Question from Evans' PDE book How do you do the second part of question 8, chapter 5, of Evans' PDE book (first edition)? I have proven the inequality for smooth, compactly supported functions using integration by parts, and I understand why approximating sequences as described in the hint exist, but I can not use the hint to extend the inequality as required.

The question is

"Integrate by parts to prove the interpolation inequality

$$\int_U |Du|^2\,dx\leqslant C\left(\int_Uu^2\, dx\right)^{\frac{1}{2}}\left(\int_U|D^2u|^2\,dx\right)^{\frac{1}{2}}$$

for $u\in C^\infty_c(U)$. By approximation, prove this inequality if $u\in H^2(U)\cap H_0^1(U)$.
"

The hint is to approximate $u$ by functions in $C^\infty_c(U)$ which converge to u in $H_0^1(U)$, and to approximate $u$ (also) by functions in $C^\infty(closure(U))$ which converge to $u$ in $H^2(U)$.

What you need is the following:

Let $v \in C^\infty_c(U)$ and $w\in C^\infty(\bar{U})$, we have

$$ \left(\int_U Dv \cdot Dw ~\mathrm{d}x\right)^2 \leq C \int_U |v|^2 \mathrm{d}x \int_U |D^2 w|^{2}\mathrm{d}x \tag{*}$$

This follows by directly integrating by parts (the boundary terms vanish as $v$ has compact support).

Now, given $u \in H^1_0(U) \cap H^2(U)$, let $v_i \to u$ in $H^1_0$ and $w_i \to u$ in $H^2(U)$ where $v_i \in C^\infty_c(U)$ and $w_i \in C^\infty(\bar{U})$.

By the strong convergence in $H^1_0$ and $H^2$ respectively, we have that for any function $f\in L^2$ we have

$$ \lim_{\ell \to \infty}\int_U \partial_{x^j} v_\ell f \mathrm{d}x = \lim_{\ell \to \infty}\int_{U} \partial_{x_j} (v_\ell - u + u) f \mathrm{d}x = \int_{U} \partial_{x^j} u f \mathrm{d}x + \lim_{\ell\to\infty}\int_{U} (\partial_{x_j}v_\ell - \partial_{x_j}u) f \mathrm{d}x $$

The second term on the RHS tends to zero using Cauchy-Schwarz and the assumed convergence of $v_\ell\to u$. Similarly we also have

$$ \lim_{\ell \to \infty}\int_U \partial_{x_j} w_\ell f \mathrm{d}x = \int_{U} \partial_{x_j} u f \mathrm{d}x $$

So we have that

$$ \int_U |Du|^2 \mathrm{d}x = \lim_{i,j\to \infty} \int_U Dv_i \cdot D w_j ~\mathrm{d}x \leq \lim_{i,j\to\infty} C \|v_i\|_{L^2} \|D^2 w_j\|_{L^2} $$

by (*). Since $v_i \to u$ in $H^1_0$, we also have $v_i \to u$ in $L^2$. Similarly as $w_j \to u$ in $H^2$ we have $D^2w \to D^2 u$ in $L^2$. So the RHS is

$$ \lim_{i,j\to\infty} C \|v_i\|_{L^2} \|D^2 w_j\|_{L^2} = C\|u\|_{L^2} \|D^2 u\|_{L^2}$$

and we have the desired result.

How about this?

Assume that $u \in H^2 (U) \cap H_0^1 (U).$ Then $$ \exists u_i \in C_c^\infty (U)\;\; s.t. \;\; \| u - u_i \|_{H_0^1} \longrightarrow 0 $$ $$ \exists u_j \in C^\infty (\overline U) \;\;s.t. \;\; \| u - u_j \|_{H^2} \longrightarrow 0.$$
We have $ \int_U |Du_i |^2 \longrightarrow  \int_U |Du|^2$ and $ \int_U u_i^2 \longrightarrow \int_U u^2 $. Thus we have \begin{eqnarray*}  \int_U |Du|^2 &=& \lim \int|Du_i |^2 \\ & \leqslant & \lim C \left( \int_U u_i^2 \right)^{1/2} \left( \int_U |D^2 u_i |^2 \right)^{1/2}  \\ &=& C \left( \int_U u^2 \right)^{1/2} \left( \lim \int_U |D^2 u_i |^2 \right)^{1/2} \\ & =& C \left( \int_U u^2 \right)^{1/2} \left( \int_U |D^2 u |^2 \right)^{1/2}.\end{eqnarray*} The last equality holds because \begin{eqnarray*}
\left| \int_U |D^2 u_i |^2 - \int_U |D^2 u |^2 \right| &\leqslant& \left| \int_U |D^2 u_i | - |D^2 u_j |  \right| + \left| \int_U |D^2 u_j | - |D^2 u |\right| \longrightarrow 0\;(i,j \to \infty).
\end{eqnarray*}


Find the least positive integer $m$ such that $m^2 - m + 11$ is a product of at least four not necessarily distinct primes. Find the least positive integer $m$ such that $m^2 - m + 11$ is a product of at least four not necessarily distinct primes.

I tried but not able to solve it.

Edit : Please note that I am looking for mathematical solution not the programming one.

Let the equality $$m(m-1)+11=N$$ It is clear that $N$ must be odd and that can be divisible by the prime $11$. However the following congruences have no solution as it is easy to prove directly checking the few possibilities with $m(m-1)$ not divisible by $3,5$ and $7$ respectively

$$\begin{cases}m(m-1)+2\equiv 0\pmod3\\m(m-1)+1\equiv 0\pmod5\\m(m-1)+4\equiv 0\pmod7\end{cases}$$

Thus $N=\prod p_i^{a_i}$ with $p_i\ge11$.

A necessary condition is that (solving the quadratic in $m$) $$\sqrt{4N-43}=x^2$$ and from this we need that $$N\equiv 1,3,7\pmod{10}$$ By chance we find the minimum $N=11^3\cdot13$ which corresponds to $\color{red}{m=132}$ and that could have been calculated by direct trials but the next solution seems to be far from $132$.


Proof by contradiction/contrapositive I'm just trying to make sure I have this right:

(b) Give a proof by contradiction of: “If n is an odd integer, then n
2
is odd.”

$n = 2k-1$

$n^2 = (2k-1)^2$

$a = n$ is odd

$b = n^2$ is odd

Since any integer $k$ multiplied by $2$ is even, and we subtract $1$, we get an odd result. That odd result, when squared, is always odd. Therefore, to assume that $n^2$ is even is actually a false statement, giving the following:

$a \rightarrow  \neg b$

$TRUE \rightarrow FALSE$

$FALSE$

(c) Give a proof by contraposition of: “If n is an odd integer, then n + 2 is odd.”

$n = 2k-1$

$n+2 = 2k+1$

$a = n$ is odd

$b = n+2$ is odd

Any integer $k$ multiplied by 2 is even, and subtracting 1 is always an odd result. Any odd result + 2 is still odd, so $a$ and $b$ are true. Assuming n+2 is even and n as even would be false statements, so we get the following:

$\neg b \rightarrow \neg a$

$FALSE \rightarrow FALSE$

$TRUE$

Given: (A) $n$ is odd.  Prove (B): $n^2$ is odd.

Your proof is not valid, since in your statement that "that odd result, when squared, is always odd" you have assumed that which was to be proven.

The steps to prove this by contradiction (which is indeed more roundabout than the simplest available proof!) would be:

Assume (B) is false.  Then $n^2$ is not odd.  By the definition of "odd",
there does not exist any integer $m$ such that $n^2 = 2m+1$.

In particular, since for all integer $r$, $2r^2+2r$ ias also an integer, there is no integer $r$ such that $n^2 =  2(2r^2+2r) + 1$.

This implies there is no integer $r$ such that $n^2=4r^2 + 4r + 1$, which implies  there is no integer $r$ such that $n=2r+1$ (for if there was, then square both sides to obtain $n^2=4r^2 + 4r + 1$).

So by the definition of "odd", using the symbol $r$ in place of the symbol $m$, $n$ is not odd.  But this contradicts (A) -- $n$ is odd -- which was given to be true.

Thus having assumed (B) is false, we have reached a contradiction.  Assuming that we accept proofs by contradiction, we have thus proved that (B) is true.

--

You have made the same mistake in the second case, using the theorem that was to be proven, as part of your proof, when you say "any odd result plus two is still odd."

Why are you using a contradiction to prove something?
$$(2k+1)(2k+1)= \text{ even+even+even}+1!$$ that's the simple way.


Let $f$ be a function $f : \mathbb{N} \to \mathbb{N}$ such that $f(2x+3y)=f(x)f(y)$, determine $f$ here what I did .

$$f(0)=f(0)^2$$
so $f(0)=1$ or $f(0)=0$
IF $f(0)=1$
we have $f(2y)=f(y)$
$$f(1)=f(2)=\ldots=f(2^n)=a$$
the equation $f(x)-a=0$ has infinitly many solutions , so $f(x)=a$

since f(0)=1 , $f(x)=1$.

i don't know how to handle the other case , i just found that $f(2x)=f(3x)=0$

A priori, there is no reason to expect $f$ to be a polynomial. So knowing that $f(x) = 0$ for infinitely many $x$ doesn't mean that $f$ is identically zero.

I consider the case when $f(0) = 1$. Then $f(2x + 0y) = f(x)$, so $f(x) = f(2x)$ for all $x$. Similarly, $f(x) = f(3x)$ for all $x$.

Let's evaluate $f(1)$.

Note that $f(5) = f(1\cdot 2 + 1\cdot 3) = f(1)^2$.

Also note that $13 = 5 \cdot 2 + 1 \cdot 3$ and $13 = 2 \cdot 2 + 3 \cdot 3$. Thus $f(13) = f(5)f(1) = f(1)^3$ and $f(13) = f(2)f(3) = f(1)^2$.

Thus $f(1)^2 = f(1)^3$, forcing $f(1) = 0$ or $f(1) = 1$.

It is a straightforward induction argument to show that $f(n)$ is a power of $f(1)$ for all $n \geq 1$. In either case, it is now apparent that $f(0) = 1$ and $f(n) = 1$ or $f(n) = 0$ for $n \geq 1$ are two possible solutions.

If $f(0) = 0$, then $f(2x + 0y) = f(x)f(0) = 0$, and similarly $f(3x) = 0$ for all $x$.

Let's evaluate $f(1)$ again.

Now note that $f(13) = f(5 \cdot 2 + 1 \cdot 3) = f(5)f(1)$, and $f(5) = f(1 \cdot 3 + 1 \cdot 2) = f(1)f(1)$, so taht $f(13) = f(1)^3$ (as above). But $f(13) = f(2 \cdot 2 + 3 \cdot 3) = f(2)f(3) = 0$, so that $f(1) = 0$.

Given any $n$ odd, we can write $n = 2x + 3y$ with $y = 1$. Then $f(n) = f(?)f(1) = 0$. Thus $f(n) = 0$ if $n$ is odd.

Given any $n$ even, as $f(2x) = 0$ for all $x$, we see that $f(n) = 0$. Thus we have shown that $f(x)$ is identically $0$ if $f(0) = 0$.

Thus in total, there are three such functions. They are

$f(n) = 0$ for all $n$.
$f(n) = 1$ for all $n$.
$f(0) = 1$, and $f(n) = 0$ for all $n \geq 1$.

If
$$
f\left(2x\right)=f\left(3y\right)=0 \Rightarrow f\left(2x+3y\right)=0
$$

Let's take a natural $n$, then can be written as $n=2x_1+3x_2$ because

first case : $n$ is even, so $n=2x_1$

second case : $n$ is odd, so $n-3$ is even so $n=2x_1+3$

So for all $n$, $n=2x_1+3x_2$ hence
$$
f=0
$$

EDIT
$$
f\left(2x\right)=0=f\left(3y\right) \Rightarrow f\left(2x\right)f\left(3y\right)=0 \Rightarrow f\left(2x+3y\right)=0
$$
because of the definition of $f$


issues with probability a man has $4$ children, given that atleast one of whom is a girl.Find the probability that he has $3$ girls and $1$ boy.

MY TRY :

probability of girl=$1/4$ and probability of boy=$3/4$ (my doubt is here, are these two probabiity correct? because it is said that atleast one of whom is a girl , not exactly 1)

=> 4c3 * (1/4)^4 * 3/4

I would imagine that the following assumptions hold: each child is either male or female; each child has the same chance of being male as of being female; and the sex of each child is independent of the sex of the other.

If at least one of the four children is a girl, then there are 15 equally-likely possibilities: BBBG, BBGB, BBGG, BGBB, BGBG, BGGB, BGGG, GBBB, GBBG, GBGB, GBGG, GGBB, GGBG, GGGB, GGGG.  Of these, there are four equally-likely cases where he has 3 girls and 1 boy (BGGG, GBGG, GGBG, GGGB).  Thus, the probability that he has 3 girls and 1 boy given that he has at least 1 girl is 4/15.

The man has four children of which have have atleast one girl so he may have 1 girl and three boys or two girls and two boys or three girls and one boy or four girls. so the probability that he has 3 girls and one boy is 1/4.


why we only need to consider $\lambda(\alpha)$ when we want to write the exact isomorphism? I am referring to this question Constructing Isomorphism between finite field

Consider $\mathbb{F}_3(\alpha)$ where $\alpha^3 - \alpha +1 = 0$ and $\mathbb{F}_3(\beta)$ where $\beta^3 - \beta^2 +1 =0$.

I know these two fields are isomorphic but I have difficulty buliding an isomorphism between them.

I know I have to determine where $\alpha$ is mapped to under the isomorphism map but I can't figure it out.

=======================================================
I know that two fields $F(\alpha)$ and $F(\beta)$ are isomorphic. My question is, why we only need to consider mapping of $\alpha$ when we write down the exact isomorphism?

If $\alpha$ exists in an extension ring of $R=\Bbb Z/n\Bbb Z$ (with $n$ possibly $0$), then any homomorphism of unital rings from $R[\alpha]$ is determined by where it sends $\alpha$. This is because we know $1\mapsto1$ hence we know where any integer multiple of $1$ is mapped to, and if we know where $\alpha$ is mapped to then we know where any positive power of $\alpha$ is mapped to, and hence know where any polynomial expression in $\alpha$ is mapped to. But every element of $R[\alpha]$ is expressible as a polynomial in $\alpha$.

Edit: Scrapping the second paragraph of my original answer because I didn't read the problem closely enough; I thought $\alpha$ and $\beta$ satisfied the same polynomial, but they're different. I almost didn't catch this issue, in which case I wouldn't have been able to fix it and readers wouldn't have been alerted.

In order to specify an isomorphism $\Bbb F_3(\alpha)\to\Bbb F_3(\beta)$, we've seen that it's enough to say where $\alpha$ goes, but wherever it goes to it must still satisfy $X^3-X+1$, so we need an element of $\Bbb F_3(\beta)$ which satisfies this polynomial. As it happens, a clever trick is possible for this task: what happens if you rewrite the given condition $\beta^3-\beta^2+1$ as a polynomial in $\beta^{-1}$ (divide it by $\beta^3$, in other words)?


Has anybody ever considered "full derivative"? When differentiating we usually take a limit and drop the infinitesimal terms.

But what if not to drop anything?

First, we extend the real numbers with an infinitesimal element $\varepsilon$ which has its own inverse $1/\varepsilon=\omega$.

And define the full derivative of a function formally as follows:

$$D_{full}[f(x)]=\frac{f(x+\varepsilon)-f(x)}{\varepsilon}$$

Now we can compute full derivatives of polynomials in closed form:

$$D_{full}[a]=0$$
$$D_{full}[ax]=a$$
$$D_{full}[x^2]=2x+\varepsilon$$
$$D_{full}[x^3]=3 x^2+3 \varepsilon  x+\varepsilon ^2$$

etc.

We also can find a function that remains invariant against full differentiation. It is not exponent with base $e$ though. To find it we solve the equation:

$$\frac{f(x+\varepsilon)-f(x)}{\varepsilon}=f(x)$$

The solution is a set of functions

$$C (\varepsilon +1)^{\frac{x}{\varepsilon }}$$

of which the most simple is

$$(\varepsilon +1)^{\frac{x}{\varepsilon }}$$

We can call it "full exponent" and re-define trigonometric and inverse trigonometric functions accordingly. For instance, full logarithm, sine and cosine become

$$\operatorname{flog}\,\,x=\frac{\varepsilon \ln(x)}{\ln(\varepsilon + 1)}$$

$$\operatorname{fsin}\,\,x=\frac{  (1+i\varepsilon)^{x/\varepsilon }-(1-i\varepsilon )^{x/\varepsilon }}{2i}$$

$$\operatorname{fcos}\,\,x=\frac{  (1+i\varepsilon)^{x/\varepsilon }+(1-i\varepsilon )^{x/\varepsilon }}{2}$$

etc (these full sine and full cosine satisfy the equation $f''=-f$ with full derivative).

The same expressions for differentiation occurs in time scale calculus with a scale parameter. I wonder whether anybody ever considered such operation of "full differentiation" either in the framework of non-standard analysis or time scales or otherwise and whether it has any established name?

Note that we can also in a similar way define its inverse operator, "full integral" that would be

$$\int_{full} f(x)dx=\varepsilon \lim_{t\to x/\varepsilon} \sum_t f(\varepsilon t)$$

where $\sum_t$ is indefinite sum.

Thus we get

$$\int_{full} a \,dx=ax$$

$$\int_{full} x \,dx=\frac{x^2}{2}-\frac{\varepsilon  x}{2}$$

$$\int_{full} x^2 \,dx=\frac{x^3}{3}-\frac{\varepsilon  x^2}{2}+\frac{\varepsilon ^2 x}{6}$$

$$\int_{full} a^x \,dx=\frac{\varepsilon  a^x}{a^{\varepsilon }-1}$$

$$\int_{full} \sin x \,dx=-\frac{1}{2} \varepsilon  \sin (x)-\frac{1}{2} \varepsilon  \cot \left(\frac{\varepsilon }{2}\right) \cos (x)$$

etc.

Note also that we can define full derivative in a more symmetric way:

$$D_{sym}[f(x)]=\frac{f(x+\varepsilon)-f(x-\varepsilon)}{2\varepsilon}$$

With this definition some formulas become simplier:

$$D_{sym}[e^x]=\frac{e^x \sinh (\varepsilon )}{\varepsilon }$$

$$D_{sym}[\sin x]=\frac{\sin (\varepsilon ) \cos (x)}{\varepsilon }$$

$$D_{sym}[1/x]=\frac{1}{\varepsilon ^2-x^2}$$

The invariant function for this operation, playing the role of exponent  will be

$$f(x)=\left(\sqrt{\varepsilon ^2+1}+\varepsilon \right)^{x/\varepsilon }$$

As I understand it, this is just the same as h-calculus. The h-derivative is defined as,

$$
D_{h} = \dfrac{f(x+h) - f(x)}{h}
$$

, where $h\ne 0$. [1] has a small chapter on it.

[1] Kac, V., & Cheung, P. (2002). Quantum calculus. Springer Science & Business Media.

Here's an example of retaining the infinitesimals - although they may be increments with a numerical value:

$$y = x^3$$

$$D(x^3) = 3x^2 + 3\epsilon x + \epsilon^2$$

$$D^2(x^3) = 6(x + \epsilon)$$

$$D^3(x^3) = 6$$

$D^n$(x) just means the nth derivative of x. This kind of thinking allows us to do something unusual - setting the slope of the linear segment. For example, set D($x^3$) to 5 and x to 2. The line with slope 5 that intersects the curve above x = 2 also intersects the curve in two other places. With those substitutions we have:

$$5 = 12 + 6\epsilon + \epsilon^2$$

Solving the quadratic for epsilon yields:

$$\epsilon = -1.586$$
$$\epsilon = -4.414$$

Meaning the horizontal distances between the right intersection and the two to its left correspond to those values. This can be confirmed by plotting the curve and line and measuring the distances.

NB The derivatives are worked out using:

$$D(f(x)) = \frac{f(x + \epsilon) - f(x)}{\epsilon}$$


Prove that if an even function $f(x)$ is $C^2$, $f(|x|)$ is also $C^2$ Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be a $C^2(\mathbb{R})$ function which is also even (ie, $f(-x) = f(x)$). Prove that the function $F: \mathbb{R^n} \rightarrow \mathbb{R}$ defined by $F(x) = f(|x|)$ (where $|x|$ denotes the Euclidean norm of the vector $x$) is also $C^2(\mathbb{R^n})$.

It feels so simple, but I'm just not getting it, so any help would be appreciated.

Thanks!

Hint $F(x)=f(x)$ for $x\ge 0$, and for $x<0$ we have $F(x)=f(|x|)=f(-x)=f(x)$, i.e. $F(x)=f(x)$ for all $x\in \mathbb{R}$


Find order of given factor group I'm trying to find the order of this factor group:
  $$(\mathbb Z_{12}\times\mathbb Z_{18}) / \langle (4,3)\rangle.$$

The order of the factor group is just the number of elements in it (aka the number of cosets).  Let $H = \langle(4,3)\rangle, G = Z_{12}\times Z_{18}$.

$$H = \{(4,3), (8,6), (0,9), (4,12), (8,15), (0,0)\}$$

So a coset of $H$ in $G$ is of the form $aH$, where $a \in G$

Let $a = (0,0)$.  Then $aH = H$.  From above, we know this coset has 6 elements.  Further, each coset must have the same number of elements.

So now we just take the order of G and divide by 6 to get the order of $G$ / $H$.

There are 12 choices for $Z_{12}$ and 18 choices for $Z_{18}$ so order of $G = 12 \cdot 18 = 216$.

$216 \div 6 = 36$.  So the order of $G / H$ is 36.

Does this look right?

Thanks for the help,
Mariogs

Yaaa.
Using Lagrange’s thm Which states that if G is a finite group and H is a subgroup of G then no of left (right) cosets of H in G is |G|/|H|


Preimage of cyclic subgroup under projection is Abelian Let $G$ be a finite non-Abelian group, let $Z(G)$ denote its centre, let $\pi: G \to G/Z(G)$ be the canonical projection, and let $H \vartriangleleft G/Z(G)$ be a cyclic normal subgroup of the quotient group.

Is it true that $\pi^{-1}(H)$ is an Abelian normal subgroup of $G$?

Suppose $s,t \in \pi^{-1}(H)$. Then $\pi(t),\pi(s) \in H$, so that $\pi(st) = \pi(s) \pi(t) \in H$, and hence $st \in \pi^{-1}(H)$. Similarly $\pi(s^{-1}) = \pi(s)^{-1} \in H$, so $s^{-1} \in \pi^{-1}(H)$. Hence $\pi^{-1}(H)$ is a subgroup of $G$.

Normality: Suppose $s \in \pi^{-1}(H)$ and $t \in G$: Then $\pi(t^{-1}st) = \pi(t)^{-1}\pi(s)\pi(t) \in H$, as $H< G/Z(G)$ is normal. Hence, $t^{-1}st \in \pi^{-1}(H)$.

I'm having some trouble proving Abelianity. Note that the above demonstrations only used the fact that $\pi$ is a homomorphism, but I think here we are really going to need some special structure of the projection.

My idea was to use the fact that, as $H$ is cyclic, it is generated by some $h \in H$, and so, if $s,t \in \pi^{-1}(H)$, then $\pi(s) = h^x$ and $\pi(t) = h^y$ for some $x$ and $y$, and then somehow use the fact that $h^{x+y} = h^{y+x}$ to show that $\pi^{-1}(H)$ is Abelian. But I seem to be getting lost in the group theoretic finesse. I think we may need to use the fact that we are quotienting by $Z(G)$ - in particular, by an Abelian group.

I would much appreciate it if someone could point me in the right direction.

Many thanks.

let $\langle h Z(G) \rangle=\overline H \vartriangleleft G/Z(G)$.

If $ \ g_1,g_2 \in \pi^{-1}(\overline H) \ $  then $\pi(g_k)=h^{x_k}Z(G)$ , ($k=1,2$).

Also $\pi (g_k)=g_kZ(G)$

Hence we have,

$g_1=h^{x_1}z_1$  And  $g_2=h^{x_2}z_2$

From here it follows, $g_1g_2=g_2g_1$

You can prove another claim: If $G/Z(G)$ is cyclic, then in fact $G$ is abelian.

Hint:

There are several ways to prove this. Note that $G/Z(G)$ is isomorphic to the group of innermorphisms on $G$. (Why?) So then $\text{Inn}(G)$ is cyclic with some generator $\gamma_a$; this means that for any $g \in G$ there is $n \in \mathbb Z$ such that $\gamma_g = \gamma_a^n$. Then evaluating $\gamma_g(a)$ tells us that $a$ commutes with every $g \in G$. (Why?) Hence $\gamma_g = \gamma_a^n$ acts trivially on $G$, which implies that $g$ also commutes with every element of $G$. But $g$ was arbitrary, so we conclude that $G$ is abelian.

(Fill the the details!)


Russian Roulette variant, option to point away Two people are forced to sit at a table with a pile of money, and a six chambered revolver that has one bullet in a randomly spun chamber. The first player is given the option of pointing the gun at his own head or at the ceiling and pulling the trigger. If the first player points the gun at the ceiling and pulls the trigger, then he must point it at his own head and pull the trigger a second time.

After the first player has pulled the trigger with the gun to his own head once, then it becomes the second players turn and he is given the same option as the first player.

The first player to shoot the bullet at the ceiling wins, alternatively if the opposing player shoots the bullet at his own head then the player left alive obviously wins as well. In either case the game ends there and the winner gets the money.

We know that the first time the trigger is pulled there is a 1/6 chance of a bullet coming out.

So its either:

OPTION A 1/6 chance of shooting the ceiling and winning followed by a 4/5 chance of aiming at his own head and pulling the trigger with no bullet coming out and surviving his turn to fire.

Or

OPTION B 1/6 chance of aiming at his own head and losing with the first pull of the trigger or a 5/6 chance of surviving his turn to fire.

Would it be better to try shooting the ceiling first, or aiming at your own head to try to end your turn faster?

Let us work backwards to see if there is a clear advance for either option.

If there is only one chamber left, the person will shoot at the ceiling and win.

If there are two chamber left, the person can shoot at his head, and survive with a $\frac{1}{2}$ chance and lose the game or lose and die with $\frac{1}{2}$ chance.
If  he shoots at the ceiling, he can win with $\frac{1}{2}$ chance or lose and die with $\frac{1}{2}$ chance.
So with 2 bullets left, you will shoot at the ceiling, chance of winning $\frac{1}{2}$, chance of losing $\frac{1}{2}$.

With three chamber left, you can shoot yourself in the head, and die with $\frac{1}{3}$ chance, or win with $\frac{2}{3}\cdot\frac{1}{2}=\frac{1}{3}$ chance.
If you shoot at the ceiling, you either win directly with $\frac{1}{3}$ chance, or you die with $\frac{2}{3}\cdot \frac{1}{2}=\frac{1}{3}$ chance, or you lose with $\frac{1}{3}$ as after this there is only one bullet left.
So with three bullets it doesn't matter what you do, unless you want the other person to do, in which case you shoot yourself, chance of winning $\frac{1}{3}$, chance of losing $\frac{2}{3}$.

With four chamber , you shoot yourself with $\frac{1}{4}$ chance of dying or $\frac{3}{4} \cdot \frac{2}{3} = \frac{1}{2}$ of winning.
Or you shoot at the ceiling, which gives you a $\frac{3}{4} \cdot \frac{1}{3} = \frac{1}{4}$ chance of dying, and a $\frac{1}{4}$ chance of winning directly, and a $\frac{3}{4}\cdot \frac{2}{3} \cdot \frac{1}{2}=\frac{1}{4}$ chance of winning by surviving the round.
So with four bullets left, you shoot at the ceiling, with $\frac{1}{2}$ chance of winning, $\frac{1}{2}$ chance of losing.

The same kind of argument shows that you want to shoot at the ceiling if there are 5 chamber left.

And the final argument shows that you have a $\frac{1}{2}$ chance of winning by shooting yourself or the ceiling with 6 chamber left.
So it doesn't matter what you do if you only want to win.

Let us see what your chance of survival is.
If there is 1 chamber left, you both will survive.

If there are 2 chambers left, you will survive with $\frac{1}{2}$ chance, and the other person will live.

For 3 chambers, you die $\frac{1}{3}$ of the time, and the other person either survives or die $\frac{2}{3} \times \frac{1}{2} = \frac{1}{3}$ of the time.

With 4 chambers left, shooting yourself results in $\frac{1}{4}$ or $\frac{1}{4} + \frac{3}{4}\cdot \frac{1}{3} = \frac{1}{2}$ chance of dying. The other person will have a $\frac{3}{4} \cdot \frac{1}{3} = \frac{1}{4}$ chance of dying.
Shooting the ceiling you die with $\frac{3}{4} \cdot \frac{1}{3} = \frac{1}{4}$  chance. And the other person has a $\frac{3}{4} \cdot \frac{2}{3} \cdot \frac{1}{2} = \frac{1}{4}$ chance of dying. So you shoot at the ceiling.

With 5 chambers left, shooting yourself result in a $\frac{1}{5}+\frac{4}{5} \cdot \frac{1}{4} = \frac{2}{5}$ chance of dying and $\frac{4}{5} \cdot \frac{1}{4} = \frac{1}{5}$ chance of dying for the other person.
The ceiling let you die with either $\frac{4}{5} \cdot \frac{1}{4}  = \frac{1}{5}$ or $\frac{4}{5} \cdot \frac{1}{4} +\frac{4}{5} \cdot  \frac{3}{4} \cdot \frac{1}{3} = \frac{2}{5}$ chance of dying. The other person dies with $\frac{4}{5} \cdot \frac{3}{4} \cdot \frac{1}{3} = \frac{1}{5}$chance.
So you shoot the ceiling and die $\frac{1}{5}$ or $\frac{2}{5}$ and the other person $\frac{1}{5}$.

For 6 chambers, shooting yourself gives a $\frac{1}{6} + \frac{5}{6} \cdot \frac{1}{5} = \frac{1}{3}$ of dying. The other person has a $\frac{5}{6} \cdot \frac{1}{5} = \frac{1}{6}$ or $\frac{5}{6} \cdot \frac{2}{5} = \frac{1}{3}$ chance of dying.
Shooting the ceiling gives you a $\frac{5}{6} \cdot \frac{1}{5} + \frac{5}{6}\cdot \frac{4}{5} \cdot \frac{1}{4} = \frac{1}{3}$ chance of dying. The other person has a $\frac{5}{6} \cdot \frac{4}{5} \cdot \frac{1}{4} = \frac{1}{6}$ chance of dying.

So for your own survival, it doesn't matter what you do. For the survival of the other person, you can double his/her chance to survive.

The answer I came up with is that if there are an odd number of chambers left in the gun to fire you want to aim at the ceiling, and if there are an even number of chambers left you want to aim at yourself. What I'm missing is the math to back up my answer, I just think its the right choice.


Combination with Repetitions Including Duplication of N Value Is a Combination with Repetition the correct term for the following problem

  N - Letters a, b, c
    R - 2

Example Result should equal

  aa
  ab
  ac
  bb
  ba
  bc
  cc
  ca
  cb

Total Results: 9

However, in using every online calculator for combinations with repetitions I receive a total result of 6.

Example calculator: Combination with Repetition Calculator

What is the correct term for the mathematical equation / formula that I am looking for the accomplish the example data set provided from the variables presented?

Thanks!

Assuming you are not limited in the number of each letter which appears and you are asking how many ways you can have a total of $r$ letters appearing (with possible repitition) in a specific order (i.e. "baa" is considered different from "aba" and different from "aab"), this problem is often worded as

"Find the total number of strings of length $n$ from an alphabet with $r$ characters available."

Alternatively, one can think of this as "Find the number of functions from the set $\{1,2,\dots,n\}\to\{1,2,\dots,r\}$"

In either case, the result will be $r^n$.

This can be seen using multiplication principle.

Pick what the character in the first slot is (equivalently pick what $f(1)$ is).
Pick what the character in the second slot is
$\vdots$

In each step there are $r$ possibilities.  There are a total of $n$ steps.  Multiplication principle says that to get the total number of possibilities, we multiply the number of options at each step, giving a total of $r^n$

The calculator you link to answers a related but different question.  How many multisets of size $r$ exist with $n$ elements available to choose from?

Equivalently, how many integer solutions are there to the system $\begin{cases} x_1+x_2+\dots+x_n=r\\x_i\geq 0\end{cases}$

Equivalently, how many strings of length $r$ where letters must appear in alphabetical order exist taking characters from a set with $n$ characters available.

Equivalently, how many strings of length $r$ taking characters from a set with $n$ characters available exist where order of characters doesn't matter.  (i.e. $ab$ is considered the same as $ba$)

This can be seen via stars and bars to be $\binom{n+r-1}{r}$

In the example in the original post, the possibilities are aa, ab, ac, bb, bc, cc for a total of $6$ possibilities.  (ba, ca, and cb were not included since they are already in the list written as ab, ac, and bc respectively)

That is a permutation without repetition, not a combination.

n P r = n! / (n-r)!

3 P 2 = 3! / (3-2)!

=3x2 / 1!

=6/1 = 6

Edit: I'm sorry didn't read your question properly, the correct term should be  "permutation with repetition" because the value A , B and C repeats eg. AA, BB, CC therefore it's n ^r.

3 ^2
=3x3 =9

It is a permutation because the order of the values matter example AB is different from BA.

If it's a combination then the value of AB and BA will be the same because in combination order of the value doesn't matter u just combine them.


$\int_0^1 \sqrt {\tan^{-1}x}\space dx=$? How to evaluate $$\int_0^1 \sqrt {\tan^{-1}x}\space dx\qquad ?$$ Is there an elementary expression or value for it? (though I know that there is no elementary expression for $\int \sqrt {\tan^{-1}x}\space dx$)  .

You can have a solution as an infinite sum

  $$\int_0^1 \sqrt {\tan^{-1}x}\space dx = \frac{\sqrt {\pi }}{2}-\frac{1}{\sqrt {\pi }}\sum _{n=0}^{\infty }{\frac {\psi^{(2\,n+1
)} \left( \frac{1}{2} \right) }{ 16^n\left( 4\,n+3 \right)  \left( 2\,
n+1 \right) !}} \sim 0.6298233443,$$

where $\psi(x)$ is the digamma function.

By using the formula http://upload.wikimedia.org/wikipedia/en/math/7/2/a/72a1058ad2087aec467af24bddcf9479.png, we have $\int_0^1\sqrt{\tan^{-1}x}~dx=\sum\limits_{n=1}^\infty\sum\limits_{m=1}^{2^n-1}\dfrac{(-1)^{m+1}}{2^n}\sqrt{\tan^{-1}\dfrac{m}{2^n}}$


Is $1$ raised to any complex power equal to $1$? I just saw a solution that says it is since,  for any complex number $z$

$$1^z = e^{z\log1} = e^{z(0)} = 1$$

However, isn't this only true for the principle branch of $1^z$, since by definiton, (letting capital L denote the princple value of log

$$\log1 = \operatorname{Log}1 +  i2\pi k$$

for any $k \in \mathbb{Z}$?

You are correct: $1$ is not the only value of $1^z$ if $z$ is not an integer.  For example, $-1$ is also one of the values of $1^{1/2}$.  In general the possible values of $1^z$ are $e^{2 \pi i n z}$ for integers $n$.


uniform distribution for probability The city needs to perform some road maintenance and will rent excavator machines from a company.   Each excavator will work for at least one hour and no more than $4$ hours in a day.   The working time is evenly distributed.
Given that a excavator has already worked for $2.5$ hours what is the probability that it will work at least $1$ more hour?

I know that its a uniform distribution, I just do not know how to solve the questions, I know I am looking for $\mathsf P(P>3.5 \mid P>2.5)$ but I do not know how to solve it.


Are total recursive functions recursively enumerable? In quite some literature I found that primitive recursive functions are recursively enumerable, but total recursive ones are not. To what set do total recursive functions belong?

I am asking this since I learned that even the halting problem is in recursively enumerable.

Summary: total recursive functions are not recursively enumerable; the proof follows from a typical diagonal argument.

“Recursively enumerable” is a property of a set of integers. “Partial recursive” or “total recursive” or “primitive recursive” are properties of functions from integers to integers. The concepts aren't quite on the same plane. Then there's also the RE complexity class, which is a class of functions from some problem domain to booleans; saying that a decision problem is in RE is equivalent to saying that the language that it recognizes is recursively enumerable, if the language is encoded in in the integers.

It is not possible to encode all functions as integers (by a diagonal argument, the set of all functions from integers to integers is uncountable and therefore you cannot find a distinct integer for every function).

On the other hand, it is possible to encode all partial recursive functions as integers, for example by writing these functions in a programming language and reading the bytes that make up the source code as the digits of some integer in base 256 (there are many programs that implement each function, of course, so pick the one that gives the smallest integer). Note that an encoding must be one-to-one, but need not be surjective. If you take a particular such encoding (any “sensible” one will give equivalent results in the end), then you can consider a class of partial recursive functions (the class of all partial recursive functions, the class of primitive recursive functions, the class of constant functions, etc.) and ask what kind of properties the set of all encodings of such functions have.

The set of encodings of partial recursive functions is recursively enumerable. As above, consider the source code of functions expressed in some idealized programming language. Program sources are easy to enumerate: sort strings by length then in lexicographic order, and skip the ones that are not syntactically well-formed (this is a decidable procedure). You have an enumeration of the partial recursive functions (each function will be reached an infinite number of times in the enumeration since there are infinitely many ways to implement it).

On the other hand, the set of encodings of total recursive functions is not recursively enumerable. As often, the proof is based on a diagonal argument. Let $E(f)$ be the encoding of a function $f$ and $F(x)$ be the preimage of $x$ if it exists (i.e. $E(F(x))=x$). Suppose there is an enumeration $r$ of all encodings of recursive functions, i.e. a recursive function $r : \mathbb{N} \mapsto \mathbb{N}$ such that for any total recursive function $f$, there is an integer $x$ such that $r(x)=E(f)$. Define a function $h : \mathbb{N} \mapsto \mathbb{N}$ by $h(x) = F(r(x))(x) + 1$. Then $h$ is not a total recursive function (otherwise there would be some $x$ such that $h = F(r(x))$). Yet because $r$ is recursive, $h$ is total recursive. The assumption that the set of encodings of total recursive functions is recursively enumerable is thus false.

If you take the class of all primitive recursive functions, then it turns out that the set of encodings is recursively enumerable. In other words, you can list all the primitive recursive functions. The idea is pretty similar to the case of the partial recursive functions above. There is no decision procedure that given a program source, can determine whether it is the source of a primitive recursive function (more generally, there is no such decision procedure for any non-trivial property — this is Rice's theorem). But recall that every function can be implemented in (infinitely) many ways: it's enough to enumerate one implementation of every function. By definition, any primitive recursive function can be expressed with projections, constants, the successor function, composition and primitive recursion. Whether an expression is of this form is easily decidable (it's a context-free grammar plus a check that variable names are used correctly). So there is a way to enumerate a set that contains the source code of every primitive recursive function, and only contains the source code of primitive recursive functions. This provides an enumeration of the set of all encodings of primitive recursive functions.

(The enumeration of primitive recursive functions above lists every function many times, for example the identity function also appears as $x \mapsto x+0$, $x \mapsto S(S(S(x) \times 1)-3))$, etc. Interestingly, it is possible, but a lot more difficult, to enumerate the primitive recursive functions without duplication, even though equality of primitive recursive functions is not decidable. See The Primitive Recursive Functions are Recursively Enumerable by Stefan Kahrs).

P.S. “We cannot say for some non-primitive non-halting function is it in the set of primitive ones”: true, but that would tend to indicate that the primitive recursive functions are no r.e.; as seen above, they are, because it's enough to find one way of expressing the function that we can tell is primitive recursive. “If total recursive functions are in non-r.e., that means their complement is in r.e.”: no, that's wrong, the non-total-recursive functions are not recursively enumerable either; the halting problem is not semi-decidable either way.

Gilles, I agree that the set of all programs (i.e., source code) that finitely express the class of total recursive functions is not recursively enumerable (or called computably enumerable). But the class of total recursive functions is itself recursively enumerable. Here is a proof:

Let a Turing machine M enumerate all partial recursive functions. For each step, it simulates program p on the i th input for j steps. If p stops with a result k, then it generates a record <p, i, k, m> and increases the number m by 1, here m serves as the index of the record. If p doesn't stop, it doesn't generate anything and moves on to the next step to continue the enumeration process.

Given a time in the future, M generates a finite set of such records, each of which has a unique index m. When we sort all the records in the syntactical order (or Godel numbers) of programs, we can obtain a set of records sharing the same program p. Such a set doesn't constitute the complete but a subset of the properties of p.

When we let M runs forever, i.e., in countably infinite steps, we would eventually generate the entire records that constitute the complete class of total recursive functions. (Consequently, at the end of the infinite time, of course that never comes, all the records sharing the same p constitute the entire properties of p.)

This set is very special. It doesn't resolve the halting problem unless we waited to the end of infinite time. It represents a strictly subset of all total recursive functions at any given time in the future. But it progressively accumulates more and more records, and eventually converges to the complete properties of all total recursive functions. This set is recursively enumerable.

By the way, the diagonal method you used only shows that we have uncountable total functions in the number-theoretical theory while all total recursive functions are countable. An uncountable set is certainly not recursively enumerable.


Binomial coefficients inequation problem Can anyone help me solve this:
$$5\binom{13}{x} < \binom{x + 2}{4}$$
After turning it to factorial I don't know what to do nothing seems to cancel out. $x$ is a positive integer.
I end up with this when trying to solve it:
$$\frac{13!}{x(x+1)(13 - x)!} < \frac{(x+2)!}{5!}$$
Edit: I found solutions in a book with this task, says solutions are {11, 12, 13} I've plugged them in and they seem to work. But can anyone explain how it was solved?

$$5\binom{13}{x} < \binom{x + 2}{4}$$
can be seen as:
\begin{align}
5 \, \frac{13!}{x! \, (13-x)!} &< \frac{(x+2)!}{4! \, (x-2)!} \\
5! \, 13! \, (x-2)! &< x! \, (13-x)! \, (x+2)! \\
5! \, 13! \, x! &< x(x-1) \, x! \, (x+2)! \, (13-x)! \\
5! \, 13! &< x(x-1) \, (x+2)! \, (13-x)!
\end{align}
From this can be stated that, for $x$ being an integer, $2 \leq x \leq 13$ for allowable values. Note that by inspection the right-hand binomial limits the value of $x$ to $x \leq 13$. The right-hand binomial provides the lower limit $x \geq 2$. By utilizing the allowable values of $x$ it is seen that the values $x \in \{11, 12, 13\}$ are valid solutions.


Rudin against Pugh for Textbook for First Course in Real Analysis So as I have said before in a previous question, I am taking a first course in Mathematical Analysis, and I'm quite excited. I just found out though that unlike the other professors at my university, my professor is using Real Mathematical Analysis by Pugh. I thought it was rather strange because I have read from so many places that Rudin's text on the topic is "the bible" of mathematical analysis, and also he is the only professor who doesn't use it. So I was wondering what some of you experienced mathemeticians thought of choosing this book over Rudin? Is this book a little easier to use than Rudin's? I have heard the Rudin is quite rigorous.

Every great book on a particular topic will have things another great book omits. Having seen both Rudin and Pugh, I would say they are both excellent choices for a rigorous course in mathematical analysis. Pugh's book may be easier to understand as Rudin is very terse.

Often what we call the bible is just what has been used for many years by many people, but newer alternatives exist and can be equally awesome too.

There are flaws s Rudin book. Taylors theorem is not handled nicely.
differential forms are rather messy. implicist inverse not covered very nicely.
i would say pugh offers lot of insights. but the exercies are quite hard and measure theory is not good. stokes theorem is not complete. implict and inverse function theorem are good but he should offered  general proof in final paragraph which he has left to exercises  dieuodennes foundations is classic
Bartles elments of realanalysis is a very good bookand easier bt implict inverse are not well handled.
langs ntroduces even distributions.book on undergraduate analysis is good and some insights are givem . some thorems are proved in two different ways at different places and comments that interchange order of integration and differentiation under integral sign are corelated. treatment of differential forms is excellent.


How can we show that $\varphi(a) \neq a +1$ for every $a \in A$? Let $A$ be a nonzero finite dimensional unital algebra over a field $F$ with
$char(F) = 0$, and let $\varphi$  be an automorphism of $A$.

  How can we show  that $\varphi(a)  \neq a +1$  for
  every $a \in A$?

How can we prove by $L_{\varphi (a)}$ through $\varphi$,$\varphi^{-1}$, and $L_{a}$?

( $L_{a}$  is  left multiplication map $x \mapsto ax$) .

Now it's on MathSE, let me post the answer as I thought of it.

Suppose $\varphi(a) = a + 1$. Since $A$ is finite-dimensional, $a$ has a minimal (monic) polynomial $P$. Since $\varphi(a)=a+1$, $\varphi$ preserves $F[a]$. Since it maps $0=P(a)$ to $P(a+1)$, we have $P(a+1)=0$, i.e. $Q(a)=0$, where $Q(t)=P(t+1)$. By uniqueness of the minimal monic polynomial, we have $Q=P$. So $P(t)=P(t+1)$.

It's standard to check that this implies, in characteristic zero, that $P$ is constant ($\star$), hence $P=1$. Clearly 1 cannot be the minimal polynomial of any element, and we have a contradiction.

($\star$) if $P(t)=P(t+1)$ in the polynomial ring $F[t]$, $n$ is a root of $P-P(0)$ for every integer $n\ge 0$, and since in characteristic zero these form infinitely many roots, we deduce $P-P(0)=0$, i.e., $P$ is constant.


Find the time to move up when the escalator is standstill given the following conditions? Mr. X takes 30 seconds to go up by using "moving up" escalator and takes 120 seconds to come down using "moving up" escalator. Find the time to move up when the escalator is standstill.

MyApproach:

My Confusion is this Can I take the speed of the man and esclator to be x and y?

Because they are acting on the same body so
I applied relative velocity concept

I was doing t= D/(x+y)=$30$

Similarly, another equation would be t= D/(x-y)=$120$

  Am i right in my approach?Please correct me if I am wrong?


Series involving Logs I'm trying to find the name of, and a good online reference to, a type of "logarithm series", e.g.

$$(1+x)^9 = \sum_{k=0}^{\infty} \frac{9^k\ln^k(1+x)}{k!} $$

I realise that this comes from $x^y \equiv \operatorname{e}^{y\ln x}$ and the Taylor series of the exponential function.

Wolfram-Alpha gives this as a potential series for $(1+x)^9$ alongside, for example, the binomial. I also remember reading about this type of "logarith series" some years ago. I've tried to Google it but, as you can probably imagine, it just lists the series of the natural log.

We do have Taylor's Series and Fourier Series and each is extremely useful.  There are some applications where one of them is of the most use; some where the other is; and a few where either would do.

Series in terms of powers of log(1+x) are a form of Taylor's series in which you expand f(x) in a Taylor's series, substitute log(x) for x in the series, and thus get an expansion for f(logx).  Lucian has pointed this out for the function in your example.

I think the reason you are not finding good references is that series like this are not used for general purposes as are the Taylor's and Fourier series.  Mathematicians develop analysis of ideas that are useful and widely applicable.  Since the Taylor's series include the kind of logarthmic series you are interested in, all the work that has been done in developing that theory applies to your interest.


When does an $LU$ decomposition not require row interchanges? I'm writing a program, and I want to decompose the matrix $A$ into the product of a lower triangular matrix $L$ and an upper triangular matrix $U$. So I am performing an $LU$ decomposition.

But there are also $PA = LU$ decompositions in which you must interchange rows to get things to work out. I do not want to do this, and I want to quit the program if this is necessary.

So my question is: Given a matrix $A$, how can I (quickly/efficiently) determine whether the $LU$ decomposition requires pivoting? I have searched online and cannot find anything.

Thanks

In a numeric case, i.e. $A \in \mathbb{R}^{n \times n}$, you never need pivoting on a computer. That is because you only need to use pivoting if you get zeros in places where you don't want them. However, with real numbers and finite precision on computers, you will never see a true zero, you will only see something like $1e-19$. Of course, you might still not want to divide by that, but formally, pivoting is not necessary.
If you want to handle real matrices, I would suggest reading up on them. The usual algorithms use pivoting even when not needed, to keep the result more stable (again, while you could in theory divide by something like $1e-19$ without pivoting, you really don't want to...).

Now assume you are in a case where we have exact precision, e.g. $A \in F^{n \times n}$ where $F$ is a finite field. Here, I think the computational fastest way is to really run the LU-algorithm and quit once pivoting gets necessary. Even things like eigenvalues, minimal polynomial, etc. usually take longer to compute than an LU decomposition; in fact computing such a decomposition is the first step in many algorithms. What you can do, if you don't already, is first compute only $L$, that might speed your things up a little bit.


sketching the graph of functions Good morning!

How to draw the graph of the following function and discuss the increasing and decreasing intervals of y.

$y = $x$  + \sqrt[]{x^2}$

Also, I would like to know free online graphing calculator

Thanks in advance.

Hint: Recall that $\sqrt{x^2} = |x|$, so

$$f(x) = x + |x| = \left\{\begin{array}{lr} 2x &: \text{ if } x \ge 0 \\ 0 &: \text{ if } x < 0 \end{array}\right.$$

For an online graphic calculator, I would suggest Wolfram Alpha.

You can try Calcpad, it is free for both online and desktop:

http://calcpad.net/Calculator

Just type the following:

f(x) = x + sqr(x^2)

$Plot{f(x)  @ x = -1 : 1}

This is the graph you will get

Plot 1

You can also define a function with three segments as follows:

f(x) = -2*(x > 0) + 0.5*(x ≡ 0) - x^3 * (x < 0)

The conditions in the brackets return 1 for "true" and 0 for "false" which makes possible to create an expression like the above one. It will evaluate correctly:

f(-2) = 8

f(0) = 0.5

f(2) = -2

However, it will not plot correctly as one function. You have to use the "multiple" plot command. It has the following syntax:

$Plot{x1 | y1 & x2 | y2 & x3 | y3 @ x = a : b}

For your example, type the following...

$Plot{x*(x > 0) | -2 & 0 | 0.5 & x*(x < 0) | -x^3*(x < 0) @ x = -1 : 1}

...and you will get this plot:

Plot 2


Is the function $xy-(\sin{y})(\sin{x} +2y)$ non-positive near $(0,0)$? As in the title is the function $f(x,y)=xy-(\sin{y})(\sin{x} +2y)$ non-positive near $(0,0)$?

The wolfram claims so - see the plot. On the other hand one can argue as follows. Choose arbitrary small $x>0$ since $x> \sin x$ one can find $y>0$ small enough to have $x> \sin x +2y$ since also $y> \sin y$ one easel gets $xy \ge(\sin{y})(\sin{x} +2y)$ and wolfram confirms that also for some concrete numbers - see this. What am I missing? Thanks in advance for Your help.

If you expand as Taylor series around $x=0$
$$A=x y-\sin (y) (\sin (x)+2 y)=-2 (y \sin (y))+x (y-\sin (y))+\frac{1}{6} x^3 \sin (y)+O\left(x^4\right)$$ Peform another expansion of the previous around $y=0$ to get for $y << x$ $$A\approx \frac 16 x^3 y$$ which matches the number in the second run.

Using, as in the second run, $x=10^{-3}$ and computing with illimited precision, what we can notice is that $A >0$ as soon as $y < 10^{-11}$.

Considering limit from all sides
You get,
$$\lim_{(x,y) \to (0,0)} \lfloor xy-\sin y (\sin x +2y)\rfloor = 0$$

Which indicates the fact that,
$$\lim_{(x,y)\to (0,0) } f(x,y) \to 0^{+}$$

Thus,
The function is positive near the given the point.


How can Bayesian and Frequentist approach be different? Let's say I am trying to add numbers, like say one to ten. I can either add them in order, or I can notice that I can group them into five groups of eleven, so I suppose which method to use depends on which gives the better answer in practice. Low and behold though, the answer is the same, and turns out to be 55. It turns out arithmetic is self-consistent, as should all branches of mathematics be.

Another example of using two different methods to solve a problem is Bayesian v.s. Frequentist. According to this article, given the information that the tanks 2, 6, 7, and 14, were observed, a frequentist would estimate there are 16.5 tanks. A bayesianist would estimate there are 19.5. Wait what?

How can they give different answers given the same, givens. Namely, is the estimated number of tanks 16.5 or 19.5 (or something else.) It doesn't make sense to ask what kind of person the analyzer is in getting the number of tanks. Is one of them an approximation, in the same way people say that $\pi = 3.14$ when it does not.

Another thing to note is that both are based on the same foundations. If I ask what $\infty + 1$ is, it is sensible to ask what number system one is use. This does not seem to apply in this case though. There is a certain expected value of tanks that is well defined.

I do not understand how a mathematical question can have contradictory answers in such a way.

Following up on OP's request to turn my loss function comment into an answer. Note that in statistics there is rarely a universally agreed-upon approach to an estimation problem, much less a notion of "correct" estimator. The rudimentary setup for parametric statistics is as follows.

We have a family of distributions $\{P_\theta\}_{\theta \in \Theta}$ indexed by a parameter $\theta \in \Theta$. For example, if we are trying to estimate the mean $\theta$ of a $N(\theta,1)$ normal random variable, we might take $\Theta = \mathbb{R}$. One could estimate the mean by
$$\hat{\theta} = \bar{X} = \frac{1}{n}\sum_{i=1}^n X_i,$$
where $X_i$ are i.i.d. $N(\theta,1)$. Notice that $\hat{\theta}$ takes values in $ \Theta$. This is an estimator OP is familiar with. Now I propose the estimator
$$\hat{\theta}_7 \equiv 7.$$
That's right: I estimate the completely unknown mean $\theta$ by 7. Is it a bad estimator? Maybe. But it is admissible in a sense that I will now endeavour to define.

Consider a positive loss function $L(\theta,\hat{\theta})$ defined on $\Theta \times \Theta$. $L(\theta,\hat{\theta})$ is a measure of the discrepancy between our estimate $\hat{\theta}$ and the true value $\theta$. Maybe we want to penalise every time $\hat{\theta}$ misses $\theta$:
$$L(\theta,\hat{\theta}) = 1_{\theta \neq \hat{\theta}}(X_1,\dots,X_n).$$
Because the normal distribution is continuous,  $\mathbb{P}_\theta(\bar{X} = \theta) = 0$, so we almost always have $L(\theta,\hat{\theta}) = 1$. On the other hand, I could take
$$L(\theta,\hat{\theta}) = (\theta-\hat{\theta})^2.$$
The average value of $L$ has a special name. It is known as the mean squared error (MSE) and is defined as
$$MSE(\theta,\hat{\theta}) = \mathbb{E}_\theta[(\theta-\hat{\theta})] = \mathbb{E}_\theta[L(\theta,\hat{\theta})^2].$$
The MSE has the following heuristic interpretation: given that $\theta$ is the actual value (hence the $\theta$ on $\mathbb{E}_\theta$), the MSE is the average square distance of the estimator from $\theta$.

Given a loss function $L$, an estimator $\hat\theta_1$ is said to be admissible if no other estimator $\hat\theta_2$ dominates it. That is, there is no $\hat\theta_2$ such that, for all $\theta \in \Theta$,
$$E_\theta[L(\theta,\hat\theta_2)] \leq E_\theta[L(\theta,\hat\theta_1)],$$
with strict equality for at least one $\theta$. One can interpret this as saying there is no estimator which performs better (in the sense of $L$) for every $\theta$. Going back to my bold estimator $\hat\theta_7 = 7$, it's clear that for $\theta \in \Theta\setminus\{7\}$,
$$MSE(\theta,\bar X) < MSE(\theta,\hat\theta).$$
However, in the event that $\theta$ is 7, my estimator is superb:
$$0 = MSE(7,\hat\theta_7) < MSE(7,\bar X).$$
In this sense, $\hat\theta_7$ is admissible.

I have written all this out because OP said in comments that $3.5$ is the correct estimate for the roll of a die. I wish to convey that such a mindset should be immediately dropped.

Finally, to return to OP's question, let us consider the estimation problem  being discussed. The frequentist approach described on the Wikipedia page is to use the UMVUE -- the uniformly minimum variance unbiased estimator. This turns out to be equivalent to minimising the MSE for an unbiased estimator. On the other hand, the Bayesian approach was essentially to find the $\theta$ for which $P_\theta(X_1 = 2, X_2 = 6, X_3 = 7, X_4 = 14)$ was maximised.

You said : I do not understand how a mathematical question can have contradictory answers in such a way.

The problem is that these are not "mathematical questions" as most people would like them to be.

If I am about to roll a six sided die once and ask you to estimate the result, you might say "5" and someone else might say "4." Which answer is right?


How many triplet primes of the form $p, p+2, p+4$ are there? Prove your conjecture. I am giving this problem to 8th grade students, and I am hoping that people can help me find elementary ways to prove this problem. I would love to find other arguments that are accessible to 8th graders, so that I can help them with their arguments. Here is one "proof" I have so far.

There is only one such prime triplet $(3,5,7)$. Either the first number is divisible by $3$, or it is not.

If the first number is divisible by $3$, then the number the first number is not prime (except the case $3,5,7$), so it is not a prime triplet.

If the first number is not divisible by $3$, then there are two cases. The remainder equals 1 or the remainder equals $2$.

If $r=1$, we have a group of 3s with 1 left over. The next prime triplet is $p+2$, so adding 2 to the remainder of 1 will create another group of 3. Therefore, $p+2$ is divisible by 3, so this is not a prime triplet.

If $r=2$, we have a group of 3s with 2 left over. The third prime triplet is $p+4$, so adding 4 to the remainder of 2 will create two more groups of 3. Therefore, $p+4$ is divisible by 3, so this is not a prime triplet.

So, in any case, one of the numbers will be divisible by 3 making $3,5,7$ the only prime triplet.

If $p+4$ isn't divisible by $3$, then neither is $p+1$. So the three consecutive integers $p$, $p+1$ and $p+2$ aren't divisible by $3$, which is impossible, since there are only two integers between successive multiples of $3$.

There is only 1, which is 3, 5, 7.

I think your method is most suitable for 8-grade students.

Suppose for the sake of contradiction that there exists another triplet, i.e. (p, p+2, p+4)

We also know that $p+4\equiv p+1 \pmod 3$ hence we can replace the current triplet with (p, p+1, p+2)

We know that if we have $n$ consecutive integers, then one of these integers is divisible by $n$. Proof: There are $n$ residue classes mod $n$. If you have $n$ consecutive integers, they will fill up all $n$ residue classes. One of them must be in the residue class of things divisible by $n$; i.e. things $=0\pmod{n}$.

Hence the result follows.


Some questions about Banach Tarski proof Banach-Tarski proof as been the topic of a video by the well-known Youtube channel VSauce but there were some parts that I didn't understand. So I went reading for the proof on Wikipedia, and I didn't understand the same parts too. It's about the countable infinity of "poles".

Here is a quick overview of the proof :

The infinite group with two genrators $\{A, B\}$ is called the free group $F$
Each element $X$ of $F$ can be bijectively mapped to a string $s(X)$ made of $a, b, a^{-1}, b^{-1}$ with no $aa^{-1}, a^{-1}a, bb^{-1}, b^{-1}b$ substring
If we note $\{x*\}$ the strings starting with $x$, and $e$ the empty string, then :

$F = s^{-1}(\{a*\} \cup \{b*\} \cup \{a^{-1}*\} \cup \{b^{-1}*\} \cup \{e\})$

$F = A^{-1}s^{-1}(\{a*\}) \cup s^{-1}(\{a^{-1}*\})$

$F = B^{-1}s^{-1}(\{b*\}) \cup s^{-1}(\{b^{-1}*\})$
One can chose two rotations $R$ and $T$ such that the action of their generated group $H$ on $\mathbb{S}_2$ is isomorphic to $F$.
If we note $M=\mathbb{S}_2/H$, then the equalities above rewrites as :

$\mathbb{S}_2 = s^{-1}(\{r*\} \cup \{r^{-1}*\} \cup \{t*\} \cup \{t^{-1}*\} \cup \{e\}) \times M$

$\mathbb{S}_2 = (R^{-1}s^{-1}(\{r*\}) \cup s^{-1}(\{r^{-1}*\})) \times M$

$\mathbb{S}_2 = (T^{-1}s^{-1}(\{t*\}) \cup s^{-1}(\{t^{-1}*\})) \times M$

What I don't understand how the poles are a problem. First the rotations of $H$ will have a countable infinity of poles $P$. Those cannot be elements of $H$ since there would then be a finite subgroup of $H$ and that's impossible given that $H$ is isomorph to $F$. So these poles are in a different equivalence class i.e. no fixed point string isn't it ? Then in the video, Michael states that there is a countable infinity of poles. But isn't the set of poles on the sphere $P \times M$ uncountable ? What are those troubles Wikipedia is talking about ?

  In Step 3, we partitioned the sphere into orbits of our group $H$. To streamline the proof, we omitted the discussion of points that are fixed by some rotation; since the paradoxical decomposition of $F$ relies on shifting certain subsets, the fact that some points are fixed might cause some trouble.

The poles are the points of $S^2$ that are left fixed by at least one nontrivial element of $H$. Since $H$ is countable and each element of $H$ has exactly two fixed points, there are countably many poles.

In the presence of fixed points, the fact "every point in $S^2$ can be reached in exactly one way by applying the proper rotation from $H$" that is used in step 3 of the proof as given in Wikipedia is wrong.

Therefore, one needs to get rid of these poles - which happens by the same trick we can use to let a single point disappear as in $\mathbb N\to\mathbb N$, $n\mapsto n+1$.


Sylow subgroup of a symmetric group Consider the symmetric group of$S_{20}$ and it's subgroup $A_{20}$ consisting of all even permutations. Let $H$ be a $7$-Sylow subgroup of$A_{20}$. Is $H$  cyclic? And is correct the statement which says that any $7$-Sylow subgroup of $S_{20}$ is subset of $A_{20}$?

I know that order of H is 49 and H is not normal subgroup of $A_{20}$. But I don't understand whether it is cyclic or not.

In $|S_{20}|$, the highest power of $7$ which divides $20!$ is $7^2$. So it is clear that the Sylow-$7$ subgroup of $S_{20}$ is of order $7^2$.

Group of order $7^2$ is either cyclic or  isomorphic to $Z_7\times Z_7$.

If it is cyclic, then $S_{20}$ will have an element of order $49$, and it should be product of disjoint cycles. Check, whether this is possible? You will reach to your answer.

For the last option
Sn has two types of subgroups: a subgroup which contains half even and half odd permutations or it contains only even permutations.


Closed form formula for repeated mods recursion I'm looking for a closed form formula for the recursion:

$$x_1=0$$
For $n\ge2$

$$x_n= x_{n-1}+k\mod n$$

i.e. for $k=5$,

$$x_1=0, x_2=1, x_3=0, x_4=1, x_5=1, x_6=0, x_7=5$$

It seems to me that finding a closed formula (in a traditional sense) for this kind of problem is quite hard, and maybe unfruitful (since the formula may not be efficiently computable).

But we can do the following, I don't know if this helps (I'm going to be sketchy and with an ordinal point of view, focused on the sequences as dynamical systems):

Suppose $n$ is already much bigger than $k$, and that $x_n=0$. Then, for a while, the following terms of the sequence are going to behave as a usual arithmetic progression (i.e., $x_m=k(m-n)$), until the sequence, due to its progression by $k$, catches again the size of $m'$, with its progression by $1$. Then the remainder operation will behave in a nontrivial manner and will "reset" the sequence, producing a number between $0$ and $k-1$ for $x_{m'}$.

So a good question is: Given $n$, can we find $m'$, i.e., can we predict when the sequence will reset? We can. Let call $next(n,k)$ to the first index $i$ such that the sequence $a_i:=k\cdot i$ equals or surpasses the sequence $b_i:=n+i$; then while $a_i$ covers the first $n$ elements in $n/k$ steps, $b_i$ covers $n/k$ elements more, which are covered by $a_i$ in $(n/k)/k$ steps, etc. Hence $$next(n,k)=\lceil n\cdot \sum_{i=0}^\infty\frac1{k^i}\rceil =  \lceil\frac{kn}{k-1}\rceil.$$

Therefore, until $next(n,k)$, which is efficiently computed, the sequence just behaves as a usual arithmetic progression, then resets (we compute the remainder of the corresponding term of the sequence modulo $next(n,k)$) and continues until the index hits either $next(next(n,k),k)$ or the same number minus $1$, depending on whether the remainder is big enough or not to skip one index (i.e., close to $0$ or close to $k$).

Update: I'm adding an specific example due to request from the OP.

Suppose $k=3$. Then $\displaystyle\sum_{i=0}^\infty\frac1{k^i}=\frac32$, which is the number we need to compute the next function.

Using the trick stated above for $n\leq k$ is just a bit confusing if we want to understand it from the scratch, so let us compute up to $n=4$ just by hand: $x_1=0, x_2=1, x_3=1, x_4=0$.

Now we have $x_4=0$, and we can say that at index 4 the arithmetic progression by 3 has been reset, conveniently to 0 (we'll analyze other cases in a moment).

We know that from $n=4$ onwards, $x_n$ will behave as a usual arithmetic progression, until $n$ hits $next(4,3)=4\cdot\frac32 = 6$, value in which it resets again. Then simply $x_5=3$ and $x_6=0$, since we started at value $0$ (in $x_4$) and actually hit an integer inside the next function (because the product by $\frac32$ is an integer, not half-integer).

Now we start the arithmetic progression from value $0$ at index 6 until index $next(6,3)=6\frac32=9$, which is integer and therefore hit, so that $x_7=3, x_8=6, x_9=0$.

The next arithmetic progression goes from $n=9$ until $next(9,3)=\lceil 9\frac32 \rceil = \lceil 13.5\rceil = 14$, meaning that we should have $x_{13.5}=0$ and so $13$ is the last index before reset (and $14$ the first one after reset). Therefore $x_{10}=3, x_{11}=6, x_{12}= 9, x_{13}= 12$ and $x_{14}=1$.

We can also compute directly the values at $n=13$ and $n=14$, so that we do not need to follow the whole progression: the reset did not happen at value 0 because the starting index was an odd number $n=2m+1$; this implies that $next(n,3)=3m+2$ (index in which the reset has already produced), and the value at index $3m+1$ is just $3m$ (as we go from $2m+1$ to $3m+1$, in $m$ steps and starting from 0, so that our arithmetic progression by 3 grows up to $3m$). Therefore we have proved that if we start at 0 with odd index $2m+1$, then the value at $next(n,3)=3m+2$ is $1$ (from $3m+3$ mod $3m+2$).

Next arithmetic progression goes from $n=14$, starting at 1, up to $next(14,3)=21$, ending also at value 1 (since $n=14$ is even). Hence $x_{14}=1, x_{15}=4, x_{16}=7, x_{17}=10, x_{18}=13, x_{19}=16, x_{20}=19$.

Next one goes from $n=21$ (odd) starting at 1, supposedly up to $next(21,3)=32$, but the fact that we start at odd index plus the fact that the starting value is 1 complot to make the reset one index before, at $n=31$:

We start at odd index $n=2m+1$ with value 1. Then $next(n,3)=3m+2$ as before and the value at $3m+1$ would be, as before, $3m$ if we started from 0, but as we started from 1, its actual value is $3m+1$, which is 0 mod $3m+1$.

The next nontrivial steps in our computations are $x_{31}=0$ (odd, start at 0), $x_{47}=1$ (odd, start at 1), $x_{70}=0$ (even, start at 0), $x_{105}=0$ (odd, start at 0), $x_{158}=1$, (even, start at 1), $x_{237}=1$, etc. We fill the holes with the respective arithmetic progressions.

It is easy to see that
$$
x_n=((\ldots(k \mod 2+k) \mod 3+k) \mod 4) \ldots)\mod (n-1)+k) \mod n.
$$
I don't sure that there is a closed formula.


Logic: Proving ~(a ↔ b) and (a ↔ ~b) are equivalent in LPL's Fitch I am working on a proof where I was able to derive this general form:

  ~(a ↔ b)

From this I would like to obtain:

  a ↔ ~b

I have made sure that the two statements are equivalent by drawing out truth-tables. The problem is that I am not really sure how to go from the first statement to the second statement in Fitch.

If it helps in answering the question, the specific proof I am working on is LPL's Exercise 8.50. It is the following:

1.| Cube(b) ↔ (Cube(a) ↔ Cube(c))
2.| | Dodec(b)                            (assumption)
3.| | | Cube(a)↔Cube(c)                   (assumption)
4.| | | Cube(b)                           ↔Elim 1, 3
5.| | | ⊥                                 AnaCon 2, 4
6.| | ~(Cube(a)↔Cube(c))                  ~Intro 3-5
∴ Dodec(b) → (Cube(a) ↔ ¬Cube(c))

Lines 2-6 are my work. I know that this argument is valid because if b was anything besides a cube (a dodec or a tet), then if Cube(a), it would necessarily follow that ~Cube(c) and vice versa, because if Cube(a) implied Cube(c) or vice versa, it would necessitate that b would have to be a cube.

0) $\lnot (a \leftrightarrow b)$ --- premise

1) $a$ --- assumed [a]

2) $b \to a$ --- from 1) by $\to$-intro

3) $b$ --- assumed [b]

4) $a \to b$ --- from 3) by $\to$-intro

5) $a \leftrightarrow b$ --- from 2) and 4)

6) $\bot$ --- from 0) and 5)

7) $\lnot b$ --- from 1) and 6) by $\lnot$-intro, discharging [b]

  8) $a \to \lnot b$ --- from 1) and 7) by $\to$-intro, discharging [a].

9) $\lnot b$ --- assumed [c]

10) $b$ --- assumed [d]

11) $\bot$ --- from 9) and 10)

12) $a$ --- from 11)

13) $b \to a$ --- from 10) and 12) by $\to$-intro, discharging [d]

14) $\lnot a$ --- assumed [e]

15) $a$ --- assumed [f]

16) $\bot$ --- from 14) and 15)

17) $b$ --- from 16)

18) $a \to b$ --- from 15) and 17) by $\to$-intro, discharging [f]

19) $a \leftrightarrow b$ --- from 13) and 18)

20) $\bot$ --- from 0) and 19)

21) $a$ --- from 14) and 20) by Double Negation, discharging [e]

  22) $\lnot b \to a$ --- from 9) and 21) by $\to$-intro, discharging [c]

    23) $a \leftrightarrow \lnot b$ --- from 8) and 22).

One direction of the equivalence is relatively easy to show: $A ↔ ¬B ∴ ¬(A ↔ B)$. The basic plan is to negate the goal, derive a contradiction which will then allow one to derive the goal.

Here is a proof using a Fitch-style natural deduction proof checker:

To show the other direction is more involved. Again, one can negate the goal, $A ↔ ¬B$, but one should rewrite the goal as a conjunction of disjunctions, $(¬A ∨ ¬B) ∧ (A ∨ B)$, so one can use other inference rules. Once one reaches a contradiction this will give one the rewritten goal from which one can derive the desired biconditional.

Here is a proof of the other direction, $¬(A ↔ B) ∴ A ↔ ¬B$:

For the proof checker and the associated textbook see the links below.

Kevin Klement's JavaScript/PHP Fitch-style natural deduction proof editor and checker http://proofs.openlogicproject.org/

P. D. Magnus, Tim Button with additions by J. Robert Loftis remixed and revised by Aaron Thomas-Bolduc, Richard Zach, forallx Calgary Remix: An Introduction to Formal Logic, Fall 2019. http://forallx.openlogicproject.org/forallxyyc.pdf


Prove properties of exponential function using a limit definition I am using ${\displaystyle \lim_{n \to \infty} \left( 1 + \frac{x}{n} \right)^{n}}=e^x$ and want to show that $e^{x+y} =e^xe^y$ and $e^{-x}=(e^x)^{-1}$ using only the given limit.

The trouble is that I believe $e^{x+y}=e^xe^y$ is not true for large values of $n$

Since
$$\left(1+\frac{x}{n}\right)^n\cdot\left(1+\frac{y}{n}\right)^n = \left(1+\frac{x+y}{n}+\frac{xy}{n^2}\right)^n=\left(1+\frac{x+y}{n}\right)^n\cdot\left(1+\frac{xy}{n^2+nx+ny}\right)^n$$
we just need to show that for any $x,y$ we have:
$$\lim_{n\to +\infty}\left(1+\frac{xy}{n^2+nx+ny}\right)^n=1\tag{1}$$
that follows from:
$$\lim_{n\to +\infty}\left(1+\frac{xy}{n^2}\right)^n=1.\tag{2}$$

At $x$ units, the value of the function is $e^x$. What happens when we move $y$ units furthur? The value then is $e^{x+y}$.

Divide the distance between $x$ and $x+y$ into $n$ intervals where $n$ is very large. The length of each interval is $\frac{y}{n}$. When we move from $x$ to $x+\frac{y}{n}$, what is the new value of the function?

You can think of $e^x$ as the principal money deposited in the bank. To get the new value, just multiply the principal money by $1+\frac{y}{n}$. This works because the exponential function is its own derivative. This is simply using the equation $e^{x+dx}=e^x+\frac{d(e^x)}{dx}dx=e^x(1+dx)$.

So the new value is $e^x(1+\frac{y}{n})$. This new value will work as our new principal money. When we furthur move by $\frac{y}{n}$ units, just multiply the new principal money by $1+\frac{y}{n}$ to get the next principal money. The next principal money is $e^x(1+\frac{y}{n})^2$

In this way, we compound our principal money $e^x$ for $n$ intervals (to get from $e^x$ to $e^{x+y}$). After compounding $n$ times, the result is obviously $e^x(1+y/n)^n$. As $n$ tends to infinity, this equals $e^xe^y$.

There's another identity that $(e^a)^x=e^{ax}$.

You can look at the derivative of $(e^a)^x$ to see why that indentity works. The derivative is $a(e^a)^x$. The base $e^a$  just means we're compounding $a$ times faster than the usual base $e$, for $x$ units of time. Rate and time are obviously multiplicative in calculating interest. At $0$, the function equals $1$. At $0+\frac{x}{n}$, it equals $1(1+a\frac{x}{n})$. After compounding $n$ times, it equals ${(1+\frac{ax}{n})^n}$, which is $e^{ax}$ as $n$ tends to infinity.


Projection expressed as a Matrix * a Vector So, let's say we want the projection of vector v [2, 1] on to L (line L) L = C[2, 1]), where vector x is a vector that lies along the Line.
I know that one way to figure out the projection would be  C*vectorX (C representing some scaled version of vector x).
Then, a second way to figure out the projection would be to normalize vector x. then C = (vector v * vector x) * vector x (or vector x as vector u (normalized).
But the final way to figure it out would be to see the projection as a linear transformation. Projection is Matrix A * vector v. For some reason when expressed as a Matrix / Vector product, it is less clear/intuitive for me.
The key point of confusion: I'm not sure for example how we get from a normalized version of vector x to expressing it as a Matrix (let's say Matrix A). Where then we would simply multiply it by vector v.
Hopefully this is clear.

A fact about projection matrices i hope you would find interesting is that they are always positive semi-definite. All eigenvalues are all ones (for a vector that is in column space) or zeros (for a vector that is not in column space). I am very new here. Sorry for my answer being improper.
I got your question better after reading it again. For $$P = A(A^T A)^{-1}A$$. $$ (A^TA)$$ does the weighting job here. If A were an orthogonal matrix it would be the identity. You know that for a vector v,$$v^Tv$$ is the square of the 2-norm of the vector. This could be the relation you are looking for.


Subgroup of order $n-1$ of a group of order $n$ Here is question 2.1.5 from Dummit and Foote : Prove that $G$ cannot have a subgroup $H$ with $|H| = n-1$, where $n = |G| > 2$.

How can one show this without using Lagrange's theorem (which is in chapter 3 of Dummit).

Thank you

We know that any subgroup must have the identity element in it. We also know that every subgroup must contain inverses for all of its elements. Suppose that $H$ is a subgroup of order $n-1$. Let $x$ designate the one element of $G$ not in $H$. Then $x$ must be its own inverse, as if $x^{-1} \not= x$ we have that $x^{-1}\in H$, yet $x$ is not in $H$ (which is a contradiction).

Now take any non-identity $y\in H$. Then if $xy$ is in $H$, then this implies that $x$ is in $H$ since we can multiply by $y^{-1}$. The only way that $xy$ is not in $H$ is if $y=1$. But we assumed otherwise.

We then reach the contradiction.

Consider an element $x$ in $G$ and not in $H$. Since $|H| = n - 1$, $x^{-1}$ must be in $H$. However, the inverse is unique so it's easy to see that $x^{-1}$ has no inverse in $H$ because $x$ is not in $H$.


Is it possible to construct an isosceles triangle by using a ruler and without using a pair of compasses? It is well known that on Euclidean plane one can construct an isosceles triangle on given straight line by using a ruler and a pair of compasses.

Also it is possible to construct straight line containing given point and parallel to given straight line by using only a ruler, with the condition that we can measure out a segment equal to any given segment.

But in this conditions there are difficulties with constructing perpendicular for given straight line. Ofcourse it is equivalent to constructing of an isosceles triangle on given straight line by using only a ruler, with the condition that we can measure out a segment equal to any given segment.

I haven't yet able to do this, also I haven't able to prove that it is impossible. Do anybody know something about this topic?

Thanks in advance.

Yes, this is possible.

It is possible to construct a pair of perpendicular lines somewhere on plane. To do this, make a rhombus. (On two lines that meet at a point A, construct segments of the same length starting from A. Then constructing parallels yields a parallelogram with two adjacent sides of equal length. This is a rhombus.) The diagonals of the rhombus are perpendicular.

Think of the two perpendicular lines as the x-axis and the y-axis, so we may draw horizontal and vertical lines through any point. This allows us to assume that the origin $(0,0)$ is on the given line. Choose some other point $(x,y)$ on the line. The horizontal and the vertical lines through $(x,y)$ meet the two axes at $(x,0)$ and $(0,y)$. So we may construct the points $(y,0)$ and $(0,−x)$. The vertical line through $(y,0)$ meets the horizontal line through $(0,−x)$ at the point $(y,−x)$. The line joining $(0,0)$ and $(y,−x)$ is perpendicular for the line line joining $(0,0)$ and $(x, y)$.

Wait, @EvgenyKuznetsovs, why are you answering your own questions? You literally answered one day after you asked. Its not like you put any real research into this if you could find an answer so soon after asking - or perhaps you already knew it when you asked?  And youre also going to award best answer to yourself, on top of it?

If you have the ability to measure length then simply extend two congruent line segments from any given point you wish, with any angle between them you wish, then connect their other endpoints.  Voila, you have an isosceles triangle. Being able to measure length is immensely powerful.

This business about parallel and perpendicular lines and rhombi is a bit unnecessary and excessive.  Don't you think? You only asked if it was possible, not how to construct a particular one in a particular way. You put no other conditions or stipulations on this, so there you go.

By "ruler" did you mean straightedge? I ask because people frequently abuse language and speak inexactly because they are ignorant of the particular meanings.

If you mean straightedge then 'no', you cannot construct an isosceles triangle. Insofar as I am aware. Not without a compass anyway, or having at least one circle already existing in the plane, or a way of measuring out length. Im no expert, per se, but Ive been studying both restricted and liberated construction methods to Euclidean geometry for a number of years now.

The answer you provided yourself in your response to yourself relies on the ability to measure lengths, which is not possible with a straightedge. But as I pointed out, there is a more straightforward way of doing it than the construction you gave. Youre not doing a Euclidean construction; youre doing, for lack of a better term, a "Cartesian construction", since you've embedded length measure into the plane by way of a ruler.

If by "ruler" you meant an actual demarcated length-measuring device, or a straightedge you're allowed to mark, then 'yes', fairly trivially. As I have outlined above, in my second paragraph.  Your method works too, but is hyper-convoluted.  I say that because I presume there is no other context you left out, and all you wanted was some arbitrary isosceles triangle in the plane.

As for constructing parallel and perpendicular lines, your method is good, although unnecessary for the intended goal. In fact it would have taken me some time to realize how to do the perpendicular.

An alternative construction for the parallel can be found on the wikipedia article for the Poncelet-Steiner theorem (linked above). Therein is an animated GIF image depicting the construction step-by-step. Just ignore the circle; what you need are the two points on the line (the line from which youre making a parallel) that passes through the circle center. These three points can be found without a compass simply be making congruent end-to-end line segments with your length measuring device.


Characteristics of a pretty ring This is a problem from a test I took today.

Definition: A pretty ring $R$ is a ring with unity 1, not a field, and each nonzero element can be written uniquely as a sum of a unit and a nonunit element of $R$.

The first problem, which is much easier, is to find such a ring. A got difficulty at first but came out with $$R = \left\{\begin{pmatrix} a & 0 \\ 0 & b \end{pmatrix}\mid a,b \in \mathbb{Z}_2\right\}$$
with usual operation done in modulo $2$.

The latter problem, which I can't solve yet is to find all possible characteristic of pretty ring.

From some experiment with almost the same ring, I came to conclusion that a pretty ring can only has characteristic 2. But, I can't prove it for the general case.

We show that a pretty ring has exactly one unit. Indeed, if $0 \neq u$ is not a unit and $e$ is a unit, then
$$ (u+e) + 0 = e + u $$
tells us that $u+e$ is not a unit (otherwise we get the contradiction $u=0$). Nowe take two units $e, \tilde{e}$ then
$$ e + (u+ \tilde{e}) = \tilde{e} + (u+ e) $$
implies $e=\tilde{e}$ and hence $1$ is the only unit.

On the other hand every ring with only one unit is a pretty ring as we can write every $x\neq 0$ as
$$ x = 1 + (x-1).$$

Thus, we have for a unital ring $R\neq \mathbb{Z}/2 \mathbb{Z}$:
$$ R \text{ is a pretty ring} \quad \Leftrightarrow \quad \vert R^\times \vert =1 $$
In particular we have $-1=1$ and thus a pretty ring has either characteristic equal to $1$ or $2$. Note that both cases are possible as the zero ring is a pretty ring.

Shorter proof: Assume that that the characteristic of the pretty ring is not $2$. Then we get from $1+0=-1+2$ must be a unit. Let $u\neq 0$ be a non-unit (exists as a pretty ring is not a field), then $1+0=(u+1) - u$ implies that $u+1$ is a non-unit. Then we get from $(u+1)+1=u +2$ that $0=1$, ie. our pretty ring is the zero ring. Therefore, a pretty ring has characteristic $1$ or $2$.

Only a partial answer. Suppose $n \geq 3$ be the characteristic of ring. Then $n.1 = 0$ implies that $ 1 = -(n-1).1 = -(n-2).1 + (-1).1$, since $-1$ is unit. If $-(n-2).1$ is not unit then by the uniqueness we get $-(n-2).1 = 0$ which contradicts that $n$ is the characteristic. So $-(n-2).1$ must be unit. Since $(n-2).1 + 2.1 = 0$, which implies that $2.1$ is a unit. Hence characteristic can not be $2$, even can not be even number. So we have proved that such $n$ must be odd.


Asymptotic variance of an estimator Suppose we have an estimator (i.e. a sequence of estimators) $T_n$ which is asymptotically normal, in the sense that $\sqrt{n}(T_n - \theta)$ converges in distribution to $\mathcal{N}(0, \sigma^2)$. The variance $\sigma^2$ is usually called the asymptotic variance of the estimator, but can we write that $\lim_{n\to\infty}\textrm{Var}[\sqrt{n}T_n]=\sigma^2$ ? If not, what additional conditions on the sequence $T_n$ we would need in order to do so ? Are consistency of $T_n$ and uniform integrability of $T_n^2$ sufficient conditions ?

I don't think you can get away with anything less than the uniform integrability of $(\sqrt{n} (T_n - \theta))^2$ and its weak convergence to $\mathcal{N}(0, \sigma^2)$. Note that if $T_n = n^{-1}\sum_{i=1}^n \xi_i$ for some iid $\xi_i$ with $E \xi_1 = 0$ and $E \xi_1^2 < \infty$, then $(\sqrt{n} T_n)^2$ is uniformly integrable (why?). There should also be a one-liner way of doing this, by appeal to some convergence theorem, or else using a trick like Skorokhod's representation theorem. I would be curious to know a shorter way; below is the "direct" analysis way.
Let $Y_n = \sqrt{n}(T_n - \theta)$ and let $Y$ be $\mathcal{N}(0, \sigma^2)$. The weak convergence of $Y_n$ to $Y$ means that, for any bounded continuous function $f$ (I write $f \in C_b$), $E[f(Y_n)] \rightarrow E[f(Y)].$ Unfortunately, the function $f(y) = y^2$ is not bounded on $\mathbb{R}$. We will have to approximate $f(y)$ by a sequence $\{f_M\} \subset C_b$ and take limits; this is where uniform integrability of $Y_n^2$ will come in.
For $0 < M < \infty$, define $f_M(y) = y^2 \wedge M$, and note that $f_M \in C_b$. We wish to show that $E[f(Y_n)] \rightarrow E[f(Y)]$, where $f(y) = y^2$. We have, for any $M$,
\begin{align*}
|E[f(Y_n)] - E[f(Y)]| &\leq |E[f(Y_n)] - E[f_M(Y_n)]| \tag{1}\\
&+ |E[f_M(Y_n)] - E[f_M(Y)]| \tag{2}\\
&+ |E[f_M(Y)] - E[f(Y)]|. \tag{3}
\end{align*}
We will use uniform integrability to pick an $M$ which bounds the first and the last term uniformly in $n$. Then, for fixed $M$, we can pick $n$ large enough to make the middle term as small as desired using the weak convergence of $Y_n$ to $Y$.
Let $\varepsilon > 0$. Since $\{Y_n^2\}_{n\geq 1}$ is uniformly integrable, so is $\{Y_n^2\}_{n \geq 1} \cup \{Y^2\}$. By uniform integrability, there is $M \in (0, \infty)$ such that
$$\sup_{n \geq 1} E[1\{Y_n^2 \geq M\} Y_n^2] < \varepsilon/8, \quad E[1\{Y^2 \geq M\} Y^2] < \varepsilon/8.\tag{4}$$
Fix such an $M$ once and for all. By the weak convergence of $Y_n$ to $Y$, for the  fixed function $f_M \in C_b$ and $\varepsilon > 0$, there is $N<\infty$ depending only on $f_M$ and $\varepsilon$ such that, for all $n \geq N$,
$$|E[f_M(Y_n)] - E[f_M(Y)]| \leq \varepsilon/2.\tag{5}$$
We have set our estimates, and what follows below holds for the given $\varepsilon > 0$, and any $n \geq N$.
To use $(4)$ in $(1)$, note that
\begin{align}
&\quad|E[f(Y_n)] - E[f_M(Y_n)]| \\
&= E[(f(Y_n) - f_M(Y_n))1\{Y_n^2 \geq M\}|] + E[(f(Y_n) - f_M(Y_n))1\{Y_n^2 < M\}|] .\tag{6}
\end{align}
The first term in $(6)$ is restricted to the event $\{Y_n^2 \geq M\}$,  and  each term  $f(Y_n)$ and $f_M(Y_n)$ contributes little to the expectation: we have for any $n \geq 1$,
$$E[f(Y_n) 1\{Y_n^2 \geq M\}] = E[Y_n^2 1\{Y_n^2 \geq M\}] < \varepsilon/8,\tag{7}$$
and also notice that the pointwise inequality $(Y_n^2 \wedge M) 1\{Y_n^2 \geq M\} \leq Y_n^2 1\{Y_n^2 \geq M\}$, which gives
$$E[f_M(Y_n) 1\{Y_n^2 \geq M\}] \leq E[Y_n^2 1\{Y_n^2 \geq M\}] < \varepsilon/8.\tag{8}$$
The second term in $(6)$ requires the cancellation of $f(Y_n)$ and $f_M(Y_n)$. However, it occurs on the event $\{Y_n^2 < M\}$, so we have the pointwise equality $(Y_n^2 \wedge M) 1\{Y_n^2 < M\} = Y_n^2 1\{Y_n^2 < M\}$, and so in fact the second term in $(6)$ is zero. Applying the triangle inequality on the first term of $(6)$ and using $(7)$ and $(8)$, we find $|E[f(Y_n) - f_M(Y_n)]| < \varepsilon/4$.
The same argument as was applied to use $(4)$ in $(1)$ can be recycled to use $(4)$ in $(3)$, and estimate $|E[f_M(Y)] - E[f(Y)]| < \varepsilon/4$. Finally, we can use $(5)$ directly in $(2)$ to deduce that, for all $n \geq N$,
$$|E[f(Y_n)] - E[f(Y)]| < \frac{\varepsilon}{4} + \frac{\varepsilon}{4} + \frac{\varepsilon}{2} = \varepsilon.$$
This is what we wanted, since for any centered random variable $Z$,
$$E[f(Z)] = E[Z^2] = \mathrm{Var}(Z).$$
Such a result must be true, and probably under milder conditions, because one can even numerically estimate the asymptotic variance in (well-converged) Markov chains. A general statement can probably be found somewhere in Meyn & Tweedie's book on stochastic stability.

Let me try the following:
$\sigma^2=\lim_{n\to\infty}\textrm{Var}[\sqrt{n}(T_n-\theta)]=\lim_{n\to\infty}(E[n(T_n-\theta)^2]-(E[\sqrt{n}(T_n-\theta)])^2)$
$=\lim_{n\to\infty}n(E[(T_n-\theta)^2]-E[T_n-\theta]^2)$
$=\lim_{n\to\infty}n(E[T_n^2]+\theta^2-2\theta E[T_n]-(E[T_n]^2+\theta^2-2\theta E[T_n]))$
$=\lim_{n\to\infty}n(E[T_n^2]-E[T_n]^2)$
$=\lim_{n\to\infty}nVar(T_n)$
$=\lim_{n\to\infty}Var(\sqrt{n}T_n)$
I only used that $\theta$ is a constant so i guess we don't need further assumptions.


Differential Equation Describing the Flowing of Water Out of a Tank I have the differential equation $${{dh} \over {dt}} =  - K\sqrt h $$

describing water flowing out of the bottom of a tank of uniform cross-sectional area under the action of gravity. $h(t)$ is the water depth at time t with ${h_0}$ being the initial depth. K is a positive constant.

I have no idea how to solve this. I only know how to separate variables to solve DEs. How would I solve this one? This is one of the starred questions in the book. I can't find another example like this. Thanks!

You can still solve this with separation of variables. divide both sides by $\sqrt{h}$ and you have a separated equation.

I present this way for solution your answer. Following the below way
                             dh/dt=-k√(h,)
You can solve this equation very easy. Divide both side of equation to √(h,). In this case, equation is changed to a separated of variable. By integrating both of them relative to separation variable and using initial condition, at which h(t=0)=h_0 , you can find integration constant, c=(h_0 )^1/2 and finally its answer:
h=(-1/2*kt+h_0^(1/2))^1/2
This equation shows the h versus t variation. By the way, you can draw it to linearity with h^2 versus t.


How to prove inequality $\frac{a}{a+bc}+\frac{b}{b+cd}+\frac{c}{c+da}+\frac{d}{d+ab}\ge 2$ Question:

Let $$a,b,c,d>0,a+b+c+d=4$$
show that
$$\dfrac{a}{a+bc}+\dfrac{b}{b+cd}+\dfrac{c}{c+da}+\dfrac{d}{d+ab}\ge 2$$

when I solved this problem, I have see following three variables inequality:
Assumming that $a,b,c>0,a+b+c=3$, show that :$$f(a,b,c)=\dfrac{a}{a+bc}+\dfrac{b}{b+ca}+\dfrac{c}{c+ab}\ge\dfrac{3}{2}$$
solution can see:inequality
I found this three answer all is not true,
1、such as  dear @Macavity, in fact
$$\sum_{cyc}\dfrac{a^2}{a^2+abc}\ge\dfrac{16}{\sum_{cyc}(a^2+abc)}$$
the Right not
$\dfrac{16}{4+\sum_{cyc}abc}$
2、and the @ante.ceperic is also not true.in fact
$$a^2+b^2+c^2+d^2+abc+bcd+cda+dab\le 8$$ is not true with $a+b+c+d=4$
such let $a=3$

Let $$a,b,c,d>0,a+b+c+d=4$$
show that
$$P=\dfrac{a}{a+bc}+\dfrac{b}{b+cd}+\dfrac{c}{c+da}+\dfrac{d}{d+ab}\ge 2$$

Now using Cauchy Schwarz:
$$\left(\dfrac{a}{a+bc}+\dfrac{b}{b+cd}+\dfrac{c}{c+da}+\dfrac{d}{d+ab}\right)((a+bc)+(b+cd)+(c+da)+(d+ab))\ge (\sqrt{a}+\sqrt{b}+\sqrt{c}+\sqrt{d})^2$$
Now:
$$P\ge\frac{4+2(\sqrt{ab}+\sqrt{bc}+\sqrt{cd}+\sqrt{da}+\sqrt{ac}+\sqrt{bd})}{4+ab+bc+cd+da}$$
Let $S_1=\sqrt{ab}+\sqrt{bc}+\sqrt{cd}+\sqrt{da}+\sqrt{ac}+\sqrt{bd},S_2=ab+bc+cd+da$
Now we need to prove:
$$\frac{4+2S_1}{4+S_2}\ge2\implies S_1-S_2\ge2$$
If we prove $S_1\ge6$ and $S_2\le4$, then $S_1-S_2\ge6-4=2$. Both inequalities can be proved easily and both hold when $a=b=c=d=1$:
$$(a+b+c+d)(b+c+d+a)\ge(ab+bc+cd+da)^2\implies S_2^2\le16\\\text{ since }a,b,c,d>0\implies S_2\le4$$

Take $1^{st}$ and $3^{rd}$ term apply the inequality, then take $2^{nd}$ and $4^{th}$ term and apply the inequality, you get

$$\ge 2\left(\frac1{1+\sqrt{bd}}+\frac1{1+\sqrt{ac}}\right)$$

again apply the inequality you get $\displaystyle \ge 2\left(\frac2{1+\sqrt{abcd}}\right)$

for $abcd =1$ you get your equality ie $a=b=c=d=1$


Proving that $\sqrt{2}+\sqrt{3}$ is irrational This is from Spivak.

Prove that $\sqrt{2}+\sqrt{3}$ is irrational.

So far, I have this:

If $\sqrt{2}+\sqrt{3}$ is rational, then it can be written as $\frac{p}{q}$ with integral $p, q$ and in lowest terms.

$$\sqrt{2}+\sqrt{3}=\frac{p}{q}$$
$$2\sqrt{6}+5 =\frac{p^2}{q^2}$$
$$(2\sqrt{6}+5)q^2=p^2$$

And that's about where I get stuck. In a similar question (prove that $\sqrt{2}+\sqrt{6}$ is irrational,) I was able to show that both $p,q$ had to be even which is impossible. I obviously can't apply this trick here. Any hints?

$(\sqrt3+\sqrt2)(\sqrt3-\sqrt2)=3-2=1$ is rational, so if the first factor were rational, so would be the second, and their difference $2\sqrt2$. But surely you know that the latter isn't rational.

From your very last step:
 $$\sqrt{6} = \frac{p^2 - 5q^2}{2q^2}$$ This a rational number- a contradiction, hence the answer.


Exponential congruence Hi All am a bit stuck on some revision that I am trying to do.

Firstly (part a) I must calculate the inverse of 11 modulo 41, which I have done and believe it to be 15.

The next part is to:
Now use your result from part (a) to find x where x^11 mod 42 = 10

I have been trying all sorts but don't seem to be getting anywhere.

Thanks

$4\cdot 11\equiv 3\pmod{41}$, hence $$8\cdot 11\equiv 6\pmod{41}$$
and by multiplying both sides by seven we get that $8\cdot 7\equiv 56\equiv 15\equiv 11^{-1}.$

By the Chinese theorem, $x^{11}\equiv 10\pmod{42}$ is equivalent to:
$$\left\{\begin{array}{rcl} x^{11} &\equiv& 0\pmod{2},\\ x^{11} &\equiv& 1\pmod{3},\\x^{11}&\equiv& 3\pmod{7},\end{array}\right.$$
or to:
$$\left\{\begin{array}{rcl} x &\equiv& 0\pmod{2},\\ x &\equiv& 1\pmod{3},\\x^{-1}&\equiv& 3\pmod{7},\end{array}\right.$$
and since the inverse of $3\pmod{7}$ is $5$, from:
$$\left\{\begin{array}{rcl} x &\equiv& 0\pmod{2},\\ x &\equiv& 1\pmod{3},\\x&\equiv& 5\pmod{7},\end{array}\right.$$
we get the solution $x\equiv 40\equiv -2\pmod{42}$.

I cannot see how the inverse of $11\pmod{41}$ is involved here: $42$ is not a prime!


Factorial series with ratio test inconclusive I have to determine the convergence of this series:

$\displaystyle\sum_{n=1}^\infty {2n! \over (n+3)! }$

The ratio test gives 1, so I don't know how to solve it. Wolfram alpha suggests me to use the comparison test, but which series should I compare it with?

Thanks for the help!

$$\frac{a_{n+1}}{a_n}=\frac{(2n+2)!}{(n+4)!}\frac{(n+3)!}{(2n)!}=\frac{(2n+1)(2n+2)}{n+4}\xrightarrow[n\to\infty]{}\infty$$

and thus the series diverges

If you menat $\;2\cdot n!\;$ , then

$$\frac{2n!}{(n+3)!}=\frac{2}{(n+1)(n+2)(n+3)}\le\frac2{n^2}$$

and by the comparison test the series converges


Is there a proof of 3x3 Magic Square General Form? Below, I have included a picture of an article I read from Martin Gardner on additional structures found within 3x3 magic squares, concerning their row (and column) products’ sums.
He casually mentions that the form of 3x3 magic squares MUST fit the general form at the top-right, but I can’t find a proof anywhere. I may be looking in the wrong places.
Does anyone have a DOI link or journal reference for a general proof? The reference page in the article was also cut off, so I couldn’t find it there, if it is there.
Thank you / sorry for not being more helpful.

This is actually a reasonable straightforward thing to work out algebraically: essentially, you are trying to solve $7$ linear systems which express that the sum of any row, column, or diagonal is equal to the sum of any other.

Let's prove this! Suppose that the following is a magic square:
$$\begin{pmatrix}A_{11} & A_{12} & A_{13} \\ A_{21} & A_{22} & A_{23} \\A_{31} & A_{32} & A_{33}\end{pmatrix}$$
The main substance of the form that Martin Gardner notes is that the sum of every row, column, or diagonal is $3A_{22}$. To prove this, let $S$ be this sum. Note that
$$S=A_{11}+A_{22}+A_{33}=A_{21}+A_{22}+A_{23}=A_{31}+A_{22}+A_{13}.$$
We note that $A_{22}$ appears three times here and that the other coefficients are laid out in two rows. Notice that adding these three forms together gives
$$3S=3A_{22}+(A_{11}+A_{21}+A_{31})+(A_{13}+A_{23}+A_{33}).$$
However, the two parenthesized terms are columns, so sum to $S$ each, giving
$$S=3A_{22}$$
after cancellation.

The rest is less insightful - just a convenient choice of variables. First, $A_{22}$ is really important, so we'll give that a name $a$ and note that every row, column, and diagonal must sum to $3a$. We know that $A_{11}$ has to be something and we can certainly write any number in the form $a+b$. Likewise, $A_{13}$ has to be something, so let's call it $a+c$. We have done nothing in this step except choose some names.

Everything else follows easily now. For instance, $A_{33}$ must be $a-b$ because $A_{11}+A_{22}+A_{33}=3a$, but we already know two of the terms so can solve for the last one. Similarly, once we note that $A_{31}$ must be $a-c$, we then can solve for the four remaining values by looking at the outer rows and columns, as we know two of the terms in each. Finally, for mathematical completeness, we check that the equations whose truth we did not force - that the middle row and column sum to $3a$ - is indeed true. Thus, the given form is a magic square, but is also the only way to make a magic square because we deduced it from the mere statement "this is a magic square."

All of the rows, columns, and diagonals sum to $3a$, so it satisfies the conditions on a magic square. It would probably still be a magic square if you replaced $a$ with $2a$.


Prove that a convolution product $(f\ast g)(x)$, with $f,g\in\mathcal{L}^2(\mathbb{R})$, goes to $0$ as $|x|$ goes to $\infty$. I want to proof that the convolution product $(f\ast g)(x)$ of two square-integrable functions $f,g\in\mathcal{L}^2(\mathbb{R})$ goes to $0$ as $|x|$ goes to $+\infty$. The convolution product here denotes
$$(f\ast g)(x)=\int_\mathbb{R}f(y)g(x-y)dy.$$
I know that, using Hölder's inequality,
$$|(f\ast g)(x)|\leq \int_\mathbb{R}|f(y)g(x-y)|dy\leq ||f||_2||g||_2$$
but I do not immediately see how this can help me to finalise my proof.

To do this question, you'd better know some density theorem of $L^p$ space. We will use a theorem which shows for an $L^2(\mathbb{R})$-function, there exists a sequence of simple measurable functions $f_k$ with bounded support such that $f_k\to f$ in $L^2(\mathbb{R})$ and $f_k$'s can be chosen to be in $L^2(\mathbb{R})$.

Since simple function with bounded support is dense in $L^2(\mathbb{R})$, there exists $f_k\to f$ in $L^2$, where $f_k$'s are simple functions with bounded support in $L^2$. This shows
$$|(f*g)(x)|\leq \lVert f-f_k\rVert_2\lVert g\rVert_2+\left|\int_{\mathbb{R}^n}f_k(x-y)g(y)\;dy\right|$$
Similarly, we can find a sequence of simple function $g_k$ with bounded support and $g_n\to g$ in $L^2$. Then,
$$\left|\int_{\mathbb{R}^n}f_k(x-y)g(y)\;dy\right|\leq \lVert f_k\rVert_2\lVert g-g_n\rVert_2+\left|\int_{\mathbb{R}^n}f_k(x-y)g_n(y)\;dy\right|$$
Note that $f_k(x-y)g_n(y)=0$ for large enough $|x|$. This is because if the radius of the support of $f_k$ is $r_k$ and the radius of support of $g_n$ is $R_n$, then if $|x|>r_k+R_n$, either $|y|>R_n$ or $|x-y|>r_k$, so either $f_k(x-y)=0$ or $g_n(y)=0$. This implies that
$$|(f*g)(x)|\leq\lVert f-f_k\rVert_2\lVert g\rVert_2+\lVert f_k\rVert_2\lVert g-g_n\rVert_2+\left|\int_{\mathbb{R}^n}f_k(x-y)g_n(y)\;dy\right|$$
First take $|x|\to\infty$ on both sides, we have
$$\lim_{|x|\to\infty}|(f*g)(x)|\leq \lVert f-f_k\rVert_2\lVert g\rVert_2+\lVert f_k\rVert_2\lVert g-g_n\rVert_2$$
Then take $n\to\infty$ on both sides, since LHS is independent of $n$, we have
$$\lim_{|x|\to\infty}|(f*g)(x)|\leq\lVert f-f_k\rVert_2\lVert g\rVert_2$$
Finally, take $k\to\infty$ on both sides, since LHS is independent of $k$, we obtain $\lim_{|x|\to\infty}|(f*g)(x)|\leq 0$, i.e., $(f*g)(x)\to 0$ as $|x|\to \infty$.

If we have:
$$I(x)=\lim_{n\to\infty}\int_{-n}^nf(y)g(x-y)dy$$
this denotes our convolution. Now we can say that we want $|x|\to\infty$ so we can write this as:
$$I_\infty=\lim_{x\to\infty}\lim_{n\to\infty}\int_{-n}^nf(y)g(x-y)dy$$
we can make the substitution $u=x-y,du=-dy$ and so:
$$I_\infty=\lim_{x\to\infty}\lim_{n\to\infty}\int_{x+n}^{x-n}f(x-u)g(u)du=-\lim_{x\to\infty}\lim_{n\to\infty}\int_{x-n}^{x+n}f(x-u)g(u)du$$
also:
$$I_{\infty}=\lim_{x\to\infty}\lim_{n\to\infty}\int_{-n}^nf(-y)g(x+y)dy$$
$$I_\infty=\lim_{x\to\infty}\lim_{n\to\infty}\int_{x-n}^{x+n}f(u-x)g(u)du=\lim_{x\to\infty}\lim_{n\to\infty}\int_{x-n}^{x+n}f(-(x-u))g(u)du$$
this suggests that $f(-(x-u))=-f(x-u)$ and since we are talking about $|x|\to\infty$ we can say that $f(-)=-f(+)$. If $g$ is an even function then we have an odd integrand over the real domain which will equal zero.