Monday, July 5, 2021

Recent Questions - Mathematics Stack Exchange

Recent Questions - Mathematics Stack Exchange


Easy to read books for Weak Topologies

Posted: 05 Jul 2021 08:10 PM PDT

The only reason that I could learn first three chapters of Conway's Functional Analysis book was that I had studied Kreyszig's Functional Analysis book before that! Unfortunately, Kreyszig's book didn't include contents of chapter 5 of Conway's book so I have find an easy to read book/online-resource for weak topology to be able to understand chapter 5 of Conway's book easily.

Content of Chapter 5 :

enter image description here

Any suggestions?

Not sure where my error is for integrating a sequence of functions

Posted: 05 Jul 2021 08:09 PM PDT

We have a sequence of functions on the domain $x\in [0,1]$ which looks like this: $f_n(x)=(1+x^2+\frac{x^4}{n})^{-1}$. From my understanding, if $f_n$ is Riemann integrable on our domain and the sequence converges uniformly then $\lim_{n\to\infty}\int_0^1f_n(x)dx=\int_0^1f(x)dx$ where $f(x)=lim_{n\to\infty}f_n(x)$. I then broke it up into cases to check for uniform convergence and got that for $x=0$ we see that the sequence of functions converges to $f_n(x) \to 1$. Then through a similar process, I checked that for $x=1$ that $lim_{n\to\infty}f_n(x)=\frac{1}{2}$. Then to check anything in between I checked $x=\frac{1}{n}$ and got that $lim_{n\to\infty} f_n(x)=1$. Did I do something wrong? Because this looks like our limit function is discontinuous so I'm not sure how I would go about integrating it? I feel like I must have done something wrong so any guidance on how to complete this integral would be great! Thank you :)

Truncation of Laurent series

Posted: 05 Jul 2021 08:08 PM PDT

Let $L$ be the $\Bbb C$-algebra consisting of Laurent series $ \sum_{i\geq m} a_i z^i $ whose principal part has only finitely many terms, and let $L'=\Bbb C[z,z^{-1}]$ be the subalgebra of Laurent polynomials. For an integer $n$, let $T_n:L\to L'$ denote the truncation map defined by $\sum_{i\geq m}a_i z^i \mapsto \sum_{i= m}^{n-1} a_iz^i$. For $f(z)=\sum_{i\geq m} a_iz^i\in L$ with $a_m\neq 0$, define $m$ to be the order of $f(z)$.

Now fix two $f_1(z)=\sum_{i\geq m_1}a_i z^i, f_2(z)=\sum_{i\geq m_2}b_iz^i \in L$ with $a_{m_1}\neq 0\neq b_{m_2}$ and $f_1+f_2\neq 0$. Thus the order of $f_i$ is $m_i$, and the order $m$ of $f_1+f_2$ is $\geq \min\{m_1,m_2\}$. Fix a Laurent polynomial $r(z)=\sum_{i= l}^n c_iz^i$. How can we show that $$ T_{n-k}\circ T_{n+1+m}((f_1+f_2)r) = T_{n-k}\circ T_{n+1+m_1} (f_1r) + T_{n-k}\circ T_{n+1+m_2} (f_2r) $$ where $k$ is a nonnegative integer? (This is needed to prove linearity of the map defined in the proof of Lemma 3.4, p.189 of Miranda's book Algebraic Curves and Riemann Surfaces. Since the truncation map is linear, it suffices to show that $$T_{n-k}\circ T_{n+1+m}(f_ir)=T_{n-k}\circ T_{n+1+m_i}(f_ir) $$ for $i=1,2$. This is certainly clear if $k$ is sufficiently large, but I can't see this in general.

for ring $A$ and prime numbers $p,q$, $p=q=0\in A$, then $p=q$?

Posted: 05 Jul 2021 08:06 PM PDT

Sorry for my bad English.

Let $A$ be commutative ring with 1, and $p,q\in \mathbb{N}$ be prime numbers.

If $p=q=0\in A$, then $p=q\in \mathbb{N}$ ?

I can't proof and can't make counter example, so please tell me solutions or hints.

$\frac{\partial ^2}{\partial x^2}\phi(x,y,k) + \frac{\partial ^2}{\partial y^2}\phi(x,y,k) = -k^2.\phi(x,y,k)$

Posted: 05 Jul 2021 08:04 PM PDT

Any idea of how can I solve the differential equation $\frac{\partial ^2}{\partial x^2}\phi(x,y,k) + \frac{\partial ^2}{\partial y^2}\phi(x,y,k) = -k^2.\phi(x,y,k)$? I thought about variable separation but I don't know what would be the solution.

Problem in a Trigonometric Series

Posted: 05 Jul 2021 07:59 PM PDT

We have to prove that $\frac{1}{cosx+cos2x}$+ $\frac{1}{cosx+cos5x}$+$\frac{1}{cosx+cos7x}$ till n terms is cosecx[tan(n+1)x-tanx]. Now ive tried solving it by writing applying cosa+cosb formula but that didnt lead me anywhere. We definitely have to form a Telescoping Series. Now if anyone has any idea how to do so I would really appreciate it.

For what primes $p$ and positive integers $k$ is this algebraic expression divisible by $3$?

Posted: 05 Jul 2021 07:59 PM PDT

My initial question is as is in the title:

For what prime $p$ and positive integers $k$ is this algebraic expression divisible by $3$? $$A(p,k):=\dfrac{p^{2k+2} - 4p^{2k+1} + 6p^{2k} + 2p^{k+1} - 8p^k + 3}{2(p - 1)^2}$$

I would like to qualify that I am specifically interested in those values for $p$ and $k$ satisfying the congruence $p \equiv k \equiv 1 \pmod 4$.

MY ATTEMPT

Here, I will evaluate my expression for $$p_1 = 5, k_1 = 1,$$ which gives $A(p_1,k_1)=9$ (which is divisible by $3$), $$p_2 = 13, k_2 = 1,$$ which gives $A(p_2,k_2)=73$ (which is not divisible by $3$), and $$p_3 = 17, k_3 = 1,$$ which gives $A(p_3,k_3)=129$ (which is divisible by $3$).

Here is my final question:

Will it be possible to consider the cases $p \equiv 1 \pmod 3$ and $p \equiv 2 \pmod 3$ separately, then use the Chinese Remainder Theorem afterwards? (I know the concept, but I would have forgotten how to do that.)

CONJECTURE: If $p \equiv 2 \pmod 3$, then $3 \mid A(p,k)$.

Alas, this is where I get stuck.

Find the shortest distance between straight line joining points A(3,2,-4) and B(1,6,-6) and straight line joining points C(-1,1,-2) and D(-3,1,-6).

Posted: 05 Jul 2021 07:56 PM PDT

The complete question is below because the world limit doesn't allow me to post full question in title.

Find the shortest distance between the straight line joining the points A(3,2,-4) and B(1,6,-6) and the straight line joining the points C(-1,1,-2) and D(-3,1,-6). Also find equation of the line of shortest distance and coordinates of the feet of the common perpendicular

I have found the shortest distance by two methods

  1. by determining that lines are skew and proceeding further on e.g. taking dot product with direction vectors and then by finding s and t. Then I found the direction vector and thus by finding its magnitude of that direction vector I got the shortest distance

  2. by finding equation of parallel planes and by solving them using distance formula between two planes

In both cases, I got shortest distance equals to sqrt(21). [is it right]

But my real query is how to find equation of the line of shortest distance and coordinates of the feet of the common perpendicular. Please help.

Evaluating $\mathrm{\int_0^1 ?(x)dx}$

Posted: 05 Jul 2021 08:10 PM PDT

This function interestingly shown as ?(x) is dubbed the Minkowski Question Mark Function. It looks very similar to x. Wolfram Alpha can even plot the derivative of this apparently smooth function. Here are some details for what I want to find without using trivial solutions like the Riemann Sum definition nor $n^{th}$ derivatives in the answer from Taylor Series.$ \ a_k$ represents a continued fraction of x=$[a_1,…,a_k]$ as seen in the first top link.

$$q=\int_0^1 ?(x)dx=-2\int_0^1\limits \sum_k \frac{(-1)^k}{2^{\sum_{n=1}^ka_n}}dk≈.5$$

This brings up the property of $$?(x^{-1})=\frac1{2^{x-1}}\implies ?(x)=2^{1-\frac1x}, x=k^n\mathop \implies ^? q\mathop =^?\int_0^1 2^{1-\frac1x}=2\ln(2)\mathrm{Ei}(-\ln(2))+1=.475050468…≈.5$$

This result seems correct for a closed form with the exponential integral function and the graph looks the same, but smoother.

Here is what the constant looks amazingly like: enter image description here

It would be very nice to find a non-integral form of this constant which is not a trivial solution like the one described in paragraph one. This function does not looks smooth, but that does not mean we cannot integrate it. What is a better, closed form or not, alternative representation or form for the area under Minkowski's Question Mark function as defined over the unit interval $x\in$[0,1]? Please correct me and give me feedback!

How to sample this subset of $\mathbb{Z}^6$ uniformly

Posted: 05 Jul 2021 07:46 PM PDT

Let $X \subset \mathbb{Z}^6$ be the set of all $(a_1, a_2, a_3, a_4, a_5, a_6)$ with all $0 \leq a_i \leq 63$ and $\sum a_i \leq 127$. The constants $63, 127$ are mostly arbitrary.

I am trying to come up with a fast method to sample this space uniformly. I have tried multiple schemes but my intuition and sample histograms tell me I haven't got it

My latest thought was taking an in progress selection, say $(a_1, a_2)$, and selecting $a_3$ from the suitable integer range, and selecting in proportion to the number of points of $X$ that start with $(a_1, a_2, a_3...$. But calculating this on the fly is difficult. The upper bounds on the $a_i$ make a closed form difficult to reach

Difference in gradient magnitude for different inputs

Posted: 05 Jul 2021 07:43 PM PDT

I have generated saliency maps of few images using the gradient of output with respect to input. I noticed that if I add noise to the images, the summation of magnitudes of saliency maps is higher than the original images. My question is if the mean or summation of magnitudes is high, what information I can gain from that?

What is the precise definition of $\lim_{x\rightarrow a^-}f(x) = -∞$?

Posted: 05 Jul 2021 07:52 PM PDT

In particular, what are the precise definitions of infinite one sided limits?

Problem in understanding the dictionary order topology

Posted: 05 Jul 2021 07:40 PM PDT

The question is from Munkres:

Find the closure of the set $A=\{1-\frac{1}{n} \times 0 \}$ in the ordered square.

I can see from the answer given herewhat the closure will be.

$p=\langle 0,1\rangle$, and suppose that $U$ is an open nbhd of $p$. Then there are $u,v\in X$ such that $p\in(u,v)\subseteq U$, where $(u,v)$ is the open interval between $u$ and $v$ in the order topology. This means that $u\prec p\prec v$. If $u=\langle a,b\rangle$ and $v=\langle c,d\rangle$, we have

$$\langle a,b\rangle\prec\langle 0,1\rangle\prec\langle c,d\rangle\;,$$

which implies that $a=0,b<1$, and $c>0$. If you make a sketch, it should be clear.

However my problem lies in understanding how will my sketch look like and how do I understand that this occurs, $$(u,v)=\big(\{0\}\times(b,1]\big)\cup\big((0,c)\times[0,1]\big)\cup\big(\{c\}\times[0,d)\big)\;.$$

Using burnside's lemma to calculate a smaller subset of unique, color-agnostic bracelets

Posted: 05 Jul 2021 08:11 PM PDT

We have a child's toy, which is a ball made of 12 colored wedges (3 Red, 3 Green, 3 Blue, 3 Yellow). Our child asked the sensible question 'how many different patterns are possible?'.

In researching the answer, I discovered it's a mathematically complex and interesting question! My math level is advanced high school/entry level university - sufficient to determine that Burnside's lemma is the right tool and to more or less understand its application to this problem. This is essentially a 'bracelet' problem, as I understand it. But I would like to solve not for the total unique bracelets, but for a smaller subset of unique arrangements that are color agnostic (I'll define this here as 'unique patterns', lacking a more accurate term). E.g. -RRY-GGG-BBB-YRY- is considered the same arrangement as -GGB-RRR-YYY-BGB-. The pattern is the same even though the colors are switched.

From my research, my starting point was a similar related problem asked at the link below and answered very helpfully by Marko Riedel:

Necklace problem with Burnside's lemma

As Marko showed using Burnside's lemma, the total arrangements of that 6 bead 'bracelet' (or a simpler toy of 6 wedges of three colors) would be 11. But the smaller subset of 'unique patterns' which are color agnostic are 5. It was easy enough to identify those 5 manually, but what I am interested in is how Burnside's lemma could be correctly applied to find the unique (color agnostic) patterns possible on this more complex toy of 12 wedges in 4 colors.

My starting point was to calculate the higher amount of all possible 'bracelets'. The identity gives 12! / 4x3! = 369,600. And unless I'm mistaken, the reflection symmetries yield no candidates and the only rotation symmetries which do are by 4 and 8 spaces (giving 24 each). This gives:

(369,600 + 24 + 24) / 24 = 15,402 possible bracelets.

This gives me an upper limit, but how would I then reduce this to the subset of 'unique patterns' which are agnostic as to color placement, as it's not feasible to calculate this manually.

Any advice would be greatly appreciated by this math curious family!

(note: resubmitted with additional context as requested by moderator to my first question)

Dynamic thresholds for EWMA - How?

Posted: 05 Jul 2021 07:32 PM PDT

The following paper talks about the benefits of adding a probabilistic component to an exponentially weighted moving average. However my confusion begins with the plotting of the dynamic thresholds.

With the algorithm being: PEWMA - Algorithm 1

With the only info provided (as far as I can see) for the dynamic thresholds being:

"We are able to estimate a local standard deviationˆσt+1 as described in Algorithm 1, and use this parameter both for the evaluation of Pt and for setting detection thresholds"

Am I fundamentally missing something?

Laplace operator on $H^{-1}$

Posted: 05 Jul 2021 07:31 PM PDT

Let $\Omega$ be a domain in $\mathbb R^n$ (say, with smooth boundary for simplicity). I am looking for a reference for the fact that if $f \in H^{-1}$ (that is, the dual of $H^1 = W^{1,2}$) then, there is a weak solution $u \in H^1$ of $\Delta u = f$ in $\Omega$, $u = 0$ on $\partial \Omega$. In order words, the operator $\Delta:H^1_0 \to H^{-1}$ is an isomorphism.

I have found in many books the equivalent statement for $\Delta:H^1_0 \cap H^{k}\to H^{k-2}$ for $k \geq 2$, but not for $k=1$. Theorem 2.44 in these lecture notes gives what I want, but I would like to have a proper reference from a book/paper that I could cite. If you know of a reference dealing with the analogous question on compact manifolds, it would be even better.

Thanks in advance!

$\int_{0}^{1}f(x)e^{nx}dx=0$ $\forall n \in N\bigcup{\{0\}} \implies f(x) = 0$

Posted: 05 Jul 2021 07:33 PM PDT

$f$ is a continuous real valued function $f: [0,1] \to R$ and $\int_{0}^{1}f(x)e^{nx}dx=0$ $\forall n \in N\bigcup{\{0\}} \implies f(x) = 0$ on the interval $[0,1]$

I am trying to prove this result. I have tried using Stone-Weirestrass with both polynomials and exponential functions; however, I cannot seem to prove it this way. Can someone offer a hint or suggestion.

Determine whether the linear transformation T is one-to-one, onto or neither.

Posted: 05 Jul 2021 07:38 PM PDT

Determine whether the linear transformation $T$ is one-to-one, onto or neither.

$T$ : P → P defined by $T(p(x)) = p'(x)$

How to prove this property about a measurable function f

Posted: 05 Jul 2021 07:42 PM PDT

Let $\mathcal{A}$ be a $\sigma$-algebra in $X$ and $f:X \to \mathbb{R}^N$ a measurable function. Prove that $G\subset \mathbb{R}^N$ open $\implies f^{-1}(G) \in\mathcal{A}$.

My attempet: If we write $f(x)=(f_1(x),\dots, f_N(x))$ where $f_i:X \to \mathbb{R}$, we have that $$ f_i=P_i \circ f$$ where $P_i:\mathbb{R}^N \to \mathbb{R}$ is the projection of the $i$-coordinate. Since $P_i$ is continuous and an open map it's true that $P_i(G)=A \subset \mathbb{R}$ open and then follows that $$f_i^{-1}(A)=f^{-1}(P^{-1}_i(A))). $$ By the description of the open sets in $\mathbb{R}$ we can write $A$ as an disjoint countable union of open intervals and maybe with some work we can prove by this that $f^{-1}_i(A) \in \mathcal{A}$. The real problem here is that it's not true that $P^{-1}_i(A)=G$. There is a way to make this argument work?

Closed kernel implies linear operator to finite-dimensional space is continuous

Posted: 05 Jul 2021 08:00 PM PDT

First I'd like to say that although I have seen similar questions here, all answers I could find used the quotient space, which we have not seen in class. Apologies if it has already been answered without using quotient spaces and I didn't find it. The result I want to prove is the following

Let $E$ and $F$ be normed spaces, with $F$ finite-dimensional, and let $T: E \to F$ be a linear operator. Show that if $\mathrm{ker}(T)$ is closed then $T$ is continuous.

We have already proved it holds for a functional $\phi: E\to \mathbb{R}$ or $\mathbb{C}$ so I thought maybe I could extend that result but couldn't quite see how.

Polynomial vector as ordered set notation

Posted: 05 Jul 2021 07:32 PM PDT

I'm wondering if it's correct to write a polynomial in linear algebra with ordered set notation.

I have seen some matrix written as:

$$ \begin{bmatrix}a & b \\c & d \end{bmatrix} = (a, b, c ,d) $$

So I thought:

$$ a x^{2}+ bx + c = (a, b, c) $$

Can I do it?

Final number after a skip-delete algorithm acting on a circular seating of numbers

Posted: 05 Jul 2021 07:53 PM PDT

There are 128 numbers 1, 2, . . . , 128 which are arranged in a circular pattern in clockwise order. We start deleting numbers from this set in a clockwise fashion as follows. First delete the number 2, then skip the next available number (which is 3) and delete 4. Continue in this manner, that is, after deleting a number, skip the next available number clockwise and delete the number available after that, till only one number remains. What is the last number left ? Source

Some examples I've calculated by considering shorter number strings: $$ 1,2 \to 1$$ $$ 1,2,3 \to 3$$ $$ 1,2,3,4 \to 1$$ $$ 1,2,3,4,5 \to 3$$ $$ 1,2,3,4,5,6 \to 5$$ $$ 1,2,3,4,5,6,7 \to 7$$

$$ 1,2,3,4,5,6,7,8 \to 1 $$ $$ 1,2,3,4,5,6,7,8,9 \to 3 $$

I can't see any pattern in the number the string gets shot down to. Any how I tried a brute-forcish approach by doing the deletion in cycles (everytime I look back to the beginning numbers, I put an arrow):

$$ 1,2,3,4....,(128) \to 1,3,5,7....(127) \to 1,7,11 ..(??)$$

Quickly I figured that this is NOT the approach.


A bit later on, I thought of a more 'generating functionish' approach:

I thought of considering the number as a polynomial $p(x) = x^1 + x^2...+x^{128}$ then I thought maybe if I could find an operation which captures the 'deletion' process as described by the question onto this polynomial, then I would have solved the question. However, I could find only find such a process for the 'first sieve' i.e: the first time we delete till the end and reach back at the starting numbers. It is given as:

$$p_1(x) = \frac{p(x)+p(-x)}{2}$$

Annnnnd I am out of ideas

Asymptotic Difference of Binomial Expansions

Posted: 05 Jul 2021 08:05 PM PDT

Given $x\rightarrow\infty$, I know that $(1-\frac{2\ln2}{x})^{x-2}-(1-\frac{\ln2}{x})^{2(x-1)}$ goes to zero because \begin{align*} \Big(1-\frac{2\ln2}{x}\Big)^{x-2}-\Big(1-\frac{\ln2}{x}\Big)^{2(x-1)} &\sim\exp\Big(-\frac{2\ln2}{x}\cdot x\Big)-\exp\Big(-\frac{\ln2}{x}\cdot 2x\Big) \\ &=0, \end{align*} where I used $\sim$ to mean asymptotic equivalence. How can I find the specific scaling (i.e., the rate of convergence) of the difference?

From the plots I made, it seems to me that it should be something like $O(\frac{1}{x})$, which can be further verified with Wolfram alpha's Big-O domination calculator -- the result is \begin{align*} \Big(1-\frac{2\ln2}{x}\Big)^{x-2}-\Big(1-\frac{\ln2}{x}\Big)^{2(x-1)} &=\frac{\ln4-\ln^22}{4x}-O\Big(\frac{1}{x^2}\Big) \\ &=O\Big(\frac{1}{x}\Big). \end{align*} I tried using Binomial expansions and Taylor expansions but had trouble simplifying. Is there a simpler way to do this (without having to resort to Laurent series which Wolfram claims)?

Inequality involving sums of exponentiated values

Posted: 05 Jul 2021 07:38 PM PDT

Given arbitrary values $x_1,\dots,x_n\ge 0$, is it necessarily the case that $$ f(p) := \frac{\left(\sum_i x_i^p\right)^2}{\sum_i x_i^{2p}} $$ is decreasing in $p$ for $p\ge 0$? As a sanity-check, $f(0)=n$, and $f(p)\le n$ for all $p$.

How to prove all eigenvalues of $AS+(I-S)$ are with positive real parts?

Posted: 05 Jul 2021 07:36 PM PDT

$A$ is a symmetric and positive definite matrix. $S$ is a diagonal matrix with elements equal to 1 or 0. $I$ is an identity matrix. I've done 10 million numerical experiments and found that this is always true.

Converse of the Sobolev Theorem (Exercise 33. from Folland)

Posted: 05 Jul 2021 07:44 PM PDT

Question 33. from Folland's book is the following:

(Converse of the Sobolev Theorem) If $H_{s}(\mathbb{R}^n) \subset C_{0}^{k}(\mathbb{R}^n)$, then $s>k+\frac{1}{2} n$. (Use the closed graph theorem to show that the inclusion map $H_{s} \rightarrow C_{0}^{k}$ is continuous and hence that $\partial^{\alpha} \delta \in\left(H_{s}\right)^{*}$ for $|\alpha| \leq k$.)

To use the closed graph theorem, we need to know how to compare convergence in $ C_{0}^{k}$ and $H_{s}$. This is the first key point that I don't see. Moreover, I don't know how to proof the other facts too. I will appreciate any help or suggestions.

Compute $\iint_ S \text{F}.\vec n\; \text{d}S$, with and without using Divergence theorem

Posted: 05 Jul 2021 07:32 PM PDT

Given a vector field $$\text{F}=(xz^2)\vec i+(x^2y-z^3)\vec j+(2xy+y^2z)\vec k$$


  • How to describe $\Sigma_1$ in cylindrical coordinates?
  • How to generally find $\vec n$?

And $S$ is the sphere half sphere $z= \sqrt {4-x^2-y^2}$ and $z=0$, then compute $$\iint_ S \text{F}\cdot\vec n\; \text{d}S$$


From Divergence theorem I know that the double integral is indeed:

$$\begin{align} \iiint_V \nabla \cdot \text{F} \;\text{d} V &= \iiint_V x^2+y^2+z^2 \;\text{d} V\\[1ex] &= \int_{0}^{2 \pi}\int_{0}^{2}\int_{0}^{\sqrt {4- \rho^2 }}(\rho^2+z^2)\rho \;\text{d}z\text{d}\rho\text{d}\theta\\[1ex] &=\frac{64\pi}{5}\end{align}$$

If I was going to compute $\iint_ S \text{F}\cdot\vec n\; \text{d}S$ directly without using Divergence theorem, then how I could proceed?


The surface $S$ can be written as $$S= \Sigma_1 \cup \Sigma_2$$

Where $$\Sigma_1= \left\{(\rho \cos\theta,\rho \sin\theta ,z): 0\le \theta \le 2\pi , 0 \le z \le \sqrt{4-\rho^2} \right\}$$ $$\Sigma_2= \left\{(2 \cos\theta,2 \sin\theta,0): 0\le \theta \le 2\pi \right\}$$ And $$\vec n_1=\frac{\partial \Sigma_1}{\partial \theta}× \frac{\partial \Sigma_1}{\partial z}=\langle \rho\cos\theta,\rho\sin\theta,0 \rangle $$

So:$$\iint_ {\Sigma_1} \text{F}\cdot\vec n_1\; \text{d}S$$$$=\int_{0}^{2 \pi}\int_{0}^{\sqrt{4-\rho^2} } z^{2}r^{2}\cos^{2}\theta+\left(r^{2}\cos^{2}\theta\sin\theta-z^{3}\right)r\sin\theta\;\text{d}z\text{d}\theta\tag{I} $$

And $$\vec n_2=\frac{\partial \Sigma_2}{\partial \theta}\times \frac{\partial \Sigma_2}{\partial z}=\langle 0,0,0 \rangle $$

So:$$\iint_ {\Sigma_2} \text{F}\cdot\vec n_2\; \text{d}S=0 \tag{II}$$

Now we need to add $(\text{I})$ and $(\text{II})$

$$\iint_ S \text{F}\cdot\vec n\; \text{d}S=\iint_ {\Sigma_1} \text{F}\cdot\vec n_1\; \text{d}S+\iint_ {\Sigma_2} \text{F}\cdot\vec n_2\; \text{d}S$$

However I doubt if $\Sigma_1$ describes the half sphere correctly.

Solutions to $A^N+B^N=C^N \pm 1$

Posted: 05 Jul 2021 08:10 PM PDT

Is there a solution to $A^N+B^N=C^N \pm 1$ where $A,B,C,N\in\Bbb{N}$, such that $A,B,C > 1,N \geq 4$ and $gcd(A,B)=gcd(B,C)=gcd(A,C)=1$?

This question was inspired by this one: $A^X+B^Y=C^Z\pm 1$ Beal's conjecture "almost" solutions


A related question was asked here before.

Noam Elkies has published a paper on this subject. The table on page 15 of the paper and his extended table here have some interesting results. If I understood his work correctly, there are no solutions for $4 \leq N \leq 20$ and $C < 2^{23}=8,388,608$.

I also found a MathWorld article on 4th order diophantine equations, but it doesn't cover equations of this form. I also checked that the OEIS has no sequences for such Fermat near misses.

finding essential prime implicants

Posted: 05 Jul 2021 08:07 PM PDT

F(w,x,y,z)=Σ(0,1,2,4,5,6,7,10,15)

which one is correct, or both wrong. i'm confused about finding prime implicants at top right and bottom right enter image description here

Finding the length of an angle bisector in a (right) triangle

Posted: 05 Jul 2021 08:08 PM PDT

We have a triangle $\triangle ABC$ such that $AB = 6,$ $BC = 8,$ and $CA = 10$. If $AD$ is an angle bisector such that $D$ is on $BC$, then find the value of $AD^2.$

So far I used the angle bisector theorem to get $CD/DB = AC/AB$ so $CD/DB = 10/6$. What should I do next?

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