No, I don't mean a quotient group. So in this exercise there are lots of S over F, where whether S over F is a vector space or not and then dimension is also to be stated e.g.: is Q over R is not a vector space (not closed under scalar multiplication), C over R is two dimensional vector space etc.
OK. then the answer is a quick no. there's no "vector space" over something that isn't a field.
it's still a no if you're asking if there's some way of defining a scalar multiplication on Q/Z that would turn it into a vector space over Q, but that's a different story.
it satisfies all of the various axioms that it would need to, with multiplication defined in the obvious way, but Z fails to satisfy the condition that it is a field. so Q is a Z-module (as is any abelian group) but not a vector space over Z.
While thinking about the relation:
$$\zeta(s)\Gamma(s)=\int_0^\infty \frac{x^{s-1}}{e^x-1}~dx $$
I started with a Mellin transform on bounded support:
$$ f(s)= \int_0^1 x^{s-1}\bigg(-\frac{1}{e^{\frac{1}{\ln(x)}}-1}-1\bigg)~dx $$
Then the integrand can be rewritten using a series:
$$ =\su...
consider B the linear map given by (x,y,z,t) -> (3x, y, z, t, 0) and A the linear map given by (a,b,c,d,e) -> (a,b,c,d). det(AB) is, as xander foretold in the prophecies, 3.
One puzzling question that I'd like to share: If A is a 4 by 5 matrix and B is a 5 by 4 matrix then what is the determinant of BA? (assume that all entries of A and B come from $\mathbb C$)
(corrected version of the question just for the record).
Quick question.... I've done the majority of this, but I realized I didn't show the orthonormal part. So after all the other stuff is done, specifically showing that it is orthogonal I'll have this:
I have shown it is orthogonal, but to show it is orthonormal, I would have to say let $v_i = \frac{v_i}{\|v_i\|^2}$. But I don't know that about the vector.
the inner product of v_1 with itself is 1 because v_1 = 1 v_1 + 0 v_2 + ... + 0 v_n and so from the definition of the inner product, v_1 innerproduct v_1 is 1*1 + 0*0 + ... + 0*0
Now to the thought I was having after I had done this.......After proving this I am saying that under any real or complex vector space with the standard inner product all bases will be orthonormal?
dc3rd: there's no "standard inner product" on an an arbitrary vector space, although if you choose a basis you can use this recipe to come up with an inner product
you see this even outside of the inner product context. you can think of elements of an n-dimensional vector space over F as n-tuples of elements of F, after choosing a basis
math language spares no space for clutter....so compact and to the point.
also thanks for pushing geometric intuition on to me. In one of my stats classes we were talking about orthogonal contrasts and even though it is not entirely crisp to me, I have been visualizing how the linear algebra ideas you discuss are being used. And it is revealing.
not fully there yet, but I see how the ideas are being used.....has got me excited.
From the moment I started working with your text every time I see a concpet I've always tried to "picture" it in some sense. It really gives you an idea of what it is you're trying to accomplish in simple terms....then you can fill in the details....
consider the set $$\{(e^t \cos(t), e^t \sin(t)) \in \mathbb{R}^2: t\leq 0\}$$
I want to show the set is not bounded.
I already showed the case of $t > 0$ not being bounded, by just letting my set be bounded by $t$ and taking a sequence of the terms which would simplify to $e^{2t} > t$. How do I state using the $t$'s properly to allow me to use the same argument? Can I "say" take the a decreasing sequence of integers $\{-1,-2,...\}$?
@AkivaWeinberger Singular means its not absolutely continuous.
Continuous means singletons have no mass.
Every $\sigma$-finite measure can be decomposed into a discrete (every point has nonzero mass), an absolutely continuous, and a singular continuous measure.
@LearningCHelpMeV2 I would approach this by working with two cases: $\sin(x-\alpha)=\sin(\beta)$ and $\sin(x-\alpha)=-\sin(\beta)$ and compute $\sin(x)$ and $\cos(x)$ from there.
If $x_i(t)$ $i=1,..,n$ ($n=2,3$) regular functions($k$ times continuously differentiable functions) and $\sum_{i=1}^n(x_i^\prime(t))>0$ $n=2,3,t\in(a,b)$ then
$x_i=(x_i(t))$ is equation for regular cuve.
Now proof starts with saying that we need only to prove that it is local bijective map.But w...
My most recent question was quite poorly received with a score of -2/0, and while I don't really care about math.se rep and I got good useful answers I would like to avoid asking bad questions in the future. However I'm not sure what was wrong, with the question. Was it that I made a mistake in my own attempt to solve the problem? Or is this some topic I shouldn't ask about because it's been done to death or something. I'm not really sure how to improve.
What is wrong with this? A=(a,$r_1,r_2,r_3), B=(b,r_1,r_2,r_3)$, where a,b and $r_i$'s are column vectors in $\mathbb R^4$ then det(A+B) is to be found. It is given that det A= 4 and det B=1. It is clear that b=$c_1a+\sum c_ir_i$, where $c_1\ne 0$. It follows that det B= det ($b,r_1,r_2,r_3)= det(c_1 a+\sum c_i r_i, r_1,r_2,r_3)=det (c_1 a, r_1,r_2,r_3)=c_1 (a,r_1,r_2,r_3)=4c_1=1$ so $c_1=\frac 14$. $det (A+B)= det((1+\frac 14)a,2r_1,2r_2,2r_3)=\frac 54 \times 2^3=10$.
@WheatWizard I don't get it, looks like a fine question to me. some people appear to downvote questions that they don't find interesting, or think are too 'easy' to be worth asking, even if they are asked well. it's weird that nobody explained their downvote in comments.
If $x_i(t)$ $i=1,..,n$ ($n=2,3$) regular functions($k$ times continuously differentiable functions) and $\sum_{i=1}^n(x_i^\prime(t))>0$ $n=2,3,t\in(a,b)$ then
$x_i=(x_i(t))$ is equation for regular cuve. Why here we don't need to prove that inverse of a map is continuous?
do you mean to have a subscript i on both sides of $x_i = (x_i(t))$? what is your definition of "regular curve"? do you have some form of the inverse function theorem?
it seems like there's a proof of something being omitted here, and you're worried about a missing step. what steps are given?
123: if you travel 4 units along a line in one direction, and then turn around and travel 9 units along the same line in the opposite direction, how many units did you travel
pretty sure the total displacement on the graph there isn't 3.9, also, but let's ignore that
my book defines regular curve as a curve when every γ(t) point has neighbourhood on which part of a curve can be regular parametrized by $γ(t)=x_i(t)$
Author mentions that we need only prove that map is locally bijective
break your time interval into intervals on which the motion is only in one direction. compute displacement on each of those intervals and add them together.
this can involve difficult computation if the equation of motion is complex, because it will depend on the zeros of the derivative of that thing (which might have really ugly formulas)
if x: [a,b] -> R describes the position at a time t and x'(t) has finitely many zeros t_1 < t_2 < t_3 < ... < t_n on [a,b], let t_0 = 0 and t_{n+1} = b and compute $\sum_{i=1}^{n+1} |x(t_i) - x(t_{i-1})|$. in your picture example, we have $t_0 = 0$, $t_1 = 2$, $t_2 = 5$ and get $|x(2)-x(0)| + |x(5)-x(2)| = |-3-1| + |6 - (-3)| = 4 + 9$ as above.
the point of a problem like this is probably to get you to read the graph and not work in terms of some general formula
the shape of the graph is completely irrelevant to this, other than that it doesn't change direction between t = 0 and t = 2, and between t = 2 and t = 5.
the real red herring would be to give you a formula for how position depends on time. then some people would apply the arc length formula and compute the arc length of the graph in the time-position plane. this is not a distance traveled (as you could see by noticing for example that if the person never went anywhere, the graph in the time-position plane is a horizontal line which has an arc length that grows with time although the person never moves)
xander: you were recommending those catci, right? also, which one of us is talking?
@leslietownes I do dislike the downvotes to an answer that I think is good and that give no explanation. It is their right to downvote, but it is not helpful if there is no comment to indicate what needs to be fixed.
There were a lot of interesting rituals around mushrooms in the New World.
The one I find most comment worthy is an Inuit tradition in which a shaman (who has eaten a lot of mushrooms in his day and therefore has some resistance) would eat shrooms, and others would drink his urine to get the hallucinogenic effect.
One of my favorite scenes in the Frisco Kid (Gene Wilder playing a rabbi, en route to California, with a guide, young Harrison Ford. He teaches a Jewish dance to a tribe gathered around a fire, after smoking/drinking some hallucinogenic.
I generally work with geometric zeta functions, but the spectral zeta function, under appropriate hypotheses, can be obtained from the geometric zeta function (via Mellin transform).
@amWhy That movie deserves to be better known. Here's a nice review: forward.com/culture/349138/… "I don't want to hurt you. I just want to make you kosher."
While thinking about the relation:
$$\zeta(s)\Gamma(s)=\int_0^\infty \frac{x^{s-1}}{e^x-1}~dx $$
I started with a Mellin transform on bounded support:
$$ f(s)= \int_0^1 x^{s-1}\bigg(-\frac{1}{e^{\frac{1}{\ln(x)}}-1}-1\bigg)~dx $$
Then the integrand can be rewritten using a series:
$$ =\su...
