Is there a good proof that all the polynomials in this family are irreducible? Writing the few lines below in PARI/GP, one easily checks that the polynomials

$$(X^3-4019680)-(a_2X^2+a_1X+a_0)$$

are all irreducible over $\mathbb Z$ when $a_0,a_1$ and $a_2$ are integers between $0$ and $20$. Is there a computer-free (or more illuminating) proof of this fact ?

ff3(a2,a1,a0)=1-polisirreducible(x^3-(a2*x+a1*x+4019680+a0))

vv1=vector(21^3,k,7)

for(a2=0,20,for(a1=0,20,for(a0=0,20,
vv1[(21^2)*a2+21*a1+a0+1]=ff3(a2,a1,a0);););)

check=sum(k=1,length(vv1),vv1[k])

Sketch. Because the constant term is so large relative to the other terms, possible integer roots of such a polynomial lie in the relatively small interval $[160, 165]$ far to the right of the origin. The fact that $X_0$ is a root is equivalent to $4019680$ having a representation in base $X_0$ with particularly large coefficients, and by writing $4019680$ in the corresponding bases we see that it doesn't have such representations.

Proof. Reducibility is equivalent to the existence of an integer root. By inspection the polynomial can have at most one real root $X_0$, somewhere past $\sqrt[3]{4019680} = 159.0000131...$. By setting $a_0 = a_1 = a_2 = 20$ we check numerically that $X_0 < 166$, so the only possible roots are $X_0 = 160, 161, 162, 163, 164, 165$. Any integer root must divide the constant term, but:
$$4019680 \equiv 0 \bmod 160$$
$$4019680 \equiv 154 \bmod 161$$
$$4019680 \equiv 136 \bmod 162$$
$$4019680 \equiv 100 \bmod 163$$
$$4019680 \equiv 40 \bmod 164$$
$$4019680 \equiv 115 \bmod 165.$$

Since $0 \le a_0 \le 20$, it is impossible for $X_0$ to divide the constant term unless $X_0 = 160, a_0 = 0$ or $X_0 = 161, a_0 = 7$. In either case, after dividing out by $X_0$, $X_0$ must divide the new constant terms, which are $-(25123+a_1), -(24967+a_1)$ respectively. But
$$25123 \equiv 3 \bmod 160$$
$$24967 \equiv 12 \bmod 161$$

and again since $0 \le a_1 \le 20$, this is impossible.

If you are really interested only in $0 \leq a_i \leq 20$, write the polynom as product of two polynoms of degree 1 and 2 with unknown coefficients:

$(A x^2 + B x + C)(Dx + E) = ADx^3 + (BD+AE)x^2+(BE+CD)x+CE$

If I did it correct, this yields for any B, C, E:

$E \cdot C = a_0 - 4019680$
$C + B\cdot E = a_1$
$E+B = a_2$

But these euqations are not solveable for $0 \leq a_i \leq 20$.


In measure theoretic probability, why is $P_X = P \circ X^{-1}$ the same as the CDF $F_X$ Suppose that we are given a random variable $X: \Omega \to \mathbb{R}$ in a probability space $(\Omega, \mathcal{A}, P)$.

Then the expectation is given by,

$$
E(X) = \int X(w) dP(w)
$$

I am trying to make sense of the "change of variable" formula,

$$
E(X) = \int_\Omega X(w) dP(w) = \int_\mathbb{R} x dP_X(x) = \int_\mathbb{R}x dF_X(x)
$$

where $P_X$ is the distribution of $X$ and $F_X(x)$ is the CDF of $X$.

I am stuck at various points:

To start, let $x = X(w) \Leftrightarrow X^{-1}(x) = w$,

Then $P(w) = P(X^{-1}(x)) = (P \circ X^{-1})(x)$

Then define $P_X := P \circ X^{-1}$.

But then we claim that $P_X = F_X$.

Recall that $F_X(x) = P(X \leq x) = P(\{w \in \Omega | X(w) \leq x)$.

But $P \circ X^{-1}(x) =  P(w) \neq F_X(x)$          $\quad \quad \quad \quad \quad $ (?)

How do I proceed from here?

Start by assuming $X$ is a simple function and show the equalities in this case.  This case just amounts to $P_X$ having only point-masses and $F_X$ having finitely many jumps (and otherwise constant), so you can evaluate all the integrals directly.

Then assume $X \geq 0$ and approximate $X$ from below by simple functions (use the monotone convergence theorem).

Finally write $X$ as the difference of two non-negative random variables and, assuming at least one is integrable, conclude the equations hold whenever they make sense.  (The only case not dealt with by this treatment is one in which the integrals don't exist)

Suppose that we are given a random variable $X: \Omega \to \mathbb{R}$ in a probability space $(\Omega, \mathcal{A}, P)$.
Then the expectation is given by,
$$ E(X) = \int X(\omega) dP(\omega)$$

No, it is:
$$\mathsf E(X)=\int_\Omega X(\omega)~\mathsf P(\mathrm d\omega)$$

I am trying to make sense of the "change of variable" formula,

$$
\mathsf E(X) = \int_\Omega X(\omega)~\mathsf P(\mathrm d\omega) = \int_\mathbb{R} x ~\mathsf P_X(\mathrm dx) = \int_{\mathbb{R}}x~\dfrac{\mathrm dF_X(x)}{\mathrm d x}\mathrm d x
$$


Matrix group isomorphic to $\mathbb Z$. The set  $G=\left\{\begin{pmatrix}1 & n  \\
        0 & 1  \\
        \end{pmatrix}\mid  n\in \Bbb Z\right\}$ with the operation of matrix multiplication is a group. Show that $$\phi:\Bbb Z \to G,$$ $$\phi(n)=\begin{pmatrix}1 & n  \\
        0 & 1  \\
        \end{pmatrix}$$
is a group isomorphism (where the operation on $\Bbb Z$ is ordinary addtion).

TO show it's isomorphism:  I know I must show one-to-one, onto and homomorphism. I've done these examples before but never with matrices.

How can I show if $\phi(a)=\phi(b)$ then $a=b$? Same question for onto and operation preserving with matrices.

Thank you!

Hint: For your first question, write down $\phi(a)$ and $\phi(b)$ (go ahead, write down the matrices on a sheet of paper). Now, if those two are equal, what does it tell you?

The other parts are obviously different, but the idea is the same: just look at the matrices involved and use what you know about matrix multiplication (for the last part).

Hint:

There are other ways to show that this map is one-to-one.

Use that the $ker(\phi)=0$.

Does there exist an $n$ such that the rank of
$\phi(n)=\begin{pmatrix}1 & n  \\
        0 & 1  \\
        \end{pmatrix}$ is not $2$?

Alternatively, construct $\phi^{-1}(n)=\begin{pmatrix}1 & -n  \\
        0 & 1  \\
        \end{pmatrix}$. The existence of an inverse with the homomorphic property implies it's an isomorphism.


Proof that $x^2+4xy+y^2=1$ has infinitely many integer solutions The question would, naturally, be very straight forward if there was a $2xy$ instead of a $4xy$. Then it would simply be a matter of doing:

$$
x^2+2xy+y^2=1\\
(x+y)^2=1\\
\sqrt{(x+y)^2}=\sqrt{1}\\
|x+y|=1
$$

Clearly, there are infinitely many pairs of integers that differ by one so there is an infinite number of integer solutions to $x^2+2xy+y^2=1$. Unfortunately, the same principle does not apply to $x^2+4xy+y^2=1$ where if I attempt to construct a similar proof all I can do is:

$$
x^2+4xy+y^2=1\\
(x+y)^2+2xy=1\\
2xy=1-(x+y)^2\\
2xy=(1+x+y)(1-x-y)
$$

And, from there, I have no idea how to proceed to complete the proof that there are an infinite number of integer solutions. I'm wondering whether I'm approaching the question entirely in the wrong way or if I am simply missing something. A nudge in the right direction would be much appreciated!

Hint:

Have you learnt about Pell's equation?

Try adding a multiple of $y^2$ (on both sides) to complete the square on the left.

It's easy...
Keep 'Y' constant and vary 'X'.
For single value of 'Y' You will get two values of 'X',
Go on varying 'Y' you will get different value of 'X' for each 'Y'.


Dimension of quotients of Noetherian rings Suppose that $A$ is a Noetherian ring containing a field $k$ and consider an ideal $I$ with $J =\sqrt{I}$ the radical of $I$. Suppose that $A/J$ is a finite-dimensional vector space over $k$. I am trying to prove that this implies that $A/I$ is also a finite-dimensional vector space over $k$.

I know that since $A$ is Noetherian there exists $m$ such that $J^m \subset I$ and so I was thinking of considering the filtration $J^m \subset \cdots \subset J^2 \subset J \subset A$ and then saying something like $\dim_k J^i/J^{i+1} = \dim_k A/J$ for all $i$ up to $i = m$. Then I would claim that $\dim_k A/I \le m \cdot \dim_k A/J$ using exact sequences like
$$
0 \rightarrow J/J^2 \rightarrow A/J^2 \rightarrow  A/J \rightarrow 0.
$$
If this works, then $A/I$ is finite-dimensional. My issue is with the statement that $\dim_k J^i/J^{i+1} = \dim_k A/J$ which I think is incorrect. Is this the right approach? It seems like something is missing.

$\dim_k J^i/J^{i+1}<\infty$ for all $i\ge 1$ for the following reason: $J^i/J^{i+1}$ is an $A/J$-module (since $J(J^i/J^{i+1})=0$), $J^i$ is finitely generated (since $A$ is noetherian), therefore $J^i/J^{i+1}$ is a finitely generated $A/J$-module. Furthermore, since $A/J$ is a finitely generated $k$-module we can deduce that $J^i/J^{i+1}$ is a finitely generated $k$-module and thus $\dim_k J^i/J^{i+1}<\infty$.

Now, step by step, one can prove that $\dim_kA/J^i<\infty$ for all $i\ge 1$.

Hint: You don't need $\dim_k J^i/J^{i + 1}$ to be independent of $i$, you just need it to be finite for each $i$ up to $m$.  Use the Nakayama lemma.


What is the method used to find the singular points of an ODE? Suppose you have an ODE, say
$$
x^{2} (x+1) y'' +  2y' + xy = 0
$$
How would you find the singular points of this?

I've looked online for an explanation of the method used to do this, but have not found any that I understand very well. Can anyone outline the method used to do this?

Re-writing
$$y''+\frac{2}{x^{2}(x+1)}y'+\frac{1}{x(x+1)}y$$
You see that the singular points are $x=0, -1$.
See http://www.math.washington.edu/~toro/Courses/99-00/135/regular.pdf
For a further discussion

Further Edit
Write the most general ODE
$$N(x)y''+P(x)y'+ Q(x) = 0$$
in standard form
$$y''+p(x)y'+q(x)y=0$$

Let $x_{0}$ be a singular point, and, multiply by $(x-x_{0})^{2}/N(x)$ one can write
$$(x-x_{0})^{2}y''+(x-x_{0})u(x)y'+v(x)y=0$$ where
\begin{align}
u(x) &= \frac{(x-x_{0})P(x)}{N(x)}\\
v(x) &= \frac{(x-x_{0})^{2}Q(x)}{N(x)}
\end{align}
$x_{0}$ is then regular if $u(x)$ and $v(x)$ have no singularity at $x_{0}$; otherwise it is an irregular point.

I have regurgitated a lot of the detail in the link I have posted, see the theory on page 1.

Cusp points when $ y''$ goes to infinity. Here at $ x=0,-1$there would be curve reversal/sudden direction change etc.


Manipulating equations question In the equation: $$T = 2\pi \sqrt {\frac lg}$$

it is for determining period of pendulum swing

If I want to solve for $g$ and I want to start by removing the root

do I square everything in the equation, on just the left hand side?
thank you

If two things are equal, for example $A=B$, then whatever we do on the left-hand side (LHS) we have to do on the right-hand side (RHS), otherwise they are not equal anymore, e.g. $A = B \rightarrow A^2 = B^2$ or $A+3=B+3$. So yes, you have to square both sides. Given this, can you give an expression for $g$?

Notice, in order to get the value of $g$ $$T = 2\pi \sqrt {\frac lg}$$ $$(T)^2 =\left( 2\pi \sqrt {\frac lg}\right)^2$$ $$T^2 =4\pi^2\frac{l}{g}$$ $$\bbox[5px, border: 2px solid #C0A000]{\color{red}{g=\frac{4\pi^2l}{T^2}}}$$
Now, in order to calculate the acceleration due to Earth's gravity at particular point, substitute the time period $T$ & length of pendulum $l$ in the above relation.


Set of infinite subsets. Is it a topology? Here is the question:

  Let $X = \mathbb{R}$ and let $\Omega$ consist of the empty set and all infinite subsets of $\mathbb{R}$. Is $\Omega$ a topological structure?

My attempt : I think the answer is No; it is not a topology.
Because we have $$ \Omega =  \lbrace \varnothing \rbrace \cup \lbrace U \subseteq \mathbb{R} \mid \text{$U$ is infinite} \rbrace $$

If we check the axiom for the finite intersection property then let $U_1, U_2 ..U_n$ be finite elements of $\Omega$ now $\cap_{i=1}^{n} U_i$ might be finite and thus doesn't belong to $\Omega$. Is this explanation correct?

EDIT: Let $U_1$ be set of all positive even numbers and $U_2$ be the set of all primes. Now we know both $U_1$ and $U_2$ are elements of $\Omega$ as they are infinite sets but $ U_1 \cap U_2 = \lbrace 2 \rbrace $ which is finite.

No, because two infinite sets may have a non-empty finite intersection (for example, the intersection of prime numbers and even numbers is $\{2\}$). This violates the condition that the intersection of two open sets be open in a topology.

However, if one requires not only that the non-empty “open” sets be infinite but also that their complements be finite (which is a stronger condition), then one has a topology. It is called the cofinite topology or finite-complement topology: $$\Omega=\{\varnothing\}\cup\{U\subseteq R\,|\,U^{\mathsf c}\text{ is finite}\}.$$

I think the answer is no,it is not a topology Because it does not satisfied the 1st condition of topology.As taa consist of empty set and all infinite subsets of $\mathbb{R}$ But it does not have $\mathbb{R}$ in it. According to the first condition of topology, empty set and $X${ground set} which is $\mathbb{R}$ belong to taa but it have all infinite subsets of $\mathbb{R}$,  not $\mathbb{R}$ itself. That's why in my view's taa is not a topology


Prove $\lim\limits_{n \to \infty} \int_0^1 x^n e^x dx = 0$ by evaluating closed form While preparing for my numerical analysis exam I encountered the following question

  Find a recursive definition $I_n = R(I_{n - 1})$ for $I_n := \int_{0}^{1} x^n e^x dx$ and prove
  $$
I_n = \sum_{k = 0}^{n} (-1)^{k} \frac{n! e}{(n - k)!} - (-1)^n n!
$$
  and give an explanation for the behaviour of $I_n$ for $n \to \infty$.

With integration by parts I found $I_n = e - n I_{n - 1}$, $I_0 = e - 1$ and proved the formula with induction.
Just for reference, is the following induction step correct?
\begin{align}
I_{n + 1} = e - (n + 1)I_n
& = e - (n + 1)\left( \sum_{k = 0}^{n} (-1)^{k} \frac{n! e}{(n - k)!} - (-1)^n n!\right) \\
& = e - \sum_{k = 0}^{n} (-1)^{k} \frac{(n + 1)! e}{(n - k)!} - (-1)^{n + 1} (n + 1)! \\
& = e\left( 1 - \sum_{k = 0}^{n} (-1)^{k} \frac{(n + 1)!}{(n - k)!} \right) - (-1)^{n + 1} (n + 1)! \\
& = e\left( 1 - \sum_{k = 1}^{n + 1} (-1)^{k - 1} \frac{(n + 1)!}{(n - k + 1)!} \right) - (-1)^{n + 1} (n + 1)! \\
& = \sum_{k = 0}^{n + 1} (-1)^{k} \frac{(n + 1)! e}{(n + 1 - k)!} - (-1)^{n + 1} (n + 1)!
\end{align}
Now I am trying to investigate the behaviour for $n \to \infty$.
First I rewrote
$$
I_n
= \sum_{k = 0}^{n} (-1)^k n! \left( \frac{e}{(n - k)!} - (-1)^{n - k}\right)
$$
Wolfram Alpha yields
$$
= (-1)^{n + 1} n! \left( n + 1 - e \sum_{k = 0}^{n} \frac{(-1)^k}{k!} \right),
$$
but I don't know how to proceed from here.
I though that the last term would cancel out since $\sum_{k = 0}^{\infty} \frac{(-1)^k}{k!} = \frac{1}{e}$ but I am not sure.

On the question paper, we are given the following numerical values:

0.2280015155 for n = 10,
0.1238038308 for n = 20,
0.08502696925 for n = 30,
0.06475689045 for n = 40 and
0.05229363830 for n = 50.

They are all positive and seem to be converging to 0.
But because of the $(-1)^{n + 1}$ term I wouldn't expect this to converge.

Intuitively the result make perfect sense because for larger $n$ the $x^n e^x$ is very small for "most of" $[0,1]$ and then "in the last moment" rises steeply to $(1,1)$.

Your induction step seems correct to me.
However, your approach is not best for the asymptotic behaviour. I suggest that you prove for instance that $0 \leq I_n \leq e\int_0^1{x^n\,dx}$.

Using integration by parts, we obtain:
$$I(n)=\int_0^1x^ne^xdx=\left[x^ne^x\right]_0^1-n\int_0^1x^{n-1}e^xdx$$
so we could say that:
$$I(n)=\left[x^ne^x\right]_0^1-nI_{n-1}=\left[x^ne^x\right]_0^1-\frac{dI}{dn}$$
You could maybe try to solve this using the differential equation (not sure on this):
$$I+I'=e$$
However if we continue with the pattern of using IBP we notice:
$$I(n)=\left[x^ne^x-nx^{n-1}e^x\right]_0^1+n(n-1)\int_0^1x^{n-2}e^xdx$$
$$=\left[x^ne^x-nx^{n-1}e^x+n(n-1)x^{n-2}e^x\right]_0^1-n(n-1)(n-2)\int_0^1x^{n-3}e^xdx$$
If we were to do this $n$ times we would get:
$$I(n)=\left[e^x\left(\sum_{i=0}^n\frac{(-1)^in!x^{n-i}}{(n-i)!}\right)\right]_0^1-n!\int_0^1e^xdx$$
$$=e\sum_{i=0}^n\frac{(-1)^in!}{(n-i)!}-en!$$
if we notice that $en!$ is just the first term of this series, we can say that:
$$I(n)=e\sum_{i=1}^n\frac{(-1)^in!}{(n-i)!}$$


Derivative of sigmoid function $\sigma (x) = \frac{1}{1+e^{-x}}$ In my AI textbook there is this paragraph, without any explanation.

  The sigmoid function is defined as follows

  $$\sigma (x) = \frac{1}{1+e^{-x}}.$$

  This function is easy to differentiate because

  $$\frac{d\sigma (x)}{d(x)} = \sigma (x)\cdot (1-\sigma(x)).$$

It has been a long time since I've taken differential equations, so could anyone tell me how they got from the first equation to the second?

Let's denote the sigmoid function as $\sigma(x) = \dfrac{1}{1 + e^{-x}}$.

The derivative of the sigmoid is $\dfrac{d}{dx}\sigma(x) = \sigma(x)(1 - \sigma(x))$.

Here's a detailed derivation:

$$
\begin{align}
\dfrac{d}{dx} \sigma(x) &= \dfrac{d}{dx} \left[ \dfrac{1}{1 + e^{-x}} \right] \\
&= \dfrac{d}{dx} \left( 1 + \mathrm{e}^{-x} \right)^{-1} \\
&= -(1 + e^{-x})^{-2}(-e^{-x}) \\
&= \dfrac{e^{-x}}{\left(1 + e^{-x}\right)^2} \\
&= \dfrac{1}{1 + e^{-x}\ } \cdot \dfrac{e^{-x}}{1 + e^{-x}}  \\
&= \dfrac{1}{1 + e^{-x}\ } \cdot \dfrac{(1 + e^{-x}) - 1}{1 + e^{-x}}  \\
&= \dfrac{1}{1 + e^{-x}\ } \cdot \left( \dfrac{1 + e^{-x}}{1 + e^{-x}} - \dfrac{1}{1 + e^{-x}} \right) \\
&= \dfrac{1}{1 + e^{-x}\ } \cdot \left( 1 - \dfrac{1}{1 + e^{-x}} \right) \\
&= \sigma(x) \cdot (1 - \sigma(x))
\end{align}
$$

Using my HP Prime, I differentiated 1/(1+exp(-x)) to get exp(-x)/(1+exp(-x))^2.  Factor out 1/(1+exp(-x), which is sigma(x), and the rest is 1-sigma(x).  That is proof by calculator. Beware!  That machine can become addictive because of the way it amplifies your capability.


What is the pre-image of $f(x) = \frac{1 }{1 + x^2}$ at $\{0\}$ I need to find the pre-image at $\{0\}$ with the funtion in the domain $$f: \mathbb{R} \to \mathbb{R}$$

$$f(x) = \frac{1 }{1 + x^2}$$

I first found the inverse function

$$f^{-1} = y = \frac{1 }{1 + x^2}$$

$$x = \frac{1 }{1 + y^2}$$
$$1 + y^2 = \frac{1 }{x}$$
$$y = (\frac{1 }{x} - 1)^{\frac{1 }{2}} = f^{-1}(x)$$

Then I replace $x$ with $0$:
$$ f^{-1}({0}) = (\frac{1 }{0} - 1)^{\frac{1}{2}} $$

But since you can't divide by $0$, their is no pre-image at $0$ for this function in the given domain and range. Am I correct?

A  bit of support to Sadek, rephrased:

$\dfrac{1}{1+x^2} = 0$

has no solution for $x \in \mathbb{R}.$

$f(x) = \dfrac{1}{1+x^2} : $

Domain: $D= \mathbb{R}.$  Range: $f(D) = (0,1],$

$0\not\in f(D).$

$\frac 1 {1+x^2}=0$  is imposible (because it gives $1=0$) therefore $f^{-1}(\{0\})=\emptyset$

again $f^{-1}$ is not the inverse function of $f$
it is the inverse image of $f$ and takes a set as an argument and gives back a set


Prove that if $U$ and $W$ are 3-dimensional subspaces of $\mathbb{R}^5$ then $U ∩ W$ is non-trivial I said that since a 3-dimensional subspace is the span of 3 linearly independent vectors then $U$ and $W$ must be linearly independent.

Since $U$ and $W$ are linearly independent then they span the entire 3-dimensional space.

Since $U$ and $W$ span the entire space, they intersect everywhere.

Therefore, $U \cap W$ is non-trivial for every vector in $\mathbb{R}^3$ except for the null vector.

This was my reasoning but can anybody tell me if my thinking is correct. I feel like I am missing something since I did not make use of $\mathbb{R}^5$.

Choose bases $A=\{u_1,u_2,u_3\}$ and $B=\{w_1,w_2,w_3\}$ for $U$ and $W$ respectively. If $A \cap B$ is nonempty we are done. Suppose not. Then $A \cup B$ contains $6$ distinct vectors in $\mathbb{R}^5$, a $5$ dimensional space, so they must be linearly dependent. This means there exist scalars $\lambda_i, \mu_i$ not all equal to zero such that
$$\sum_i \lambda_i u_i +\sum_i \mu_i w_i=0.$$
Taking the right sum to the other side we see that $\sum_i \lambda_i u_i \in W$. Since it is also in $U$, it is in $U \cap W$. It is nonzero because if it would be zero then the $\lambda_i$ would all be zero and by linear independence of the $u_i$ and the $\mu_i$ would be zero as well by linear independence of the $w_i$.  This shows $U \cap W$ contains a nonzero element.

Recall that $\dim(U\cap V) + \dim(U+V) = \dim U + \dim V=6$. If the intersection was trivial, $U+V$ would have dimension $6$, which of course is impossible.


What, if anything, is a square radian? My 1st year Mathematics BSc course notes on circular motion use
\begin{align}
\frac{d}{dt}(\sin\theta)
&=\frac{d}{dt}(\sin(\omega t))\tag{1.1}\\
&=\omega\cos(\omega t),\tag{1.2}\\
\frac{d^2}{dt^2}(\sin\theta)
&=\frac{d^2}{dt^2}(\sin(\omega t))\tag{2.1}\\
&=\frac{d}{dt}(\omega\cos(\omega t))\tag{2.2}\\
&=-\omega^2\cos(\omega t),\tag{2.3}
\end{align}
where

$\omega$ is angular speed, a scalar, with units $\text{rad}\cdot s^{-1}$;
$t$ is time, a scalar, with units $s$.

From this it seems to me to follow that $\omega^2$ has units $(\text{rad}\cdot s^{-1})^2=\text{rad}^2\cdot s^{-2}$. Is that right, and, if so, does a square radian have a physical meaning?

"rad(ians)" is just another word for the number $1$, a unit-less unit. An angle is some part of a full rotation at $2\pi$, a unitless number. However, to tell that any number stands for an angle and not some amount of apples or cake, there is the habit to add "radians" to it.

OP answering own question.

Remarks by Hagen von Eitzen and Lutz Lehmann have prompted me to think of the following answer.

A radian is a dimensionless fraction with value
$$\frac{1}{2\pi}.$$

It seems to me to follow that a square radian is a dimensionless fraction with value
$$\left(\frac{1}{2\pi}\right)^2=\frac{1}{4\pi^2}.$$

This can be thought of as the angle by which the angular speed increases each second.


Finding a Coordinate on a circle using radius, angle, and origin I am trying to calculate a point on a circle using an angle and a different point.

With this picture, I know the origin O, the radius r, the angle A, and the point B. Now I want to find the point C. How could I do this? (Note: the origin is not at (0,0))

There are several different approaches you can take.  Since you're asking about trigonometry and polar coordinates, here's an approach:

Find the angle to B from the horizontal.
Find the angle to C by subtracting A.
Find C using its angle with r

The angle to B can be found with $\theta_B = \sin^{-1}(B_y/B_x)$ where $B_y$ is the $y$ coordinate of B.

Then the angle to C is $\theta_C = \theta_B - A$

Finally, $C = (C_x,C_y) = (r \cos(\theta_C), r \sin(\theta_C)) $

If your origin $O$ is not at $(0,0)$, you can shift the whole thing down so that $O$ is temporarily at $(0,0)$, then you just shift your result back by the opposite amount.

I will use $\alpha$ instead of $A$. suppose first that O(0,0). Note that every point on the circle has the coordinates $(R\cos\theta,R\sin\theta)$. If $\theta_B$ is the angle of point B, then the angle $\theta_C$ of point C is $\theta_C=\theta_B-\alpha$. Hence, the coordinates of $C$ are
$$
(R\cos(\theta_B-\alpha),R\sin(\theta_B-\alpha))
$$
that is
$$
(R\cos\theta_B\cos\alpha+R\sin\theta_B\sin\alpha,R\sin\theta_B\cos\alpha-R\cos\theta_B\sin\alpha)
$$
By setting $x_B=R\cos\theta_B$, $y_B=R\sin\theta_B$ we get
$$
(x_B\cos\alpha+y_B\sin\alpha,y_B\cos\alpha-x_B\sin\alpha)
$$
that is
$$
\begin{pmatrix}
 \cos\alpha  & \sin\alpha \\
 -\sin\alpha & \cos\alpha \\
\end{pmatrix}
\begin{pmatrix}
   x_B \\ y_B
\end{pmatrix}
$$
in matrix form. Now, if $O\neq(0,0)$, say $O(o_1,o_2)$, then
$$
\begin{pmatrix}
   o_1 \\ o_2
\end{pmatrix}+\begin{pmatrix}
 \cos\alpha  & \sin\alpha \\
 -\sin\alpha & \cos\alpha \\
\end{pmatrix}
\begin{pmatrix}
   x_B-o_1 \\ y_B-o_2
\end{pmatrix}
$$
is the desired coordinates.


Probability of a sequence of random letters Suppose an the letters of  a  random sequence  are chosen independently and uniformly  from the set of 26 English alphabet .If the sequence contains n letters what is the probability that it contains  the word "Love" as a  subsequence?Thanks a lot for nay help in advance

The way of dealing with such questions is to think clearly about what is involved. Suppose I am looking for the word "the". Let the probablity that it appears in the first $n$ letters be $p_n$.

Now if "the" appears, either it appears in the first $n-1$ letters, or it appears for the first time at the $n^{th}$ letter. In this case the last three letters chosen are "the" with probability $\frac 1{26^3}$, and the first $n-3$ letters do not contain the word "the".

I am left with $$p_n=p_{n-1}+\frac 1{26^3}(1-p_{n-3})$$ and $p_0=p_1=p_2=0$. This is now a recurrence which can be explicitly solved.

If I were looking for the single letter "a" the same approach would give me $p_n=p_{n-1}+\frac 1{26}(1-p_{n-1})$ or $p_n = \frac {25}{26}p_{n-1}+\frac 1{26}$

The solution to this is $p_n=1+A\left(\frac{25}{26}\right)^n$ and $p_0=0$ gives $A=-1$, which checks with simpler ways of computing, which are available for a single letter.

The total number of combinations is $n!$

The total number of combinations including the word 'love' is $(n-3)!$
This is because we can treat the sequence of the four letters 'love' as one object, so there are only n-3 to arrange.

so $P($String contains 'love'$) = (n-3)!/(n)! = 1/(n*(n-1)*(n-2))$


The indefinite integration of $\frac{1}{\sqrt{n^4-1}}$ I need the indefinite integral:

$$\int\frac{1}{\sqrt{n^4-1}}dn $$

I know it has a relation with the $\tanh^{-1}$ function, but can't find a proper substitution.

Introduce variables $c, s$ and $\theta$ such that

$$\cos\theta = c = \frac{1}{n}\quad\text{ and }\quad \sin\theta = s = \sqrt{1-c^2} = \frac{\sqrt{n^2-1}}{n}.$$
We have
$$\begin{align}
\int \frac{dn}{\sqrt{n^4-1}}
&= - \int \frac{dc}{c^2\sqrt{\frac{1}{c^4}-1}}
= - \int \frac{dc}{\sqrt{1-c^4}}\\
&= - \frac{1}{\sqrt{2}} \int \frac{dc}{\sqrt{(1-c^2)(1 - \frac12(1-c^2))}}\\
&= \frac{1}{\sqrt{2}}\int \frac{ds}{\sqrt{(1-s^2)(1-\frac12 s^2)}}\\
&= \frac{1}{\sqrt{2}}\int \frac{d\theta}{\sqrt{1-\frac12 \sin^2\theta}}\\
\end{align}$$
The integrals in last two lines are in the Jacobi's form and regular form of the incomplete elliptic integral of the first kind for modulus $m = \frac12$:

$$
\begin{align}
F(x;m)     &= \int_0^x \frac{dt}{\sqrt{(1-t^2)(1-mt^2)}}\\
F(\phi\mid m) &= \int_0^\phi \frac{d\theta}{\sqrt{1-m\sin^2\theta}} = F(\sin\phi;m)
\end{align}
$$

As a result

$$\int_1^x \frac{dn}{\sqrt{n^4-1}}
= \frac{1}{\sqrt{2}}F\left(\frac{\sqrt{x^2-1}}{x};\frac12\right)
= \frac{1}{\sqrt{2}}F\left(\cos^{-1}\frac{1}{x}\bigg|\frac12\right)
$$
On WolframAlpha, the regular form of the incomplete elliptic integral can be accessed with the function EllipticF. The integral can be evaluated using following expression (for real $x > 1$):
$$\bf 1/\text{Sqrt}[2]*\text{EllipticF}[\text{ArcCos}[1/x],1/2]$$

Put n=tanx, after making substitution you will be left with a cosine function which is easy to integrate


Good final exam bonus problem for calculus students? I'm making a final exam for a first course in calculus at a university.  I need some suggestions for a good bonus problem.  By "good" I mean an interesting problem that some of the brighter students could solve in about 15 minutes.

Hopefully it should involve limits, derivatives, or integrals.

The problem should NOT involve trig functions.

The problem should NOT be theoretical. These students can't do real analysis proofs, for instance.

Here are a few interesting problems:

Let $f^n(x)$ denote the nth iterate of a function (for example, $f^2(x)=(f\circ f)(x)$). If $f$ has a fixed point at $x=x_0$ (meaning that $f(x_0)=x_0$) then what is $\frac{d}{dx}f^n(x_0)$ in terms of $f'(x)$?
Prove that the intersection of the parabola and the rectangle in the picture is $\frac{2}{3}$ the area of the rectangle:

A triangle starts is an equilateral triangle $ABC$ at $t=0$. However, each second, both $BC$ and the length of the altitude to $BC$ increase by $1$ unit. What does the measure of angle $BAC$ approach as $t$ approaches $\infty$?

Browse old problems from the Putnam exam. http://kskedlaya.org/putnam-archive/


Denest $\sqrt{\left(2+\sqrt{2}\right) \left(10-2 \sqrt{5}\right)}$ Is it possible to denest following radical to sum of terms with smaller root count inside?

$\sqrt{20+10 \sqrt{2}-4 \sqrt{5}-2 \sqrt{10}}=\sqrt{\left(2+\sqrt{2}\right) \left(10-2 \sqrt{5}\right)}$

I've found that it cannot be denested into $a+b\sqrt{2}+c\sqrt{5}+d\sqrt{10}$ by squaring and equating coefficients. Resulting system didn't have rational solutions.

Maybe it can be denested to something as sum of $\sqrt{a+b\sqrt{c}}$ or $\sqrt{a+b\sqrt{c}+d\sqrt{e}}$ or even $\sqrt[4]{a+b\sqrt{c}}$?

For example, following can be denested:
$$\sqrt{8+2 \sqrt{2}-2 \sqrt{6}}=\sqrt{6+3\sqrt{2}}-\sqrt{2-\sqrt{2}}$$

$$\sqrt{110-60 \sqrt{3}+46 \sqrt{5}-28 \sqrt{15}}=2\sqrt{15+6\sqrt{5}}-\sqrt{50+22\sqrt{5}}$$

$$\sqrt{48+12 \sqrt{2}+16 \sqrt{5}-12 \sqrt{6}+4\sqrt{10}-4\sqrt{30}}=$$
$$=\sqrt{6+3\sqrt{2}}-\sqrt{2-\sqrt{2}}+\sqrt{30+15\sqrt{2}}-\sqrt{10-5\sqrt{2}}$$

Well if you're still looking for something:

$$\sqrt{2+\sqrt{2}}=\sqrt{\frac{2+\sqrt{2}}{2}}+\sqrt{\frac{2-\sqrt{2}}{2}}$$
and

$$\sqrt{10-2\sqrt{5}}=2 \cdot \sqrt{\frac{5-\sqrt{5}}{2}}$$

so
$$\sqrt{\left(2+\sqrt{2}\right) \cdot \left(10-2\sqrt{5}\right)}=2\cdot \sqrt{\frac{5-\sqrt{5}}{2}} \cdot \left(\sqrt{\frac{2+\sqrt{2}}{2}}+\sqrt{\frac{2-\sqrt{2}}{2}}\right)$$

Try factoring $2\sqrt{10}$ as $2\sqrt{5}\sqrt{2}$, then the expression under the radical can be factored.


How to count the number of elements in the given set I am stuck at this counting.
I am explaining my question with an example.

Suppose we take $n=2\times 3\times 5\times 7$.
I  want to find those elements in the set
$$X=\{7,14,21,28,35,42,49,56,\ldots ,203\}=\{7k\,|\, 1\leqslant 7k<n\}$$
which  have $2,3,5$ as one of their prime factors and count their number.
I call this number $x$.

The numbers in the set $X$  which  have $2,3,5$ as one of their prime factors are $$\{14,21,28,35,42,56,63,70,84,98,105,112,126,140,147,154,175,182,189,196\}.$$

I took the subgroup generated by $2\times 7=14$ which is $\{0,14,28,42,\ldots,196\}$.
I took the subgroup generated by $3 \times 7=21$ which is $\{0,21,42,63,\dots,189\}$.
I took the subgroup generated by $5\times 7=35$ which is $\{0,35,70,105,\dots,175\}$.

I found $x=$ number of elements in subgroup generated by $14$ + number of elements in subgroup generated by $21$+number of elements in subgroup generated by $35$  -number of elements in subgroup generated by $42$-number of elements in subgroup generated by $70$-number of elements in subgroup generated by $105=15+10+6-6-3-2=20$.

My question is:

  If $n=p_1^{\alpha_1}p_2^{\alpha_2}\cdots p_k^{\alpha_k}$
  and if we consider the cyclic subgroup generated by $p_k$ denoted by $\langle p_k\rangle$, what will be the number of those elements in $\langle p_k\rangle$ which have $p_i$ where $1\le i\le k-1$ as one of their prime factors?

I have provided as much detail as I could.
If I need to provide any other details please ask in the comment box

Regarding the example: Let's allow $0$ in $X$, then $X = 7\mathbb{Z}/n\mathbb{Z} \cong \mathbb{Z}/m\mathbb{Z} =: R$, where $m = 2\cdot 3\cdot 5$. You are searching for the number $x$ of Elements in $X$ which share a factor with $|R| = m$. Below I show that the number of $k\in\mathbb{N}, k\leq m$ which are coprime to $m$ is equal to the number of units (invertible elements) in $R$, which is given by $\varphi(m)$, where $\varphi$ is Euler's totient function.
Finally you get
$$x = m - \varphi(m) - 1 = 30 - 8 - 1 = 21 $$
(I'm subtracting $1$, because $X$ contains $0$ now. I don't know why our counts are off by one, though)

Edit: I just noticed I haven't really answered your generalized question, but I'd say it works just like in the special case from above.

Proof of the claim from above.

Let $\varphi : \mathbb{N} \longrightarrow \mathbb{N}, n \longmapsto |E(\mathbb{Z}_n)|$ and $M_n := \{k\in\mathbb{N} \mid k \leq n,\; \gcd(k,n) = 1 \}$.
We show $|M_n| = \varphi(n)$. Choose an arbitrary $n\in\mathbb{N}$.

''$\leq$'': Let $m \in M_n$. Then the Euclidian algorithm gives $k,\ell \in \mathbb{Z}$ such that $1 = km + \ell n$. I.e., $\bar 1 = \overline{km} + \overline{\ell n} = \overline{km}$ in $\mathbb{Z}_n$, thus $(\bar m)^{-1} = \bar k$ and thus $\bar m \in E(Z_n)$. Because $m \leq n$ for all $m \in M_n$, we get $\bar m' \neq \bar m$ for $m'\neq m \in M_n$.

''$\geq$'': Let $\bar m \in E(\mathbb{Z}_n)$. Then there is $k \in \mathbb{Z}$ such that $1 = \overline{mk}$, hence there is $\ell \in \mathbb{Z}$ such that $1 = km + \ell n$. Thus $1 \in m\mathbb{Z} + n\mathbb{Z} = d\mathbb{Z}$. Such a  $d\in\mathbb{Z}$ exists, because $\mathbb{Z}$ is a principle ideal domain. Then $d$ is a gcd of $n$ and $m$. From $1 \in d\mathbb{Z}$ follows $d\mathbb{Z} = \mathbb{Z}$ and therefore $d\in \{1,-1\}$.

If you are talking about arithmetic sequence , you can easily count the number of its elements using this formula :

                  $a_{n} = a_{1} + (n-1)d$

where, $a_{n}$ is the last number in the arithmetic sequence, $a_{1}$ is the first number, $n$ is number of elements ,and $d$ is the difference between any two successive elemets

So in your question  the number of elements in the set is : $n=29$ using the above formula


$\sum\limits_{k=1}^n \frac{x^k}{k}=H_n+\sum\limits_{k=1}^n \binom{n}{k} \frac{(x-1)^k}{k}$ Prove:
$$\sum\limits_{k=1}^n \frac{x^k}{k}=H_n+\sum\limits_{k=1}^n \binom{n}{k} \frac{(x-1)^k}{k}$$
where $H_n=\sum\limits_{k=1}^n\frac1k$
For the $x^k$ i tried  $x^k=((x-1)+1)^k$ and decompose it into binomial expansion but I got dual sum at the one side one sum at the other side
For the second method:
$$\sum\limits_{k=1}^n \binom{n}{k} \frac{(x-1)^k}{k}-\sum\limits_{k=1}^n \frac{x^k}{k}=H_n$$
and try to show
$$\binom{n}{k}(x-1)^k-x^k=1$$
Either the problem is wrong or I cannot do it.

I think the identity does not hold. Using
$$ x^k=[1+(x-1)]^k=\sum_{i=0}^k\binom{k}{i}(1-x)^i=1+\sum_{i=1}^k\binom{k}{i}(1-x)^i $$
one has
\begin{eqnarray}
&&\sum\limits_{k=1}^n \frac{x^k}{k}\\
&=&\sum\limits_{k=1}^n\frac{1}{k}\bigg(1+\sum_{i=1}^k\binom{k}{i}(1-x)^i\bigg)\\
&=&H_n+\sum\limits_{k=1}^n\sum_{i=1}^k \binom{k}{i} \frac{(x-1)^i}{k}\\
&\neq&H_n+\sum\limits_{k=1}^n \binom{n}{k} \frac{(x-1)^k}{k}.
\end{eqnarray}

This identity does hold true
express both the RHS and LHS as integrals
then interchange the integrals and summations.
Hope this helps


Is 0.9 repeating = 1 disproved by asymptotes? I'm discussing proofs that 0.9 repeating equals 1 with some friends, and they use asymptotes to disprove this. One says if we had the function $y=x/0.000\ldots1$ (and he's only using that impossible number for theoretical purposes), the slope of the asymptote would be as close to undefined as you can get, but the value would be 0.9 repeating (if you let 1 represent infinity).

I don't know that much about asymptotes yet, but I'm comparing it to an infinite series, because he says that the function's graph forever approaches the number that is the value of the asymptote, but effectively never reaches it. The infinite series $1/2 + 1/4 + 1/8+\cdots$ forever approaches $1$, and you could say that it effectively never does, but even still the answer is exactly $1$. Is there a better argument against the asymptote argument?

EDIT: I know about the many valid proofs for this, but my problem is that they all refuse to accept those proofs. They're throwing invalid arguments my way, and since they reject mine, I'm stuck having to disprove theirs. I know those arguments can be disproved, just not how, and my question is how to disprove this one. Just clarifying.

$1/2 + 1/4 + 1/8 + \ldots$ is just a number. It doesn't "approach" anything: it simply is $1$.

It is the sequence of partial sums that approaches $1$: that is, the list of numbers

$1/2 = 1/2$
$3/4 = 1/2 + 1/4$
$7/8 = 1/2 + 1/4+ 1/8$
$15/16 = 1/2 + 1/4+ 1/8 + 1/16$
...

is a sequence of numbers whose limit is $1$.

$1/2 + 1/4 + 1/8 + \ldots$ is not this sequence: it is the number that is equal to the limit of this sequence.

$0.9$ repeating is equivalent to $1$ because the difference of them is $0.0000...1$ or $0.1$ to the power of infinity, but if there is an infinite amount of zeros there's no room for the one. You might say that it will always have enough room for one because it's infinite, but if you can add a one you can add a zero, making the number invalid for the difference, so if there can't be a one at the end it means it can't exist, therefore saying that there is no difference between $0.9$ repeating and $1$ because the difference between them cannot exist.


Find all functions $f(x)$ such that $f\left(x^2+f(y)\right)$=$y+(f(x))^2$ Let $\mathbb R$ denote the set of all real numbers. Find all function $f: R\to \ R$ such that   $$f\left(x^2+f(y)\right)=y+(f(x))^2$$   It is the problem. I tried to it by putting many at the place of $x$ and $y$ but I can't proceed. Please somebody help me.

Let me summarize the proposed ideas and give a self-contained answer: Let $f : \Bbb{R} \to \Bbb{R}$ satisfy the functional equation

$$ f(x^2 + f(y)) = y + f(x)^2 \tag{*}$$

for all $x, y \in \Bbb{R}$.

  Step 1. $f(0) = 0$.

(This solution is due to @Leo163.) Let $a \in \Bbb{R}$ be such that $f(a) = 0$. For instance, plugging $y = -f(x)^2$ to $\text{(*)}$ confirms the existence of such $a$. Then plugging $(x, y) = (a, a)$ to $\text{(*)}$ gives

$$ f(a^2) = f(a^2 + f(a)) = a + f(a)^2 = a.$$

Finally, plugging $(x, y) = (0, a^2)$ to $\text{(*)}$ gives

$$ 0 = f(a) = f(f(a^2)) = a^2 + f(0)^2. $$

This shows that $f(0) = 0$.

  Step 2. $f$ satisfies various identities:

  $f(x^2) = f(x)^2$.
  $f(f(x)) = x$. In particular, $f$ is bijective.
  $f(-x) = -f(x)$.
  $f(x+y) = f(x)+f(y)$.

The identity 1 (resp. 2) follow by plugging $y = 0$ (resp. $x = 0$) to $\text{(*)}$. For 3, we may assume that $x \geq 0$. Then replacing $(x, y)$ in $\text{(*)}$ by $(x^{1/2}, f(-x))$, we have

$$ 0 = f(x - x) = f(-x) + f(x^{1/2})^2 = f(-x) + f(x) $$

and 3 follows. Finally, replacing $(x, y)$ in $\text{(*)}$ by $(x^{1/2}, f(y))$ gives

$$ f(x + y) = f(x) + f(y), \qquad \forall x \geq 0, \ y \in \Bbb{R}. $$

The restriction that $x \geq 0$ can be removed by combining this with 3: if $x < 0$, then

$$ f(x+y) = -f(-x-y) = -(f(-x) + f(-y)) = f(x) + f(y). $$

  Step 3. $f(x) = x$.

If $x > 0$ then by the first identity of the previous step,

$$ f(x) = f((x^{1/2})^2) = f(x^{1/2})^2 > 0. $$

(We can exclude the equality because $f$ is bijective and $f(0) = 0$.) Combining this with the fact that $f$ is an odd function, we have:

$$ f(x) \geq 0 \quad \Longleftrightarrow \quad x \geq 0.$$

Now for any $x \in \Bbb{R}$,

$$ f(x) -x  = f(x) - f(f(x)) = f(x - f(x)). $$

Therefore $f(x) - x \geq 0$ if and only if $x - f(x) \geq 0$, which implies that $f(x) = x$.

Remark. The Cauchy functional equation has pathological solutions which are also involution, and we need to use the extra structure given by the equation $\text{(*)}$ to exclude such possibilities. The first identity of Step 2 was essential in our solution.

Observe

$f(x^2+f(0)) = [f(x)]^2$
$f(f(x)) = x+[f(0)]^2$.

Plugging $0$ into expression 1 yields
\begin{align}
f_2(0):=f(f(0)) = [f(0)]^2
\end{align}
and plugging $f(0)$ into expression 1 yields
\begin{align}
f(f(0)^2+f(0)) =[f_2(0)]^2 = [f(0)]^4.
\end{align}
Moreover, by expression 2, we have
\begin{align}
f(f(0)^2+f(0))= f([f(0)]^2)=f(f_2(0)) = f_3(0).
\end{align}
Hence we have
\begin{align}
f_3(0) = [f(0)]^4.
\end{align}

Next, using both 1 and 2, we have
\begin{align}
f(f(x^2+f(0)))=x^2+f(0)+[f(0)]^2. \ \ (\ast)
\end{align}
Plugging $0$ into $(\ast)$ yields
\begin{align}
f_3(0) = f(0)+[f(0)]^2.
\end{align}
Combining everything yields
\begin{align}
[f(0)]^4= f_3(0) = f(0)+[f(0)]^2 \ \ \Rightarrow \ \ f(0)[f(0)^3-f(0)-1] =0.
\end{align}
Solving for $f(0)$ yields that either $f(0) = 0$ or $f(0)$ is a root of the polynomial $p(x) = x^3-x-1$, which only has one real root.

Moreover, observe
\begin{align}
f_4(0) = [f(0)]^4
\end{align}
and
\begin{align}
f_4(0) = f(f(f_2(0))) = f_2(0) + [f(0)]^2 = 2[f(0)]^2
\end{align}
which means
\begin{align}
[f(0)]^4-2[f(0)]^2 = 0.
\end{align}
Thus, $f(0)$ is also a root of $x^4-2x^2 = x^2(x^2-2)$.

In conclusion, $f(0)$ has to be $0$. Thus, $f(f(x)) = x$ and $f(x^2) = [f(x)]^2$.


Conditions under which pointwise limits of functions are Riemann integrable. A linchpin of Lebesgue theory is that limits of measurable functions are measurable.

Is there any similar condition for Riemann integrability? Can we create a class of functions for Riemann integrability that is closed under limits?

One class could be "the set of continuous functions under uniform limits". But can we modify this in some way to allow pointwise limits?

From what I remember there is a proof possible using classical (Riemann integrable arguments) that you can do which say if your limit function is Riemann integrable, then the integral and the limit commute (this is obviously true as it's a consequence of Lebesgue theory).

I think one set would be equicontinuous functions using some fixed delta, any others which are more relaxed?

It is definitely not true that the pointwise limit of Riemann integrable functions is Riemann integrable.

Let $\mathbb{Q}= \{q_n: n\in\mathbb{N}\}$ be the rationals. Then, define;
\begin{equation}
f_N(x)= \sum_{n=1}^n\chi_{q_n}(x),
\end{equation}
where,
\begin{equation}
\chi_{q_n}(x)= \begin{cases}
1, x= q_n\\
0, x\neq q_n.
\end{cases}
\end{equation}

It is clear that each $f_N$ is Riemann integrable (use definition) with $\int f_N= 0$. However, the pointwise limit, $f_N\to \chi_{\mathbb{Q}}$ is not Riemann integrable.

$y=sinx$ between $0$ and $\pi$ is Riemann integrable. So are the partial sums of Fourier series between the same limits. A Fourier series can be found for a step function. The step function is Riemann integrable, but the limit of a sequence of step functions might not be. So even "nice" functions can generate functions that are not Riemann integrable. This lack of "closure" of Riemann integrals showed the need for Lebesgue theory.


Show that $a_n=(n+1)^{\frac{1}{\ln(n+1)}} \to e$ when $n \to \infty$ Show that $a_n=(n+1)^{\frac{1}{\ln(n+1)}} \to e$ when $n \to \infty$ Can you give me a hint? I can't use L'Hopital. Thanks.

Hint 1: It's a trick question. The solution is really, really, really easy.

Hint 2: Take a spreadsheet program and calculate the values for n = 1 to 10.


What is the period of a periodic function |f(x)| , if period of f(x) is T? If period of the function $f(x)$  is $T$ then what is the period of the function $|f(x)|$.

It can be any divisor of $T$. For example, $T/5$

This is going to depend on whether the function $f(x)$ becomes negative or not and on where those zeroes are (if they exist).
Example 1:
If $f(x)=\sin x + 3$ then both $f(x)$ and $|f(x)|$ have the same period.
Example 2:
If $f(x)=\sin x + 0.5$ then both $f(x)$ and $|f(x)|$ have the same period.
Example 3:
If $f(x)=\sin x$ then $|f(x)|$ has half the period of $f(x)$.
Plot the three different situations and try to see why this is so.


Dimensions for the cheapest possible rectangular box What are the dimensions for the cheapest possible rectangular box with a volume of 504 cm3  if the material for the bottom costs \$20/cm2, material for the sides costs \$6/cm2, and material for the top costs \$8/cm2 ?

I not sure how to use the information that I was given to solve this question. I know that I should use more 6 dollar pieces but I don't know how many of each I should use without doing guess and check.

For length $l$ cm, width $w$ cm and height $h$ cm, the material costs $c$ in dollars are
$c= 20lw+ 6(2lh+2wh)+ 8lw$ $=28lw + 12h(l+w)$
Note that the set volume of the box means that $h=504/lw$, so
$c=28lw + 6048(l+w)/lw$ $=28lw+6048/l + 6048/w$.

Then the partial derivatives on $l$ and $w$ are
$\frac{\delta c}{\delta l} = 28w-6048/l^2  $ and
$\frac{\delta c}{\delta w} = 28l-6048/w^2  $

Solving for zero derivative we get
$l^2w = 216$ and $w^2l= 216 $

and thus $l=w$ at the minimum and we can solve for the dimensions.

Initial answer

It's relatively obvious that the top and bottom should be square, since we're looking to get minimum perimeter for the area.

So given that $s$ is the side length for the square base, the height $h(s)=504/s^2$ and the cost $c(s) = 20s^2+ 6\cdot 4sh(s)+ 8s^2$ $=  28s^2 + 12096/s$.

Then $\frac{dc}{ds} = 56s-12096/s^2$ and we can solve for the minimum at $\frac{dc}{ds} =0$.


Combinations and Permutations: Number of ways of taking out 1 $ bills Can A has N 1 $ bills. Can B is empty. At each step you can either take a bill from can A or put a bill you already have into can B. You can choose to keep some bills in your hand and take some more bills from can A. At the end all the bills should be in can B.

For e.g. Let N = 10, I can do a sequence:

take 1, take 1, take 1, put 1, put 1, take 1 take 1, put 1, take 1, put 1, put 1, take 1, take 1, put 1, take 1, take 1, put 1, put 1, put 1, put 1

In how many possible ways can I do this?

I am trying to reduce this to a problem of solving number of non-negative solutions of equations
     x1 - y1 >=0;
     x1 + x2 - (y1 + y2) >= 0;
     x1 + x2 + x3 - (y1 + y2 + y3) >= 0

where xis correspond to taking a xi bills from can A and yi to putting yi bills in can B.
But, I am stuck here and not able to go any further.

Consider the case when all the bills are distinct. Now, how many permutations are there in which I can do the above process?

We can associate each dollar bill with a letter, and then consider a word of length $2n$ (where $n$ is the number of dollar bills) with the requirement that each associated letter appears exactly twice in the word.  Each such word will give us a unique (up to permutation) way to draw the dollar bills and put them into can B by moving along the word from left to right, and when we encounter a letter the first time we move the associated dollar bill from can A to our hand, and when we encounter it the second time we move it from our hand to can B.

For example, for three dollar bills we could associate the letters a,b,c, and the word 'abbcca' would tell us how to move the dollar bills; first moving a to our hand, then b, then moving b to can B, and so forth.

It then suffices to count the number of such words which are distinct, which is $\dfrac{(2n)!}{2^n}$.

It's a simple 2D path problem with a square grid. With N rights and N downs to go, the number is
$${2N \choose N}$$


Is there any example of a non-measurable set whose proof of existence doesn't appeal to the Axiom of choice? Is there any example of a non-measurable set whose proof of existence doesn't appeal to the Axiom of choice?

What would it imply if there was such an example?

EDIT: For instance, maybe this will help understand the kind of example I had in mind, it is known that an important feature to determine that the AC is needed in the Banach-Tarski case is the non-transitivity of rotations on Euclidean space, one might find an example of a transitive group for some given space keeping the rest equal and make the AC unnecessary?

I guess then it might be said that this transitivity will be in this particular case equivalent to the AC or some amount of it, but I guess that it would be important to show this if it hadn't been realized before.

No, there is not - it is consistent with ZF that every set of reals is Lebesgue measurable, so some amount of the axiom of choice is needed.

This is even true if we assume dependent choice, a certain reasonably strong choice property (enough to do most useful theorems from analysis): this was proved by Solovay. He needed an additional assumption, though: that the theory ZFC+"There is an inaccessible cardinal" is consistent. It was later shown by Shelah to be necessary: we can prove that ZF+"There is an inaccessible cardinal" is consistent if and only if ZF+DC+"All sets of reals are measurable" is consistent.

In lieu of choice, I believe the consistency of ZF+"All sets of reals are measurable" from merely the consistency of ZF (the weakest possible assumption here) was proved by Truss. However, it's worth pointing out that in ZF alone, "measurable" doesn't necessarily mean what you think. For example, it's consistent with ZF that $\mathbb{R}$ is a countable union of countable sets, and in that context basic analytic properties like measurability go absolutely bonkers. So really the right base theory to look at, in my opinion, is ZF+DC; and then it's a very interesting fact that an inaccessible is needed.

I think bof's interpretation of my original question is reasonable. I'm compelled to the following argument as an anwer to it in those terms but I'm certainly no set theorist so any correction is welcome: It seems to me that an affirmative answer would imply that ZF is not powerful enough to produce proofs about set measurability without the aid of the AC, probably because the AC (or the relavant part of it that bears on measurability) may be a direct consequence of the primitive notion of set that is needed to construct the ZF axioms and therefore ZF without an explicit extra axiom like AC cannot prove the existence of such an unmeasurable set. Unfortunately this would mean this affirmative answer could not be proved within ZF so it ultimately has to appeal to the AC wich would actually answer the original answer in the negative,  leading to a situation of undecidability of the question within ZF to avoid contradiction.


When considering a finite-type scheme as a ringed space, is it enough to look at its $k$-points? I am reading a set of notes by Michel Brion about automorphism groups of projective varieties.
The following claim appears in the proof of a theorem stating that if G is a connected group scheme, $X$ a $G$-scheme, and $\pi:X \to Y$ a proper morphism such that $\pi_*(\mathcal{O}_X)=\mathcal{O}_Y$, then there is a unique $G$-action on $Y$ such that $\pi$ is $G$-equivariant.
We will consider a finite-type scheme $Z$, as a ringed space $(Z(k), \mathcal{O}_Z)$, where the set $Z(k)$ is equipped with the Zariski topology, this makes sense as $Z$ is of finite type.
What about considering $Z$ as a ringed space makes it ok to consider only its $k$-points, and why is it necessary that $Z$ be of finite-type?

When $k$ is algebraically closed and $Z,Z'$ are of finite type over $k$:

the data of $(Z(k),\mathcal{O}_Z|_{Z(k)})$ is enough to recover $Z$, and
given a morphism $\varphi:(Z(k),\mathcal{O}_Z|_{Z(k)})\to (Z'(k),\mathcal{O}_{Z'}|_{Z'(k)})$, there is a unique morphism $\psi:Z\to Z'$ which has $\varphi$ as its restriction to the $k$-points. (Additionally, no two distinct morphisms $Z\to Z'$ restrict to the same morphism $Z(k)\to Z'(k)$.)

The recipe for recovering $Z$ from $Z(k)$ is straightforward: we need to freely adjoin generic points, which we can do by applying a nifty functor. Define a functor $t(-):Top\to Top$ which associates to any topological space $X$ the set of its irreducible closed subsets with the topology given by declaring that for every closed $Y\subset X$ we have that $t(Y)\subset t(X)$ is a closed set. This functor acts on a morphism $f:X\to X'$ by sending an irreducible closed subset $Y\subset X$ to $\overline{f(Y)}\subset X'$. For any topological space $X$, there is an induced morphism $\alpha:X\to t(X)$ by sending $x\mapsto \overline{\{x\}}$, and this map induces a bijection between open subsets of $X$ and $t(X)$.
To verify that $(t(Z(k)),\alpha_*(\mathcal{O}_Z|_{Z(k)}))\cong Z$, we'll need to make some observations about $Z(k)$. First, $Z(k)$ can be identified with the set of closed points of $Z$: by Zariski's lemma, the closed points of a scheme of finite type over a field $F$ are exactly those points which have as their residue field a finite extension of $F$. If $F$ is algebraically closed, it has no nontrivial finite extensions, so the closed points of $Z$ are exactly $Z(k)$, and they have the induced topology from $Z$. Next, for any scheme of finite type over a field, the closed points are very dense: they're dense in every nonempty closed subset.
Now we may define a map of topological spaces $t(Z(k))\to Z$ by sending any irreducible closed subset $Y\subset Z(k)$ to the generic point of its closure in $Z$. It's not difficult to see that this is a homeomorphism by our previous observations, and as $\alpha: Z(k)\to t(Z(k))$ gives a bijection on open sets, we have that $\alpha_*(\mathcal{O}_Z|_{Z(k)})\cong \mathcal{O}_Z$.
This rather quickly settles the second part, too: given $\varphi$, we can let $\psi$ be the morphism $(t(\varphi),\alpha_*\varphi^\sharp)$, and checking uniqueness is not difficult.
Essentially what we've proven here is that there's an equivalence of categories between schemes of finite type over an algebraically closed field $k$ and the $k$-points of schemes of finite type over an algebraically closed field equipped with the restriction of the structure sheaf. The upshot of this for you is that if you can construct the map in the statement you wish to prove in this restricted setting, you'll automatically get the map of schemes you want.

Both algebraically closed and finite type are necessary conditions: if either of these are violated, there are $k$-schemes which cannot be distinguished from the empty $k$-scheme by looking at their $k$-points, and our proof above breaks. In the non-algebraically closed case, consider the spectrum of the algebraic closure, and in the non-finite type case, consider the spectrum of the transcendental extension $k(x)$. In both cases, a $k$-point is equivalent to a series of $k$-algebra map $k\to F\to k$ which is the identity. But this can't happen: the map $F\to k$ must have nontrivial kernel by dimension reasons, which means that $F\to k$ is the zero map because $F$ is a field.

Question: "When considering a finite-type scheme as a ringed space, is it enough to look at its k-points?"
Answer: If $k$ is a field and $A:=k[x,y]/(y-x)^{n+1}$, $S:=Spec(A)$ it follows
$$S(k):=Hom_{k-alg}(A,k)$$
is the set of maps of $k$-algebras $\phi: A \rightarrow k$. The map $\phi$ is determined by its values $\phi(x)=a, \phi(y)=b$ with $(b-a)^{n+1}=0$, and since $k$ is a field it follows $b=a$. Hence $S$ has the same $k$-rational points as the affine line $\mathbb{A}^1_k:=Spec(k[t])$. There is no isomorphism $S \cong \mathbb{A}^1_k$. Hence $S$ is not determined by $S(k)$
For simplicity, Let $S:=Spec(A)$ be an integral affine scheme of finite type over an algebraically closed field $k$, with $A:=k[x_1,..,x_n]/I$ where $I$ is a prime ideal. You may consider the set $S(k)$ of $k$-rational points of $S$ and embed this set into $k^n$ - in the sense of Hartshorne, Chapter I, it follows $S(k)\subseteq k^n$ is an algebraic variety, and $I(S(k))=I$ by the Nullstellensatz. Hence you recover the ring $A$ from $S(k)$: There is an isomorphism of $k$-algebras
$$R \cong S_k:=k[x_1,..,x_n]/I(S(k))$$
and
$$S\cong Spec(S_k).$$
Hence if $k$ is algebraically closed and $S$ is an affine integral scheme of finite type over $k$  with the above construction you recover $S$ from $S(k)$. There are group schemes that are non-reduced.
Question: "What is the underlying topological space of Z if not Z(k)?"
Answer: Is $A:=\mathbb{R}[x]$ and $S:=Spec(A)$ it follows the underlying topological space of $S$ is the set of prime ideals in $A$. The prime ideals $I$ in $A$ have - by the nullstellensatz - a residue field that is a finite extension of $k:=\mathbb{R}$. Hence they are on the form
$$I:=(x-r)$$
for a real number $r$ or $I:=(p_z(x))$
where $p_z(x):=(x-z)(x-\overline{z}):=x^2-2ax+a^2+b^2$ where $b\in \mathbb{R}^*$. Hence there is no equality between the topological space of $S$ and the set $S(k) \cong k$.
Question: "We will consider a finite-type scheme $Z$, as a ringed space $(Z(k),O_Z)$, where the set $Z(k)$ is equipped with the Zariski topology, this makes sense as $Z$ is of finite type."
Answer: Note: You must make sense of $\mathcal{O}_Z$ - $\mathcal{O}_Z$ is not a sheaf of rings on $Z(k) \subseteq Z$ which is a topological subspace.
A finitely generated $k$-algebra $A$ is a Hilbert-Jacobson ring. In such a ring any prime ideal is the intersection ot the maximal ideals containing it. Let $X:=Spec(A)$ and let $X^m \subseteq X$ be the set of closed points in $X$ with the induced topology. Let $i: X^m \rightarrow X$ be the inclusion map. The map $i$ is continuous and you may take the topological inverse image $\mathcal{O}_{X^m}:=i^{-1}(\mathcal{O}_X)$ to get a ringed topological space $F(A):=(X^m, \mathcal{O}_{X^m})$. This defines in a functorial way a ringed topological space $F(A)$ to any finitely generated $k$-algebra, and for any morphism $\phi: A \rightarrow B$ you get a canonical map of ringed spaces
$$F(\phi): F(B) \rightarrow F(A).$$
With this construction any map of ringed spaces $f: F(B) \rightarrow F(A)$
comes from a map of rings $\phi: A \rightarrow B$ and if $F(\phi)=F(\psi)$ it follows $\phi=\psi$. If $k$ is algebraically closed it follows $X^m=X(k)$ is the space of $k$-rational points of $X$.
More generally if $X$ is of finite type over a field or Dedekind domain $k$, you let $X^m \subseteq X$ be the topological subspace of closed points of $X$ and define $F(X):=(X^m, \mathcal{O}_{X^m})$ similarly. You get to any scheme $X$ of finite type over $k$ a ringed space $F(X)$ which is functorial in $X$.
Any morphism $F(X) \rightarrow F(Y)$ comes from a map of schemes $X \rightarrow Y$ and if $F(\phi)=F(\psi)$ it follows $\phi=\psi$. Hence the ringed space $F(X)$ - which only depends on the closed points of $X$ - captures all properties of the scheme $X$. If $k$ is an algebraically closed field it follows $F(X)=X(k)$ is the space of $k$-rational points of $X$. This construction works over any Dedekind domain. The construction is "straightforward" and you avoid too many abstract "category theoretical" results that may lead to set theoretical problems: Since $X$ is a set it follows $X^m$ is also a set since it is a subset of $X$.
You find an explanation of this construction at the following link:
https://mathoverflow.net/questions/377922/building-algebraic-geometry-without-prime-ideals/378961#378961


Prove that the maximum area of a rectangle inscribed in an ellipse is $2ab$ Prove that the maximum area of a rectangle inscribed in an ellipse $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$ is $2ab$.

My attempt:
Equation of ellipse: $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$.
Assume that $a>b$ & let $l$ & $w$ are length and width of rectangle then area will be $$A=l\times w \tag{1}$$
Now, I substituted the point $\left(\dfrac {l}{2}, \dfrac {w}{2}\right)$ in the equation of ellipse
$$\frac{(l/2)^2}{a^2}+\frac{(w/2)^2}{b^2}=1,$$
$$b^2l^2+a^2w^2=4a^2b^2.\tag{2}$$
I am not sure how to proceed from here.

Consider the coordinate transformation $$(u,v) = (x/a, y/b),$$ which maps the ellipse $$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$$ to the unit circle $$u^2 + v^2 = 1.$$  Such a transformation preserves the ratios of areas; in particular, if a quadrilateral inscribed in the ellipse has maximal area among all such quadrilaterals, then the transformed quadrilateral under this mapping will also have maximal area; and vice versa.  This is not difficult to show.
Therefore, it suffices to consider all quadrilaterals inscribed in the unit circle in the transformed coordinate system.  Since we know that the quadrilateral of maximal area in a circle is a square (proof below), it follows that the family of quadrilaterals of maximal area inscribed in the ellipse are those parallelograms that arise from the inverse transformation.  These squares all have diagonal with length $2$, thus area of $2$, and upon the inverse transformation, the corresponding parallelograms all have area $2ab$.  Of these, there is only one that is a rectangle, corresponding to the configuration where the sides are parallel to the coordinate axes; although merely observing existence is sufficient (uniqueness is irrelevant).

For the proof that a quadrilateral of maximal area inscribed in a circle is a square, let $ABCD$ be any quadrilateral inscribed in the circle.  Consider a diagonal, say $AC$.  Then $\triangle ABC$ is an inscribed triangle whose area is maximized when the height $h_B$ perpendicular to $AC$ is maximized, since $AC$ is fixed.  This corresponds to choosing the midpoint of $AC$ as the foot of the perpendicular.  Similarly, the area of $\triangle ADC$ is maximized in the same manner.  But this means $BD$ must be a diameter, since the feet of the perpendiculars coincide, both being the midpoint of $AC$, and the perpendicular to any chord through its midpoint is a diameter.  Thus among all quadrilaterals with fixed diagonal $AC$, the one that maximizes the area is one where the other diagonal $BD$ is a diameter.  Then among all such quadrilaterals with $BD$ as diameter, the one that maximizes the area is the one where $AC$ is also a diameter.  Since $AC \perp BD$, it follows $ABCD$ is a square.

Let one vertex of the rectangle be $(a\cos t, b\sin t)$. Then, the other three are known as well and the area is
$$A= 4ab|\sin t \cos t |\le 2ab (\cos^2t+\sin^2t)=2ab$$
where the inequality $2uv\le u^2+v^2 $ is used.


Number of integral solutions of the equation $x+y+z \le 25$ I came up with this equation
$x+y+z≤25$, such that $x≥-3$, $y≥-2$ and $z≥0$.

I introduced a dummy variable $w$ and the equation became $x+y+z+w=25$ where $w≤30$.

I used the Multinomial Theorom and I proceeded in this way,
Solutions are equivalent to finding coefficient of $x^{25}$
$(x^{-3}+x^{-2}+x^{-1}+1...x^{25})(x^{-2}+x^{-1}+1+...x^{25})(1+x+x^2+...x^{25})(1+x+x^2+x^3...x^{30})$

This can be converted to
$\frac{(x^{-3})(1-x^{29})(x^{-2})(1-x^{28})(1-x^{26})(1-x^{31})}{(1-x)^4}$

Now When I found the coefficients of $x^{25}$ I found $^{23}C_3- ^4C_3 - ^7C_3 - ^5C_3$ Which is not the answer written in the textbook.

Can someone explain What did I do wrong?

Finding the number of integer solutions to $\begin{cases}x+y+z\leq 25\\x\geq -3\\y\geq -2\\ z\geq 0\end{cases}$ is equivalent to finding the number of integer solutions to

$$\begin{cases} x'+y'+z + w = 30\\x'\geq 0\\y'\geq 0\\z\geq 0\\w\geq 0\end{cases}$$

seen by a change of variables $x'=x+3, y'=y+2$ and introducing a dummy variable $w = 25-x-y-z$

Your attempt and the subtraction of binomial coefficients looks like you were trying to deal with upper bounds on one or more of the variables, like if it were $\begin{cases}x'+y'+z+w=30\\0\leq x'\leq 10\\0\leq y'\leq 15\\ 0\leq z\\0\leq w\end{cases}$ or similar...

Note in particular that the condition $-3\leq x$ is different than the condition $-3\leq x\color{red}{\leq 25}$.  If we were to be using a generating function here to describe $-3\leq x$, it would have involved $(x^{-3}+x^{-2}+x^{-1}+1+x+x^2+\dots+x^{25}+x^{26}+\dots+\dots)$ and would not have ended at the $x^{25}$ term.

Here, we have no upper bounds to worry about... and the answer is a straightforward textbook example of stars-and-bars

$$\binom{30+4-1}{4-1}$$

Nice question

$x+y+z ≤ 25$

Where $ x≥ -3$, $y ≥ -2$, $z ≥ 0$
Where $(x,y,z)€Z$

My intention here is to create a table of values, but since i can't type &&&

Let's say the equation is
$x+y+z ≤ w$
Where $w$, their sum is such that $w ≤ 25$
The range of values of $x$, $y$ and $z$ are $:$

$x → (-3,-2,-1,0,1,2,3,............)$

$y → (-2,-1,0,1,2,3,4,.......…....)$

$z → (0,1,2,3,4,5,6,7,............)$

Now the maximum value of $x$ will occur when both $y$ and $z$ are minimum, because they all have a fixed range $w$

$w → (-5,-4,-3,-2,-1,0,1,2...........,23,24,25)$

So also the maximum of $y$ will occur when $x$ and $z$ are minimum and vice versa

$x → (-3,-2,-1,0,1,2,..........25,26,27)$

$y → (-2,-1,0,1,2,.................25,26,,27,28)$

$z → (0,1,2,3,4,..............27,28,29,30)$

Since we are trying to find the number of integer solutions to the inequality, then number of integers in each set of its variables $x$, $y$ and $z$ will  hold the key

The number of elements in set $x$ is 31, same with $y$ and $z$
Therefore number of integer solutions $31×31×31$

$29,791$


Methods for determining thresholds for future use based on historical data (in R or Excel) Let's say I have some data about the amounts of reports to Help Desk (about technical problems) which were monitored and registered every five minutes for the whole day (even during the night). Is there any way to efficiently calculate the "alarming" thresholds:

top - above this means there are too many reports in the system! Time to get to work! (most important).
bottom - lower than this means there are too few of them (report system got broken?)
The problems I face:

System is fairly new, so at the start of the registering process there were extreme outliers. I want to have thresholds for a more stable system in the future.
People obviously don't work during the night, so there's a lot of observations (let's call them "moments" - those "five minutes") where the number of reports stays the same.

An "alarming" threshold indicates that something becomes "abnormally" high (or low). This could be interpreted in statistical terms as "the phenomenon under consideration exceeded its mean value plus its standard deviation" (or went under).

So you need to calculate sample means and standard deviations to begin with. Of what metrics? You can easily find all around the web the Key Performance Indicators (KPI's) for Support Call Centers -no need to re-invent the wheel. As I happen to have some professional experience in the field, I would especially pay close attention to how the following two metrics evolve : 1) evolution of backlog (the amount of pending/open/unresolved reports) 2) evolution of average time it takes to resolve and close a report.

But to calculate various statistics of the metrics you will eventually select, and do this "efficiently" and also not-misleadingly, you have to understand the special characteristics of your data (and study the corresponding areas of statistics).
In your case, these special characteristics are : your data is time series, your data is high frequency, your data exhibits all sorts of seasonality (through the day, daily, weekly, monthly). These together create the following complex situation : your data depends on the immediate past (eg, backlog at beginning of each, say, hour, depends on what happened the previous hour (high frequency time series), but it also co-varies with what happened the same hour the previous day (high-frequency seasonality).
You need to study these aspects to understand what kind of model you must set up in order for the various statistics like mean and standard deviation to eventually help you and not fool you.

You should look into Statistical Process Control it uses values derived from a baseline period to determine whether a system is stable (experiencing normal variation) or if there is something that needs to be investigated.


Are these two sentences logically equivalent? What i'm essentially asking is if the following statement is true:

$ \forall x \exists y (R(x) \lor Q(y)) :\Leftrightarrow  \exists y \forall x (R(x) \lor Q(y)) $

where $:\Leftrightarrow $ means that the two statements are logically equivalent.
(sometimes $\vDash$ and it's mirror are used for logical equivalence but can't find the mirrored symbol in mathjax

Edit: I suspect that it is not true but i'm trying to prove it and seems it should be.

Yes, the sentence is true — the two sides are equivalent, because both are equivalent to
$$\forall x \, R(x) \lor \exists y \, Q(y).
$$
If either predicate used both variables this wouldn't be so: in general, $\exists\forall \to \forall\exists$ but not conversely. But here, you can rearrange the quantifiers because:

  If $v$ is not free in $p$ then
  $$\exists v\,(p \lor \varphi(v)) \equiv (p \lor \exists v\,\varphi(v))$$
  and
  $$\forall v\,(p \lor \varphi(v)) \equiv (p \lor \forall v\,\varphi(v)).$$

Nope; but one implication is true:
\begin{equation}
\exists y\forall x(R(x)\vee Q(y))\to \forall x\exists y(R(x)\vee Q(y)).
\end{equation}
It's more or less intuitive that this holds: when you have "$\forall x\exists y\dots$" this defines a function: given $x$, exists $x(y)$ such that $\dots$ (You need Choice to define this to be an actual function, so you can well-order the universe and take the least $x$ such that "$\dots$"). So if you have "$\exists y\forall x\dots$" you can define a constant function with output $y$. Therefore you have "$\forall x\exists y\dots$". With this you can see that the other implication is false beacause the function $x(y)$ can take different values.


Find the volume common to two circular cylinders, each with radius r, if the axes of the cylinders intersect at right angles. (using disk/washer) Find the volume common to two circular cylinders, each with radius r, if the axes of the cylinders intersect at right angles. (using disk/washer)

I saw no example of this problem anywhere.. I saw an example how to solve it without calculus but I'm afraid that'll be deductions on my test.. I have the volume of a cylinder and the method of area between two curves but this is out of my reach.... I don't know what I can try...

The main challenge in this problem is to predict the solid itself. Take a look at the image below

As we increase the height $z$ of the intersecting plane the section of the solid is a square with a side $a$. Since the solid is created with two cylinders of radius $r$, $a$ and $z$ are related by $$a^2+z^2=r^2$$
The area of the square is $A=a^2$ or in terms of $z$: $$A(z)=a^2=r^2-z^2$$ The method of washers tells us $$V=8\int_0^rA(z)dz=8\int_0^r(r^2-z^2)dz=8\left(r^3-\frac{r^3}{3}\right)=\frac{16}{3}r^3$$
Note that the coefficient $8$ is required since we are considering a $1/8$th part of the solid.

The volume of a Steinmetz solid $(SS)$, formed by the intersection of $2$ cylinders (radii $r$) at right angles is:

$$V = 4r \cdot r \cdot r \left(\frac{\pi}{2} - \frac{1}{3}\right) = 4r^3\left(\frac{\pi}{2} - \frac{1}{3}\right)$$

As the volume of $SS$ is formed by sections of a cylinder, the answer must contain $\pi$.


Catenoid is a minimal surface i want to show that the catenoid is a minimal surface. I have given

$f:I \times (0,2\pi)\longrightarrow \mathbb{R}^3$ with $f(r,\phi)=\left( \begin{array}{c}\cosh(r) \;\cos(\phi)\\\cosh(r) \;\sin(\phi)\\r\end{array} \right)$.

I know that:

$f$ is minimal surface $\Longleftrightarrow$ $\Delta f=0$.

$f$ is given in polar coordinates so i have to calculate the following:

$\Delta f= \frac{\partial^2f}{\partial r^2}+\frac{1}{r}\frac{\partial f}{\partial r}+\frac{1}{r^2}\frac{\partial^2f}{\partial \phi^2}$

$\frac{\partial f}{\partial r}= \left( \begin{array}{c}\sinh(r) \;\cos(\phi)\\\sinh(r) \;\sin(\phi)\\1\end{array} \right)$ , $\frac{\partial^2f}{\partial r^2}=\left( \begin{array}{c}\cosh(r) \;\cos(\phi)\\\cosh(r) \;\sin(\phi)\\0\end{array} \right)$

$\frac{\partial f}{\partial \phi}=\left( \begin{array}{c}-\cosh(r) \;\sin(\phi)\\\cosh(r) \;\cos(\phi)\\0\end{array} \right)$ ,  $\frac{\partial^2 f}{\partial \phi^2}=\left( \begin{array}{c}-\cosh(r) \;\cos(\phi)\\-\cosh(r) \;\sin(\phi)\\0\end{array} \right)$.

But when I put all together I can not show that $\Delta f$ is 0. What did I do wrong?
Can someone help me please?

Thanks in advance.

I think you are a little bit confused about the harmonic characterization of (conformally immersed) minimal surfaces.

We know that every regular $2$-dimensional surface can be described locally in isothermal coordinates (i.e. for neighborhood of the surface, there's a coordinate map that preserves angles aka is conformal aka has 1st fundamental form satisfying $E=G$, $F=0$).

So we can cover the surface by a family of coordinate maps $$\vec{x_\alpha}(u,v) = \big(x_1(u,v), x_2(u,v), x_3(u,v)\big)$$

with each $\vec{x_\alpha}$ conformally mapping an open subset of $\mathbb{R}^2$ to $\mathbb{R}^3$.

The harmonic characterization says that the surface is minimal iff for each $\vec{x}_\alpha$ in such a family, the coordinates $x_i(u,v)$ are harmonic functions with respect to the coordinates (u,v).  I would advise going back to look at the proof of this characterization for clarification, and thinking about geometrically what it means to be conformal (preserve angles).

I think where you have been misled is in thinking of this as a polar parametrization and using the so-called "polar form of the laplacian."

In your case you have a conformal coordinate map describing the entire catenoid in coordinates $r, \varphi$.  What does $$\big(\cosh(r)\cos(\varphi)\big)_{rr} + \cosh(r)\cos(\varphi)\big)_{\varphi \varphi}$$ look like?

This not a complete answer but that's too long for me to post it in comment.

Preliminary for differential geometry of surfaces

\begin{align*}
  \mathbf{x}(u,v)
  &= \begin{pmatrix} x(u,v) \\ y(u,v) \\ z(u,v) \end{pmatrix} \\
  \mathbf{x}_u &= \frac{\partial \mathbf{x}}{\partial u} \\
  \mathbf{x}_v &= \frac{\partial \mathbf{x}}{\partial v} \\
  \mathbf{N} &=
  \frac{\mathbf{x}_u \times \mathbf{x}_v}{|\mathbf{x}_u \times \mathbf{x}_v|}
  \tag{unit normal vector} \\
\end{align*}

First fundamental form
$$\mathbb{I}=
\begin{pmatrix} E & F \\ F & G \end{pmatrix}=
\begin{pmatrix} \mathbf{x}_{u} \\ \mathbf{x}_{v} \end{pmatrix}
\begin{pmatrix} \mathbf{x}_{u} & \mathbf{x}_{v} \end{pmatrix} $$

Second fundamental form
$$\mathbb{II}=
 \begin{pmatrix} e & f \\ f & g \end{pmatrix}=
-\begin{pmatrix} \mathbf{N}_{u} \\ \mathbf{N}_{v} \end{pmatrix}
 \begin{pmatrix} \mathbf{x}_{u} & \mathbf{x}_{v} \end{pmatrix}$$

Metric
$$ds^2=E\, du^2+2F\, du\, dv+G\, dv^2$$

Element of area
$$dA=|\det \mathbb{I}| \, du \, dv
=|\mathbf{x}_u \times \mathbf{x}_v| \, du\, dv
=\sqrt{EG-F^2} \, du\, dv$$

Principal curvatures

Let $\begin{pmatrix} \mathbf{N}_{u} \\ \mathbf{N}_{v} \end{pmatrix}=
\mathbb{A} \begin{pmatrix} \mathbf{x}_{u} \\ \mathbf{x}_{v} \end{pmatrix}$
where $\mathbb{A}=
\begin{pmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{pmatrix}$.

Now
\begin{align*}
   \begin{pmatrix} \mathbf{N}_{u} \\ \mathbf{N}_{v} \end{pmatrix}
   \begin{pmatrix} \mathbf{x}_{u} & \mathbf{x}_{v} \end{pmatrix} &=
   \mathbb{A} \begin{pmatrix} \mathbf{x}_{u} \\ \mathbf{x}_{v} \end{pmatrix}
   \begin{pmatrix} \mathbf{x}_{u} & \mathbf{x}_{v} \end{pmatrix} \\
  -\begin{pmatrix} e & f \\ f & g \end{pmatrix} &=
   \mathbb{A} \begin{pmatrix} E & F \\ F & G \end{pmatrix} \\
   \mathbb{A} &=
  -\begin{pmatrix} e & f \\ f & g \end{pmatrix}
   \begin{pmatrix} E & F \\ F & G \end{pmatrix}^{-1}
\end{align*}

The principal curvatures $k_{1}, k_{2}$ are the eigenvalues of $-\mathbb{A}$.

Mean curvature

$$H=\frac{k_{1}+k_{2}}{2}
=-\frac{1}{2} \operatorname{tr} \mathbb{A}
=\frac{eg-f^2}{EG-F^2}$$

Gaussian curvature

$$K=k_{1} k_{2}
=(-1)^{2} \det \mathbb{A}
=\frac{eG-2fF+gE}{2(EG-F^2)}$$

  For minimal surface, $$H=0$$


Show that no line with a y-int of 10 will ever be tangential to the curve with $y=3x^2+7x-2$ Show that no line with a y-int of 10 will ever be tangential to the curve with $y=3x^2+7x-2$.

Having trouble in showing this. So far these are my process.

Let line be $y=mx+10$
$mx+10 = 3x^2+7x-2$
$3x^2+(7-m)x-12=0$
Apply quadratic formula
$\frac{(7-m\pm\sqrt{(m^2-14m+193)}}6$

A bit stuck here. Maybe I've missed the whole point and complicated this. Any help is appreciated! Thanks in advance :)

You tagged calculus so with derivatives: the slope of a tangent to the given function is

$$y'=6x+7\implies\;\text{for any point on the graph }\;\;(a, 3a^2+7a-2)$$

the tangent line to the function at that point is

$$y-(3a^2+7a-2)=(6a+7)(x-a)\implies y=(6a+7)x-3a^2-2$$

and thus the $\;y\,-$ intercept is $\;-3a^2-2\;$ , and this is $\;10\;$ iff

$$-3a^2-2=10\iff a^2=-4$$

and this last equality can't be true in the real numbers

If you assume that the line is tangent to the parabola then
$$
mx+10=3x^2+7x-2
$$
at some point. Simplifying the previous equation we have
$$
3x^2+(7-m)x-12=0.
$$
Since we are assuming they are tangential, the discriminant of the last equation should be equal to zero. That is,
$$
m^2-14m+193=0.
$$
Solving for $m$ using the quadratic formula you get
$x=7+12i$ or $x=7-12i$, which are not solutions in the real numbers. Therefore the line and the parabola are not tangential.


Second derivative of a matrix quartic form I need to compute the second derivative of the following quartic expression: $$x^H A^H x x^H A x$$ where is Hermitian. I have tried to compute the first derivative, and if I am not wrong, it should be: $$(A+A^H) x x^H (A+A^H) x$$
But then, I do not know how to proceed to calculate the second derivative. Could someone sketch the steps I need to follow? Thank you.

Define the scalar variables
$$\eqalign{
&\phi &= x^HAx = (A^Tx^*)^Tx \cr
&\phi^* &= x^HA^Hx = (A^*x^*)^Tx \cr
&\psi &= \phi^*\phi \cr
}$$
Find the gradient of your function $(\psi)$  with respect to $x$, treating $x^*$ as an independent variable.
$$\eqalign{
d\phi &= (A^Tx^*)^T\,dx \cr
d\phi^* &= (A^*x^*)^T\,dx \cr
d\psi
 &= \phi\,d\phi^* + \phi^*\,d\phi \cr
 &= (\phi A^*x^* + \phi^*A^Tx^*)^Tdx \cr
g = \frac{\partial\psi}{\partial x}
 &= A^*x^*\phi + A^Tx^*\phi^* \cr
g^* = \frac{\partial\psi}{\partial x^*}
 &= Ax\phi^* + A^Hx\phi \cr
}$$
Not that we need it, but the last equation is a consequence of the fact that $\psi=\psi^*\,$ (it's real).

Now the Hessian is just the gradient of the gradient, so
$$\eqalign{
dg
 &= (A^*x^*)\,d\phi + (A^Tx^*)\,d\phi^* \cr
 &= \Big((A^*x^*)(A^Tx^*)^T + (A^Tx^*)(A^*x^*)^T\Big)\,dx \cr
H = \frac{\partial g}{\partial x}
  &= A^*x^*x^HA + A^Tx^*x^HA^H \cr
}$$
Note that the Hessian is symmetric, but it is not Hermitian.

The second derivative of a multi-variable function $f=f(x_1,\ldots,x_n)$ is usually expressed as its Hessian matrix
$$
Hf(x_1,\ldots,x_n)=\left(\frac{\partial^2 f}{\partial x_i\partial x_j}\right)_{i,j=1}^n
$$
Here
$$
f(x)=x^tA^txx^tAx=(x^tBx)^2, \quad \text{where $B=\frac{1}{2}(A+A^t)=(b_{ii})$}.
$$
Hence
$$
\frac{\partial f}{\partial x_i}=4(x^tBx)(Bx)_i
$$
and
$$
\frac{\partial^2 f}{\partial x_i\partial x_j}=8(Bx)_i(Bx)_j+4(x^tBx)b_{ij},
$$
and thus
$$
Hf(x)=8Bx(Bx)^t+4(x^tBx)B
$$


How can I find $\Gamma\cap \Omega $ if $\Gamma$ and $\Omega $ are parametrized surface? Let consider two surfaces $$\Gamma=\{\gamma (u,v)\mid u,v\in \mathbb R\}$$
and $$\Omega =\{\omega (u,v)\mid u,v\in \mathbb R\}.$$

How can I find $\Gamma\cap \Omega $ if $\Gamma$ and $\Omega $ are parametrized surface ? Do I have to pass by cartesian coordinates or I can do it directly ? Let take an example $$\Gamma=\{\gamma (u,v)=(\cos u\sin v,\sin u \sin v, \cos v)\mid v\in [0,\pi], u\in [0,2\pi]\}$$
and $$\Omega =\{\omega (u,v)=(u,v,0)\mid u,v\in\mathbb R\}.$$

Let $(x,y,z)\in \Gamma\cap \Omega $. The only thing I can say is that there are $u,v\in[0,2\pi]\times [0,\pi]$ and $s,t\in \mathbb R$ s.t. $$(x,y,z)=\gamma (u,v)=\omega (s,t),$$
but it won't allow me to the circle $$\theta \longmapsto (\cos \theta ,\sin \theta ,0),\quad \theta \in [0,2\pi].$$
What I did is that I find the cartesian equation of $\Gamma$ and $\Omega $ and then parametrized it as following
$$\Gamma: x^2+y^2+z^2=1\quad \text{and}\quad \Omega : z=0,$$
and thus we find $$\Gamma\cap \Omega :\begin{cases}x^2+y^2=1\\ z=0\end{cases},$$
and thus, after parametrization one finds $$\Gamma\cap \Omega =\{c(\theta )=(\cos\theta ,\sin\theta ,0)\mid \theta \in [0,2\pi]\}.$$

Question : First of all, I know that my method using cartesian coordinate is correct, and my question is not to know if the way I solved the problem is correct or not since I know it's correct. My question is : Is there a direct method from
$$\Gamma=\{\gamma (u,v)\mid u,v\in \mathbb R\}\quad \text{and}\quad \Omega =\{\omega (u,v)\mid u,v\in \mathbb R\},$$
to find $\Gamma\cap \Omega $ (i.e. Without Using Cartesian Coordinate) or I must use them ?

I would thing about something as : finding $u,v$ s.t. $\gamma (u,v)=\omega (u,v)$ but this doesn't work. Other idea would be to fine $u,v,s,t$ s.t. $\gamma (u,v)=\omega (s,t)$, but still doesn't work, so How can I do it if it's possible ?

So I recall the question : How would you find $\Gamma\cap \Omega $ without using coordinate coordinate ?

Calling $\gamma(u,v)=\Gamma$ and $\omega(u,v)=\Omega$ the two intersecting surfaces we have

$$
p_1 = p_1(u_1,v_1) \leftarrow \gamma(u_1,v_1) = \Gamma\\
p_2 = p_2(u_2,v_2) \leftarrow \omega(u_2,v_2) = \Omega
$$

where $p_i = (x_i,y_i,z_i), \;\; i \in \{1,2\}$

For generic surfaces, the intersection can be obtained by solving by an iterative procedure like Newton-Raphson methods.

So our problem is to solve

$$
p_1(u_1,v_1)-p_2(u_2,v_2) = 0
$$

which gives three equations and four unknowns. The missing equation is used as initial point locator and then as a step constraint.

This problem can also be handled as a minimization problem or as

$$
\min_{u_1,v_1,u_2,v_2}||p_1-p_2||
$$

I hope to be helpful.
Attached the result for the Newton-Raphson iterative process application to the intersection curve determination. The algorithm used is a simpler one without step uniformity correction.


Prove that $s_n\leq 2-\frac{1}{n}$ for all $n\geq 1$ We have the sequence $(s_n)_{n\geq 1}$ given by $s_n=\sum^n_{k=1}\frac{1}{k^2}$. Prove that  $s_n\leq 2-\frac{1}{n}$ for all $n\geq 1$. Thanks in advance!

Hint

$$\frac1{k^2}\le\frac1{k(k-1)}=\frac1{k-1}-\frac1k\quad \forall k\ge2$$
and telescope.

Look at the Riemann zeta function.

$\zeta(2)=\pi^2 / 6$ which is smaller than $1.7$.

Since every value of your function is smaller than $\zeta(2)$ your claim follows.

See : http://en.wikipedia.org/wiki/Riemann_zeta_function#Specific_values


How do you define the derivative of a function without an argument? So the derivative of $f: x\mapsto f(x)$ is defined by $f':x \mapsto \lim_{h\to0}\dfrac{f(x+h)-f(x)}{\phantom{f}(x+h)\,-\,(x)}$.  But is there a way to define $f'$ solely in terms of $f$, without explicit reference to $x$?

Here's an attempt.  For any $h \in \mathbb{R}$, we define a function $\tau_h$ by $\tau_h(x)=x+h$.  (Think of $\tau_h$ as "translation by $h$", if you like.)  (Edit: If you don't like the explicit use of variables in the definition of $\tau_h$, you can define $\tau_h = id +\hat{h}$, where $id$ denotes the identity function and $\hat{h}$ denotes the constant function with value $h$.)  Now for any function $f$ and any real $h$ we can define a difference quotient as
$$\Delta_h(f)=\frac{ f \circ \tau_h  - f}{h}$$

Note that $\Delta_h(f)$ is a function defined without explicit reference to variables (although it does of course require an explicit reference to the parameter $h$).

Now to finish the story we can define $f' = \lim_{h \to 0} \Delta_h(f)$, where the limit here is understood in the sense of pointwise-convergence.

Edited to add:  One nice advantage of this formulation is that it lends itself naturally to generalizations.  Let $\sigma = \{\sigma_h\}_{h \in \mathbb{R}}$ be any parametrized family of functions $\sigma_h: \mathbb{R} \to \mathbb{R}$.  Then we can define difference quotients relative to $\sigma_h$ as
$$\Delta_h^{\sigma}(f)=\frac{ f \circ \sigma_h  - f}{h}$$
and then define the "derivative of $f$ relative to $\sigma$" as $f^{\sigma} = \lim_{h \to 0} \Delta_h^{\sigma}(f)$.  In this notation, the usual derivative $f'$ is just $f^{\tau}$, where $\tau = \{\tau_h\}_{h \in \mathbb{R}} $ is the family of translations defined earlier.

Note your definition of $f'$ has an argument-you wrote it as $f'(x)$, which is correct.   You can apply the same definition:  $$f''(x) = \lim_{h\to0}\dfrac{f'(x+h)-f'(x)}{\phantom{f}(x+h)\,-\,(x)}$$  If you want it in terms of $f$, you can substitute in and have $$f''(x) = \lim_{h\to0}\dfrac{f(x+2h)-2f(x+h)+f(x)}{\phantom{f}((x+h)\,-\,(x))^2}$$


How to demonstrate that there is no all-prime generating polynomial with rational cofficents? It seems like there is no polynomial with finite variables known, which could generate all prime numbers, by integer assignments. Is there a proof that such polynomial can not exist and does anyone have one in his/her stack?

In fact there does not even exist a non-constant polynomial $f$ (I assume you want integer coefficients) which only takes prime values with integer inputs. It suffices to prove this for polynomials in one variable. By the hypothesis that $f$ is non-constant, it takes arbitrarily large values, so without loss of generality $|f(0)| > 1$; in particular, $f(0)$ is divisible by some prime $p$. Then $f(kp)$ is always divisible by $p$, hence cannot be prime for sufficiently large $k$.

However, remarkably there do exist polynomials in more than one variable all of whose positive values are prime.

Consider a polynomial P(n) with integer coefficients;
also, assume that the constant term (coefficient of n^0), is a0.

Case 1) a0 != 1 :
        Then, putting n=c*a0, where c is a non-zero integer, shall yield a multiple of a0.

In general, P(c*P(1)+1) shall always be divisible by P(1). Hence, the given polynomial cannot generate only primes!!


Lebesgue measure of a set $A \subset [0;1]$. A set consists of numbers (in interval $[0;1]$) which, when written in a decimal form, have digit $2$ occurring "before" digit $3$ can be found. It sounds awkwardly, but I hope you get what I mean:

$0.1213 \in A \\
0.1231 \in A \\
0.1211 \in A \\
0.1111 \notin A \\
0.1321 \notin A \\
0.1312 \notin A$

I saw a few somewhat similar examples, so I suspect that the set is indeed Lebesgue measurable, but I'm struggling to find a union of (hopefully distinct) intervals for it...

Longish Hint: It is probably easier to compute the Lebesgue measure of the complement $[0,1] \setminus A$ of this set.  A real number in $[0,1]$ will be in the complement iff it has a decimal exapansion in which $3$ occurs before any occurrence of $2$[*].  So here are pairwise disjoint intervals which make up the set:

$[0.3,0.4]$;
$[0.03,0.04]$, $[0.13,0.14]$, $[0.43,0.44]$, $[0.53,0.54]$, $[0.63,0.64]$, $[0.73,0.74]$, $[0.83,0.84]$, $[0.93,0.94]$,  (the interval $[0.33,0.34]$ is a subset of the interval from the previous level, so we omit it here);
$[0.ij3,0.ij4]$ where $i,j \in \{ 0 , 1 , 4, 5, 6, 7, 8, 9 \}$.
$\vdots$

In general, for the $n$th level there will be $8^{n-1}$ intervals each of length $10^{-n}$, given a total measure of $8^{n-1} 10^{-n}$.  From here it should be easy to compute the Lebesgue measure of $[0,1] \setminus A$, and thus also the Lebesgue measure of $A$.

[*] There might be a slight quibble about this, depending on whether you see $0.4000\ldots$ as belonging to $A$, or $0.3999\ldots$ as belonging to $[0,1] \setminus A$.  However, as only rational numbers can have two different decimal expansions the Lebesgue measure will be the same with either choice.  (I'm opting for $0.4000\ldots = 0.3999\ldots \notin A$.)

Addendum: Thanks to comments from the OP, I have realised that the intended definition of the set $A$ is as follows: $x \in [0,1]$ belongs to $A$ iff if has a decimal expansion in which $2$ occurs, and there are no occurrences of $3$ before the first occurrence of a $2$.  However the analysis done above will work with only minor modifications to determine the measure of the set $A$ (and this we will compute directly).  A collection of pairwise disjoint intervals making up $A$ are:

$[0.2,0.3]$;
$[0.02,0.03]$, $[0.12,0.13]$, $[0.42,0.43]$, $[0.52,0.53]$, $[0.62,0.63]$, $[0.72,0.73]$, $[0.82,0.83]$, $[0.92,0.93]$,  (the interval $[0.22,0.23]$ is a subset of the interval from the previous level, so we omit it here);
$[0.ij2,0.ij3]$ where $i,j \in \{ 0 , 1 , 4, 5, 6, 7, 8, 9 \}$.
$\vdots$

The connection between this and the above should be clear.

So, the numbers 00-99 are 100 combinations,
and there are 19 numbers that contains 3 somewhere. (3 first, 10 combinations, 3 last, 10 combinations, but one combination is counted twice).

Therefore, 19/100 is the measure of your set.

Finding the intervals should be straightforward.


The distance between two sets inside euclidean space Let $A,B \subseteq \mathbb{R}^d$ be non-empty sets. Define their distance to be
$$ d(A,B) = \inf \{ \|x-y\| : x \in A, \; \; y \in B \} $$
For any $A,B$, I want to prove that $d(A,B) = d( \overline{A}, \overline{B} )$.
My Attempt
Put $\alpha = d(A,B)$. Therefore, $\alpha = \|x_0 - y_0 \|$ for some $x_0 \in A$ and some $y_0 \in B$. But, notice $x_0 \in \overline{A}$ and $y_0 \in \overline{B}$ by definition. hence
$$ \| x_0 - y_0\| \geq \inf\{ \|x' - y'\| : x' \in \overline{A}, \; \; y' \in \overline{B} \} =  d( \overline{A}, \overline{B} )$$
$$\therefore d(A,B) \geq d( \overline{A}, \overline{B} )$$
I am stuck trying to show the other direction: $d(A,B) \leq d( \overline{A}, \overline{B} )$
Can someone help me? thanks a lot

Note that the function $\Phi:\mathbb R^d\times \mathbb R^d\to\mathbb R^+$ defined by $\Phi(x,y)=\Vert x-y\Vert$ is continuous, and that $\overline A\times \overline B=\overline{A\times B}$. So, in order to prove what you want, it is enough to show that if $\Phi:X\to\mathbb R^+$ is a continuous function on a topological space $X$, then $\inf\, \Phi(M)=\inf\, \Phi (\overline M)$ for any (nonempty) set $M\subset X$. Now, you have $\Phi(M)\subset \Phi(\overline M)\subset \overline{\Phi(M)}$, the second inclusion being true because $\Phi$ is continuous. So $\inf\, \overline{\Phi(M)}\leq \inf\, \Phi(\overline M)\leq \inf\, \Phi(M)$. To conclude, it remains to observe that for any (nonempty) set $E\subset\mathbb R^+$ you have $\inf\, E\in \overline E$, so that $\inf \, E=\inf\overline E$. Aplying this with $E=\Phi(M)$, this gives $\inf\, \Phi(\overline M)=\inf \,\Phi(M)$ as required.

Note, why do you always have $d(A,B)\geq d(\bar{A},\bar{B})$?

if $d(\bar{A},\bar{B}) = d$ then $\exists x_1,x_2,...\in \bar{A}$ and $y_1,y_2,...\in \bar{B}$, for $\epsilon>0$, $\exists N$ such that $d(x_n,y_n)\leq d+\epsilon$ for $n>N$

now find a sequence $x_1',x_2',...\in A$ and $y_1',y_2',...\in B$ which also does the job, i.e.
for $\epsilon>0$, $\exists N$ such that $d(x_n',y_n')\leq d+\epsilon$ for $n>N$ (how? use definition of closure)

For a proper proof: Distance between two sets in a metric space is equal to the distance between their closures


Representing the area of a circle as the sum of circumferences I had this idea of calculating the area of a circle as the sum of circumferences where the radius is being shortened by an infinitesimal amount until it reaches 0. I was told that this was impossible, because an area can only be represented as a sum of smaller areas, but not lines.
I tried to make a program that calculates the area of a circle by using this method, here's the algorithm:

This works, I've tested it with numerous examples. The bigger the n, the greater the precision.

My 2 questions are:

Why is this working if an area cannot be represented as a sum of lines?
Why must I divide the result given by this method by the reciprocal value of the value I shorten the radius with? As I said, the greater the n, the greater the precision. Thus this will be the exact area when n approaches infinity, but wouldn't I be dividing by infinity later? How does this give me an exact result?

Your algorithm computes the following:
$$
P = \frac{1}{10^n} \sum_{\substack{i \ge 0 \\ r - 10^{-n} i >  0 }} 2 \pi (r  - 10^{-n} i)
$$
For simplicity, let's assume that $r$ is an integer multiple of $10^{-n}$; for example, $r$ is an integer. Then we can write this as
$$
P = \frac{1}{10^n} \sum_{j = 1}^{10^n r}2 \pi j
$$
This is a Riemann sum (and thus a good approximation for) for the integral
$$
\int_{0}^r 2 \pi r' \; dr' = \pi r^2,
$$
and the approximation gets better the more subintervals you have, i.e. the higher $n$ is.

You have good intuition -- the area of the circle can be thought of as the sum of the (infinitely many) lengths of all of the circumferences you can draw inside it with smaller radius. This is what the above says. In calculus terms, the integral of circumference is area. This is why your algorithm works.

I’ve been thinking about the same thing recently. I haven’t taken calculus yet, but I have alternative way of looking at it as multiplying length by width. You have an arithmetic series representing the varying lengths of circumferences within the circle. The sum of an arithmetic series is equal to its average multiplied by the number of terms. In this way, you can multiply the average circumference(length) by the the width that the length occurs along(the radius). That way, you multiply length by width to find area.


List the elements of G So we are asked to write out the elements of G and H where G= $ \mathbb{Z}/ <20> $ and

H = $<4, 20>$ . I understand how to do H and I got:
{$0 + <20>, 4 + <20> , 8 + <20>, 12 + <20>, 16 + <20>$ } from here through, would G=
{$1+ <20>, 2+ <20>, 3+ <20>$, etc....} or how do I find this???

Your $G$ is correct... But you are mentioning $H=<4,20>$, now see $<20>$ $\subset <4>$, so $<4,20>=<4>$ = {$0, \pm4, \pm 8 \ldots \ldots$}.


Let $V$ be the space of $n\times n$ matrices, $B$ a fix matrix, then what is matrix of operator $L_B(A) = BA$ on $V$? I guess the answer is the $n^2\times n^2$ block matrix $diag(B, B, ..., B)$. Where shall I start to prove it? I tried to find the matrix by figuring out the image of the standard basis of $V$, but it gets complicated.

Any idea would be appreciated.

One more question

  $\bullet$ Any idea about how to compute $det(L_B)$? I am sure that the answer is $det(B)^n$ if $B$ is invertible.

I found the answer of question $\bullet$ here :)

If we use a standard basis in which the matrix
$$
A=\begin{pmatrix}
a_{11}&a_{12}&\cdots &a_{1n}\\
a_{21}&a_{22}&\cdots &a_{2n}\\
\cdots\\
a_{n1}&a_{n2}&\cdots &a_{nn}\\
\end{pmatrix}
$$
is represented as a vector of components
$$
A=\begin{pmatrix}
a_{11}\\\cdot\\\cdot \\a_{1n}\\
a_{21}\\\cdot\\\cdot \\a_{2n}\\
\cdots\\\cdots\\
a_{n1}\\\cdot \\\cdot\\a_{nn}\\
\end{pmatrix}
$$
than the matrix that represents the transformation $L_B(A)=BA$ is a block matrix of the form
$$
\begin{pmatrix}
B_{11}&B_{12}&\cdots &B_{1n}\\
B_{21}&B_{22}&\cdots &B_{2n}\\
\cdots\\
B_{n1}&B_{n2}&\cdots &B_{nn}\\
\end{pmatrix}
$$
where $B_{ij}$ is a diagonal matrix with as diagonal  value the element $b_{ij}$ of the matrix $B$

You can prove this result starting from the case $n=2$ and using induction.

Hint If you use the standard basis $E_{ij}$m, then
$$L_B(E_{ij})=BE_{ij}=\begin{bmatrix} 0&0 &...& b_{1i} & 0 & ..0 \\
0&0 &...& b_{1i} & 0 & ..&0 \\
0&0 &...& b_{2i} & 0 & ..&0 \\
...& ...&...&...&...&...&...\\
0&0 &...& b_{ni} & 0 & ..0 \\
\end{bmatrix}$$
where the non-zero elements appear in column $j$. Write this as a linear combination of elementary matrices.


Metric spaces and openness I am asked to prove two things. I would like to know if the proof was elaborate and concise.

I would also like to know if proving reductio ad absurdum is looked down upon. I have heard from my professor say it's not recommended, but my analysis book says it is an invaluable tool.

a) Prove that in a metric space the complement of a point is open.

Suppose that the point $x$ is contained in a closed interval [x]. Now suppose for contradiction the complement of the point is closed, that is, $a<=x$ and $x<= b$. Now, the absurdity is at hand: $[x] \cup [a, b] \leftrightarrow [a, b]$ when they were originally meant to be disjoint. So the complement of the point must be open, that is, $a < x < b$.

b) Prove that any set in a metric space is an intersection of open sets.

No clue here. I would gladly appreciate any clues. I am self-learning this.

Let $(X,d)$ be a metric space.

Let $x \in X$.

We wish to show that $X\setminus \{x\}$ is open.

By definition, this means showing that for each $y \in X \setminus \{x\}$ there exists an $\epsilon\in(0,\infty)$ such that ….

Fix a specific $y$. Then the distance between $x$ and $y$ is $d(x,y)$.

Let $\epsilon = \text{something or other to do with }d(x,y)$.

Then …

a) Prove that in a metric space the complement of a point is open.

Suppose there exists a metric space $x\in X$. We wish to show $X - {x}$ is open. Thus, define

$$\forall y \in M : D(x, y) < \epsilon$$

By definition, $ \forall y \in M, S_{\epsilon}(x)$ is an open ball.  Because open balls are open, it has been shown the complement of a point is open.

b) Prove that any set in a metric space is an intersection of open sets.

Suppose there exists a set $E \subset M, E = E_{1} \cup E_{n}$. Next, suppose $e \in E$. Then $e \in E_{n} \forall n$. Since $E_{n}$ is open we have $B_{r_{i}}(e) \subset E_{n}$. If we take $r = min (r_1, r_2, ... r_n)$, then $B_{r}(e) \subset E$. QED.


does $\lim_{N\to\infty}\frac{\sum_{i=1}^N a_i}{\sum_{i=1}^N b_i}$ converge to $\lim_{N\to\infty}\frac{1}{N}\sum_{i=1}^N\frac{a_i}{b_i}$ Can this ever be the case?
$$\lim\limits_{N\to\infty}\frac{\sum\limits_{i=1}^N a_i}{\sum\limits_{i=1}^N b_i} = \lim\limits_{N\to\infty}\frac{1}{N}\sum_{i=1}^N\frac{a_i}{b_i}$$ with $a_i>0$, $b_i>0$, $a_i<b_i$. As others pointed out simulations indicate convergence, but is there formal ground to it?

For a simple counterexample, take $a_i=\dfrac1{3^i}$ and $b_i=\dfrac1{2^i}$.

Then the LHS ratio tends to a finite value ($\frac12$), while the RHS tends to a finite value ($2$) over $N$, i.e. $0$.

We want to show
\begin{equation}
  \lim\limits_{n\to\infty}\frac{\sum\limits_{i=1}^{n}{a_i}}{\sum\limits_{i=1}^{n}{b_i}}=\lim\limits_{n\to\infty}\frac{1}{n}\sum\limits_{i=1}^{n}\frac{a_i}{b_i}
\end{equation}
Assuming right hand side is convergent, take
\begin{equation}
\lim\limits_{n\to\infty}\frac{1}{n}\sum\limits_{i=1}^{n}\frac{a_i}{b_i}\to\bar{\varepsilon}
\end{equation}
rewrite as
\begin{equation}
  \frac{\sum\limits_{i=1}^{\infty}{a_i}}{\sum\limits_{i=1}^{\infty}{b_i}}=\bar\varepsilon
\end{equation}
let
\begin{equation}
\varepsilon_i = \frac{a_i}{b_i}
\end{equation}
multiply right hand side by 1 and substitute $\varepsilon_i$
\begin{equation}
  \frac{\sum\limits_{i=1}^{\infty}{\varepsilon_ib_i}}{\sum\limits_{i=1}^{\infty}{b_i}}=\frac{\sum\limits_{i=1}^{\infty}\bar\varepsilon{b_i}}{\sum\limits_{i=1}^{\infty}{b_i}}
\end{equation}
substitute $\varepsilon_i=\bar\varepsilon+\Delta\varepsilon_i$
\begin{equation}
  \frac{\sum\limits_{i=1}^{\infty}{\bar\varepsilon b_i}+\sum\limits_{i=1}^{\infty}{\Delta\varepsilon_i b_i}}{\sum\limits_{i=1}^{\infty}{b_i}}=\frac{\sum\limits_{i=1}^{\infty}\bar\varepsilon{b_i}}{\sum\limits_{i=1}^{\infty}{b_i}}
\end{equation}
Then it suffices to show
\begin{equation}
\frac{\sum\limits_{i=1}^{\infty}{\Delta\varepsilon_i b_i}}{\sum\limits_{i=1}^{\infty}{b_i}}\to 0
\end{equation}
Since $\forall i:a_i>0, b_i>0$ we obtain inequality
\begin{equation}
\frac{\sum\limits_{i=1}^{\infty}{\Delta\varepsilon_i b_i}}{\sum\limits_{i=1}^{\infty}{b_i}}<\frac{\left(\sum\limits_{i=1}^{\infty}{\Delta\varepsilon_i}\right)\left(\sum\limits_{i=1}^{\infty}{b_i}\right)}{\sum\limits_{i=1}^{\infty}{b_i}}
\end{equation}
\begin{equation}
\frac{\sum\limits_{i=1}^{\infty}{\Delta\varepsilon_i b_i}}{\sum\limits_{i=1}^{\infty}{b_i}}<{\sum\limits_{i=1}^{\infty}{\Delta\varepsilon_i}}
\end{equation}
where
\begin{equation}
{\sum\limits_{i=1}^{\infty}{\Delta\varepsilon_i}}\to 0
\end{equation}
due to expectation $\mathrm{E}[\Delta\varepsilon_i]=0$ by its definition