6

The formula $v = v_0 + at$ assumes that the acceleration is constant. The formula $v = v_0 + \int_{t_0}^{t_f} a(t) \, dt$ allows for the possibility that the acceleration changes with time.


3

The issue is that you assume that the small semicircular integral is $0$. If you integrate over an arc segment with radius $r$ and angle $\alpha$ around a simple pole $z_0$, $$\lim_{r\to 0}\int_{C(r,\alpha)}f(z) dz=\alpha i \mathrm{Res}(f,z_0)$$ See for example this answer. Then if you go around the pole on a semicircle counterclockwise your integral is $\pi ...


3

The modules $126$ and $126^*$ are different. The one has highest weight $2\lambda_4$, the other $2\lambda_5$. The graph automorphism interchanges them. $16$ is different from $16^*$, which are $\lambda_4$ and $\lambda_5$. Indeed, we see $L(2\lambda_4)$ in the symmetric square of $L(\lambda_4)$. The tensor square of a module breaks up as the exterior square ...


2

Yes, there is. Let me start with a very general notion from metric geometry. Let $(X,d)$ be a metric space, $\Gamma$ a discrete group of isometries of $X$ by which I mean that all elements of $\Gamma$ preserve the metric $d$ and if $\gamma_i\in \Gamma$ is any infinite sequence of distinct elements then for all $x\in X$ $$ \lim_{i\to\infty} d(x, \gamma_i x)=\...


2

The second displayed formula you write is correct and standard in exponentiating Lie algebra coproducts into tensor products of group elements: you use it in composing rotation group representations all the time. The first one is nonsense, and you should desist from mindless general manipulations until you are comfortable with what they actually mean. I am ...


2

You integral diverges. Consider the region $u_k \in [(2k-1)\epsilon, 2k\epsilon]$. Explicitly $$ (u_1, u_2, u_3, u_4, u_5) \in G_\epsilon = [\epsilon, 2\epsilon] \times [3\epsilon, 4\epsilon] \times [5\epsilon, 6\epsilon] \times [7\epsilon, 8\epsilon] \times [9\epsilon, 10\epsilon]. $$ Each term in the product is bounded from below $$ \left(2 |k - m| - 1\...


2

In OP's derivation we only have a small beef with OP's order of arguments at one step. It seems a bit unwarranted to a priori assume in eq. (3) that $F_u$ does not depend on ${\bf v}$. After all, the change $\Delta L$ in the Lagrangian under infinitesimal boosts only needs to be a total time-derivative $dF_u/dt$ (so that the infinitesimal boost is a quasi-...


2

The first formula $v=at$ is valid for a constant whereas the second one is valid also for not constant acceleration that is $v=\int_0^t a(u)du$.


1

hint Sans perdre la généralité, we can Assume that $ a<b $ and put $$a=b\cos(t)$$ then $$|a+b|=|b|(1+\cos(t))$$ $$|a-b|=|b|(1-\cos(t))$$ So $$|a+b|+|a-b|=2|b|\le 2c$$


1

Let’s start with the standard sine function: $$f(t)=\sin t$$ If the radius of the wheel is $r$, then to adjust the amplitude, i.e. the farthest the wheel can go from its middle position, you need to multiply with $r$: $$f(t)=r\sin t $$ Now, you need to adjust the time period. Note, the period for $\sin (nx)$ is $\frac{2\pi}{n}$. Set $\frac{2\pi}{n} =60 \...


1

Note that 2 revolutions in 30 seconds works out to 1 revolution every 15 seconds. The question asks the distance traveled in 15 seconds, which due to the previous calculation, works out to exactly 1 revolution. The distance traveled in one revolution is the circumference of the wheel, which is $2 \times \pi \times 10$ m, that is $20\pi$ meter, nearly 62 m.


1

It is known that $$v(t)=f'(t)$$ Assuming that the conditions of FTC are satisfied, we have $$\int_a^bf'(t)dt=f(b)-f(a)$$ or $$\int_a^bv(t)dt=f(b)-f(a)$$ and $$f(b)=\int_\color{red}{a}^bv(t)dt+f(\color{red}{a})$$ we look for $ f(t)$ and we know $ f(0)$, so we take $ b=t $ and $ a=0$. If we have given $ f(1)$, we would get $$f(t)=\int_1^tv(t)dt+f(1)$$ In your ...


1

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