Are there automorphisms of $H$ which are not restriction of an automorphism of $G$? Let $G$ be a group and $H$ a characteristic subgroup of $G$ (that is, invariant under all automorphisms of $G$).
Let $\phi \in \mathcal{Aut}(H)$, we'll call $\widetilde{\phi}$ an automorphism of $G$ such that

$$\widetilde{\phi}(h) = \phi(h) \quad \forall h \in H$$

Or in other words, such that $\widetilde{\phi}_H = \phi$, where $\widetilde{\phi}_H$ denotes its restriction to $H$.

Can we construct a $\widetilde{\phi}$ for every $\phi \in \mathcal{Aut}(H)$?

Or in other words, are there automorphisms of $H$ which are not restrictions to $H$ of an automorphism of $G$?

It's easy to show that if the answer is no, then there is a bijection $\mathfrak{A}: \mathcal{Aut}(H) \xrightarrow{\sim} \mathcal{Aut}(G)$ such that $\mathfrak{A}(\phi) = \widetilde{\phi}$. Could this fact be used to disprove the claim above by contradiction?

The answer is no. One counterexample is the group $G = {\rm AGL}(1,8)$, which has order $56$. It has a normal (and therefor characteristic) Sylow $2$-subgroup $P$, which is elementary abelian of order $8$.

Now ${\rm Aut}(G)$ is the group ${\rm A \Gamma L}(1,8)$, which is a solvable group of order $168$ and has $G$ as a normal subgroup of index $3$. However, ${\rm Aut}(P)$ is the group ${\rm PSL}(3,2)$, which coincidentally also has order $168$, but it is a nonabelian simple group. In fact only $21$ of the $168$ automorphisms of $P$ extend to automorphisms of $G$.

If $H$ is characteristic in $G$ then every automorphism of $G$ is an automorphism of $H$. Thus, the question reduces to answering whether $A(H) \subset A(G)$ is true. Consider $Gh_1$ and $Gh_2$ where $h_1$ and $h_2$ are in $H$. Then $g_1h_1g_2h_2=g_1g_2'h_1h_2$. This shows that the right-regular representation of $H$ is a subgroup of the right-regular representation of $G$. Thus the holomorph of $H$ is a subgroup of the holomorph of $G$. Since $A(H)$ and $A(G)$ fix the identity in the respective holomorphs, $A(H) \subset A(G)$.

EDIT: The claim that the holomorph of $H$ is a subgroup of the holomorph of $G$ is indeed incorrect, for the normalizer of the right regular representation of $H$ in $S_G$(The symmetric group on elements of $G$) need not necessarily be contained in the normalizer of the right regular representation of $G$ in $S_G$.


How do I find the upper and lower sum of the area of the graph by summation? Solve for the area $y=\sqrt{x}$

I have the following formulas for lower and upper sum.

$$LowerSum = \sum_{i=1} ^n f(m_i)\Delta x$$

$$Upper Sum = \sum_{i=1} ^n f(M_i)\Delta x$$

I know that the interval for $[a, b]$ is $[0, 1]$ and that there are $n$ rectangles. The change in x is $\Delta x = \frac{b-a}{n}$ so:

$$\Delta x = \frac{1}{n}$$

For the left endpoint, $f(m_i)$, the book has something like $f(m_i) = a + (i - 1)\frac{b-a}{n}$ so:

$$f(m_i) = \frac{i - 1}{n}$$

Now I have:

$$LowerSum = \sum_{i=1} ^n f\left(\frac{i - 1}{n}\right)\left(\frac{1}{n}\right)$$

Moving the change in x to outside and plug into the function:

$$\left(\frac{1}{n}\right)\sum_{i=1} ^n \sqrt{\frac{i - 1}{n}}$$

Now what do I do? The book doesn't have any example with a square root so in their example, they are left with a single $i$ or $i^2$ which there are summation formulas for those; what about a square root? Am I even doing it right?

Reman Sums can be applied in this case

For the function $f(x)=\sqrt x$:

$\lim\limits_{ n\to\infty} \sum\limits_{i=1}^{n} f(x_i)\Delta X$

let $x_i$ be the right end point, $x_i=\frac{i^2}{n^2}$

$\Delta x=\frac{1^2}{n^2}-\frac{(i-1)^2}{n^2}=\frac{2i-1}{n^2}$

Thus: $\lim\limits_{ n\to\infty} \sum\limits_{i=1}^n \sqrt{\frac{1^2}{n^2}}\frac{2i-1}{n^2}=\frac23$


Must heredity be explicilty stated in addtion to Peano's axioms when defining natural numbers? My question is stated in bold text.

This question pertains to the following formulation of Peano's axioms
used to formalize an introduction of the natural numbers (beginning
with 1) consisting of successions of vertical strokes drawn on paper
in horizontal rows.

I. $1$ is a number.
II. To every number $a$ there corresponds a unique number $a^{\prime},$
called its successor.
III. If $a^{\prime}=b^{\prime},$ then $a=b.$
IV. $a^{\prime}\ne1$ for every number $a.$
V. Let $A\left(x\right)$ be a proposition containing the variable
$x.$ If $A\left(1\right)$ holds and if $A\left(n^{\prime}\right)$
follows from $A\left(n\right)$ for every number $n,$ then $A\left(x\right)$
holds for every number $x.$

Are these axioms sufficient to determine the natural numbers $\mathbb{N}\equiv\left\{ 1,2,3,\dots\right\} ?$
Or is it necessary to add a further requirement that a number must
inherit its membership in $\mathbb{N}$ from $1?$

For example, it seems that the set $\mathcal{N}=\mathbb{N}\cup\left\{ h\vert i\in\mathbb{Z}\land h=i+\frac{1}{2}\right\} $
is consistent with these axioms, with $a^{\prime}\equiv a+1,$ in
the standard sense.

I propose that the following statement would fulfill such a requirement:
The set of natural numbers contains the number $1$ as well as every
number reached successively beginning with $1,$ with succession conforming
to Peano's axioms as stated.

That avoids explicit reference to a "successor function", which
would require additional definitions. Nonetheless, any such requirement
stated in addition to Peano's axioms seems to presuppose ordering.
Terms such as beginning with, repeated application, successively, inherited from, etc., appeal to
our a priori notions of ordering. So I'm left wondering to what extent
our subsequent theorems pertaining to ordering are actually derived
results.

Yes, the (first-order) theory PA is not sufficient for characterizing $\mathbb{N}$ up to isomorphism.

In fact, no first-order theory can do the job for any infinite structure at all! This is a consequence of the compactness theorem:

Suppose $\mathcal{A}$ is an infinite structure in a language $\Sigma$ and $\Gamma$ is a set of sentences true in $\mathcal{A}$.
Fix a cardinal $\kappa>\vert\Gamma\vert$, and let $\Sigma^*$ be the language gotten by adding $\kappa$-new constant symbols $\{c_\eta:\eta<\kappa\}$ to $\Sigma$.
Let $\Gamma^*=\Gamma\cup\{c_\eta\not=c_\theta:\eta<\theta<\kappa\}$.
By the compactness theorem, $\Gamma^*$ has a model - any finite subset of $\Gamma^*$ can only mention finitely many of the new constants, so an appropriate expansion of $\mathcal{A}$ will do the job.
But any model of $\Gamma^*$ has cardinality $\ge\kappa$ - hence can't be $\mathcal{A}$ itself. But the reduct of such a model to the original language $\Sigma$ is a model of $\Gamma$, so there are $\Sigma$-structures satisfying $\Gamma$ which aren't (isomorphic to) $\mathcal{A}$.

(Note that the original formulation of PA was as a second-order theory; second-order logic can indeed capture $\mathbb{N}$ up to isomorphism.)

The answer is no.  It is fairly easy to show that the axioms as stated are sufficient to characterize the natural numbers.

Proof of the following theorem suffices:  $\forall_n\left[ n=1\lor \exists_m n=m^\prime\right].$  This is shown as follows: $P\left[1\right]$ is given, and $P\left[n\right]\implies P\left[n^\prime\right]$ is axiomatic.

But this really isn't the entire proposition which I wanted to demonstrate.

A more complete statement such as: every number is either $1$ or succeeds $1$ does appear to require second order logic.  By $n$ succeeds $1$, I mean that the number $n$ is either $1^\prime$, or is the successor of a number which succeeds $1$.

$$S\left[x\right]\text{:=}\left(x=1'\lor \exists _y\left[S\left[y\right]\land x=y'\right]\right).$$

The proposition that $n$ is either $1$ or $n$ succeeds $1$ is now:

$$P\left[x\right]\text{:=}\left(x=1\lor S\left[x\right]\right).$$

The inclusion of the existential quantifier in $S$ raises this to logic of the second order.

Part of the reason for my confusion is that the authors of BBFSK seem to take the existence of the natural numbers as a-priori.  This is especially true the in first chapter where they omit the first two of Peano's axioms in their second-order logic statement of them.


Does one need to learn set theory before learning category theory? I am having a course in Algebraic Topology and learning some basic category theory. But I only have a very limited understanding of basic set theory. I have no idea what is ZFC, and stuff like that. Thus, I find it hard to understand some motivation of the category theory.

My question is: does one need to learn some rigorous basic set theory before learning about category theory? If yes, do you have any recommendations?

I don't think you really need to go to formal set theory in order to understand category theory. What you need of course is basic understanding of what sets and functions are, nothing more then it is needed to understand algebra and topology (which I assume you know otherwise why taking a course in algebraic topology) :D

That said, you have to understand that category theory was born in algebraic topology and homological algebra, not for the sake of set theory.

Category theory is an abstract framework where is possible to study with the right level of generality(abstraction) many phenomena occurring in different context like topology, algebra and geometry.
In particular in category theory you can define in a uniform way what are products, coproducts and many other general constructions which arise in different fields in mathematics.

In order to understand category theory can be good having some background knowledge in algebra, geometry, topology (etc) in order to have a good list of example one can use when learning categorical concepts, although not strictly necessary. Take a look to Awodey's Category theory for an introduction on category theory with very low prerequisites.

Having some knowledge in set theory can help in finding some other interesting examples (and applications of category theory) but it is not really fundamental (at least if you ignore size issues, something many mathematicians do).

If I were you, I will "learn" set theory as much as I can. Take a course in set theory at your university and do well in that class, then take the second course before learning category. The set theory knowledge you need to master should allow you to do higher level analysis, higher modern algebra, and even topology which are required for you to become a mathematician. Cathegory theory is used to study higher modern algebra and higher number theory and that is where it will take you.


Solving Higher-Order Non-Homogeneous ODE $y'''-3y''+3y'-y=24(x+1)e^{x}$ Given equation:

$$y'''-3y''+3y'-y=24(x+1)e^x$$ with initial values $y(0)=1$ and $y'(0)=2$ and $y''(0)=7$

I get the homogeneous equation as $(c_1+c_2x+c_3x^{2})e^{x}$

But I don't get how to get the particular solution as it is written $24(x+1)e^{x}$

Easy way:
$$y'''-3y''+3y'-y=24(x+1)e^x$$
$$(ye^{-x})'''=24(x+1)$$
Integrate three times.

Another way:
$$y'''-3y''+3y'-y=24(x+1)e^x$$
The poynomial characteristic is:
$$r^3-3r^2+3r-1=0 \implies (r-1)^3$$
The solution to the homogeneous equation is
$$y_h=c_1e^x+c_2xe^x+c_3x^2e^x$$
For the particular solution try:
$$y_p=x^3(Ax+B)e^x$$

$$\frac{1}{f(D)} \left ( e^{ax} \phi(x) \right ) = e^{ax} \left ( \frac{1}{f(D+a)} \phi(x) \right )$$

The formula above would be of help to solve the particular integral $$\int 24 x(e^x)\,dx.$$


What is wrong with the following proof? This originates from the problem $\#3.8$ of Rudin, the problem is as followed:

  If $ \sum a_{n} $ converges and if ${b_{n}}$ is monotonic and bounded,
  prove $\sum a_{n}b_{n}$converges.

I know that if we remove the assumption that $b_{n}$ is monotonic, then the conclusion won't hold. That is, if we have $ \sum a_{n} $ converges and  ${b_{n}}$ bounded, we can not reach the conclusion that $\sum a_{n}b_{n}$converges. I think of a counter example as $a_{n}=\frac{(-1)^{n}}{n}$ and $b_{n}=(-1)^{n}$,then  $ \sum a_{n} $ converges and  ${b_{n}}$ bounded, however, $\sum a_{n}b_{n}$ diverges.

Then my question is, however, I read of a proof  the problem without using the assumption that $b_{n}$ is monotonic, and hardly can I find where is wrong in the proof. So can anyone tell my what is wrong in the following proof? Thanks in advance guys!

  Exercise $\mathbf{3.8}$

  We're told that $\{b_n\}$ is bounded. Let $\alpha$ be the upper bound of $\{|b_n|\}$. We're also told that $\sum a_n$ converges: so for any arbitrarily small $\epsilon$, we can find an integer $N$ such that
  $$\left|\,\sum_{k=m}^n a_k\right|\le\dfrac{\epsilon}{\alpha}\,\,\text{for all $n,m$ such that $n\ge m\ge N$}$$
  which is algebraically equivalent to
  $$\left|\,\sum_{k=m}^n a_k\alpha\right|\le \epsilon\,\,\text{for all $n,m$ such that $n\ge m\ge N$}$$
  which since $|b_k|\le \alpha$ for every $k$, means that
  $$\left|\,\sum_{k=m}^n a_k b_k\right|\le\left|\,\sum_{k=m}^n a_k \alpha\right|\le\epsilon\,\,\text{for all $n,m$ such that $n\ge m\ge N$}$$
  By theorem $3.22$, this is sufficient to prove that $\sum a_n b_n$ converges.

The last inequality is wrong. Just take $n=m+1$ and $a_n = -a_m$. Then $|\sum_{k=m}^na_k \alpha| = 0$, but it certainly doesn't follow that $|\sum_{k=m}^na_kb_k| = 0$ for all possible choices of $b_m, b_n$.

The first inequality is wrong.  It assumes that $\sum a_n$ converges absolutely, when the sum was only stated to converge.


Logic puzzle: Which octopus is telling the truth? King Octopus has servants with six, seven, or eight legs. The servants with seven legs always lie, but the servants with either six or eight legs always tell the truth.

One day, four servants met.
The blue one says, “Altogether, we have 28 legs.”
The green one says, “Altogether, we have 27 legs.”
The yellow one says, “Altogether, we have 26 legs.”
The red one says, “Altogether, we have 25 legs.”

What is the colour of the servant who tells the truth?

Hint:  since they all disagree, at least three are lying.  How many legs does that account for?

Since at least 3 octopuses must be telling the lie, then it must be the green servant (with 6 legs) who is telling truth and 3 others (with 7 legs) are telling lies.

Other combinations listed here will not make sense (unless I missed out any).

blue = 28
8,8,6,6
7,7,7,7
8,7,7,6

green = 27
8,7,6,6
7,7,7,6 -> here it is!

yellow = 26
8,6,6,6
7,7,6,6

red = 25
7,6,6,6


Unramified subextension of the Galois closure of a totaly ramified $p$-adic field Let $L/L'/K/\mathbb{Q}_p$ be a tower of finite field extensions such that $L'/K$ is totally ramified and $L/K$ is its Galois closure. We suppose that $L/L'$ is unramified. Let $M/K$ be a Galois subextension of $L/K$ such that $L/M$ is unramified. Is necessarily $L=M$ ?

The answer seems to be yes if the ramification of $L'/K$ is tame. We then have $L'=K(\sqrt[e]{\pi_K})$, so the residual degree $f(L/K)=[L:L']$ is equal to the multiplicative order of $|k_K|\bmod e$, where $|k_K|$ is the order of the residue field. If we write $M/M_0/K$ with $M_0/K$ maximal unramified, then $M_0$ must contain the $e$-th roots of unity, so $|k_{M_0}|\equiv 1\bmod e$. We conclude that $f(L/K)$ divides $f(M_0/K)$, so they must we equal. Then $M=L$ by minimality of $L$.

In general, if $Gal(L/L')$ is generated by a Frobenius lift $\phi$, then $Gal(L/M)$ is generated by an element $i\phi^k$ with some $i\in I(L/K)$ and some $k$ dividing $f(L/K)$. The group $Gal(L/M)$ is normal in $Gal(L/K)$, so one can prove that $\phi$ commutes with $i$. Then, $L'\cap M$ is the subfield of $L$ fixed by $\phi$ and $i$ and $$Gal(L/L'\cap M)=\langle\phi\rangle\times\langle i\rangle.$$ Furtherfore, one can prove that the quotient $$Gal(M/L'\cap M)=(\langle\phi\rangle\times\langle i\rangle)/\langle i\phi^k\rangle$$ is cyclic generated by the image of the Frobenius $\phi$.

My idea was to try to prove that $M/L'\cap M$ is unramified, thus leading to a conclusion that $i$ is trivial, and then that $k=1$ by minimality of $L$, but I am not even sure if that's possible.

(edit: in hindsight, I think there is a flaw in my argument, so read if you please.)

Obviously if we know $M$ contains $L'$, by the definition of $L$ as the Galois closure of $L'$ we immediately get $L=M$.

Since your statement is about a bunch of finite extensions over $\mathbb{Q}_p$ and about unramifiedness, I figure it's best to think about it using finite fields.

Given this slogan, we know that $k_{L} / (k_{L'} = k_K) / \mathbb{F}_p$ are finite extensions. Let's think about where $k_M$ should sit in. Since $L/M/K$, we know that $k_L/k_M/k_K$, so now our tower of finite fields becomes $k_L/k_M/(k_K = k_{L'})$.

Theory about unramified extensions (and along with the fact that there is only one finite extension of a given degree of a finite field) now forces $M/L'$, so we are done.


Let $\varphi:\left(a,b \right)\rightarrow R$ twice derivatible with $\varphi\left(a \right)=\varphi\left(b \right)=0$ for all $x\in[a,b]$ Let $\varphi:\left(a,b \right)\rightarrow R$ twice derivatible with $\varphi\left(a \right)=\varphi\left(b \right)=0$ for all $x\in[a,b]$ and ${\varphi}^{(2)}(x)>0$ Prove that $\varphi\left(x \right)<0$ for all $x\in(a,b)$.

how ${\varphi}^{(2)}(x)>0$ for all $x\in[a,b]$ we have ${\varphi}^{(1)}$ is strictly decreasing, by the rolle theorem, there exists $c\in(a,b)$ such that  ${\varphi}^{(1)}(c)=0$,  As ${\varphi}^{(1)}$ is strictly decreasing in $(a, b)$ and ${\varphi}^{(1)}(c)=0$, so ${\varphi}^{(1)}<0$ in $[a,c)$ and ${\varphi}^{(1)}>0$ in $(c,b]$.

Is this conclusion right?

As ${\varphi}^{(1)}$ is strictly decreasing in $(a, b)$ and ${\varphi}^{(1)}(c)=0$, so ${\varphi}^{(1)}<0$ in $[a,c)$ and ${\varphi}^{(1)}>0$ in $(c,b]$.

You can conclude like this: $\phi'<0$ in $[a,c)$ implies that $\phi$ strictly decreases on $[a,c]$ for every $x\in [a,c], \phi(x)<\phi(a)=0$. On the other hand, you have $\phi$ strictly increases on $[c,b]$, and if $x\in[c,b], \phi(x)<\phi(b)=0$.

If $\varphi''(x)>0$ $\forall x\in (a,b)$ then $\varphi'(x)$ is strictly incresing on $(a,b).$ Now:

Is it possible to have $\varphi'>0?$ In such a case $\varphi$ will be sctrictly increasing and thus $\varphi(b)>\varphi(a).$ Contradiction.
Is it possible to have $\varphi'<0?$ In such a case $\varphi$ will be sctrictly decreasing and thus $\varphi(b)<\varphi(a).$ Contradiction.

Thus, we have that there exists $c\in (a,b)$ such that $\varphi'<0$ on $(a,c),$ $\varphi'(c)=0$ and $\varphi'>0$ on $(c,b).$

Can you conclude?


Uniform convergence of $\sum \frac{nx}{1+n^2x^2}$ I know that $\frac{nx}{1+n^2x^2}$ isn't uniformly convergent.
However I am not sure about the series. None of the tests for series convergence, namely wierstrass m test, Abel test or Dirichlet test give anything about proving non convergence.
Does non-uniform convergence of individual terms imply non-uniform convergence of series?

It's overkill in this case since the series
$$
\sum_{n=1}^\infty \frac{nx}{1+n^2x^2}
$$
is not even pointwise convergent for $x\neq 0$, but in general:

If you have uniform convergence over $E$ of a series with partial sum $S_n=\sum_{k=1}^n f_n$ to a limit function $S$, then
$$\begin{align*}
\sup_{x\in E}|f_n(x)| &= \sup_{x\in E} |S_{n}(x)-S_{n-1}(x)| \leq \sup_{x\in E} |S_{n}(x)-S(x)|+ \sup_{x\in E} |S_{n-1}(x)-S(x)| \\&\xrightarrow[n\to\infty]{} 0+0=0
\end{align*}$$
so $(f_n)_n$ must converge uniformly (on $E$) to the zero function.

Let $f_n(x)=\frac{n x}{1+n^2 x}$, the series $\sum_{n=1}^{\infty} \frac{n x}{1+n^2 x}$ is not defined on $x=\frac{-1}{n^2},\,\, n=1,2,3,...$. and for all other values of $x$, we have $\frac{n x}{1+n^2 x} \to 0,\,\, as\,\, n \to 0$.
According to M-test theorem, the series convergent uniformly on any bounded interval $(a,b],\,\, b >a>0$ for
$$\frac{n x}{1+n^2 x} \le \frac{b} {a n^2} $$
and it is clear that
$$\frac{b}{a}\sum_{n=1}^{\infty}\frac{1}{n^2} = \frac{b \pi^2}{6 a}.$$
The series diverges at $x= \frac{1}{1-n^2},\,\,\, n=1,2,3,...$ because the expression $\frac{n x}{1+n^2 x} =1$ at $x=\frac{1}{1-n^2}$ and the infinite summation for the last diverges.


Is group division similar to number division? The below written lines are taken from book title " Visual Group Theory" by Natham Carter

  The most general way to deconstruct a large group into two factors is called taking quotients. It reveals not only direct products but also semi direct products. It gives us deep insight into groups structure.

Question : Why quotients in groups needed? I have heard about division in numbers but in groups how it make sense? 8/4 = 2, but $C_4 / C_2$ ? How researcher in mathematics comes up with the idea of quotient?

I have seen this question Why the term and the concept of quotient group?

my opinion and self experience is that the quotient groups are famous and very important for group theorists since the have important behave and they built by normal subgroups which are very very important in group theory.somehow quotient groups play the role of fraction in numbers(that is if G\N is quotient group, we called G without considering N inside) but the whole story is about algebraic theorist like normal groups and normal subgroups so the quotient groups are important to them.and be careful u never can compare quotient with divisions but in some groups. for example u can study $\frac{z}{n{z}}$ which z is integer numbers.


$k$-rational points of the automorphism functor of a scheme Let $X$ be a scheme and let $\operatorname{Aut}_X$ denote the functor sending a scheme $T$ to the set of $T$-automorphisms of $X \times T$.
Assume that $\operatorname{Aut}_X$ is representable by a group scheme. I will continue to denote this group scheme by $\operatorname{Aut}_X$.
I am reading "Notes on automorphism groups of projective varieties" by Michel Brion.
In it he defines $\operatorname{Aut}(X):=\operatorname{Aut}_X(k)$ and then writes that $\operatorname{Aut}_X$ is algebraic if and only if $\operatorname{Aut}(X)$ is.
The property of being algebraic is something assigned to a group scheme, but, as far as I understand,  $\operatorname{Aut}_X(k)$  is just a group.
Does $\operatorname{Aut}_X(k)$ also have a group scheme structure?

There are several definitions of the notion "group scheme". If a group scheme
$G$ (over a commutative ring $k$) is a representable functor (let $G:=Spec(R)$)
$$h_G: Aff(k) \rightarrow Groups$$
where $C:=Aff(k)$ is the category of affine schemes over $k$ and $Groups$ is the category of groups, it follows
$$G(k):=h_G(Spec(k)):=Hom_C(Spec(k), G) \cong Hom_{k-alg}(R,k).$$
Note: When considering a group valued functor $h_G$ as above, there are problems since the category $Aff(k)$ of arbitrary $k$-algebras is not a small category - the objects $Ob(Aff(k))$ of $Aff(k)$ is not a set. You could view a group scheme as a triple $(G,m,i)$ where $G$ is a scheme over $k$,
$m:G\times_k G \rightarrow G$ and $i:G \rightarrow G$ are morphisms satisfying a set of criteria. Here $m$ is the group multiplication and $i$ is the inversion. Then you avoid such set theoretical problems. If you are not careful when introducing methods from category theory, you may find you are unknowingly working with "non standard set theory". On page 5 in "Representations of algebraic groups" (Jantzen) you find the following:
"Let us assume $k$ to be arbitrary again. In the definitions to follow, we shall be rather careless about the foundations of mathematics. Instead of working with "all" $k$-algebras, we should (as in [DG]) take only those in some "universe". We shall leave the appropriate modifications to the interested reader."
If you care about foundations you should either read about the "universe" mentioned in Jatzen's book or instead work with a triple $(G,m,i)$ as explained above.
If you dont care, many mathematicians will accept the following method of proof: When giving a lecture on your results you use the "hand-waving method" to prove results. You "wave your hands" and speak about "categories" and "universes".
One of the problems with being a student in modern algebraic geometry is that much of the results presented in the EGA book series is formulated in the functorial language and hence it assumes the notion "universe" as mentioned in Jantzens book. The EGA book series consists of several tousands of pages and as a student you will not have the time to read everything and verify every detail. This is why you - as a student - will be forced to do some "hand-waving" every now and then.
Let us from now "assume all $k$-algebras live in a fixed universe".
Question: "Does AutX(k) also have a group scheme structure?"
Answer:  If $G:=GL(V)$ where $V:=k\{e_1,..,e_n\}$ it follows $R:=k[x_{ij}][t]/(tdet(x_{ij})-1)$ and you may identify $G(k)$ with the set of matrices
$$\{A:=(a_{ij})\in Mat(n,k)\text{ with }det(A)\in k^*\},$$
hence the set $G(k):=GL(V)(k)$ may be interpreted as a set of invertible $n\times n$ matrices with coefficients in $k$ - this set has a group operation.
Hence if a "group scheme" in your language is a representable functor
it follows $Aut(X):=Aut_X(k)$ is not a group scheme - it is the group of $k$-rational points of the group scheme $Aut_X$ - this is by definition a "set" and not a "representable functor". If a group scheme is a triple $(G,m,i)$ as defined above, it follows $Aut_X(k)$ is not a group scheme since $Aut_X(k)$ is a set and not such a triple $(G,m,i)$.
It seems to me that a group scheme $G$ in this setting is "algebraic" iff $k$ is an algebraically closed field and $R$ is a finitely generated $k$-algebra which is an integral domain.
It follows $R$ is a Hilbert-Jacobson ring and that any prime ideal in $R$ equals the intersection of the maximal ideals containing it. The set
$$G(k)\cong Hom_{k-alg}(R,k)$$
may be identified with the maximal ideals in $R$ and when you choose a set of generators $R\cong k[x_1,..,x_n]/\mathfrak{p}$ you choose an embedding $V(\mathfrak{p}) \subseteq \mathbb{A}^n_k$ as a closed subvariety.
Here - since $k$ is algebraically closed we use the language of Hartshorne Chapter I where an "algebraic variety" is an irreducible algebraic subset of $k^n$.
If $V(\mathfrak{p})$ is  irreducible it follows there is - by the Nullstellensatz - an equality of ideals
$$I(V(\mathfrak{p})) = \mathfrak{p},$$
hence in a sense you recover the ring $R$ from the set of maximal ideals in $R$. There is an isomorphism
$$R \cong k[x_1,..,x_n]/I(V(\mathfrak{p})) .$$
This may explain the statement
Proposition:"$Aut_X$ is algebraic if and only if $Aut(X)$ is."
Proof: If $Aut_X:=Spec(R)$ it follows that if the $k$-rational points $Aut(X) \subseteq k^n$ is an algebraic subvariety of $k^n$ in the above sense, it follows $R$ is a finitely generated $k$-algebra which is an integral domain, hence $Aut_X$ is algebraic. Conversely we define a group scheme $G:=Spec(R)$ to be algebraic iff $R$ is a finitely generated $k$-algebra which is a domain. It follows the "set of $k$-rational points" $G(k) \subseteq k^n$ may be realized as an irreducible algebraic subset for some $n\geq 1$.
We also observe that if we choose an embedding $G(k) \subseteq k^n$ as an algebraic variety and an ideal $I(G(k)) \subseteq k[x_1,..,x_n]$ we may define $R(G(k)):=k[x_1,..,x_n]/I(G(k))$ and then since $\mathfrak{p}$ is prime we get
$$R \cong R(G(k))$$
hence we recover $G:=Spec(R)$ from $G(k)$: There is an isomorphism of affine schemes
$$G \cong  G_k:=Spec(R(G(k)))$$
and an isomorphism of group schemes
$$h_G \cong h_{G_k}.$$
Question: "Does AutX(k) also have a group scheme structure?"
Answer: No, $G(k):=Aut_X(k)$ does not have a group scheme structure, but there is a way to construct a group scheme $h_{G_k}$ from $G(k)$ and an isomorphism $h_G \cong h_{G_k}$ of group schemes. Hence if the base field $k$ is algebraically closed and if the group scheme is reduced, you may recover the group scheme $Aut_X$ from the group $Aut_X(k)$.


Non-real numbers in system of equations Given $a^2+b^2=1$,
$c^2+d^2=1$,
$ac+bd=0$
To prove
$a^2+c^2=1$,$b^2+d^2=1$,$ab+cd=0$
Now this can be easily done by trigonometric substitution if it was given that $a,b,c,d$ are real numbers.
I have a solution using matrix which I think is valid even if the given numbers are non real.
https://artofproblemsolving.com/community/c6h475289p2661966
I am curious to know if there other ways to it Algebraical or with any other method.

This says that
$$\pmatrix{a&b\\c&d}\pmatrix{a&c\\b&d}=\pmatrix{1&0\\0&1}$$
implies
$$\pmatrix{a&c\\b&d}\pmatrix{a&b\\c&d}=\pmatrix{1&0\\0&1}.$$
This follows from the fact that (over any commutative ring) any square matrix
with a left inverse has a right inverse (necessarily the same as the left inverse).
But if you want an "equational" proof avoiding concepts from linear algebra
observe that
$$a^2+c^2=1+a^2c^2-(1-a^2)(1-c^2)=a+a^2c^2-b^2d^2=1+(ac+bd)(ac-bd)=1$$
and similarly for $b^2+d^2=1$. Also
$$ab+cd=ab(c^2+d^2)+(a^2+b^2)cd=(ac+bd)(ad+bc)=0.$$

The vectors $(a,b)$ and $(c,d)$ are unit and orthogonal.
A reflection around the bissector of the first quadrant swaps the abscissas and the ordinates, and creates the unit orthogonal vectors $(a,c)$ and $(b,d)$.

By trigonometry, the general solution of the first system is
$$a=\cos\theta,b=\sin\theta,c=\pm\sin\theta,d=\mp\cos\theta.$$


Prove: $A × (B ∩ C) = (A × B) ∩ (A × C).$ What I know:

-I need to prove the sets are equal, that is, prove the LH side is a subset of the RH side, and vice versa.
-$A × (B ∩ C)$ is the set of all ordered pairs with first entry an element of $A$, and second entry an element of $(B ∩ C)$.

-$(A × B)$ is the set of all ordered pairs with first entry an element of $A$, and second entry an element of $B$.
-$(A × C)$ is the set of all ordered pairs with first entry an element of $A$, and second entry an element of $B$.
So $(A × B) ∩ (A × C)$ is the set of ordered pairs that are both in $(A × B)$ and $(A × C)$.

How can I link these concepts together? I don't need a complete solution; getting started is the hardest part!

You need to "element chase"; that is, suppose $(x,y)\in A\times(B\cap C)$ and then show you must have $(x,y)\in (A\times B)\cap(A\times C)$, then do the other direction as well. I'll do one direction for you and you can try the other yourself.

Suppose $(x,y)\in A\times(B\cap C)$. Then $x\in A$ and $y\in B\cap C$, meaning $y\in B$ and $y\in C$. Because $y\in B$, $(x,y)\in A\times B$, and because $y\in C$, $(x,y)\in A\times C$. Therefore $(x,y)\in(A\times B)\cap(A\times C)$. This tells us $A\times(B\cap C)\subseteq (A\times B)\cap(A\times C)$.

Proof:
$$\begin{align}
A\times(B\cup C)&=\{(x,y):x\in A,y\in B\cup C\}\\
&=\{(x,y):x\in A,y\in B\text{ or }x\in A,y\in C\}\\
&=\{(x,y):(x,y)\in A\times B\text{ or }(x,y)\in A\times C\}\\
&=(A\times B)\cup(A\times C)
\end{align}$$
Therefore,  $A\times (B\cup C) = (A\times B)\cup(A\times C)$.


Prove that the following is not true or true. Prove that the following is not true or true. If $x,y,z \geq 1$ and they are all integers then $z^2=9xy+6y+6x+3$. If true, find $x,y,z$. If false, prove it.

I first tried some values but the fact that I am being asked to prove it if it is false, then I guess the above is false.

I tried some values like

$x=1,y=1$ then $z^2=24$ but that violates $z$ being an integer.

So if the above is false, how would i proceed in proving it.

EDIT: Sorry for all the confusion but I'm not trying to approach this problem by counterexamples. I want to know why does the equation $z^2=9xy+6y+6x+3$ not work? Why wont the left hand side ever provide a solution on the right hand side

So, there are two possible questions here:

$1)$ As stated, your question states, prove or disprove that $\forall x,y,z\in \mathbb{Z}_{\ge 1}$
$$ z^2=9xy+6y+6x+3.$$
Here is the counterexample: consider $x=y=z=1$. This implies
$$ 1=24.$$
This is blatantly false, and so this can not hold for all $x,y,z,\in \mathbb{Z}_{\ge 1}$.

$2)$ The alternative question states, prove or disprove that $\exists x,y,z\in\mathbb{Z}_{\ge1}$ such that
$$z^2=9xy+6y+6x+3. $$
Through software, or some sort of exhaustive calculation, we can see that $x=5,y=8\implies z=21$, a solution to the equation. So there exists such a $3-$tuple in $\mathbb{Z}_{\ge1}$ which solves the equation.

You haven't quite understood the question.

z^2 is defined as the result got by adding together 9xy + 6y + 6x +3

you're being asked whether 9xy + 6y + 6x + 3 can be a perfect square,
when x and y are both positive integers.

Many integers are not squares; they want you to show 9xy + 6y  + 6x + 3 can be a square, presumably for just one set of values, though this seems unclear.

One approach might be to try to find x and y so that 3xy + 2y + 2x +1 is 3 x a square, such as 3, 12, 27, 48... maybe by factorising.

(x + 1)(y+1) = xy + y + x +1 umm..

or let's try x=y, so we get 9X^2 + 12x + 3 = 3(3x+1)(x+1).
Then say x+1 = 3M to allow for the 3
= 3(3M)(3x + 1)  to get a square, M (3x+1) must be square (where M=(x+1)/3)
so M( 3 (3M -1) + 1) has to be a square.
so M(9M -2) has to be square. So ... ummm...
Anyway, give it a go....


Separation of variables on a second order ODE I am attempting to grasp the basics of separation of variables for a second order separable differential equation, and am failing to do so:

Given the equation:

$$ x=\frac{d^2y}{dx^2}$$

I know from calculus class intuitively that the solution is:

$$ y=\frac{x^3}{6}$$

But if I am to use separation of variables, why don't I get:

$$ \frac{y^2}{2}=\frac{x^3}{6} $$

Since I should have had to integrate the $y$ side twice as well.

You don't have something like $f(x)dx dx=g(y)dy dy$. You have $f(x) dx =g(y)d\left(\frac{dy}{dx}\right)$. Remember that the second deriviative is $\frac{d \left(\frac{dy}{dx}\right)}{dx}$. The actual solution is $\frac{x^3}6+cx+k$, where $c$ and $k$ are constants. This should be your process:

$x=\frac{d^2y}{dx^2}$

$x dx=d\left(\frac{dy}{dx}\right)$

$\int xdx=\int d\left(\frac{dy}{dx}\right)$

$\frac{x^2}2=\frac{dy}{dx}$

You have treated $\frac{d^2y}{dx^2}$ as $\frac{d^2y}{dx^2} = (\frac{dy}{dx})^2 = \frac{dydy}{dxdx}$. This of course isn't true. The second derivative of $y$ wrt $x$ is $\frac{d}{dx}(\frac{dy}{dx})$ which if you want to use the fraction analogy can be thought of as $\frac{dy}{dxdx}$. Now, using separation of variable on the equation you've supplied would yield $$\iint  xdxdx = \int  dy$$
Solving this would produce the correct solution. This isn't a mathematically rigorous explanation, but it works to show the difference between $\frac{d^2y}{dx^2}$ and the first derivative squared.


Prove a square is homeomorphic to a circle $s:=\{|x|\le 1,|y|\le 1\} $
$c:=\{{x}^{2}+{y}^{2}\le1\}$
Prove $\overset{\circ}{s}  \cong \overset{\circ}{c}$

ok... not to sure what to do.
I think $\overset{\circ}{s}  \to\overset{\circ}{c}$ is something like:
$$(x,y)\rightarrow\left(\frac{x}{\sqrt{{x}^{2}+{y}^{2}}},\frac{y}{\sqrt{{x}^{2}+{y}^{2}}}\right)$$
What is the inverse for this? Do i need an inverse? Do i just prove the function is continous and the inverse is continuous?
Please help...

I think the following more general result is true and the argument is also quite simple:

  Let $V$ be a real normed vector space and $||\cdot ||_1$, $||\cdot ||_2$ two equivalent norms on $V$, i.e. there are positive constants $C_1, C_2$ such that $||x||_1\le C_1 ||x||_2$ and $||x||_2\le C_2 ||x||_1$. Note that each norm gives rise to the same topology in $V$ because they are equivalent. Define the balls unit balls with respect to each norm as $B_1=\{x\in V:||x||_1\le 1\}$ and $B_2=\{x\in V:||x||_2\le 1\}$. Then $B_1\approx B_2$.

Define $\phi:D_1\to D_2$ as $\phi(x)=\frac{||x||_1}{||x||_2} x$ if $x\neq 0$ and $\phi(0)=0$. It's clear that $\phi$ is continuous in $V-0$ because the norms and the scalar product are continuous functions and continuity at $0$ is also easy to verify using the fact that $||\cdot ||_1$ and $||\cdot ||_2$ are equivalent norms. The inverse is given by $\psi:D_2\to D_1,\psi(x)=\frac{||x||_2}{||x||_1} x$ if $x\neq 0$ and $\psi(0)=0$. So $\phi$ is the desired homeomorphism.

The problem here is a special case of this taking $V=R^2$, the norms $||\cdot||_2$ and $||\cdot||_\infty$ (remembering that all norms in $R^2$, a finite dimensional vector space, are equivalent) and noting that my homeomorphism $\phi$ maps the boundary of the circle to the boundary of the square and so you get an homeomorphism from the interiors by restricting the domain.

You don't need to find the inverse because these two spaces are compact.


Evaluating binomial coefficients $$\binom{1}{0},\binom{1}{1},\binom{1}{2}$$
 What does this mean, and how do I achieve an numerical value when trying to solve a proof or problem in this form?

As you know, in general, $\binom{n}{k}$  is a number with value = the count of ways $k$ objects can be selected out of $n$ objects (when we are not interested in the order of selection, but just in the set of elements that are chosen).

Respectively:

$\binom{1}{0} = 1$ since there is only one way to select 0 elements out of 1 - just do not select anything.
$\binom{1}{1} = 1$ too, since we must just select the only element we have. We know also that $\binom{1}{1} = \binom{1}{0}$, because $1=1-0$.
$\binom{1}{2} = 0$ since we cannot select 2 elements out of 1, at all. In general, $\binom{n}{k} = 0$, if $k>n$ or $k<0$.

An expression for ${n}\choose{k}$ that you can use to compute it is $\frac{n}{k!(n-k)!}$.

The actual interpretation of the values is:

$\binom{1}{0} = 1$ means there is only one way to select 0 elements out of 1element
$\binom{1}{1} = 1$ means we one element and select the only element we have.
$\binom{1}{2} = 0$ means we cannot select 2 elements out of 1 so 0


Prove $ \oint_{\partial V} (\mathbf{\hat{n}} \times \mathbf{A}) \; \mathrm{d}S = \int_V (\nabla \times \mathbf{A}) \; \mathrm{d}V   $ I need help proving the following vector calculus identity:

$$
\oint_{\partial V} (\mathbf{\hat{n}} \times \mathbf{A}) \; \mathrm{d}S = \int_V (\nabla \times \mathbf{A}) \; \mathrm{d}V
$$

the identity is also found on this link, under "Surface–volume integrals".

Thank you very much

Since $\mathbf{\hat{n}}$ and $\mathbf{dS}$ are parallel, $\mathbf{dS} \times \mathbf{\hat{n}} = 0$, so
\begin{align*}
\mathbf{\hat{n}} \times (\mathbf{A} \times \mathbf{dS}) & = - \mathbf{A} \times (\mathbf{dS} \times \mathbf{\hat{n}}) - \mathbf{dS} \times (\mathbf{\hat{n}} \times \mathbf{A}) \\
 & = - \mathbf{dS} \times (\mathbf{\hat{n}} \times \mathbf{A}) \\
 & = (\mathbf{\hat{n}} \times \mathbf{A}) \times \mathbf{dS}
\end{align*}
by the the Jacobi Identity and divergence theorem.
$$\mathbf{A} ~dS = (\mathbf{\hat{n}} \times \mathbf{A}) \times \mathbf{dS} + \mathbf{\hat{n}} \cdot \mathbf{A} ~\mathbf{dS}.$$
Then use the corollaries to show that
$$\iint_{\partial V} \mathbf{A} ~dS = \iiint_V \left(\nabla(\mathbf{\hat{n}} \cdot \mathbf{A})- \nabla \times (\mathbf{\hat{n}} \times \mathbf{A})\right) ~dV.$$

here this guy explains it nicely
http://www.youtube.com/watch?v=1qLb0B40YnA


How to show a limit exist in maple? Im trying to show $ \lim_{x \to 0}\frac{\sin(\sin(2x)^2}{x^2} = 4 $ in maple, but I am not getting it. This is what I have

It shouldn't be that hard.  Maple has a "limit" command  Try something like limit((sin(sin(2*x)))^2/x^2,x=0);

You miscounted your parentheses so I'm guessing what you meant and I don't guarantee I counted them right, but that's how you do it.

You need to modify the condition, it should be $|f - 4| > \epsilon $


Conjectured closed form of $G^{2~2}_{3~3}\left(1\middle|\begin{array}c1,1;b+1\\b,b;0\end{array}\right)$ In my answer to this question, I come across the following case of the Meijer G-function:

$$F(b)=G^{2~2}_{3~3}\left(1\middle|\begin{array}c1,1;b+1\\b,b;0\end{array}\right), b>0$$

and based on my experiments, $F(b)$ have the following closed form:

$$F(b)
\stackrel?=\frac{\Gamma(b)}{b}\left(-\gamma-\psi(b)+\frac{2^{1-b}}{b}{_2F_1}\left(\begin{array}c1,1\\b+1\end{array}\middle|-1\right)+b{_3F_2}\left(\begin{array}c1,1,b+1\\2,2\end{array}\middle|-1\right)\right)$$

Is there any chance of proving this?

Edit: Using the definition of the Meijer G-Function, $F(b)$ have the integral representation:
$$F(b)=\frac{1}{2\pi}\int^{+\infty}_{-\infty}\frac{\Gamma(\tfrac{b}{2}+ix)\Gamma(\tfrac{b}{2}-ix)}{\tfrac{b^2}{4}+x^2}dx.$$

Edit 2: I've found a further generalization:
$$F(b,z)=G^{2~2}_{3~3}\left(z\middle|\begin{array}c1,1;b+1\\b,b;0\end{array}\right)=\frac{1}{2\pi}\int^{+\infty}_{-\infty}\frac{\Gamma(\tfrac{b}{2}+ix)\Gamma(\tfrac{b}{2}-ix)}{\tfrac{b^2}{4}+x^2}e^{(b/2+ix)\log z}dx\\
\stackrel?=\frac{\Gamma(b)z^b}{b}\left(-\log z-\gamma-\psi(b)+\frac{(z+1)^{1-b}}{b}{_2F_1}\left(\begin{array}c1,1\\b+1\end{array}\middle|-z\right)+bz~{_3F_2}\left(\begin{array}c1,1,b+1\\2,2\end{array}\middle|-z\right)\right), b\not\in\{0,-1,-2,\dots\},z\neq0,-1.$$
Edit 3: Further simplified the ${_2F_1}$ part.

Case $F(1,1)$ is this claim:
$$
2 \int_{-\infty}^\infty \frac{dx}{(4 x^2+1)\cosh(\pi x)} = 2\log 2 .
$$
Maybe start with a proof of that.

added March 24

We will attempt this $F(1,1)$ case using residues.  I changed variables slightly to make the problem
$$
\int_{-\infty}^{+\infty} \frac{dx}{(x^2+1)\cosh(\pi x/2)} = 2\log 2 .
$$
The function
$$
F(z) := \frac{1}{(z^2+1)\cosh(\pi z/2)}
$$
is analytic in the complex plane except for poles at $(2n+1)i$, $n \in \mathbb Z$.  Let $M$ be a large integer, and consider the contour integral of $F$ around the square with vertices
$$
-4 M ,\quad 4 M ,\quad 4M + 8 M i,\quad -4 M\pi + 8 M i .
$$
The integral along the bottom side is
$$
\int_{-4M}^{4M} \frac{dx}{(x^2+1)\cosh(\pi x/2)}
$$
so the limit as $M \to \infty$ is the integral to be computed.  We can do estimates for the other three sides to show their integrals go to zero as $M \to \infty$.  (Postponed...)

The poles for $F(z)$ inside the square are at $z=(2n+1) i$, for $n=0,1,\dots, 4M-1$.  Now $z=i$ is a double pole, the residue there is
$$
\mathrm{Res}_{z=i} F(z) = \frac{-i}{2\pi} .
$$
For $n>0$, the residue is
$$
\mathrm{Res}_{z=(2n+1) i} F(z) = \frac{(-1)^n i}{2 n (n+1)\pi}
$$
Now as $M \to \infty$, the square encloses more and more of these poles, so in the limit the sum is the sum of all the residues in the upper half-plane:
$$
\frac{-i}{2\pi}+\sum_{n=1}^\infty \frac{(-1)^n i}{2n(n+1)\pi}
=\frac{-i\log 2}{\pi}
$$
so finally the integral around the contour (which converges to the integral we want) also converges to $2\pi i$ times this sum of residues,
$$
\int_{-\infty}^{+\infty} F(z)\,dz = 2 \log 2 .
$$

Inspired by GEdgar's answer for the $F(1,1)$ case, I give an answer (to my own question!) for the general case.

We start from the integral representation $$
F(b,z)=\frac{1}{2\pi}\int^{+\infty}_{-\infty}\frac{\Gamma(\tfrac{b}{2}+ix)\Gamma(\tfrac{b}{2}-ix)}{\tfrac{b^2}{4}+x^2}z^{\frac{b}{2}+ix}dx.
$$

We consider the function $G(w)=\frac{\Gamma(\tfrac{b}{2}+iw)\Gamma(\tfrac{b}{2}-iw)}{\tfrac{b^2}{4}+w^2}z^{\frac{b}{2}+iw}dw$. $G(w)$ have double poles at $w=\pm\frac{b}{2}i$, and single poles at $w=\pm(\frac{b}{2}+n)i$ for all positive integer $n$.

Now let $N$ be a large integer, and consider the integral path around the rectancle with vertices $-N,N,N-(N+\frac{b+1}{2})i,-N-(N+\frac{b+1}{2})i$. It is possible to prove that the integrals on the three sides go to zero, and therefore we have
$$F(b,z)=-i(\text{sum of residues of $G(w)$ in the lower half-plane}).$$

It remains to calculate the residues at each pole.

In fact, we have $$
\begin{align*}
-i\operatorname{Res}_{w=-\frac{b}{2}i}G(w)&=\frac{z^b\Gamma(b)}{b}\left(\frac1b-\log z-\gamma-\psi(b)\right)\\
-i\operatorname{Res}_{w=-(\frac{b}{2}+n)i}G(w)&=\frac{(-1)^{n+1}z^{b+n}\Gamma(b+n)}{n~n!(b+n)}.
\end{align*}
$$

Comparing with the conjectured closed form, we only need to prove that $$
\frac1b-\sum_{n=1}^{\infty}\frac{(-z)^nb\Gamma(b+n)}{n~n!\Gamma(b)(b+n)}\stackrel?=\frac{(z+1)^{1-b}}{b}{_2F_1}\left(\begin{array}c1,1\\b+1\end{array}\middle|-z\right)+bz~{_3F_2}\left(\begin{array}c1,1,b+1\\2,2\end{array}\middle|-z\right).$$

Now note that $$
\frac{(z+1)^{1-b}}{b}{_2F_1}\left(\begin{array}c1,1\\b+1\end{array}\middle|-z\right)+bz~{_3F_2}\left(\begin{array}c1,1,b+1\\2,2\end{array}\middle|-z\right)\\
=b^{-1}{_2F_1}\left(\begin{array}cb,b\\b+1\end{array}\middle|-z\right)+bz~{_3F_2}\left(\begin{array}c1,1,b+1\\2,2\end{array}\middle|-z\right)\\
=\underbrace{\sum_{m=0}^{\infty}\frac{\Gamma(b+m)^2\Gamma(b+1)}{b\Gamma(b)^2\Gamma(b+m+1)m!}(-z)^m}_{n=m}-\underbrace{\sum_{m=0}^{\infty}\frac{b\Gamma(m+1)^2\Gamma(b+m+1)\Gamma(2)^2}{\Gamma(1)^2\Gamma(b+1)\Gamma(m+2)^2m!}(-z)^{m+1}}_{n=m+1}\\
=\sum_{n=0}^{\infty}\frac{\Gamma(b+n)}{\Gamma(b)(b+n)n!}(-z)^n-\sum_{n=1}^{\infty}\frac{\Gamma(b+n)}{n~n!\Gamma(b)}(-z)^{n}\\
=\frac1b+\sum_{n=1}^{\infty}\left(\frac{\Gamma(b+n)}{\Gamma(b)(b+n)n!}(-z)^n-\frac{\Gamma(b+n)}{n~n!\Gamma(b)}(-z)^{n}\right)\\
=\frac1b+\sum_{n=1}^{\infty}\frac{\Gamma(b+n)(-z)^n}{n!\Gamma(b)}\left(\frac{1}{b+n}-\frac{1}{b}\right)\\
=\frac1b-\sum_{n=1}^{\infty}\frac{b\Gamma(b+n)(-z)^n}{n(b+n)n!\Gamma(b)}.
$$

Edit: Ninja'd by 2 minute.


Solve without using L'Hôpital's rule? Is it possible to solve this without using L'Hôpital's rule?

$$
\lim_{x\to 0} \Big(\frac{3x+1}{x}-\frac{1}{\sin x}\Big)
$$

I tried to solve it but I got stuck at the $\frac{1}{\sin x}$ part.

$$
=\lim_{x\to 0}\frac{3+x^{-1}}{1}-\lim_{x\to 0}\frac{1}{\sin x}
$$

since $\lim_{x\to 0}\frac{1}{\sin x}$  does not exist.

alright so lets break up the first term to get

lim(x->0) [3x/x + 1/x - 1/sinx]
now lets use properties of limits

lim(x->0) 3 + lim(x->0) (1/x) - lim(x->0) 1/sinx
now knowing that lim(x->0) sinx/x = 1 lets divide the third term by x at the numerator as well as the denominator to obtain

lim(x->0) 3 + lim(x->0) (1/x) - lim(x->0) (1/x)/sinx/x

and since the denominator is just 1 we are left with

lim(x->0) 3 + lim(x->0) (1/x) - lim(x->0) (1/x)

BUT notice lim x--> 0 1/x does not exist. so this direct method of finding limit would not work. another use is power series as one of the answer is posted :)


Jordan form exercise What am I doing wrong?

I've been learning how to put matrices into Jordan canonical form and it was going fine until I encountered this $4 \times 4$ matrix:

$A=\begin{bmatrix}
2 & 2 & 0 & -1 \\
0 & 0 & 0 & 1 \\
1 & 5 & 2 & -1 \\
0 & -4 & 0 & 4 \\
\end{bmatrix}
$

Which has as only eigenvalue $\lambda_1=\lambda_2=\lambda_3=\lambda_4=2$ with 2 corresponding eigenvectors, which I will for now call $v_1$ and $v_2$:

$v_1 = \pmatrix{0\\0\\1\\0}, v_2=\pmatrix{-3 \\ 1 \\ 0 \\ 2}
$

2 eigenvectors means 2 Jordan blocks so I have 2 possibilities:

$J= \pmatrix{2 & 1 & 0 & 0 \\
0 & 2 & 1 & 0 \\
0 & 0 & 2 & 0 \\
0 & 0 & 0 & 2} $ or $ J= \pmatrix{2 & 1 & 0 & 0 \\
0 & 2 & 0 & 0 \\
0 & 0 & 2 & 1 \\
0 & 0 & 0 & 2}
$

I consider the first possibility. This gives me the relations:

$Ax_1=2x_1 \\
Ax_2=x_1+2x_2 \\
Ax_3=2x_3+x_2 \\
Ax_4=2x_4 \\
$

where $x_1$ and $x_4$ should be $v_1$ and $v_2$. From the second relation $(A-2I)x_2=x_1$ I see

$\pmatrix{0 & 2 & 0 & -1 \\
0 & -2 & 0 & 1 \\
1 & 5 & 0 & -1 \\
0 & -4 & 0 & 2} \pmatrix{a \\ b \\ c \\ d} =\pmatrix{0 \\ 0 \\ 1 \\ 0} $

( $v_2= \pmatrix{ -3 \\ 1 \\ 0 \\ 2}
$ will give an inconsistent system)

Now I get that $x_2 = \pmatrix{-2 \\ 1 \\ 0 \\ 2}
$

From the third relation $(A-2I)x_3=x_2$:

$\pmatrix{0 & 2 & 0 & -1 \\
0 & -2 & 0 & 1 \\
1 & 5 & 0 & -1 \\
0 & -4 & 0 & 2} \pmatrix{e \\ f \\ g \\ h} =\pmatrix{-2 \\ 1 \\ 0 \\ 2} $

But this system is inconsistent as well! No matter which vectors I try in which places, when I try to generalize eigenvectors I seem to always end up with some inconsistency.

Is there something staring me in the face that I am overlooking? Or am I doing it completely wrong (even though this method worked fine for me before)?

Sorry for the lengthiness and thank you in advance.

Take a look at
$$
       (A-2I)^2 = \begin{bmatrix}
0 & 2 & 0 & -1 \\
0 & -2 & 0 & 1 \\
1 & 5 & 0 & -1 \\
0 & -4 & 0 & 2 \\
\end{bmatrix}^2
$$
The $a_{1,2}$ entry is non-zero. What does that tell you?

Look at the image space (column-wise) of $S = ( A - 2I )$, which yields two vectors  $S(v), v = (0,0,2,0) , (-1,1,-1,2)$ [SPECIFICALLY IN THIS ORDER]
Now find $u$ such that $S(u) = v$      say $u = (0,0,0,1)$
Now add any vector $w$ such that $w$ is in $Ker(S)$ and altogether completes a basis for $\Bbb R^4$; clearly $w = (-3,1,0,2)$

You can check that $\mathcal B = \{ (0,0,2,0), (-1,1,-1,2), (0,0,0,1), (-3,1,0,2)\}$does the job for JCF transform.

P.s not experienced in this typefont, but the maths works!:)


Why is finding all subsets of a set (power set) an exponential problem? I know it's function of $2^n$ where $n$ is the number of elements in the original set. But what is so doubling (100% growth) about finding all subset of a set?

Usually for $2^n$ function we can represent it as a binary tree but I can't work my way to understand why it's exponential in this case. It's a property but I can't link it to usual methods of understanding $2^n$.

You can represent the process of identifying all subsets of $\{1,\ldots,n\}$ as a binary tree. Here’s a rough sketch for $n=3$:

At the first stage we ask whether $1$ is in the subset; at the second whether $2$ is in the subset; and so on. Since the question always has $2$ possible answers, each level of the tree has twice as many nodes as the previous level, and after $n$ questions we have

$$1\cdot\underbrace{2\cdot 2\cdot\ldots\cdot 2}_{n\text{ doublings}}=2^n$$

nodes. Each subset of $\{1,\ldots,n\}$ is completely determined by the answers to these $n$ questions, and each string of yes/no answers completely determines a subset of $\{1,\ldots,n\}$, so at the last level there is exactly one node for each subset of $\{1,\ldots,n\}$. Thus, $\{1,\ldots,n\}$ must have $2^n$ subsets.

It is sufficient that given any subset of $\{0,...,n-2\}$ (with $n-1$ elements), there are two ways to extend it to a subset of $\{0,....,n-1\}$ (with $n$ elements). But the extension either includes or excludes $n-1$, so we are done.


Dimension theory "based on $\mathbb R^n$" This question is somewhat vague, so please be gentle with me. I want to know if there is some definition of topological dimension that has $\mathbb R^n$ as a "paradigm", something like 'A nice (connected, normal, whatever) topological space $X$ has dimension $n$ if $X$ can be "related" to $\mathbb R^n$ in some topological way'. Of course, the core of this tentative definition is what we mean by "related". An (almost certainly doomed) attempt of mine, illustrating my point of view (and my intentions) is the following:

  $X$ has topological dimension $n$ if $X$ is homeomorphic to some subspace of $\mathbb R^n$, and it is not homeomorphic to any subspace of $\mathbb R^{n-1}$.

EDIT

Murphy's Law applied in this case. I refrained to ask for a definition not limited to topological manifolds, because I thought that it was clear that my intention was to get an answer with enough generality, for example for spaces like $\mathbb Q^n$. Unfortunately this was not the case. Summarizing, I am asking for a definition not limited to topological manifolds.  Also, please don't forget that the tentative definition above is just for illustration, I am not taking it seriously.

Perhaps what you are aiming at is the more general Urysohn dimension theory, see http://en.wikipedia.org/wiki/Inductive_dimension

The path to connect the space $\mathbb R^n$ with the euclidean space of $n$ dimensions, is through the cartesian product of real number lines.  For this kind of space, one can directly used established 'analytical geometry' formulae (eg a line as $y=ax+b$), to do this.

The relation to non-euclidean space is that it is possible to map the coordinate system $\mathbb R^n$ onto something like hyperbolic space, and then show that there is a topological bending of euclidean into hyperbolic space, and thus some kind of local connectivity applies.

One can consider other sets like $\mathbb C^n$ onto the 'unitary space'.  The 'unitary line' is the argand diagram $\mathbb C^1$.  The same formulae, like the line as $y=ax+b$ still work but the real-line as line is now replaced by the argand-diagram as line.  It's very useful for exploring even dimensions, like $n=4$.

For example, in 2D, one can represent rotation around a point by the relation $X(t)=X(0) \operatorname{cis}(\omega t)$.  In the 'four-dimensional' $\mathbb C^2$, the same product, will cause a rotation that rotates everything around the point $(0,0)$, specifically:

$$X(t),Y(t) = X(0) \operatorname{cis}(\omega t), Y(0) \operatorname{cis}(\omega t)$$

The cute thing to note here is that ratio $Y(t)/X(t) = Y(0)/X(0)$, and must stay on the same 'argand diagram' or complex line through $(0,0)$.  This equation then describes rotation of all four-dimensional space around a point, and it's relatively easily shown.  Likewise, introducing $Z(), \dots $, show that this is true of all even dimensions.

Mapping things like $\mathbb Q^n$, also is useful.  $\mathbb Q$ is what i call a discrete set, in as much as there are numbers in its scope that don't belong to it: for example, $\sqrt{2}$ is not rational.  A set of this nature is rather like a sparse lattice formed by $\mathbb Z^n$.  For example, the set $CQ$ contains units on the unit circle like $0.6+0.8i$.

More usefully, I use it of the sets of the span of chords of a polygon $\mathbb Zp$, and their corresponding cyclotomic numbers, $\mathbb {CZ}n$.  This leads to spaces like $\mathbb Zp^n$.

The uniform polytope 'bi-truncated 24choron', or 'octagonny', or $o3x4x3o$, tiles space in a piecewise discrete manner.  This means that we can completely make all of the figures that are attatched to it.  One can evaluate the space that can be reached by edge-traversal, rather like the cubic gives $\mathbb Z^4$.  In this case, it's $\mathbb Z4^4$, the set of coordinates of the form $z_1 + z_2 \sqrt{2}$.  This space is infinitely dense, but is itself a double-projection of something in 8-dimensions, like $E_8$.  There's a set of quaterions based on octagonny.


Betting game with numbers Someone is challenged to play the following game: there are 36 marbles in an urn. 21 of them are red, 9 are blue and 6 are green. Each red is worth 0 rupees, each blue 100 rupees and each green 1000 rupees. By betting 1000 rupees only once, you must pick as many marbles as you want (obviously without seeing its color), one by one (you draw each marble, look at its color and either stop or continue) and without replacement and you can stop at any time, and you earn the product of the values of the marbles you have picked so far.
Assuming that the player plays optimally, what is the average percentage of the average percentage of profit for the game?
My understanding is that we must calculate the probability (not) to draw a 0 in a series of consecutive draws from 1 to 36.
For the 1st draw, the probability to get a 0 is $P_1 = \frac {21}{36}$ and the game ends, so with probability $P'_1 = \frac {15}{36}$ the player continues ("optimal" playing refers to this, otherwise if he continues, he will get 0 anyway).
Up to now, the player has earned 100 rupees with probability $Pw'_1 = \frac {15*9}{36*15}$ and 1000 rupees with probability $Pww'_1 = \frac {15*6}{36*15}$.
For the 2nd draw, the probability not to get a 0 is $P'_2 = \frac {14}{35}$ and so on.
The player will earn again the product of either 100 or 1000 of the first earning, with probability equal to the product of $P'_2$ by $\frac {8}{14}$ or $\frac {5}{14}$.
Can you please help me continue? It is getting really confusing from this point forward!
Thank you very much!

As long as there is at least one blue or green marble in the urn, it makes sense to keep playing, since you have a better than 1 in 36 chance of multiplying your winnings by at least 100. So the optimal strategy is to draw 15 marbles. Your expected earnings will be the product of: (1) the probability that every one of the first 15 marbles drawn is non-red, times (2) the number of rupees you will earn if you draw all 15 non-red marbles and no red marbles. Can you compute these two quantities? The result is kind of funny, in that you wind up with a minuscule probability of winning an astronomically large amount of money.
Edit: There are
$$
{36\choose 15}=\frac{36!}{15!\, 21!}=5567902560
$$
possible sets of 15 marbles you could draw. Exactly one of them has no red marbles, so using the optimal strategy, the probability you win money is $1/5567902560$. In the event you draw 15 non-red marbles, the amount of money you win is $100^9\cdot 1000^6 = 10^{36}$. Your expected winnings are then
$$
\frac{10^{36}}{5567902560}\approx 1.796\cdot 10^{26},
$$
and your expected percent profit is
$$
\frac{\frac{10^{36}}{5567902560}-1000}{1000}\cdot 100\%\approx 17960084416419816082413526\%.
$$


Curve enclosing the maximum area the curve of fixed length $l$ that joins the points $(0,0)$ and $(1,0)$ lies above the $x-axis$ and encloses the maximum area between itself and the $x-axis$, is a segment of

A straight line
A parabola
An ellipse
A circle

I don't know exactly how to solve it but it seems Circle is the right Answer

Here l is the fixed perimeter of plane curve passing through two given points (0,0) and (1,0). Let S be the area enclosed by the plane curve and x-axis

Then, we are maximize $S=\int _0^1 y\;dx$ with boundary conditions y(0)=y(1)=0 subject to constraint $\int_0^1(1+y'^2)^\frac{1}{2}\;dx=l$

Let $F(x,y,y')=y+ \lambda(1+y'^2)^\frac{1}{2} $

Where $\lambda$ is the lagrange multiplier . Then the required extremal satiesfied the Euler's equation

$\frac{\partial F}{\partial y}-\frac{d}{dx}(\frac{\partial F}{\partial y'})=0\\
\Rightarrow 1-\frac{d}{dx}\{\frac{\lambda y'}{(1+y'^2)^\frac{1}{2}}\}=0\\
Integrating \;w\;r\;to\;x, x-\{\frac{\lambda y'}{(1+y'^2)^\frac{1}{2}}\}=a\\
\Rightarrow (x-a)^2=\{\frac{\lambda^2 y'^2}{(1+y'^2)}\}\\
\Rightarrow \frac{1+y'^2}{y'^2}=\frac{\lambda^2}{(x-a)^2}\\
\Rightarrow y'^2=\frac{(x-a)^2}{\lambda^2-(x-a)^2}\\
\frac{dy}{dx}=\pm \frac{(x-a)}{(\lambda^2-(x-a)^2)^\frac{1}{2}}\\
Integrating, \; y=b\pm\{\lambda^2-(x-a)^2\}^\frac{1}{2}\\
\Rightarrow (y-b)^2=\lambda^2-(x-a)^2\\
\Rightarrow (x-a)^2+(y-b)^2=\lambda^2$
this is equation of a circle with radius $\lambda$

By reflecting the curve across the $x$-axis we can make it in to a closed loop, which doubles both the length and the area. Thus this is equivalent to the usual isoperimetric problem (maximize area inside a loop of fixed length) with the constraint of reflection symmetry. Since the circle is the unconstrained optimum and also has the reflection symmetry, it provides the solution to your problem: the curve is a semicircle.


If $(A-\lambda I)^r q(A) =0$ then $q(V) \subset \text{ker}(A-\lambda I)^r$ I was reading this answer to a question about the multiplicity of the roots in the minimal polynomial. The hypothesis of the original question is as follows:

  Let $ V\neq \{0 \}$ be a finite-dimensional vector space over a field $F$ and let $A \in \text{End}(V)$

At the beginning of the answer, I was reading the author establishes the following:

  We have $(A-\lambda I)^r q(A) =0$, which is to say that $q(V) \subset \text{ker}(A-\lambda I)^r$

I understand that the first equation is $0$ because by definition the minimal polynomial annihilates the transformation to which it's related to, but I don't see how this implies the second part. It is my understanding that $ \text{ker}(A-\lambda I)^r$ is the set of all generalized eigenvectors associated to the eigenvalue $\lambda$, so is this saying that any vector in $V$ applied to the polynomial $q$ is a generalized eigenvector?

Can someone tell me why this implication holds or tell me what I'm missing? Thank you in advance!

EDIT: The original author of the question edited his answer to have $q(A)V$ instead of $q(A)$, which is what some people already said they suspected in the comments of this question.

Given a linear operator $A : V \to V$ and a polynomial $q(t)$ such that $(A - \lambda I)^r q(A) = 0,$ it follows that $(A - \lambda I)^r q(A)(v) = 0(v) = 0$ for all $v$ in $V,$ hence we have $\operatorname{range}(q(A)) \subseteq \ker(A - \lambda I)^r.$ Considering that $\operatorname{range}(q(A)) = q(A)(V)$ by definition, we conclude that $q(A)(V) \subseteq \ker(A - \lambda I)^r.$

The kernel of an operator $T\in \Bbb L(V$) is defined by $\{v\in V \mid T(v)=0\}$. Hence $(A-\lambda I)^r q(A) =0$ is exactly the same as saying every  $q(A)(v) \subset \text{ker}(A-\lambda I)^r\; \forall v\in V$

(I believe the V was supposed to be A. I've learned this part in Linear Algebra but we never used q(V) with V a vector space.)


How is rearranging $56\times 100\div 8$ into $56\div 8\times 100$ allowed by the commutative property? So according to the commutative property for multiplication:

$a \times b = b \times a$

However this does not hold for division

$a \div b \neq b \div a$

Why is it that in the following case:

$56 \times 100 \div 8 = 56 \div 8\times 100$

It seems like division is breaking the rule. There is something I am misunderstanding here.

Is it because $a\times b\div c=a\div c\times b$ is allowed since $b\div c$ are not being rearranged so that $c\div b$?

If this is the case are you allowed to rearrange values in equations so long as no values have the form $a \div b = b \div a$ and $a - b = b -a$ ?

Notice that you have always $\div 8$, no matter the order of the other terms. You don't divide by a different number. It might help to think $\div c$ as a multiplication with $d=1/c$. Then everything would look easier:
$$a\times b\div c=a\times b\times d=a\times d\times b=a\div c\times b$$

With multiplication and division alone (no addition or subtraction) they are associative with respect to one another, BUT division itself is NOT commutative.

So basically you can do the multiplication or division in either order, BUT you must respect which way you interpret the inputs to the left or right of the operator's sign. If you flip this it has the effect of flipping the inputs which is equal precisely when the operator is commutative. As you noted, with division, it is not commutative.


How do I use the Lagrange multiplier in this problem? Suppose I have this problem:

max     f(x,y)
s.t    2x + 4y = 3

If the function $f(x,y) = x^{0.7}y^{0.3}$ then how can I use Lagrange multiplier to find the maximum input choices? I found out that $\nabla$$g(x,y) = 2i + 4j$ and $\nabla$$f(x,y) = 0.7(y/x)^{0.3} + 0.3(x/y)^{0.7}$. So I know that $\nabla$f = λ$\nabla$g but how do I use these numbers to find the maximum input choices?

Consider $$F=x^{7/10} y^{3/10}+\lambda  (2 x+4 y-3)$$ compute the derivatives and set them equal to zero $$F'_x=2 \lambda +\frac{7 y^{3/10}}{10 x^{3/10}}=0$$ $$F'_y=4 \lambda +\frac{3 x^{7/10}}{10 y^{7/10}}=0$$ $$F'_\lambda=2 x+4 y-3=0$$ From $F'_x=0$, you can extract $$\lambda=-\frac{7 y^{3/10}}{20 x^{3/10}}$$ Replacing $\lambda$ in $F'_y$, you get $$F'_y=\frac{3 x^{7/10}}{10 y^{7/10}}-\frac{7 y^{3/10}}{5 x^{3/10}}=\frac{3 x-14 y}{10 x^{3/10} y^{7/10}}=0$$ which makes $$3x-14y=0$$ Combine this with $F'_\lambda$ : two linear equation for two unknowns $x,y$. Solve and compute the value of $x^{7/10} y^{3/10}$.

At first glance, I thought that the problem is simple because the sum of the exponents is equal to $1$. Effectively, making it more general such as $$F=x^{a} y^{1-a}+\lambda  (2 x+4 y-3)$$ just doing the same, you should arrive to $$(1-a)x-2ay=0$$

But, the problem stays simple if set as $$F=x^{a} y^{b}+\lambda  (2 x+4 y-3)$$ Repeat the same steps to arrive to $b x-2ay=0$.

I am sure than you can take it from here.

during the step
Replacing λ in F'y,
you divided the denominator by 4 because of 4λ. Can you actually do this? It seems like it would change the value.


How to verify that $\frac{68}{25}$ is an upper bound for the set $A=\{(1+\frac{1}{n})^n|n\in\Bbb N^*\}$ I've tried to prove it aiming for a contradiction:If $\frac{68}{25}$ is not an upper bound of A, then, $\exists n\in\Bbb N^*$ such that $$\frac{68}{25}\lt(1+\frac{1}{n})^n\Rightarrow\frac{(68)^{1/n}}{(25)^{1/n}}-\frac{1}{n}<1\Rightarrow n(68)^{1/n}-(25)^{1/n}\lt n(25)^{1/n}\Rightarrow n(68)^{1/n}\lt(25)^{1/n}(n+1)$$ At this part i get stuck, because, we see that $n<n+1$ and instinctively i'd say that

$(68)^{1/n}<(25)^{1/n}$, apparently, i´ve found a contradiction: $68<25$. But i've found that in general if

we have $a<b$ and $ac<bd$ it´s not always true that $c<d$, for instance, $2<3$ and $2\sqrt2<3(1)$

don't imply that
$\sqrt2<1$. What can i do?, Please, help me. THANKS

The expression $\left(1 + \frac{1}{n}\right)^n$ is increasing, and its limit as $n \to \infty$ is $$e = \sum_{k = 0}^{\infty} \frac{1}{k!} ,$$
so as hinted in the comments, the problem is equivalent to showing that $$ e \leq \frac{68}{25}.$$

Method 1: Taylor expansion The above series converges rapidly, suggesting we can produce tight bounds by keeping the first several terms and then bounding the tail by a small quantity.

Expanding the series, we have
$$e = 1 + 1 + \frac{1}{2} + \frac{1}{6} + \frac{1}{24} + \frac{1}{120} + \frac{1}{720} + R ,$$
for a remainder series $R$. But the $\ell$th term of $R$ is a product of $\frac{1}{720}$ with a product of $\ell$ integers $> 2$, so $$R < \frac{1}{720} \sum_{\ell = 1}^{\infty} 2^{-\ell} = \frac{1}{720}.$$ Combining this with the previous inequality gives $e < \frac{979}{630}$, and cross-multipliying gives $\frac{979}{630} < \frac{68}{25}$. Of course, a similar method could be applied to other series with positive terms converging to $e$.

Method 2: Continued fraction expansion From the series characterization of $e$, its continued fraction expansion is $[2; 1, 2, 1, 1, 4, 1, 1, 6, \ldots, 1, 1, 2n, \ldots]$. The successive convergents of a continued fraction expansions (1) approximate the value more and more closely and (2) alternate being underestimates and overestimates. Computing from the expansion gives that the convergents of this series are: $2, 3, \frac{8}{3}, \frac{11}{4}, \frac{19}{7}, \frac{87}{32}, \ldots .$ Since $\frac{19}{7} < \frac{87}{32}$, the latter is an overestimate for $e$, and again cross-multiplying gives $e < \frac{87}{32} < \frac{68}{25}$.

$$
\left(1+\frac1n\right)^n < \left(1+\frac1n\right)^{n+1}
$$
and the right hand side is decreasing.
From this you get $\sup A < \left(1+\frac1{1000}\right)^{1000+1}<\frac{68}{25}$.

Edit for the non-believers: the last inequality is a true fact which can be verified with exact arithmetic: $25\times1001^{1001}<68\times1000^{1001}$.

I still don't get how this correct answer can get downvoted (but still get also copied by later answers...).


Evaluate the complex logarithmic function $\log(1-i)$ Evaluate $\log(1-i)$.

I have $\log(1-i) = \operatorname{Log}|1-i|+i\operatorname{Arg}(1-i)+i2k\pi$

How do I compute $\operatorname{Log}|1-i|$?

First, find $|1-i|$. That's positive real ($\left|a+bi\right|=\sqrt{a^2+b^2}$). Then find its log.

Firstly, you compute the modulus.  Second, determine the argument.     The format is $\log(r)+i
arg(2\pi n)$.
This is how complex logarithm is determined.


Does “arbitrarily small” mean very close to zero or very negative? In mathematical writing, does “arbitrarily small” mean very close to zero (like $0.000001$) or very negative (like $-1000000$)?

Are there better phrases to distinguish these two cases?

Traditionally, it's the first one: "arbitrarily small" means "small in magnitude", close to zero.

I see the logic behind the second interpretation, but numbers like $-10^9$ are normally thought of as large negative numbers, so using "small" to describe them is bound to be confusing.

Also on English.SE: What is an adjective for a very large negative number?

Usually, "Arbitrarily Close" is used to reference a proximity or distance between two points.  If two points are arbitrarily close, the difference is such that they are all but the same point; they are so close the difference is insignificant - except that they are different.


Prove that polynomial is always greater than zero Let $p(x)$ be a polynomial such that $p(1) = 0$ and $p'(x) > p(x)$ for all $x \geq 1$ , then we have to prove $p(x) > 0$, for all $x >1$.

I got it when I thought it logically . But how can we prove it mathematically.

According to me $p'(1)$ should be positive .

So after $ x= 1$ the polynomial will be positive and hence the $p'(x)$ cannot be negative i.e the polynomial is decreasing as polynomial is positive. Hence polynomial is always positive.

But how can we prove it mathematically that the polynomial is always positive . I am not getting any idea  .

Consider a function $h(x) = p(x) e^{-x} ,$ we have $h' (x) =e^{-x} (p'(x) -p(x) )>0 $ for $x>1$ thus $h$ is increasing and therefore $h(x)>h(1) $ for $x>1.$

I have done at last . I don't know which all polynomial are satisfying , but I have proved it .


Group with a property that $x^{-1}=x$ Suppose that we have some group G in which the following holds for all $x $ belongs to G: $x^{-1} =x$. Prove that G is abelian.

How would you guys solve this one?

Well, we have
$$xy = (xy)^{-1} = y^{-1}x^{-1} = yx$$ for all $x,y \in G$. Hence $G$ is abelian.

Take $x,y \in G$ , you want to show that $xy = yx$.

Or , the commutator of $x,y$ is the identity , that is $[x,y] = xyx^{-1}y^{-1}=e$.

Can you take this from here?


Definitions 4.32 and 4.33 and Theorem 4.34 in Baby Rudin: Are these proofs the limit statements based on these definitions correct? Here's Theorem 4.34 in the book Principles of Mathematical Analysis by Walter Rudin, 3rd edition:

Let $f$ and $g$ be defined on $E \subset \mathbb{R}$. Suppose $$ f(t) \to A, \ \ \ g(t) \to B, \ \mbox{ as } \ t \to x.$$ Then
(a) $f(t) \to A^\prime$ implies $A^\prime = A$.
(b) $\left( f+g \right)(t) \to A+B$.
(c) $\left( fg \right)(t) \to AB$.
(d) $\left( f/g \right)(t) \to A/B$.
provided the right members of (b), (c), and (d) are defined.
Note that $\infty - \infty$, $0 \cdot \infty$, $\infty / \infty$, $A/0$ are not defined.

Now here is Definition 4.33 in Rudin:

Let $f$ be a real function defined on $E \subset \mathbb{R}$. We say that $$ f(t) \to A \ \mbox{ as } \ t \to x,$$ where $A$ and $x$ are in the extended real number system, if for every neighborhood $U$ of $A$ there is a neighborhood $V$ of $x$ such that $V \cap E$ is not empty, and such that $f(t) \in U$ for all $t \in V \cap E$, $t \neq x$.

And, finally here is Definition 4.32:

For any real $c$, the set of real numbers $x$ such that $x > c$ is called a neighborhood of $+\infty$ and is written $(c, +\infty)$. Similarly, the set $(-\infty, c)$ is a neighborhood of $-\infty$.

How to prove Theorem 4.34 using Definitions 4.33 and 4.32?
My effort:
Theorem 4.34(a):

If $f(t) \to A$ and $f(t) \to A^\prime$ as $t \to x$, then, for every neighborhood $U$ of $A$, we can find a neighborhood $V$ of $x$ such that $V \cap E$ is not empty, and such that $f(t) \in U$ for all $t \in V \cap E$, $t \neq x$, and for every neighborhood $U^\prime$ of $A^\prime$, we can find a neighborhood $V^\prime$ of $x$ such that $V^\prime \cap E$ is not empty, and such that $f(t) \in U^\prime$ for all $t \in V^\prime \cap E$, $t \neq x$.
Now if $A \neq A^\prime$, then we can find neighborhoods $U$ and $U^\prime$ of $A$ and $A^\prime$, respectively, such that $U \cap U^\prime$ is empty. Now $V \cap V^\prime$ is not empty and is a neighborhood of $x$. What next?

Theorem 4.34(b), (c), and (d):

If $f(t) \to A$ and $g(t) \to B$ as $t \to x$, then, for every neighborhood $U$ of $A$ and for every neighborhood $V$ of $B$, we can find neighborhoods $W_1$ and $W_2$, respectively, of $x$ such that $W_1 \cap E$ and $W_2 \cap E$ are not empty, and such that $f(t) \in U$ for all $t \in W_1 \cap E$, $t \neq x$, and such that $g(t) \in V$ for all $t \in W_2 \cap E$, $t \neq x$. What next?

4.34(a) is not necessarily right. Consider a function f defined on E={1,2,3} such that f(j)=j for all j, and x=2. Then according to 4.33, you can show both f(t) -> 1 and f(t)-> 3 when t -> 2.
Specifically, for this function f, you can pick any neighborhood of 1, and of course you can always choose V=(0, 2.5), so V intersects E with {1,2}, and f(1)=1 is of course in the neighborhood of 1. A similar story can be told for any neighborhood of 3.
One way to fix the problem is to add an assumption in 4.33. For we to be able to say that f(t)-> A when t-> x, we add the assumption that x is a limit point of E. After that, (a) is not a problem. You can try (b,c,d).


How prove this Polynomial $g(x)=\sum_{i=1}^{n}a^m_{i}x^i$have only real roots? Question 1:
let Polynomial $f(x)=\displaystyle\sum_{i=0}^{3}a_{i}x^i,$ have three real numbers roots,where $a_{i}>0,i=1,2,3$.

show that:
$$g(x)=\sum_{i=0}^{3}a^m_{i}x^i$$ have only real roots,where $m\in R,m\ge 1$

My try:
case (1):
suppose that $f(x)$  has  a zero of multiplicity 3, then we assume

$$f(x)=(x+p)^3=x^3+3x^2p+3xp^2+p^3$$
then
$$g(x)=x^3+(3p)^mx^2+(3p^2)^mx+(p^3)^{m}=(x+p^m)[x^2+(3^mp^m-p^m)x+p^{2m}]$$

then

$$h(x)=x^2+p^m(3^m-1)x+p^{2m}\Longrightarrow \Delta =(p^m(3^m-1))^2-4p^{2m}>0$$
so this case $g(x)$ have only three real roots.

for case (2):
let $f(x)=(x+p)^2(x+q)$,
I can't prove it,
and the case (3):
$$f(x)=(x+p)(x+q)(x+r)$$
and this case I can't prove it too.
I hope someone can help solve this nice problem ;
Thank you very much!

As I explained in my comment, it is enough to do case (2).

Let $x,y \ge 0$. Then $(X+x)(X+x)(X+y) = X^3+(2x+y)X^2 + (x^2+2xy)X + x^2y$.
The discriminant of $X^3+bX^2+cX+d$ is, according to wikipedia, $\Delta(b,c,d) = b^2c^2-4c^3-4b^3d-27d^2+18bcd$.

The derivative of $m \mapsto \Delta(b^m,c^m,d^m)$ at $m=1$ is
$$\delta(b,c,d) \\
= (2b^2c^2-12b^3d+18bcd)\log b + (2b^2c^2-12b^3d+18bcd)\log c + (18bcd-4b^3d-54d^2)\log d
$$

Plugging $b = 2x+y,c = x^2+2xy,d = x^2y$ we compute and obtain
$$ \delta(x,y) =
 4x^6(\frac yx -1)^3\left(-(2+ \frac yx)\log \frac {(2x+y)^2}{x^2+2xy} + \frac yx \log \frac {(2x+y)^3} {x^2y}\right)$$

$\delta(x,y)/4x^6$ is actually a function of $t = \frac yx$ and has the same sign as $\delta(x,y)$ so it is enough to study the sign of the function $$g(t) = (t-1)\left(-(2+t)\log \frac{(2+t)^2}{1+2t} + t \log \frac {(2+t)^3}t\right) \\
 = (t-1)\left((t-4)\log(2+t) + (2+t)\log(1+2t) - t\log t\right)
$$

Letting $g(t) = (t-1)h(t)$, we compute $h'(t) = \log(9+(t- t^{-1})^2) + 2\frac{(t-1)^2}{2t^2+5t+2} \ge \log 9 > 0 $. Since $h(1) = -3\log3+3\log3-\log1 = 0$, we have $h(t)>0$ for $t>1$, $h(t)<0$ for $t<1$, and $h(1)=0$.

Going back to $g$ we learn that $g(t)>0$ except when $t=1$, which means that $\delta(x,y) > 0$ except when $x=y$, which is what we wanted.

For the first question, this is really a question about the derivative. $f$ has three real zeros if and only if its derivative has two zeros, and $f$ is positive at the smaller critical point, and negative at the bigger critical point [by the way, I am dividing through by $a_3$ to make the polynomial monic -- this does not change the question). This can be checked for $g$ by tedious computation, which I leave to you...


Riemann hypothesis: is Bender-Brody-Müller Hamiltonian a new line of attack? There is a beautiful paper in Physical Review Letters [PRL 118, 130201 (2017), DOI:10.1103/PhysRevLett.118.130201] by Carl Bender, Dorje Brody, and Markus Müller (BBM) on a Hamiltonian approach to the Riemann Hypothesis. The paper  is surprisingly easy to follow
for a physicist.
BBM define a Hamiltonian
$$\hat{H} = (1- e^{-i \hat{p}} )^{-1} ( \hat{x} \hat{p} + \hat{p} \hat{x} ) (1- e^{-i \hat{p}})$$
where $p=- i \partial_x$ is the momentum operator in $\hbar=1$ units.
The authors show that eigenfunctions of $\hat{H}$ vanishing at infinity must be be in the form of Hurwitz theta function, $\psi_z= - \zeta(z,x+1)$, so that $$\hat{H} \psi_z = i (2 z -1)\psi_z $$
Imposing a boundary condition $\psi_z(0)=0$, by the virtue of $\zeta(z,1)=\zeta(z)$, they show that all non-trivial zeros of the Riemann $\zeta$ function must be eigenvalues of $\hat{H}$ with the imposed boundary conditions.
BBM call this result a "a complex extended
version of the Berry-Keating conjecture" and go on to provide heuristic arguments that all eigenvalues of $\hat{H}$ are real.
How promising is this new development in the context of solving the Riemann Hypothesis?
Steven Strogatz seems optimistic.
Update [19.10.2020]: The authors of [1] have published additional comments [3
, 4], including a response [4] to @Jean Bellissard's comment [5] that grew out of his answer below.

I tried to put my own argument on this blog recently but I failed. Let me try again.
1)- The original definition of $\zeta(z,x) $ is

$$\zeta(z,x) = \sum_{n=1}^\infty \frac{1}{(n+x)^z}
$$

Then $\zeta(z,x)-\zeta(z,x-1)=-1/x^z$. Then, using the integral representation, this relation extends by analiticity to the maximal domain in which both sides are defined and holomorphic. For our purpose, it is enough to consider $x>0$ and $\Re{z}>0$. Let $\zeta_z$ denote the map $x\to \zeta(z,x)$. Then, using the generator $A$ of the dilation operator, it is easy to check that the eigenvalues have the form $1/2+\imath E$. For $A=(pq+qp)/2=-\imath xd/dx-\imath 1/2$. This gives $A(1/x^z)=\imath (z-1/2)(1/x^z)$. Going from $A$ to $H$ is done using the operator $\Delta$, leading to the formal equation $H\zeta_{1/2+\imath E}=E\zeta_z$. The Dirichlet b.c. at the origin forces E to satisfy $\zeta(1/2+\imath E)=0$.

2)- The first problem I see is that $\zeta_z$ does not seem to belong to the Hilbert space ${\mathcal H}=L^2(0,\infty)$. Using a classical method by Hadamard

$$\int_0^\infty |\zeta(z,x)|^2 dx=\frac{1}{1-2\Re{z}}\zeta(2\Re{z}-1)$$

which converges for $\Re{z}>1$. Using the integral representation, a similar firmula can be obtained, but I failed to show that square integrability hold for $\Re{z}=1/2$. Hence I see a problem here.

3)- The other problem is the definition of $\Delta$. Let $S$ denote the translation operator, defined non rigorously by $Sf(x)=f(x-1)$. If restricted to $\mathcal H$, it is not unitary, because it is defined only for $x>1$. We can defined it by imposing $Sf(x)=0$ for $x\in [0,1]$. Then it is only a partial isometry. For $S^\ast f(x)=f(x+1)$ leading to $S^\ast S=I$, but $S S^\ast =P$ is the projection onto $L^2(1,\infty)$. Using these notations $\Delta= I-S=(S^\ast-I)S$. But we get a problem here: the function $\zeta_z$ is defined for $x>-1$, and the extension to the interval $(-1,0]$ is explicitly used in 1)-. So we cannot use
$S$. But then what is the operator $e^{-\imath \hat{p}}$ used by the authors?

4)- If $\hat{p}$ is the usual operator

$$\hat{p}=-\imath \frac{d}{dx}$$

then there is a problem with its domain of definition. On $L^2(\mathbb R)$, it is selfadjoint as can be seen by using Fourier transform. But on $\mathcal H$, it is not. This is a classical exercise found in the very old book of Courant-Hilbert. Namely one can always define it in the set of $L^2$ functions with $L^2$ derivative, vanishing at $x=0$. Then it is symmetric. If so its adjoint is defined on the same space but without the vanishing at $x=0$. Not only the adjoint is not symmetric but its set of eigenvalues is the open lower half plane. This is because if $f_z(x)=e^{zx}$, then $\hat{p}^\ast f_z=-\imath z f_z$, while $f_z\in \mathcal{H}$ for $\Re{z}<0$. The same argument shows that $+\imath$ cannot be an eigenvalue. This means that the "defect indices", namely the dimension of the eigenspaces with eigenvalues $\pm\imath$, are not equal. Then, the von Neumann theorem show that the operator $\hat{p}$ has no selfajoint extension. If not selfadjoint, the definition of its exponential becomes a problem, because the functional calculus is nit defined in general.

5)- The previous argument can be rephrased in terms of the operator $S$. Its adjoint admits a lot of eigenvalues, namely the points inside the unit disk.

In conclusion, the sloppyness of the definitions used but the authors leads to a complete mess. Nothing is correct in this paper.

As long as physicists use algebra, or algorithmic arguments, they can find outstanding results. But when it comes to analysis, they may loose their judgment, and grave mistakes show up at the corner. Analysis is not easily amenable to algoritmic descriptions. And this is precisely where the power of Mathematics lies: by manipulating infinities, Mathematics goes way beyond the Church-Turing definition of computability. And what is Analysis if not manipulating infinities, through limits, convergence and the likes?

The current situation is as follows. The zetagrid project no longer exists. The Z function has a pole in Re (z) = 1, but is analytic for Re (z) > 1,
and divergent for Re (z) < 1. Riemann analytically extended the function so that it could be analytic in the region 0 < Re (z) < 1, which is what is known
as the critical strip or research interval. Where we have that for Re (z) = 1/2 there are infinite zeros (Hardy, 1914) and it is assumed (RH) that all the zeros
of the Z are there. However, in November of 2017, this server showed that there are infinite zeros of the Z in the critical interval distributed in
different families of lines, with which the Riemann Hypothesis is false! In addition, the existence of the zeroes by reflection predicted by the
Riemann functional equation was confirmed and according to which if in the interval 0 < Re (z) < 1/2 there is some zero, this zero must be reflected
in the interval 1/2 < Re (z) < 1
enter image description here
0.44 + 1977.19-> 0.56 + 1977.19
0.45 + 1329.09-> 0.55 + 1329.09
0.45 + 1415.59-> 0.55 + 1415.59
0.46 + 25.0-> 0.54 + 25.0
0.46 + 37.6-> 0.54 + 37.6
0.47 + 14.09-> 0.53 + 14.09
0.47 + 21.0-> 0.53 + 21.0
0.48 + 14.09-> 0.52 + 14.09
0.48 + 21.0-> 0.52 + 21.0
0.49 + 14.09-> 0.51 + 14.09
0.49 + 21.0-> 0.51 + 21.0

all of them are zeros of the Z (by reflection), which also are not on the line Re (z) = 1/2, as it is easy to see and check using my expressions
for the Z function. I leave that task to you.


Solve the binomial equation Solve the binomial equation $$z^4 = -8$$

Below is the steps i have done

1: I have taken  |-8| that is 8 and then done 8^(1/4) which is 2^(1/4).

2: Since $z=r(cos\alpha+isin\alpha)$ leads me to

$r^4(cos4\alpha+isin4\alpha)=-8(cos\pi/2+isin\pi/2)$

Divide by 4 since the z term is raised by four gives $2^{1/4}(cos\pi/8 + k * \pi/2 +sin\pi/8 + k * \pi/2) $

Is this the correct way to solve this problem ? I am asking since i just started with binomic equations and been stuck for some hours with the question.

Another approach that is an efficient way forward is to exploit Euler's Identity and  write

$$z^4=r^4e^{i4\theta}=-8=8e^{i(2n+1)\pi}$$

for all integer $n$.  Thus, upon inverting we have for $z$

$$z=re^{i\theta}=2^{3/4}e^{i(2n+1)\pi/4}$$

for $n=\pm 1, \pm 3$.  Therefore, the $4$ roots of $z$ are

$$\bbox[5px,border:2px solid #C0A000]{z=2^{3/4}e^{\pm i\pi/4}\,\,\text{and}\,\,2^{3/4}e^{\pm i3\pi/4}}$$

or in rectangular form

$$\bbox[5px,border:2px solid #C0A000]{z=2^{1/4}(1\pm i)\,\,\text{and}\,\,2^{1/4}(-1\pm i)}$$


Does GL(2,R) contain cyclic subgroup of order n ? Does GL(2,R) contain cyclic subgroup of order n ?  GL(2,R) is a General Linear group of order 2.

I just can not figure out this. Can you tell me the answer with explanation?

I know that the group contains a Infinite cyclic subgroup generated by a matrix whose (2,1) element is 0 and others are 1, and a cyclic subgroup of order 2.

Your question is equivalent to finding an element of order $n$ in $GL(2,\Bbb{R})$.

Try the matrix \begin{pmatrix}\cos 2\pi/n &-\sin2\pi/n \\ \sin2\pi/n & \cos 2\pi/n\end{pmatrix}

You can put the $n^{\rm{th}}$ root of unity and/or its powers on the diagonal. For example
$$\begin{pmatrix}\omega&0\\0&\omega^\alpha\end{pmatrix},$$ where $\omega^n=1$ and $\gcd(\alpha,n)=1$.

Moreover, as order of the generator is finite, using basic linear algebra arguments you can also prove that any such cyclic subgroup will have a generator which is a diagonal matrix. Further any such diagonal generator will be of the form as given above.


$y= 2x^2+8-1$: why does not touch the $y$ axis? If $y=0$ then  $2x^2+7=0$; $2x^2=-7$; $x^{2}= -\frac{7}{2}.$; if $x=0$ then $x =-\frac{7}{2}$ right?
Please someone explain to me this concept, although it seems a dumb question?

Since $x\geq 0$ for all $x\in \mathbb R$ we have that $y=2x^{2}+7\geq7>0.$
So the minimum value of $y$ is $7$ and this occurs when $x=0$. Indeed when $x=0$ we have $y=2(0)^{2}+7=7$, which is where the quadratic crosses the $y$-axis.
If you set $y=0$, then this gives you the values where the graph crosses the $x$-axis, but $y\geq7$, so it will never cross the $x$-axis. Usually you would compute the discriminant to find the number of roots.
Below is a sketch:

In fact it touches the y axis (x=0). The intersection of the curve and the y axis is at the point (0,7).
The curve does not touch the x axis (y=0) because in this case the resulting polynomial has two imaginary solutions (one is the complex conjugate of the other).


If $|f'(z)| \leq |z|$ for all $z\in \mathbb{C}$ then $f(z) = a + bz^2$ for arbitrary $a,b\in\mathbb{C}$ with $|b| \leq 1$. Is there something wrong with this problem?

  Let $f$ be entire function and $|f'(z)| \leq |z|$ for all $z\in
\mathbb{C}$. Show that $f(z) = a + bz^2$ for arbitrary
  $a,b\in\mathbb{C}$ with $|b| \leq 1$.

I was using Liouville theorem, but the result is not the same.

By writing $$\left|\frac{f'(z)}{z}\right| \leq 1$$ we have that $f'(z)/z$ (*here where I'm not sure) is entire and bounded by $1$. So by Liouville's theorem $f'(z)/z$ is constant, that is $$f'(z)/z = c$$ hence $f'(z) = cz$ and therefore $f(z) = \frac{1}{2}cz^2 + a$. Letting $\frac{1}{2}c = b$ yields the result, but with $|b| = \frac{1}{2}|c| \leq \frac{1}{2}\times1 = \frac{1}{2}$ instead of $|b| \leq 1$.

Here are the questions

*How to show that $f'(z)/z$ also analytic at $z = 0$?
Is there an error in my proof above? Or with the problem?

If $f'(z)/z$ need not be analytic at $z = 0$ then I can't use Liouville theorem. All I know is that $f'(0) = 0$, so I think that $f'(z)/z$ is also analytic at $z = 0$.

You know that your function is entire because
$$
|\frac{f'(z)}{z}| \leq 1
$$
is bounded. Since it is bounded and holomorphic around 0, you can extend it to $z=0$ and therefore you have an entire function. It is a "removable singularity", you can find characterisations of it in any textbook or script.
As for your other question:
It is trivially true that $|b|\leq \frac{1}{2}\leq 1$. You can just plug in the derivative and check that
$|bz| \leq |z|$ implies $|b| \leq 1$. In your proof, you had a statement on the derivate, i.e. $f'(z)=cz$. And then you integrated both sides. In the process, you got a factor of $\frac{1}{2}$ which, however, vanishes when you differentiate with respect to $z$.
The problem here is that you take a statement/restriction on the derivative and by integrating the statement changed. Look at it that way, just ignore the $|.|$ for a second:
$$
f'(z) \leq cz
$$
becomes:
$$
\int f'(z)=f(z)\leq  \frac{1}{2}cz^2=\int cz
$$

Let $z_0\in\mathbb{C}$ be arbitrary. Then for $|z-z_0|\leq R$, we have
$$|f'(z)|\leq |z| = |z-z_0+z_0|\leq R+|z_0|.$$
By cauchy's inequality,
$$\left|\frac{f'''(z_0)}{2!}\right|\leq \frac{R+|z_0|}{R^2}\rightarrow 0 \mbox{ as } R\rightarrow \infty.$$
So, $f'''(z_0)=0$. Since $z_0$ was arbitrary, $f'''(z)=0$ on $\mathbb{C}$. Now, integrate it from $0$ to $z$, we get
$$f''(z)=f''(0).$$
Again, integrate from $0$ to $z$,
$$f'(z)=f''(0)z+f'(0).$$
As $f'(0)=0$, $f'(z)=f''(0)z$.

Therefore, $f(z)=a+\frac{f''(0)}{2}z^2$ where $a$ is arbitrary. Take $b=\frac{f''(0)}{2}$. So, $|b|=|\frac{f''(0)}{2}|\leq 1$, as $|f''(0)|\leq 1$.

Observe that, $f''(0)=\lim\limits_{z\rightarrow 0}\frac{f'(z)}{z}$, as $f'(0)=0$.


How to formally prove that $\gcd(a,1) = 1$ when $a \in Z$ I was trying to prove that $$\gcd(2a+1,9a+4) = 1 , a\in Z$$
My proof went like this:
$$ (9a+4) = 4(2a+1) + a$$
$$ (2a+1) = 2(a) + 1 $$
$$ \Rightarrow \gcd(9a+4,2a+1) = \gcd(2a+1,a)=\gcd(a,1) = 1$$
The last line of reasoning seems intuitive for me; The greatest common divisor of any integer with 1 is definitely one because the only divisor of 1 is itself. However, the question of how to prove it formally arose.
Thanks for any suggestions!

The formal definition for the greatest common divisor function is:
$$\gcd(n,m)=\max\{d\in\mathbb{N}:d\mid n \,\wedge\,d\mid m\}.$$
Set $m=1$. The only number $d$ that satisfies $d\mid n$ and $d\mid1$ is $1$. So we have
$$\gcd(n,1)=\max\{1\}=1.$$

using Euclid's Lemma, GCD (a,b) = GCD(a,a-b), , You can continue like this ,by subtracting smaller number from larger number.

${(2a+1 , (9a+4-2a-1))=(2a+1,7a+3)...  (a,a-1)=1}$

-|  but there is a faster way to do this.

${ GCD(2a+1,9a+4)= (2a+1, 4(2a+1)+a)= (2a+1,a)= (a,2a+1)=(1,a)=1 }$


Probability of getting three zeros when sampling a set Say we have a set of $12$ numbers - $(0,1,2,0,3,0,4,0,5,6,0).$
We go thru this set one number at a time, and choose a number with a probability $1/3$; we sample. What is the probability we end up with $3$ $0$'s out of the $4$ numbers we will likely choose(on average)?
Is there a closed-form solution for this?

If I understand the problem, you have a list of 12 objects, 5 of which are special (the 0s).  (I use the word "list" instead of "set" since the latter implies there is no repetition among the objects).  Each of the 12 objects will be independently considered, and "chosen" with probability 1/3.  You want to know the probability of 3 of the special objects being chosen.

In that case, the 7 non-special objects are irrelevant.  You just want to know the probability that exactly 3 out of 5 independent events occur, when each one occurs with probability 1/3.  This is the probability of a binomial random variable with $n=5$ and $p=1/3$ taking on the value $3$.  The linked Wikipedia page has a formula for computing this.

The probability that you want is $(1/3)^4(2/3)^8 N$, where $N$ is the number of subsets of length $4$ containing exactly $3$ zeroes.


Distribution function (CDF) of the sum of two random variables + law of iterated expectations I'm taking my first probability class, and we're studying sums of independent random variables. We're using Ross's First Course in Probability. It states the definition of a convolution, but doesn't give much more depth into the subject. I checked out statlect.com's page on the sums of independent random variables, but I'm struggling with the first proof on the page.

It states the following theorem: Let $X$ and $Y$ be two independent random variables and denote $F_{X}(x)$ and $F_{Y}(y)$ their distribution functions. Let: $Z=X+Y$ and denote the distribution of $Z$ by $F_{Z}(z)$. The following holds: $F_{Z}(z) = E[F_{X}(z-Y)] = E[F_{Y}(z-X)]$

Proof:

$F_{Z}(z)=P(Z\leq z)\\\;\;\;\;\;\;\;\;\;=P(X+Y\leq z)\\\;\;\;\;\;\;\;\;\;=P(X\leq z-Y)\\\;\;\;\;\;\;\;\;\;=*E[P(X\leq z-Y|Y=y)]* \text{by the law of iterated sums}\\\;\;\;\;\;\;\;\;\;=E[F_{X}(z-Y)]$

The law of iterated sums states: $E[E[X|Y]]=E[X]$. I follow the proof of that theorem just fine. However, I'm struggling to see how it allows you to make the jump from $P(X\leq z-Y)$ to $E[P(X\leq z-Y|Y=y)]$.

In Ross's A First Course in Probability, the proof for $F_{Z}(z)=E[F_{X}(z-Y)]$ makes more sense. But I can't link the steps in his proof to see where iterated expectations come in to play. Any help would be greatly appreciated!

I will denote $P(X\le z-Y)$ as $F_X(z-Y)$, the pdf as $f()$, and integrals using $\int_{lb}^{ub}$:

First, write CDF of $z-Y$ as conditional probability on the value of $Y$.

$$F_X(z-Y) = \int_{y=-\infty}^{\infty} F_X(z-y|Y=y)f_Y(Y=y)dy   \qquad (1)$$

In other words, you are finding the cumulative probability that $X$ is less than $z-Y$ by conditioning on the value of the r.v. $Y$ and then integrating over all possible values of $Y$.

Recall the definition of "expectation"

$$E_Y[Y] = \int_{y=-\infty}^{y=\infty} yf_{Y}(y)dy     \qquad (2)$$

You will see that you can rewrite eqn $(1)$ as

$$E_Y[F_X(z-y|Y=y)]$$

In general, when working with conditional probability/expectation proofs I find it helpful to write things out in terms of integrals.


prove that for every integer,$ n$, greater than $1$,$ (3^{2n+1}) + (5^{2n-1})$ is divisible by $16$ Could someone please help me out with this proof?

Prove that for every integer $n≥1$, $3^{2n+1} + 5^{2n-1}$, is divisible by $16$.

I get to a point where I have...

$$3^{2k+1} \cdot 3^2 + 5^{2k-1} \cdot 5^2 $$

Thanks!

The order of $3$ and $5$ in $(\mathbb{Z}/16\mathbb{Z})^*$ is four, hence you need to check the claim only for two cases: $n=1$ and $n=2$. For the remaining cases, just use the fact that:
$$3^{2(n+2)+1}\equiv 3^{2n+1}\pmod{16},\qquad 5^{2(n+2)-1}\equiv 5^{2n-1}\pmod{16}.$$

I generalized this here:
When is there an $m$ that divides $u^{an+b}+v^{cn+d}$ for all $n$

I asked:
Find conditions on
 positive integers
$u, a, v, c$
and non-negative integers
$b, d$,
such that
there exists an $m$
which divides
$f(n) = u^{an+b}+v^{cn+d}$
for all $n$.
Determine $m$
in terms of
$u, a, b, v, c, d$.

One conclusion was
that if
$(u^a-v^c)|(u^b+v^d)$,
it divides $f(n)$
for all $n$.

In this case,
putting $n+1$ for $n$
(so $d$ is not negative),
$(u, a, v, c, b, d)
=
(3, 2, 5, 2, 3, 1)
$,
so
$u^a-v^c
=3^2-5^2
=-16
$
and
$u^b+v^d
=3^3+5^1
=27+5
=32
$.
Since $-16$ divides $32$,
$16$ divide all terms.


Inductive Proof that $k!<k^k$, for $k\geq 2$. Call $P(k): k!<k^k$, for $k\geq 2$
Test it out with 2, and it's true ($2<4$).

Assume that $P(k)$ is true for some $k\geq2$.

Then show that $P(k+1)$ is true.

$P(k+1): (k+1)!<(k+1)^{k+1}$

Ministep: $(k+1)!=k!(k+1)$
Ministep: $(k+1)^{k+1}=(k+1)^{k}(k+1)$

$k!(k+1)<(k+1)^{k}(k+1)$

Can I just pull out that (k+1) and call it a day? That will make $k!<(k+1)^{k}$.

We're already assuming that $k!<k^k$, for some integer $k\geq 2$, and I don't think I need to prove that $k^{k}<(k+1)^{k}$. I mean, I guess I could go prove that as well, but let's assume for the moment that I don't have to. Since we're assuming the original $P(k)$ is true, I should be able to write $k!<k^{k}<(k+1)^{k}$ right? That would prove the $k+1$ case. Is this the correct way to go about things? It seems too easy, but makes sense.

Thanks for confirmation and/or pointing out mistakes!

I should be able to write $k!<k^{k}<(k+1)^{k}$ right?

Sort of. The key part, which it sounds like you called "too easy," is realizing that you can use $k^k<(k+1)^k$ in the context of the proof; that is, this realization makes it possible to easily get from the left-hand side of what you need to prove to the right-hand side. Nonetheless, I would encourage you to take a look at this post on how to write a clear induction proof. Your proof is not as clear as it could be with the sidebar "ministeps," the fact that you never once mentioned where you used the inductive hypothesis, etc. If you were my student, then I would probably give you 8/10. My point in answering is to provide what I think, in my humble opinion, is a nice way of writing up the proof. I hope it helps.

Claim: For $n\geq 2, n\in\mathbb{Z}, n!<n^n$.

Proof. For $n\geq 2, n\in\mathbb{Z}$, let $P(n)$ denote the statement
$$
P(n) : n!<n^n.
$$
Base case ($n=2$): $P(2)$ says that $2!=2<4=2^2$, and this is true.

Inductive step $P(k)\to P(k+1)$: Fix some $k\geq 2$, and assume that
$$
P(k) : \color{blue}{k!<k^k}
$$
holds. To be shown is that $P(k+1)$ follows where
$$
P(k+1) : \color{green}{(k+1)!<(k+1)^{k+1}}.
$$
Starting with the left-hand side of $P(k+1)$,
\begin{align}
\color{green}{(k+1)!} &= (k+1)\cdot \color{blue}{k!}\tag{by definition}\\[0.5em]
&< (k+1)\cdot\color{blue}{k^k}\tag{by $P(k)$, the ind. hyp.}\\[0.5em]
&< (k+1)(k+1)^k\tag{since $k<k+1$}\\[0.5em]
&= \color{green}{(k+1)^{k+1}},\tag{by definition}
\end{align}
we end up at the right-hand side of $P(k+1)$, completing the inductive step.

By mathematical induction, the statement $P(n)$ is true for all integers $n$ where $n\geq 2$. $\blacksquare$

Both expressions are a product of $k$ factors. It's easy to see that in $k^k$, each of the factors is larger than the factors from $k!$.

I get that you are looking for an inductive proof, but any proof is trivially inductive.

Assume $(k-1)!<(k-1)^{k-1}$. Now I need that to imply $k!<k^k$. Since $k!<k^k$, as explained above, I've done the induction step. It doesn't matter that you don't see the logical connection. It's still formally and logically true that $P(k-1)\implies P(k)$, since $P(k)$'s truth can be established independently, quickly.