System of differential equations $x'=f(y-x),\, y'=f(x-y)$ I have a problem regarding differential equations, $f$ is increasing. $x,y$ are dependent on $t$
$$ f(0)=0 $$
$$x'=f(y-x)$$
$$y'=f(x-y)$$
$$x(0)=1$$
$$y(0)=0$$
prove that when $x,y$ are the solutions of this equation then $$ \lim_{t \to \infty} x = \lim_{t \to\infty}y $$
So far I have managed to see that $x'(0)=f(-1)$ and $y'(0)=f(1)$ which implies that $x'(0)<0$ and $y'(0)>0$ and basically $x'(t)<0$, $y'(t)>0$ as $x>y$ so as $f$ is increasing and $x$ is declining with $t$ and $y$ is increasing with $t$, the derivatives of $x$ and $y$ will be declining and increasing respectivly and they will have the boundary. Can anyone help from this on?

First we note, that $\lim_{t\rightarrow \infty} x = x_\infty$ indeed exists. This is because $x$ is decreasing monotone. On the other hand $y_\infty$ exists aswell. We have $x_\infty \geq y_\infty$, because if there is a $t_0$ with $x(t_0)=y(t_0)$ we have $x'(t_0)=y'(t_0)$ and this will be $x_\infty$.

Now assume that $x_\infty >y_\infty$. The genral solution of a differential equation is:
$$x(t) = x(0)+ \int_0^t f(x(t)-y(t)dt$$
But we then would have
$$\lim_{t\rightarrow \infty} x(t) =x(0) +\int_0^t f(y-x)dt \leq\lim_{t\rightarrow \infty} x(0) +\int_0^t f(y_\infty-x_\infty)dt =-\infty $$

This however contradicts that $x_\infty>y_\infty$

Let $u=y-x$. Then
$$
\frac{du}{dt}=\frac{dy}{dt}-\frac{dx}{dt}=f(-u)-f(u)
$$

Which is automatic with an equilibrium solution $u=0$. Note that your initial conditions become $u(0)=-1$

For $u<0$ since $f$ is increasing $f(-u)-f(u)>0$ giving $u$ is increasing (for all $u<0$). Assuming $u$ is a continuous solution it would need to pass through $u=0$ but cannot actually exceed this value because of the equilibrium there. so $\lim_{t\to \infty} u=0$ Thus $\lim_{t\to \infty} y-x =0$. So if either of the limits $\lim_{t \to 0} x$ or $\lim_{t \to 0} y$ exist (and are real) they must be equal. If they're $\pm \infty$ they also must be the same one.


Equivalent definition of the measurable sets of a finite outer measure Usually if I have a outer measure $\mu^*:\mathbb R^n \to \mathbb R$  induced by a measure $\mu:\mathcal R \to \mathbb R$ where $\mathcal R$ denotes a ring over $\mathbb R^n$ we define the $\mu^*$-measurable sets as those sets $A$ for which $Q \subset \mathbb R^n \Rightarrow \mu^*(Q)= \mu^*(Q\cap A)+ \mu^*(Q\cap A^c)$. But I just read that if $\mu^*$ is finite, a set A is measurable if and only if $\mu^*(\mathbb R^n)=\mu^*(A)+\mu^*(A^c)$. How could I show this? (Any hints would also be welcome)

Because R^n is of sigma-finite measure.

This definitions works on any space X on which there is an outer measure \mu^* induced from a sigma-finte premeasure.

For any E\subset X
if \mu^(X) = \mu^(E) + \mu^*(E^C)

then E is \mu^* measurable


Maximising a Ratio - Relationship of Marginal Numerator and Denominator I am trying to maximize a ratio with respect to $x$, where both numerator and denominator are dependent on variable $x.$

The problem is a follows:

$$ S(x) = \frac {E - C(x)} {V - R(x)} $$

$$ C(x) = \frac {\frac{O w p}{x^2} + \frac{O w \gamma G k}{x}} {QA} $$

$$ R(x) = M - h x $$

Where my goal is to maximize $S(x).$ All the other variables are independent of $x.$

I have been thinking about the conditions that must be met, if the ratio is maximized. My current guess is that the maximum $S(x)$ is where following holds:

$$ \frac{E} {V} = \frac {C'(x)} {R'(x)} $$

Where $C'(x)$ and $R'(x)$ are the first derivatives respectively.

However, I am not entirely sure that this is the case and I can't prove this mathematically. Is this conditions right? How can I prove this?

Many thanks in advance.

The function $S(x)$ is maximized for $x\rightarrow 0,$ where $S(x)\rightarrow \infty$ (depending on the sign of the constants and from which side you approach), which can be seen by considering $C(x).$

Disregarding the singularity, the maximum occurs when
$$S'(x)=0 \implies \frac{C'(x)}{R'(x)}=\frac{E-C(x)}{V-R(x)},$$
which can be shown using standard differentiation techniques (product and chain rule). Note that you have to check the limits $x\rightarrow \pm \infty$ (where $S(x)\rightarrow 0$) and that this equation has multiple solutions, all of which needs to be checked.

Maxima may occur when $\frac{d}{dx}S(x)=0\implies \frac{C(x)}{R(x)}=\frac{C'(x)}{R'(x)}$. I am saying "may" because this is not a necessary condition. You need to check further.


A continuous, nowhere differentiable but invertible function? I am aware of a few example of continuous, nowhere differentiable functions. The most famous is perhaps the Weierstrass functions

$$W(t)=\sum_k^{\infty} a^k\cos\left(b^k t\right)$$

but there are other examples, like the van der Waerden functions, or the Faber functions. Most of these "look like" some variation of:

(Weierstrass functions from Wolfram)

Specifically, they are clearly not invertible. Since these functions are generally self-similar at many scales, this non-invertability would seem to hold essentially everywhere.

I'm wondering if it's possible to construct such a function which is invertible. Intuitively, maybe this would be "jittery" in the same way as the Weierstrass function, but if it had a slope which always increased, it would be invertible. Or perhaps there is at least an example in which the function is invertible over some segment of the range.

Interestingly, there are no such examples! For a continuous function $f : \mathbb{R} \rightarrow \mathbb{R}$ to be invertible, it must be either monotone increasing or decreasing. A famous classical result in analysis, Lebesgue's Monotone Function Theorem, states that any monotone function on an open interval is differentiable almost everywhere. Hence, there are no continuous functions that are invertible and nowhere differentiable.

I will assume by invertible you mean injective, i.e. bijective on its image. As others have mentioned, there are no such examples among functions from $\mathbb{R}$ to $\mathbb{R}$.

However you might find such examples in other topological spaces.

An example $\mathbb{R} \rightarrow \mathbb{R}^2$ is given by $g(t) = (W(t), t)$ with $W$ the Weierstrass function; it is obviously injective and continuous but nowhere differentiable, considering its projection to the first coordinate.

You can also build an example from $\mathbb{R}^2 \rightarrow \mathbb{R}^2$ by considering $g(t_1, t_2) = (t_1, W(t_1) + t_2)$ or its inverse, given by $g^{-1}(x_1, x_2) = (x_1, x_2 - W(x_1))$.

An injection $\mathbb{R}^2 \rightarrow \mathbb{R}$ cannot be continuous so you won't find a counterexample there.


How is $ \lim_{z \to z_o} (z-z_o)\frac{f(z)}{g(z)} = \lim_{z \to z_o} \frac{f(z)}{g(z)-g(z_o)/(z-z_o)}= \frac{f(z_o)}{g'(z_o)}$? I was reading this proof in Gamelin Complex Analysis (page 196):

If $ f(z) $ and $ g(z) $ are analytic at $ z_o $ and if $ g(z) $ has a simple zero at $ z_o $

$$ Res[ \frac{f(z)}{g(z)},z_o ] = \frac{f(z_o)}{g'(z_o)} $$
In this case $ f(z)/g(z) $ has a simple pole at $ z_o$

$$  \lim_{z \to z_o} (z-z_o)\frac{f(z)}{g(z)} = \lim_{z \to z_o} \frac{f(z)}{g(z)-g(z_o)/(z-z_o)}= \frac{f(z_o)}{g'(z_o)}$$

I'm confused where does $g(z)-g(z_o)$ in the denominator come from?

Since $g(z)$ has a simple zero in $z_0$ you can freely add or subtract $g(z_0)$ in the denominator, because $g(z_0)=0$:
$$
(z-z_0) \frac{f(z)}{g(z)}=(z-z_0)\frac{f(z)}{g(z)-g(z_0)}=\frac{f(z)}{\frac{g(z)-g(z_0)}{z-z_0}}\to_{z\to z_0} \frac{f(z_0)}{g'(z_0)}.
$$

both numerator and denominator are of the form 0/0. So use L'Hospital rule.


Understanding an implication involving symetric groups, alternating groups, and normal subgroups I am trying to understand the following statement:

Let's suppose that $\forall n, H \triangleleft A_n$ contains a $3$-cycle.

$A_n \triangleleft S_n$ has and index of $2$.

If  $(abc) \in H \implies \exists (de) \in S_n/A_n$ s.t  $(de)$ and $(abc)$ are disjoint and s.t. $(abc)(de) = (de)(abc)$

Here's what I don't understand

Why does there exist $(ed) \in S_n/A_n$?
Why are $(de)$ and $(abc)$ disjoint?
Why do they commute?

The mapping $S_n\rightarrow\{\pm 1\}:\pi\mapsto {\rm sgn}(\pi)$ is a group homomorphism, surjective if $n\geq 2$. The kernel is $A_n$ and so $A_n$ has index 2 in $S_n$.


Probability - Dice Toss When rolling two dice, what is the probability that the roll will be $5$, $2$ on two consecutive rolls? I'm pretty sure it's $\frac{2}{36}$ for one roll, but I'm not sure how to figure the second. The answer in the book says it's $\frac{1}{36}$, but I can't figure out how it works.

Thanks for your help.

Have you quoted the question exactly ?

My hunch is that the question was asking :

When rolling a die twice, what is the probability of getting 5,2.

The book answer would then be correct, Pr = $\dfrac16\times\dfrac16 = \dfrac{1}{36}$

Because its two consecutive rolls , it takes place one after the other.. so, prob to get either 5 or 2 in first roll is (2/36) and prob to get either 5 or 2 in secd roll is (2/36). so the total prob will be 4/36=1/9.. so my answer is 1/9.


Cellular Boundary Formula In Hatcher's book we find, when computing the boundary maps of cellular homology,

Cellular Boundary Formula: $d_n(e^{n}_\alpha)=\sum_\beta d_{\alpha\beta}e^{n-1}_{\beta} $ where $d_{\alpha\beta}$ is the degree of the map $S^{n-1}_\alpha \rightarrow X^{n-1}\rightarrow S^{n-1}_\beta$ that is the composition of the attaching map of $e^{n}_\alpha$ with the quotient map collapsing $X^{n-1}- e^{n-1}_\beta $ to a point.

My question is:

When we identify $X^{n-1}/(X^{n-1}- e^{n-1}_\beta)$ with $S^{n-1}_\beta$, we are choosing a homeomorphism. Doesn't this choice affect the degree of the map in question?

Essentially (as I see it), we must choose a quotient map from $D^{n-1}$ to $S^{n-1}$ which identifies $S^{n-2}$ to a point.

There is indeed a choice of homeomorphism $D^n/S^{n-1} \cong S^n$ but as others have pointed out this won't affect any homology calculations as long as you consistently use this homeomorphism for each cell.

However, the degree is affected by the choice of homeomorphism, so the cellular boundary formula as found in Hatcher is only right to a sign. In order for the cellular boundary formula to make sense it is necessary that the generators of $H_n(X^n, X^{n-1})$ are  of the form $e_{\alpha}^n =\Phi_{\alpha *}^n([D^n])$ for a fixed generator $[D^n]$ of $H_n(D^n, S^{n-1})$. Now in order for the cellular boundary formula to hold we need that these generators fit together according to the sequence of isomorphisms

$H_n(D^n, S^{n-1}) \cong H_{n-1}(S^{n-1}) \cong H_{n-1}(D^{n-1}/S^{n-2}) \cong H_{n-1}(D^{n-1}, S^{n-1})$

where the first isomorphism is the boundary map coming from the long exact sequence for the pair $(D^n, S^{n-1})$ and the second isomorphism comes from the chosen homeomorphism $S^{n-1} \cong D^{n-1}/S^{n-2}$. If the homeomorphisms make the chosen generators incompatible then the formula may fail to hold.

This compatibility ensures the last two isomorphisms followed by $\Phi_{\beta *}^{n-1}$ sends $\partial [D^n]$ to $[D^{n-1}]$ and then to $e_{\beta}^{n-1}$. Thus it sends $\Delta_{\alpha \beta} \partial [D^n] = d_{\alpha \beta} \partial [D^n]$ to $d_{\alpha \beta} e_{\beta}^{n-1}$. The standard diagram chase shows $d_n(e_{\alpha}^n) = \sum_{\beta} (H_{n-1}(S^{n-1})\to H_{n-1}(D^{n-1}/S^{n-2}) \to H_{n-1}(D^{n-1}, S^{n-1}))\Delta_{\alpha \beta} \partial [D^n]$ so the theorem is proved.

Notice that any homeomorphism has degree $1$, as it induces an isomorphism in homology. Thus there is absolutely no problem related to such a choice in the construction of cellular homology.


Linear operators on the functions $f:\mathbb{R}\rightarrow\mathbb{R}$ that distribute over multiplication Let $V$ denote the vector space of all functions $f:\mathbb{R}\rightarrow\mathbb{R}$. What are the linear operators $L:V\rightarrow V$ such that $L[fg]=L[f]L[g]$ for all $f,g\in V$?

I made a bit of progress by considering the functions
$$\chi_t(x) = \begin{cases}
1 & x=t \\
0 & x\neq t
\end{cases}.$$
For fixed $x\in\mathbb{R}$, the value of $L[\chi_t](x)$ is either $0$ or $1$ for each $t\in\mathbb{R}$. If there exists some $t$ such that $L[\chi_t](x)=1$, then $L[f](x)=f(t)$. I was unable to do the case in which $L[\chi_t](x)=0$ for all $t$.

$\newcommand{\IR}{\mathbb{R}}$You are asking about endomorphisms of the algebra $A:=\IR^I := \prod_{i\in I} \IR$. All of them are of the form $x \mapsto (f_i(x))$ where $f_i$ are arbitrary homorphisms $A\to\IR$.

In particular we have to find certain maximal ideals of $A$. Now luckily there is a nice(ish) classification of all ideals of products of fields: They are all of the form $V_\phi := \{ x \mid \{i | x_i=0\} \in\phi\}$ where $\phi$ is a filter on $I$. The maximal ideals correspond to ultrafilters. (This is a nice exercise)

There are two cases of ultrafilters: Principal ultrafilters, i.e. $\phi=\phi_x=\{P\subseteq I \mid x\in P\}$ for some fixed $x\in I$, and non-principal ultrafilters.

The principal ones are easy: They are exactly the kernels of the projection maps. Also there are no non-identity ringhomomorphisms $\IR\to\IR$. Therefore any $f: A\to\IR$ with kernel $\phi_x$ is equal to the $x$-th projection map.

The non-principal ultrafilters are more tricky. The quotients $A/V_\phi$ are called ultraproducts of $\IR$. These are ordered fields as well, but interestingly they are not isomorphic to $\IR$, they don't even embed into $\IR$, because they have infinitesimal and infinite elements. Therefore they cannot occur as kernels of our maps $f_i: A\to\IR$.

Summary: All algebra homomorphisms $A\to A$ are of the form $(x_i)\mapsto (x_{g(i)})$ for some map $g:I\to I$.

This paper says that they are all evaluation functions: http://projecteuclid.org/download/pdf_1/euclid.pja/1195524949

(Provided that you consider only continuous functions.)


The existence of group isomorphism between Euclidean space. Is there any group isomorphism for addition $\mathbb{R}^n$ to $\mathbb{R}^m$ where $n\neq m$? I could prove that there exists any vector space isomorphism or smooth map, but I still could not know that if we consider only abelian group structure for the addition of them, then the group isomorphism between them exits or not.

Hint: For vector spaces over $\mathbb{Q}$, you can see the underlying group homomorphism as a $\mathbb{Q}$-linear map. This reduces the problem to seeing if dimensions of the vector spaces over $\mathbb{Q}$ are equal as this is equivalent to the underlying groups being isomorphic. In turn, it is known (facts section) that the dimension of an infinite dimensional vector space over  $\mathbb{Q}$ is the cardinality of the vector set.

Partial answer:
Assume $n \neq m$ and that $f : \mathbb R^n \rightarrow \mathbb R^m$ is a group isomorphism for addition, i.e. for all $u,v \in \mathbb R^n$, $f(u+v) = f(u) + f(v)$. Then $f$ cannot be continuous.
Indeed, you can easily successively show that for all $u \in \mathbb R^n$, $f(nu) = nf(u)$ for all $n \in \mathbb N$, then for all $n \in \mathbb Z$, then $f(qu) = qf(u)$ for all $q \in \mathbb Q$. Now if $f$ is continuous you have $f(\lambda u) = \lambda f(u)$ for all $\lambda \in \mathbb R$. So $f$ is linear and therefore $n = m$, which is a contradiction.
Here we have also shown that any group isomorphism for addition from $\mathbb R^n$ to $\mathbb R^m$ is a $\mathbb Q$-linear map.
I didn't manage to disprove the statement for non-continuous applications though but I will look into it!


Let $m$ and $n$ be odd integers. Prove $ \sum_{i=1}^{m}i^n≡ 0 \pmod{m}$ Let $m$ and $n$ be odd integers. Prove $ \sum_{i=1}^{m}i^n≡ 0 \pmod{m}$

Here's my attempt at an Induction Proof:

Let this
$$ \sum_{i=1}^{m}i^n≡ 0  \pmod{m} $$
be correct for some $m$. Now let's prove that's correct also for $m+1$:
$$ \sum_{i=1}^{m+1}i^{n} = i^1+\dots+i^m+i^{m+1} ≡ 0  \pmod{m} $$
and then this $i^1+\dots+i^m$ is divisible by $m$ because the first part of induction.

What next? Can anyone help me?

Let $m,n$ be odd integers, then we want to show $$\sum_{i=1}^{m}i^n \equiv 0  \pmod{m}.$$

Lets look at an example $$1^7 + 2^7 + 3^7 + 4^7 + 5^7 \equiv 0 \pmod 5.$$

Clearly we can throw away the last term: $$1^7 + 2^7 + 3^7 + 4^7 \equiv 0 \pmod 5$$

Now notice that half the terms may be written as negatives $$1^7 + 2^7 + (-1 \cdot 2)^7 + (-1)^7 \equiv 0 \pmod 5$$

and we can pull that sign out when we have an odd exponent $$1^7 + 2^7 - (2^7 + 1^7) \equiv 0 \pmod 5$$ which is obviously true.

this inspires a proof: $$\sum_{i=1}^{m}i^n \equiv \sum_{i=1}^{m-1} i^n \equiv \left(\sum_{i=1}^{(m-1)/2} i^n\right) + \left(\sum_{i=(m+1)/2}^{m-1} i^n\right)$$ $$\equiv \left(\sum_{i=1}^{(m-1)/2} i^n\right) + \left(\sum_{i=1}^{(m-1)/2} (m-i)^n\right)\equiv \left(\sum_{i=1}^{(m-1)/2} i^n\right) - \left(\sum_{i=1}^{(m-1)/2} i^n\right)\equiv 0 .$$

$\begin{array}\\
i^n+(m-i)^n
&=i^n+\sum_{j=0}^n \binom{n}{j}m^j(-1)^{n-j}i^{n-j}\\
&=i^n+(-1)^ni^n+\sum_{j=1}^n \binom{n}{j}m^j(-1)^{n-j}i^{n-j}\\
&=\sum_{j=1}^n \binom{n}{j}m^j(-1)^{n-j}i^{n-j}
\qquad\text{since }n\text{ is odd}\\
&=m\sum_{j=1}^n \binom{n}{j}m^{j-1}(-1)^{n-j}i^{n-j}\\
&≡ 0  \pmod{m}\\
\end{array}
$

Since
$\sum_{i=1}^{m-1}i^n
=\sum_{i=1}^{m-1}(m-i)^n
$,
$2\sum_{i=1}^{m-1}i^n
≡ 0  \pmod{m}
$.

Since $m$ is odd,
this implies that
$\sum_{i=1}^{m-1}i^n
≡ 0  \pmod{m}
$.

(I know you can skip this
by pairing symmetric terms,
but,
what the heck.)

Note that the term
$m^n$ can be ignored.


Struggling to understand principal ideals in Ring Theory I studied group theory last year and the concept of $<g>$ or $<g,h>$ for elements $g,h$ of a group $G$ made complete sense as it was just the set of all the products of elements within $< >$ and their inverses just like how span works.
Now when I get to Ring Theory (old edition with NO UNITAL ELEMENT requirement) the definition of $<r>$ and $<r,s>$ for $r,s$ in a ring $R$ just doesn’t make sense anymore as it’s no longer just the products (and addition) of the elements within $< >$ but instead involves elements outside of $< >$ as well.

I just can’t picture in my head what’s going on here and it’s making harder following my classes harder. For example one line in a proof we were doing was:

“Clearly $<a>^3 \subseteq RaR$“ ($R$ is a ring and $a$ is an element in $R$) but this isn’t at all clear for me.

Since you are specifying that your ring doesn't have a unit, you need to be careful when defining ideals: $\left<a\right>$ is not the set of all multiples of $a$, since otherwise $a$ would not necessarily be an element of $\left<a\right>$ as there is no unit.

$\left<a\right>$ is the smallest subset of $R$ that

contains $a$
is an additive subgroup
is stable by external multiplication with elements of $R$.

This last requirement, which is the one that bothers you, is actually quite natural. You can think of how an even number is still an even number when you multiply it by any number, odd or even (here we are looking at the set of even numbers, $\left<2\right>$, inside $\Bbb Z$).

See this question regarding why ideals play a more important role than subrings in ring theory.


Good examples of Ansätze Frequently, when talking to mathematicians, I have some trouble when I mention, use, or try to explain what an Ansatz is. (Apparently it is more of a physics term than a maths one, for some reason.) The Wikipedia page on it has what I think is a good definition:

  an educated guess that is verified later by its results.

It also has one example which really resonates with the way I normally see the term being used, namely exponential Ansätze for the solutions of a differential equation. There one has a problem such as
$$y''(x)+ay'(x)+by(x)=0,$$
and one quite nonchalantly assumes that the solution is $y(x)=e^{kx}$, leaving some leeway into not specifying $k$. This (Ansatz) is of course unjustified, and the only rigorous explanation for what one is doing is that one is blindly testing to see if a function of that form can be one particular solution.

One then, of course, goes on to show that this is indeed de case when $k$ satisfies $k^2+ak+b=0$, and this usually yields two distinct roots $k_1$ and $k_2$ with associated linearly independent solutions. The upshot of this is that one can now something very general about any solution of the original problem - i.e. that it is of the form
$$y=A e^{k_1 x}+B e^{k_2x}$$
for unspecified complex coefficients $A$ and $B$ - from the original, very limited Ansatz.

While this example is nice, I can't think of other simple, strong examples of this sort of argument, where a simple and limited educated guess turns out, at the end, to encapsulate the whole generality of the problem; I would like to see more of those.

I am not a great fan of the notion of ansatz, simply because it seems magical.

One can but imagine that most ansätze are the result of whole series of unsuccessful tries, and that we only hear of the successful ones. So whenever I read «let us try the ansatz such and such» I understand «ok, I spent a couple of weeks trying stuff that did not go anywhere but, who knows how, finally managed to make it work, so let me just tell you the short story and, by the by, make myself look like I came up with this stuff out of the blue»

How about the integral
$$
\int \sec x dx = \int \sec x \frac{\tan x + \sec x}{\tan x + \sec x}dx,
$$
with the next step being the u-substitution $u = \tan x + \sec x$?


Rod cutting: requesting a proof for $2^{n-1}$ possible ways The rod cutting problem states that: Given a rod of length $n$ inches and a table of prices $P_i$ for $i=1,\dots,n$, determine the maximum revenue $R_n$.

In CLRS chapter 15 Rod cutting page 360:

  We can cut up a rod of length $n$ in $2^{n-1}$ different ways

I can intuitively understand why this is so; basically we can construct a binary tree of height $\log N$. But can someone provide me a proof of why this is so?

A rod of length $n$ can be cut at any of the positions $1$ inch, $2$ inches, up to $n-1$ inches. Therefore for each way of cutting the rod, there is a corresponding sequence of $0$s and $1$s of length $n-1$, with a $1$ in the $i$th place meaning that the rod is cut at this position and a $0$ meaning that it is not cut.

Since there are $2^{n-1}$ such sequences, there must be $2^{n-1}$ ways of cutting the rod.

This can be proved using Binomial theorem $(1 + x)^n$.

$(1 + x)^n = nC_0*(x^0) + nC_1*(x^1) + nC_2*(x^2) + .... + nC_{n-1}*(x^n-1) + nC_n*(x^n)$

Total number of Cuts can be made = $(n - 1)$
    So, $n ==> (n - 1)$ & $x = 1$

0 Cut NO CUT - $(n-1)C0$
1 Cut can be made in different ways - $(n-1)C1$
2 Cut can be made in different ways - $(n-1)C2$
.....
n-1 Cut can be made in different ways - $(n-1)Cn-1$

Hence, Total possible ways to cut $= (n-1)C_0 + (n-1)C_1 + (n-1)C_2 + ... + (n-1)C_{n-1}$

That will be equals to -> $(1 + 1)^n-1 ==> 2^(n-1)$

Hence Proved.


Homomorphisms from $(\Bbb Q,+)$ to $(\Bbb Q,+)$ and from $(\Bbb Q^*,\cdot)$ to $(\Bbb Q^*,\cdot)$ Hope this isn't a duplicate.

Considering homomorphisms from $(\Bbb Q,+)$ to itself, I notice that all the homomorphisms are of the form $\phi(x) = ax$ , where $a=\phi(1)$ . So there are exactly $\aleph_0$ number of homomorphisms from $(\Bbb Q,+)$ to itself and except the trivial 0-homomorphism, all others are onto-homomorphism . Is it correct?

On the other hand, considering homomorphisms from $(\Bbb Q^*,\cdot)$ to itself, I get that $\phi(1) =1$ and also get some additional restrictions like if $x \in \Bbb Q^*$ and $x$ is a square number then $\phi(x) \neq y$ , where y is either (negative) or (positive but does not admit a rational square root) . Are there some other restrictions on a general homomorphism  from $(\Bbb Q^*,\cdot)$ to itself ?

The following maps,(i) $ x \mapsto x$ and (ii) $x \mapsto x^{-1}$ (since, $(\Bbb Q^*,\cdot)$ is abelian) are automorphisms of $(\Bbb Q^*,\cdot)$ . How can one find the complete list of $Aut(\Bbb Q^*,\cdot)$ ?

Thanks in advance for help.

There are lot more automorphisms of $\Bbb Q^*$. By the fundamental theorem
of arithmetic, $\Bbb Q^*\cong (\Bbb Z/ 2\Bbb Z)\times\bigoplus_p\Bbb Z$
where the $p$ range over all primes. The automorphism group of $\Bbb Q^*$
is uncountable, as for instance we can consider the maps fixing $-1$
and taking $p$ to $\epsilon_p p$ with $\epsilon_p=\pm1$.

$$
\mathbb{Q}^*\cong gr(-1)\times\prod_{p\in P} gr(p).
$$
Here, P is the set of primes.
From this it follows that
$$
\prod_{p\in P} gr(-1)<Aut(\mathbb{Q}^*).
$$


Finding arc length parametrization of a cycloid Find an arc length parametrization of the cycloid with parametrization
r(t)= .
I took the derivative and found the speed to be sqrt(2(1-cost))but now I'm unsure how to integrate that to get s. How do you integrate sqrt(2(1-cost))?

Try this one

$(1-\cos t) = 2 \sin ^2(\frac{t}{2})$

Hint: If ${\bf r}: I \to \Bbb R^n$ is a curve, it's arclength function is given by: $$s(t) = \int_{t_0}^t \|{\bf r}'(\xi)\|\,{\rm d}\xi.$$
If $0 \in I$, we usually take $t_0 = 0$, for simplicity. Solve the integral, and you will have $s$ as a function of $t$. Solve for $t$ in terms of $s$.


Confirming that an integral equals 1 I was looking at Wolfram Alpha's interpretation of the integral $\int_{-\infty}^{\infty} e^{-e^{-x} - x}dx = 1$ and I'm trying to explicitly compute that now. My first attempt was to substitute with $u = -e^{-x}$, but I ended up getting a $u^u$ in the resulting integral, which I don't think can be derived in elementary functions. After several more unsuccessful attempts at solving this integral, I think that either I'm missing something obvious or that this integral can't be evaluated with elementary functions. I am curious to read your htoughts.

( http://www.wolframalpha.com/input/?i=integrate+-infinity+to+infinity+e%5E%28-x-e%5E%28-x%29%29 )

This integral cannot be solved in terms of elementary functions. Try using Maclaurin series!


Same order of element in group... If we have two elements $a$ and $b$ in a group such that $a^3 = e$ and $b^3 = e$ does this imply that $a=b$? I suspect so, but I think im missing something trivial here?

No: Another counterexample:

Take $(1, 2, 3),\;(1, 3, 2) \in S_3$. Each of these permutations is of order $3$ (confirm that, if you need to), yet clearly $(1, 2, 3) \neq (1, 3, 2)$.

No. Consider the group of integers mod 3.


A question about probability (conflicting solutions) There is a square of side $n$ units. Join the diagonals. Now the square is divided into 4 regions of equal area. Each of them is coloured differently. Given 2 points that can lie within any of the four regions, what is the probability that two points are in the same coloured region?

The problem:

I and my friend get two different solutions using two different lines of reasoning. It has to do with the order of the points being placed:

Solution 1 (mine): The order in which the points are placed does not matter. The number of ways 2 identical objects can be placed in 4 distinct regions is 10 (stars and bars). Out of those, there are 4 cases which satisfy the condition. Hence the probability is $2/5$.

Solution 2 (my friend's): The order of the points being placed does matter. That means there are 16 ways to place the points (4 orientations × 4 regions). Out of those, 4 cases satisfy the condition. Hence the probability is $1/4$.

Now of course, one line of reasoning must be wrong here. Which one is wrong? An explanation is appreciated.

The first solution is wrong and the second is correct. Whichever region the first point lies in, there is a $\frac14$ chance the second point will be in that same region.

The first solution assumes that the probabilities of having the two points in one given region and those of having them in two separate, given regions are the same, when in fact the second is twice the first.


Show that $a^2 \bmod b = (a \bmod b)^2 \bmod b$ Show that
$$a^2 \mathrm{\ mod \ } b = (a \mathrm{\ mod \ } b)^2 \mathrm{\ mod\ } b$$
for $ a, b \in \mathbb{Z}^+ $.

this was derived from an Informatics olympiad question.

$$a=kb+n\\a^2=k^2b^2+2kbn+n^2\\a^2=n^2\mod{b}$$
Now since $a=n \mod{b}$ this is the proof

The argument of @kingW3 is sufficient.

By definition of integer division, $a = b\ (a \ div\ b)+(a\ mod\ b)$.

Then $$a^2\ mod\ b=\left(b(a\ div\ b)+(a\ mod\ b)\right)^2\ mod\ b=\left(b^2(a\ div\ b)^2 + 2 b(a\ div\ b)(a\ mod\ b)+(a\ mod\ b)^2\right)\ mod\ b = (a\ mod\ b)^2\ mod\ b.$$


Limit of indeteminate form $1^{(∞)}$ If we consider the function $f(x)=[(ax+1)/(bx+2)]^{x}$ where $a,\,b >0$ and $a<b,\,$ then we have to find $\lim_{x\to \infty} f(x)$

I tried as follows]1

But at end I got stuck .

$$\lim_{x\to \infty}\left(\frac{ax+1}{bx+2}\right)^x=\lim_{x\to \infty}\left(\frac ab\right)^x\left(\frac{1+\frac{1/a}{x}}{1+\frac{2/b}{x}}\right)^x \tag 1$$

Using the limit definition of the exponential function, we see that

$$\lim_{x\to \infty}\left(\frac{1+\frac{1/a}{x}}{1+\frac{2/b}{x}}\right)^x=e^{1/a-2/b}$$

However, we have

$$\lim_{x\to \infty} \left(\frac ab\right)^x=\begin{cases}\infty&,a>b>0\\\\0&,b>a>0\\\\1&,a=b>0\end{cases}$$

Therefore, we conclude that if $a>b>0$, then the limit is $\infty$.  If $b>a>0$, then the limit is $0$.  And if $a=b>0$, then the limit is $e^{-1/a}$.

In any situation with this indeterminant form, I always use log and L'Hopital. First, consider $\lim\log (f (x)) $ instead. We're now looking at $\lim x(\log (ax + 1) - \log (bx + 2))$, after applying a couple of log properties. Next, rephrase it as $\lim\frac {\log (ax +1) - \log (bx +2)}{1/x} $. This has form 0/0, so L'Hopital's rule applies.


How to read a book in mathematics? How is it that you read a mathematics book?
Do you keep a notebook of definitions? What about theorems?
Do you do all the exercises? Focus on or ignore the proofs?

I have been reading Munkres, Artin, Halmos, etc. but I get a bit lost usually around the middle. Also, about how fast should you be reading it? Any advice is wanted, I just reached the upper division level.

This method has worked well for me (but what works well for one person won't necessarily work well for everyone).  I take it in several passes:

Read 0: Don't read the book, read the Wikipedia article or ask a friend what the subject is about.  Learn about the big questions asked in the subject, and the basics of the theorems that answer them.  Often the most important ideas are those that can be stated concisely, so you should be able to remember them once you are engaging the book.

Read 1: Let your eyes jump from definition to lemma to theorem without reading the proofs in between unless something grabs your attention or bothers you. If the book has exercises, see if you can do the first one of each chapter or section as you go.

Read 2: Read the book but this time read the proofs.  But don't worry if you don't get all the details.  If some logical jump doesn't make complete sense, feel free to ignore it at your discretion as long as you understand the overall flow of reasoning.

Read 3: Read through the lens of a skeptic.  Work through all of the proofs with a fine toothed comb, and ask yourself every question you think of.  You should never have to ask yourself "why" you are proving what you are proving at this point, but you have a chance to get the details down.

This approach is well suited to many math textbooks, which seem to be written to read well to people who already understand the subject.  Most of the "classic" textbooks are labeled as such because they are comprehensive or well organized, not because they present challenging abstract ideas well to the uninitiated.

(Steps 1-3 are based on a three step heuristic method for writing proofs: convince yourself, convince a friend, convince a skeptic)

This is a seven-year post but I am putting here my experience learning mathematics as an undergraduate.

First, I would explain a bit what's usually inside a mathematics book. For every chapter or sections, there are usually two parts, the contents and the exercises. The purpose of contents is to let you expose yourself to the knowledge while the exercises are to let you improve until you really understand the material well. So, first and foremost, do not feel disappointed if you cannot solve some of the exercises as you have met the exercise at your level! If you can solve all exercises easily or it's so hard that you can't solve any of them, that book is not for you. For me, a book suitable for you would be the one you can solve at least 60% of the questions most of the time. Also, do not try to solve every single exercise in a book as it leads to diminishing returns. There are a lot of fun subjects that you should go for it. This paragraph is to let you have a correct attitude towards exercises in mathematics. Now I will share my way to study a mathematics book.

The first level - Read the chapters briefly. If you are following a course, lecture's class would be at this level. If you are self-studying, read a chapter that you can understand at least half of the content including proofs. If at this point, it takes so long for you to understand (even the first chapter), consider changing to an easier book.

The second level - Read the chapters in detail. Jot down the concept or part of proofs that you can't understand. Copy down all the theorem statements and definitions on a piece of paper. This is extremely helpful when you attempt the exercises.

The third level - Do the exercises in each chapter. After struggling, do not feel embarrassed to look the answer to the questions you cannot solve. Make sure to understand how they solve it.

The fourth level - After you have completed half of the book. It's time for revision! Try to prove everything in the text once again. Revise the exercise that you had done. If you can't prove some of them, reread again and try to memorise the key parts of the proofs. Then reprove them after a few days.

The fifth level - First you do the same as in the fourth level to the last half of the book. Then, write a revision sheet for all the chapters. This revision sheet is for your future reference, so it should be short and concise.

Usually, you can read 4 mathematics book in 3 months by following this process provided the level of the book is suitable for you. After this, you may study another subject of your interest or further sharpen your skills by reading a more advanced text of the same subject. For me, it's the fastest way. Of course, to do this, you need to invest at least 8 hours per day on it.


How to confirm $\phi(F_1(x,y))＝F_2(\phi(x),\phi(y))$,where $F_1$ and $F_2$ are formal group law of elliptic curve $E_1$, $E_2$. This question is from Silverman's 'the arithmetic of elliptic curves',$p134$.
Let $K$ be a field of characteristic $p > 0$, let $E_1/ K$ and $ E_2/K$ be elliptic curves, and let $\phi : E_1 \to E_2$ be a nonzero isogeny defined over $K$. Further, let $f: \hat{E_1} \to \hat{E_2} $ be the homomorphism of formal groups induced by $\phi$.
My question:
How does an isogeny $\phi$ on elliptic curves induces a homomorphism of corresponding formal groups?
I guess $f(T)＝\phi(T)$,but I cannot check this is actually homomorphism.
My question is, I would like to know the confirming process
$\phi(F_1(x,y))＝F_2(\phi(x),\phi(y))$,where $F_1$ and $F_2$ are formal group law of elliptic curve $E_1$, $E_2$.

If you have an elliptic curve $E_j/k:y_j^2=x_j^3+a_j x+b_j$ then

the group law is given by a rational function in 4 variables $x_j,y_j,x_j',y_j'$  which can be rewritten as a rational function in $x_j/y_j,1/y_j,x_j'/y_j',1/y_j'$.

$1/y_j$ is in $k[[x_j/y_j]]$

whence the group law is given by a formal series $F_j\in k[[x_j/y_j, x_j'/y_j']]$,

the power series expansion of the algebraic-rational function giving $x_j(A+B)/y_j(A+B)$ from $x_j(A)/y_j(A)$ and $x_j(B)/y_j(B)$. This is the formal group law. When $k=\Bbb{C}$ this power series is an analytic function.

Your isogeny $f:E_1\to E_2$ is given by a pair of rational functions in $x_1,y_1$, can be rewritten as a pair of rational functions in the $x_1/y_1,1/y_1$ coordinates. Replacing $1/y_1$ by its formal series $\in k[[x_1/y_1]$ this pair of rational functions is given by one formal series $\phi(x_1/y_1)\in k[[x_1/y_1]]$, the power series expansion of the algebraic-rational function giving the $x_2/y_2$ coordinate from $x_1/y_1$.

$T\to \phi(T)$ is your homomorphism of formal group law.
We have just made a change of coordinate and embedded the rational-algebraic functions defining the group law and the isogeny in some larger ring of formal series. The group law and the isogeny commute, so do their formal series representation.


Why is variance squared rather than cubed (or any other exponent)? Variance is

$$\dfrac{\sum_{i=1}^{n}(x_i-\bar x)^2}{n-1}$$

But why square the difference? Why not cube it, or any other exponent?

Related question.

You can also use other exponents, but then it is not called variance anymore. See this wikipedia page, for example.

Variance comes up quite naturally in various contexts, and carries a lot of information about the distribution.

Still, I am not convinced by this answer.
The naming convention can be separate topics to discuss.
Here what is the issue if we take cubes or quad etc?


What is the free abelian group on $M \times N$ where $M,N$ are modules. The free abelian group (equivalently the free $\mathbb{Z}$-module) $F(M \times N)$ is
defined as the set of all linear combinations of elements of $M \times N$,

$$F(M \times N) = \{n_1x_1 + \cdots +n_kx_k : n_i \in \mathbb{Z}, x_i \in M \times N, k \in \mathbb{Z}_{\ge0}\}$$

Wouldn't these linear combinations already be in $M \times N$?  How is the free abelian group different from $M \times N$?

No. For instance, suppose you took the $\mathbb{Z}$-module $\mathbb{Z}/2\mathbb{Z}=M$. Then, in $M^2$ you have that $2(1,0)=(0,0)$ but in $F(M^2)$ one has that $2(1,0)\ne (0,0)$. The relations that hold amongst elements of $M^2$ DO NOT hold for elements of $F[M^2]$.

A right $\mathbb Z$-module is also a left $\mathbb Z$-module, if that helps.


maximum and number of distinct real roots If $f(x)=x^6+2x^5+bx^4+x^3+bx^2+ax+1$ and $f(1)=0$ and $f(-1)\neq 0.$ Then maximum number of distinct real roots of $f(x)=0,$ is

what i try: $x^6+2x^5+bx^4+x^3+bx^2+ax+1=0$

$x=0$ in not a solution

$$x^3+\frac{1}{x^3}+\bigg(2x^2+\frac{a}{x^2}\bigg)+b\bigg(x+\frac{1}{x}\bigg)+1=0$$

How do i solve it Help me please

Hint:

If $f(1)=0$ then the sum of the even and odd coefficients differ only by sign. That is, $2b+2=-(a+3)\to a=-2b-5$


How to find $\lim_{n \to \infty}|\left( 1+\frac{z}{n}\right)^{n}|$? This is a step on the way to proving
$$
\lim_{n \to \infty}\left(1 + \frac{z}{n}\right)^{n} = e^{z}.$$

I'm looking for an answer without
1) a summation or
2) a logarithmic function.

So far, I have that
$$
\lim_{n \to \infty} |z_{n}| = \lim_{n \to \infty} \vert\left(1+\frac{z}{n}\right)^{n}\vert = \lim_{n \to \infty} \left(\sqrt{\left(1+\frac{x}{n}\right)^{2} +\left( \frac{y}{n}\right)^{2} }\right)^{n} = \lim_{n \to \infty}\left(\left( 1+\frac{x}{n}\right)^{2} + \left(\frac{y}{n} \right)^{2} \right)^{n/2} = \lim_{n \to \infty} \left( 1 + \frac{2x}{n} + \left(\frac{x}{n}\right)^{2} + \left( \frac{y}{n}\right)^{2}\right)^{n/2}.
$$
Now, I was given the following hint, and I would like to use it, but I don't know how:

  Hint: Argue that you can discard terms with $1/n^{2}$; for example, by showing that the limit $\displaystyle \lim_{n \to \infty} \frac{\left(\left( 1 + \frac{x}{n}\right)^{2} + \left(\frac{y}{n}\right)^{2}\right)^{n/2}}{\left(1 + \frac{2x}{n}\right)^{n/2}} = 1$.

How would I show this ($\epsilon$-definition?), and how would it help me here with what I've already done?

I think the best approach is the one which I found on MSE (will add link when I find it). It is based on the following general theorem.

  Theorem: If $a_{n}$ is a sequence of real or complex terms such that $n(a_{n} - 1) \to 0$ as $n \to \infty$ then $a_{n}^{n}  \to 1$ as $n \to \infty$.

The proof is done by writing $a_{n} = 1 + a_{n} - 1 = 1 + b_{n}$ (say) so that $nb_{n} \to 0$ as $n \to \infty$. Clearly we have
\begin{align}
a_{n}^{n} &= (1 + b_{n})^{n}\notag\\
&= 1 + nb_{n} + \dfrac{1 - \dfrac{1}{n}}{2!}(nb_{n})^{2} + \cdots\notag
\end{align}
and hence
\begin{align}
|a_{n}^{n} - 1| &\leq |nb_{n}| + \frac{1}{2!}|nb_{n}|^{2} + \cdots + \frac{1}{n!}|nb_{n}|^{n}\notag\\
&\leq|nb_{n}| + \frac{1}{2}|nb_{n}| + \frac{1}{2^{2}}|nb_{n}|^{2} + \cdots\notag\\
&= \dfrac{|nb_{n}|}{1 - \dfrac{|nb_{n}|}{2}}\notag\\
&\to 0\text{ as }n \to \infty\notag
\end{align}
Hence $a_{n}^{n} \to 1$ as $n \to \infty$.

It is now easy to show that if $z = x + iy$ then $$\left(1 + \frac{z}{n}\right)^{n} \to e^{x}(\cos y + i\sin y)$$ for all real $x, y$ where symbol $e^{x}$ is defined by the limit $$e^{x} = \lim_{n \to \infty}\left(1 + \frac{x}{n}\right)^{n}\tag{1}$$ I assume that you know the proof that the limit in $(1)$ exists for all real $x$ and this limit is non-zero.

I show that if $$f(z) = \lim_{n \to \infty}\left(1 + \frac{z}{n}\right)^{n}\tag{2}$$ then $$f(z_{1} + z_{2}) = f(z_{1})f(z_{2})\tag{3}$$ for all $z_{1}, z_{2} \in \mathbb{C}$. Consider the sequence $$a_{n} = \dfrac{1 + \dfrac{z_{1} + z_{2}}{n}}{\left(1 + \dfrac{z_{1}}{n}\right)\left(1 + \dfrac{z_{2}}{n}\right)}\tag{4}$$ and then $$n(a_{n} - 1) = -\left(\dfrac{\dfrac{z_{1}z_{2}}{n}}{\left(1 + \dfrac{z_{1}}{n}\right)\left(1 + \dfrac{z_{2}}{n}\right)}\right)$$ so that $n(a_{n} - 1) \to 0$ as $n \to \infty$. Hence $a_{n}^{n} \to 1$ as $n \to \infty$. From definition of $a_{n}$ in equation $(4)$ it follows that $f(z_{1} + z_{2}) = f(z_{1})f(z_{2})$ provided that both the limits $f(z_{1}), f(z_{2})$ exist and are non-zero.

Next I show that the limit $f(z)$ defined by equation $(2)$ exists and is non-zero for all $z \in \mathbb{C}$. If $z$ is real then by assumption mentioned after equation $(1)$, the limit $f(z)$ exists and is non-zero. Next let $z = iy$ be totally imaginary. I will show that in this case $f(z) = \cos y + i\sin y$ and hence it is non-zero. Consider again the sequence $$c_{n} = \dfrac{1 + \dfrac{iy}{n}}{\cos\left(\dfrac{y}{n}\right) + i\sin\left(\dfrac{y}{n}\right)} = \cos\left(\frac{y}{n}\right) + \left(\frac{y}{n}\right)\sin\left(\frac{y}{n}\right) + i\left\{\left(\frac{y}{n}\right)\cos\left(\frac{y}{n}\right) -\sin\left(\frac{y}{n}\right)\right\}$$ and it is easy to see that $n(c_{n} - 1) \to 0$ as $n \to \infty$ and hence $c_{n}^{n} \to 1$ and from defining equation for $c_{n}$ we can see that denominator of $c_{n}^{n}$ is $(\cos y + i\sin y)$. It follows that $$(1 + iy/n)^{n} \to (\cos y + i\sin y)$$ Thus we have established that $f(z)$ exists and is non-zero in both the cases when $z$ is purely real and when $z$ is purely imaginary.

From equation $(3)$ it follows that if $z = x + iy$ then $f(z) = f(x)f(iy)$ and $f(z)$ is non-zero because $f(x), f(iy)$ are non-zero. We thus have $f(z) = e^{x}(\cos y + i\sin y)$.

It is easy to use the theorem mentioned in the beginning of my post to prove the result which OP is seeking in order to develop a proper theory of the limit of $(1 + z/n)^{n}$. Thus I show directly that if $z_{n} = (1 + z/n)^{n}$ then $|z_{n}|^{2} \to e^{2x}$ where $z = x + iy$ and hence $|z_{n}| \to e^{x}$. Clearly as mentioned by OP we have $$|z_{n}|^{2} = \left\{\left(1 + \frac{x}{n}\right)^{2} + \frac{y^{2}}{n^{2}}\right\}^{n}$$ Taking $$a_{n} = \dfrac{\left(1 + \dfrac{x}{n}\right)^{2} + \dfrac{y^{2}}{n^{2}}}{1 + \dfrac{2x}{n}}$$ we can see that $n(a_{n} - 1) \to 0$ and hence $a_{n}^{n} \to 1$ so that $|z_{n}|^{2} \to e^{2x}$.

OP has asked in the comments about choosing an appropriate sequence $a_{n}$ to make use of the theorem mentioned in my post. The basic idea is that the theorem is supposed to be used in situations where you want to find the limit of an expression whose exponent is $n$. Thus say you want to find limit of a sequence of type $c_{n}^{n}$. Moreover you intuitively know the answer as to what the limit will be. The idea is to find another sequence of type $d_{n}^{n}$ whose limit you already know and then set $a_{n} = c_{n}/d_{n}$.

Thus for example in OP's question there is a hint that the limit of $\left(\left(1 + \dfrac{x}{n}\right)^{2} + \dfrac{y^{2}}{n^{2}}\right)^{n/2}$ is related to that of the sequence $(1 + 2x/n)^{n/2}$. Squaring these sequences we get the exponents as $n$ and then it is a simple matter to figure out that $$c_{n} = \left(1 + \dfrac{x}{n}\right)^{2} + \dfrac{y^{2}}{n^{2}}, d_{n} = 1 + \frac{2x}{n}, a_{n} = \frac{c_{n}}{d_{n}}$$ A better case can be made for the limit $(1 + iy/n)^{n}$. We intuitively know that the limit should be $e^{iy} = \cos y + i\sin y$. Let's then try to find another expression whose $n^{\text{th}}$ power tends to $\cos y + i\sin y$. Here we are damn lucky because of De Moivre's Theorem $$\left(\cos\frac{y}{n} + i\sin\frac{y}{n}\right)^{n} = \cos y + i\sin y$$ and therefore we set $$c_{n} = 1 + \frac{iy}{n}, d_{n} = \cos\frac{y}{n} + i\sin\frac{y}{n}, a_{n} = \frac{c_{n}}{d_{n}}$$

Just show that

$$\left(1 + \frac{2x}{n} + \frac{D}{n^2}\right)^n = \sum_{k=0}^n \left(1+\frac{2x}{n}\right)^{n-k} C_{n}^k \frac{D^{k}}{n^{2k}} = (1 + \frac{2x}{n})^n + \sum_{k=1}^{n} (1 + \frac{2x}{n})^{n-k}C_n^k \frac{D^k}{n^{2k}}$$

the latter term take $n\to \infty$, there is a constant $M$ that $(1 + \frac{2x}{n})^{n-k} \le M$ for all $k$. thus we just need to show that

$$\sum_{k=1}^{n} C_n^k \frac{D^k}{n^{2k}}\to 0$$

$$\sum_{k=1}^{n} C_n^k \frac{D^k}{n^{2k}} = \sum_{k=1}^{n} \frac{n(n-1)\dots (n-k+1)}{k(k-1)\dots 1} \frac{D^k}{n^{2k}}$$

and

$$\frac{n}{k}\le \frac{n-1}{k-1}\le \dots \le \frac{n-k+1}{1}$$

we just have to show

$$\sum_{k=1}^{n}\left(n-k+1\right)^k\frac{D^k}{n^{2k}}\to 0$$

while

$$(n-k+1)\le n$$

$$\sum_{k=1}^{n}\left(n-k+1\right)^k\frac{D^k}{n^{2k}}\le \sum_{k=1}^{n} \frac{D^kn^k}{n^{2k}} = \sum_{k=1}^{n}\frac{D^k}{n^k} \le \sum_{k=1}^{\infty} \frac{D^k}{n^k} = \frac{D}{n-D}\to 0$$


Solve the functional equation $f(x)=f\left({x\over 3}\right)+f\left({2x\over 3}\right)$ with $f : [0,\infty) \to \mathbb R$ continuous Solve the functional equation

  $$f(x)=f\left({x\over 3}\right)+f\left({2x\over 3}\right)\qquad \forall x\geq 0$$
  with $f : [0,\infty) \to \mathbb R$ continuous.

I can't manage to get this one to the form of Cauchy's functional equation, but I imagine that's how it's maybe done.

To my surprise, there are non-linear solutions to this functional equation.
Consider functions of the form:

$$F(x;\alpha,k) = x^\alpha \sin( k\log x )$$

If one can choose $\alpha \in (0,1)$ and $k \in (0,\infty)$ such that:

$$\begin{align}
\left(\frac13\right)^\alpha \cos\left(k \log\frac13\right) + \left(\frac23\right)^\alpha \cos\left(k\log\frac23\right) &= 1\\
\left(\frac13\right)^\alpha \sin\left(k \log\frac13\right) + \left(\frac23\right)^\alpha \sin\left(k\log\frac23\right) &= 0\tag{*}
\end{align}$$

then $F$ is continuous over $[0,\infty)$ and satisfies:
$$ F(x;\alpha,k) =  F(\frac{x}{3};\alpha,k) + F(\frac{2x}{3};\alpha,k)$$

By trial and error, I find at least one solution of $(*)$ with $(\alpha,k) \sim ( 0.7586093, 16.4941542 )$ and the corresponding $f(x)$ looks like this:

Observe that : $$f(x)=f\left(\frac{x}{3}+\frac{2x}{3}\right)=f\left(\frac{x}{3}\right)+f\left(\frac{2x}{3}\right)$$

you can make a substitution :

$$\frac{x}{3}=y;$$
$$\frac{2x}{3}=z.$$

$$f(y+z)=f(y)+f(z)$$ which is a Cauchy functional equation with  the solution:
$$f(y)=Ky, K \in \mathbb{R}.$$

or $$f\left(\frac{x}{3}\right)=\frac{x}{3}$$
$$\frac{x}{3} \mapsto X $$

so $$f(X)=kX$$ with $k \in \mathbb{R}$.


Asymptotic Expansion of $\ f(x)=(1-\beta \frac{ log(log(x))}{log(x)})^{\beta}$ So I got this function and I'm looking for an asymptotic expansion for different values of$\ \beta > 1 $
$\ f(x)=\left(1- \beta \frac{\log \left( \log(x) \right)}{\log(x)} \right)^{\beta}$ as $\ x \to \infty $

Does anyone have any idea on how to go about this? Obviously for very large $\ x$, we go to $\ 1$ very quickly and the first order is obviously $\ 1$, but what about the other orders?

Let the function $f(x)$ be given by

$$f(x)=\left(1-\beta\frac{\log \log x}{\log x}\right)^{\beta}$$

where $x>1$.  We wish to find the Taylor series for $f(x)$.

To that end, let $g(t)$ be the function given by

$$g(t)=(1-t)^{\beta}$$

and note that $f(x)=g\left(\beta\frac{\log \log x}{\log x}\right)$.

Straightforward calculation shows that the $n$'th order derivative $g^{(n)}(0)$ of $g(t)$ at $t=0$ can be written as

$$\begin{align}
g^{(n)}(0)&=(-1)^n\beta(\beta-1)(\beta-2)\cdots(\beta-n+1)\\\\
&=(-1)^n\frac{\Gamma(\beta+1)}{\Gamma(\beta-n+1)} \tag 1
\end{align}$$

where $\Gamma(\beta)$ is the Gamma Function.  Note that we exploited the identity $\Gamma(\beta+1)=\beta\Gamma(\beta)$ repeatedly ($n\,\,$times) to arrive at $(1)$.

Now, the Taylor series for $g(t)$ is

$$g(t)=\sum_{n=0}^{\infty}\frac{\Gamma(\beta+1)}{\Gamma(\beta-n+1)} \frac{(-1)^nt^n}{n!}$$

Simple application of the ratio test show that the series is convergent for all $t$.

Finally, letting $t=\beta\,\frac{\log \log x}{\log x}$, $x>1$ reveals

$$\bbox[5px,border:2px solid #C0A000]{f(x)=\sum_{n=0}^{\infty}\frac{\Gamma(\beta+1)}{\Gamma(\beta-n+1)}\frac{(-1)^n\beta^n}{n!}\left(\frac{\log\log x}{\log x}\right)^n} \tag 2$$

For convenience, we provide the first $4$ terms of $(2)$ as

$$\bbox[5px,border:2px solid #C0A000]{f(x)\sim 1-\beta^2\left(\frac{\log\log x}{\log x}\right)+\frac{\beta^3(\beta-1)}{2!}\left(\frac{\log\log x}{\log x}\right)^2-\frac{\beta^4(\beta-1)(\beta-2)}{3!}\left(\frac{\log\log x}{\log x}\right)^3}$$

Hint

$$
\lim_{x \rightarrow \infty}
\left( 1 - \beta \frac{\log(\log(x))}{\log(x)} \right)^\beta
=
\lim_{y \rightarrow \infty}
\left( 1 - \beta  \frac{\log(y)}{y} \right)^\beta
=
\lim_{z \rightarrow \infty}
\Big( 1 - \beta  z \exp(-z) \Big)^\beta
$$


Why $\cos(n^2x) \not\to 0$ for any real $x$? I'd like to show, as simply as possible, that $\sum_{n=1}^\infty\cos(n^2 x)$ diverges for every real $x$.

(I know how to prove it for rational $x$. For irrational $x$, I don't know if $(n^2 x)$ is equidistributed mod 1. But if it is, perhaps this fact is not needed for a simple proof.)

Edit: Wikipedia says that Weyl proved in 1916 that $(n^2 x)$ is equidistributed mod 1 for irrational $x$, so this answers my question. The question remains if there's a simple proof not using this fact.

Assume that this sequence is convergent to $0$ so the subsequence $(\cos(4n^2x))$ is also convergent to $0$. Moreover by the identity

$$\cos^2 y+\sin^2y=1$$

the sequence $(\sin^2(n^2x))$ is convergent to $1$ but in this case

$$\cos(4n^2x)=\cos^2(2n^2x)-\sin^2(2n^2x)\\=(\cos^2(n^2x)-\sin^2(n^2x))^2-4\sin^2(n^2x)\cos^2(n^2x)\xrightarrow{n\to\infty}1$$
contradiction!

A necessary condition for convergence of series $\displaystyle \sum_{1}^{\infty}a_n\,\,\, is: \lim_{n \to \infty} a_n = 0$.

The series $\sum_{n=1}^\infty\cos(n^2 x)$ not met for any x --> the series diverge.


How is Faulhaber's formula derived? I have been wanting to understand how to find the sum of this series.

$$1^p + 2^p + 3^p +{\dots} + n^p$$

I am familiar with Gauss' diagonalised adding trick for the sum of the first $n$ natural numbers.

I can prove the formulas for

$$\begin{align}
 \sum_{1}^{n} k^2 &= \frac{n(n+1)(2n+1)}{6}\\
 \sum_{1}^{n} k^3 &= \frac{n^2(n+1)^2}{4}\\
\sum_{1}^{n} k^4 &= \frac{n(n+1)(2n+1)(3n^2+3n-1)}{30} \\
\end{align}$$

With mathematical induction. But, beyond that even proofs with mathematical induction are difficult.

I'm interested in learning the theory and the proof behind Faulhaber's formula. What is the knowledge required to understand this proof ?

The following interesting derivation is from Aigner's "A Course in Enumeration" (Springer, 2007).

Remember that if we have the following exponential generating functions:

$\begin{align}
   \widehat{A}(z)
     &= \sum_{n \ge 0} a_n \frac{z^n}{n!} \\
   \widehat{B}(z)
     &= \sum_{n \ge 0} b_n \frac{z^n}{n!}
\end{align}$

then also:

$\begin{align}
   \widehat{A}(z) \cdot \widehat{B}(z)
     &= \sum_{n \ge 0}
          \left(
            \sum_{0 \le k \le n} \frac{a_k}{k!} \frac{b_{n - k}}{(n - k)!}
          \right) z^n \\
     &= \sum_{n \ge 0}
          \left(
            \sum_{0 \le k \le n} \frac{n!}{k! (n - k)!} a_k b_{n - k}
          \right) \frac{z^n}{n!} \\
     &= \sum_{n \ge 0}
          \left(
            \sum_{0 \le k \le n} \binom{n}{k} a_k b_{n - k}
          \right) \frac{z^n}{n!}
\end{align}$

Let's define:

$\begin{align}
S_m(n)
      = \sum_{1 \le k \le n - 1} k^m
\end{align}$

We can define the exponential generating function:

$\begin{align}
\widehat{S}_n(z)
      &= \sum_{m \ge 0} S_m(n) \frac{z^m}{m!} \\
      &= \sum_{1 \le k \le n - 1} \sum_{m \ge 0} \frac{k^m z^m}{m!} \\
      &= \sum_{1 \le k \le n - 1} \mathrm{e}^{k z} \\
      &= \frac{\mathrm{e}^{n z} - 1}{\mathrm{e}^z - 1} \\
\end{align}$

This is almost the exponential generating function of the powers of $n$:

$\begin{align}
  \widehat{P}_n(z)
      &= \sum_{m \ge 0} n^m \frac{z^m}{m!} \\
      &= \mathrm{e}^{n z}
\end{align}$

We can write:

$\begin{align}
   (\widehat{P}_n(z) - 1) \widehat{B}(z)
     = z \widehat{S}_n(z) \tag{1}
\end{align}$

where we have the exponential generating function of the Bernoulli numbers:

$\begin{align}
   \widehat{B}(z)
      &= \frac{z}{\mathrm{e}^z - 1} \\
      &= \sum_{n \ge 0} B_n \frac{z^n}{n!}
\end{align}$

Comparing coefficients of $z^{m + 1}$ in (1):

$\begin{align}
 \sum_{m \ge 1} z^m
   \sum_{0 \le k \le m}
      \binom{m}{k} \frac{(n z)^{m - k}}{(m - k)!} B_k
      = \sum_{m \ge 0} S_m(n) \frac{z^{m + 1}}{m!}
\end{align}$

we get after simpĺifying:

$\begin{align}
   S_m(n)
      = \frac{1}{m + 1} \,
           \sum_{0 \le k \le m} \binom{m + 1}{k} B_k n^{m + 1 - k}
\end{align}$

Note: The formula given is often associated with Faulhaber, but Faulhaber's formulas where quite different (and computationally more efficient). This formula is due to Bernoulli.

Here I give an induction formula: from which all sums can be derived:

Let $$ S_n(p)=\sum_{k=1}^{n} k^p\qquad n, p\in\mathbb N ~~~~~\text{called Cavalieri sum of oder p}$$

then,
We know the following Binomial formula

$$ (k+1)^p = k^p+ \sum_{i=0}^{p-1}\binom{p}{i} k^i$$
where $\binom{p}{i}= \frac{p!}{i!(p-i)!}$.
Which implies that,

$$\sum_{k=1}^{n} (k+1)^p =\sum_{k=1}^{n} k^p+\sum_{i=0}^{p-1}\binom{p}{i} \sum_{k=1}^{n} k^i = S_n(p) +\sum_{i=0}^{p-1}\binom{p}{i} S_n(i) $$

But $$\sum_{k=1}^{n} (k+1)^p = \sum_{k=2}^{n+1} k^p = S_{n+1}(p) -1 = S_n(p) +(n+1)^p -1$$

Hence finally we get the formula :

$$\color{red}{(n+1)^p -1  =\sum_{i=0}^{p-1}\binom{p}{i} S_n(i)} $$

  From this it is possible to compute the sum for any $p\ge 1 $ in $ \mathbb N $.


Approximate a positive Sobolev function by positive smooth functions Here is a problem that I have encountered in PDE book several times. But I have never seen a proof of it. I will be very grateful if someone could give me a proof.

Question: Let $B$ be the unit ball in $\mathbb{R}^n$, $f$ a non-negative function in $H_0^1(B)$, prove that there exists a sequence of non-negative functions $\varphi_k\in C_c^\infty(B)$ such that $\varphi_k\rightarrow f$ in $H_0^1(B)$.

Edit: What if we replace $B$ by a general domain $\Omega$?

Edit II: Thanks to Hans's idea (which should work for any star-shaped domain), if the boundary of $\Omega$ suitably good (for example, it admits finite covering of star-shaped open sets), then using partition of unity we should be able to construct the desired approximation.

Edit III: If I didn't make any mistake. L.C. Evans's Partial Differential Equations (First Edition) page 260 gives a proof for $C^1$ domain. Although he was actually proving something else, the key ingredient works in our situation!

Let $\psi$ be a standard non-negative positive mollifier, $C^\infty$, supported on the unit ball in $\mathbb{R}^n$, with $\int_{\mathbb{R}^n} \psi = 1$. Define $\psi_k(x) = k^n \psi(kx)$ as usual. Extend $f$ trivially outside the unit ball and define $\tilde f_k(x) = f(kx/(k-1))$ for all $x$, for $K > 1$. This function is now supported on the ball with radius $1 - k^{-1}$ and is non-negative and in $H^1_0$. Show that $\tilde f_k \to f$ in $H^1_0$. (It is enough to show weak convergence in $L^2$ and convergence of the norm, since we are working in a Hilbert space.)

Then define
$$
\varphi_k(x) = \int_{\mathbb{R}^n} \psi_k(x-y)\tilde f_k(y) dy
$$
as usual. Then $\varphi_k$ is $C^\infty$, supported in the unit ball, and non-negative. Apply a triangle inequality argument to show that $\varphi_k \to f$ in $H^1_0$.

Are you getting mixed up with the definitions?

Let $\Omega\subset\mathbb{R}^n$ be open.  $H^{1}_0(\Omega)=H^{1,2}_0(\Omega)$ is $\mathbf{defined}$ to be the closure of $C^{\infty}_c(\Omega)$ in the $W^{1,2}(\Omega)$ norm, which is the norm that $H^{1,2}_0(\Omega)$ inherits.

Hence, $f\in H^{1,2}_0(\Omega)$ if and only if there exists a sequence $(\varphi_n)$ in $C^{\infty}_c(\Omega)$ such that $\varphi_n\longrightarrow f$ in $H^{1,2}_0(\Omega)$, by definition.

I.e. every function in $H^{1,2}_0(\Omega)$ is either a function in $C^{\infty}_c(\Omega)$, or is the limit of some sequence in $C^{\infty}_c(\Omega)$, under the $W^{1,2}(\Omega)$ norm.

There is nothing to prove.


norm of non-square matrix I have the question about the non-square matrix.
If $A$ is $m \times n$-matrix, the $m$ is not equal $n$. $x$ is a vector which the dimension is $n \times 1$.
Is it correct that
   $$\left\Vert Ax \right\Vert \leq \left\Vert A\right\Vert \left\Vert x\right\Vert ?$$
If $A$ is $m \times n$-matrix and $B$ is $n \times m$-matrix, is it correct that
   $$\left\Vert AB \right\Vert \leq \left\Vert A\right\Vert \left \Vert B \right\Vert?$$

Note: I'm mainly ask about norms like $\|\cdot\|_1$, $\|\cdot\|_2$, $\|\cdot\|_\infty$

For a vector norm $\|\cdot\|_\star$ (e.g. $\star=1,2,\infty$) one define an induced matrix norm by
$$\|A\|_\star=\max_{\|x\|_\star\ne 0}\frac{\|A x\|_\star}{\|x\|_\star}$$
for any finit dimentional matrix $A$.
it can be proved from that the following is equivalent definition
$$\|A\|_\star=\max_{\|x\|_\star=1}{\|A x\|_\star}$$

Let $x^*$ be a vector such that $\|x^*\|_\star=1$ and
$$\max_{\|x\|_\star=1}{\|A x\|_\star}=\|A x^*\|_\star,$$
then for any other $\|x\|_\star=1$ we have
$$\|A x\|_\star\leq\|A x^*\|_\star=\|A\|_\star=\|A\|_\star\|x\|_\star$$

Now $\|ABx\|_\star\leq \|A\|_\star\|Bx\|_\star\leq \|A\|_\star\|B\|_\star\|x\|_\star$ , therefore $$\|AB\|=\max_{\|x\|_\star=1}{\|A B x\|_\star}\leq
\max_{\|x\|_\star=1}{\|A\|\|B\|\|x\|_\star}=
\|A\|_\star\|B\|_\star$$

Note that nothing in the above required that the matrices needs to be square, so it works for any $n\times m$ matrix.

By the way, for the Frobeneous Norm which isn't an induced norm these properties still holds. I guess for a general non induced norm this may and may not work.

Shortly, yes, the inequality holds. In details...

Generally, let $A: V \to W$ be linear map between linear spaces $V$ with norm $\Vert \cdot \Vert_V$ and $W$ with norm $\Vert \cdot \Vert_W$. Then you can define operator norm (though the name may be different in different books) as
$$
\Vert A \Vert_{W,V} = \sup_{x  \neq 0 , \in V} \frac{ \Vert Ax \Vert_W }{ \Vert x \Vert_V }.
$$
From here you can easily prove assuming  all norms to be finite that for all $x\in V$
$$
\Vert Ax \Vert_W \leq \Vert A \Vert_{W,V} \cdot \Vert x\Vert_V
$$
and then prove that for $B:W \to V$
$$
\Vert AB \Vert_W \leq \Vert A \Vert_{W,V} \cdot \Vert B \Vert_{V,W}
$$
which is what you need.


Sum $\sum_{n=0}^{5}{n}{(n+1)}$ Since this series seems to be 0,2,6,12,20 and 30 there isn't a common difference that is identical between each term - it in itself seems to increase by 2 each time.

I can't use $S_n=\frac{n[2a+(n-1)d]}{2}$ since I don't know what to do about the common difference.

Answer is supposed to be 112.

$$\sum_{n=0}^{5}{n}{(n+1)}$$

You could use the formula
$$\sum_{n=0}^N n^2=\frac{N(N+1)(2N+1)}{6}$$
together with the AP formula to get
$$\sum_{n=0}^5n(n+1)=\sum_{n=0}^5n^2+\sum_{n=0}^5n=\frac{5\times6\times11}{6}+\frac{5\times6}{2}=55+15=70\ .$$
On the other hand, since you have just worked out all the terms and there are not many of them, you may as well just add them up...

The answer would be $112$ if the upper limit of summation was $6$ instead of $5$.

Another way to do this involves converting the sum to a sum of a telescoping series and a trivial series.  Consider

$$(n+1)^3-n^3 = 3 n^2+3 n+1 = 3 n (n+1)+1 \implies n (n+1) = \frac13 [(n+1)^3-n^3] - \frac13 $$

Thus,

$$\sum_{n=1}^{N} n (n+1) = \frac13 \sum_{n=1}^{N} [(n+1)^3-n^3] - \frac13 \sum_{n=1}^{N} 1 = \frac13 [(N+1)^3 - 1] - \frac13 N$$

You can simplify this to get

$$\sum_{n=1}^{N} n (n+1) = \frac13 (N^3 + 3 N^2 +2 N) = \frac13 N (N+1)(N+2)$$

Plug in $N=5$ to get your answer.


How can I find the common axis of 2 cones in space that have the same base radius but different heights? How do I find the 3D vector describing the axis of 2 overlapping cones, like this:

If I have only the following information:

Coordinates of the common tip
Coordinates of a point on the yellow cone's base
Coordinates of a point on the blue cone's base
The base radius (same for both)
The separation between the tip and both bases, along the common axis

Given that the axis can be an arbitrary vector.

Let $\mathbf{v}$ be a unit vector along the common axis of the cones, let $\mathbf{a}, \mathbf{b}$ be the vectors joining the vertex to the points on the yellow and blue cones' bases. Since we know the radii and the height of these cones, we can compute $\theta_1, \theta_2$ the angles between $\mathbf{v}, \mathbf{a}$ and $\mathbf{v}, \mathbf{b}$ respectively. So $\mathbf{v} \cdot \mathbf{a} = |\mathbf{a}|\cos{\theta_1} $, $\mathbf{v} \cdot \mathbf{b} = |\mathbf{b}|\cos{\theta_2}$ and $|\mathbf{v}| = 1$ - Three equations with three unknowns so we can solve for $\mathbf{v}$.


How to solve $\int_0^{2\pi} $ $(\int_1^\infty$ $\sin(s-t) \over t^3$ $dt$ $)ds$? I already tried to solve it myself and to type it into the calculator, but it gives out some weird stuff with an endless long solution.

Since $\left|\sin\right|\leq 1$, Fubini's theorem applies, hence:
$$ I=\int_{0}^{2\pi}\int_{1}^{+\infty}\frac{\sin(s-t)}{t^3}\,dt\,ds = \int_{1}^{+\infty}\frac{1}{t^3}\int_{0}^{2\pi}\sin(s-t)\,ds\,dt = 0. $$
But you don't even need Fubini's theorem to state the same:
$$ \int_{1}^{+\infty}\frac{\sin(s-t)}{t^3}\,dt = \sin(s)\int_{1}^{+\infty}\frac{\cos(t)}{t^3}\,dt - \cos(s)\int_{1}^{+\infty}\frac{\sin(t)}{t^3}\,dt $$
where both $\int_{1}^{+\infty}\frac{\cos(t)}{t^3}\,dt$ and $\int_{1}^{+\infty}\frac{\sin(t)}{t^3}\,dt$ are converging integrals, bounded by $\int_{1}^{+\infty}\frac{dt}{t^3}=\frac{1}{2}$ in absolute value. It follows that:
$$ I = \int_{0}^{2\pi}\left(A \sin(s)+B\cos(s)\right)\,ds = 0.$$

Zero is the correct result.

Switching the integrals you have

$$\int_1^{+\infty}\frac{\text{d}t}{t^3}\int_0^{+2\pi} \sin(s-t)\ \text{d}s = \int_1^{+\infty}\frac{\text{d}t}{t^3}\left(-\cos(s-t)\bigg|_{0}^{+2\pi}\right)$$

But

$$\left(-\cos(s-t)\bigg|_{0}^{+2\pi}\right) = 0$$

Hence

$$\int_1^{+\infty} \frac{\text{d}t}{t^3}\cdot 0 = 0$$


modular arithmetics - remainder I need to find the remainder of
${1011}^{10}+{10}^{11}$
when divided by $101$.

According to this website it is 55 but I fail to see how.

For 1011, we can write it as $1$ because $1011=10*101+1$ so ${1011}^{10}$ is $1$.

For ${10}^{11}$, I wrote the $11$ like $8+2+1$ or $2^3+2^1+2^0$

Then multiplying the results $10\cdot100\cdot1$ and taking the remainder of 101 I got 91.
$$1000=9\cdot101+91$$
So overall I have 92 for the answer, but the website says 55.
Was my method or understaning wrong or is there is a problem or limitation on large numbers in that calculator?

Also is there any website or mehtod to check myself with this kind of big numbers?

I don't think it's the web-site's fault.  The input on that page was 1.1156078e30.  You're not giving it the correct number $1011^{10} + 10^{11}$, which is 1115607835569227940375059334601, just the first $8$ decimal digits.  All the digits matter here!

Try codepad.org

python works with the big number correctly

http://codepad.org/mZ9HW4zW


Solving the differential equation $y'=(y^2-1)e^{ty}$ In a practice exam, we're asked to solve the differential equation $y'=(y^2-1)e^{ty}$, with $y(1)=0$.

However, the only techniques we've covered are separable differential equations, first order linear differential equations with integration factors or variation of parameters, and differential forms.

This looks like a nonlinear differential equation because I can't separate the $ty$ in the exponent. I wrote it as a form
$$
(1-y^2)e^{ty}dt+1dy=0
$$
but this is not exact, and I can't find an integration factor, since the usual formulas don't end up being expressions only in $t$ or $y$. It's not homogeneous either. I think it might be a typo since it's not something we covered, but is there a way to solve it I'm overlooking?

Hint:

Let $u=ty$ ,

Then $y=\dfrac{u}{t}$

$\dfrac{dy}{dt}=\dfrac{1}{t}\dfrac{du}{dt}-\dfrac{u}{t^2}$

$\therefore\dfrac{1}{t}\dfrac{du}{dt}-\dfrac{u}{t^2}=\left(\dfrac{u^2}{t^2}-1\right)e^u$ with $u(t=1)=0$

$\dfrac{1}{t}\dfrac{du}{dt}=\dfrac{u^2e^u+u}{t^2}-e^u$ with $u(t=1)=0$

$(u^2e^u+u-e^ut^2)\dfrac{dt}{du}=t$ with $t(u=0)=1$

Let $v=t^2$ ,

Then $\dfrac{dv}{du}=2t\dfrac{dt}{du}$

$\therefore\dfrac{u^2e^u+u-e^ut^2}{2t}\dfrac{dv}{du}=t$ with $v(u=0)=1$

$(u^2+ue^{-u}-t^2)\dfrac{dv}{du}=2e^{-u}t^2$ with $v(u=0)=1$

$(u^2+ue^{-u}-v)\dfrac{dv}{du}=2e^{-u}v$ with $v(u=0)=1$

Let $w=u^2+ue^{-u}-v$ ,

Then $v=u^2+ue^{-u}-w$

$\dfrac{dv}{du}=2u+(1-u)e^{-u}-\dfrac{dw}{du}$

$\therefore w\left(2u+(1-u)e^{-u}-\dfrac{dw}{du}\right)=2e^{-u}(u^2+ue^{-u}-w)$ with $w(u=0)=1$

$(2u+(1-u)e^{-u})w-w\dfrac{dw}{du}=2ue^{-u}(u+e^{-u})-2e^{-u}w$ with $w(u=0)=1$

$w\dfrac{dw}{du}=(2u+(3-u)e^{-u})w-2ue^{-u}(u+e^{-u})$ with $w(u=0)=1$

This belongs to an Abel equation of the second kind.

In fact all Abel equation of the second kind can be transformed into Abel equation of the first kind.

Let $w=\dfrac{1}{z}$ ,

Then $\dfrac{dw}{du}=-\dfrac{1}{z^2}\dfrac{dz}{du}$

$\therefore-\dfrac{1}{z^3}\dfrac{dz}{du}=\dfrac{2u+(3-u)e^{-u}}{z}-2ue^{-u}(u+e^{-u})$ with $z(u=0)=1$

$\dfrac{dz}{du}=2ue^{-u}(u+e^{-u})z^3+((u-3)e^{-u}-2u)z^2$ with $z(u=0)=1$

Please follow the method in http://www.hindawi.com/journals/ijmms/2011/387429/#sec2


Why do we define $e^x$ as the $\lim_{x \to \infty} (1 + \frac{x}{n})^n$? I am making my way through an analysis textbook and we have defined $e$ as the  $\lim_{n \to \infty} (1 + \frac{1}{n})^n$. We also defined $e^x$ as  $\lim_{n \to \infty} (1 + \frac{x}{n})^n$.

I am quite uncertain as to where this exponentiation definition comes from. For example, in the book we have shown how $\sqrt{2}$ can be represented as an infinite decimal. The method we used was to take increasing decimal expansions that were just under $2$, adding one decimal at a time. So for example:

$$1^2 < 2 < 2^2$$
$$1.4^2 < 2 < 1.5^2$$
$$1.41^2 < 2 < 1.42^2$$

and so on. We then define $\sqrt{2}$ to be the supremum of the set of improving approximations of $\sqrt{2}$, sup $\{1, 1.4^2, 1.41^2, 1.414^2,...\}$ and we showed that $2$ is indeed the least upper bound.

So it seems to me that we have already established a way to take irrational numbers to integer powers. So now my question is, why not just do the same for $e^x$? I don't understand why we introduce the limit definition to define exponentiation. Once we are given the definition of $e$, which is just a constant, why can't we use the same procedure as described above to define exponents of $e$? (By same procedure I mean taking increasingly accurate finite decimal approximations of $e$ and multiplying them together $x$ times).

For example, my confusion is along these lines: Say we have defined exponentiation of natural numbers to natural numbers. We define it in the usual intuitive sense, so $3^5 = 3\cdot3\cdot3\cdot3\cdot3$. Then all of a sudden we say introduce a limit to define $7^8$, instead of using the established exponentiation.

I know that for any (integer at least) value of $x$, $\lim_{n \to \infty} (1 + \frac{x}{n})^n$ converges to the same value that we would get if using the other method, so obviously the definition is consistent. However, where exactly does it come from? Do we use it simply because it 'works', and converges to $e^x$ had we used the other method?

Note: I know a caveat to the other method presented here is that in the described form, it can only be used for integer powers. In my book it does say this can be generalised to real powers, but using $e^x$, which doesn't give me much insight.

The provided digit-by-digit process you've provided for computing $\sqrt2$ has several drawbacks.

It only allows you to define $a^b$ for rational $b$. To get irrational $b$, you would need to go through the whole process again within the exponent.
It is hard to work with. It doesn't directly give you what $a^b$ is to work with.
It does not obviously have desired exponentiation properties. Does $(a^b)^c=a^{bc}$ and $a^b\cdot a^c=a^{b+c}$? Is it monotone? Continuous? Differentiable?

By playing around with the limit, we can see that this should intuitively give us exponentiation as we know it:

$$\lim_{n\to\infty}\left(1+\frac xn\right)^n\stackrel?=\lim_{n\to\infty}\left[\left(1+\frac xn\right)^{n/x}\right]^x\stackrel?=\lim_{n\to\infty}\left[\left(1+\frac1n\right)^n\right]^x=e^x$$

Of course this is not rigorous seeing as we don't have exponentiation to work with yet, as it is what we are trying to define, but this should help your intuition. It remains to be shown that the listed properties are easily shown, and from them we can prove exponentiation as you know it works.

It's hard to say exactly what you want clarification about. There are at least two things your why may mean (not mutually exclusive, recognise):

(1) What originally motivated this definition of $e^x$? Well, the short answer to this strand is, the problem of continuous compounding of interest. This was originally investigated by one of the Bernoullis -- I've forgotten precisely which at the moment.

(2) Or perhaps you meant to ask why this definition is equivalent to other definitions of $e^x.$ Well, again, a short answer is that all these different definitions satisfy properties that everyone agrees the function $e^x$ should have.

But another possible source of confusion is apparent from your explanation:

  So it seems to me that we have already established a way to take irrational numbers to integer powers. So now my question is, why not just do the same for $e^x$? I don't understand why we introduce the limit definition to define exponentiation. Once we are given the definition of $e,$ which is just a constant, why can't we use the same procedure as described above to define exponents of $e$? (By same procedure I mean taking increasingly accurate finite decimal approximations of $e$ and multiplying them together $x$ times).

Well, the short answer to this would be that in the definition you're asking about, $x$ can be any real number whatsoever, not just positive integers as you're suggesting.


Boolean algebra: can it be simplified even further? Can it be simplified even further?

$$\overline{b_2} \overline{b_3} + b_0 \overline{b_1} b_2 + \overline{b_0} \overline{b_1} \overline{b_2} $$


Computing marginal PMF The problem is as follows:

Let  N  be a positive integer random variable with PMF of the form

$$P_N(n)=\frac{1}2*n*2^{-n},n=1,2,…. $$

Once we determine the numerical value of  N , we then draw a random variable  K  whose (conditional) PMF is uniform on the set  {1,2,…,2n}.

The joint PMF is  $$P(K=k,N=n)= \begin{cases}\frac{1}{4*2^{-n}} & n \in \{1,2, ...\},k\in \{1,2,...,2n\}\\
0 & \text{ otherwise }
\end{cases}$$

Now I would like to find the marginal PMF $P_K(k)$ as a function of $k$.We only need to answer only for the case when  $k$  is an even number. I use simple example to draw out the conditional PMF $ P(K = k | N = n)$ table.

$$\begin{array}{c|c|c|}
 & \text{k=1} & \text{k=2} \\ \hline
\text{n=1} & 1/2 & 1/2 \\ \hline
\end{array}$$

But having trouble understanding how to compute the marginal PMF function. Any guidance is appreciated!

Assume $k$ is even.

\begin{align*}
P(K = k) &= \sum_{n = 1}^\infty P(N = n, K = k)\\
&= \sum_{n = 1}^{\frac{k}{2} - 1} P(N = n, K = k) + \sum_{n=\frac{k}{2}}^\infty P(N = n, K = k)\\
&= \sum_{n = 1}^{\frac{k}{2} - 1} 0 + \sum_{n=\frac{k}{2}}^\infty \frac{1}{4 * 2^n}\\
&= \sum_{n=\frac{k}{2}}^\infty \frac{1}{4 * 2^n}\\
&= \frac{1}{2^{\frac{k}{2}+1}}
\end{align*}

Why is $P(N = n, K = k) = 0$ for $1 \leq n \leq \frac{k}{2} - 1$?

We know that $P(N = n, K = k) = 0$ for $k \geq 2n+1$ which is just $n \leq \frac{k}{2} - \frac{1}{2}$.

But if we assume $k$ is even and use the fact that $n$ is an integer, $n \leq \frac{k}{2} - \frac{1}{2}$ $\implies $ $n \leq \frac{k}{2} - 1$.

How about making a table in which you have the joint distribution and then you check, what is the marginal probability for $k=2, k=4$, and the like.
For $k=2$, the marginal probability should be $1/4$
For $k=4$, the marginal probability should be $1/8$
...
This can be expressed in terms of $k$ as:
$$\dfrac  1 { 2^{1 + k/2}}$$


Why other definitions of convergence fail to be correct? The following is an exercise from the book Advanced Calculus:

Well obviously the b. is not correct since for $\epsilon= \frac19$, ${\{a_n}\}={\{\frac1n}\}$ and $N=2$ it fails to be correct. But the a. and the c. are exactly as same as the accepted definition of convergence which is given in the book, i.e.

  Definition. A sequence ${\{a_n}\}$ is said to converge to the number a provided that for every positive number $\epsilon$ there is an index $N$ such that $|a_n - a| < \epsilon$ for all indices $n \ge N$.

In fact, the a. and the c. are just re-phrasing the above mentioned definition.

Please help!

No, a and c are not the same! Those changes you call "rephrasings" totally change the meaning - that's the point to the exercise.

Let $\epsilon=3$. Let $N=1$. Then $|1/n-1| < \epsilon$ for every $n>N$.

So it is true that for some $\epsilon>0$ there is $N$ so $|1/n-1|<\epsilon$ for every $n>N$. Which is what a says - if a is the definition of limit we've just proved that $$\lim_{n\to\infty}\frac1n=1.$$

And if c were the same as the actual definition that it would not be true that $\lim 1/n = 0$. Because c does not hold here - it is not true that there is an $N$ such that for every $\epsilon>0$ we have $|1/n-0|<\epsilon$ for every $n>N$. Given $N$, let $\epsilon=N/3$ and $n=2N$. Then $n>N$ but $|1/n-0|>\epsilon$.

For $a$, the problem is the phrase for all indicies. Choose small number $\epsilon$ and go to work. For $c$, the problem is the phrase There is an index N. If you fix $N$ in advance and then choose epsilon later, you will run into a problem. So given $N$, how do you choose $\epsilon$ to make your counterexample $a_n = \frac{1}{n}$ work?


The longest string of none consecutive repeated pattern Construct a string by using only $n$ characters, the string should do not contain any consecutive repeated pattern, in other words, the string must have not any substring matching /(.+)\1/, what is the length l(n) of the longest string S(n)?

n    S(n)    l(n)
1    a       1
2    aba     3
3    ?       ?
4    ?       ?

Edit:
abcabacbabcabacabcac has none consecutive  repeated pattern,so l(3)>=20

A theorem of Thue's (not Thue's theorem!) states that for $n = 3$ (and therefore, also for larger $n$) an infinite sequence of characters (chosen out of $n$) without repeated sub-sequences (a.k.a square-free) exists.

Edit: here's a slightly related reference: J.D. Currie's paper 'There are ternary circular square-free words of length
n for n  18.', which at least has a reference to Thue's paper from 1906.

Edit 2: Another reference, with explicit construction of sequences of length 3, 9, 27 and so on: http://arxiv.org/abs/0712.0139.
That paper gives the following sequence of length 27: 123213231213123132312132123

To find a maximal string for n letters, add a new letter in every other place of a maximal string for n-1 letters.  It is still the case that no consecutive identical substrings exist.  No more letters can be added because two of them would have to appear consecutively, so this is the best that is possible.  The length of the string is 2^n-1.

a
bab
cbcacbc
dcdbdcdadcdbdcd


Does this $a_n=(1-\frac{1}{2!})^{(\frac{1}{2!}-\frac{1}{3!})^{\ldots^{(\frac{1}{n!}-\frac{1}{(n+1)!})}}}$ have a finit limit? My question here is related to telescopic sum using factorial and it is related to my question here, I have computed some values of $a_n=(1-\frac{1}{2!})^{(\frac{1}{2!}-\frac{1}{3!})^{\ldots^{(\frac{1}{n!}-\frac{1}{(n+1)!})}}}$ for odd parity and even parity but it is not fixed for example for $n=2$ we have $0.793700$ and it decreases for $n=4$ to $0.77982$, now for $n=3$ we have $0.5465$ and it increases for $n=5$ to $0.54876$ , it seems increasing for odd parity and decreasing for even parity iteration. Now I have looked to all given answers here but I can't juge whether that sequence converges or not by means it has a limit or not?

    My question here is:
    Is this $a_n=(1-\frac{1}{2!})^{(\frac{1}{2!}-\frac{1}{3!})^{\ldots^{(\frac{1}{n!}-\frac{1}{(n+1)!})}}}$ have a finit limit ?

Note The motivation of this question is looking to the behavior of the Gamma function in the power telescoping sum.

We have similar behavior to the sequence that you linked to, except the limiting values are different:  For even $n$, it is $$a_n \to 0.77954333600168773503298455024204190801488463615921\ldots,$$ and for odd $n$, it is $$a_n \to 0.54877354704085687513069922740691455562600046738030\ldots.$$  The number of correct decimal places increases slightly faster than quadratically in $n$; i.e., if $\epsilon(n)$ is the absolute error as a function of $n$, then $$-\log \epsilon(n) \sim O(n^{k}),$$ where I estimate $k$ to be approximately $2.3$, certainly greater than $2$ but less than $2.4$.  I do not recognize these constants as having a closed form.

I don't think this really counts as a proof, but I made some code in Matlab code to investigate this.

n=input('please input a value for n ');

for m=1:n;

for I=1:m;

    b(m+1)=1;

    b(m-I+1)=(1/factorial(m-I+1)-1/factorial(m-I+2))^b(m-I+2);

    a(m)=b(1);

end

end

for I=1:n;

fprintf('a(%.f)=%.8f \n',I,a(I));

end

Inputting n=12 outputs
a(1)=0.50000000
a(2)=0.79370053
a(3)=0.54650798
a(4)=0.77982290
a(5)=0.54876028
a(6)=0.77954370
a(7)=0.54877354
a(8)=0.77954334
a(9)=0.54877355
a(10)=0.77954334
a(11)=0.54877355
a(12)=0.77954334
So while it seems that $a_{2n}$ and $a_{2n-1}$ each converge, they don't converge to the sam value.

The thing is that as n tends to infinity $(\frac{1}{(n)!}-\frac{1}{(n+1)!})$ rapidly tends to 0, so $(\frac{1}{(m-n)!}-\frac{1}{(m+1-n)!})^{(\frac{1}{(n+1)!}-\frac{1}{(n+2)!})}$ quickly tends to 1.


Extensions of a field by a root of a monic irreducible polynomial Let $K$ be a field and $f(x) \in K[x]$ a monic irreducible polynomial of degree $n$ and consider $\frac{K[X]}{(f(x))}$.

Is $\frac{K[X]}{(f(x))} \cong K(a)$, $\forall a \in K$ st. $f(a)=0$ and $[K(a):K] = n$?

Let $a$ be a root of $f$. Since $f$ is irreducible, it is the minimal polynomial of $a$ (the polynomial of minimum degree with coefficients in $K$ having $a$ as a root). So no nonzero linear combination of $1, a, a^2, \dots a^{n-1}$ can equal $0$, so $K(a)$ has $K$-basis $1, a, a^2, \dots a^{n-1}$. Thus $[K(a) : K] = n$.

Similarly $K[X]/(f(X))$ is a vector space with basis $1, X, \dots X^{n-1}$ over $K$. The natural map $K[X]/(f(X)) \to K(a)$ sending $X^i \to a^i$ is therefore an isomorphism.


three conjectures in statistics of iid variates Consider $N$ different i.i.d. variates $x_i$ (for $1\leq i\leq N$). Let $\hat{x}$'s be the ordered variates, in other words $\hat{x}_j$ is the $j$th largest variate.

The density function for $\hat{x}_n$ may be computed as
$$\hat{f}_n(\alpha)=\frac{N!}{(n-1)!(N-n)!}F^{N-n}(\alpha)\Big(1-F(\alpha)\Big)^{n-1}f(\alpha)$$
in obvious notations.

What may be said about the following propositions?

The limit
$$\lim_{N\rightarrow\infty}Nf(\langle\hat{x}_1\rangle)$$
is finite (and nonzero).
$$\text{Var}(\hat{x}_1)\geq\text{Var}(\hat{x}_2)\geq\text{Var}(\hat{x}_3)\geq\cdots\geq\text{Var}(\hat{x}_{\lfloor\frac{N}{2}\rfloor})$$
$$\lim_{N\rightarrow\infty}\hat{f}_{\lfloor rN\rfloor}=\delta(.-F^{-1}(r))\hspace{5mm}\forall r\in(0,1)$$

P.S. You may (if you want) further assume that $f$ is nonzero on $(0,+\infty)$ and is monotonously decreasing.

A counterexample to the first propositionis afforded by a continuous uniform distribution, where $f$ is constant and thus the expression under the limit is proportional to $N$.

A counterexample to the second proposition is affored by a discrete uniform distribution on three values, where for large $N$ the variance of the median goes to zero (since the median is almost certainly the middle value) whereas the variance of the tertiles remains finite in the limit (since they have finite probabilities of being the middle value or one of the outer values).

The third proposition doesn't seems implausible on the face of it, as the left-hand side is independent of $r$ and the left-hand side doesn't seem to be.


Proof of calculating MinHash I'm reading about MinHash technique to estimate the similarity between 2 sets: Given set A and B, h is the hash function and $h_\min(S)$ is the minimum hash of set S, i.e. $h_\min(S) = \min(h(s))$ for s in S. We have the equation:
$$
p(h_\min(A) = h_\min(B)) = \frac{|A \cap B|}{|A \cup B|}
$$
Which means the probability that minimum hash of A equals to minimum hash of B is the Jaccard similarity of $A$ and $B$.

I am trying to prove above equation and come up with a proof, which is:
for $a \in A$ and $b \in B$ such that $h(a) = h_\min(A)$ and $h(b) = h_\min(B)$. So, if $h_\min(A) = h_\min(B)$ then $h(a) = h(b)$. Assume that hash function h can hash keys to distinct hash value, so $h(a) = h(b)$ if and only if $a = b$, which is $\frac{|A \cap B|}{|A \cup B|}$. However, my proof is not complete since hash function can return the same value for different keys. So, I'm asking for your help to find a proof which can be applied regardless the hash function. Thanks.

L

It seems like it is an underlying assumption in this formula that different elements map to distinct hashes. Otherwise, just have your hash function map everyone to 0. The probability you are looking at would be 1 regardless of any sets.

There is a remarkable connection between minhashing and Jaccard similarity
of the sets that are minhashed.

• The probability that the minhash function for a random permutation of
rows produces the same value for two sets equals the Jaccard similarity
of those sets.

To see why, we need to picture the columns for those two sets. If we restrict
ourselves to the columns for sets S1 and S2, then rows can be divided into three
classes:

Type X rows have 1 in both columns.
Type Y rows have 1 in one of the columns and 0 in the other.
Type Z rows have 0 in both columns.

Since the matrix is sparse, most rows are of type Z. However, it is the ratio
of the numbers of type X and type Y rows that determine both SIM(S1, S2)
and the probability that h(S1) = h(S2). Let there be x rows of type X and y
rows of type Y . Then SIM(S1, S2) = x/(x + y). The reason is that x is the size
of S1 ∩ S2 and x + y is the size of S1 ∪ S2.

Now, consider the probability that h(S1) = h(S2). If we imagine the rows
permuted randomly, and we proceed from the top, the probability that we shall
meet a type X row before we meet a type Y row is x/(x + y). But if the
first row from the top other than type Z rows is a type X row, then surely
h(S1) = h(S2). On the other hand, if the first row other than a type Z row
that we meet is a type Y row, then the set with a 1 gets that row as its minhash
value. However the set with a 0 in that row surely gets some row further down
the permuted list. Thus, we know h(S1) 6= h(S2) if we first meet a type Y row.
We conclude the probability that h(S1) = h(S2) is x/(x + y), which is also the
Jaccard similarity of S1 and S2. ( from Mining of Massive Datasets book )


How can I compare two vectors if their inner product is zero? I have the following:

$\vec{w} = \begin{bmatrix}1 \\ \\ 2\end{bmatrix},$ where $\vec{w}$ is a position vector.

If I do a vector multiplication with $\vec{w}$ and two coordinates $x_1$ and $x_2$,

$\begin{bmatrix}1 \\ \\ 2\end{bmatrix}^T \begin{bmatrix}x_1 \\ \\ x_2\end{bmatrix} = 3$, where $3$ is the intercept.

The above matrix equation becomes

$x_1 + 2x_2 - 3 = 0$

or

$x_2 = -\dfrac{1}{2}x_1 + \dfrac{3}{2}$

The graphical presentation of the relationship between $\vec{w}$ and the above equation $x_2 = -\dfrac{1}{2}x_1 + \dfrac{3}{2}$ becomes perpendicular as follows:

But how can I show the equation and $\vec{w}$ is perpendicular?

The line $OW$ has its equation $x_2=2x_1$ with slope $2$ and the other line has slope $\dfrac{-1}{2}$ so their product is $-1$.

One way is to calculate the $w$ which is closest from origo. All points $w$ closest to a differentiable manifold will be perpendicular to the surface tangent space (and parallell to the normal) thanks to a result in differential geometry. A line is maybe the simplest differentiable manifold that exists.

So we can solve $$w_O = \min_w \{\|w\|_2 + \|w_1+2w_2-3\|_2\}$$

Which is a linear least squares problem.


Properties of matrix equation CA=C I'm interested in the properties of the matric equation $CA=C$, where $C$ is an $m\times n$ matrix with real entries, and $A$ is a square $n\times n$ matrix with real entries, with $m\leq n$. What can I say about $A$, when $C$ is assigned? I know the following holds true:

Let $A=C^*C$, where $C^*$ is any generalized inverse of $C$. Then
the system above always admits at least one solution.
Whichever $C$, $A$ must be an orthogonal projection mapping the
rows of $C$ into themselves.

As for (1), can I go the other way round, i.e. $CA=C$ implies that $A$ be some linear function of (one of the) generalized inverses of $C$?

As for (2), does it imply that $A$ be symmetric (since real) and idempotent?

Thanks

See solving matrix equations $AX=B$ in

F.Gantmacher. Theory of matrices, AMS Chelsea publishing, 1959.

In fact you can only solve this question by this approach:

$CA=C$

$CA-C=0$

$C(A-I)=0$

Let $B=A-I$ ,

Then $CB=0$

Suppose $C=\left(\begin{array}{rrrr}c_{11}&c_{12}&\cdots&c_{1n}\\c_{21}&c_{22}&\cdots&c_{2n}\\\vdots&\vdots&\ddots&\vdots\\c_{m1}&c_{m2}&\cdots&c_{mn}\end{array}\right)$ and let $B=\left(\begin{array}{rrrr}b_{11}&b_{12}&\cdots&b_{1n}\\b_{21}&b_{22}&\cdots&b_{2n}\\\vdots&\vdots&\ddots&\vdots\\b_{n1}&b_{n2}&\cdots&b_{nn}\end{array}\right)$ ,

Then $\left(\begin{array}{rrrr}c_{11}&c_{12}&\cdots&c_{1n}\\c_{21}&c_{22}&\cdots&c_{2n}\\\vdots&\vdots&\ddots&\vdots\\c_{m1}&c_{m2}&\cdots&c_{mn}\end{array}\right)\left(\begin{array}{rrrr}b_{11}&b_{12}&\cdots&b_{1n}\\b_{21}&b_{22}&\cdots&b_{2n}\\\vdots&\vdots&\ddots&\vdots\\b_{n1}&b_{n2}&\cdots&b_{nn}\end{array}\right)=\left(\begin{array}{rrrr}0&0&\cdots&0\\0&0&\cdots&0\\\vdots&\vdots&\ddots&\vdots\\0&0&\cdots&0\end{array}\right)$

$\left(\begin{array}{rrrr}\sum_{k=1}^nc_{1k}b_{k1}&\sum_{k=1}^nc_{1k}b_{k2}&\cdots&\sum_{k=1}^nc_{1k}b_{kn}\\\sum_{k=1}^nc_{2k}b_{k1}&\sum_{k=1}^nc_{2k}b_{k2}&\cdots&\sum_{k=1}^nc_{2k}b_{kn}\\\vdots&\vdots&\ddots&\vdots\\\sum_{k=1}^nc_{mk}b_{k1}&\sum_{k=1}^nc_{mk}b_{k2}&\cdots&\sum_{k=1}^nc_{mk}b_{kn}\end{array}\right)=\left(\begin{array}{rrrr}0&0&\cdots&0\\0&0&\cdots&0\\\vdots&\vdots&\ddots&\vdots\\0&0&\cdots&0\end{array}\right)$

$\therefore\begin{cases}\sum_{k=1}^nc_{1k}b_{k1}=0\\\sum_{k=1}^nc_{2k}b_{k1}=0\\\vdots\\\sum_{k=1}^nc_{mk}b_{k1}=0\\\sum_{k=1}^nc_{1k}b_{k2}=0\\\sum_{k=1}^nc_{2k}b_{k2}=0\\\vdots\\\sum_{k=1}^nc_{mk}b_{k2}=0\\\vdots\\\sum_{k=1}^nc_{1k}b_{kn}=0\\\sum_{k=1}^nc_{2k}b_{kn}=0\\\vdots\\\sum_{k=1}^nc_{mk}b_{kn}=0\end{cases}$

You should handle this system of linear equations.


Find a (16,6,1) Balanced incomplete block design (BIBD) I'm trying to find a balanced incomplete block design with the 16 items and $\lambda= 1$. I've calculated (using these defenitions)  that a BIBD with 8 blocks and 6 items per block should be possible. However, when trying to contruct this with r, it only finds not-balanced sets.

A balanced set with $\lambda=2$ (and 16 blocks) can be found for example, but as the goal is to let participants rank sets of items, the target is to keep the effort to a minimum.

My question: is such a BIBD possible, and if yes, how do I construct a valid one?

Thanks in advance,
Simon

Such a design cannot exist for it will violate Fisher's inequality.

For those interested, it's not possible. The first possibility with 16,6 is a lambda of 2.


For groups elements $a$ and $b$, how do you prove that $(a^{-1}ba)^n = a^{-1}b^na$? For any elements $a$ and $b$ from a group and any integer $n$, prove that $(a^{-1}ba)^n = a^{-1}b^na$.

Suppose $P(n):$ $(a^{-1}ba)^n = a^{-1}b^na$.

$1.$ $n=1$ $(a^{-1}ba)^1 =a^{-1}ba =a^{-1}b^1a$

$2.$ Suppose $P(n)$ is true we want to show $P(n+1)$ is true.

$(a^{-1}ba)^{n+1}=(a^{-1}ba)^{n}(a^{-1}ba)=(a^{-1}b^{n}a)(a^{-1}ba)$

Then using associative and inverse property this reduces to
$a^{-1}b^{n+1}a$

Thus $P(n+1)$ is true, therefore $P(n)$ is true for every natural number.

I'm not sure how to show it for the negative numbers or could I just use this result to conclude it holds for the negative numbers also.

I suggest that you prove
(for $n\geq 1$)
$(a^{-1} ba)^n (a^{-1} b^{-n} a)$
is equal to the neutral element of the group,
thus calculating the inverse of $(a^{-1} ba)^n$.

This way, you dont need to do another induction.

Induction is overkill. Just write it down:

$$
(a^{-1}ba)^n =
\underbrace{(a^{-1}ba)(a^{-1}ba)\dotsb(a^{-1}ba)}_{n \text{ copies}} =
\underbrace{a^{-1}baa^{-1}ba\dotsb a^{-1}ba}_{n \text{ copies}} =
a^{-1}\underbrace{bb\dotsb b}_{n \text{ copies}}a =
a^{-1}b^na
$$


Upper bound for $\frac{\sum_{i=0}^{k} \binom{n-2}{i}}{\sum_{i=0}^{k} \binom{n}{i}}$ How to simplify $P = \frac{\sum_{i=0}^{k} \binom{n-2}{i}}{\sum_{i=0}^{k} \binom{n}{i}}$ to get an upper bound in terms of $n$ and $k$. Here $k \le n$ and $\binom{n}{r}$ is the binomial coefficient that denotes the number of ways to choose $r$ objects from a set of $n$ objects.

$P$ is the probability that 2 fixed objects $u,v \in S$ where $S$ is a set of $n$ distinct objects, are both not present in a random subset $C$ of $S$ where $|C| \le k$ so trivially $P \le 1$

Interpretation:

Numerator = Number of subsets of size $\le k$ without $u,v$

Denominator = All possible subsets with size $\le k$

Let $F_n$ and $f_n$ be respectively CDF and pmf of $B(n,\frac 1 2)$. Then $$P=\frac 1 4\frac {F_{n-2}(k)}{F_n(k)}=\frac 1 4\frac {F_{n-2}(k)}{F_{n-2}(k)-\frac 3 4f_{n-2}(k)-\frac 1 4f_{n-2}(k-1)}$$ From CLT we get that for $k\ge \frac n 2-O(n^{\frac 1 2})$ $$P= \frac 1 4+O(n^{-\frac 1 2})\tag 1$$
For $k=o(n)$, we easily get $P\to 1$
and if $k= an$ for $0<a<\frac 1 2$, let $r=\frac{a}{1-a}$ to get $$P\to \frac 1 {(1+r)^2}=(1-a)^2\tag 2$$

which agrees with the other two regimes. This result is very easy to explain - when $k=an<\frac n 2$, most of the sets in consideration are very close in size to $k$ and chances that two elements don't belong to a specific such set are $(1-a)^2$.


How to integrate $\int_{-\pi}^\pi|\cos(x)|dx$ I need to calculate the average value $\mu$ with the formula:
$$
\mu = \frac{1}{b-a} \int_a^bf(x)\,dx
$$
in my case:
$$
\mu = \dfrac{1}{2\pi}\int_{-\pi}^\pi |\cos(x)|\,dx =\dfrac{1}{\pi}\int_{0}^\pi |\cos(x)|\,dx
$$
but the problem is that I don't know how to integrate the absolute value of $\cos$: '$|\cos(x)|$'
How do I integrate it?

Something to consider is that between $0$ and $\pi/2$, $cos$ is positive, and so $|cos| = cos$. And then between $\pi/2$ and $\pi$, $cos$ is negative, so $|cos| = -cos$.

All you need to do at that point is split your integral up:

$$
\int_0^\pi |\cos(x)| \,dx = \int_0^{\pi/2} |\cos(x)|\, dx + \int_{\pi/2}^\pi |\cos(x)| \,dx = \int_0^{\pi/2} \cos(x) \,dx - \int_{\pi/2}^\pi \cos(x)\, dx.
$$

You can go a step further to solving then by noting that $\int_{\pi/2}^\pi \cos(x)\, dx = -\int_0^{\pi/2} \cos(x)\, dx$, so all in all your integral resolves down to:

$$
\int_0^\pi |cos(x)| \,dx = 2\int_0^{\pi/2} cos(x) \,dx.
$$

But that last bit is extra to your original question :) Long story short, in the general case, when looking at integrals of absolute values you need to break it into the separate cases:

$$
\int_x |f(x)| = \int_{x: f(x) \geq 0} f(x) - \int_{x: f(x) < 0} f(x)
$$

Hope that helps you!


Is there a short name for series which satisfy the hypothesis of the alternating series test? The alternating series test says that an alternating series whose terms are monotonically decreasing in absolute value and which approach zero in the limit converges.

Of course this is not an if-and-only-if statement -- most convergent alternating series do not have monotonically decreasing terms -- but if $a_n$ is an alternating series satisfying the hypothesis of the test, is there any term shorter than "alternating series with terms monotonically decreasing in absolute value which approach zero in the limit?"

For the past decade or so I thought that there was such a term and that the term was "telescoping series."  This would seem to make sense, because the partial sums of such a series telescope in on themselves.  Apparently, however, "telescoping series" has something to do with the formulas for the partial sums, and not for the partial sums themselves.

Typically we refer to an alternating series that converges as "conditionally-convergent". Of course if the series converges absolutely then it is also absolutely convergent. Naturally, there are some of the former that aren't the latter.

Telescoping refers to when the partial sum terms all cancel except for perhaps a couple of the "end-terms". For example $\sum(\frac{1}{n}-\frac{1}{n+1})$ is telescoping.


How do I adjust model parameters to get stable oscillations with Arditi-Ginzburg equations? I am using the predator-prey equations from the Arditi-Ginzburg to model the interactions of 14 species in a video game.  The equations for two species are:
$$
dx=x(A-Bx)-Cxy/(x+y)\\
dy=-Gy+Mxy/(x+y)
$$
with careful selection of parameters A,B,C,G,M I get the expected out-of-phase oscillations between the two species, but there is a damping factor.  See the plot.

A,B: controls the growth of the prey w/o predator

C controls how fast the predator consumes the prey

G controls how fast the predator dies w/o prey

M controls how fast the predator grows with maximum prey

How can I adjust the parameters to make the oscillations continue indefinitely?  Stable systems are very common, and damped oscillations are common.  Where are the infinite oscillations?

P.S. Most choices for parameters quickly result in a stable system.. why?

Edit: With lots of tweaking, here are repeating cycles, with growing amplitude, see plot.

Increasing parameter A from 0.21 to 0.23 and then increasing parameter C from 0.7 to 0.82 made the difference.  The slightly increasing amplitude can be adjusted by increasing Parameter G very slightly.

Now the peaks are fine, but the valleys are too low and too long.  I am using LibreOffice to synthesize and visualize this.  Can you guys recommend better open source tools for hacking functions?

UPDATE

I added one more predator to the mix.

$$
dx=x(A-Bx)-Cxy/(x+y)\\
dy=-Gy+Mxy/(x+y)-C1yz/(y+z)\\
dz=-G1y+M1yz/(y+z)
$$
dx= change in grass population

dy= change in mice population

dz= change in fox population

Starting Parameters:

X=1000 Y=1000 Z=100
A= 0.23 B=0.00001 C=0.82 G=0.53 M=0.71 C1=0.42 G1=0.4 M1=0.71

The plot shows the basic out-of-phase relationships are starting to emerge, and the populations are headed towards a steady state.  A little more tweaking should get these three into steady oscillations, and then I will continue adding to the food chain...


Relation between rad(G) , Δ(G) and |G| in a graph Is there a formula for resolving the following question : "Given a graph G with rad(G) = 3 and Δ(G) = 5, can we have |G| = 2006?"

rad(G) is  is the minimum graph eccentricity of any vertex in G.(The eccentricity is the maximum graph between v and any other vertex u of G)

Δ(G) is the maximum degree in G(a degree of a vertex is the nr. of edges at it)

|G| is the nr. of vertices

I was thinking using the theorem "The number of vertices of odd degree in any graph G is even". By this, it can be |G| = 2006 if Δ(G) = 5, but how can I use the rad(G) as well?

Note that in a graph of maximum degree $d$, for a fixed vertex $v$ there are

at most $d$ vertexes at distance $1$ from $v$,
at most $d(d-1)$ vertexes at distance $2$ from $v$,
at most $d(d-1)^2$ vertexes at distance $3$ from $v$,
and so on.

Take $v$ a vertex that is a "center" so has distance at most $3$ to each other vertex and do the calculation. (Do not forget $v$ itself.)

If it is radius = 3  and diameter = 5 according to

http://www.sciencedirect.com/science/article/pii/0012365X73901167

there is no reason why such a graph could not be constructed.


Does a linear map of normed vector spaces have a unique extension? Let $ V$
  and $ W$
  be finite dimensional normed vector spaces and let $ U$
  be an open subset of $ V$
  (the topology induced by the norms). Let $ L:U\rightarrow W$
  be a linear map. Does there exist a unique linear map $ T:V\rightarrow W$
  such that the restriction of $  T$
  to $ U$
  is $ L$
 ? If this is not true, then are there any additional conditions which can be imposed on $ V$
  and $ W$
  such that the extension exists? In particular, I am interested in the case where $ V$
  and $ W$
  are Euclidean.

The domain $U$ of the linear map $L$ shoud be a subspace of $V$ !!
If $U$ is a non- empty open subset of $V$ and is a subspace of $V$, then we have $U=V$ ! Try a proof