Hi. How can I find the Stirling transform of $(k-1)!$ i.e. $\displaystyle\sum_{k=1}^{n}\left(\frac{1}{k!}\sum_{j=0}^{k}(-1)^{k-j}\binom{k}{j}j^n(k-1)!\right)$ ?
@Semiclassical trying to learn probability. I didn't pay attention when they taught product measures in school so I'm doing a lot of "nod and smile like you know what's going on"
i'm not sure what you're getting at. that definition is fine for computing the moment of inertia for the rod. but it's hardly convenient for computing moments of inertia of the circle (besides the case when the axis goes through its center perpendicular)
or, to frame it a little differently: the whole point of the parallel axis theorem is that, if you want to figure out a moment of inertia relative to a given axis $A$, it's enough to know about the moment of inertia relative to another parallel axis $A'$.
2) or you could just take the answer you get for axis 2, and use the parallel axis theorem to turn that into the answer for the axis being at the bottom of the handle
for a good practice problem in moments of inertia: take a sphere of known mass and volume, and drill a hole through it parallel to the $z$-axis. then try to figure out the moment of inertia relative to the $z$-axis.
if that's too complicated, use a flat disk instead of a sphere.
ok I think I am gonna also use marion and thornton as something with me like more refrence I guess I got 90 in the midterm,because of algebra mistake but final I would like to crush it so it would be nice if I understand stuff inside out.
physicists use it without realizing it in some places (thermodynamics, for example). in other places it'd be a lot better if they did (electromagnetism)
If you think of $I$ as a linear map, you apply it to the vector. But it's really a bilinear form, not a linear map, and then you have to turn it into a linear map to feed it a vector. This is also why physicists distinguish between vectors and pseudovectors.
the whole vector/pseudo-vector point is one which i haven't had a whole lot of contact with myself. but that's largely due to my avoiding a particle physics route
A device is statically balanced if the center of mass lies on the axis of rotation so if it that is not the case then there will be gravity acting on it so it will cause outward force if it is statically balanced
@Semiclassic, pseudovector (like a cross product) is something that's really a 2-vector and then you use $\Lambda^2\Bbb R^3\cong \Bbb R^3$ to think of it as a vector.
what's confusing to me is that "if [a device is not statically balanced] then there will be gravity acting on it so it will cause outward force if it is statically balanced"
It was never mentioned in the physics courses I took, @Semiclassic, but it comes up in our junior/senior level E&M class, I'm told by my friend who teaches it.
ok let us say that a device is not statically balanced then that means center of mass doesn't lie on the axis of rotation what movements will happen on the object ? @Semiclassical
i didn't see it in undergrad physics. did in grad, though
here's the dfn Wikipedia gives: "Static balance occurs when the centre of gravity of an object is on the axis of rotation. The object can therefore remain stationary, with the axis horizontal, without the application of any braking force. It has no tendency to rotate due to the force of gravity."
there's a line after that: " This is seen in bike wheels where the reflective plate is plated opposite the valve to distribute the centre of mass to the centre of the wheel."
I'll admit I'm not the best at mechanics, of course, but at least in my limited experience, $L$ and $\omega$ have always been parallel whenever I've dealt with them.
@TedShifrin I haven't seen the Hessian used in that quadratic manner before as in lecture 48. That's really cool! I'm not used to the quadratic form approach yet, but I am starting to get the hang of it. Nifty stuff!
@anon: I edited my answer to clarify what's wrong with your attempted proof. It's (unless I missed something) correct. The problem is just that $\text{Homeo}^+(\mathbb R^2)$ should not be simply connected!
@TedShifrin In lecture 50, you mention the defintion that two vectors are parallel when you can obtain the other via scalar multiplication. In particular, $\vec{\nabla}f = \lambda \vec{\nabla} g$. I don't quite follow. Suppose the vectors in question differ by an affine transformation? They can still be parallel but they can't be equated by scalar multiplication...how does your definition of parallel account for this?
@TedShifrin: I just did it by slicing horizontally instead of vertically and it works. But the calculation is a lot more complicated because then I need to know the parabolas along the triangle. I'll try another way in a bit.
@StanShunpike Because it is standard to view vectors as equivalence classes of "arrow thingies" in affine space, where two are equivalent iff they have the same length and direction (when defined).
Thus, the translation component of any affine transformation has no effect on vectors
@BalarkaSen, why do we prove that well ordering implies principle of induction for $\mathbb N$, but in Peano Axioms principle of induction is already assumed?
@TobiasKildetoft, i mean by induction A subset of the natural numbers with 0 in it, such that it has the successor of every number in it, is the same set as the natural numbers.
For example, most of the comments on my post there which just say something like "nice post" are really just there because the name of the commenter is a link to a website hoping to boost their search rankings
@Soham $\mathcal{C}([0, 1])$ be the set of all continuous functions from $[0, 1]$ to $\Bbb R$. Prove that this a metric space equipped with $d(f, g) = ||f - g||$, where $||f|| = \sup_{x \in [0, 1]}\{|f(x)|\} < \infty$ (prove that it even make sense).
@SohamChowdhury I don't know a lot of combinatorics. I am only vaguely familiar with combinatorial number theory, and some Ramsey theory I picked up from lectures of Kaj.
