Proving $\sum\limits_{l=1}^n \sum\limits _{k=1}^{n-1}\tan \frac {lk\pi }{2n+1}\tan \frac {l(k+1)\pi }{2n+1}=0$ Prove that $$\sum _{l=1}^{n}\sum _{k=1}^{n-1}\tan \frac {lk\pi } {2n+1}\tan \frac {l( k+1) \pi } {2n+1}=0$$

It is very easy to prove this identity for each fixed $n$ . For example let $n = 6$; writing out all terms in a $5 \times 6$ matrix, we obtain:

$\begin{matrix}
\tan \dfrac {\pi } {13}\tan \dfrac {2\pi } {13}
 &
\tan \dfrac {2\pi } {13}\tan \dfrac {3\pi } {13}
&
\tan \dfrac {3\pi } {13}\tan \dfrac {4\pi } {13}
&
\tan \dfrac {4\pi } {13}\tan \dfrac {5\pi } {13}
&
\tan \dfrac {5\pi } {13}\tan \dfrac {6\pi } {13}
 \\
\tan \dfrac {2\pi } {13}\tan \dfrac {4\pi } {13}
 &
 \tan \dfrac {4\pi } {13}\tan \dfrac {6\pi } {13}
&
\tan \dfrac {6\pi } {13}\tan \dfrac {8\pi } {13}
&
\tan \dfrac {8\pi } {13}\tan \dfrac {10\pi } {13}
&
\tan \dfrac {10\pi } {13}\tan \dfrac {12\pi } {13}
\\
\tan \dfrac {3\pi } {13}\tan \dfrac {6\pi } {13}
&
\tan \dfrac {6\pi } {13}\tan \dfrac {9\pi } {13}
&
\tan \dfrac {9\pi } {13}\tan \dfrac {12\pi } {13}
&
\tan \dfrac {12\pi } {13}\tan \dfrac {15\pi } {13}
&
\tan \dfrac {15\pi } {13}\tan \dfrac {18\pi } {13}
\\
\tan \dfrac {4\pi } {13}\tan \dfrac {8\pi } {13}
&
\tan \dfrac {8\pi } {13}\tan \dfrac {12\pi } {13}
&
\tan \dfrac {12\pi } {13}\tan \dfrac {16\pi } {13}
&
\tan \dfrac {16\pi } {13}\tan \dfrac {20\pi } {13}
&
\tan \dfrac {20\pi } {13}\tan \dfrac {24\pi } {13}
\\
\tan \dfrac {5\pi } {13}\tan \dfrac {10\pi } {13}
&
\tan \dfrac {10\pi } {13}\tan \dfrac {15\pi } {13}
&
\tan \dfrac {15\pi } {13}\tan \dfrac {20\pi } {13}
&
\tan \dfrac {20\pi } {13}\tan \dfrac {25\pi } {13}
&
\tan \dfrac {25\pi } {13}\tan \dfrac {30\pi } {13}
\\
\tan \dfrac {6\pi } {13}\tan \dfrac {12\pi } {13}
&
\tan \dfrac {12\pi } {13}\tan \dfrac {18\pi } {13}
&
\tan \dfrac {18\pi } {13}\tan \dfrac {24\pi } {13}
&
\tan \dfrac {24\pi } {13}\tan \dfrac {30\pi } {13}
&
\tan \dfrac {30\pi } {13}\tan \dfrac {36\pi } {13}
 \end{matrix}$

one can notice then, that the first column vanish the fourth one :

$\tan \dfrac {\pi } {13}\tan \dfrac {2\pi } {13}=-\tan \dfrac {12\pi } {13}\tan \dfrac {15\pi } {13}$

$\tan \dfrac {2\pi } {13}\tan \dfrac {4\pi } {13}=-\tan \dfrac {24\pi } {13}\tan \dfrac {30\pi } {13}$

$\tan \dfrac {3\pi } {13}\tan \dfrac {6\pi } {13}=-\tan \dfrac {16\pi } {13}\tan \dfrac {20\pi } {13}$

$\tan \dfrac {4\pi } {13}\tan \dfrac {8\pi } {13}=-\tan \dfrac {4\pi } {13}\tan \dfrac {5\pi } {13}$

$\tan \dfrac {5\pi } {13}\tan \dfrac {10\pi } {13}=-\tan \dfrac {8\pi } {13}\tan \dfrac {10\pi } {13}$

$\tan \dfrac {6\pi } {13}\tan \dfrac {12\pi } {13}=-\tan \dfrac {20\pi } {13}\tan \dfrac {25\pi } {13}$

and the third column vanish the fifth one :

$\tan \dfrac {3\pi } {13}\tan \dfrac {4\pi } {13}=-\tan \dfrac {30\pi } {13}\tan \dfrac {36\pi } {13}$

$\tan \dfrac {6\pi } {13}\tan \dfrac {8\pi } {13}=-\tan \dfrac {5\pi } {13}\tan \dfrac {6\pi } {13}$

$\tan \dfrac {9\pi } {13}\tan \dfrac {12\pi } {13}=-\tan \dfrac {25\pi } {13}\tan \dfrac {30\pi } {13}$

$\tan \dfrac {12\pi } {13}\tan \dfrac {16\pi } {13}=-\tan \dfrac {10\pi } {13}\tan \dfrac {12\pi } {13}$

$\tan \dfrac {15\pi } {13}\tan \dfrac {20\pi } {13}=-\tan \dfrac {20\pi } {13}\tan \dfrac {24\pi } {13}$

$\tan \dfrac {18\pi } {13}\tan \dfrac {24\pi } {13}=-\tan \dfrac {15\pi } {13}\tan \dfrac {18\pi } {13}$

while the second column is self-vanishing:

$\tan \dfrac {2\pi } {13}\tan \dfrac {3\pi } {13}=-\tan \dfrac {10\pi } {13}\tan \dfrac {15\pi } {13}$

$\tan \dfrac {4\pi } {13}\tan \dfrac {6\pi } {13}=-\tan \dfrac {6\pi } {13}\tan \dfrac {9\pi } {13}$

$\tan \dfrac {8\pi } {13}\tan \dfrac {12\pi } {13}=-\tan \dfrac {12\pi } {13}\tan \dfrac {18\pi } {13}$ .

So the equality occurs.
But how to generalize the proof?

Micah already pointed the way in a comment: The identity

$$\tan a \tan b = \frac{\tan a - \tan b}{\tan(a-b)} - 1$$

causes the inner sum to telescope. To make full use of this, let's extend the inner sum to $k=2n$:

$$
\begin{align}
\sum _{l=1}^{n}\sum _{k=1}^{2n}\tan \frac {lk\pi } {2n+1}\tan \frac {l(k+1) \pi } {2n+1}
&=
\sum _{l=1}^{n}\sum _{k=1}^{2n}\left(\frac{\tan \frac {l(k+1)\pi } {2n+1}-\tan \frac {lk \pi } {2n+1}}{\tan\frac {l\pi}{2n+1}}-1\right)
\\
&=
\sum _{l=1}^{n}\left(\frac{\tan \frac {l(2n+1)\pi } {2n+1}-\tan \frac {l\pi } {2n+1}}{\tan\frac {l\pi}{2n+1}}-2n\right)
\\
&=
\sum _{l=1}^{n}(0-1-2n)
\\
&=
-n(2n+1)\;.
\end{align}
$$

This sum contains each of the terms we want to sum twice, with mirror symmetry, and it contains one additional term in the middle for $k=n$. Thus, for our sum to vanish, we need

$$
\sum _{l=1}^{n}\tan \frac {ln\pi } {2n+1}\tan \frac {l(n+1) \pi } {2n+1}=-n(2n+1)\;.
$$

The arguments of the two factors add to $l\pi$, so they're negatives of each other, so we're looking for

$$
-\sum _{l=1}^{n}\tan^2 \frac {ln\pi } {2n+1}\;.
$$

We can again extend the sum to $2n$ to double it, since the arguments form pairs that add up to $n\pi$; then, since $n$ and $2n+1$ are coprime, we can replace $ln$ by $l$ while traversing the same arguments; and then we can set the upper limit back to $n$, since the arguments still add up to $\pi$ in pairs. Thus, what we need is

$$
\sum _{l=1}^{n}\tan^2 \frac {l\pi } {2n+1}=n(2n+1)\;.
$$

How to find this sum is shown at Prove that $\sum\limits_{k=1}^{n-1}\tan^{2}\frac{k \pi}{2n} = \frac{(n-1)(2n-1)}{3}$. Adapting the argument in the accepted answer there for our odd denominator, we obtain

$$
\left(\cos\frac{k\pi}{2n+1}+\mathrm i\sin\frac{k\pi}{2n+1}\right)^{2n+1}=(-1)^k\;,
$$

taking the imaginary part,

$$
\sum_{r=0}^n\binom{2n+1}{2r+1}\left(\cos\frac{k\pi}{2n+1}\right)^{2n-2r}\left(\mathrm i\sin\frac{k\pi}{2n+1}\right)^{2r+1}=0\;,
$$

dividing by $\left(\cos\frac{k\pi}{2n+1}\right)^{2n+1}$,

$$
\sum_{r=0}^n\binom{2n+1}{2r+1}\left(\mathrm i\tan\frac{k\pi}{2n+1}\right)^{2r+1}=0\;,
$$

and dividing by $\tan\frac{k\pi}{2n+1}$ and letting $x=\tan^2\frac{k\pi}{2n+1}$,

$$
\sum_{r=0}^n\binom{2n+1}{2r+1}(-x)^r=0\;.
$$

Then Vieta's formula shows that the sum of the roots of this equation is

$$\frac{\binom{2n+1}2}{\binom{2n+1}0}=n(2n+1)\;,$$

as required.

[Update:]

This answer suggests an alternative, more elementary way to calculate the sum of the squares of the tangents: On the grid $\frac{l\pi}{2n+1}$, the tangent decomposes into $n$ mutually orthogonal sines, each of whose dot product with itself is $2n+1$, so the dot product of the tangent with itself is $n(2n+1)$.

To prove that,

$$\sum _{l=1}^{n}\sum _{k=1}^{n-1}\tan \frac {lk\pi } {2n+1}\tan \frac {l( k+1) \pi } {2n+1}=0$$

Consider the following,

$$\tan ( \frac {l(k+1)\pi } {2n+1} - \frac {l(k)\pi } {2n+1} ) = \frac { \tan \frac {l(k+1)\pi } {2n+1} - \tan \frac {l(k)\pi } {2n+1} } {1 + \tan \frac {l(k+1))\pi } {2n+1}\tan \frac {l(k) \pi } {2n+1}}$$

i.e. $$\tan ( \frac {l\pi } {2n+1} ) = \frac { \tan \frac {l(k+1)\pi } {2n+1} - \tan \frac {l(k)\pi } {2n+1} } {1 + \tan \frac {l(k+1))\pi } {2n+1}\tan \frac {l(k) \pi } {2n+1}}$$

i.e. $$\ ( 1 + \tan \frac {l(k+1))\pi } {2n+1}\tan \frac {l(k) \pi } {2n+1}) = \frac { \tan \frac {l(k+1)\pi } {2n+1} - \tan \frac {l(k)\pi } {2n+1} } {\tan \frac {l\pi } {2n+1}}$$

i.e. $$\tan \frac {l(k+1))\pi } {2n+1}\tan \frac {l(k) \pi } {2n+1} = (\frac { \tan \frac {l(k+1)\pi } {2n+1} - \tan \frac {l(k)\pi } {2n+1} } {\tan \frac {l\pi } {2n+1}} - 1)$$

Hold l = 1 and take $$\sum _{k=1}^{n-1}$$

Next hold l = 2 and take $$\sum _{k=1}^{n-1}$$

.......

Continue this till l exhausts for l = n

Now sum all the terms you get and we see that the sum is indeed equal to 0

i.e. $$\sum _{l=1}^{n}\sum _{k=1}^{n-1}\tan \frac {lk\pi } {2n+1}\tan \frac {l( k+1) \pi } {2n+1}=0$$


Number of connected components after removing $t$ edges from K-edge-connected graph $k,s,t$ are positive integers such that $k(s+1)>2t$.

prove: if $G$ k-edge-connected graph then removing any $t$ edges from $G$ will yield a graph with at most $s$ connected components.

question from old exam i'm trying to solve .... have no idea where to start .
the graph is k-connected so i guess i'm expected to use "Whitney theorem" or "Manger theorem" but i don't see how i can derive something about number of connected components using mentioned theorems and the given inequality

after some thought i think i need to use induction here .

some trail and error led me to this solution hope its a valid one .

given $s$ number of connected components after the removal of $t$ edges .
each component had at least $k$ edges what were connecting vertices in it to the other vertices of the original graph $G$ , moreover each removed edge were connected to 2 vertices
so it was connecting 2 components before it was removed. from this we can derive the wanted inequality

$$sk < 2t < sk + k$$

where:

$sk$ the minimal number of edges you need to remove to yield $s$ connected components
$2t$ the number of removed edges
$sk + k = k(s+1)$ the number of edges you would need to remove if there were $s+1$ connected components


Calculus Leibniz' notation I'm currently doing integration by parts, and I'm finding that the notation is what makes it tough for me. So I looked it up and found that:

$$\int u(x)v'(x)dx= u(x)v(x) - \int u'(x)v(x)dx$$

But the wikipedia said that this above equality is the same as:

$$ \int u dv = uv - \int v du $$

I think I have located my problem: I don't see how these 2 are equal at all. How could for example know that $$\int u(x)v'(x) dx = \int udv$$

I can't really see the equivalence here.

The point is that when Leibniz conceived the derivative, he thought of it as the quotient of two infinitely small numbers $dy$ and $dx$ if $y = f(x)$, for example. In that case, the derivative was really written as $dy/dx$, and since $f'(x)=dy/dx$, we have $dy=f'(x)dx$ as the infinitely small quantity that $y$ varies at a rate $f'(x)$ when $x$ varies the infinitely small quantity $dx$.

However, this stuff isn't rigorous. Indeed, in standard analysis, it is impossible to conceive a number like an infinitesimal, and the use of this even as mere notation may lead to confusion. That's why in the modern language, we simply use the prime notation to denote the derivative. The next best thing to replace the infinitesimals $dy$ and $dx$ is the notion of a differential form; there's so much about them to be said that I won't explain here.

So, in truth, if you use this stuff that's not rigorous you have $dv=v'(x)dx$ and $du=u'(x)dx$ so that we can write:

$$\int u(x)v'(x)dx = u(x)v(x)-\int v(x)u'(x)dx$$

Simply as:

$$\int udv = uv - \int v du$$

Now, understand that this last thing is just a mnemonic rule so that people remember what to do when find an integral like that. The rigorous version is the first formula that's derived directly from the product rule for derivatives.

Indeed, many of the notations regarding one-dimensional integrals that refer to "let $u$ be that, then $du$ is that other thing" and so forth are just rules for you to remember easier what to do. As I've said, those $dx$, $du$ and everything else can have a rigorous meaning as differential forms when you study differential geometry.

The "answer" lies in the proof of Integration by Parts. It also helps to recall that $v^\prime(x)=\frac{d v}{d x}$.


Are there journals related to mathematical writing - by this I mean styles of writing, organization, formatting etc... Are there journals where one can submit ideas on mathematical writing style eg: how to improve style, clarity, formatting etc. The closest I can find is journals related to mathematical education, but I'm not sure this would be a good fit.

Typically, if you have your own mathematical style and want to share it with the world, you start by employing it in a peer reviewed paper that you publish in a scientific journal. Preface the text with an explanation of the new style that you are using. If others read your paper and like your style, they may use it in their papers as well. If enough people use it, it may be used as a standard--the style may even be named after you.


Show that $A \lor B ⊢ B \lor A$ Prove the following derivability claim using only our primitive rules: $A \lor B ⊢ B \lor A$

I know this can be attributed to the commutative property, but I'm not exactly sure how to prove this using only the primitive rules of sentential logic.

Edit: Sorry about that guys - I am working with the formal system from Teller's Primer. Also using contradiction is fine. However, primitives rules only means no derived rules so De Morgan's is out of the question.

I assume the natural deduction rules for disjunction here.

Unlike conjunction, that has only one introduction and elimination rule, the disjunction has two different intros and one elim - which is why it needs some further clarification here:

Elim: $ \qquad \dfrac{\Gamma, p \vdash  r \qquad \Gamma, q \vdash r \qquad \Gamma \vdash p \lor q}{\Gamma \vdash r}$
Intro 0: $\qquad \dfrac{ \Gamma \vdash p}{ \Gamma \vdash p \lor q}$
Intro 1: $\qquad \dfrac{\Gamma \vdash q}{\Gamma \vdash p \lor q}$

Then:

1) $A \lor B$, Premise

$\qquad$ 2) $B$, Assumption

$\qquad$ 3) $B \lor A$, 2, $\lor$ Intro 0

4) $B \rightarrow (B \lor A)$, 2-3, $\rightarrow$ Intro

$\qquad$ 5) $A$, Assumption

$\qquad$ 6) $B \lor A$, 2, $\lor$ Intro 1

7) $A \rightarrow (B \lor A)$, 2-3, $\rightarrow$ Intro

8) $B \lor A$, 1, 4, 7, $\lor$ Elim

I am using rules I think are the ones assumed in most treatments of Sentence logic.
How about:

$A  ∨ B ⊢ B  ∨ A $

Start by assuming:
1) $ A ∨ B $

2)Assume -$(B∨ A) $

Conclude $-B \land  -A$

Conclude $- A$

From  $ -A $ conclude $ -A \lor -B $

From $ -A \land -B $ conclude

-$( A ∨ B)$ , a contradiction , conclude

$(B ∨ A) $

Discharging assumptions :

$( A ∨ B) ⊢ (B∨A)   $


Proof verification: $x_n $diverges to $\infty\leftrightarrow$ lim inf $x_n$ = lim sup $x_n$ = $\infty$ Here's how I did it:

Suppose $x_n \rightarrow \infty$. Hence, for any $M \in \mathbb{R}$, we can find some $N \in \mathbb{N}$ such that $n \geq N$ implies $x_n \geq M$. Then, for such $N$, we take $N^{th}$ tail of that sequence to have:

inf ($T_n) \geq M$ and sup $(T_n) \geq M$.

Since we can do this for any $M,$ it is implied that

lim inf $x_n$ = lim sup $x_n$ = $\infty$.

The proof for the second part is very similar.

Your proof is correct. If you want to be a little more precise:

Suppose $x_n\to\infty$.

For $n\in\mathbb{N}$, define
$$
T_n:=\{x_m:m\geq n\}\\
A_n:=\inf\left(T_n\right)
$$

Note that if $m\geq n$, then $T_m\subseteq T_n$ so
$$
A_m\geq A_n\tag{1}
$$

Let $M\in\mathbb{R}$. Since $x_n\to\infty$, there exists $N\in\mathbb{N}$ such that $x_n\geq M$ if $n\geq N$. By definition of infimum as the greatest lower bound, it follows that $A_{N}\geq M$. By $(1)$, it follows that if $n\geq N$ then $A_n\geq M$.

This shows that
$$
\lim_{n\to\infty}A_n=\infty
$$

Hence, by definition of limit inferior, $\liminf x_n=\infty$.

Note. If $x_n\to\infty$, then $\{x_n\}$ is not bounded above and $\sup T_n$ is always infinite. So if $\{x_n\}$ is not bounded above, then we usually define $\limsup x_n$ to be $\infty$.

Just come back to definitions, when you hesitate.

Let's have $v_n=inf\,\{x_k\,|\,k\ge n\}$ and $w_n=sup\,\{x_k\,|\,k\ge n\}$.

$w_n\ge v_n$ so we only have to proove that $v_n\to+\infty$ and $w_n\to+\infty$ is automatic.

From $(x_n)_n$ divergence we have $\forall M\in\mathbb R\quad v_N\ge M$

Thus since $(v_n)_n$ is increasing we get $\forall n\ge N\quad v_n\ge v_N\ge M$, which is by definition $v_n\to+\infty$.

And since $\underline\lim x_n=\lim v_n$ and $\overline\lim x_n=\lim w_n$ the demonstration is done.


How to prove a specific sequence is Fibonacci's with no prior knowledge nor trial and error? Let $n$ be a positive integer and let $s_n$ be the number of increasing sequences of integers, alternatingly even and odd, starting with $0$ and ending with $n$. E.g. for $n=3$ we only have the two sequences $$ 0,1,2,3 \ \  \text{and} \ \ 0,3.$$ hence $s_3=2$. Prove that the $s_n$ are the Fibonacci numbers.

Here's the deal. We already know that the sequence is Fibonacci's, so it suffices to show that $s_n$ satisfies the recursive definition of $F_n$. And it is in fact straightforward to put the sequences for $n=k,k+1$ into one-to-one correspondence with those for $n=k+2$.
So what I'm looking for is an approach able to find  that $s_n=F_n$, without knowing it beforehand or guessing.

So the number of sequences $s_{n+1}$ ending in $n+1$ is obtained by adding $n+1$ to sequences ending in $n$ or ...

you can't add $n+1$ to a sequence ending in $n-1$ because they have the same parity. But you can substitute $n+1$ for $n-1$ at the end of a sequence because they have the same parity.

Every sequence ending in $n+1$ can be obtained in this way, because either the penultimate element is $n$ or it is less than $n-1$ (parity again).

Hence $s_{n+1}=s_n+s_{n-1}$ and it suffices to show that $s_1=s_2=1$

If you did not know beforehand that the result is Fibonacci then start by trying to find a relation for $s_n$  in terms of its previous terms.


Integration of $\log(\sin x)$ In trying to integrate $\log(\sin x)$ and I ended up looking for a solution and found the one at this link: http://www.meritnation.com/ask-answer/question/how-to-integrate-f-log-sin-x-dx/math/766517

However, there is one step in the process I don't understand. How do the limits from $0$ to $\pi$ change to be $0$ to $\pi/2$ in the bottom part of the image and the $t$ remains but the $1/2$ disappears? I'm relatively new to this process...

  From equation (3) Now,

  $$\int_0^{\pi/2}\left[\log(\sin 2x)dx\right] = \int_0^\pi\left[\frac{1}{2}\log(\sin t) \right]dt=\int_0^{\pi/2}\left[\log(\sin t) \right]dt=\int_0^{\pi/2}\log(\sin t)dt = I$$

Here's a quick proof i could come up with:

$2I=I'=\int^{\pi/2}_{0} \ln(\sin t) dt+\int_{\pi/2}^{\pi} \ln(\sin t) dt $

Consider second integral,

Put $y=-\frac{\pi}{2}+t$

$I_2=\int^{\pi/2}_{0}\ln (\sin \left(y+\frac{\pi}{2}\right)) dy$

$I_2=\int^{\pi/2}_{0}\ln (\sin \left(t+\frac{\pi}{2}\right)) dt$

$I_2=\int^{\pi/2}_{0}\ln \sin (t) dt$

This is applicable to all possible function. Depending on whether odd or even function.

in property $6$ of tbk, after applying upper limit changes from $2a$ to a which implies $2a=\pi$. After applying the property $6, a = \frac\pi2.$


Definition of conditional probability for the intersection of 2 events - am I right? Am I right saying that (assuming that $P(A)>0, P(B)>0, P(C)>0$)

$P(A\cap B|C)=\frac{P(A\cap B\cap C)}{P(C)}?$

Yes, you are correct. To help see for yourself, let's call the event $A \cap B$ by a new name, $E$. The definition of conditional probability gives us that
\begin{equation*}
P(E \mid C) = \frac{P(E \cap C)}{P(C)}
\end{equation*}
We also know that $E \cap C = (A \cap B) \cap C = A \cap B \cap C$, as events, so we get the result you ask about.

(Also, note that we only need to assume $P(C) > 0$. The result is perfectly valid when $P(A) = 0$ and/or $P(B) = 0$.)

This is comes from defination of conditional probability for two sets, you can also try for more than three and any finite as well infinite numbers of sets.as i think well?


"Logically equivalent formulae express the same _______." <- What word do logicians use for the blank? Meaning denotes the truth conditions of a sentence: what would have to be the case for the interpreted formula to be true. Nevertheless, without an interpretation, two logically equivalent formulae express something that makes them equivalent. What is the name for that?

I think that this one is a "slippery" question.
We can approach it from different points of view, with increasing degree of "difficulty" and, so to say, "philosophical committment".
First approach : is it possible to avoid to be "involved with" meaning in the elucidation ?
Stay, for simplicity, with propositional logic; we have that :

$\varphi \leftrightarrow \psi$ iff $\vdash \varphi \rightarrow \psi$ and $\vdash \psi \rightarrow \varphi$.

If we assume modus ponens, this amount to "interderivability", i.e. :

$\varphi \vdash \psi$ and $\psi \vdash  \varphi$.

There is no "smell" of meaning here. But, without meaning, i.e. without interpretation of the above formulae, what is "expressed" by their equivalence ?
Second approach : there is a long tradition in modern logic regarding "extensionality". From Frege's Begriffsschrift onwards, mathematical logic has been concerned mainly with truth-functional contexts.
According to Frege, the Bedeutung of an entire proposition is its truth-value, either the True or the False. For Frge, complete propositions, like names, have objects as their Bedeutungen, and in particular, the truth-values the True or the False. In this way, he is able to transcribe sentential connectives such as “and” and “or,” etc., as truth functions in the strictest sense — functions that take truth-values as argument and yield truth-values as value. [see Kevin Klement, Frege and the Logic of Sense and Reference (2002)]
According to W&R [see Alfred North Whitehead & Bertrand Russell, Principia Mathematica to 56 (2nd ed - 1927), page 115] :

It is obvious that two propositions are equivalent when, and only when, both are true or both are false. Following Frege, we shall call the truthvalue of a proposition truth if it is true, and falsehood if it is false. Thus two propositions are equivalent when they have the same truth-value.

AIt should be observed that, if $p \equiv q$, $q$ may be substituted for $p$ without altering the truth-value of any function of $p$ which involves no primitive ideas except those enumerated in *1 [the truth-functional connectives].

We shall give the name of a truth-function to a function $f(p)$ whose argument is a proposition, and whose truth-value depends only upon the truth-value of its argument. All the functions of propositions with which we shall be specially concerned will be truth-functions, i.e. we shall have

$p \equiv q . \supset .f(p) \equiv f(q)$.

The reason of this is, that the functions of propositions with which we deal are all built up by means of the primitive ideas of *1. But it is not a universal characteristic of functions of propositions to be truth-functions. For example, "A believes $p$" may be true for one true value of $p$ and false for another.

In "modern" terms, this is the so-called Replacement theorem [see S.C.Kleene, Introduction to Metamathematics (1952), page 116] :

If $A$ and $B$ are formulas, $C_A$ is a formula constructed from a specified occurnce of $A$ using only the [truth-functional connectives], and $C_B$ results from $C_A$ by replacing this occurrence of $A$ by $B$, then $A \leftrightarrow B \vdash C_A \leftrightarrow C_B$.

And equivalently : $A \leftrightarrow B, C_A \vdash C_B$.
In a "universe" of "extensional" contexts, that considered by the  "simplified" semantics of logical languages, it is easy to "equate" the condition for two equivalent formulae of having the same "truth conditions" with that of expressing "the same thing".
It is likely then, that "that thing" is their meaning.
Third approach : see Michael Dummett and his The Logical Basis of Metaphysics (Harvard UP, 1991) and Proof-Theoretic Semantics.
The basic idea - I think - is to build a theory of meaning without truth. But I'm not prepared to discuss it.

Usually, two logically equivalent formulas express interderivability, or one could say that the formulas are interderivable.  In other words, in a given calculus with axioms, if A is logically equivalent to B, then if A gets assumed, then we can derive B using only the rules of inference and axioms of the calculus, and if we assume B, then we can derive A using only the rules of inference and axioms of the calculus.

So one might say that logically equivalent formulae express the same derivability.

For instance, in Lukasiewicz three-valued logic (p$\rightarrow$q) is equivalent with ($\lnot$q$\rightarrow$$\lnot$p) implying that if (p$\rightarrow$q) is interderivable with ($\lnot$q$\rightarrow$$\lnot$p) (note that this does NOT imply that [(p$\rightarrow$q)$\rightarrow$($\lnot$q$\rightarrow$$\lnot$p)] is a theorem NOR that [($\lnot$q$\rightarrow$$\lnot$p)$\rightarrow$(p$\rightarrow$q)] is a theorem here (though in fact they are theorems here), since Lukasiewicz three-valued logic does not have a deduction theorem which allows for reasoning to those theorems in this way).

Interderivability in some ways comes as a more general notion than logical equivalence, since logical equivalence only applies to well-formed formulas.  In contradistinction, interderivability applies to sets of well-formed formulas under a fixed rule.  For example, the axiom set {CpCqp, CCpCqrCCpqCpr, CCNpNqCqp} under detachment and uniform substitution is interderivable with the axiom set {CCpqCCqrCpr, CpCNpq, CCNppp} under detachment and substitution.  But, the set {CpCqp, CCpCqrCCpqCpr, CCNpNqCqp} under detachment and substitution is not equivalent to {CCpqCCqrCpr, CpCNpq, CCNppp} under detachment and substitution since neither set with its rules is a well-formed formula.


Closed Form involving double factorial Is there a closed form of
$$1+\frac{x}{3}+\frac{x^2}{5\cdot 3\cdot 1}+\frac{x^3}{7\cdot 5\cdot 3\cdot 1}+ \cdots = \sum_{n=0}^{\infty}\frac{x^n}{(2n+1)!!}$$
where $n!!$ is the double factorial?
Thank you very much!

Since 1, $1*3$, $1*3*5$  is the double factorial definition,
$\sum_{n=0}^{\infty}\frac{x^n}{(2n+1)!!}$ this is the right form....the equation you wrote is correct.
From wikipedia about the double factorial

and for odd n it is ${\displaystyle n!!=\prod _{k=1}^{\frac
> {n+1}{2}}(2k-1)=n(n-2)(n-4)\cdots 3\cdot 1\,.}$

However, I hardly see how the double factorial form help you


Deriving Black Scholes using CAPM I am referring to http://www.frouah.com/finance%20notes/Black%20Scholes%20PDE.pdf Section 3, which is a bit more detailed version of the original derivation from https://www.cs.princeton.edu/courses/archive/fall09/cos323/papers/black_scholes73.pdf

There are several things that are not clear to me, and I would like to ask someone to explain them:

(1) on page 7 we read "Drop dt from both sides and take the covariance of rV and rM, noting that only the second term on the right-hand side of Equation (10) is stochastic" : how on Earth do we see this? the first term also contains Vt and St just like the second one, so what would make it less stochastic?

(2) first we find E[dV_t/V_t] and then we multiply the equation by V_t... Without arguing at all we assume that E[dV_t/V_t]=E[dV_t]/V_t. Why is that so? Isn't Vt random as well, it's not a constant, is it?

(3) after all, what is the precise mathematical meaning of the expression E[dV_t]?

Hope someone can clarify this. Thanks.

Just seen this - probably too late, but if not..:

re. (1): You are in $t$, i.e. $t$ is now, and $V_t$ and $S_t$ are real market prices observable right in this moment ($t$).

re (2): same answer as to (1). $S_t$ and $V_t$ are here not symbols for random variables, but current realizations which you see now (in time $t$).  Of course, you don't know future prices, so from your current perspective, $V_{t+1}$ and $S_{t+1}$ (and hence $dV_t$ and $dS_t$) are random (note that I use infinitesimal small units here for $t$...)

re (3)  $E$ is the expectation operator, $d$ indicates change in an infinitesimal small time interval, $V_t$ is the option value. $E(dV_t)$ is the expected change of the option value over the next infinitesimal small time interval.

It should be noted that the original paper by Black & Scholes is not mathematically rigorous. There is no rigorous meaning to the expression $\mathrm{d}t$. Nor is there any rigorous meaning to the expression $\mathrm{d}V_t$. Actions  such as 'dropping $\mathrm{d}t$' have no mathematical meaning. Nevertheless they hold  because beneath this heuristic garbage lies  a deep mathematical theory pinning what you are allowed to do. If you are looking for a decent, non rigorous derivation to the Black Scholes equation, then Wilmott - The Mathematics of Financial Derivatives is a good book to look at.

That paper has no meaning in a mathematical sense. It is impossible to answer your question.

(too long to be a comment)


Prove that $\frac1{x}-x$ is multiplicatively subadditive in $(0, 1)$ If
$f(x) = \dfrac1{x}-x$,
prove that
$f(x)$ is multiplicatively subadditive in
$(0, 1)$,
that is,

if $a, b \in (0, 1)$,
then
$f(a)+f(b) < f(ab)$.

To show one expression is larger than another, it is often helpful to consider the difference:

$$\begin{align}
f(ab) - f(a) - f(b) &= \frac{1}{ab} - \frac{1}{a} - \frac{1}{b} + a + b - ab\\
&= \left(\frac{1}{ab} - \frac{1}{a} - \frac{1}{b} + 1\right) - \left(1 - a - b + ab\right)\\
&= \left(\frac{1}{a} - 1\right)\left(\frac{1}{b} - 1\right) - (1-a)(1-b)\\
&= \frac{(1-a)(1-b)}{ab} - (1-a)(1-b)\\
&= (1-a)(1-b)\left(\frac{1}{ab} - 1\right)\\
&> 0.
\end{align}$$

Suppose we take $f(a)$ as $a$ approaches $0$. Then $f(a)$ approaches infinity. If we take $f(b)$ approaching $1$, then $f(b)$ approaches $0$. If we take $f(ab)$ with the same conditions, then ab should approach $0$ - but it is less than $f(a)$, which is important to note, because that means $f(ab)$ approaches infinity faster than $f(a)$ does.


How find this maximum of the $\sin^2{\theta_{1}}+\sin^2{\theta_{2}}+\cdots+\sin^2{\theta_{n}}$ Question:

let $\theta_{1},\theta_{2},\cdots,\theta_{n}\ge 0$,and such
$$\theta_{1}+\theta_{2}+\theta_{3}+\cdots+\theta_{n}=\pi$$
find the $P$ the maximum of value $P(n)$
$$P=\sin^2{\theta_{1}}+\sin^2{\theta_{2}}+\cdots+\sin^2{\theta_{n}}$$

Find the closed $P(n)$
I found this
when $n=2$
then
$$P=\sin^2{\theta_{1}}+\sin^2{\theta_{2}}=1-\dfrac{1}{2}(\cos{2\theta_{1}}+\cos{2\theta_{2}})=1-\cos{(\theta_{1}+\theta_{2})}\cos{(\theta_{1}-\theta_{2})}=1+\cos{(\theta_{1}-\theta_{2})}\le 2$$
when $\theta_{1}=\theta_{2}=\dfrac{\pi}{2}$.
so
$$P(2)=2$$
and for general $n$,maybe have use other methods,Thank you

I guess we can prove
$$P(3)=P(4)=P(5)=\cdots=P(n)?$$

If you restrict in the interior of $ \tilde {\theta} = \{  ( \theta _1 , \dots , \theta _n)\mid \theta _1 + \dots + \theta _n=\pi\}$ the critical point is $ \theta _1= \dots =\theta _n$ where you get $P= n \sin^2 ( \pi/n)$.

For case $n=2$ you have found the maximum. For $n=3$, you compare the value of $P$ at the critical point and the maximum at the boundary. The maximum at the boundary is the same as the previous case namely for $n=2$. So you compare $3 \sin^2 \pi /3=9/4 >2$ and you conclude the the maximum for $n=3$ is $ 9/4$.

Now you can proceed inductively comparing the critical value you get from Lagrange multipliers with the max at the boundary, which is known.

Use a Lagrange multiplier to fix the constraint. It then falls out that all $\theta_i$ are the same and the rest is easy.


Convex hull of $3$ dimensional set reduced to $2$ dimensional set Let $S = \{(f_1(t), f_2(t), f_3(t)) : t \in \mathbb{R}\}$ and suppose $f_3(t) \geq 1$ for all $t \in \mathbb{R}$. Is finding the convex hull of $S$ in some way equivalent to find the convex hull of $T = \{f_1(t)/f_3(t), f_2(t)/f_3(t)\}$?

In particular, $S = \{(a^3 + a, a^2, a^4 + 1) : a \in \mathbb{R}\}$ here.

Not at all. If $S$  is a $3-$dimensional it must be the subset of its convex hull. Then the dimension of Convex hull cannot be smaller than of the set itself.


Simplify the Complex Fraction I am having trouble with the following complex fraction. I have simplified everything for the most part, but I am stuck on the last part and need to know what I have to do next.

$$\frac{\frac{2}{3}+\frac{4}{5}}{\frac{5}{6}-\frac{1}{2}}$$

(I hope this converts over to the way it needs to be displayed.

I have simplified it to
    $\frac{\frac{10}{15}+\frac{12}{15}}{\frac{10}{12}-\frac{6}{12}}$

which equals to
    $\frac{\frac{22}{15}}{\frac{4}{12}}$

simplified to

$\frac{4}{12}×\frac{15}{22}$

At this point do I cross multiply and then add the two results? I can't figure out what to do next. Please help me.

(5/6)-(1/2) <- for this one you could just multiply the numerator and denominator of the second fraction by 3 instead of 6 and it'll be (5/6)-(3/6)

Then you'll get (22/15)/(2/6) , Then, to divide two fractions you take the second one and flip it then multiply. So, this (22/15)/(2/6) becomes this (22/15)*(6/2)

To make your life easier see if you can simplify before multiplying. Well, in this case we certainly can. 22 is divisible by 2 and 2 is divisible by 2. Also, 6 is divisible by 3 and 15 is divisible by 3. After simplifying it'll be (11/5)*(2/1) at this point there isn't anymore simplification you can make so the answer is:

22/5


Finding a primitive root of a prime number How would you find a primitive root of a prime number such as 761? How do you pick the primitive roots to test? Randomly?

Thanks

There is no general formula to find a primitive root. Typically, what you do is you pick a number and test. Once you find one primitive root, you find all the others.

How you test

To test that $a$ is a primitive root of $p$ you need to do the following. First, let $s=\phi(p)$ where $\phi()$ is the Euler's totient function. If $p$ is prime, then $s=p-1$. Then you need to determine all the prime factors of $s$: $p_1,\ldots,p_k$. Finally, calculate $a^{s/p_i}\mod p$ for all $i=1\ldots k$, and if you find $1$ among residuals then it is NOT a primitive root, otherwise it is.

So, basically you need to calculate and check $k$ numbers where $k$ is the number of different prime factors in $\phi(p)$.

Let us find the lowest primitive root of $761$:

$s=\phi(761)=760=2^3\times5\times19$
the powers to test are: $760/2=380$, $760/5=152$ and $760/19=40$ (just 3 instead of testing all of them)
test 2:

$2^{380}\equiv 1\mod 761$ oops

test 3:

$3^{380}\equiv -1\mod 761$ OK
$3^{152}\equiv 1\mod 761$ oops

test 5 (skip 4 because it is $2^2$):

$5^{380}\equiv 1\mod 761$ oops

test 6:

$6^{380}\equiv -1\mod 761$ OK
$6^{152}\equiv 67\mod 761$ OK
$6^{40}\equiv -263\mod 761$ hooray!

So, the least primitive root of 761 is 6.

How you pick

Typically, you either pick at random, or starting from 2 and going up (when looking for the least primitive root, for example), or in any other sequence depending on your needs.

Note that when you choose at random, the more prime factors are there in $\phi(p)$, the less, in general, is the probability of finding one at random. Also, there will be more powers to test.

For example, if you pick a random number to test for being a primitive root of $761$, then the probability of finding one is roughly $\frac{1}{2}\times\frac{4}{5}\times\frac{18}{19}$ or 38%, and there are 3 powers to test. But if you are looking for primitive roots of, say, $2311$ then the probability of finding one at random is about 20% and there are 5 powers to test.

How you find all the other primitive roots

Once you have found one primitive root, you can easily find all the others. Indeed, if $a$ is a primitive root mod $p$, and $p$ is prime (for simplicity), then $a$ can generate all other remainders $1\ldots(p-1)$ as powers: $a^1\equiv a,a^2,\ldots,a^{p-1}\equiv 1$. And $a^m \mod p$ is another primitive root if and only if $m$ and $p-1$ are coprime (if $\gcd(m,p-1)=d$ then $(a^m)^{(p-1)/d}\equiv (a^{p-1})^{m/d}\equiv 1\mod p$, so we need $d=1$). By the way, this is exactly why you have $\phi(p-1)$ primitive roots when $p$ is prime.

For example, $6^2=36$ or $6^{15}\equiv 686$ are not primitive roots of $761$ because $\gcd(2,760)=2>1$ and $\gcd(15,760)=5>1$, but, for example, $6^3=216$ is another primitive root of 761.

Finding a primitive root of unity for $\text{modulo-}761$.
For computing purposes, create/calculate the $\text{powers of } 2 \text{ table}$ with 95 rows and 4 columns shown in the second section, corresponding to the subgroup generated by $[2]$; the table contains $378$ elements, missing only $[1]$ and $[760]=[-1]$ from the subgroup.
From the last row of table , where duplicates (two, $[722]$ and $[39]$) are first detected by the computing algorithm,
$\tag 1 \Large2^{-95} \equiv -2^{95} \pmod{761}$
and therefore the order of $[2]$ is $380$ and $[2]$ is not a primitive root of unity.
The number $[3]$ is not in the table but (from the $65^{th}$ row),
$\tag 2 \Large2^{65} \equiv 3^{2} \pmod{761}$
The order of $[2^{65}]$ is $76$ and the order of $[3]$ is $152$ (see this) and
$[3]$ is not a primitive root of unity.
The number $[4]$ is in the table (on the $2^{nd}$ row) and therefore can't be a primitive root of unity.
The number $[5]$ is in the table (on the $50^{th}$ row, $\,\large 2^{-50} = 5$) and therefore
can't be a primitive root of unity.
The number $[6]$ is not in the table but (from the $67^{th}$ row),
$\tag 3 \Large2^{67} \equiv 6^2 \pmod{761}$
The order of $[2^{67}]$ is $380$ and the order of $[6]$ is $760$ and $[6]$ is a primitive root of unity.

$\quad \small 2^n \;\,\, 2^{-n} \quad\; -2^n \, -2^{-n}$
002 381     759 380
004 571     757 190
008 666     753 095
016 333     745 428
032 547     729 214
064 654     697 107
128 327     633 434
256 544     505 217
512 272     249 489
263 136     498 625
526 068     235 693
291 034     470 727
582 017     179 744
403 389     358 372
045 575     716 186
090 668     671 093
180 334     581 427
360 167     401 594
720 464     041 297
679 232     082 529
597 116     164 645
433 058     328 703
105 029     656 732
210 395     551 366
420 578     341 183
079 289     682 472
158 525     603 236
316 643     445 118
632 702     129 059
503 351     258 410
245 556     516 205
490 278     271 483
219 139     542 622
438 450     323 311
115 225     646 536
230 493     531 268
460 627     301 134
159 694     602 067
318 347     443 414
636 554     125 207
511 277     250 484
261 519     500 242
522 640     239 121
283 320     478 441
566 160     195 601
371 080     390 681
742 040     019 721
723 020     038 741
685 010     076 751
609 005     152 756
457 383     304 378
153 572     608 189
306 286     455 475
612 143     149 618
463 452     298 309
165 226     596 535
330 113     431 648
660 437     101 324
559 599     202 162
357 680     404 081
714 340     047 421
667 170     094 591
573 085     188 676
385 423     376 338
009 592     752 169
018 296     743 465
036 148     725 613
072 074     689 687
144 037     617 724
288 399     473 362
576 580     185 181
391 290     370 471
021 145     740 616
042 453     719 308
084 607     677 154
168 684     593 077
336 342     425 419
672 171     089 590
583 466     178 295
405 233     356 528
049 497     712 264
098 629     663 132
196 695     565 066
392 728     369 033
023 364     738 397
046 182     715 579
092 091     669 670
184 426     577 335
368 213     393 548
736 487     025 274
711 624     050 137
661 312     100 449
561 156     200 605
361 078     400 683
722 039     039 722


Big $\mathcal{O}$ Notation $f(N)=\mathcal{O}\left(\frac{1}{\sqrt{N}}\right)$ If $f(N)=\mathcal{O}\left(\frac{1}{\sqrt{N}}\right)$ (the big-oh) means that $\frac{1}{\sqrt{N}}$ grows faster than $f(N)$ as $N \to \infty$. So it is possible to take $f(N)=\frac{1}{N}$ or  $f(N)=\frac{1}{N^2}$.

Am I correct ?

When you say $f(N) = \mathcal{O} \left( g(N) \right)$ this means that $f(N)$ is equal to some function that has the property

$$
f(N) \leq c g(N)
$$

for all $N \geq N_0$, and $c$ some constant. In your case this means that there's an $N_0$ and constant $c$ for which

$$f(N) \leq \frac{c}{\sqrt{N}}.$$

Therefore your choices are both valid as candidates for $f(N)$, but trivially even $f(N) = 0$ is a suitable candidate.

By definition

  $$f(x) = \mathcal{O}(g(x)) \iff f(x) \leq cg(x)$$

In your case let $f(N)=\dfrac{1}{N^a}$ with $a \in \mathbb{N} $  and $\,g(N)=\dfrac{1}{\sqrt{N}}$, then
$$\lim_{N\to\infty}\frac{f(N)}{cg(N)} = \frac{\frac{1}{N^a}}{\frac{c}{\sqrt{N}}} = \frac{\sqrt{N}}{cN^a} = \sqrt{\frac{N}{(cN^a)^2}}= \frac{1}{cN^{\frac{a2-1}{2}}} =0 $$
Therefore
$$f(N) \leq \dfrac{c}{\sqrt{N}}\implies f(N) = \mathcal{O}\left(g(N)\right) $$
$\tag*{$\Box$}$
So you're right.


How to formulate a data fitting problem as a least squares problem Formulate the data fitting problem as a least squares problem

$\frac {1}{2} \Vert Ax-b \Vert_2^2 $

I thought I was supposed to wrote it like this:

$ \frac {1}{2} x^THx + g^T+ \gamma$ but actually that's an unconstrained
quadratic program; any help?

In data fitting, we are interested to solve:

$$\boldsymbol \theta = \underset{{\boldsymbol{\theta}} \in \mathbb{R} ^{M+1}}{\text{minimize}}J(\boldsymbol{\theta})$$

The error function $J\colon \mathbb{R^{M+1}}\to\mathbb{R}$ is given by $$J(\boldsymbol{\theta}) = \frac{1}{2N} \sum\limits_{n=1}^N \{h_{\boldsymbol{\theta}}(\boldsymbol{\phi}^{(n)})-t^{(n)}\}^2$$

where $\boldsymbol{\phi}:\mathbf{R}^D \rightarrow \mathbf{R}^{\mathcal{H}}$ a map/transformation function. The hypothesis $h_{\boldsymbol{\theta}}(\mathbf{\phi}^{(n)})$ we want to fit is given by:

$$h_{\boldsymbol{\theta}}(\boldsymbol{\phi}^{(n)}) = h(\boldsymbol{\phi}^{(n)},\boldsymbol{\theta}) =  \theta_0 + \theta_1 \phi^{(n)}_1 + \theta_2 \phi^{(n)}_2 + \dots + \theta_D \phi^{(n)}_D = \sum\limits_{d=0}^D \theta_d \phi^{(n)}_d, \quad \phi^{(n)}_0 = 1$$

If we define the parameters vector $\boldsymbol{\theta} = [\theta_0, \theta_2, \dots, \theta_D]^T \in \mathbb{R^{D+1}}$ the vectorized form of the hypothesis and the error functions respectively is $$h_{\boldsymbol{\theta}}(\boldsymbol{\phi}^{(n)}) = \boldsymbol{\theta}^T \boldsymbol{\phi}^{(n)}$$ and
$$J(\boldsymbol{\theta}) = \frac{1}{2N} (\boldsymbol{\Phi}\boldsymbol{\theta}- \mathbf{t})^T(\boldsymbol{\Phi}\boldsymbol{\theta}- \mathbf{t}) = \frac{1}{2N}||\boldsymbol{\Phi}\boldsymbol{\theta}- \mathbf{t}||^2$$ with

$$\boldsymbol{\Phi} = \begin{bmatrix}
      (\boldsymbol{\phi}^{(1)})^T \\[0.3em]
      (\boldsymbol{\phi}^{(2)})^T  \\[0.3em]
      \vdots \\[0.3em]
      (\boldsymbol{\phi}^{(n)})^T
\end{bmatrix}= \begin{bmatrix}
      \phi^{(1)}_1 & \phi^{(1)}_2 & \dots & \phi^{(1)}_D \\[0.3em]
      \phi^{(2)}_1 & \phi^{(2)}_2 & \dots & \phi^{(2)}_D \\[0.3em]
      \vdots \\[0.3em]
      \phi^{(N)}_1 & \phi^{(N)}_2 & \dots & \phi^{(N)}_D \\[0.3em]
\end{bmatrix}$$

Finally, the quadratic form of the error function is:

$$J(\boldsymbol{\theta}) = \frac{1}{2N} \Bigg\{ \boldsymbol{\theta}^T \boldsymbol{\Phi}^T\boldsymbol{\Phi}\boldsymbol{\theta} -2 \boldsymbol{t}^T \boldsymbol{\Phi} \boldsymbol{\theta} + \boldsymbol{t}^T\boldsymbol{t} \Bigg\}$$

PS: This methodology is also called multivariable linear regression.

Formulate the data fitting problem:

$\hat y(\theta)=\hat y(\theta,t)={\theta}_0+{\theta}_1e^{-t}+{\theta}_2t^2 $
$\theta=[{\theta}_0,{\theta}_1,{\theta}_2]$

$y_i=\hat y(\theta,t_i)+e_i$ $i=1,2,..,m$

as a least squares problem

$\frac {1}{2} \Vert Ax-b \Vert_2^2 $

I did this way:

A=$\begin{bmatrix}1& e^{-t}& t^2\\1& e^{-t}& t^2 \\ ..&..&..\\1& e^{-t}& t^2\end{bmatrix} $ $\in \mathbb{R}^{m \times3}$

x=$\begin{bmatrix} {\theta}_0 \\ {\theta}_1 \\ {\theta}_2 \end{bmatrix} $

$b=y_i \in \mathbb{R}^m$

$n=3$

Is it right?


Prove multiplication in fields is commutative This is Problem $16$ from Halmos' Linear Algebra Problem Book.

The problem asks whether or not multiplication must be commutative in a field.

The solution uses the distributive properties $a(b+c)=ab+ac$ and $(a+b)c=ac+bc$ with $(0+1)x$ and $x(0+1)$ to show that both $0x=x0=0$.

From here it states that this implies that multiplication is both commutative and associative. I can't seem to grasp the underlying logic of this. How does this imply that $xy=yx$ for every $x$ and $y$ in the field?

Thanks for any help!

-Tusike

From the pages preceeding the problem:

The author is intending to ask if all pairs multiplicatively commute, since we've been given that all nonzero products commute.

Thus the only thing left to prove is that $0$ commutes with everything.

As an aside, keep in mind that some older books, especially ones translated from French abstract algebra, use the word field to mean division ring (a "noncommutative field"). Nowadays fields are mostly assumed to be commutative rings, though.

Start here:

a*0 + a
= a*0 + a*1   (Because it's a group, it has a multiplicative identity element 1)

= a*(0 + 1)   (Using the distributive law)

= a*1         (0 + 1 = 1)

= a + 0     (0 is the additive identity element, which exists because it's a group)

So a*0 + a = a + 0   (Now add the inverses of a to both sides)

=> a*0 + a - a = a - a + 0

=> a*0 = 0 (Because he element + inverse equals the additive inverse, which is 0; and then an element + the additive inverse, is the element)

Do the same thing, except use 0 * a first. You've proved multiplication with 0 is commutative, which is all that matters, because we know that multiplication with between every other element is commutative


How one can obtain roots at the negative even integers of the Zeta function? The Riemann zeta function $ζ(s)$ is defined for all complex numbers $s ≠ 1$ with a simple pole at $s = 1$. It has zeros at the negative even integers, i.e., at $s = −2, −4, −6, ...$.

My question: How one can obtain these roots.

These roots arise from that the Riemann zeta function satisfies the following functional equation

$$
\zeta(s)=2^s \pi^{s-1} sin(\frac{\pi s}{2}) \Gamma(1-s)\zeta(1-s)
$$

Thus, when s=-2n, we get that

$$
\zeta(-2n)=2^{-2n} \pi^{-(2n+1)} sin(-n\pi ) \Gamma(1+2n)\zeta(1+2n)
$$

which gives us

$$
\zeta(-2n) = 0
$$
as $sin(-n \pi)=0 $ and $\Gamma(1+2n) $ is finite for all n $\in \mathbb{Z}^+ $.

$$\zeta(-2)=\sum_{k=1}^\infty \frac{1}{k^{-2}}$$
Consider the example: Zeta at negative integer $-2$. To me it is never zero, it is ever growing.


PDF of 6 six-sided fair dice sum I am trying to create a pdf for the sum of 6 six-sided fair dice. The sums range from 6 to 36 (either have all 1s or all 6s).

My attempt at the solution:

$x$ = sum of the 6 dice

$P(x=6) = C(6,0) (1/6)^6$  <--- all 1s

$P(x=7) = C(6,1) (1/6)^5 (1/6)$ <--- five 1s and one 2

$P(x=8) = C(6,1) (1/6)^5 (1/6) + C(6,2) (1/6)^4 (1/6)^2$  <----- five 1s and one 3 or four 1s and two 2s

$P(x=9) = C(6,1) (1/6)^5 (1/6) + 2\cdot C(6,2) (1/6)^4 (1/6) (1/6)$ <----- five 1s and one 4 or four 1s, one 3 and one 2

$P(x=10) = C(6,1) (1/6)^5 (1/6) + 2\cdot C(6,2) (1/6)^4 (1/6) (1/6) + C(6,2) (1/6)^4 (1/6)^2$ <----- five 1s and one 6 or four 1s, one 4 and one 2 or four 1s and two 3s

As you can see this process is extremely tedious to count all the possible combinations as the sums get larger, so I was wondering if anyone has a more concise approach/solution to this particular problem.

Thanks!

Yes considering that there are 6^6 = 46,656 combinations, counting them one-by-one isn't very practical. There is some good info here:

http://mathworld.wolfram.com/Dice.html


When do positive integers $0<q_1<k_1$ and $0<q_2<k_2$ exist such that $q_1k_1+q_2k_2=k_1k_2$? Let $k_1$ and $k_2$ be positive integers. When do positive integers $0<q_1<k_1$ and $0<q_2<k_2$ exist such that $q_1k_1+q_2k_2=k_1k_2$?
I believe that these exist precisely when gcd$(k_1,k_2) \neq 1$. Is this correct? Does somebody have a proof?
Examples: for $k_1=3, k_2=6$, one has $4 \cdot 3 + 1 \cdot 6 = 18 $ and $2 \cdot 3 + 2 \cdot 6 = 18 $. However, for $k_1=2$, $k_2=3$, this never seems to happen.

Your assumption is indeed correct.

First, we show that it is a necessary condition:

Assume $GCD(k_1,k_2)=1$. If you take a look at your equation, we gain $k_2|q_1, k_1|q_2$ (why?).

Since $q$ cannot be $0$, this leaves us with $q_1k_1+q_2k_2\geq k_1k_2+k_2k_1=2k_1k_2>k_1k_2$.

Now, to prove that our condition is sufficient, we say $k_1=l_1d, k_2=l_2d$, where $GCD(l_1,l_2)=1$.

Now the equation writes $d(q_1l_1+q_2l_2)=d^2l_1l_2 \Leftrightarrow q_1l_1+q_2l_2=dl_1l_2$.

If we say $q_1=l_2r_1, q_2=l_1r_2$, putting that in leaves us with $r_1+r_2=d$, which can be fulfilled due to $d>0$ and leaves us with a solution.


these functions are analytic, then why integration is zero? for both integrals below, consider c to be ccw unit circle.I used Cauchy integral formula and I calculated the integral below:
$$\int \frac{{e^z}^2}{z^2} dz = 0$$
I used unit circle parametrization and I calculated the integral below:
$$\int Re(z).\bar{z} dz = 0$$
But does the results of integrals being zero mean, functions inside integrals are analytic on and in c? ( bc, Cauchy Goursat says integration would yield a zero if function inside integral is analytic every where in and on c) . They are not analytic I believe, so why are the integrals become zero?

if f is continuous on G (G is simply connected)and the integral of f = 0
then f is analytic on G.

(converse of Cauchy theorem is true if f is continuous)


A simple question on position vectors. Is there any unit vector normal to all the following $3$ position vectors? $$a=\langle 2,-3,1\rangle,\quad b=\langle3,2,-4\rangle,\quad c=\langle-1,2,-2\rangle.$$ I know the answer to this. But the answer I have is apparently not the right one. I spent a whole hour arguing with my teacher on this trivial question as he seemed to have a different understanding of position vectors.  What I did: $$a\times b=\langle10,11,13\rangle=p \\ p\cdot c\neq0.$$ Therefore, position vectors $a,b,c$ do not lie on the same plane and do not have a unit normal vector common to all of them. What he says we should do: Shift the origin to '$a$' and take the position vectors $ab$ and $ac$. Then take a cross product and divide it by its magnitude to get the unit vector perpendicular to position vectors $a,b$ and $c$. I told him the vectors $ab$ and $ac$ weren't really the original position vectors and so that was wrong. But he kept stressing the point that they are position vectors and hence that can be done. He didn't seem to get my question. So who is right? I searched a lot online and nowhere did they seem to use his way. This is my last resort as I didn't really want to put up such a question here unless there was no other way.

You are absolutely right. If he does not agree ask him to take dot product of what he found out with $\vec a,\vec b ,\vec c$.

And if he still does not agree, ignore him.

your teacher is right: any three points lie on a plane. Your solution assumes a plane which goes through the origin.

What your teacher does is intuitively clear: when three points lie on a plane, it does not matter whether you move all of them by the same amount. They should still be on a plane (albeit a different one). Thus when your teacher subtracts a from all points, he ends up with (0,0), ab and ac. Next, he constructs a plane through the origin so (0,0) is in there. the normal of the plane is defined by ab x ac. and thus ab and ac are on the plane.


Calculate circle radius with given triangle The following information is given:

Given the legs of red triangle, find R.
Little 'r' is the length of the arc between 2 red intersections with circle.
It is unknown.
Is that possible or the task is not defined correctly ?

If we know the lengths of chord and arc then it is possible. The length of the chord is $d=\sqrt{8^2+6,5^2} \approx 10.31$. Now if we connect ends of the chord with the center of the circle, we will get a sector which has a measure $\theta$. We know that $\theta=\frac{r}{R}$. Thus we have an isosceles triangle formed by two radii and the chord which has vertex angle $\theta$. Draw a height from the center of the circle, the height will be a median and a bisector. This gives us another equation: $\sin \frac{\theta}{2}=\frac{d}{2R}$ or $$\color{blue}{\sin \frac{r}{2R}=\frac{d}{2R}}$$ This equation has one unknown and can be solved using numerical methods.

Assuming $r$ is the length of that arc and that it's given, then it is possible. Otherwise it isn't, since then you can make $R$ arbitrarily large.


Need help finding the fundamental set of solutions to the differential equation: $y^{(6)}+y=cos(t)$ I factor it to $(r^2+1)(r^4-r^2+1)$. Then I sub $x=r^2$ to get $(x^2-x+1)$. The roots I am getting aren't correct and I'm not sure if I should be subbing the second polynomial like how I did or not.


What does "induced operations" means in congruence operations It says: prove that R is a congruence (it means it's a relation of equivalence and it preserves operations) ith respect to sum and multiplication in $\mathbb{R}$.

$a,b\in \mathbb{R}: aRb \iff a-b \in \mathbb{Z}$.

This is the problem of my question -> "Describe the induced operations in ${\mathbb{R}}/{R}$. Also what are their properties?"

I have a counterexample for multiplication, meaning this isn't a congruence respect to multiplication in $\mathbb{R}$, am I right?

What is the meaning of "induced operations" and how can I find them all, does square a number counts as operation?, how can I be sure i've found them all?.

For congruence or an equivalence relation, you need 3 properties:

reflexive ($aRa$)
symmetric ($aRb \Leftrightarrow bRa$)
transitive ($aRb, bRc \Rightarrow aRc$)

In your case, $a-a=0 \forall a \in \mathbb{R}$, so $R$ is reflexive.

If $a,b \in \mathbb{R}$ and $a - b \in \mathbb{Z}$ then $b-a = -(a-b) \in \mathbb{Z}$, so $R$ is symmetric.

Can you prove $R$ is transitive?


Complex Measures: Variation I highly doubt a proof of mine for complex measures...

First, a complex measure can be decomposed into positive measures:
$$\mu=(\Re\mu)_+-(\Re\mu)_-+i(\Im\mu)_+-i(\Im\mu)_-=:\sum_{\alpha=0\ldots3}i^\alpha\mu_\alpha$$

Next, each of these has a Radon-Nikodym derivative:
$$\mu_\alpha(E)=\int_Eu_\alpha d|\mu|$$
giving rise to the complex Radon-Nikodym derivative:
$$\mu(E)=\sum_{\alpha=0\ldots3}i^\alpha\mu_\alpha=\int_E\sum_{\alpha=0\ldots3}i^\alpha u_\alpha d|\mu|=:\int_Eud|\mu|$$

Finally, the Radon-Nikodym derivative decomposes too:
$$u=(\Re u)_+-(\Re u)_-+i(\Im u)_+-i(\Im u)_-=:\sum_{\alpha=0\ldots3}i^\alpha u'_\alpha$$

  How to check that a fortiori: $u'_\alpha=u_\alpha$

(Put in words, wether the variation of measure agrees with the variation of derivative.)

Meanwhile, I got it by some lemma...

It respects real and imaginary variation by linearity:
$$\Re\int h\mathrm{d}\lambda=\int\Re h\mathrm{d}\lambda$$
$$\Im\int h\mathrm{d}\lambda=\int\Im h\mathrm{d}\lambda$$
as well as positive and negative variation by variation of integrals:
$$\left(\int h\mathrm{d}\lambda\right)_\pm=\int h_\pm\mathrm{d}\lambda$$
(For a rigorous check see: Integral Measures: Variation)

Therefore one has for the real and similarly imaginary part:
$$\int_Eu_\alpha\mathrm{d}|\mu|=(\Re\mu)_\pm(E)=\int_E(\Re u)_\pm\mathrm{d}|\mu|=\int_Eu'_\alpha\mathrm{d}|\mu|\quad(\alpha=0,2)$$

Concluding that they agree pointwise: $u'_\alpha=u_\alpha\mod|\mu|$


Question related with Polya urn model The Polya urn model for contagion is as follow.

We start with an urn which contains one white ball and one black ball. At each second we choose a ball at random from the urn and replace it together with one more ball of the same colour.

Calculate the probability that when $n$ balls are in the urn, $i$ of them are white.

How do you prove this probability is equal to $1/(n-1)$?

Are you sure that the probability must be that value?

I'm asking you this, because, if $w(n)$ is the amount of white balls and $b(n)$ the black balls at stage $n$, then the proportion $X(n):=\frac{w(n)}{w(n)+b(n)}$ is a martingale, respect to the natural filtration, then $\mathbb{E}[X(n)]=\mathbb{E}[X(0)]=1/2$, and therefore
$\mathbb{P}[w(n)=n/2]=1\neq \frac{1}{n-1}$.

I think what you are really want to prove is that

$\mathbb{P}[w(n)=i]=\frac{i}{n-i}$,

Of course if $i=1$, you get what you put before.


How do I calculate values of series? Possible Duplicate:
  How to find the sum of the following series

Hey there,

How do I (generally) calculate values of series? What tricks and theorems are there to make my life easier? E.g.:
$$\sum_{n=1}^\infty \frac{n^2}{2^n} = ?$$

I know perfectly well how to show whether a series converges or diverges, but (at least in this and a lot of other cases) I have no idea how to calculate the value to which it converges. I'm looking forward to your help.

Consider the more general problem $\sum {n^2}{z^n}$. To handle this, start with $\sum {z^n}$ and $\sum {n}{z^n}$. The first of these should be easy.


Consider the linear function$ f : \mathbb{R}^{2\times2}\to \mathbb{R}^{3\times2}$ defined as follows: Consider the linear function $f : \mathbb{R}^{2\times2} → \mathbb{R}^{3\times2}$ defined as follows:

$$\begin{pmatrix} r_1 & r_2 \\  r_3 & r_4 \\  \end{pmatrix} \in \mathbb{R}^{2\times2}\mapsto f\begin{pmatrix} r_1 & r_2 \\  r_3 & r_4 \\  \end{pmatrix} :=\begin{pmatrix} 3 & 2 \\  -2 & 1 \\ 0 & 4 \end{pmatrix} \begin{pmatrix} r_1 & r_2 \\  r_3 & r_4 \\  \end{pmatrix}$$

Check whether $f$ is injective and/or surjective. If it is bijective, find its inverse function. Finally, find bases for its Kernel and its Range.

Attempted solution:

This matrix multiplication evaluates to:

$$\begin{pmatrix} 3r_1+2r_3 & 3r_2+2r_4 \\  -2r_1+r_3 & -2r_2+r_4 \\ 4r_3 & 4r_4 \end{pmatrix}$$

$f$ cannot be surjective since $m>n$. However, it can be injective if $n=\text{rank}$. Therefore, it is not bijective.

Now I get stuck. I don't understand the technique to find the bases for its kernel or range nor can I calculate its rank. I know we can row reduce this but it isn't in the proper form so I'm unsure of how to proceed.

Let's calculate the kernel first. We are searching for all matrices $R=\begin{pmatrix} r_1 & r_2 \\ r_3 & r_4 \end{pmatrix}$ satisfying
$$f(R)=\begin{pmatrix} 3r_1+2r_3 & 3r_2+2r_4 \\  -2r_1+r_3 & -2r_2+r_4 \\ 4r_3 & 4r_4 \end{pmatrix} =0.
$$

So, from the last line of the above matrix we obtain $r_3=r_4=0$, and from the first line we obtain $r_1=r_2=0$. So the only matrix on the kernel is the zero matrix. Therefore, your map is injective.

Now, let's calculate the range.
$$f(R) = \begin{pmatrix} 3r_1+2r_3 & 3r_2+2r_4 \\  -2r_1+r_3 & -2r_2+r_4 \\ 4r_3 & 4r_4 \end{pmatrix}.
$$
This matrix can be written as:
$$f(R) = r_1\begin{pmatrix} 3 & 0 \\  -2 & 0 \\ 0 & 0 \end{pmatrix}
+r_2\begin{pmatrix} 0 & 3 \\  0 & -2 \\ 0 & 0 \end{pmatrix}
+r_3\begin{pmatrix} 2 & 0 \\  1 & 0 \\ 4 & 0 \end{pmatrix}
+r_4\begin{pmatrix} 0 & 2 \\  0 & 1 \\ 0 & 4 \end{pmatrix}.
$$

Therefore, the range of $f$ is generated by the matrices
$$\begin{pmatrix} 3 & 0 \\  -2 & 0 \\ 0 & 0 \end{pmatrix}
,\quad
\begin{pmatrix} 0 & 3 \\  0 & -2 \\ 0 & 0 \end{pmatrix}
,\quad
\begin{pmatrix} 2 & 0 \\  1 & 0 \\ 4 & 0 \end{pmatrix}
,\quad
\begin{pmatrix} 0 & 2 \\  0 & 1 \\ 0 & 4 \end{pmatrix}.
$$

Furthermore, from the injectivity of $f$ you actually obtain these matrices are linearly independent, and therefore they form a basis for the range.

Hint: Consider bases for $\mathbb{R}^{2 \times 2}$ and $\mathbb{R}^{3 \times 2}$.

Write the matrix representation of $f$ with respect to this pair of bases (since $f$ is a linear transformation).

Now we reduced the problem to calculating the kernel and range of a matrix, which is a straightforward task (this can always be done, for any linear transformation between finite dimensional vector spaces).


Function $f: \Bbb{R} \to \Bbb{R}$ that's not continuous using the definition that the preimage is not open. Let $f: \Bbb{R} \to \Bbb{R}$, $$f(x)=\begin{cases} 0, &x \leqslant 0, \\ 1, &x>0.\end{cases}$$ $f$ is not continuous, but by definition there exists an open set $V \subset \Bbb{R}$ whose preimage $f^{-1}(V)$ is not open in $\Bbb{R}$. Give an example of such an open set $V$.

I'm not sure  I understand the definition correctly I've tried plotting two real lines and finding a closed set in the domain such that the image would be open, but cannot seem to find it. I feel like $f^{-1}(V)=[-\frac12, \frac12]$ would have worked, but I'm not sure what will the image of this be since $f$ was defined piecewise. Any hints would be appreciated.

$f$ takes only two values. For $V=(-1/2,1/2)$ which is open in the codomain, $f^{-1}(V)=(-\infty,0]$ which is not open in the domain.
Note that for judging continuity of $f$, we need to look at preimages of open sets in the codomain. You may build a non-open set $S$ in the domain whose image $f(S)$ in the codomain is open, but the inverse image of $f(S)$ is not necessarily equal to $S:~S\subseteq f^{-1}(f(S))$. This is what precisely happens when you take $S=[-1/2,1/2],f(S)=\{0,1\}$ but $f^{-1}(f(S))=\Bbb R$ which is open in the domain.

Hint: Try and calculate $F^{-1}(V)$ where $V$ is any open set that includes $0$, but does not include $1$.


Find the smallest $n$ such that an integer lies between $nx$ and $ny$ for real numbers $x$ & $y$. Say I'm given two real numbers as inputs: $x$ and $y$, with $x < y$. I want to find the smallest natural number n such that there's at least one integer between $nx$ and $ny$ (inclusive of $nx$ and $ny$).

The largest $n$ can possibly be is $\lceil{\frac{1}{y-x}}\rceil$ because at that point the gap between $nx$ and $ny$ is at least $1$.

In general we should expect $x$ and $y$ to be 'very close' so checking all the possibilities is undesirable.

How can I determine $n$ efficiently?

I think $n$ will be the smallest denominator $d$ of any rational number $\frac{m}{d}$ in the range $[x,y]$.  We know $n \leq d$ because multiplying a fraction $\frac{m}{d} \in [x,y]$ by $d$ will give you an integer in the range $[dx, dy]$, namely $m$.  We know $n \geq d$ because if $n$ were smaller than $d$ you could construct a fraction in the range $[x,y]$ with a smaller denominator than $d$ (the fraction's denominator would be $n$ and its numerator would be the integer in range $[nx, ny]$)

So you want to find the "simplest" (smallest denominator) rational number in the range $[x,y]$.  This can be done in something like $O[\log(\frac{1}{y - x})]$ time by dividing an conquering with mediants.

The mediant of $\frac{a}{b}$ and $\frac{c}{d}$ is $\frac{a + b}{c + d}$.  The mediant of $\frac{a}{b}$ and $\frac{c}{d}$ will lie between $\frac{a}{b}$ and $\frac{c}{d}$, and (if $ad - bc = 1$) the mediant will be the rational in that range with the smallest possible denominator.  You can use this to binary search for lowest denominator fraction in the desired range

Here's some python code to search a Stern–Brocot tree to give the answer.  It starts with the range $[\frac{0}{1}, \frac{1}{1}]$ and uses the simplest fraction within that range as a pivot to binary search until it finds a fraction within $[x, y]$.

from fractions import Fraction

def simplest_fraction(x, y):
    """Finds the rational number with smallest denominator in the given range.
    Range is inclusive"""
    if x > y:
        raise ValueError('{} > {}'.format(x, y))
    if int(x) != int(y):
        return 1
    assert int(x) == int(y)
    int_part = int(x)
    x -= int_part
    y -= int_part
    left_frac, right_frac = Fraction(0), Fraction(1)
    assert left_frac < x < y < right_frac
    while left_frac < x:
        determinant = (left_frac.denominator * right_frac.numerator -
                    (left_frac.numerator * right_frac.denominator))
        assert determinant == 1, ('Mediant not guaranteed '
                                'to have smallest denominator in the range. '
                                'left: {}, right: {}, determinant: {}').format(
                                left_frac, right_frac, determinant)
        mediant = Fraction(upper.right_frac + left_frac.numerator,
                    left_frac.denominator + right_frac.denominator)
        if x <= mediant <= y:
            return mediant + int_part
        elif y < mediant:
            right_frac = bisection
        elif mediant < x:
            left_frac = mediant

def smallest_n(x, y):
    return simplest_fraction.denominator

You know that:
$1\leq n\leq \big\lceil\dfrac{1}{x-y}\big\rceil$
You can create a pretty fast algorithm for finding the lowest $n$ by splitting possibilities in half repeatedly. Here's some Python code.

from math import ceil
from math import floor

def findN(x,y):
    lower = 1
    upper = ceil(1/(x-y))

    while upper - lower >= 1:
        if floor(x * floor((upper+lower)/2)) - floor(y * floor((upper+lower)/2))     >= 1:
           upper = ceil((upper+lower)/2)
        else:
           lower = ceil((upper+lower)/2)
    return upper

If the average of the lower and upper bounds is a possibillity  for $n$. Then the average is the new upperbound and otherwise it's the new upperbound, until the difference between the two is no longer greater than 1. In the beginning, upper is a value for $n$ and upper only gets replaced with numbers that also work for $n$.

I just realised this won't always give you the lowest possible value of $n$, though often it does.


Locally symmetric spaces and the curvature tensor Let $M$ be a Riemannian manifold. Suppose $\nabla R=0$ where $R$ is the curvature tensor (we then say $M$ is locally symmetric). Then if $\gamma$ is a geodesic of $M$ and $X,Y,Z$ are parallel vector fields along $\gamma$ then $R(X,Y)Z$ is a parallel field along $\gamma$.

However, I do not fully understand what does it mean $\nabla R=0$. Is it to say that $\nabla_ZR(X,Y,W,T)=0$ for all vector fields $Z,X,Y,W,T$?

How can we use this to prove $\dfrac{D}{dt}R(X,Y)Z=0$?

The only thing I can think is maybe we can use that if $V(t)=Y(c(t))$ (i.e. $V$ is induced by a vector field) then $\dfrac{DV}{dt}=\nabla_{c'}Y$.

Any hints would be appreciated.

$R$ is a $2$-tensor, $\nabla R=0$ is equivalent to saying that for every vector field $\nabla_ZR=0$.

$\nabla_ZR(X,Y)=\nabla_Z(R(X,Y))-R(\nabla_ZX,Y)-R(X,\nabla_ZY)$.


How do i prove this inequality? Let $a,b,c,d,e,f$ be positive reals.

Then how do i prove the below inequality?

  If $ac-b^2≧0$ and $df-e^2≧0$, then $(a+d)(c+f)-(b+e)^2 ≧0$.

I hope someone answers this specifically.. Thank you in advance. :)

Another way:

consider quadratic polynomials $f(x)=ax^2+b/2 x+c$ and $g(x)=dx^2+e/2x+f$. Then your inequalities are equivalent to the negativity of discriminants of $f$ and $g$. And this is equivalent  to $f(x) \geq 0$ and $g(x) \geq 0$ for all $x$. But then $(f+g)(x) \geq 0$ for all x. So the discriminant of $f+g$ is negative. This is precisely $(a+d)(c+f)-(b+e)^2 \geq 0$

Given that $ac - b^2 \geq 0$ and $df - e^2 \geq 0$
.We can rewrite it as $ac\geq b^2$ and $df\geq e^2\quad\quad (1)$

We have $(a+d)(c+f) = ac+df+dc+af$.
The first two terms are respectively $\geq b^2$ and $e^2$ using $(1)$.

So, we have $ac + df \geq b^2 + e^2\quad \quad (2)$.

Since $a,b,c,d,e,f$ are positive reals, we can apply the AM-GM inequality as below:
$$\frac{af+dc}{2}\geq \sqrt{acdf}$$
In other words $$af + cd \geq 2\sqrt{acdf}\quad\quad (3)$$

Also, $\sqrt{acdf}\geq be$ using $(1)$ again .......$(4)$.

$(3)$ and $(4)$ gives $$af+dc\geq 2\sqrt{acdf}\geq 2be\quad\quad (5)$$

$(5)$ and $(2)$ gives your required result.


Convergence $I=\int_0^\infty \frac{\sin x}{x^s}dx$ Hi I am trying to find out for what values of the real parameter does the integral
$$
I=\int_0^\infty \frac{\sin x}{x^s}dx
$$
(a) convergent and (b) absolutely convergent.

I know that the integral is convergent if $s=1$ since
$$
\int_0^\infty \frac{\sin x}{x}dx=\frac{\pi}{2}.
$$
For $s=0$ it is easy to see divergent integral since $\int_0^\infty \sin x\, dx$ is divergent.  However I am stuck on figuring out when it is convergent AND or absolutely convergent.

I know to check for absolute convergence I can determine for an arbitrary series $\sum_{n=0}^\infty a_n$ by considering
$$
\sum_{n=0}^\infty |a_n|.
$$
If it helps also $$\sin x=\sum_{n=0}^\infty \frac{(-1)^{2n+1}}{(2n+1)!} {x^{2n+1}}$$.
Thank you all

This is a good problem to analyze.  We can solve it by just series methods and careful thought.

Given the following integral
\begin{equation}
\int_{0}^\infty \frac{\sin x}{x^s}dx, \tag1
\end{equation}
for what values of the real parameter s is the integral convergent and absolutely convergent.

(a) In order to solve this problem we break (1) into two pieces
\begin{equation}
\int_{0}^\infty \frac{\sin x}{x^s}dx=\int_{0}^1 \frac{\sin x}{x^s}dx + \int_{1}^\infty \frac{\sin x}{x^s}dx \tag2
\end{equation}
We can analyze each term separately.  It is easy to see that the term
$$
\int_{1}^\infty \frac{\sin x}{x^s}dx
$$
is divergent for $ s \leq 0$ since integral is proportional to $x^s$ which diverges as $x \to \infty$.  For $ s > 0$, the series is convergent since $x^{-s} \downarrow 0 \  \text{as}\  x \to \infty$.  We now consider the other term in (2) and write it explicitly in terms of a sum
$$
\int_{0}^1 \frac{\sin x}{x^s}dx=\int_{0}^1 \sum_{n=0}^{\infty}\frac{(-1)^n x^{2n+1}}{(2n+1)!}{x^{-s}}dx= \int_{0}^{1}\sum_{n=0}^{\infty}\frac{(-1)^n x^{2n+1-s}}{(2n+1)!}dx.
$$
We can evaluate if this integral is convergent by analyzing the series inside which is
\begin{equation}
\sum_{n=0}^{\infty}\frac{(-1)^n x^{2n+1-s}}{(2n+1)!}\equiv \xi
\end{equation}
Using the ratio test on $\xi$, we have
$$
\lim_{n\to \infty}\bigg| \frac{(-1)^{n+1} x^{2n+3-s} \cdot (2n+1)}{(2n+3)! \cdot (-1)^n x^{2n+1-s}} \bigg|=\lim_{n\to \infty} \frac{x^2}{4n^2+10n+6}=0.
$$
By the definition of the ratio test, this series is absolutely convergent since
$$
\lim_{n\to \infty} \bigg|\frac{\xi_{n+1}}{\xi_n}\bigg| =0 <1.
$$
We now check for uniform convergence by swapping the order of summation and integration, that is doing the integral first which yields
$$
\sum_{n=0}^{\infty} \frac{(-1)^n} {(2n+1)!}\int_{0}^{1} x^{2n+1-s} dx=\sum_{n=0}^{\infty} \frac{(-1)^n} {(2n+1)! \cdot (2n+2-s)}.
$$
Note, the $(2n+2-s) >0$ to be defined.  Computing the sum for $n=0$ we have the condition $2 -s > 0$, or $ 2>s$.  Evaluating the integral at $n=0, s=2$ we have
$$
\int_{0}^{1} x^{2n+1-s} dx=\int_{0}^{1} {x^{-1}} dx
$$
which diverges as the logarithm.

We can  conclude that  (1) is convergent for $s \in (0,2)$.

(b):For absolute convergence we check the convergence of
$$
\int_{0}^{\infty} \bigg|\frac{\sin x}{x^s}\bigg| dx.
$$
Once again, we break the integral into two parts
$$
\int_{0}^{\infty} \bigg|\frac{\sin x}{x^s}\bigg| dx=\int_{0}^{1} \bigg|\frac{\sin x}{x^s}\bigg| dx + \int_{1}^{\infty} \bigg|\frac{\sin x}{x^s}\bigg| dx.
$$
The second term on the right converges for $s > 1$ and is seen easily since
$$
\int_{1}^{\infty} \bigg|\frac{\sin x}{x^s}\bigg| dx < \int_{1}^{\infty} \bigg|\frac{1}{x^s}\bigg| dx
$$
which is convergent for $s > 1$.  We check the other term for convergence by noting that
$$
\bigg|\frac{\sin x}{x^s}\bigg|=\frac{\sin x}{x^s}
$$
for $ x \in [0,1]$.  Thus we conclude that
$$
\int_0^1 \frac{\sin x}{x^s}
$$
is absolutely convergent for $s \in (0,2)$.

Therefore, the integral in (1) is absolutely convergent for $s \in (1,2)$.

It is absolutely convergent for $1<s<2$.  First, write  the integral as

$\int_{0}^{\infty} \left|{\frac{\sin(x)}{x^s}}\right| \;dx = \int_{0}^{1} \left|\frac{\sin(x)}{x^s}\right| \;dx  + \int_{1}^{\infty} \left|\frac{\sin(x)}{x^s} \right|\;dx$.

Then $ \int_{1}^{\infty} \left|\frac{\sin(x)}{x^s} \right|\;dx$ converges for any $s>1$ since  $\int_{1}^{\infty} \left|\frac{1}{x^s} \right|\;dx$ converges for such $s$ (using the obvious fact that $|\sin(x)| \leq 1$).  Moreover, this integral diverges for $s\leq 1$.

For the other summand, recall that $\frac{\sin(x)}{x} < 1$ on $(0,1]$, so we bound the integrand:

$\left|\frac{sin(x)}{x^s}\right| \leq \frac{1}{x^{s-1}}$

for $x\in (0,1].$

It is well known that the integral $\int_0^1 \frac{1}{x^p}\;dx$ converges for $0<p<1$, so $\int_0^1 \frac{1}{x^{s-1}}\;dx$ converges for $1<s<2$.  It remains to show that this integral diverges for $s\geq 2$, but this could be accomplished by using your taylor series for $\sin(x)$.


A Math function that draws water droplet shape? I just need a quick reference. What is the function for this kind of shape?

Thanks.

Luckily for you, though the admonition of "there's no real reason to expect that an arbitrary shape admits a nice description" is true in general, there is such a curve that might suit your needs: the piriform of Longchamps (piriform - "pear-formed"), with the parametric equation (changed a bit from the equations in the links; I'll leave the proof of equivalence to you):

$$\begin{align*}x&=a(1-\sin\,t)\cos\,t\\y&=b(\sin\,t-1)\end{align*}$$

Here for instance is the case $a=1,\quad b=\frac52$:

On the other hand, remember that in a genuine physical drop, surface tension compels the lower bulbous portion of the drop to be (more or less) spherical; this is of course not the case with the piriform.

This can be considered as a trapezium of very small base and a arc of circle below it.


How to determine in polynomial time if a number is a product of two consecutive primes? How to determine in polynomial time if a number is a product of two consecutive primes?

All I can figure out is that if Cramér's conjecture is true, then we can use the AKS primality test to find $p_i < \sqrt n < p_{i+1}$, then check if $p_i * p_{i+1} = n$.  Is there some way to determine if a number is of this form in polynomial time that doesn't rely on any unproven assumptions?

Also, given that a number is of this form, what is the quickest way to factor it?  How fast will modern general-purpose factoring algorithms such as the quadratic sieve factor a product of consecutive primes?

If $\rm\ n = p\:q\ $ is a product of two "close" primes, i.e.  $\rm\:|p-q| < n^{1/3},\:$ then $\rm\:n\:$ can be factored in polynomial time, see Robert Erra; Christophe Grenier. The Fermat factorization method revisited. 2009. See also their slides How to compute RSA keys? The Art of RSA: Past, Present, Future.

The fastest factorization method is Number Field Sieve (NFS). To factor a 1024 bit integer $N$, it takes around $2^{86}$ many steps.


How is the value of $\pi$ ( Pi ) actually calculated? When I was a child I was taught $\pi$ (Circumference/Diameter) is an irrational number  and can be approximated to $22/7$ but $= 3.(142857)(\ldots)$.

But where does this value comes from?

In short How do I derive $\pi$?

There is the famous formula
$$\dfrac{\pi}{4} = 1 - \dfrac{1}{3} + \dfrac{1}{5} - \dfrac{1}{7} \pm \cdots.$$
Unfortunately, this converges rather slowly.  (If you compute to $n$ terms and multiply by $4$,
you get within roughly $1/n$ of the value of $\pi$; so to get 3 decimal places of accuracy, you'll need to sum something like the first thousand terms of the series.)

There is the Machin formula
$$\dfrac{\pi}{4} = 4\arctan \dfrac{1}{5} - \arctan \dfrac{1}{239},$$
which can be combined with an infinite series formula for $\arctan$
to provide a much more rapidly convergent infinite series which can be used to compute
$\pi$ to many digits of accuracy.

The Wikipedia entry provides more details about these and other methods, both historical and contemporary, for computing $\pi$.

$$\pi \;\approx\; 180 \cdot \frac{\sin 10^{-n}}{10^{-n}}$$

where $n$ is a positive real number.
The greater the value of $n$ the more exact
the value of π becomes.


Number of elements in a set of nth roots of unity I am confused about number of nth roots of unity because I know that set of nth roots of unity is a finite abelian group but somewhere I read for some n ,this is infinite group. Please tell me about these cases and whether n belongs to $\Bbb N$ or $\Bbb Z$

It is not quite clear from your phrasing which of these two cases you are talking about.

Case 1. For a particular $n \in \mathbb{N}$, consider the set of $n^\text{th}$ roots of unity under the operation of multiplication. This forms an abelian group with $n$ elements. For example, if $n = 4$, then you have the finite (four element) abelian group $\{1, -1, i, -i\}$ under multiplication.

Case 2. Consider the set of all roots of unity under the operation of multiplication. Again, this forms a group, but it has infinitely many elements. (It includes the first root of unity, the second roots of unity, the third roots of unity, etc.; essentially, it is the union of all such groups as described in the first case.)

The n-th roots of unity are best understood using the polar form of complex numbers and the exponential function. The n-th roots are $e^{2\pi ik/n}$ with $k\in\mathbb{N}$ and $0\leq k<n$. Geometrically these are the points of a regular n-gon on the unit circle in the complex plane with $1$ as a fixed point. From the polar form we can see that multiplying two of them together will add the $k_1$ and $k_2$ components together, and reducing them modulo $n$ will give us the polar form of another n-th root of unity. Because the multiplication of complex numbers is commutative, $1$ is a member of the n-th roots and $e^{2\pi i/n}$ generates the group it is an cyclic group of order $n$. There is no $n$ for which this is an infinite group.

However the entire unit circle in the complex plane forms a group and the n-th roots are a subgroup. I can also construct a group isomorphic to $\mathbb{Z}$ by taking an irrational number $q$ and generating the group with $e^{2\pi iq}$ and it's inverse. Maybe these are the infinite cases you've heard about.


Find inflection point of parametric cubic I'm familiar with turning points and inflection points for "normal" graphs (i.e. those that relate y and x) but how would I get an inflection point for a graph where each dimension has a separate equation?
In my case, given a parametric cubic curve $$\vec r(t) = \vec At^3 + \vec Bt^2 + \vec Ct + \vec D$$ , how would I determine the inflection point? Also, what does this point actually represent? Change in concavity? I am struggling to conceptualize this in 3D Vector space.
Thanks!

A parametrized curve has a tangent vector $$\overrightarrow{T} = \frac{\overrightarrow{dr}/dt}{|\overrightarrow{dr}/dt|} $$ divided by its length.
The curvature $\kappa$ (a scalar) is defined as $$\kappa =\frac{|\overrightarrow{dT}/dt|}{|\overrightarrow{dr}/dt|}$$
The unit normal vector is $$\overrightarrow{N} = \frac{\overrightarrow{dT}/dt}{\vert{\overrightarrow{dT}/dt}\vert} $$
An inflection point would be where the curvature is zero.


Subtract infinite unions, is it possible? $\cup_{n=1}^\infty F_n - \cup_{n=k}^\infty F_n$

Here, $k \ge 1$.

What is the result of this calculation? I am thinking for $k = 1$, it must be the null set, but what about $k = 2, 3 ...$? For example, for $k = 2$, is it then $F_1$?

Note that the subtraction sign means everything in the LHS set after removal of everything in the RHS set.

We have that

$$A\setminus B=A\cap B^c$$ and

$$(\cup_{i\in I}A_i)^c=\cap_{i\in I}A_i^c.$$ So

\begin{align}\cup_{n=1}^{\infty} F_n \setminus (\cup_{n=k}^{\infty} F_n)&=(\cup_{n=1}^{\infty} F_n) \cap (\cup_{n=k}^{\infty} F_n)^c\\ &=(\cup_{n=1}^{\infty} F_n) \cap (\cap_{n=k}^{\infty} F_n^c.)\end{align}

Note that if $k=2$ the result is not $F_1.$ For example, $F_1=[0,2], F_2=[1,3], F_n=\emptyset, n\ge 3.$ Then

$$\cup_{n=1}^{\infty} F_n \setminus (\cup_{n=2}^{\infty} F_n)=[0,1)\ne F_1.$$

Take all $F_i$ equal a fixed set. Then you get null. However, take $F_1$ to be arbitrary, then you can get any other set. The question is not well-posed.


Dividing by zero and imaginary numbers If we decide to call the square root of -1 its own number, then why not do the same for dividing by zero? Just make an entirely new type of number?

If you introduce a division by zero, you can only retain a subset of the usual rules of arithmetic or you run into contradictions. If $0\times x=1$ then associativity would give $$2\times 0 \times x = 0 \times x = 3 \times 0 \times x,\,2 \times 1 = 3\times 1.$$

Let's put this in a more helpful way: rather than saying that $\sqrt{-1}=i$, say that $i$ is a (particular consistent choice of) thing so that $i^2=-1$. Phrased this way, your question does have a sensible interpretation: we suppose there is a symbol $\varepsilon$ so that $\varepsilon^2=0$. We then form the ring of things of the form $a+b\varepsilon$, where $a$ and $b$ are reals; these are called dual numbers.

It is easy to show that the dual numbers are not a field: it has elements that multiply to zero (indeed, it has lots of them), so they cannot have inverses ("$1/\varepsilon$" would be the element that you multiply $\varepsilon$ by to get $1$, but there is no such element). For a fuller discussion of which dual numbers have inverses, have a look at the appropriate section of the Wikipedia article.

A concrete way to represent $i$ is the matrix $ J = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}$, while $1$ is represented by the identity matrix $I=\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$; one can show that the collection of matrices of the form $aI+bJ$ with $a,b \in \mathbb{R}$ form a field. Similarly, a concrete way to represent $\varepsilon$ is $\begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}$.

What do functions of dual numbers look like? Quite boring, actually. Let $p(x)$ be a real polynomial. Then by using the binomial theorem, we find
$$p(x+y\varepsilon)=p(x) + \varepsilon y p'(x).$$
This extends to any function with a Taylor expansion. This simplicity does actually have applications in computer algebra, as one way of implementing automatic differentiation.


If $2^{2k}-x^2\bigm|2^{2k}-1$ then $x=1$ This is the $y=2^k$ case of this question.
Suppose that $k\geq1$ and $0<x<2^k$ and $2^{2k}-x^2\bigm|2^{2k}-1$. Is it necessarily the case that $x=1$?
Equivalently: Suppose that there are two positive divisors of $2^{2k}-1$ which average to $2^k$. Is it necessarily the case that these two divisors are $2^k-1$ and $2^k+1$?

I continue from Thomas Browning's (the author of the question) answer. We desire to show that
$$nx^2-4(n-1)y^2=1$$
has no solutions. Note that any solution must satisfy $\gcd(nx,y)=1$. We can rewrite the equation as
$$(nx)^2-4n(n-1)y^2=n,$$
so if
$$x^2-4n(n-1)y^2=n$$
has no solutions with $\gcd(x,y)=1$ then we're done. I'm going to prove that using the fact that
$$\frac xy\approx \sqrt{4n(n-1)}\approx 2n$$
and then squeezing the inequalities together and proving that they're too tight to hold. This corner of number theory is called Diophantine Approximation, and I happen to know about it. Start with
$$\sqrt{4n(n-1)}=[2(n-1);\overline{1,4(n-1)}]$$
This is easier to prove backwards. Let
$$t=2(n-1)+\frac 1{1+\frac 1{t+2(n-1)}}$$
and then it's easy to find that the positive solution is $t=\sqrt{4n(n-1)}$.
Also if
$$x^2-dy^2=n$$
then
$$\frac xy=\sqrt{d+\frac n{y^2}}=\sqrt{d}\sqrt{1+\frac n{dy^2}}$$
$$\frac xy-\sqrt{d}<\frac n{2\sqrt{d}y^2}$$
In our case $n>0$ and $d=4n(n-1)$ so
$$0<\frac xy-\sqrt{4n(n-1)}<\frac 1{4y^2\sqrt{1-1/n}}$$
Now from Hardy and Wright intro to number theory page 153:
Theorem 184. If
$$\left|\frac pq -x\right|<\frac 1{2q^2}$$
then $p/q$ is a convergent.
Note that when H&W say convergent they require it to be in lowest terms. Which is true of our previous expression, so $x/y$ is a convergent of $\sqrt{4n(n-1)}$. But the residues $x^2-dy^2$ left by a convergent $\frac xy$ to the continued fraction of $\sqrt d$ are periodic with the same period as the continued fraction itself. You can verify that when $d=4n(n-1)$ the residues are $1$ and $-4(n-1)$.
\begin{align*}
[2(n-1)]&=\frac{2(n-1)}1                   &(2(n-1))^2-4n(n-1)1^2&=-4(n-1)\\
[2(n-1);1]&=\frac{2n-1}1                   &(2n-1)^2-4n(n-1)1^2&=1\\
[2(n-1);1,4(n-1)]&=\frac{8n^2-10n+2}{4n-3} &(8n^2-10n+2)^2-4n(n-1)(4n-3)^2&=-4(n-1)\\
[2(n-1);1,4(n-1),1]&=\frac{8n^2-8n+1}{4n-2}&(8n^2-8n+1)^2-4n(n-1)(4n-2)^2&=1\\
[2(n-1);1,4(n-1),1,4(n-1)]&=\frac{32n^3-56n^2+26n-2}{16n^2-20n+5}&(\dots)^2-4n(n-1)(\dots)^2&=-4(n-1)
\end{align*}
So $n$ can never be a residue, therefore our equation has no solution.

The equation resolves to $(2^k+x)(2^k-x) \mid (2^k+1)(2^k-1)$.  Since we have the first factor larger than the rest, we would look for a common factor in $(x- 1)$ and $(2^k-1)$ or $(x+1)$ and $2^k+1$,
We could suppose that the first is a product of say $(ab)(cd)$, and that the second is a product of $(ac)(bd)e$, where $ac$ divides $2^k+1$ and $bd$ divides $2^k-1$.  The common divisor between the first two and the last two factors, is $x+1$, and between the second and third, and the first and fourth, $x-1$.
But this common factor must also divide $2^k+1$ and $2^k-1$, and so must divide $2$.
So there is no other number which divides pairs of divisors (ie $2^k \pm 1$ and $2^k\pm x$).


Formula for $1/f(x)$ where $f$ is a polynomial Let $f$ be a polynomial having $n$ distinct real roots:
  $$f(x)=(x-x_1)(x-x_2)\dots(x-x_n)$$
  Prove that $$\frac{1}{f(x)}=\sum_{k=1}^n \frac{1}{f'(x_k)(x-x_k)}, \: \forall x \in \mathbb{R} - \{x_1,x_2,\dots,x_n \} $$

I don't know much about partial fractions, but this looks very much like them. I tried to use induction for this, but I couldn't really make the jump from $n-1$ to $n$ and I honestly wouldn't have been satisfied even if I had solved it that way, because I would really like to see where this formula comes from. I'm sorry if this is actually very easy, I simply don't know how to approach it

Let $s(x)$ be the summation and notice that

$$s(x)f(x)$$ is the Lagrangian interpolation polynomial* on the points $\left(x_k,y_k\right)$ where $$y_k:=\frac{\prod\limits_{\substack{i\ne k}}(x_k-x_i)}{f'(x_k)}.$$

But we precisely have

$$f'(x_k)=\prod\limits_{\substack{i\ne k}}(x_k-x_i)$$

so that $$s(x)f(x)$$ is identically $1$.

*Identify with

$$\sum_{k} y_k\frac{\prod\limits_{\substack{i\ne k}}(x-x_i)}{\prod\limits_{\substack{i\ne k}}(x_k-x_i)}.$$

For conciseness the ranges $1, n$ have been left implicit.

Have you tried decomposing $\frac{1}{f(x)}$ into partial fractions. By that i mean writing $\frac{1}{f(x)} = \frac{A_1}{x-x_1}+...+\frac{A_n}{x-x_n}$ and solving for each $A_i$? Then i think the first fraction would look something like $\frac{1}{(x-x_1)(x_1-x_2)(x_1-x_3)...(x_1-x_n)}$. Then calculating $f'(x_1)$ should get you the result you want.


How to get a lower bound on $|Mv|$ (matrix-vector product)? Suppose $M$ is a square matrix (with elements that are continuous functions which are bounded above and below) and $v$ is a vector. I want a lower bound like $$|Mv| \geq C|v|$$ for constant $C$.

Do I have any luck here?

There can't be a better lower bound than $0$, because it is possible to have $Mv = 0$ with $v \ne 0$.

If $M$ is not zero, then there is a unique nonzero greatest lower bound.

If $M$ is nonsingular, then that best lower bound is $1 / | M^{-1}|$.

In general, see https://doi.org/10.1016/j.laa.2010.02.014

Joseph Grcar, "A matrix lower bound", Linear Algebra and Its Applications, Volume 433, Issue 1, 15 July 2010, Pages 203-220


placing balls inside ball Is it possible to put pairwise disjoint open 3d-balls with radii $\frac{1}{2},\frac{1}{3},\frac{1}{4},\dots$ inside a unit ball?

not an original question,  I found it somewhere in the internet once, but without any answer.

As pointed out by @mjqxxxx, this is possible.

In fact, it is possible to pack two families of open balls of radii $\frac12, \frac13, \ldots$ into unit sphere.
The picture below illustrate one possible configuration (with $n$ up to $300$):

$\hspace1in$

In above picture, the two families are related to each other by a reflection with respect to origin.

Let us concentrate on the families of brown balls.
The brown ball with radius $\frac12$ is centered at $(0,0,-\frac12)$.
For $n \ge 3$, the brown ball with radius $\frac1n$ is centered at

$$(x_n,y_n,z_n) = (r_n\sin\theta_n\cos\phi_n,r_n\sin\theta_n\sin\phi_n, r_n\cos\theta_n)$$

where $\displaystyle\;r_n = 1 - \frac1n,\;\theta_n = \frac{\pi}{4}\left(\frac{n+1}{n-1}\right)\;$ and $\phi_n$ is defined recursively by

$$
\phi_n =
\begin{cases}
0, & n = 3\\
\phi_{n-1} + \rho_n + \rho_{n-1}, &n > 3
\end{cases}
\quad\text{ where }\quad
\rho_n = \sin^{-1}\left(\frac{1}{(n-1)\sin\theta_n}\right)
$$
If one orthogonal project the ball with radius $\frac1n$ onto $xy$-plane,
$\rho_n$ will be the half-angle subtended by the ball's image with result to origin. The condition
$\phi_{n} - \phi_{n-1} = \rho_{n} + \rho_{n-1}$ guarantee the balls with radii $\frac{1}{n}$ and $\frac{1}{n-1}$ are disjoint from each other.

For $n = 2$, the brown ball with radius $\frac12$ is touching
the blue ball with radius $\frac12$ and the brown ball with radius $\frac13$.
However, their interiors are disjoint.
For small $n$, one can build a 3d model of above configuration
and verify by eye the balls are disjoint from each other.
Above picture contains balls with $n$ upto $300$, balls with $2 < n \le 300$ have been inspected and they are disjoint from each other.
For larger $n$, the balls form a spiral converging to some sort of limit cycle at $\theta = \frac{\pi}{4}$.
For a typical ball there, its projection onto $xy$-plane will subtend an angle $2\rho_n \sim \frac{2\sqrt{2}}{n}$ with respect to the origin. Furthermore,
the balls nearby will be arranged info "arms". Let $\frac{1}{Kn}$ be the radius of nearest ball on the next arm of spiral (i.e. the arm closer to the limit cycle immediately above the current arm). The "angles" between the ball and its nearest neighbor on next arm will be around $2\pi$. This means
$$\int_{n}^{Kn} \frac{2\sqrt{2}}{n} dn = 2\sqrt{2}\log K \approx 2\pi \quad\implies\quad
K \approx e^{\pi/\sqrt{2}}$$

The distance between this pair of balls is around
$\theta_n - \theta_{Kn} \approx \frac{\pi}{2Kn} - \frac{\pi}{2n} = \frac{\pi}{2n}\left(\frac1K-1\right)$.
We can compare it with the required separation
$\frac{1}{n} + \frac{1}{Kn} = \frac{1}{n}\left(1 + \frac{1}{K}\right)$ between them.
Since the ratio $\frac{\pi}{2}\left(\frac{K-1}{K+1}\right) = \frac{\pi}{2}\left(\frac{e^{\pi/\sqrt{2}}-1}{e^{\pi/\sqrt{2}}+1}\right) \approx 1.263418 > 1$, when $n$ is large, the ball with radius $\frac1n$ is disjoint from
the one on next arm of spiral.

To bridge the analysis for small and large $n$, we need to check by the time $n$ reaches $300$, whether we can switch to use the result for large $n$ or not.

I don't have a rigorous proof. However, when we increase $n$ to around $100$, the relative error between $2\rho_n$ and its approximation $\frac{2\sqrt{2}}{n}$ already falls below $0.55\%$. The relative error in $K$ should be around $\frac{\pi}{\sqrt{2}} \cdot 0.55\% \sim 1.3\%$. As far as I can see, other approximations should generator relative errors of same order. Since the ratio $1.263418$ above is tens of percents larger than $1$. It seems $n = 100$ is already large enough.

Combining all these analysis, a safe bet is all the balls in above two families are disjoint from each other.

See if I did understand it right:

Given any two 3D-balls (and they are open), none can interset with each other. That is $$ B_n \cap B_k = \oslash$$ for any $k$ and $n$.

It is wanted to know if you can fit them all in a ball of radius $1$, while each $B_n$ has radius $\frac 1 n$ with $n >1$.

If the answer is yes, then you can fit them all without intersections. That is, the sum of their volume must be lesser than $\frac {4 \pi} 3$.

$$ \sum V_{olume} < \frac {4 \pi} 3 $$

We have to check if the above inequality is true.

Does the serie belows converge? To what number?

$$ \sum V_{olume} = \sum_{n=2}^\infty \frac {4 \pi} 3 \frac 1 {n^3}$$

And yes, it does converge to (approximately) $$\sum_{n=2}^\infty \frac {4 \pi} 3 \frac 1 {n^3} \approx 0.202157 \ \frac {4 \pi} 3$$

Which confirms that the inequality is true.

Hence, yes. You can fit all those 3D-balls inside a ball of radius $1$.

Based on Post new answer or significantly edit old wrong answer?

I deleted the old wrong answer and am posting this new one.


Volume of spheres in higher dimensions? What is the volume of spheres in higher dimensions?

In the real world, spheres are measured by their diameter, not their radius.  Were one to buy tyres, or plates, or ironmongery (bolts, nuts, drills, that sort of thing), and holes of all sorts, are measured by their diameter.

Moreover, there are named units corresponding to the area of an inch (diameter) circle, and the volume of an inch diameter sphere - circular inch, and spheric inch.  This is because the three regular solids that are the Nth power of a line, all define valid products, and coherent units.

Thus, we have the conversion factor in rexx style.

   C2P(N) = PPI( 1/FACT(N, 2),  N // 2)
       PPI(a, b) = a * (PI/2)^2
       FACT(a, b) =  a * (a-b) * (a-2b) ... while a-nb >0

In mathematical terms, one could write $c2p(n) = (\pi/2)^{n//2} / (n-1)!!$, and the volume of a sphere is then $vol(sph) = 2^N * c2p(n)$


Evaluation the integral in complex analysis $$\int_0^{2\pi}\frac{\cos^23\theta\, d\theta}{5-4\cos2\theta}.$$
In complex analysis, I want to solve the below problem. I know already the answer. But it is not easy to verify the answer. The function in the integral has a pole of order $5$ at $z=0$. it is strange.... The calculation is too hard. Please, help me...

Hint: Using the change of variable $z=\mathrm e^{\mathrm i\theta}$, this is, up to multiplicative constants I will let you discover, the complex integral on the unit circle of the function $f$ defined by
$$
f(z)=\frac{(z^3+z^{-3})^2}{5-2z^2-2z^{-2}}\frac1z\propto\frac1{z^5}g(z),
$$
where
$$
g(z)=\frac{(1+z^6)^2}{1-\frac52z^2+z^4},
$$
and the question is to determine the coefficient of $z^4$ in the expansion of $g(z)$.

To do so, first note that the numerator is $1+O(z^6)$ hence it does not contribute. Using $w=\frac52z^2-z^4$, the expansion $\frac1{1-w}=1+w+w^2+O(w^3)$ and the fact that $O(w^3)=O(z^6)$, one sees that
$$
g(z)=\frac1{1-w}+O(z^6)=1+\left(\frac52z^2-z^4\right)+\left(\frac52z^2\right)^2+O(z^6),
$$
hence the $z^4$ term in $g(z)$ is $-z^4+\left(\frac52z^2\right)^2$, from which the answer follows.


Why we need topology to define these new form open sets? By the definition of topology, I feel topology is just a principle to define "open sets" on a space(in other words, just a tool to expand the conception of open sets so that we can get some new forms of open sets.) But I think in the practical cases, we just considered Euclidean space most and the traditional form open set in Euclidean space works pretty well. And these new form open sets seem to have no common use. So why we need to expand the conception of the open sets? What's the motivation behind?Thanks.

I think your idea that topology is just to define open sets misses the point. I interpret the point of topology as attempting to generalize the concept of a continuous function primarily, and open sets are often the most direct pedagogical route there.

Further, one often does want to leave euclidean space. For example, we often want to consider spaces of continuous functions where the topology (say, pointwise convergence) may not even be metrizable! Topological spaces that are not first countable, and thus not metrizable, also arise quite naturally in functional analysis.

Even at times when one can fundamentally imbedd the space being worked on into some euclidean space, this is often a cumbersome process adding on unnecessary baggage.

We need topology so we can work limits. And limits are important.


Simplified method for symmetric matrices I am trying to find the determinant of the following  $4\times 4$ symmetric  matrix $A$ given as
$$A= \begin{pmatrix}
a & b & c & d \\
  b & c & d & a \\
  c & d & a & b \\
  d & a & b & c \end{pmatrix}$$
Any solution and help would be appreciated. Thanks.

I'll explain the situation for $3 \times 3$ matrices, an you will have an idea of how it will work for larger matrices. It's a nice technique, though I don't know about the constant at the end.

The key:

  Try to look for conditions on variables, that will make the determinant zero. That is, when we substitute these relations of variables into the determinant, maybe two rows become the same, or two columns become the same etc. Then, the remainder theorem will tell you that this determinant contains that relation, expressed as as a polynomial, as a factor.

I'll give you an example, by evaluating the $3 \times 3$ equivalent of this matrix:
$$
\begin{bmatrix}
a & b & c \\
b & c & a \\
c & a & b
\end{bmatrix}
$$

First of all, note that by doing $R_1 \to R_1 + R_2 + R_3$ :
$$
\begin{bmatrix}
a+b+c & a+b+c & a+b+c \\
b & c & a \\
c & a & b
\end{bmatrix}
$$

Now, if $a+b+c=0$, then the determinant will be zero. Hence, you see that $a+b+c$ is a factor of the given determinant.

Similarly, look at the original determinant. If $a=b=c$, then you can see that the determinant is zero. Hence, the determinant will be a multiple of $(a-b)^2 + (b-c)^2 + (c-a)^2$ also.

Now, since the degrees ($3$ and $3$) match,you can check that up to a constant, the determinant of the $3 \times 3$ case is $(a+b+c)((a-b)^2 + (b-c)^2 + (c-a)^2)$.

Similarly, in the four times four case:

1) Add all other rows to the first row, and note that if $a+b+c+d=0$ then the determinant is zero.

2) Similarly, if you let $c=a$ and $d=b$, then the first and third rows are the same, so you can check that $(c-a)^2 + (b-d)^2$ will divide the determinant.

3) Do $R_1 = R_1 -R_2 + R_3 -R_4$. This will give the first row as $[1,-1,1,-1] \times (a-b+c-d)$. Hence, if this quantity is zero, you get the determinant is zero, So this is another factor.

Finally, by comparing degrees, the product of these factors is up to a constant the determinant.

Last but not the least, since you want generalizations, I should compute this for a $5 \times 5$ determinant of that kind, and see where we can go.

So,  the $5 \times 5$ determinant will look like:
$$
\begin{bmatrix}
a & b & c & d & e \\
b & c & d & e & a \\
c & d & e & a & b \\
d & e & a & b & c \\
e & a &  b & c & d
\end{bmatrix}
$$

A look at the above matrix should tell you at least the following. Unfortunately Wolfram is hanging when I try to figure out the answer, but I want you to check at least the following:

1 ) $a+b+c+d+e$ is a factor of the determinant.

2) If $a=b=c=d=e$, then the determinant is zero, so $\frac 12 \sum_{x\neq y, (x,y) \in \{ a,b,c,d,e\}^2} (x-y)^2$ is a divisor of the given polynomial.

If not completely solve the question, at least this technique sheds light on the manner in which you can possibly approach larger matrices of this kind.

This technique is immensely useful in evaluating many symmetric, or patterned determinants up to a constant, especially the Cauchy formula for determinants of the form $\frac{1}{X_i + Y_j}$.

Some suggests a Row Reduction to create zeros in rows or columns, then use expansion by minors along that row/column. But I find this method easier to understand.

I hope it helps! :)


Write $\cos(x)$ in terms of $\sin(x)$ Write $\cos(x)$ in terms of $\sin(x)$ if the terminal point $x$ is in quadrant IV.

I know $\cos^2(x)$ $= 1-\sin^2(x)$. And I know that cos is positive in quadrant IV. I am

guessing that the answer is $\sqrt{1-\sin^2(x)}$  Can anyone help verify this answer?

Thank you!!

As written, it's not quite correct. Let's start with the identity

$$\sin^2{x} + \cos^2{x} = 1 \implies \cos^2{x} = 1 - \sin^2{x}$$

In quadrant IV, $\cos{x}$ is positive, so taking a square root gives us

$$\cos{x} = \sqrt{1 - \sin^2{x}}$$