I have come up with for this property are reasonable.
it doesn't seem like an abuse of terms to me. i might call it just "symmetry" (with the precise sense spelled out as you have done here). i say this only because in the abstract, "consistency" sounds like it's something that it is very bad not to have, and if you're willing to consider models that don't have the property, "this does not have the symmetry property" sounds more neutral than that it lacks a form of "consistency."
but this is in the realm of what sounds nice to my ear, not what contravenes established conventions. i don't think there are any.
@leslietownes It's bad to not to have that property. The model needs to respect it to be scientifically valid. However, they don't have to be perfect consistency; some slight deviations are expected but it should be more or less consistent.
okay, you do you. i don't know what is or isn't scientifically valid.
i have the same terminological quibble with 'bias,' in the sense of 'biased estimator.' people treat biased estimators like hot coals. when in many instances they can be just fine. but that terminology is definitely here to stay
i don't know of one. i'm not sure that it's even known whether there is a complexity gap between determining primality and fully factoring an integer as a product of primes. i am also not familiar with the best known methods of each.
how's that for a non answer of an answer.
the literal answer to your question is 'yes,' but all i can think of is "factor, and then collect the even powers of primes," and presumably you want to know if you can skip "factor"
Prove that $a^2x^2+ (b^2 + a^2 - c^2)x + b^2 = 0$ does not have real roots if $a+b>c$ and $|a-b|<c$. $a,b,c \in \mathbb{R}$.
Solution I found online:
The discriminant is given as:
$$D = (b^2 + a^2 - c^2)x - 4a^2b^2$$
$$(a^2 + b^2)^2 - 2c^2(a^2 + b^2) + c^4 - 4a^2b^2$$
Let $c^2= t$.
You get : ...
