This question has two parts. On the other hand I’d like to get a clarification what kind of theoretical and logical framework is required to implement a system that has propositional and predicate functionality, but also can determine, if formal sentence is not boolean type at all. Related to that what is needed to determine teoretical wise if given sentence is well formed at all? Hope, this makes sense.
Tag: formula
Cartan’s magic formula for diffferential graded algebra
Algebra $ A$ is called graded algebra if it has a direct sum decomposition $ A=\bigoplus_{k\in\Bbb Z} A^k$ s.t. product satisfies $ (A^k)(A^l)\subseteq(A^{k+l}) \text{ for each } k, l.$
A differential graded algebra is graded algebra with chain complex structure $ d \circ d = 0$ .
Derivation of degree $ k$ on $ A$ means a linear map $ D:A \to A$ s.t. $ $ D(A_j)\subset A_{j+k} \text{ and } D(ab)=(Da)b + (1)^{ik}a(Db), a\in A_i$ $
All smooth forms on $ n$ manifold $ M$ is a differential graded algebra $ \Omega^{\bullet}(M)=\bigoplus_{k=0}^{n} \Omega^k(M)$ , with wedge product and exterior derivative.
In proving Cartan’s magic formula $ \mathcal{L}_X=i_X \circ d + d\circ i_X$ holds for $ \Omega^{\bullet}(M)$ , we can use the following steps:

Show that two degree $ 0$ derivations on $ \Omega^{\bullet}(M)$ commuting with $ d$ are equal iff they agree on $ \Omega^0(M)$ .

Show that $ \mathcal{L}_X$ and $ i_X \circ d + d \circ i_X$ are derivations on $ \Omega^{\bullet}(M)$ commuting with $ d$ .

Show that $ \mathcal{L}_X f = Xf = i_Xdf+ d i_Xf$ for all $ f \in C^{\infty}(M)=\Omega^0(M)$ .
My question:

Why step 1 is ture? Why commuting with $ d$ is so important?

Can step 1 be extended to any derivations without restriction on degree.
Thank you.
How to solve $l^319l^2+96l144=0$ ; without using vieta’s formula
$ l^3 – 19l^2 + 96l – 144=0$ Then??. Without using vieta’s formula
needHelp with Dice rolling formula to determine odds [on hold]
I am creating a game and would like to know the formula to determine the odds of rolling a particular result:
if rolling a 12 sided dice(112) with a +4 to the roll, and a 6 sided dice (that has a different symbol on each of the six sides). What are the odds of rolling a 10 or greater AND rolling on the 6 sided dice a specific symbol that is only on 1 of the 6 sides?
I am hoping to get a formula so I can adjust any of the numbers to determine other odds that use the same mechanics.
something tells me this isn’t that hard of a formula to figure out but I am just pretty poor at math. ðŸ™‚
thank you.
Coarealike formula for BV function (not its derivative)
Let $ f \in BV(\Omega)$ . The coarea formula states that
$ $ Df = \int_{\mathbb R} D \chi_{\{f >h\}} \, dh.$ $
Do we also have that $ $ f = \int_{\mathbb R} \chi_{\{f >h\}} \, dh$ $ holds?
Coarealike formula for BV function (not its derivative)
Let $ f \in BV(\Omega)$ . The coarea formula states that
$ $ Df = \int_{\mathbb R} D \chi_{\{f >h\}} \, dh.$ $
Do we also have that $ $ f = \int_{\mathbb R} \chi_{\{f >h\}} \, dh$ $ holds?
Can Riemann’s explicit formula be generalized to semiprimes?
Following Isometry group of an integer I wonder if one can define a “mock zeta function” $ \zeta_{V}$ (where $ V:=(\mathbb{Z}/2\mathbb{Z})^{2}$ stands for “Vierergruppe”, the German word for the Klein group) whose nontrivial zeros would give rise to the analogue of Riemann’s explicit formula (relating the nontrivial zeros of $ \zeta_{\mathbb{Z}/2\mathbb{Z}}:=\zeta$ to $ \pi(x)=\pi_{\mathbb{Z}/2\mathbb{Z}}(x)$ ) for the counting function of the semi primes $ \pi_{V}(x)$ . If so could the analogue of the Riemann hypothesis for $ \zeta_{V}$ be proven false?
I suspect there might be a relationship between the isometry group of the multi set of non trivial zeros of $ \zeta_{G}$ for $ G\in\{V,\mathbb{Z}/2\mathbb{Z}\}$ (this isometry group being equal to $ V$ if and only the relevant RH is false) and the isometry group of the integers as defined in the link above counted by $ \pi_{G}(x)$ .
To state it differently, would the socalled parity problem be overcome by a proof of RH?
I need to come up with a formula that I could use and replace values
The problem is as follows:
”Letâ€™s say a company ran a prelaunch and generated 5,000 leads on their level 1 over a period of 30 days. Meaning, all of those leads came into the company directly from their own marketing efforts.
But throughout the wellexecuted prelaunch, there were multiple viral components and content released that those 5,000 people shared.
That 5,000 could balloon to 6,000â€¦ then 9,000â€¦ then 13,000, then 20,000â€¦. throughout the launch.
In this example, using the 20,000 number, thatâ€™s 15,000 extra leads that the company had zero acquisition cost.
To drill deeper, letâ€™s say the average CPA (costperacquisition) was $ 4.00 per enrollment into the prelaunch for the company.
The company spent $ 20,000.00 on marketing (5,000 X $ 4.00) to obtain those 5,000 leads.
But thanks to the viral nature of a prelaunch, they had an extra 15,000 leads that came in.
So that brings down their overall cost per lead to just $ 1.33.”
My question is… how did they come up with the result of $ 1.33 CPA.
Thanks in advance for the help!
Hector
Summation formula for this?
I have found the following summation formula based on a recurrence. It supposes $ n = 2^k$ where k is an integer. I’ve intuitively discovered that the following closed form may be true (following the constraint on n), but I’m not sure why.
$ \sum_{\textstyle i=0}^{\textstyle \lg n} {\frac n{2^i}}\ = n\sum_{\textstyle i=0}^{\textstyle \lg n} \frac 1{2^i}\ = n(1+ \frac 12 + \frac 14 +…+ \frac 1n)\ = 2n1\$
I’ve reasoned that the last line should be true because if I plug in n=32 the solution is 63, and if we think about the numbers being added as $ 1$ s in a long bit string, we will end up with lg$ n+1$ ones in a row. I’m wondering if there is a summation formula or inductive proof that can show that this is true? I’m just waving my hands thinking this must be true, but I can’t be sure.
Wu formula for manifolds with boundary
The classical Wu formula claims that if $ M$ is a smooth closed $ n$ manifold with fundamental class $ z\in H_n(M;\mathbb{Z}_2)$ , then the total StiefelWhitney class $ w(M)$ is equal to $ Sq(v)$ , where $ v=\sum v_i\in H^*(M;\mathbb{Z}_2)$ is the unique cohomology class such that $ $ \langle v\cup x,z\rangle=\langle Sq(x),z\rangle$ $ for all $ x\in H^*(M;\mathbb{Z}_2)$ . Thus, for $ k\ge0$ , $ v_k\cup x=Sq^k(x)$ for all $ x\in H^{nk}(M;\mathbb{Z}_2)$ , and $ $ w_k(M)=\sum_{i+j=k}Sq^i(v_j).$ $ Here the Poincare duality guarantees the existence and uniqueness of $ v$ .
My question: if $ M$ is a smooth compact $ n$ manifold with boundary, is there a similar Wu formula? In this case, there is a fundamental class $ z\in H_n(M,\partial M;\mathbb{Z}_2)$ and the relative Poincare duality claims that capping with $ z$ yields duality isomorphisms $ $ D:H^p(M,\partial M;\mathbb{Z}_2)\to H_{np}(M;\mathbb{Z}_2)$ $ and $ $ D:H^p(M;\mathbb{Z}_2)\to H_{np}(M,\partial M;\mathbb{Z}_2).$ $
Thank you!