$I(n)=∫_1^n(1+2^{-[x] })^x dx $ Prove: $n-11$ Let
$ I(n)=∫_1^n(1+2^{-[x] })^x dx $ Prove: $n-1<I(n)<n+1,nϵN,n>1$.

    To solve this problem I broke the iintegral and find each integral,looking for q better solu

$\newcommand{\d}[1]{\text{ d}#1}$
Before I present my solutions, I would first highlight that your inequality on the left is false for $n = 2$, as:
$$
I(2) = \int_1^2 (1 + 2^{-\lfloor{x}\rfloor})^x \d{x} = \int_1^2 \left(1 + \frac{1}{2}\right)^x \d{x} \approx 1.84973 < 2
$$
However, it is true that $n - 1 < I(n) < n + 1$.

This is probably what you've done so far:
$$
I(n) = \int_1^n (1 + 2^{-\lfloor{x}\rfloor})^x \d{x} = \sum_{k=1}^{n-1} \int_{k}^{k+1} (1 + 2^{-\lfloor{x}\rfloor})^x \d{x} = \sum_{k=1}^{n-1}\int_{k}^{k+1} (1 + 2^{-k})^x \d{x}
$$
Observe that for any $a > 1$, $a^x$ is strictly convex. For any convex function $f : [c,d] \to \mathbb{R}$,  we have $f$ lying below the line joining $(c, f(c))$ and $(d, f(d))$. This means that the area under $f$ is $\leq$ the area of the trapezium with the vertices $(c,0)$, $(c, f(c))$, $(d, f(d))$ and $(d,0)$. Thus:
\begin{align*}
\int_{k}^{k+1} (1 + 2^{-k})^x \d{x} &< \frac{1}{2}((1+2^{-k})^k + (1+2^{-k})^{k+1}) \\
&= \frac{1}{2}(1+2^{-k})(2 + 2^{-k}) \\
&= 1 + 2^{-k} + 2^{-k-1} + 2^{-2k-1}
\end{align*}
Therefore:
\begin{align*}
\sum_{k=1}^{n-1}\int_{k}^{k+1} (1 + 2^{-k})^x \d{x} &< \sum_{k=1}^{n-1}(1 + 2^{-k} + 2^{-k-1} + 2^{-2k-1}) \\
&= n - \frac{4^{1-n}}{3} - 3(2^{-n}) + \frac{5}{6} \\
&< n + 1
\end{align*}
To show that $I(n) > n - 1$, simply observe that $(1 + 2^{-k})^x > 1$ for all $k \geq 1$, so:
$$
I(n) = \sum_{k=1}^{n-1}\int_{k}^{k+1} (1 + 2^{-k})^x \d{x} > \sum_{k=1}^{n-1}\int_{k}^{k+1} 1 \d{x} = n - 1
$$

Split your I tegral up between when floor(x) changes. So we get sum_k=1 to n-1 of integral from k to k+1 (1+2^-k)^x dx.

The integral will be 1/ln(1+2^-k)*[(1+2^-k)^k+1 - (1+2^-k)^k] = 2^-k * (1+2^-k)^(k+1) /ln(1+2^-k).

I don't know how you can show that the sum of these integrals will be between n a d n+1, so I will stop here.


would it be equivalent to find this remainder? I was given this exercise: Suppose that  $N$ is written in the decimal system in the form $N=a_m a_{m-1} \dots a_1 a_0 $.

Prove that $11 \mid N \Leftrightarrow 11|a_0-a_1+ \dots+(-1)^{m-1}a_{m-1}+(-1)^m a_{m}$.

So,if I would have to find the remainder of the division of $N$ by $11$,could I find the remainder of the division of $a_0-a_1+ \dots+(-1)^{m-1}a_{m-1}+(-1)^m a_{m}$ by $11$ ??

The key of your problem is $11|(10^{k}-1)$ for all even $k$ and $11|(10^k+1)$ for all odd $k$. Since
$$
N=a_m a_{m-1} \dots a_1 a_0=\sum_{i=0}^ma_i10^i,
$$
then
\begin{align}
11|N&\iff 11|\sum_{i=0}^ma_i10^i\\
&\iff 11|\left(\sum_{i\textrm{ odd}}^ma_i10^i+\sum_{i\textrm{ even}}^ma_i10^i\right)\\
&\iff 11|\left(\sum_{i\textrm{ odd}}^ma_i(10^i+1)+\sum_{i\textrm{ even}}^ma_i(10^i-1)-\sum_{i\textrm{ odd}}^ma_i+\sum_{i\textrm{ even}}^ma_i\right).\\
\end{align}
Now using the two facts above, you will lead to what you wanted.

Mo you can not find it. If you have seen its proof this strategy works because because 9999...uptil even number of terms is removed as out will always be facor and in odd digits 10000...1 is divisible so we remove that factor
  sorry I mis understood your question I gave the explanation  for divisor . We can find the remainder it will be either number yor are getting let say n so it will be n or 11-n for n>0  and -n or n+11


Prove that $\lim_{n\rightarrow\infty} \frac{1}{n}\sum_{k = 1}^{n}{f(\frac{k}{n}) }$ $=\int_0^1 f(x)dx.$ Question:

Let $f$ be continuous on $[0,1]$. Prove that

$\lim_{n\rightarrow\infty} \frac{1}{n}\sum_{k = 1}^{n}{f(\frac{k}{n}) }$ $=\int_0^1 f(x)dx.$

where $k=0,1,...,n.$

Attempt:

I don't even know where to start. It makes sense reading the sum, as $k\rightarrow n$, and dividing it by the number of partitions, I should reach the definition of the integral. Hoping for a little push to get started.

A continuous function on a compact set is uniformly continuous.

Given an $\epsilon\gt0$, find an $N$ so that if $|x-y|\le\frac1N$ we have $|f(x)-f(y)|\le\epsilon$. Thus, for any $n\ge N$,
$$
\int_{\large\frac{k-1}n}^{\large\frac kn}\left|\,f\left(\frac kn\right)-f(x)\,\right|\,\mathrm{d}x\le\frac\epsilon{n}
$$
Therefore, for $n\ge N$,
$$
\begin{align}
\left|\,\sum_{k=1}^nf\left(\frac kn\right)\frac1n-\int_0^1f(x)\,\mathrm{d}x\,\right|
&=\left|\,\sum_{k=1}^n\left(\int_{\large\frac{k-1}n}^{\large\frac kn}f\left(\frac kn\right)\,\mathrm{d}x-\int_{\large\frac{k-1}n}^{\large\frac kn}f(x)\,\mathrm{d}x\right)\,\right|\\
&\le\sum_{k=1}^n\int_{\large\frac{k-1}n}^{\large\frac kn}\left|\,f\left(\frac kn\right)-f(x)\,\right|\,\mathrm{d}x\\[6pt]
&\le\epsilon
\end{align}
$$

The first thing I notice is that we will be multiplying by zero as the current formula stands (constant divided by infinity is zero). But, intuitively, we can not say the integral of the two parts will be zero. This leads me to thinking of rewriting the sum using some formula. In fact, if we prove this formula, we have our answer!


Fit a equilateral triangle on three arbitrary parallel lines with an edge and compass How can you fit a equilateral triangle on three arbitrary parallel lines with an edge and compass?

Well, I'm late to this, but I've been obsessing over this far too much to back out now.  So: Label your parallel lines $a$, $b$, and $c$ from bottom to top.  Construct a line $d$ with a "positive slope" that crosses $b$ such that the top right angle of their intersection is $60^\circ$.  Extend the line so that is crosses $c$.  Call the intersection of $d$ and $c$ point $A$.  Construct another line $e$ such that it lies to the left of $d$, is parallel to $d$, and is the same distance from $d$ as $a$ is from $b$.  Call the intersection of $e$ and $b$ point $B$.  Draw an arc centered at $B$ and having radius $AB$, and crossing $a$.  Call the intersection of the arc with $a$ point $C$.  $\triangle ABC$ is your triangle.

Why does this work?  Here's a hint: Start with an equilateral triangle $\triangle ABC$.  Draw any line through $A$, and then draw a parallel line through $B$.  Now identify the center $O$ of $\triangle ABC$.  You have now have a triangle and two lines.  Rotate the entire configuration $120^\circ$ around $O$.  What do you get?  Rotate another $120^\circ$.  What do you get now?

Given 3 parallel lines A B C (top to bottom), construct a large equilateral triangle between lines A and C where the bottom triangle leg is on line C and the top triangle vertex is on line A. Mark the bottom right point of the triangle point c1.  Also mark the intersecting points of lines B and the left leg of the triangle point as b1.  Set the compass point on point c1 and the compass lead on point b1.  Now draw an arc that intersects line A, at new point a1. Use the straight edge and draw the new equilateral triangle between points b1, c1, and a1.

I figured this out in 1965 and made a proof to confirm validity, but have long forgotten the proof.

Regards, Ray McKaig


how many non equivalent $\mathbb{T}^n$ bundles are there over $\mathbb{S}^n$? Intuitively, maybe you have $\mathbb{Z}$  in inequivalent bundles for each $\mathbb{S}^1$ component as you can imagine taking the torus product an interval and gluing together the ends of the interval with a "twist". But this might not be true and I don't know how to go about proving it if it is....

You have the exact sequence $0\rightarrow \mathbb{Z}^n\rightarrow \mathbb{C}^n\rightarrow\mathbb{T}^n\rightarrow 1$ which induces an isomorphism (Cech cohomology) $H^2(M,\mathbb{Z}^n)\rightarrow H^1(M,\mathbb{T}^n)$. Take $M=S^n$, $H^2(M,\mathbb{Z}^n)=0$ if $n\neq 2$ so for $n\neq 2$ the bundle are trivial.


Does the sequence converge, and to what? We have a sequence $\{a_n\}$

$$a_0 = 0$$

$$a_{n+1} = \frac{a_{n}}{2} + 1$$

Does it converge? And to what?

Hint: show by induction that the sequence $a_n$ increase. Trick to do this is to see that

$a_{n+1}-a_n=a_n/2-a_{n-1}/2$.

Now this shows that the sequence is increasing. What else do you need to see it converges?

Further, suppose that the limit is $L$. Then $L=L/2+1$, as $\lim a_{n+1}=\lim a_n $


Use of the concept of subgroup vs field extension Why is it popular to use the idea of subgroups in cases of groups and field extensions in case of fields? In both case one set is the subset of the other along with the restriction of some additional properties.

Edit: As per suggestion of @Lubin, I would rephrase the question as - why do we emphasize on the substructures in case of groups, while for fields, we emphasize the overlying structures.

My answer must be partial and idiosyncratic. People may, should, and I hope will, object to my opinions. And correct my misstatement of facts too!

When we look at the smallest possible group, namely the trivial group, to ask for extensions is to ask about all groups! Not an interesting viewpoint, I think. Even when we look at a group like $C_2$, the cyclic group of order $2$, this group is contained in (more exactly, has an injective homomorphism into) every finite group of even order. Even if you demand that $C_2$ should be a normal subgroup of the extended group, you can always look at $C_2\times G$ for any group at all, and many others, containing the group of order $2$.

On the other hand, when we look at a smallest possible field, it’s definitely an interesting question to ask what extensions it has. If your “smallest possible” is $\mathbb F_p=\mathbb Z/(p)$, it’s interesting that there’s only one extension of each finite degree, and these all are normal over $\mathbb F_p$ with cyclic Galois group. If your “smallest” is $\mathbb Q$, then there are extensions of all possible finite degrees, some normal, some not, and yet we don’t yet know whether among the normal extensions, all finite groups occur as Galois group. Interesting facts and questions all.

For substructures, when we look at a group, we can ask about all sorts of properties: abelian or not, soluble or not, nilpotent, etc. And of course we have techniques, extremely well developed. In a way, by looking at the substructures rather than the overlying structures, we have restricted our investigations to more manageable questions.

For substructures of fields, there’s the fact that some fields that are important outside the narrow subject of field theory have no proper subfields, and even for fields that aren’t prime fields, such as $\mathbb C$, there certainly are subfields, but we deal with them in most cases as extensions of $\mathbb Q$ not subfields of $\mathbb C$. A notable exception is the theorem that the only fields $K$ with $[\mathbb C\colon K]<\infty$ have this degree equal to $2$.

The split is not absolute. There are ways of studying situations in which, given a group $G$, we can make statements about groups $X$ that have $G$ as a normal subgroup. Similarly, there are statements about subfields of a given field.

I think it's due to history. Groups were discovered/invented as a complete unit and sub groups were found later(Starting with Lagrange).First field existed intuitively as rational numbers,Later Galois extended them with roots looking for 5th degree equation solutions.


Proof if $f$ is continuous then the preimage is closed. I want to show that if $f: \mathbb{R}\to\mathbb{R}$ is continuous then $f^{-1}(l) = \{x\in\mathbb{R}|f(x) = l\}$ is closed.
I am not sure of my proof as I feel like I am missing a step:
I decided to use sequential continuity and consider a convergent sequence $(x_n)\subset f^{-1}(l)$.
By sequential continuity if $x_n \to L \implies f(x_n) \to f(L) = L'$
I want to say that I am now done - but I think I need to prove that $L \in f^{-1}(l)$, or $L' \in Imf$.

[Note] : As Robert mentioned, it is easier to show that continuity implies inverse of an open set is open which I will use as a Lemma.

Lemma $(1)$ : Consider $(X,d_{X})$ and $(Y,d_{Y})$ metric spaces and let $f : X\longrightarrow Y$. We claim that $f$ is continuous in $X$ if and only if for every open set $G\subseteq Y$, $f^{-1}(G)$ is open in $X$.

Proof :
$(\implies)$ :
Let $G$ be open in $Y$, we have to show that $f^{-1}(G)$ is open in $X$. Let $x_{0}\in f^{-1}(G)\implies f(x)\in G$ and since $G$ is open, then $f(x_{0})$ is an interior point of $G$ that is to say $\exists\varepsilon>0$ such that $N_{\varepsilon}(f(x_{0}))\subseteq G$. By continuity of $f$ at $x_{0}$, $\exists\delta>0$ such that $\forall x\in N_{\delta}(x_{0})\implies f(x)\in N_{\varepsilon}(f(x_{0}))$. We now show that $x\in f^{-1}(G)$, notice that :
\begin{align*}
\text{If}\;x\in N_{\delta}(x_{0})&\implies f(x)\in N_{\varepsilon}(f(x_{0}))\subseteq G \\
&\implies x\in f^{-1}(G),\;\forall x\in N_{\delta}(x_{0})\\
&\implies N_{\delta}(x_{0})\subseteq f^{-1}(G)
\end{align*}
Hence, $x_{0}$ is an interior point of $f^{-1}(G)$ implying that $f^{-1}(G)$ is open in $X$.
$(\impliedby)$ :
We have that $\forall G\subseteq Y$ open, we have that $f^{-1}(G)$ is open in $X$. Let $x_{0}\in X$, we want to show that $f$ is continuous at $x_{0}$. Let $N_{\varepsilon}(f(x_{0}))$ be a neighborhood in $Y$ with $\varepsilon>0$, then since $N_{\varepsilon}(f(x_{0}))$ is open in $Y$, then $f^{-1}(N_{\varepsilon}(f(x_{0}))$ is open in $X$. Now $x_{0}\in f^{-1}(N_{\varepsilon}(f(x_{0}))$ because $f(x_{0})\in N_{\varepsilon}(f(x_{0}))$. Therefore, $x_{0}$ is an interior point of $f^{-1}(N_{\varepsilon}(f(x_{0}))$ which means there exist $\delta>0$ such that $N_{\delta}(x_{0})\subseteq f^{-1}(N_{\varepsilon}(f(x_{0}))$. Therefore, $\forall x\in N_{\delta}(x_{0})$ we have that $x\in f^{-1}(N_{\varepsilon}(f(x_{0}))\implies f(x)\in N_{\varepsilon}(f(x_{0}))$. Hence, $f$ is continuous at $x_{0}$.

Corollary : Consider $(X,d_{X})$ and $(Y,d_{Y})$ metric spaces and let $f : X\longrightarrow Y$. We claim that for every open set $G\subseteq Y$, we have that $f^{-1}(G)$ is open in $X$ if and only if for every closed set $F\subseteq Y$, we have $f^{-1}(F)$ is closed in $X$

Proof : The proof is trivial since it follows from the fact that $f^{-1}(E^{c})=(f^{-1}(E))^{c}$. Thus, if $f : X\longrightarrow Y$ is continuous and $F\subseteq Y$ is closed, then $F^{c}$ is open which implies $f^{-1}(F^{c})$ is open which means $(f^{-1}(F))^{c}$ is open by the $\text{(Lemma)}$ and thus $f^{-1}(F)$ is closed. The reverse implication also follows from that basis.

I think it's easiest to show that $U= \Bbb R \setminus f^{-1}(l)$ is open.  Assume $f(x) = y \neq l$.  Choose $\varepsilon= \vert y - l \vert$.  Then because $f$ is continuous $\exists \delta \gt 0$ such that $\vert z - x \vert \lt \delta \Rightarrow \vert f(z)-f(x) \vert = \vert f(z)-y \vert \lt \varepsilon= \vert y - l \vert$, which means $f(z) \neq l$.  Thus, $\{ z \in \Bbb R \mid \vert z-x \vert \lt \delta \}$ is an open set containing $y$ that is contained in $U$, so $y$ is in the interior of $U$.  But $y \in U$ was arbitrary so $U$ is open.


Proving that $L^p \cap L^{\infty} \supset L^r$ where $1 \leq p <r < \infty$ Suppose that $1 \leq p <r < \infty$.

Prove that $L^p \cap L^{\infty} \supset L^r$

I can prove this theorem for cases other than $\infty$ but not for those with $\infty$ .
Could you please guide me on how to proceed in the infinity case ?

This is not true in general. For instance, take your space to be $X=[1,\infty]$ with the standard measure, $r=2$, $p=1$, and $f(x)=\frac{1}{x}$, then $f$ is in $L^2(X)=L^r(X)$, but $f$ is not in $L^1(X)=L^p(X)$.


How to respond to "solve this equation" in a basic algebra class If it's acceptable practice on math.se, I'd like to really only ask this question of math educators as opposed to students or mathematical researchers.  Some researchers will undoubtedly think the whole question is not worthy of consideration, and some students will not understand the issue. But if you really have something to contribute, please do.

Imagine yourself teaching a basic algebra class: maybe to grade schoolers or to high schoolers, or in my case, to adults ages 18 and up in a community college. You will encounter "problems" like the following, where for now I am intentionally leaving out words:

$$2x+3=6-x$$

The "answer" to a question like this somehow communicates that $1$ is the only solution, that $x$ needs to equal $1$, that the solution set is $\{1\}$, or $\{x\mid x=1\}$, etc.

Some of my colleagues feel that if the task was to "solve this equation", that "$x=1$" is not an acceptable final response from a student. They say that "$x=1$" is an "equivalent equation" to the original equation, because it has the same solution set. They say that to "solve this equation", to the exclusion of other ways a student might respond, is to write a set as part of an English statement. They are happy with: "The solution is $1$", "The solution set is $\{1\}$", or "The solution set is $\{x\mid x=1\}$". But they are emphatic that "$x=1$" or "The solution to the equation is that $x=1$" cannot count as appropriate responses. Answers like these don't get full credit on their exams.

This matters because we are creating a library of higher quality online problems (for WeBWorK) and we need to set a standard for how the solutions should be entered. I am opposed to having answer blanks where the student merely enters a number. So for example I am opposed to having something like "The solution is __." appear on screen and the student only fills in the number. This trains the student that those words don't matter; they don't have to write them and they won't generally pay them any attention. We could program the question to understand a whole sentence, but there are too many issues with alternatively worded correct sentences, not to mention poor spelling. Remember, these responses are to be automatically evaluated.

I support making the students have to enter "$x=1$", because it is a whole statement. And if they enter something simpler, like "$1$", it is an easy matter to shoot back an automated message that their response is not the form we are looking for. And I counter the idea that "$x=1$" has to be interpreted as an "equivalent equation" by saying that sometimes "$x=1$" is an assertion (aka an assignment) rather than an equation to solve; I'm asserting that $x$ has to equal $1$ for the equation to be true.

So my question to this community, assuming you support my position, is: can you help me make better arguments for my case? I suppose to stay within math.se guidelines, I will say that I am open to being just plain wrong, and definitive mathematical vocabulary expertise can be used to prove it.

To put it bluntly, those of your colleagues who don’t accept ‘$x=1$’ as a solution don’t speak English. That sort of hyperpedantry accomplishes nothing beyond making students think that mathematics is all about invoking the right (incomprehensible) magic formula(s). I’ll go so far as to say that I think it excessive pedantry to object to $x=\frac12\left(1\pm\sqrt5\right)$ as an answer to a question asking for the solution(s) to $x^2-x-1=0$.

It would be another matter if the question were written The solution set is __: that question clearly calls for a set. In that case, however, the script should accept $\{1\}$, $\{x\in\Bbb R:x=1\}$, $\{x:x=1\}$, $[1,1]$, and any other reasonably straightforward variant, but not $x=1$, $1$, or the like.

This question reminds me of how I could be given the exact same word problem in Grade 2 and again in Grade 7.  The difference in getting the same problem is that what is required to justify my answer is wildly different between these grades.  In Grade 2, it is enough to just work out the number that is the right answer that a teacher accepts without challenge while in Grade 7 the same problem requires stating a variable, formulating an equation and solving it, not unlike what you have as a given problem here.  There is something to be said for understanding how to justify an answer which could be a proof, counter-example, or something else to demonstrate an understanding of a concept and its use on some level.

Thus, I'd be tempted to consider what kind of formality are you expecting and would you want to be feeding someone the ideas of how to go through these kinds of problems.  Thus, I'd suggest bringing in Mathematical constructs and setting a bar for what kinds of answers are reasonable.  Otherwise you may run into people that want to claim what may otherwise seem ridiculous.


Most ambiguous and inconsistent phrases and notations in maths What are some examples of notations and words in maths which have been overused or abused to the point of them being almost completely ambiguous when presented in new contexts?

For instance, a function $f$:

  $f^{-1}(x)$ can be an inverse and a preimage and sometimes even $\frac{1}{f(x)}$.

  $f^2(x)$ can be $(f\circ f)(x)$ and $(f(x))^2$.

  $f^{(2)}(x)$ on the other hand, is the second derivative, even though adding parentheses to a number usually does nothing.

  And for some functions the parentheses for the argument are omitted: $f\:x = f(x)$.

  So how should $f^{(2-3)}(x)$ be interpreted? $f^{(-1)}$, an integral of $f$? or a composition, $\left(f^{(2)}\circ f^{(-3)}\right)(x)$? Or just $f^{2-3}(x) = f^{-1}(x)$?

Another example is mathematicians notorious use of the word normal to describe... normal things?

Using similar symbols and expressions for different things is unavoidable, but it can create some ambiguity when first introduced to their other uses.

'The function $f(x)$'. No, the function is $f$.
Let $f$ and $g$ be real differentiable functions defined in $\mathbb R$. Some people denote $(f\circ g)'$ by $\dfrac{\mathrm df(g(x))}{\mathrm dx}$. Contrast with the above. I discuss this in greater detail here.
The differential equation $y'=x^2y+y^3$. Just a minor variant of 1. Correct would be $y'=fy+y^3$ where $f\colon I\to \mathbb R, x\mapsto x^2$, for some interval $I$.
This is one I find particularly disgusting. "If $t(s)$ is a function of $s$ and it is invertible, then $s(t)$ is the inverse", lol what? The concept of 'function of a variable' isn't even definable in a satisfiable way in $\sf ZFC$. Also $\left(\frac{\mathrm dy}{\mathrm dx}\right)^{-1}=\frac{\mathrm dx}{\mathrm dy}$. Contrast with 1.
In algebra it's common to denote the algebraic structure by the underlying set.
When $\langle \,\cdot\,\rangle$ is a function which takes sets as their inputs, it's common to abuse $\langle\{x\} \rangle$ as $\langle x\rangle$. More generally it's common to look at at finite set $\{x_1, \ldots ,x_n\}$ as the finite sequence $x_1, \ldots ,x_n$. This happens for instance in logic. Also in linear algebra and it's usual to go even further and talk about 'linearly independent vectors' instead of 'linearly independent set' - this only an abuse when linear (in)dependence is defined for sets instead of 'lists'.
'Consider the set $A=\{x\in \mathbb R\colon P(x)\}$'. I'm probably the only person who reads this as the set being the whole equality $A=\{x\in \mathbb R\colon P(x)\}$ instead of $A$ or $\{x\in \mathbb R\colon P(x)\}$, in any case it is an abuse. Another example of this is 'multiply by $1=\frac 2 2$'.
Denoting by $+$ both scalar addition and function addition.
Instead of $((\varphi\land \psi)\to \rho)$ people first abandon the out parentheses and use $(\varphi\land \psi)\to \rho$ and then $\land$ is given precedence over $\to$, yielding the much more common (though formally incorrect) $\varphi\land \psi\to \rho$.
Even ignoring the problem in 1., the symbol $\int x\,\mathrm dx=\frac {x^2}2$ is ambiguous as it can mean a number of things. Under one of the common interpretations the equal sign doesn't even denote an equality. I allude to that meaning here, (it is the same issue as with $f=O(g)$).
There's also the very common '$\ldots$' mentioned by Lucian in the comments.
Lucian also mentions $\mathbb C=\mathbb R^2$ which is an abuse sometimes, but not all the time, depending on how you define things.
Given a linear map $L$ and $x$ on its domain, it's not unusual to write $Lx$ instead of $L(x)$. I'm not sure if this can even be considered an abuse of notation because $Lx$ is meaningless and we should be free to define $Lx:=L(x)$, there's no ambiguity. Unless, of course, you equate linear maps with matrices and this is an abuse. On the topic of matrices, it's common to look at $1\times 1$ matrices as scalars.
Geometers like to say $\mathbb R\subseteq \mathbb R^2\subseteq \mathbb R^3$.
Using $\mathcal M_{m\times n}(\mathbb F)$ and $\mathbb F^{m\times n}$ interchangeably. On the same note, $A^{m+ n}=A^m\times A^n$ and $\left(A^m\right)^n=A^{m\times n}$.
I don't know how I forgot this one. The omission of quantifiers.
Calling 'well formed formulas' by 'formulas'.
Saying $\forall x(P(x)\to Q(x))$ is a conditional statement instead of a universal conditional statement.
Stuff like $\exists yP(x,y)\forall x$ instead of (most likely, but not certainly) $\exists y\forall xP(x,y)$.
The classic $u=x^2\implies \mathrm du=2x\mathrm dx$.
This one disturbs me deeply. Sometimes people want to say "If $A$, then $B$" or "$A\implies B$" and they say "If $A\implies B$". "If $A\implies B$" isn't even a statement, it's part of an incomplete conditional statement whose antecedent is $A\implies B$. Again: mathematics is to be parsed with priority over natural language.
Saying that $x=y\implies f(x)=f(y)$ proves that $f$ is a function.
Using $f(A)$ to denote $\{f(x)\colon x\in A\}$. Why not stick to $f[A]$ which is so standard? Another possibility is $f^\to(A)$ (or should it be square brackets?) which I learned from egreg in this comment.

$$\bigcup \{ A ,B, C\} = A \cup B \cup C $$ or $$\bigcup \mathcal A $$ where $\mathcal A $ is a set of sets. It should be

$$\bigcup_{X \in \mathcal A} X. $$ This can come up in topology sometimes (unioning over a collection of open sets), and can get rather confusing if you start doing things (edit:) like $\left(\bigcup \mathcal A\right) \cup U \cup V$ where $U$ and $V$ are just sets (eg. when adding open sets to a cover).

Edit: It's like defining $\sum S = \sum_{x\in S} x$ where $S \subset \mathbb R$ is finite (or ordered). Then usually you expect in something like $(\sum S) + x + y$ for $S$, $x$, and, $y$ to be the same sort of thing.


Hard math contest trigonometry type problem How to solve this problem:

Also, most people would use trigonometry, but is there a way to use derivative to solve this too?

Above, $O$ is the centre of the sphere, $x$ is its radius. Writing the area of trapezoid $ABCD$ as a sum of three triangles, we can easily solve for $x$ to get:

$$
x=\frac{1200\sqrt{3}}{54}=38.490017\ldots
$$

No trig or calculus needed, only geometry.

Image of the work

Why don't take the radius and subtract from the height to to make a new triangle. With the new triangle, then use half of the angle of the gap between three or just the two triangles(cones). Solve for the opposite side of the the angle, which would be the radius of the sphere. An if the the opposite side is equal to the value you reduced the height by, then it confirms that it is the radius that fits the space, thereby making the perfect circle. I found that 22.6 is the angle. I use tangent to find the side and I was typically five numbers off. The approach seems to be perfect, but it's still off.


Finding $(a, b, c)$ with $ab-c$, $bc-a$, and $ca-b$ being powers of $2$ This is a 2015 IMO problem.  It seems difficult to solve.

  Find all triples of positive integers $(a, b, c)$ such that each of the numbers $ab-c$, $bc-a$, and $ca-b$ is a power of $2$.

Four such triples are $(a,b,c)=(2,2,2),(2,3,2),(3,5,7),(2,6,11)$.

Suppose $a,b,c\in\mathbb{N}$ and $\alpha,\beta,\gamma\in\mathbb{N}_0$ are such that $bc-a=2^\alpha$, $ca-b=2^\beta$, and $ab-c=2^\gamma$. It is easily seen that either all of $a,b,c$ are odd, or at least two of them are even.  We claim that all solutions $(a,b,c)$ are permutations of $(2,2,2)$, $(2,2,3)$, $(2,6,11)$, and $(3,5,7)$.

If two of $a$, $b$, and $c$ are equal, say $a=b$, then we have $2^\alpha=bc-a=a(c-1)$, or $c-1=2^\nu$ and $a=2^{\alpha-\nu}$ for some $\nu\in\mathbb{N}_0$ such that $\nu\leq\alpha$.  Hence, $2^{2\alpha-2\nu}=a^2=ab=2^\gamma+c=2^\gamma+2^\nu+1$, which means $a=2$ and $\{\gamma,\nu\}=\{0,1\}$.  If $\nu=0$, then $(a,b,c)=(2,2,2)$; if $\nu=1$, then $(a,b,c)=(2,2,3)$.  From now on, we assume that $a$, $b$, and $c$ are pairwise distinct.  Without loss of generality, assume that $a<b<c$.  It follows immediately that $\alpha>\beta>\gamma$.

If $\gamma=0$, we have $ab-c=2^\gamma=1$, or $c=ab-1$.  Plugging this in, we have $$2^\alpha=bc-a=b(ab-1)-a=ab^2-(a+b)$$ and $$2^\beta=ca-b=(ab-1)a-b=a^2b-(a+b)\,.$$  If both $a$ and $b$ are odd, then $c$ is even, and we have a contradiction.  If $a$ or $b$ is even, then both are even.  Suppose that $k$ and $l$ are the largest positive integers such that $2^k\mid a$ and $2^l\mid b$.  We claim that $k=l=1$.  If $k\neq l$, then from $2^\alpha-2^\beta=ab(a-b)$, we obtain $\beta=k+l+\min\{k,l\}$.  Nonetheless, from $2^\alpha+2^\beta=(a+b)(ab-2)$, we get $\beta=1+\min\{k,l\}<k+l+\min\{k,l\}$, which is a contradiction.  Thus, $k=l$.  Write $a=2^ks$ and $b=2^kt$, where $s$ and $t$ are odd integers.  We see that $$2^\alpha+2^\beta=(a+b)(ab-2)=2^{k+1}(s+t)\left(2^{2k-1}st-1\right)$$ and $$2^\alpha-2^\beta=2^{3k+1}st\left(\frac{s-t}{2}\right)\,.$$  This means $\beta\geq 2^{3k+1}$.  That is, $2^{2k}\mid s+t$, whence $4\mid s+t$.  This would then mean that $\frac{s-t}{2}$ is odd.  Hence, $\beta=3k+1$.  Now, we have $$2^{3k+1}=2^\beta=a^2b-(a+b)=2^{3k}s^2t-2^k(s+t)\geq2^k\left(2^{2k}-2\right)s^2t\,.$$  If $k>1$, then $s^2t\leq 2$, implying that $s=t=1$, but this violates the assumption that $a<b$.  Consequently, $k=1$, so $s^2t\leq 4$, which gives $s=1$ and $t=3$.  Thence, $(a,b,c)=(2,6,11)$.

From now on, we assume that $\gamma>0$.  If $a$ is even, then $$2^\beta-2^\gamma=(ca-b)-(ab-c)=(c-b)(a+1)$$ leads to $2^\gamma\mid c-b$, and $$2^\beta+2^\gamma=(ca-b)+(ab-c)=(c+b)(a-1)$$ leads to $2^\gamma \mid c+b$.  Ergo, $2^\gamma\mid (c+b)-(c-b)=2b$ and $2^\gamma\mid (c+b)+(c-b)=2c$, or $2^{\gamma-1}\mid b$ and $2^{\gamma-1}\mid c$.  Now, as $2^\gamma=ab+c$, $2\mid a$, and $2^{\gamma-1}\mid b$, we conclude that $2^\gamma\mid c$.  Therefore, $2^\gamma\mid b$ as well, whence $c=2^\gamma w$ for some odd $w\in\mathbb{N}$.  Similarly, $2^\beta\mid b$ and $2^\beta\mid a$ (by considering $2^\alpha+2^\beta$ and $2^\alpha-2^\beta$), with $b=2^\beta v$ and $a=2^\delta u$ for some odd $v,u\in\mathbb{N}$ and for some $\delta\in\mathbb{N}$ such that $\delta>\beta$.  If $b$ or $c$ is even, we can use a similar method to achieve the same result.

Consequently, in the case where $\gamma>0$, if any of $a$, $b$, and $c$ is even, then $a=2^\delta u$, $b=2^\beta v$, and $c=2^\gamma w$ for some odd $u,v,w\in\mathbb{N}$ and $\delta \in \mathbb{N}$ with $\delta>\beta$.  From $2^\beta=ca-b$ and $2^\gamma=ab-c$, we get $$2^\beta=\left(ab-2^\gamma\right)a-b=a^2b-2^\gamma a-b=2^{2\delta+\beta}u^2v-2^{\gamma+\delta} u - 2^\beta v\geq \left(2^{2\delta+\beta}-2^{\gamma+\delta}-2^\beta\right)u^2v\,.$$
Therefore, $2\leq 2^{2\delta-1}u^2v\leq \left(2^{2\delta}-2^{\gamma+\delta-\beta}-1\right)u^2v \leq 1$, which is absurd.  That is, $a$, $b$, and $c$ must all be odd.

Now, $\gamma>0$ and all $a$, $b$, and $c$ are odd.  From the equality $$2^{2\beta}-2^{2\gamma}=\left(2^\beta-2^\gamma\right)\left(2^\beta+2^\gamma\right)=\big((c-b)(a+1)\big)\big((c+b)(a-1)\big)=\left(c^2-b^2\right)\left(a^2-1\right)\,,$$
we get $2^6\mid 2^{2\beta}-2^{2\gamma}$, making $\gamma \geq 3$.  Plugging $c=ab-2^\gamma$ into $bc-a=2^\alpha$ and $ca-b=2^\beta$ to get $\left(\frac{b^2-1}{2^\gamma}\right)a-b=2^{\alpha-\gamma}$ and $\left(\frac{a^2-1}{2^\gamma}\right)b-a=2^{\beta-\gamma}$.  Since $\alpha>\beta>\gamma$, we deduce that $\frac{b^2-1}{2^\gamma}$ and $\frac{a^2-1}{2^\gamma}$ are odd integers.  Hence, there exist $i,j\in\{-1,+1\}$ and odd $u,v\in\mathbb{N}$ such that $a=2^{\gamma-1}u+i$ and $b=2^{\gamma-1}v+j$.  As $c=ab-2^\gamma$, $c=2^\gamma w+k$ for some $w\in\mathbb{N}$ and $k:=ij$.  From $$2^{\gamma}w\left(2^{\gamma-1}(iu+jv)+2\right)=(c-k)(ia+jb)=(ca-b)i+(bc-a)j=2^\beta i+2^\alpha j\,,$$
we conclude that $2^{\beta-\gamma-1}\mid w$ but $2^{\beta-\gamma}\nmid w$, or $c=2^{\beta-1}z+k$ for some odd $z\in\mathbb{N}$.  Moreover, $$2^\beta=ca-b> ca-c=c(a-1)=\left(2^{\beta-1}z+k\right)(a-1) \geq \left(2^{\beta-1}-1\right)(a-1)$$ yields $a-1\leq 2$.  Therefore, $a=3$, and this forces $z=1$ and $k=-1$.  That is, $c=2^{\beta-1}-1$; i.e.,
$$2^\beta=ca-b=3\left(2^{\beta-1}-1\right)-b\,,$$
or $b=2^{\beta-1}-3$. Since $2^{\gamma-1}u+i=a$, we must have $\gamma=3$, $u=1$, and $i=-1$.  Finally, $2^\gamma=ab-c$ implies that $8=3\left(2^{\beta-1}-3\right)-\left(2^{\beta-1}-1\right)=2^\beta-8$.  That is, $\beta=4$, so $(a,b,c)=(3,5,7)$.

Without loss of generality: replace $b$ with $a+k1$ and $c$ with $a+ k2$

$(a, a+k1, a +k2) $

$ab- c = a2 + k1a -a -k2 $

$bc- a = a2 + k1k2 + (k1+k2 -1)a$

$ca- b = a2+ (k2-1)a - k1$

All these must by $2N$ for some $n$ in $I > 0$


Show $|x^2 + y^2 - 7|_7 \geq \frac{1}{7}$ in p-adics Working in the $7$-adic numbers, I want to show that for any $x, y \in \mathbb{Q}$, we can guarantee that $|x^2 + y^2 - 7|_7 \geq c$ for some $c$.

By some amount of trial and error, I've guessed that $c = 1/7$. I showed this must be true for any $|x^2 - 7|_7$, but I'm having trouble extending the insight for one square to the sum of two squares.

To simplify notation a bit, put $p = 7$. Let $\nu : \mathbb{Q} \to \mathbb{Z}$ denote the $p$-adic valuation, so that $|x|_p = p^{-\nu(x)}$. Then $\nu(x + y) = \min\{\nu(x), \nu(y)\}$ for $\nu(x)\not= \nu(y)$. Since $(-1/p) = -1$, we have $\nu(x^2 + y^2)\not = 1$, and thus $|x^2 + y^2 - 7|_p \geq |7|_p = \frac{1}{7}$.

There are a number of cases to consider, depending on $|x|_7$ and $|y|_7$.
I'll just do the most interesting one.  Suppose $|x|_7 = |y|_7 = 1$.
Write $x = x_1 + 7 x_2 + \ldots$ and $y = y_1 + 7 y_2 + \ldots$, with $x_1, y_1 \in \{1,\ldots,6\}$.  Since the nonzero squares mod $7$ are $1$, $2$, and $4$, $x_1^2+y_1^2$ can't be divisible by $7$, so
$|x^2 + y^2 - 7|_7 = |x_1^2 + y_1^2|_7 = 1$.


What are the odds of hatching 11 roosters and 1 hen? I bought one dozen fertililized eggs, incubated them and hatched 11 roosters and 1 hen! This was just the opposite of what I'd hoped.  What were the odds of this happening?

Depends on how hot the incubator was:

  There was no difference in sex ratio between chickens hatched from eggs stored at 60 and 80°F, neither departing significantly from 50 per cent. pullets. However, storage at 40°F produced significantly more pullets than cockerels, 54.6 per cent. of the chickens being pullets

Taken from this study: http://www.publish.csiro.au/paper/AR9600664.htm

So for 50% it's $${12 \choose {1}}*2*(1/2)^{12} =  3/512 \approx 0.006%$$

For 54% pullets or female chickens,

$${12 \choose {1}}*2*(.546)^{11} *(.454) \approx 0.01%$$

The likely reason you got 11 roosters to 1 hen is that you chose 11 eggs containing roosters. Their gender is estalished before laying. The big breeders want hens to keep up their egg production and because female meat chickens fetch a higher price. So they keep the female eggs -those they think are female - and sell the suspected males. I found a website suggesting definite male eggs have a pointier end, whereas definite females  do not, and are more perfectly oval. The ones in between are apparently harder to determine. I'm guessing you bought your dozen fertilised eggs from a breeder who knew how to retain as many female eggs as possible - or sell them to more knowledgable buyers.
I'm sure there is a mathematical ratio once you know how many eggs, on average, contain hens or roosters. But as I used to say to my (long-suffering) maths teachers: "How is this ever going to help me in life?!" Finally, on this occasion, I think lateral thinking wins the day. Not that I would encourage you to buy more eggs. Your experience serves to remind me of how many potentially characterful roosters are killed just to allow a human to indulge a culinary habit or preference. You can eat very well and be healthier without other species flesh, eggs or breast milk in your diet. However, it may be morally justifiable to buy/rescue 18mo layer hens when they are being sent to slaughter. Hens lay for around 5 years, slowing from around 18months, and can live 10-15 happy years when well-cared for.


Intemporal budget by lagrange Assume that a representative agent lives forever and receives an endowment, denoted yt, in each period. The entire endowment sequence is known
with certainty on date 0. The representative agent discounts future utility at
rate $\beta$ and has a period utility function given by
$u(c_{t}) = (c_{t})^{1-\sigma}/(1-\sigma)  $

(d)write the intertemporal budget constraint of the representative
agent by aggregating the flow budget constraints.
Write down the Lagrangian using the constraint in (d). How many
Lagrange multipliers are there? Find the first order conditions and write down
the Euler equation.

I have found the equation
$ L = \beta^tU(c_{t}) + (\lambda)_{t}(1/(1+r)^t)(c_{t}-y_{t}) $

then I have found the equation by taking derivative of L with respect to $c_{t}$ and $c_{t+1}$ but since there are no savings(b) I could not take derivative of L with respect to b so cannot find the ratio of lamdas :
$ \frac{U^{'}(c_{t})}{U^{'}(c_{t+1})\beta} = \frac{(\lambda)_{t}}{(\lambda)_{t+1}(r+1)}$

The lagrange-function should be look like this:

$\mathcal L=\sum_{t=0}^{n} \left( \beta ^t\cdot \frac{1}{1-\sigma}\cdot c_t^{1-\sigma}+ \lambda_t \left(  \frac{y_t}{(1+r)^t}- \frac{c_t}{(1+r)^t} \right) \right) $

$ y_t- c_t  =0\Rightarrow \text{No savings in period t}$

First order condition:

$\frac{d L}{d c_t}=\beta ^t\cdot \frac{1}{1-\sigma} \cdot (1-\sigma) c_t^{1-\sigma}-\frac{\lambda _t}{(1+r)^t}=0 \Rightarrow \lambda_t =\beta ^t\cdot \frac{(1+r)^t}{1-\sigma} \cdot (1-\sigma) c_t^{1-\sigma} \quad (1) $

$\frac{d L}{d c_{t+1}}=\beta ^{t+1}\cdot \frac{1}{1-\sigma} \cdot (1-\sigma) c_{t+1}^{1-\sigma}-\frac{\lambda _{t+1}}{(1+r)^{t+1}}=0 \Rightarrow \lambda_{t+1} =\beta ^{t+1}\cdot \frac{(1+r)^{t+1}}{1-\sigma} \cdot (1-\sigma) c_{t+1}^{1-\sigma} \quad (2) $

Dividing (2) by (1) gives $\frac{\lambda _{t+1} }{\lambda _{t}}=\beta \cdot (1+r) \frac{c_{t+1}^{1-\sigma}}{c_{t}^{1-\sigma}}$

It is the same as $\frac{\lambda _{t+1} }{\lambda _{t}}=\beta \cdot (1+r) \frac{U'(c_{t+1})}{U'(c_{t})}$

My relation is a little bit different from yours.


limit of $\sqrt{12n+20}-\sqrt{12n}$ Compute the following:

$\lim\limits_{n\to \infty}\sqrt{12n+20}-\sqrt{12n}$

$\sum\limits_{n=0}^{\infty}\frac{1}{n^2+2n}$

In both I wasn't able to find anything. I tried $\lim\limits_{n\to\infty} a_n =\lim\limits_{n\to\infty} a_{n+1}$ for the first one. And I was able to show second one converge but i need the actual value.

for the first one multiply and divide by $$ \sqrt {12n+20} +\sqrt {12n}$$

For the second one use partial fraction $$\frac {1}{n^2+2n} = (1/2)[(1/n) -1/{(n+2)}]$$

and you get a telescoping series.

You should be able to handle both of them.

$$\lim\limits_{n \to \infty} \left( \sqrt{12 n + 20} - \sqrt{12 n} \right)$$

$$= 2 \lim\limits_{n \to \infty} \left( \sqrt{3 n + 5} - \sqrt{3 n} \right)$$

$$= 2 \lim\limits_{n^\prime \to \infty} \left( \sqrt{n^\prime + 5} - \sqrt{n^\prime} \right)$$

Proof:  For any $0<\epsilon$ we can find an $n$ such that the left-hand side is less than $\epsilon$:

$$\frac{5-\epsilon^2}{2 \epsilon} < \sqrt{n}$$.

Thus the limit is $0$.


In how many ways can the letters of the word EQUATION be arranged if vowels and consonants occur together? How many words with or without meaning can be formed using all the letters of the word "EQUATION" at a time if vowels and consonants occur together.
My answer is $5!3!2!=1440$.  Am I right?

In the word "EQUATION" there are 3 consonants (Q,T,N) so there are $3!$ ways to arrange and 5 vowels (E,U,A,I,O) so $5!$ ways to arrange. In whole for both first consonant and then vowels or vice-versa, there are $2!$ ways .
so in total $2! \cdot 5! \cdot  3!=1440$ ways.


Show $\lim_{n \rightarrow \infty} \frac{n}{e^n} = 0$ $$\lim_{n \rightarrow \infty} \frac{n}{e^n}$$
$$= \lim_{n \rightarrow \infty} \exp(\ln(\frac{n}{e^n})) $$
$$= \lim_{n \rightarrow \infty} \exp(\ln(ne^{-n})) $$
$$= \lim_{n \rightarrow \infty} \exp(\ln((n^{\frac{-1}{n}}e)^{-n})) $$
$$= \lim_{n \rightarrow \infty} \exp(-n \ln((n^{\frac{-1}{n}}e))) = 0 $$

Is my argument valid?

Notice, your last step is hard to understand & justify. Although it is easy to apply L'Hospital's rule for $\frac{\infty}{\infty}$ form as follows $$\lim_{n\to \infty}\frac{n}{e^n}=\lim_{n\to \infty}\frac{\underbrace{n}_{\large\to \infty}}{\underbrace{e^n}_{\large \to \infty}}$$ $$=\lim_{n\to \infty}\frac{\frac{d}{dn}(n)}{\frac{d}{dn}(e^n)}$$ $$=\lim_{n\to \infty}\frac{1}{e^n}$$$$=\frac{1}{\infty}=0$$

Yes, it is; although you might want to read a bit about L'Hospital's Rule.


The square root of a variable is negative? If the square root of a variable is negative, as shown below:

$$\sqrt x = -1$$

Then what is $x$ equal to? The closest answer I can think of is $i^4$.

$$\sqrt{i^4}=i^{\frac42}=i^2=-1$$

But if $i^4$ is evaluated first, then it doesn't work:

$$\sqrt{i^4}=\sqrt{(-1)(-1)}=\sqrt1=1$$

If a negative number is a square root of $x$, then $x$ is the square of that negative number, and thus $x$ is a positive number.  Every nonzero complex number has two square roots.  However, when you talk about the square root of a positive number, you generally mean the positive square root, not the negative square root.

Using the Quaternion definition, $ x $ can be any of $i^2j^2$, $j^2k^2$ or $i^2k^2$, without using any arbitrairy operations order.

If $\sqrt{x} = -1$, $ x = -1 * -1$ and $-1$ can be defined by $i^2$, $j^2$ and $k^2$


Can a set be uncountable if it is the union of an uncountable quantity of finite objects? We denote by $\mathbb {iN}$ the set of all infinite subsets of $\mathbb N$ and by  $f(\mathbb S)$ the set of all finite consecutive subsets of $\mathbb S$. Ex, if $\mathbb S = \{2,4,6,8,\dots \} $ then $f(\mathbb S) =  \{ \{2\}, \{2,4\},\{2,4,6\},\{2,4,6,8\},\dots \} $

Then $\mathbb {iN}$ is NOT countable and $\mathbb N,\mathbb S,f(\mathbb S) $are countable

$$Define: \mathbb T =  \{ f(\mathbb S) : \mathbb {S} \in \mathbb {iN} \}   $$

Here $\mathbb T$ is NOT countable, it has $\mathbb {iN}$ elements

$$Define:  \mathbb {U} = \bigcup_{\mathbb {S}\in \mathbb{T}}   \mathbb {S} $$

$ \mathbb {U}$ is countable.

$ \mathbb {U}$ is the union of an uncountable number of countable sets.

This would mean that some elements of set $\mathbb T$ contribute information to the union, but most don't.

Why would this be the case if the elements of $\mathbb T$ are homogeneous?

In order to ensure that an uncountable union of non-empty sets is uncountable, you need to ensure that there is an uncountable number of sets which actually adds new information to each other.

Look at the function $f(r)=\{0\}$. Then $\bigcup_{r\in\Bbb R}f(r)=\{0\}$. An uncountable union, but a very finite set. How does that happen? Well, we have repetitions.

Given any infinite set $\mathbb S_1$ where $ f(\mathbb S_1) = \{ a_1, a_2, a_3, \dots , a_n, \dots \} $

then for any $ a_n $ there will exist an infinite family of infinite sets  $\mathbb S_{x \in \mathbb N}$ where $a_n \in f(\mathbb S_x)$

In other words, there would be no member  $ a_n $ that is unique to one set only.

Therefore the union of an uncountable number of finite sets would simply be the union of the finite sets, which is countable.


Good book about differential forms I'm a looking for a good book to self-study differential forms. Particularly, I'm looking for a book that is as similar as possible to Bert Mendelson's "Introduction to topology" (i.e. a book that procede by  following a: "Definition, theorem, proof" style). In addition, the book that I'm looking for should be as much self consistent as possibile. I'm a first year graduate student in nuclear engineering. My prerequisites are a good understanding of (multivariate and vector) calculus,  linear algebra, and a little of functional analysis, Lebesgue integration theory, PDE. I know nothing about differential geometry, but to my (very) poor understanding  differential forms and concept like manifolds and so on are linked to each other.

Differential forms are things that live on manifolds. So, to learn about differential forms, you should really also learn about manifolds. To this end, the best recommendation I can give is Loring Tu's An Introduction to Manifolds. Tu develops the basic theory of manifolds and differential forms and closes with a exposition of de Rham cohomology, which allows one to extract topological information about a manifold from the behavior of the differential forms on it.

Look no further than Bernard Schutz "Geometrical Methods of Mathematical Physics" (CUP 1980) for an intro which starts at the beginning. Differential Forms are merely infinitesimal increments of coordinate axes - which transform inversely to tangents to these axes - and linear combinations thereof. In an x,y,z coordinate system an example would be 3dx+dy+4dz. Their inverse transformation nature is signified by writing their indices UPSTAIRS rather than downstairs. A usual 1st example of a differential form is the differential of a function.


Find the least non negative residues mod 7,11 and 13 of 12345678. Find the least non negative residues mod 7,11 and 13 of 12345678.

my attempt: A number N is congruent modulo 7, 11, or 13, to the alternating sum of its digits in base 1000. (For example, 123456789 ≡ 789 − 456 + 123 ≡ 456 mod 7, 11, or 13.)

What  I can do if I have understood well the question is the clear

$$12345678=949667\cdot13+\color{red}7=1122334\cdot11+\color{red}4=1763668\cdot7+\color{red}2$$ where the asked residues are in red.


If through row reduction of a square matrix you can produce a row of zeros does that automatically make that matrix singular? If through simple row operations such as adding one row to another I can produce a row of entirely zero does this mean that the matrix will be singular?

For example:

$$\left\{ \begin{matrix} 1 & 2 & 3 & 4 & 5 \\ 0 & 1 & 2 & 3 & 4 \\
2 & 3 & 4 & 4 & 0 \\ 5 & 4 & 3 & 2 & 1 \\ 1 & 5 & 4 & 2 & 3\end{matrix} \right\}$$

Can be turned into the following by adding the first row to the fourth row

$$\left\{ \begin{matrix} 1 & 2 & 3 & 4 & 5 \\ 0 & 1 & 2 & 3 & 4 \\
2 & 3 & 4 & 4 & 0 \\ 6 & 6 & 6 & 6 & 6 \\ 1 & 5 & 4 & 2 & 3\end{matrix} \right\}$$

Which can then be turned into the following by subtracting the second row from the first

$$\left\{ \begin{matrix} 1 & 1 & 1 & 1 & 1 \\ 0 & 1 & 2 & 3 & 4 \\
2 & 3 & 4 & 4 & 0 \\ 6 & 6 & 6 & 6 & 6 \\ 1 & 5 & 4 & 2 & 3\end{matrix} \right\}$$

Now by subtracting the first row from the fourth 6 times we will produce a row of entirely $0$'s like so

$$\left\{ \begin{matrix} 1 & 1 & 1 & 1 & 1 \\ 0 & 1 & 2 & 3 & 4 \\
2 & 3 & 4 & 4 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 1 & 5 & 4 & 2 & 3\end{matrix} \right\}$$

Since we have managed to produce a row of $0$'s does this mean that the rank of this matrix after further row operations is guaranteed to be less than 5 so according to the Invertible Matrix Theorem it should not be invertible/will be singular?

Is this logic sound or is their some other way we can determine that a large matrix is non invertible without doing intensive calculations?

You're making this too complicated if your question is just about invertibility. Just compute the determinant. If either a row or a column of a matrix is entirely made of 0's, then the determinant will be 0 and by the equivalence theorem of linear algebra, the matrix is singular(noninvertible) . Of course, this will only work by hand if the matrix is relatively small-otherwise, you'll need a computer algebra program.

Interestingly, the same result is obtained if your algorithm produces 2 identical rows in a square matrix. Indeed, the result is equivalent to the previous one-just add the negative of one of the 2 rows to the other! You might think that this is just a computational quirk of square matrices, but it has far reaching consequences. To give just one-it's where the Pauli Exclusion principle of quantum mechanics comes from. When one expresses the wavefunction of fermionic particles via the Heisenberg matrix formulation, one can express it via the Slater determinant. If 2 fermionic particles-like electrons in the outer d and higher shells of an atom--occupy the same quantum states,then  2 rows or columns of the resulting matrix are identical. But then when one applies the Slater determinant to compute the probability value of the square of a given wavefunction, the deteminant vanishes. This means the value of the probability of finding an electron in a given state is 0 and this means it's impossible!

To find out more about this relationship, see any basic book on quantum mechanics or chemistry, like Ira Levine's very good book.


Examples of uncountable sets with zero Lebesgue measure I would like examples of uncountable subsets of $\mathbb{R}$ that have zero Lebesgue measure and are not obtained by the Cantor set.

Thanks.

Let $(r_n)_{n\in\mathbb{N}}$ be a dense sequence in $\mathbb{R}$ (it could, for example, be an enumeration of the rationals). For $k \in \mathbb{N}$, let

$$U_k = \bigcup_{n\in\mathbb{N}} (r_n - 2^{-(n+k)},\, r_n + 2^{-(n+k)}).$$

$U_k$ is a dense open set with Lebesgue measure $\leqslant 2^{2-k}$, thus

$$N = \bigcap_{k\in\mathbb{N}} U_k$$

is a set of the second category (Baire), hence uncountable, and has Lebesgue measure $0$.

Denote $E_2^c =$ the set of real numbers $x$ on the interval $[0, 1]$ such that in the decimal form of $x = 0.i_1 i_2 i_3 \dots$ there is no digits $2$.

Write $x \in E_2^c$ in decimal form
    $$
 x = 0. i_1 i_2 i_3 \dots, \ \
 i_k = 0, 1, 3, 4, 5, 6, 7, 8, 9, \ \
 x = \sum_{k=1}^{\infty} \frac{i_k}{10^k},
 $$
    and assume that the first digit $2$ appeals at $k$th decimal of $x$.

Denote $E_{2,1}$ as the set of real numbers $x$ on the interval $[0, 1]$ such that the first digit $2$ appeals at $1$th decimal of $x$:
    $E_{2,1}
 = \{x | x = 0. 2 \dots\}
 = [0.2, 0.3)$,
    then $m(E_{2,1}) = 10^{-1}$;

Denote $E_{2,2}$ as the set of real numbers $x$ on the interval $[0, 1]$ such that the first digit $2$ appeals at $2$th decimal of $x$:
    $E_{2,2}
 = \{x | x = 0. i_1 2 \dots\}
 = \cup_{i_1} [0.i_1 2, 0.i_1 3)$,
    $i_1 \in \{0, 1, 3, 4, 5, 6, 7, 8, 9\}$,
    then $m(E_{2,2}) = 9 \times 10^{-2}$;

Inductively,
    \begin{align}
 E_{2,k}
 &= \{x | x = 0. i_1 \dots i_{k-1} 2 \dots\} \\
 &= \cup_{i_1, \dots, i_{k-1}} [0. i_1 \dots i_{k-1} 2, 0. i_1 \dots i_{k-1} 3),
 \end{align}
    where $i_1, \dots, i_{k-1} \in \{0, 1, 3, 4, 5, 6, 7, 8, 9\}$,
    then $m(E_{2,k}) = 9^{k-1} \times 10^{-k}$;

Define $E_2 =$ the set of real numbers $x$ on the interval $[0, 1]$ such that in the decimal form of $x = 0.i_1 i_2 i_3 \dots$ there is digits $2$.

Then the measure of $E_2$ is
    $$
 m(E_2)
 = m(\cup_{k=1}^{\infty} E_{2,k})
 = \sum_{k=1}^{\infty} m(E_{2,k})
 = \sum_{k=1}^{\infty}  9^{k-1} \times 10^{-k}
 = 1.
 $$

Thus, the measure of $E_2^c$ is
    $$
 m(E_2^c)
 = m([0, 1] \setminus E_2)
 = m([0, 1]) - m(E_2)
 = 1 - 1
 = 0.
 $$


Like 1-D, is the force always directed towards the minimum of the potential in 3-D? Assertion Consider the motion of a particle of mass $m$ under the influence of a potential $V(x)$ in 1-dimension. The force on that particle is given by $\textbf{F}(x)=-\frac{dV}{dx}\hat{\textbf{x}}$. From this, it is clear that the $\textbf{F}(x)$ is always directed downhill at any point on $x$-axis i.e., towards a local minimum of the potential. This is true independent of the form of the potential $V(x)$.

Question In 3-D, $\textbf{F}(x,y,z)=-\boldsymbol{\nabla} V(x,y,z)$. What can we say in this case about the direction of the force at any point $(x,y,z)$? Again I'm looking for the most general conclusion that can be drawn for an arbitrary V(x,y,z) but reasonably well-behaved.

Let $V: \mathbb{R}^3 \rightarrow \mathbb{R}$ be a potential function.

The force is now $F = -\nabla V$, that is, it is minus the gradient of $V$. The gradient is the direction of steepest ascent. Hence, minus the gradient is the direction of steepest descent.

Now there are two claims:

Following the force (or minus the gradient) infinitesimally will give you a local minimum, provided that the potential is bounded below$^1$.
The gradient need not point in the direction of a global or local minimum.

For the second point I invite you to think about a hilly landscape, here we can define a function $h: \mathbb{R}^2 \rightarrow \mathbb{R}$ to be the height of the current position. Now, a river will flow along the direction of steepest descent (that is $-\nabla h$), but if you are standing on the bank of a river and looking downstream there is no guarantee that you are looking in the direction of the lake where the river ends up. Indeed the river may meander back and forth, and may even stream in the opposite direction before doubling back on itself!

The first point is the justification for the method of gradient descent for finding minima/maxima of functions.

A proof would presumably use the gradient theorem, but I don't have time to write anything up right now.

(1): A local minimum could still be located "at infinity", if this bothers you, you could ask for the domain of $V$ to be compact.

$$F_x=-\dfrac{\partial V(x,y,z)}{\partial(x)}$$
$$F_y=-\dfrac{\partial V(x,y,z)}{\partial(y)}$$
$$F_z=-\dfrac{\partial V(x,y,z)}{\partial(z)}$$
So the total force is:
$F=F_x+F_y+F_z$
Obviously this is a vector sum. As you said the $V(x,y,z)$
must be a reasonably well behaved potential. In that case the direction of the force is directed in a direction orthogonal to the tangent plane of $V(x,y,z)$
in any point $P(x,y,z)$


What does $dx$ mean? $dx$ appears in differential equations, such us derivatives and integrals.

For example, a function $f(x)$ its first derivative is $\dfrac{d}{dx}f(x)$ and its integral $\displaystyle\int f(x)dx$. But I don't really understand what $dx$ is.

The formal definition of an expression such as
$$
\int_0^1 x^2\,dx
$$
will depend on the setting. This is because there is not just one "theory of integration" - there are several different theories in different areas.

I like the presentation at the beginning of this note by Terence Tao. The key point is that there are really at least three different viewpoints on integration in elementary calculus:

Indefinite integration, which computes antiderivatives
An "unsigned definite integral" for finding areas under curves and masses of objects
A "signed definite integral" for computing work and other "net change" calculations.

The value of an expression such as $\int_0^1 x^2\,dx$ comes out the same under all these interpretations, of course.

In more general settings, the three interpretations generalize in different ways, so that the "dx" comes to mean different things. In the setting of measure theory, "dx" is interpreted as a measure; in  the context of differential geometry, it is interpreted as a 1-form.

But, for the purposes of elementary calculus, the only role of the "dx" is to tell which variable is the variable of integration. In other words, it lets us distinguish
$$
\int_0^1 uv\,du = v/2
$$
from
$$
\int_0^1 uv\,dv = u/2
$$

lhf's answer is fine if you just need to solve calculus problem but is incorrect.  Assuming x is an independant variable, dx is the real that satisfies for all x > 0, dx < x where x is real.


Probability of 2 consecutive heads in 5 tosses of a fair coin I'm somehow getting the answer 1. Total number of possible events is $2^5$.

Now the number of favorable events:

2 consecutive heads can happen in following ways:
$$H\: H\: (H\:or \: T)\:\:(H\:or \: T)\:\:(H\:or \: T)$$ $$ (H\:or \: T)\:H\: H\:(H\:or \: T)\:(H\:or \: T)$$$$ (H\:or \: T)\:(H\:or \: T)\:H\: H\:(H\:or \: T)$$$$ (H\:or \: T)\:(H\:or \: T)\:(H\:or \: T)\:H\: H\:$$

Each of these $4$ possibilities can happen in $2^3$ ways corresponding to whether it's H or T in the remaining 3 tosses.

So, favorable number of events = $4\cdot 2^3$

Probability$=\frac{4\cdot 2^3}{2^5}=1$

What did I miss?

The possibilities you mention aren't mutually exclusive so you cannot just add them. $HHHHHH$ is in all of them e.g.

It might be easier to count the unfavorable ones instead?

The only ways to not get two consecutive heads are
$$\sf TTTTT\\ TTTTH, TTTHT, TTHTT, THTTT, HTTTT\\ TTHTH, THTTH,HTTTH, THTHT, HTTHT, HTHTT\\ HTHTH$$

So the probability you seek is $1-13/2^5$.

That is $19/32$, or $0.59375$

The trouble with your method is the massive overcounting of common outcomes.

$$\require{cancel}\sf \phantom{[}H\phantom{\vert T]~[}H\phantom{\vert T]}~[H\cancel{\vert T}]~[H\vert T]~[H\vert T] \\ [H\cancel{\vert T}]~\phantom{[}H\phantom{\vert T]~[}H\phantom{\vert T]}~[H\vert T]~[H\vert T]$$


If $(a,b,M)$ is a Pythagorean triple, can $(b,b+a,N)$ be another triple? Does anyone know of a pair of Pythagorean triples of the form
$$(a, b, M) \quad\text{and}\quad(b, b+a, N)$$
Is such a pair possible?

Partial solution follows:

First note that if $a$ and $b$ share a common factor, then you can divide both triples by that common factor and the property still holds (e.g. if $a$ and $b$ are both even, then so is $a+b$, and hence the triplets $(a/2, b/2, M/2)$ and $(b/2, \frac{a+b}{2}, N/2)$ should also be Pythaogorean). Hence, without loss of generality, assume that $\gcd(a, b) = 1$, and hence $(a, b, M)$ is a primitive triple. Then $(b, a+b, N)$ is also a primitive triple.

Then assume that the generators of $(a, b, M)$ and $(b, a+b, N)$ are $(x_1, y_1)$ and $(x_2, y_2)$ respectively, i.e. that $a$ and $b$ equal $2x_1y_1$ and $x_1^2-y_1^2$ in some order, and similarly for $b$ and $a+b$. That gives four possible cases to test against, but three of them can very quickly be eliminated:

$a+b$ is always odd (because exactly one of $a$ and $b$ is even), so it must be that $b = 2x_2y_2$ and $a+b = x_2^2 - y_2^2$. Hence $b$ is even, so $a$ is odd, hence $a = x_1^2 - y_1^2$ and $b = 2x_1y_1$, implying that $x_1y_1 = x_2y_2$.

We can then rearrange all of that to eliminate $a$, $b$ and $y_2$, and hence have an expression relating $x_1$, $y_1$ and $x_2$:

$x_2^4 - (x_1^2 - y_1^2 + 2x_1y_1)x_2^2 - x_1^2y_1^2 = 0$

And we note that in this form, it is a quadratic in $x_2^2$, so the triples exist only if that quadratic has at least one solution that is a perfect square. An equivalent equation can be written for $y_2$. So for any given Pythagorean triple, there is a means of testing if it will form a second triple, but I don't know of any way to simplify that or prove that it will never have a solution.

Using Wolfram Alpha to ensure accurate algegra, we start with Euclid's formula (modified) for the OP.

$$A=m^2-n^2\qquad B=2mn+(m^2-n^2)\qquad C=m^2+n^2$$

$$A^2=m^4 - 2 m^2 n^2 + n^4$$

$$B^2=m^4 + 4 m^3 n + 2 m^2 n^2 - 4 m n^3 + n^4$$

$$C^2=m^4 + 2 m^2 n^2 + n^4$$

$$A^2+B^2-C^2=m^4 + 4 m^3 n - 2 m^2 n^2 - 4 m n^3 + n^4=0$$

The only solutions to the last equation are:

$$n =\pm m\qquad n = 2 m \pm \sqrt{5} m\qquad n=m=0$$

$\therefore$ there are no integer solutions except $A,B,C=(0,0,0), (0,2,2), (0,8,8), (0,18,18), ...$


Throwing $k$ balls into $n$ bins. I have the following question:

Throwing $k$ balls into $n$ bins. What is the probability that exactly $z$ bins are not empty?

I thought about something like:
$$\Pr(z)=\frac{n! z^{k-z}}{n^k (n-z)!},$$
but this is not correct.

Another idea is: $\Pr(bin ~empty)=(1-1/n)^k$, $\Pr(bin ~Not ~empty)=1-(1-1/n)^k$, so:

$$\Pr(z)=(\Pr(bin ~empty))^{n-z}(\Pr(bin ~Not ~empty))^{z}\cdot A$$

So, how to obtain the $A$?

Thanks!

The probability that at most $r$ particular bins are not empty is $(r/n)^k$. Thus by inclusion–exclusion the probability that exactly $z$ particular bins are not empty is

$$
\sum_{j=0}^z(-1)^j\binom zj\left(\frac{z-j}n\right)^k=\frac{z!}{n^k}\left\{k\atop z\right\}\;,
$$

where $\displaystyle\left\{k\atop z\right\}$ is a Stirling number of the second kind. Since there are $\displaystyle\binom nz$ ways of selecting $z$ particular bins, the desired probability is

$$
\binom nz\frac{z!}{n^k}\left\{k\atop z\right\}=\frac{n!}{(n-z)!n^k}\left\{k\atop z\right\}\;.
$$

You can also derive this result by noting that there are $n^k$ outcomes in total, and the favourable outcomes are characterized by a partition of the set of $k$ balls into $z$ non-empty subsets (of which there are $\displaystyle\left\{k\atop z\right\}$ ) and then $n(n-1)\cdots(n-z+1)$ choices where to place each subset.

I wonder why it is not the simple answer shown below.
Choose z bins(any), put 1 ball each in those bins, then randomly put balls in those z bins.

The required prob. is nCz*(z/n)^(k-z)


Trigonometric Identity Involving Hyperbolic Tangents and Inverse Tangents I am given the following question:
Prove that $$ \tan^{-1}\left(\frac{\tan 2\alpha +\tanh2\beta}{\tan 2\alpha -\tanh2\beta} \right) +\tan^{-1}\left(\frac{\tan\alpha -\tanh\beta}{\tan\alpha +\tanh\beta}\right)=\tan^{-1}(\cot\alpha\coth\beta)$$

How do I solve this?

I believe the expression on the RHS should be $\tan^{-1}(\cot\alpha\coth\beta)$. As for the proof, since $\tan^{-1}(a)+\tan^{-1}(b)=\tan^{-1}\left(\frac{a+b}{1-ab}\right)$,
the LHS of your equation equals $$\tan^{-1}\left(\frac{(\tan2\alpha+\tanh2\beta)(\tan\alpha+\tanh\beta) +(\tan2\alpha-\tanh2\beta)(\tan\alpha-\tanh\beta)}{(\tan2\alpha-\tanh2\beta)(\tan\alpha+\tanh\beta) - (\tan2\alpha+\tanh2\beta)(\tan\alpha-\tanh\beta)}  \right)$$
$$=\tan^{-1}\left(\frac{\tan\alpha\tan2\alpha+\tanh\beta\tanh2\beta}{\tanh\beta\tan2\alpha-\tan\alpha\tanh2\beta} \right).$$
Recall that $\tan2\alpha=\frac{2\tan\alpha}{1-\tan^2\alpha}$ and $\tanh2\beta=\frac{2\tanh\beta}{1+\tanh^2\beta}$, so substituting into the above expression yields
$$\tan^{-1}\left(\frac{\frac{2\tan^2\alpha}{1-\tan^2\alpha}+\frac{2\tanh^2\beta}{1+\tanh^2\beta}}{\frac{2\tan\alpha\tanh\beta}{1-\tan^2\alpha}-\frac{2\tan\alpha\tanh\beta}{1+\tanh^2\beta}}\right)$$
$$=\tan^{-1}\left(\frac{2\tan^2\alpha(1+\tanh^2\beta)+2\tanh^2\beta(1-\tan^2\alpha)}{2\tan\alpha\tanh\beta(1+\tanh^2\beta)-2\tan\alpha\tanh\beta(1-\tan^2\alpha)}\right)$$
$$=\tan^{-1}\left(\frac{\tan^2\alpha+\tanh^2\beta}{\tan\alpha\tanh^3\beta+\tan^3\alpha\tanh\beta}\right)$$
$$=\tan^{-1}\left(\frac{1}{\tan\alpha\tanh\beta}\right)$$
as required.


A regular expression for all strings over $\{a,b\}$ which do not have both substrings $bba$ and $abb$ I am looking at the answer in a solution manual to the following question.

  Let the alphabet $= \{a,b\}$.
    Write regular expressions for:

  all the words that don't have both substrings $bba$ and $abb$.

The answer given was

  $a^*(baa^*)^*b+b^*(a^*ab)^*a^*$

I don't think this is correct, since it can't make $abb$, which is in the language. So it's got to be wrong.

So I came up with my own. Would this be all the words that don't have both substring $bba$ and $abb$?

$(a+ba)^*(bb+b+\lambda)$

Or maybe this one should work

$(b+\lambda)(ab+a)^*+b^*$

(In the above $\lambda$ is the empty string.)

The regex a*(baa*)*b+b*(a*ab)*a* does match abb:

a* matches a
(baa*)* matches empty string
b+ matches bb,
(a*ab)*a* matches empty string.

Also, both of your regexes match bbabb which shouldn't be matched according to the task. And what is the point of bb+b+, isn't it the same as bbb+?

Not sure what are you trying to accomplish with "^ is an empty string by the way", but usually ^ matches the beginning of the string

These are incorrect regular expressions. We generate $bba$ from the second part.

I think $b^*+a^*(ab)a+a^*(ba)a+a^*ba(a+ab)a+a^*ab(a+ab)a$ may be a correct regular expression.

The correct regular expression is
$$b^*+a^*(a^*aba^*)^*+a^*(a^*ba)^*a^*(V+b)+a^*(ba)^*(a+ab)^*a+a^*(ab)^*(a+ab)*.$$

$ababaab$ accepted
$abababaab$ accepted
$baababaab$ accepted
$ababba$ Rejected


Why does the series $\sum_{n=1}^\infty\frac1n$ not converge? Can someone give a simple explanation as to why the harmonic series

  $$\sum_{n=1}^\infty\frac1n=\frac 1 1 + \frac 12 + \frac 13 + \cdots $$

doesn't converge, on the other hand it grows very slowly?

I'd prefer an easily comprehensible explanation rather than a rigorous proof regularly found in undergraduate textbooks.

Let's group the terms as follows:

Group $1$ : $\displaystyle\frac11\qquad$           ($1$ term)

Group $2$ : $\displaystyle\frac12+\frac13\qquad$($2$ terms)

Group $3$ : $\displaystyle\frac14+\frac15+\frac16+\frac17\qquad$($4$ terms)

Group $4$ : $\displaystyle\frac18+\frac19+\cdots+\frac1{15}\qquad$   ($8$ terms)

$\quad\vdots$

In general, group $n$ contains $2^{n-1}$ terms.  But also, notice that the smallest element in group $n$ is larger than $\dfrac1{2^n}$.  For example all elements in group $2$ are larger than $\dfrac1{2^2}$.  So the sum of the terms in each group is larger than $2^{n-1} \cdot \dfrac1{2^n} =  \dfrac1{2}$.  Since there are infinitely many groups, and the sum in each group is larger than $\dfrac1{2}$, it follows that the total sum is infinite.

This proof is often attributed to Nicole Oresme.

Given that the harmonic series of primes $ \sum_{p \, prime} \frac{1}{p}$ diverges, since

$$ \sum_{n=1}^{\infty} \frac{1}{n} > \sum_{p \, prime} \frac{1}{p}$$

it then follows that the harmonic series $\sum_{n=1}^{\infty} \frac{1}{n}$ also diverges.


Sum of finite ordinals: $\lambda_1+\lambda_2+\dots+\lambda_n+\dots=\omega$ Prove: $\lambda_1+\lambda_2+\dots+\lambda_n+\dots=\omega$, where $\lambda_i$ are finite ordinal nonzero numbers.

I tried like this.

$A_i$ set and ord($A_i$)$=\lambda_i, i=\{1,2,...\}$ and ord($\mathbb{N}$)$=\omega$.

$\lambda_1+\lambda_2+\dots+\lambda_n+\dots=$ord($\bigcup (A_i$x$\{i\})$)

$\bigcup (A_i$x$\{i\})$ is countable union of finite set. And that is countable.
So ord($\bigcup (A_i$x$\{i\})$)=ord($\mathbb{N}$)

Is this correct?

EDIT:
Definition: $\{\lambda_i\}_{i\in I}$ is family of ordinal numbers and $\lambda_i=$ord($A_i$). Then $\sum_{i\in I}=$ord($\bigcup_{i\in I} (A_i$x$\{i\})$)

I think that I need to find order isomorphism from $\bigcup_{i\in I} (A_i$x$\{i\})$ to $\mathbb{N}$ but I'm not sure how.

Given two posets $(S,\le_S)$ and $(T,\le_T)$, and order isomorphism from $(S,\le_S)$ to $(T,\le_T)$ is bijective function $f$ from $S$ to $T$ with the property that, for every $x,y\in S$, $x\le_S y$ iff $f(x)\le_T f(y)$

It is easy to show the sum is not finite. The sum is a limit of finite sums however. the only candidate is $\omega$.

Note $\lambda_{i} = 1 + 1 + 1 \cdots$ ($\lambda_{i}$ times). Replace $\lambda_{i}$ accordingly in the sum. By the associative property we can remove parentheses and get $(1 + 1 + 1 + \cdots)$ which is obviously $\omega$.


why double Sigma summation? The lecture slides is about covariance. One page it is

on the other page, it is

Why there are double Sigma summation on the first one?Is it just a typo?

I think a very clear definition of covariance is:

$Cov(X,Y) = \sum\limits_{x \in X}\sum\limits_{y \in Y} (x- \mu_X)(y-\mu_Y)f(x,y)$

If you define $S = X \times Y = \{(x,y) | x \in X, y \in Y\}$ (cartesian product).

Then you can rewrite the definition as:

$Cov(X,Y) = \sum\limits_{(x,y) \in S} (x- \mu_X)(y-\mu_Y)f(x,y)$

I think the first equation in your post is a confusion between the two definitions written in this answer.
It is not even clear to me what summation sign $(x,y) \in S$ belongs to.
However, a lot of notation abuse like this is made, if it is clear what is being meant. Which was obviously not the case here since you were confused.

It is a double summation, one over $x$ and one over $y$, so strictly speaking you need two $\Sigma$ signs on both slides.

Sometimes people leave only one just for easier notation.


Numerically Invert the Wakeby Percent Point Function I am looking at the possibility of using the Wakeby Distribution to attempt to model color components in image rows and columns (it is a very silly idea for "compressing" images that I want to see explode).  I chose this distribution because I like how it seems possible to model a PDF that seems to respect skewness (naturally, any comments about distributions that can be parameterized using moments like mean, variance, skewness, and kurtosis would be appreciated).  I think I can ramp up on the MLE method to estimate $\alpha$, $\beta$, $\gamma$, and $\delta$ if I can just get past this one problem.

Both the Wikipedia page and the NIST page on the Wakeby Distribution state that finding the PDF for the distribution requires numerically inverting the "percent point function" without any further explanation.  From other research, I have found that "percent point function" is another name for the cumulative density function that is referred to as $U$ on both those pages.  In trying to invert the formula to solve for $U$, I have gotten as far as

$\beta\delta(x-\xi)+\beta\gamma-\alpha\delta=\frac{\beta\gamma-\alpha\delta(1-U)^{\beta+\delta}}{(1-U)^\delta}$

but I can get no further.  I have tried looking for any sites or papers that address this issue, but it seems that everyone uses external tools to perform this work.

Could someone help me understand how to find the cumulative distribution function $F(x)$ from the Wakeby Distribution formula?

I am a layman, but it seems that you have to solve the integral (your function in the integral boundaries (in the given interval)). A possible way to do it, is numerical integration (e.g. simpson's rule).


Understanding Variable of Integration I am currently trying to get a deeper understanding of multivariable calculus and I am finding my one stumbling block to be the variable of integration. In single variable calculus, I have always considered the variable of integration (in antiderivatives) as the "what do I have to differentiate the answer with respect to in order to get the integrand." For instance, in the following:

$$\int 2x dx=x^2$$

I have always interpreted the dx as meaning I have to differentiate my answer ($x^2$) with respect to x in order to get my integrand ($2x$). In other words, I have to take $\frac{d}{dx}$ of $x^2$ to get $2x$.

I am struggling to understand a certain concept though: multiplication of variables of integration. Say, for instance, that I have the following:

$$\int (x -6)^2 dx$$

If I define $u = x - 6$, I can write the following equivalent integral

$$\int u^2 \frac{du}{dx}dx$$

It seems to me that now there are "multiple" variables of integration. Am I looking for an answer that I have to differentiate with respect to $u$ or $x$ in order to get the integrand? This can be simplified to:

$$\int u^2 du$$

Now, it is clear that I am looking for a function that I have to differentiate with respect to $u$ in order to get my integrand. However, I don't understand why the variables of integration ($dx$) can just cancel. My understanding is that they must mean more than "what do I have to differentiate my answer...." In the case of definite integrals, it is relatively clear (I can imagine them as small changes along a certain axis). But, with indefinite integrals, I cannot think of what they represent. Any help would be very much appreciated. Thank you!

What you are talking about is a technique called integration by substitution it is a techique to integrate functions involving chain rule in their derivative.In fact it comes to help the power rule to find the anti-derivative, however, the power rule to find the anti-derivative becomes useless if the derivation contains applying the chain rule, that is, in order to apply the power rule you have to guarantee that the integration will stop at applying the power rule so we integrate with respect to the derivative of the inner function(the function we were going to apply the chain rule at i.e u in your example.


Fitch natural deduction proof of $\vdash (\neg P \lor Q) \to (P \to Q)$ I have been working on a problem for hours now - trying to prove a statement which has no premises. This is what the problem looks like:

You can see that so far I've tried to work my way backwards, starting with the answer, but without premises I am finding it very difficult to move forward. I've gone through my textbook back and forth and the examples in there are much less complex than this one. I'm pretty sure this is a rather obvious one, but I just can't seem to figure it out.

Thanks so much.

I'd go about it a bit differently: we need a rule to eliminate the $\lor$, in the system I learnt there was a $\lor$-elimination rule of the form (see also here:

  If $P \lor Q$ holds, and we can prove $R$ from the assumption $P$ in some subproof (so $P \to R$ from the introduction of $\to$ rule) and also $Q \to R$ (via the same rule and some subproof), then we can conclude $R$; if $R$ follows from both members of a disjunction and the disjunction holds we conclude $R$.

So

$\lnot P \lor Q$ (base assumption we will eliminate at the end)
$P$ (we want to show $P \to Q$ eventually so we assume this, so a start of a subproof).
$\lnot P$ (assumption 1 for the elimination of $\lor$).
$\bot$ ((2+3: introduction of falsum $\bot$), or maybe in your system you need  to go via $P \land \lnot P$ first?)
$Q$ (from 4: Ex Falso Quodlibet, or whatever you call the rule).
$\lnot P \to Q$ (from 3-5), end of subproof that started at 3).
$Q$ (assumption, new subproof).
$Q$ (trivial conclusion/copy from 7., I used to have to do this explicitly)
$Q \to Q$ (intro $\to$) from 7-8 end of subproof 2.
$Q$ ( from elimination of $\lor$, we have 1. holding, and combining with 6. and 9. we apply the rule I described above.)
$P \to Q$ from the subproof that started at 2. and ended at 10. asumption 2. can be dropped now).
$\lnot P \lor Q) \to (P \to Q)$ (from 1. - 11., assumption 1 is now dropped and we have an assumptionless proof of the statement.

On line 3 what you appear to be wanting to use is disjunctive syllogism (DS). This rule is available in this proof checker but it may require you to first construct $\lnot\lnot P$.  However, that is easy to do as shown on lines 3-5 below. Once that is done, you can use disjunctive syllogism on lines 1 and 5 to derive $Q$. Then follow your approach to complete the proof.

Here is what the proof might look like:

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

P. D. Magnus, Tim Button with additions by J. Robert Loftis remixed and revised by Aaron Thomas-Bolduc, Richard Zach, forallx Calgary Remix: An Introduction to Formal Logic, Winter 2018. http://forallx.openlogicproject.org/


Is the ideal $(X^2-3)$ proper in $\mathbb{F}[[X]]$? Let $\mathbb{F}$ be a field and $R=\mathbb{F}[[X]]$ be the ring of formal power series over $\mathbb{F}$. Is the ideal $(X^2-3)$ proper in $R$? Does the answer depend upon $\mathbb{F}$?

Clearly $X^2-3=(X+\sqrt3)(X-\sqrt3)$ and hence $X^2-3$ is not zero.

I have no idea whether the ideal is proper or not.

So far, I didn't learn any theorem to prove an ideal is proper. Perhaps I should start with definition of proper ideal? Find one element in $R$ but not in the ideal ?

The element $\sum _{i \geq 0}  a_i X^i \in R$ is invertible in $R$ if and only if $a_0\neq 0$.
The key to the proof of that relatively easy result is the identity $(1-X)^{-1}=\sum_{i \geq 0}  1. X^i \in R $
In your question $a_0=-3$ so that the element $X^2-3$ is invertible in $R=F[[X]]$ (which is equivalent to the ideal $(X^2-3)\subset R$ being proper) if and only if the characteristic of the field $F$ is not $3$:   $$  (X^2-3)\subset R \;\text {proper} \iff \operatorname{char} F\neq 3.$$

Since $\mathbb{F}$ is a field then R as you have defined it is an integral domain.  Therefore if you can prove that the ideal does not contain 1 then the ideal is proper.


Independence, conditioning, and correlations Suppose $X$ and $Y$ are independent random variables uniformly distributed on $[0,1]$. Suppose we consider a conditional distribution of $X$ and $Y$ on some event $C$. Is it possible that these conditional distributions are correlated?

Yes. Try the event $C:=\{Y>X\}$. The joint density of $(X,Y)$ conditional on $C$ is uniform on the 'upper triangle' of the unit square, so $Y$ and $X$ are positively correlated given $C$.

An image generated with Python function seaborn.jointplot http://stanford.edu/~mwaskom/software/seaborn/generated/seaborn.jointplot.html, illustrating the answer given by @grand.chat.
EDITED The graph of a joint probability density function of $(X,Y)$ is a surface $z=f_{X,Y}(x,y)$. Usually to illustrate such a distribution one displays its contour plot, that is some level lines projected onto the $x-y$ plane.
In our case the joint p.d.f. $f_{X,Y|((X,Y)\in C)}$ is equal to $2$ for $(x,y)\in C$ and zero otherwise. Hence the theoretical contour plot consists in just one contour line (the triangle of vertices $(0,0), (0,1), (1,1)$). But in the figure posted here it is drawn the  contour plot of the estimated from data joint distribution function (I simulated $(X,Y)|((X,Y)\in C)$). Since this experimental joint pdf is not a constant at the points of $C$, you can see more contours lines, not just one.


Find the values of $a$ and $b$ when the line tangent to the ellipse in the first quadrant forms a triangle whose area is a minimum given $a+b=1$. The problem has 3 parts to it. First I was asked to find the tangent line to the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ in the first quadrant where the area of the triangle formed is a minimum. I found it to be $$\frac{x}{a\sqrt{2}}+\frac{y}{b\sqrt{2}}=1$$ The second part asks me to find the minimum area based from the first part of the question when $a+b=1$. This I found to be $\mbox{Area}=ab=b(1-b)$ since $a+b=1$ implies $a=1-b$. What I'm really puzzled about is the one that asks me to find the values of $a$ and $b$ for which this minimum area occurs. I'm not sure what to do with this. Please help.

P.S. The triangle formed here is the one bounded by the tangent line and the coordinate axes.

Let's write the parametric equations of the ellipse as $P(t) = (a\cos t, b \sin t)$. The tangent line at the point $P(t)$ has equation:
$$
(b \cos t)x + (a \sin t)y = ab
$$
This line intersects the $y$-axis where $y = b/ \sin t$ and it intersects the $x$-axis where $x = a/ \cos t$. So, the area of the enclosed triangle is
$$
A(t) = \frac{ab}{2\sin t \cos t} = \frac{ab}{\sin (2t)}
$$
Clearly this has a minimum value of $ab$ when $t = \pi/4$. Setting $t = \pi/4$ in the tangent line equation above gives the same result that you found.

If $a + b = 1$, then the minimum area is $b(1-b)$. If you graph the function $f(b) = b(1-b)$, you will see that it's a "hump" shaped parabola that crosses the horizontal axis at $b=0$ and $b=1$. So, by setting $b$ to a value that's slightly larger than $0$ or slightly less than $1$, you can can make the area as small as you like. So, in some sense, the minimum area is zero (or, more precisely, it can be made arbitrarily small).

Note that the second part of the question is pretty much unrelated to the first part: it just deals with understanding the function $f(b) = b(1-b)$.

I  think area asked is between tangent and coordinate axis.

The tangent in 1st quadrant is $y= -mx + \sqrt{(a^2*m^2)+b^2}$  where m>0.
Y intercept of tangent is $\sqrt{(a^2*m^2) +b^2}$ ;
X intercept is   $\sqrt{(a^2*m^2)+b^2}/m$
Area of triangle with coordinate Axes is $= ((a^2*m^2)+b^2)/2m$ ;

  Area = (1/2)(a^2*m+(b^2)/m)>=a*b>=1/4   I.e when m^2 = b^2/a^2
  and a is just greater than 1/2 and his just less than 1/2. In limiting case a and b are approaching towards 1/2.


calculatimg $\lim_{z\to0}\left(\frac{\Re(z^{3})}{|z|^{2}}\right)$ I'm trying to solve the limit of $z$ where $z \in \mathbb{C} $

$$
\lim_{z\to0}\left(\dfrac{\Re(z^{3})}{|z|^{2}}\right)
$$

What you would use is that $|\Re z|\le|z|$ . Therefore

$$|\Re (z^3)| \le |z^3| = |z|^3 $$

So you have that:

$$\left|{\Re (z^3)\over |z|^2}\right|\le |z|$$

which will imply that $\lim_{z\to0}\Re(z^3)/|z|^2 = 0$.

You are adviced to remember the definition of limit, that for every $\epsilon>0$ there exists an $\delta$ such that $|f(z)-L|<\epsilon$ whenever $0<|z-a|<\delta$, here we have that $|\Re(z^3)/|z|^2-0|<\epsilon$ whenever $0<|z-0|<\delta=\epsilon$.

The approach of using limits along lines is bad because limit along straigh lines doesn't imply that the limit exists as defined above. For example $f(x+iy) = x/y$ (except on $y$ axis where $f(x) = 0)$. Along straight lines it aproaches zero, but not as a limit in $\mathbb C$ (for example look at what happens if you aproach zero along the parabola $y=x^2$).

first of all we know that : $z^{n}=r^{n}cis(n\theta)$ also $|z|=r=\sqrt{x^2+y^2}$
and $\Re(z)=rcos\theta ,\Im(z)=rsin\theta$.

depending on what is above we have :

$\lim_{z\to0}(\dfrac{\Re(z^{3})}{|z|^{2}}) = \lim_{r\to0}(\dfrac{r^3cos3\theta}{r^{2}})=\lim_{r\to0}(rcos3\theta)=0
$


$f(x) = f(x-1)*b  \to  f(x) = c_1 b^{x-1}$ I'm trying to teach myself mathematics. I only started last year with a previous knowledge of basic arithmetic. I often make a detour into regions featuring problems that are a bit too deep for me. I'm often too curious to wait until I was able to catch up on the necessary bedding. I hope you'll excuse me :-)

Wolfram|Alpha says:

$$f(x) = f(x-1)*b \to f(x) = c_1 b^{x-1}$$

This is how I would compute the original recursive algorithm(?):

$$f(x) = f(x-1)*b$$

$$\begin{align} \\
f(4) &= f(4-1)*b \\
&= f(f(3-1)*b)*b \\
&= f(f(f(2-1)*b)*b)*b \\
&= f(f(f(f(1-1)*b)*b)*b)*b \\
&= \qquad ...
\end{align}$$

But how do I compute the recurrence relation $f(x) = c_1 b^{x-1}$?

I would compute the Fibonacci numbers like this:

$$\begin{align}
F_n &= F_{n-1} + F_{n-2}\\
    F_0 &= 0\\
    F_1 &= 1\\

   F_5 &= F_{5-1} + F_{5-2}\\
       &= F_4 + F_3\\
       &= (F_3 + F_2) + (F_2 + F_1)\\
       &= ((F_2 + F_1) + (F_1 + F_0)) + ((F_1 + F_0) + F_1)\\
       &= (((F_1 + F_0) + F_1) + (F_1 + F_0)) + ((F_1 + F_0) + F_1) \\
       &= 5
\end{align}$$

I'm also not sure about the difference between a recursive algorithm and its recurrence relation. I'd be grateful if someone could enlighten me or simply link me up.

I believe that your way to compute $f(n)$ is not correct, as you have outlined in the question. Observe that $f(4) = b * f(3) = b^2 * f(2)$ and not $f(4) = b * f(f(2)*b)$.

That said, you have $$f(n) = b * f(n-1) = b^2 * f(n-2) = \ldots = b^{n-1} * f(1).$$
Say that your initial condition states that $f(1) = c_1$, you can conclude that $f(n) = c_1 * b^{n-1}$.

A recurrence relation defines a function depending on its values on points of lower "rank". Since you want to be able to calculate the function, you also need some bases cases that are just given (like axioms). For example, the recurrence relation defining the Fibonacci numbers defines $F_n$ in terms of $F_{n-1},F_{n-2}$; the base cases $F_1=1,F_0=0$ are not defined in terms of previous elements.

We can have more complicated definitions of "rank". Here are two examples. One can define the binomial coefficients using Pascal's identity $$\binom{n}{k} = \binom{n-1}{k-1} + \binom{n-1}{k}$$ with appropriate bases cases. Written out, that would be $$C(n,k) = C(n-1,k-1) + C(n-1,k), \quad C(0,k) = 1_{k=0}.$$ So in that cases $(n-1,\cdot)$ is of lower rank than $(n,\cdot)$.

Another example is the determinant - it's a function on $n \times n$ matrices (arrays of numbers) which is defined as the single value in case $n=1$, and using some formula involving $(n-1)\times(n-1)$ matrices otherwise.

In general, for the definition to work, the notation of rank must be a well ordering.

A recursive algorithm is a procedure which calls itself. For example, you can use the following pseudocode to calculate the Fibonacci numbers:

if ((n == 0) || (n==1)) return n;
else return F(n-1) + F(n-2);

Every recurrence relation can be computed this way. In this particular case, the algorithm for calculating $F_n$ runs in exponential time (in other words, it's very inefficient), and there are better ways to calculate $F_n$; for example, you could calculate the values iteratively:

F[0] = 0; F[1] = 1;
for (int i = 2; i <= n; i++) F[i] = F[i-1] + F[i-2];

There are even better algorithms.

Conversely, recursive algorithms have other uses, for example the best algorithms for sorting a list of numbers are recursive and use the "divide and conquer" paradigm.


Is there a simple function that generates the series; $1,1,2,1,1,2,1,1,2...$ or $-1,-1,1,-1,-1,1...$ I'm thinking about this question in the sense that we often have a term $(-1)^n$ for an integer $n$, so that we get a sequence $1,-1,1,-1...$ but I'm trying to find an expression that only gives every 3rd term as positive, thus it would read;
$-1,-1,1,-1,-1,1,-1,-1...$

Alternatively a sequence yielding $1,1,2,1,1,2,1,1,2...$ could also work, as $n$ could just be substituted by it in $(-1)^n$

Let $F_n$ be the Fibonacci sequence $0,1,1,2,3,5,8,13,21,34,...$ then a possible sequence is $$(-1)^{F_{n+1}}$$

Here is my proposal, for $n \ge 1$:

$$1+\left\lfloor \frac{1}{n^{(n \pmod {3})}} \right\rfloor - \left\lfloor \frac{1}{n}\right\rfloor$$

It will provide the sequence: $\{1,1,2,1,1,2,...\}$


How to interpret scientific notation? I'm having a problem understanding scientific notation. What is the difference between the following:

$$\text{5e2, 5e-2, -5e2, -5e-2}$$


Is there possibly a largest prime number? Prime numbers are numbers with no factors other than one and itself.

Factors of a number are always lower or equal to than a given number; so, the larger the number is, the larger the pool of "possible factors" that number might have.

So the larger the number, it seems like the less likely the number is to be a prime.

Surely there must be a number where, simply, every number above it has some other factors.  A "critical point" where every number larger than it simply will always have some factors other than one and itself.

Has there been any research as to finding this critical point, or has it been proven not to exist?  That for any $n$ there is always guaranteed to be a number higher than $n$ that has no factors other than one and itself?

Euclid's famous proof is as follows: Suppose there is a finite number of primes.  Let $x$ be the product of all of these primes.  Then look at $x+1$.  It is clear that $x$ is coprime to $x+1$.  Therefore, no nontrivial factor of $x$ is a factor of $x+1$, but every prime is a factor of $x$.  By the fundamental theorem of arithmetic, $x+1$ admits a prime factorization, and by the above remark, none of these prime factors can be a factor of $x$, but $x$ is the product of all primes.  This is a contradiction.

Proof that there is always a higher prime number and therfore an infinite amount of prime numbers:

Mathematicians often hide behind an array of jargonish symbology. We need more proofs like the one I invent here (took me 90 minutes after watching “An infinite Mind” thanks Patel for channeling Ramanujan):

Easiest layman’s proof that there is always a higher prime number than any prime you are given:

Let Y be a prime number of any positive value.

Y!-1 cannot be divided by Y or any other integer lower than Y because those numbers are already factors of Y! and therefore cannot be factors of Y!-1; Therefore any factor of Y!-1 must have a value higher than Y, and yet not be divisible by any other factor lower than Y, which means it must be a prime number higher than Y. So in any case, either Y!-1 is a prime number OR Y!-1 has at least one factor which is prime numbers higher than Y: in either case, there is a higher prime number than Y.

[Y! means 1x2x3x4x5x … all the way up to … Y]


Calculating the residues of $f(z)=\frac{ze^{iaz}}{z^4+4}$ I have $f(z)=\frac{ze^{iaz}}{z^4+4}$. The poles of $f$ are {$\pm1\pm i$}.
How do I calculate $2\pi i \sum \operatorname{Res}\limits_{z = \pm 1 + i} f(z)$ ? (I'm only looking at positive $i$'s.)
I've tried using the property $\operatorname{Res}f(z) = \frac1{g'(p)}$ and that gives me $\frac1 4 e^{ia(\pm1+i)}$. But then I realised the property only works for one pole. How do I do this?

Consider, for instance, the residue at $1+i$. Since$$z^4+4=(z-1-i)(1-1+i)(z+1-i)(z+1+i),$$define$$g(z)=\frac{ze^{iaz}}{(z-1+i)(z+1-i)(z+1+i)},$$and you have $f(z)=\frac{g(z)}{z-1-i}$. Therefore,\begin{align}\operatorname{res}_{z=1+i}\bigl(f(z)\bigr)&=\operatorname{res}_{z=1+i}\left(\frac{g(z)}{z-1-i}\right)\\&=g(1+i)\\&=-\frac{ie^{(-1+i)a}}8.\end{align}


Find the limit and show that it converges to that limit. Consider the sequence $\{(\sqrt{n+1}-\sqrt{n})\sqrt{n}\}$. Find the limit and prove it converges to that limit or prove that it does not converge.
My attempt:
$(\sqrt{n+1}-\sqrt{n})\sqrt{n}=\sqrt{n}\sqrt{n+1}-n=\sqrt{n^2+n}-n$
Multiplying by the conjugate then dividing by it gives:
$\frac{n}{\sqrt{n^2+n}+n}=\frac{1}{\sqrt{\frac{n+1}{n}}+1}$.
Then taking the limit of this gives 1/2.
To prove that the sequence converges to 1/2, I have to show that $|(\sqrt{n+1}-\sqrt{n})\sqrt{n}-\frac{1}{2}|<\epsilon$ for some $\epsilon>0$. I know that I have to find an index $N<\epsilon$ but that is where I am stuck.

Just work with the equivalent expression which you found: $\frac{1}{\sqrt{\frac{n+1}{n}}+1}$
Take an arbitrary $\epsilon$
Find N such that
$|\frac{1}{\sqrt{\frac{n+1}{n}}+1} - \frac{1}{2}|<\epsilon$
when $n \gt N$

Hint:  Can you use the binomial theorem?
\begin{aligned}\sqrt{n}\big(\sqrt{n+1}-\sqrt n\big) &= n\Bigg(\Big(1 + \frac{1}{n}\Big)^{1/2}-1\Bigg) \\
&=\Bigg(\frac{1}{2}-\frac{1}{8n}+\cdots\Bigg)
\end{aligned}
Gives a handle on the error term and therefore a route to find $N$ needed to make the error less than $\varepsilon$


What is $\langle 3u+2v , -u+4v\rangle$ Given $\lVert u\rVert$ , $\lVert v\rVert$ , and $\langle u,v\rangle$? What is $\langle 3u+2v , -u+4v\rangle$ Given $\lVert u\rVert$ , $\lVert v\rVert$ , and $\langle u,v\rangle$?
\begin{align}
    \lVert u\rVert &= 3\\
    \lVert v\rVert &= 5\\
    \langle u,v\rangle &= -4
\end{align}
Well, to solve this I thought I'd break down certain elements.

I know that...
\begin{gather}
 \langle u,v\rangle = \mathbf{u}\cdot\mathbf{v} = u_1v_1 + u_2v_2 + u_3v_3 + \cdots + u_nv_n = -4\\
    \langle u+v , u+v\rangle = \langle u,u\rangle + \langle u,v\rangle + \langle v,u\rangle + \langle v,v\rangle
\end{gather}
But I don't know what $\langle 3u+2v , -u+4v\rangle$ actually is...

Is it
$$
    \langle 3u,-u\rangle + \langle 3u,4v\rangle + \langle 2v,-u\rangle + \langle 2v,4v\rangle \mbox{?}
$$

Hints.$\def\v#1{{\bf#1}}$

Expand $(3\v u+2\v v , -\v u+4\v v)$ pretty much as you did for $(\v u+\v v,\v u+\v v)$.
You are given $(\v u,\v v)$: how does this help you to find $(\v v,\v u)$?
You are given $\|\v u\|$ and $\|\v v\|$: what does this tell you about the other terms in your expansion?

$\langle 3u+2v,-u+4v\rangle    =\langle 3u,-u\rangle +\langle 3u,4v\rangle +\langle 2v,-u\rangle +\langle 2v,4v\rangle$
\begin{align}
          &=-3\langle u,u\rangle +12\langle u,v\rangle -2\langle v,u\rangle +8\langle v,v\rangle\\
              &=-3\cdot 9 + 12\cdot -4 - 2\cdot 4 + 8\cdot 25
\end{align}
since $\lVert u\rVert = 3$, $\lVert v\rVert = 5$, $\langle u,v\rangle = -4$ and  $\lVert u\rVert^2 =\langle u,u\rangle$


Need help to solving the inequality term of $\frac{|2x-4|}{5x-15} \geq \frac{2+\sqrt{x-8}}{x^2+3x-130}$ $$\frac{|2x-4|}{5x-15} \geq \frac{2+\sqrt{x-8}}{x^2+3x-130}$$
My work so far is :

Finding the definition term for the left hand side.
$$5x-15 \neq 0 $$
$$5(x-3)\neq 0 $$
$$x \neq 3$$

Finding the definition term for the right hand side.
$$x^2+3x-130 \neq 0$$
$$(x+13)(x-10) \neq 0$$
$$x\neq -13 \lor x\neq 10$$

How to finding the inequality term and simplify this form ?

I also try raising both side into power of two, but i fail here
$$ (2+\sqrt{x-8} )^2$$
This still contain radical form.

Please help me to solving this inequality. Thank you so much.

Clearly $x\geq 8$. Let $t=\sqrt{x-8}$, so $t\geq 0$. Then $x= t^2+8$ and we can rewrite the inequality as:

$${2(t^2+6)\over 3(t^2+5)}\geq {2+t\over (t^2+21)(t^2-2)}$$

If $t<\sqrt{2}$, i.e. $x<10$ then this is always true (since the left side is positive and right is negative).

So we have to consider only the case when $t>\sqrt{2}$. In this case we get a following polynomial inequality we have to solve:

$$2(t^2+6)(t^2+21)(t^2-2)\geq 3(t^2+5) (2+t)$$


maximize $M+N+P$ with the restraint $5(MN+MP+PN)=7MNP$ and $M,N,P$ are positive integers Mal, Num, and Pin each have distinct number of marbles. Five times the sum of
the product of the number of marbles of any two of them equals to seven times
the product of the number of the marbles the three of them have. Find the largest
possible sum of their marbles.

the question is essentially maximize $M+N+P$ with the restraint $5(MN+MP+PN)=7MNP$ and $M,N,P$ are positive integers

I tried using lagrange multipliers but it didn't seem to work, I think because the degree of the expression is 1.

the  expression shows that
atleast one of the variables are a multiple of 5

hints, suggestions and solutions would all be appreciated

taken from the 2017 InImc https://chiuchang.org/wp-content/uploads/sites/2/2018/01/2017_IWYMIC_answer-2.x17381.pdf

Actually, we can do this question in a simple way.
WLOG, let $M<N<P$.

$\dfrac{1}{M}+\dfrac{1}{N}+\dfrac{1}{P}=\dfrac{MN+MP+NP}{MNP}=\dfrac{7}{5}$

If $M=2$, then $\dfrac{1}{M}+\dfrac{1}{N}+\dfrac{1}{P}<\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}=\dfrac{13}{12}<\dfrac{7}{5}$, therefore $M=1, \dfrac{1}{N}+\dfrac{1}{P}=\dfrac{2}{5}$

Rearrange it and we get $\left(2N-5\right)\left(2P-5\right)=25$

$\because N<P \\ \therefore \begin{cases} 2N-5=1 \\ 2P-5=25 \end{cases} \rightarrow \begin{cases} N=3 \\ P=15 \end{cases}$

$M+N+P=1+3+15=\boxed{19}$

Was the constraint $5(MN+ MP+ PN)= 7MNP$?  What you have "$5(MN+ MP+ PM)= 7MNP$"  doesn't make sense (MP+ PM= 2MP).. Using Lagrange multipliers write $f(M,N,P)= M+ N+ P$  and $g(M,N,P)= 5(MN+ MP+ PN)- 7MNP$. Then $\nabla f= <1, 1, 1>$ and $\nabla g= <5N+ 5P- 7NP, 5M+ 5P- 7MP, 5M+ 5N- 7MN>$. So we must have $5N+ 5P- 7NP= \lambda$, $5M+ 5P- 7MP= \lambda$, and $5M+ 5N- 7MN=\lambda$.  Since those are all equal to $\lambda$, they are all equal to each other:
5N+ 5P- 7NP= 5M+ 5P- 7MP so 5N- 7NP= 5M- 7MP or 5(N-M)= 7P(N-M) or, as long as N is not equal to M, 5= 7P and P= 5/7.  And 5M+ 5P- 7MP= 5M+ 5N- 7MN so 5P- 7MP= 5N- 7MN or 5(P- M)= 7M(P- M) so, as long as P is not equal to M, M= 5/7.  Also, of course, M, N, and P must satisfy the restraint 5MN+ 5MP+ 5PN= 7MNP.  We also have the constraint, 5(MN+ MP+ PN)= 7MNP.


Two questions on the existence of sets associated with LaSalle invariance principle I am copying in verbatim a standard reference on Lasalle's theorem (which is Hassan Khalil's textbook, page 128)
Theorem:

Let $\Omega \subset D$ be a compact set that is positively invariant
with respect to an ODE $\dot x = f(x)$. Let $V: D \to \mathbb{R}$ be a
continuously differentiable function such that $\dot V(x) \leq 0$ in
$\Omega$. Let $E$ be the set of all points in $\Omega$ where $\dot
 V(x) = 0$. Let $M$ be the largest invariant set in $E$. Then every
solution trajectory starting in $\Omega$ approaches $M$ as $t \to
 \infty$.

Here are my two questions about this theorem:

How would you know if there exists such a positively invariant set $\Omega$ in the first place? And how would you construct this set $\Omega$?

How would you know there exists an invariant set $M$ in $E$ in the first place? And how would you construct this set $M$?

Thank you.

For added context, this question involves an extremely well studied/known theorem in dynamical system theory. The text in reference is cited 40,000 times, hence not some obscure result. However, very little is said about the existence (or construction) of certain sets used in this theorem and most examples do not verify the existence of such sets, or does so in a case-by-case basis with no universal method of constructing these sets.

LaSalle's principle assumes the existence of $\Omega$, so you don't need to construct it to prove the theorem. If you are trying to apply the theorem, then usually there's an obvious candidate for $\Omega$, such as an equilibrium point, bounded set containing the solution, etc.

The null set is (trivially) always an invariant set in $E$. To construct $M$, let $\mathcal{M}$ be the set of all invariant subsets of $E$ (nonempty because $\varnothing\in \mathcal{M}$) and set $M = \cup \mathcal{M}$. Again, it is usually easy to tell what $M$ is (e.g. an equilibrium) from the specific problem.