Determine the point on the plane $4x-2y+z=1$ that is closest to the point $(-2, -1, 5)$ Determine the point on the plane $4x-2y+z=1$ that is closest to the point $(-2, -1, 5)$. This question is from Pauls's Online Math Notes. He starts by defining a distance function:

$z = 1 - 4x + 2y$

$d(x, y) = \sqrt{(x + 2)^2 + (y + 1)^2 + (-4 -4x + 2y)^2}$

However, at this point, to make the calculus simpler he finds the partial derivatives of $d^2$ instead of $d$. Why does this give you the same answer?

In general, if $$d(x,y)=\sqrt {f(x,y)}$$ for some positive function $f(x,y)$ then the minima of d will correspond to the minima of f.  So, if $f(x,y)$ is differentiable, it makes sense to search for solutions of $$ \frac{\partial f}{\partial x}=0, \frac{\partial f}{\partial y}=0$$ where $f=d^2$ rather than calculating partial derivatives of $d$.

In any case, $$ \frac{\partial d}{\partial x} = 0$$ is the same as $$\frac {1}{2} \frac {1}{\sqrt {f(x,y)}} \frac{\partial f}{\partial x}=0 $$ by applying the chain rule which requires $$\frac{\partial f}{\partial x}=0$$ which is where we started. (Similar conclusion for derivative with respect to $y$)

Here is  a simple solution using tools from middle school for the computation:

Denote $x,y,z$ the coordinates of the orthogonal projection of the point $(-2,-1,5)$
 It satisfies the equations of  proportionality:
$$\frac{x+2}4=\frac{y+1}{-2}=\frac{z-5}1$$
This common ratio is also equal to
$$\frac{4(x+2)-2(y+1)+1(z-5)}{4^2+(-2)^2+1^2}=\frac{(4x-2y+z)+1}{21}=\frac{2}{21}$$
whence the solution
\begin{cases}
x=-2+\dfrac 8{21}=-\dfrac{34}{21},\\[1ex]
y=-1-\dfrac 4{21}=-\dfrac{25}{21}, \\[1ex]
z= 5+\dfrac 2{21}=\dfrac{105}{21}.
\end{cases}
With these elements, the distance squared is
$$d^2=(x+2)^2+(y+1)^2+(z-5)^2=\frac{8^2+(-4)^2+2^2}{21^2}=\frac{84}{21^2}=\frac{4}{21}$$
and finally $\;d=\dfrac 2{\sqrt{21}}.$

As to your exact question, the differential of $d^2$ is $2d D(d)$, hence the critical values are obtained at the same points, and as $d>0$, $d^2$ and $d$ both increase or both decrease.


Should I care if some system forms a group? I am currently studying a discrete dynamical system where the function composition on the set of periodic points forms a finite group.

From a dynamical systems perspective, is there any reason to care about this property?  Would it be worth continuing to study the properties of this group?

Assuming that you have checked that each element does have an inverse, note that:

  The set if iterates on the set of periodic points is finite if and only if there is a maximal period.

So it is fair to say that nothing else can be learned.


Does this choice require Axiom of Choice? This theorem is from Munkres Topology 2nd ed.:

In the last para, it is said that 'choose an element' $n$ of $D$. Why we do not need axiom of choice here? (Here $D$ may not be finite.)

The fact that we can choose an $n\in D$ is just the fact that $D$ is assumed nonempty.  This doesn't require choice.  As a heuristic, you don't need the axioms of choice to choose a sock from a pair of socks (or one sock from each pair of finitely many pairs of socks), but if I give you an infinite collection of pairs, you need some form of the axiom of choice to select one from each pair simultaneously.  You can grab one element from each set in a finite collection of sets from the axioms of set theory.  If you want to grab one from infinitely many sets, you need a stronger axiom in general.

We use axiom of choice while choosing an element from infinitely many sets 'at the the same time'. Otherwise of course we can choose an element one by one. Here i think it is the case that we choose an element from infinitely many sets at the same time.


$(\mathbb{Z}/2\mathbb{Z})[x] /\langle x^2 -2\rangle \ncong (\mathbb{Z}/2\mathbb{Z})[x] /\langle x^2 -3\rangle$ today I have a problem.

Let $R_1=\mathbb{Z}_2[x] /\langle x^2 -2\rangle$ and $R_2=\mathbb{Z}_2[x] /\langle x^2 -3\rangle$

prove or disprove $R_1$ and $R_2$ are isomorphic.

I felt confuse because $x^2 =2$ and $x^2=3$ have solution in $\mathbb{Z}/2\mathbb{Z}$

I don't know what to do.

Both sets have four elements, namely the equivalence classes represented by all polynomials over $\mathbb{Z}_2$ of up to first order. Now make multiplication tables for both sets. If they are essentially the same (up to relabelling entries and reordering rows and columns), the rings are isomorphic. Otherwise they aren't.

I'm not good at editing.I'm sorry:(
First of all, 2 in Z2 is O, because 2 mod 2=0
(the remainder of the division 2 / 2 is 0), then 3 in Z2 is 1 (3/2= 1 , remainder=1).
=>
            X^2 = X ^2 - 1
            -1 is 2-1=1 in Z2

You should know that -m in Zn is n-m.
=> X^2= X^2 + 1 | +1
   X^2 + 1=x^2+2 (2 is 0 in Z2)=>

=> X^2 +1= X^2 | + X ^2

X^2+X^2+1=X^2+ X^2

 X^2 (1+1) +1=X^2(1+1)

1+1 is 0 =>

0+1=0

1=0
False.
Your ecuation  doesn't have a solution. x ^2=2 means x^2=0, with the only solution x=0, and x^2=3 it means x^2=1, x=1. We have a contradiction: x=1=0 False


Prove every odd integer is the difference of two squares I know that I should use the definition of an odd integer ($2k+1$), but that's about it.

Thanks in advance!

Step 1: pick an odd number (like $n=13$ here)

Step 2: bend it in "half" (any odd number $n$ can be written as $2k+1$, and $13=2\cdot 6 + 1$)

Step 3: fill in the blank space

Step 4: Count squares.  (Here, the blue square has area $36=6^2$, while the whole square has area $49=7^2$)

You could add.

For mod ((2x+1),4)= 1, 2x+1 has at least one more solution than (k^2+1)-k^2=2x+1 and k^2 will be odd and (k-n)^2 will be even. Opposite if  Mod((2x+1),4)= 3. n is always odd.

k^2-(k+n)^2= 2x+1, if 2x+1 is a composite number, if not, 2x+1 only has the solution (k^2+1)-k^2=2x+1 and 2x+1 is a prime number.

Sorry Mathjax is not working on my computer.


Find the number of ways of arranging the letters Find the number of ways of arranging the letters $\text{AAAAA, BBB, CCC, D, EE & F}$ in a row if no two $\text{C's}$ are together?

My Attempt:

Well, I should I arrive at the answer if I subtract the cases where 3 $\text{C's}$ and 2 $\text{C's}$ appear together from the total.

Total possibilities $= \frac{15!}{5!3!3!2!}$

Total possibilities where 3 $\text{C's}$ appear $=\frac{13!}{5!3!2!}$

However, I am not able to find the possibilities for 2 $\text{C's}$ being together and get to the answer.

Any help would be appreciated.

Ignore the C's for the moment and arrange the remaining letters. There are $\frac{12!}{5!3!2!}=332640$ ways to do this.

Now consider the thirteen spaces between and beyond the letters – at most one C may be inserted into each space, so given an arrangement of the other letters there are $\binom{13}3=286$ ways to add C's.

The total number of admissible combinations is thus $332640×286=95135040$.

First arrange all the letters without $C$'s, i.e. $$\frac{12!}{5!3!1!2!1!}.$$
Now we have to fill inside the gaps between arranged letters.
As they given no two $C$'s come together, means $3$ $C$'s also doesn't come together.
The answer:
$$
\frac{12!}{5!3!1!2!1!}*\frac{P^{13}_3}{3!}.
$$
For example I have $3$ chocolates, that doesn't mean that I don't have $2$.


Does $\operatorname{div}\left(\nabla G +xG\right)=0\Longleftrightarrow \nabla G +xG=0$? Let $G$ be a smooth function defined on $\textbf{R}^d$. What are the assumptions I should use to assume that
$$\operatorname{div}\left(\nabla G(x) +xG(x)\right)=0 \quad (\forall x\in \textbf{R}^d)$$
implies that $G$ is a Gaussian? (Several answers are possible I guess...)

Many thanks !

I'm going to use an argument along the lines of what @hOff proposed.  In addition to the smoothness of $G$ I am going to suppose that it decays rapidly at infinity so that all of the following calculations are justified.  For example, we could assume that $G$ is Schwartz class.

As the first order of business we expand your equation as
$$
\Delta G(x) + x \cdot \nabla G(x) + d G(x) =0.
$$
To attack this problem we're going to use the Fourier transform, which we define as
$$
  \hat{f}(\xi) = \int_{\mathbb{R}^d} f(x) e^{-2\pi i x\cdot \xi} dx.
$$

We have the following useful identities:
$$
\widehat{\Delta f}(\xi) = -4 \pi^2 |\xi|^2 \hat{f}(\xi) \\
\widehat{x \cdot \nabla f}(\xi) = -\xi \cdot \nabla \hat{f}(x) - d \hat{f}(\xi).
$$
These both follow by integrating by parts in the formula for the Fourier transform.  For example,
$$
\widehat{x \cdot \nabla f}(\xi) = \sum_{j=1}^d \int_{\mathbb{R}^d} x_j \partial_j f(x) e^{-2\pi i x\cdot \xi} dx \\
=  -\sum_{j=1}^d \int_{\mathbb{R}^d}  f(x) \partial_j( x_j e^{-2\pi i x\cdot \xi})  dx  \\
= -\sum_{j=1}^d \int_{\mathbb{R}^d} f(x) [x_j(-2\pi i \xi_j) + 1] e^{-2\pi i x\cdot \xi} dx  \\
= - d \hat{f}(\xi) -  \sum_{j=1}^d \xi_j \int_{\mathbb{R}^d} f(x)(-2\pi i x_j)e^{-2\pi i x\cdot \xi} dx \\
=  - d \hat{f}(\xi) -  \sum_{j=1}^d \xi_j \partial_{\xi_j} \int_{\mathbb{R}^d} f(x)e^{-2\pi i x\cdot \xi} dx \\
= - d \hat{f}(\xi)  - \xi \cdot \nabla \hat{f}(\xi).
$$

Now we apply the Fourier transform to our equation for $G$ to find that
$$
0= - 4\pi^2 |\xi|^2 \hat{G}(\xi) -\xi \cdot \nabla \hat{G}(\xi) - d \hat{G}(\xi) +  d \hat{G}(\xi) \\
= - 4\pi^2 |\xi|^2 \hat{G}(\xi) -\xi \cdot \nabla \hat{G}(\xi).
$$

This is a first order PDE that we can solve using the method of characteristics.  Suppose for the moment that $\xi_0 \in \mathbb{R}^d$ is such that $|\xi_0|=1$.  Consider the function $g(t) = \hat{G}(e^t \xi_0)$.  We compute
$$
\frac{d}{dt} g(t) = \nabla \hat{G}(e^t \xi_0) \cdot e^t \xi_0 = -4\pi^2 |e^t \xi_0|^2 \hat{G}(e^t \xi_0) = -4\pi^2 e^{2t} g(t),
$$
which implies that (using standard ODE theory)
$$
g(t) = g(0) \exp\left(-2\pi^2(e^{2t} -1) \right)
$$
and hence that
$$
\hat{G}(e^t \xi_0) = \hat{G}(\xi_0) \exp\left(-2\pi^2(e^{2t} -1) \right).
$$
Now, for any $\xi \neq 0$ we write $\xi = e^t \xi_0$ for $\xi_0 = \xi/|\xi|$ and $|\xi| = e^t$.  Plugging in above then shows that
$$
\hat{G}(\xi) = \hat{G}\left(\frac{\xi}{|\xi|} \right)\exp\left(-2\pi^2(|\xi|^2 -1) \right).
$$
Since we have assumed that $G$ is integrable, we know that $\hat{G}$ is continuous, and so we have derived the most general form for $\hat{G}$, namely
$$
\hat{G}(\xi) = K\left(\frac{\xi}{|\xi|} \right) e^{-2\pi^2 |\xi|^2}
$$
for some continuous function $K$.  However, the exponential term is unity at $\xi=0$, so the continuity of $\hat{G}$ requires that actually $K$ is constant.

Hence $\hat{G}(\xi) = Ke^{-2\pi^2 |\xi|^2}$, which then implies that
$$
G(x) = \frac{K}{(2\pi)^{d/2}} e^{-|x|^2/2}.
$$
Let's now go back to your question of what assumptions have to be made on $G$.  To make the above analysis valid we need that $G$, $\Delta G$, and $x\cdot \nabla G$ all decay fast enough to justify applying the Fourier transform.  For example, we can assume that
$$
G, \Delta G, x \cdot \nabla G \in L^1(\mathbb{R}^d).
$$

A last remark: You should also be able to prove that $G$ is Gaussian if and only if $G$ is radial.  One direction is trivial.  For the other you assume that $G$ is radial and then rewrite the PDE as a second order ODE in $r$ and the resulting solution should be a Gaussian.

For the functions from Schwartz functional space $S_{1,1}$ where $S_{\alpha,\beta} = |x^{\alpha}\frac{\partial^{\beta}}{\partial x^{\beta}}G| \lt \infty$ where norm means existence if integral over $R^d$ we can use multidimensional Fourier transform.

This end up in: $$0 = \digamma[\nabla(\nabla G + xG)] = i<k \cdot \hat{F}>$$ where $\hat{F} = ik\hat{G} +i\nabla\hat{G}$ and $\hat{G} = \digamma[G]$

Now, $\hat{F}$ vector is parallel to $k$, i.e. we do not have any perpendicular components that vanish in scalar product $<k \cdot \hat{F}>$.

It is known that Schwartz space is dense in $L^p$, so we can approximate function from $L^p$ by the sequence of function from $S$ to converge weakly by $L^p$ norm or a.e. approximation.


Probability of three events occurring given correlation? I am facing a problem that I cannot find the answer to. I have three variables, A, B and C. There are only two possibilities for each of these, A either happens or it does not, B happens or it does not and C happens or it does not. I know that if these events are independent that the probability of them all occurring is simply $P(A)\cdot P(B)\cdot P(C)$. So if the probability of each happening is 10% then all three have a $10\%·10\%·10\% = 0.1\%$ probability of occurring. But how would this formula change if the events were not independent but were instead positively correlated.

I can solve this for just two variables with the formula:
$P(A \cap B) = P(A)\cdot P(B) + \rho_{AB}\cdot \sqrt{P(A)\cdot (1-P(A))\cdot P(B)\cdot (1-P(B))} $, where $\rho_{AB}$ is the correlation coefficient between A and B.

How would I change this formula to calculate the probability that A, B and C all occur? I.e. calculating $P(A \cap B \cap C)$ knowing $P(A)$, $P(B)$, $P(C)$, $\rho_{AB}$, $\rho_{AC}$, $\rho_{BC}$.

Thanks in advance for the help!

As I found out in this presentation, where the above-mentioned issue is dealt with in the context of default correlations, the problem is not possible to be solved with the inputs given.

I.e., when 3 events have defined probabilities of happening, the pairwise correlations are not enough to construct the joint probability of all 3 events happening.

In the context of default correlations, complex problems of correlations of events are solved using other methods such as Copula functions.

I believe you can take the average correlation and plug it in while adding the addition P(C) & P(C)⋅(1−P(C)) to each side of the equation.

This at least works for my use case.


Solving for $x$ when $\sqrt{\sqrt{3} - \sqrt{\sqrt{3} + x}} = x$ Suppose $x$ is a real number such that $\sqrt{\sqrt{3} - \sqrt{\sqrt{3} + x}} = x.$ Find $x.$

I first squared to get rid of the first square root, which gave me $\sqrt{3} - \sqrt{\sqrt{3} + x} = x^2.$ However, I'm not sure how to move on from there. Can someone give me a hint?

From where you left off:
$$\begin{align}
\sqrt{3}-\sqrt{\sqrt3+x}
&=x^2\\
\sqrt{3}-x^2
&=\sqrt{\sqrt3+x}\\
3-2\sqrt{3}x^2+x^4
&=\sqrt{3}+x\\
x^4-2\sqrt{3}x^2-x+\left(3-\sqrt3\right)&=0\\
\left(x^2+x+(1-\sqrt{3})\right)\left(x^2-x-\sqrt{3}\right)&=0
\end{align}$$
So if there is a solution to the original equation, it is a root of this 4th degree polynomial. It's easy to find its four roots since it factors. But some roots of this polynomial might not solve the original equation, since we squared a few times earlier. So each one should be checked.
Note that the original left-hand side is not real unless $x$ is in $\left[-\sqrt{3},3-\sqrt{3}\right]$. That should help eliminate several of the four polynomial roots.

Comment:
If you put $x=\sqrt{\sqrt{3} - \sqrt{\sqrt{3} + x}} $ under the third radical you get:
$$\sqrt{\sqrt{3} - \sqrt{\sqrt{3} + \sqrt{\sqrt{3} - \sqrt{\sqrt{3} + \sqrt{\sqrt{3} - \sqrt{\sqrt{3} + }} . . .}} }} = x$$
Suppose there is no negative sign then by squaring both sides you get:
$$X^2=\sqrt 3 +X$$
Which gives:
$$X=\frac{-1 ±\sqrt{1+4\sqrt 3}}{2}≈0.9≈\frac{\sqrt3}{2}$$
So $x<X≈\frac{\sqrt 3}2$


By applying the second version of the Fundamental Theorem of Calculus find the integral: The second version of the Fundamental Theorem of Calculus states that if $F'(x)=f(x)$ then $\int_{a}^{b} f(x) dx = F(b)-F(a)$.

I need to use this to find a) $\int_{-2}^{-1} \frac{1}{x^3} dx $ and b) $\int_{0}^{x} t \cdot cos(t^2) dt $

Now I can solve both integrals easily enough and get a) $-\frac{3}{8}$ and b) $\frac{1}{2}sin(x^2)$ as my answers. However I don't know what qualifies as using the Fundamental Theorem of Calculus. I've just integrated as I normally would and got my answers. Do I need to be particularly rigorous or what?

Computing these integrals "easily enough" means using the FTC. That's what the FTC is all about.

The "hard way" would be to rewrite them as limits of Riemann sums and computing these limits "directly". The latter can only be done in special cases, like the following: $\int_0^1 e^x\ dx$.

we have $\left(-\frac{1}{2}x^{-2}\right)'=\frac{1}{x^3}$ thus we get
$\int_{-2}^{-1}\frac{1}{x^3}dx=\frac{1}{2}(-2)^{-2}-\frac{1}{2}(-1)^{-2}=-\frac{3}{8}$
is it this what you mean?
sorry for my mistake, further we have
$\left(\frac{1}{2}\sin(x^2)\right)'=\frac{1}{2}\cos(x^2)2x=x\cos(x^2)$ and we get
$\int_{0}^{x}t\cos(t^2)dt=\frac{1}{2}\sin(x^2)$


Deriving the $Var(X) = E(X^2) - \mu^2$ from the moment around the mean formula. I want to be able to derive  $Var(X) = E(X^2) - \mu^2$  from the moment around the mean formula $Var(X) = E((X-\mu)^2)$. Is there a way to do this?

Thanks in advance.

Please find the solution, let me know if you have questions.


Stochastic process with delta correlation in time I am trying to learn stochastic calculus and when they talk about the Langevin equation they say that the correlation of the gaussian white noise (which i believe is the covariance between two random variables in the stochastic process) has a "$\delta$ correlation in time":

  $\langle\xi(t)\xi(\tau)\rangle=\delta(t-\tau)$

Where $\delta$ is "the delta function". Now I wonder which one they mean, is it the generalised function (does that mean the correlation is $\infty$?) or is it the indicator function? Or something entirely different?

I think that when something like this is said, it is meant, that the processe $\xi(t)$ is correlated only at one single moment of time: that's the meaning of delta-function term. :) So the autocorrelation function vanishes in other moments. In some way it is the indicator function (as your have said) of the moment when it is not zero.

The sharp braces indicate the ensemble average, using the Kronecker delta means its only fully positively correlated with itself when $t = \tau$. For any $t \neq \tau$ it is uncorrelated.


If $x\not\leq y$, then is $x>y$, or $x\geq y$? I'm currently reading about surreal numbers from here.

At multiple points in this paper, the author has stated that if $x\not\leq y$, then $x\geq y$.

Shoudn't the relation be "if $x\not\leq y$, then $x>y$"?

Hasn't the possibility of $x=y$ already been negated when we said $x\not\leq y$?

Thanks in advance.

The paper defines $\leq$ and it is not necessarily the "standard" $\leq$ that you and I are used to. It is often better to substitute some other symbol like $x\preceq y$, and realize we have to prove stuff about this symbol before we can use it.

In particular, in that paper, it is proved in Theorem 1.4 that $x\not\leq y$ implies $y\leq x$. It does not prove that $y<x$.

So trying to reason about $\leq$ as it if means "less than or equal" is a mistake. There is no definition of "less than" to use here.  There is only $\leq$.

There are examples where $x\leq y$ and $y\leq x$. This means that if we define $x<y$ in terms of $x\leq y$ and $x\neq y$ then we get $x<y$ and $y<x$ and therefore we find that $<$ defined this way is non-transitive.

I think the name "surreal" should give a strong hint at the non-standard nature of these "numbers."

Here's another way of looking at it.

$$(x\le{y})\leftrightarrow (x<y \lor x=y)$$

$$(x\not\le{y})\leftrightarrow(x>y\land{x\ne{y}})$$

$$x\not\le{y}\rightarrow{x\ne{y}}$$

Therefore:
$$x\not\le{y}\not\rightarrow{{(x\ge{y})}}$$


AVL tree to find kth smallest key I'm working on this problem and having difficulty with it

Search trees such as BST, AVL, 2-3 trees are used to search a set of n keys for a key value. We can augment the tree to allow searching by rank, that is, we can ask for the kth smallest key in the tree. On way to do thus for AVL trees is to store a field L(x) at each node(in addition to the balance information for x) such that L(x) is 1+ the number of nodes in x's left subtree. Design an algorithm that inputs an AVL tree T with rank fields (L(x)) and n nodes and an integer k such that 1<=k<=n and outputs the kth smallest key in T. Analyze worst case complexity of the algorithm Explain how the rank information of T should be updated after the insertion of a new key.

First of all, I'm not sure why L(x) <-rank(x) is 1 + the number of nodes in x's left subtree. And for the time complexity, I think it will remain O(logn) as the insertion will rebalance the tree and it taks O(logn)

Can anyone please clarify this?

The #(number) nodes in the left subtree of $x$ is the # of nodes which have value less than that of the $x$ in the tree rooted at $x$.

The rank of an element is # of elements which are less than itself +1.

If you trying to search the rank of some element $y$ less than $x$, then rank of $y$ in tree rooted at $x$ is rank of $y$ in tree rooted at left subtree of $x$.

If you are trying to search the rank of some element $y>x$, then the rank of $y$ in tree rooted at $x$ =(rank of $x$ + rank of $y$ in right subtree of $x$).

Lastly if you are looking for rank of $x$, you just query at the root.

As far as updating the rank information is concerned.
Every time a new element is added, check the root.
If it is less than the root increment the root's rank field by one, and recurse on left-subtree, else dont increment the root's rank field and recurse on the right subtree.

Time to update the rank information is same as the time to insert an element in the tree.


Find the only f=vt that has f(2t)=4f(t) I'm just starting calculus 1 and I don't know how to solve this.  Can someone please help?
The problem below involves linear functions $f(t) = vt + C$. Find the constants v and C.
Find the only $f=vt$ that has $f(2t)=4f(t)$.  show that every $f=\frac{1}{2}at^2$ has this property.

Assuming $v$ is constant: $f(2t) = 4f(t)$ implies $v(2t) = 4(vt)$ and therefore $2vt = 0$.
This yields either $t$ or $v$ must be $0$, but since the equation has to hold for all $t$ this means $v$ must be $0$.
Now let $f(t) = \frac{1}{2}at^2$. $f(2t) = \frac{1}{2}a(2t)^2 = \frac{1}{2}a(4t^2) = 4(\frac{1}{2}at^2) = 4f(t)$, regardless of $a$.
Hope this helps!

Let $t=2^s$.
Then
$$f(2t)=4f(t)$$ is
$$f(2\cdot2^s)=f(2^{s+1})=4f(2^s)$$ or $$g(s+1)=4g(s).$$
As no other information is given, $g$ is an arbitrary function in a unit range, and its value in other ranges are derived from the above recurrence, as
$$g(s+n)=4^ng(s)$$
or
$$f(2^nt)=4^nf(t)$$ where $f$ is arbitrary in, say, $[1,2)$.

$vt$ cannot work, because $v\,2^nt\ne 4^nv\,t.$
$at^2$ does, because $a\,(2^nt)^2=4^na\,t^2.$


The space $C^0([0;1],\mathbb{R})$ of all continuous, real-valued functions on $[0,1]$ is not reflexive. How does one prove that $C^0([0;1],\mathbb{R})$ equipped with the supremum norm is not reflexive?

I don't understand how to show that the $J$ mapping is not surjective.

As we have seen in a previous question, if $E$ is a reflexive Banach space then each linear continuous functional attains its norm. So, in order to show that $E:=C^0([0,1],\Bbb R)$ endowed with the supremum norm is not reflexive, it is enough to find a linear functional which is not norm attaining. We can define
$$x'(f):=\int_0^{1/2}f(t)\mathrm dt-\int_{1/2}^1f(t)\mathrm dt;$$
$x'$ is linear and $|x'(f)|\leqslant \lVert f\rVert_{\infty}$, hence $x'$ is continuous and its norm is $\leqslant 1$.
To see that the norm is indeed $1$, for $n$ integer, consider a function $f_n$ which is $1$ on $[0,1/2-1/n)$ and $-1$ on $(1/2+1/n,1)$, and linear on $(1/2-1/n,1/2+1/n)$. We can see that $\lVert f_n\rVert=1$ and $x'(f_n)=1-2/n$.
Now, we have to show that we cannot find $f\in E$ such that $x'(f)=1$ and $\lVert f\rVert=1$. Let $f$ be continuous of norm $1$. We have to show that $x'(f)\neq 1$, for a fixed $\varepsilon>0$ we can find $\delta>0$ such that $|f(t)-f(1/2)|\leqslant \varepsilon$ whenever $|t-1/2|\leqslant \delta$. Since
$$\tag{*}  x'(f)=\int_0^{1/2-\delta}f(t)\mathrm dt+\int_{1/2-\delta}^{1/2+\delta}f(t)\mathrm dt -\int_{1/2+\delta}^1f(t)\mathrm dt,$$
we can assume that $f(x)=1$ on $[0,1/2-\delta]$ and $f(x)=-1$ on $[1/2+\delta,1]$, otherwise it is clear that $x'(f)\neq 1$.
By (*), it follows that $|x'(f)|\leqslant 1-2\delta+2\delta\varepsilon+\delta| f(1/2)|$. Taking $\varepsilon\lt 1-|f(1/2)|$, we get that $|x'(f)|\lt 1$.

you could also use the following facts:

$BV[a,b]$ is not separable,
$C[a,b]$ is separable,
if $X$ is a Banach Space and $X^*$ is its dual and is separable, then $X$ is separable.

Since $BV[a,b]$ is the dual of $C[a,b]$, reflexivity would imply $BV[a,b]$ is separable.


Car travelling with lowest fuel Suppose a car's fuel tank can carry at most 500 unit fuel, but it need to travel from A to B where the distance between them is 1000 km. Suppose that the fuel consumption rate is 1 unit/km, and between A and B, the car can store any amount of fuel at any point. The question is, what is the minimum unit of fuel needed at the starting point A?

To make the problem clearer, consider a simple problem of traveling 5 km, with the car's max fuel 4 unit. Of course, if you just take 4 unit fuel at the starting point, it cannot finish 5km. So you can pick a position, say 1 km from the start, when your full fueled car reach here, you have 3 unit left, so you store 1 unit here and then go back to start with 4 unit again. This time, when you reach this point, you have 3 unit left because you traveled 1km, but you can take 1 unit from your previous storage to make your fuel tank full again. And now you have 4 unit fuel with your distance to destination 4km, so you can finish your journey smoothly.

For the original problem, assume that I can only store the fuel at integral km distance from the start, I obtain the solution that I need 3848 unit at the start, and I will store the fuel at 24 km, 62 km, 107 km, 163 km, 234 km, 334 km and 500 km such that when I reach 500 km, I can guarantee that I have 500 unit for the rest of the journey. But in the original problem, there is no such limit which means that you can store at, for example, 24.675 km if only it can reduce the total fuel needed. So how can we obtain the optimal solution for this?

Wouldn't it be more optimal to store $\frac{500}3$ units at $\frac{500}3$ km, go back. Now your fuel budget is $500\cdot (1 + 1/3)$, so store $500\cdot \frac{1 + \frac 13}3$ units at $500\cdot \frac{1 + 1/3}3$ km, go back. Now your fuel supply is $500\cdot (1 + (1 + \frac 13)\cdot \frac 13)$, so store $500\cdot (1 + (1 + \frac 13)\cdot \frac 13)$ etc...it's a series you need to get to 2.

1

$\frac 13 + 1 = \frac{1+3}3$

$\frac{4/3}3 + 1 = \frac{1+3+9}9$

$\frac{10/9}3 + 1 = \frac{1+3+9+27}{27}$

etc. When does $(3^0 + 3^1 + 3^2 + 3^3 +\cdots  + 3^n)/3^n$ reach 2?


Prove $2\cos\frac{2\pi}{7}$ is root of $x^3 + x^2 - 2x -1$ To prove that a regular septagon cannot be constructed by a straightedge and compass, it suffices to prove that $2\cos(\frac{2\pi}{7})$ is not constructible.

Several other answers to this question follow with the fact that $2\cos(\frac{2\pi}{7})$ is a root of

$$x^3 + x^2 - 2x -1$$

How do I derive that?

I know that since the cubic polynomial is irreducible in $\mathbb{Q}$, it has no real roots, implying that $cos(\frac{2\pi}{7})$ is not constructible.

Let $a = \frac{2\pi}7$. Then, $3a + 4a = 2\pi$ and

$$\cos4a = \cos3a$$

Use the identities $\cos 2t = 2\cos^2t -1$ and $\cos3t = 4\cos^3 t-3\cos t$ to expand the two sides,

$$2(2\cos^2 a -1)^2 -1 = 4\cos^3a -3\cos a $$

Rearrage,

$$8\cos^4a -4\cos^3a -8\cos^2a +3\cos a+1=0$$

which factorizes as

$$(\cos a -1) (8\cos^3a + 4\cos^2a-4\cos a-1)=0$$

Since $\cos a \ne 1$, the equation above shows that $2\cos \frac{2\pi}7 $ satisfies

$$x^3+x^2 -2x-1=0$$


Weak convergence, $f\in L^p(\Bbb R^d) $ and $g\in L^q(\Bbb R^d) $ then, $\lim_{j\to\infty} \int_{\Bbb R^d}f_j(x)g(x)dx = 0$ Assume $f\in L^p(\Bbb R^d) $ and  $g\in L^q(\Bbb R^d) $
Where, $1<p<\infty$ and $1<q<\infty$ are dual ecxponents namely, $$\frac1p+\frac1q =1$$
Then for every $s\in\Bbb R$ such that, $sp\le d$ show that,
$$\lim_{j\to\infty} \int_{\Bbb R^d}f_j(x)g(x)dx = 0$$

Where, $$f_j(x) = j^sf(jx)~~~$$

My attempt,

I Applied Holder inequality, and use the change of variables $u =jx$ to get the following,

\begin{align}\left|\int_{\Bbb R^d}f_j(x)g(x)dx\right| &\le \left(\int_{\Bbb R^d}|f_j(x)|^pdx\right)^{1/p} \left(\int_{\Bbb R^d}|g(x)|^qdx\right)^{1/q}\\&=\frac{1}{j^{\frac dp-s}}\left(\int_{\Bbb R^d}|f(x)|^pdx\right)^{1/p} \left(\int_{\Bbb R^d}|g(x)|^qdx\right)^{1/q} \\&=\frac{1}{j^{\frac dp-s}}\|f\|_{p}\|f\|_{q}\end{align}

If I suppose that, $\color{blue}{d>sp}$ then $\frac{1}{j^{\frac dp-s}}\to 0$
and the result follows.

  Question: How do I prove the case where $sp= d$?

Note it's not true for $p=1$. Assume $p>1$. Come to think of it, it's also false for $p=\infty$. Assume $p<\infty$.

Let $\epsilon>0$. Choose $A>0$ so that $$\left(\int_{|x|\le A}|g(x)|^q\right)^{1/q}<\epsilon.$$(This is where $q<\infty$ is needed.)
An explicit calculation shows that $$\left(\int_{|x|>A}|f_j(x)|^p\right)^{1/p}\to0.$$(This uses $p<\infty$.)

Edit: What's above will look like a solution to many readers. Others may be wondering how those two lines prove what's asserted. The (standard) argument goes like so: Assume $\epsilon$ and $A$ are as above. Choose $N$ so that $$\left(\int_{|x|>A}|f_j(x)|^p\right)^{1/p}<\epsilon\quad(j>N).$$
Noting that $||f_j||_p=||f||_p$ (since $sp=d$), for $j>N$ we have $$\left|\int f_jg\right|\le\left|\int_{|x\le A}f_jg\right|+\left|\int_{|x|>A}f_jg\right|\le||f||_p\epsilon+\epsilon||g||_q.$$

This is just a reminiscent of the this:
How to prove that, $\lim\limits_{k \rightarrow \infty} \int_{\Bbb{R}} f_k(x) g(x) dm(x) = 0$.

If $sp =d$ then, Proceed as follows:

  Lemma: Continuous function of compact support are dense in $L^r(\Bbb R^d)$ for, $1<r<\infty$.

For,  $\varepsilon>0$ we know that there exist a $f_\varepsilon,g_\varepsilon\in C_c(\Bbb R)$ a continuous function of compact support such that,
$$ \|g-g_\varepsilon\|_q\le \varepsilon \|f\|_p^{-1} $$
and
$$ \|f-f_\varepsilon\|_p\le \varepsilon \|g_\varepsilon\|_q^{-1} $$

Proceeding like here with   Cauchy Schwartz inequality replaced by Holder inequality, one easily arrives at

$$\left|\int_\mathbb R f_j(x) g(x) dx\right| \le\left[\varepsilon+\varepsilon j^{s-d}+ j^{s-d}\left|\int_{\text{supp }f_\varepsilon}f_\varepsilon(x) g_\varepsilon(\frac{x}{j}) dm(x)\right| \right]\\= \left[\varepsilon+\varepsilon j^{s(1-p)}+ j^{s(1-p)}\left|\int_{\text{supp }f_\varepsilon}f_\varepsilon(x) g_\varepsilon(\frac{x}{j}) dm(x)\right| \right]$$

Moreover, $f_\varepsilon$ $h_\varepsilon$ are continuous functions  with compact support, we get
 $$\color{blue}{\lim_{j\to \infty}\int_{\text{supp }f_\varepsilon}f_\varepsilon(x) g_\varepsilon(\frac{x}{j}) dx=\int_{\text{supp }f_\varepsilon}f_\varepsilon(x) g_\varepsilon(0) dx}$$

Whence,
$$ \lim_{j\to \infty} \left|\int_\mathbb R f_j(x) g(x) dx\right| \le\lim_{j\to \infty} \left[\varepsilon+\varepsilon j^{s-d}+ j^{s-d}\left|\int_{\text{supp }f_\varepsilon}f_\varepsilon(x) g_\varepsilon(\frac{x}{j}) dx\right| \right]=\varepsilon $$

Thus, for all $\varepsilon>0,$ we have, $$ \limsup_{j\to \infty} \left|\int_\mathbb R f_j(x) g(x) dx\right| \le \varepsilon $$
letting  $\varepsilon\to 0$  yields the require result.

$$ \color{red}{\lim_{j\to \infty}  \int_\mathbb R f_j(x) g(x) dx =0.} $$


$\int_0^1|f_n|^3\leq 1\Rightarrow \int_E |f_n|<\varepsilon$ when $|E|$ is small Let $f_n\colon [0,1]\to\mathbb{R}$ be Lebesgue measurable with $$\int_0^1|f_n|^3\leq 1 \mbox{ for all } n.$$ Show that for all $\varepsilon>0$ there exists $\delta>0$ so that if $E\subset[0,1]$ is Lebesgue measurable with $|E|<\delta$ then $$\int_E |f_n|<\varepsilon\mbox{ for all } n$$

I don't want to give away the whole answer, but use that $\displaystyle \int_E |f_n|=\int_{[0,1]}\chi_E|f_n|$ and try to use the Holder's inequality hint on the right side (what should $q$ be?).

I would complete proof of JLA. Suppose $ p=3 $ and then Using the equality $ \frac{1}{p} + \frac{1}{q} =1 $ we have $ q= \frac{3}{2}$. We can write
$$ \int_E |f_n|=\int_{[0,1]}\chi_E|f_n| \leq \left(\int_{[0,1]}{|f_n|^3}\right)^\frac{1}{3} \left(\int_{[0,1]}{\chi_E}^{\frac{3}{2}}\right)^{\frac{2}{3}}$$


Function mapping of $\sqrt{x}$ Can $\sqrt{x}$ be considered a function for x (set of real numbers greater than or equal to 0)?
Moreover, why is the graph of $\sqrt{x}$ only maps to positive y values if for example $\sqrt{9}$ = 3 and -3 shouldn’t it be symmetrical over the x axis?

The domain and range of sqrt(x) is always non negative number.
1.) Domain must be non negative otherwise the number would become imaginary. E.g sqrt(-1) = iota.
2.) As square root of non negative number is also non negative, range must be toward positive side.


Solutions that oscillate more rapidly Which one of the following two equations has solutions that oscillate more rapidly?
$$x''+\sqrt{t^6+3t^5+1} x =0$$
$$ x''+2t^3 x =0$$
I was trying to use sturm separation theorem but not getting exact approach towards it!!
Is there any direct way to do it?
Edit: The book from where I have given this question has answer as first equation in the hints provided at the back of book.
Book name: "A Textbook  on Ordinary  Differential  Equations" by Shair Ahmad, Antonio Ambrosetti(Second Edition)-Chapter5-Exercise-D2

If we assume that $t$ is confined in an interval with small amplitude we may assume that $t^3$ is approximately constant. By recalling that a solution of $x''+ K x=0$ (with $K>0$) has a frequency of $\frac{\sqrt{K}}{2\pi}$ we immediately get that a solution of $x''(t)+2t^3 x(t)=0$ oscillates faster than a solution of $x''(t)+\sqrt{t^6+3t^5+1}\,x(t)=0$.

Just solve them and plot the two:

The full forms have constants, that must be set for initial conditions.  For example $c_1 \cos t + c_2 \sin t$.  In the above, I've set all constants to $1$, just to reveal the periods... the focus of the question.


Why is integration the inverse of differentiation Why is integration the inverse of differentiation, I mean why do I get the same function when I integrate and then differentiate the result?

I'm going to assume that you want an intuitive explanation rather than the mathematical details. An intuitive explanation is that

integration is like saying "give me a running total of all the values this function has had up to a certain point"
differentiation is like saying "how much is the value of this function going to increase from a given point to the next point"

So you can see if I ask for a running total, I'm adding up values the function takes one by one, and integration can be roughly read as accumulation. On the other hand if I ask how much a function is going to increase from one point to the next I'm asking for the difference between one value and the next.

The intuitive relationship that differentiation undoes integration can then be seen as follows. If you take the difference between one value of a running total and the next value of the running total, you get what you added to the current value of the running total to get the next value of the running total, which is the function you had in the first place before you integrated it.

This is illustrated pictorially below where the simple function $y=1$ at the top of the diagram is accumulated over an interval size (or step size) of 1 to give the step function at the bottom left. You can see that each time you go from $x$ to $x+1$ we accumulate a yellow box on the bottom left graph. Conversely to go from the bottom left graph to the top graph we just take the difference between the total at $x$ and the total at $x+1$, which is a yellow box.

What happens in calculus is that the points are moved closer and closer together until the intervals over which functions are accumulated (integration) or differenced (differentiation) are infinitesimally small.  As the interval size is made smaller and smaller the steps gradually get closer and closer together until in our example you get the smooth function $y=x$ at the bottom right of the diagram. This is equivalent to taking the area under the curve to get to $y=x$, and taking the gradient of the curve to go back from $y=x$ to $y=1$.

The relationship between integration and differentiation is a very important relationship in calculus, so important that it is called the Fundamental Theorem of Calculus.

I will suggest you to go and see the definition of integration (indefinite).
There you will see that integration is a method to find the function when at any point in domain, its differentiation is provided to you. And so it becomes the inverse of differentiation.


What is probability of language knowledge in company? In a company are 100 workers.

50 of them (100) speak ENG
30 of them (100) speak GER
15 of them (100) speak ENG and GER

What is the probability of randomly selected worker speaks GER on condition he does NOT speaks ENG?

I have two suggestions:

a) 3/10
b) lower than 3/10

As described in the problem, a worker speaking German and a worker speaking English are independent events. As such, the fact that a worker speaks English doesn't actually give you any information as to whether or not a worker speaks German. Thus, the odds that a worker speaks German given that he speaks English is the same as the odds that a worker speaks German: $\frac{3}{10}$

ENG          Non-ENG
☺☺☺☺☺☺☺☺☺☺ | ☺☺☺☺☺☺☺☺☺☺
☺☺☺☺☺☺☺☺☺☺ | ☺☺☺☺☺☺☺☺☺☺
☺☺☺☺☺☺☺☺☺☺ | ☺☺☺☺☺☺☺☺☺☺
☻☻☻☻☻☺☺☺☺☺ | ☻☻☻☻☻☺☺☺☺☺
☻☻☻☻☻☻☻☻☻☻ | ☻☻☻☻☻☻☻☻☻☻

☻ =  GER

The odds of randomly selecting a GER speaker out of the total are $3/10$ (30 GER out of 100 workers) and half of the workers also speak ENG.

Therefore, the odds of a randomly selected worker speaking GER and not ENG are $1/2$(Fraction of workers that don't speak ENG) times 3/10(fraction of workers that speak GER).

$.15$ or $3/20$


Odds of winning at minesweeper with perfect play How would someone go about doing this? Assume that the first "click" will never be a bomb, and that the number of mines and the area are both known. Rather hoping there is a clever way to do this, but I will not be so surprised if there isn't.

EDIT:
I would assume (though without any real proof) that a program could be written that could solve minesweeper in linear time (as the board gets bigger linearly, if the mines/area ratio stays the same).

It would seem to me that in general no more than 9 blocks need to be considered (the high end of what i've see playing minesweeper at expert) to determine if

its a mine
its a safe square
the odds that its a mine

That would support my earlier assertion.

EDIT 2: This would also seem to contradict the fact that minesweeper is NP complete, and with probably not so much work one (maybe even I, but probably not) could write an algorithm that can play a perfect game of minesweeper that would have a linearly increasing runtime which would contradict (summery of) the paper here. So I guess this raises the next question which is: where is the flaw in my logic?

EDIT 3: I really am more interesting in the odds than in the algorithm to solve minesweeper. And it would be helpful to me if someone could explain why the number of checks/tests/calculations one has to do does not rise linearly with respect to area.

I am a very good minesweeper player, and I can say that perfect play can get you to win in $99\%$ of the easy ($8\times 8$ with $10 $mines) or intermediate ($16 \times 16$ with $40$ mines) levels. In the expert level ($16 \times 30$ with $99$ mines) it becomes harder to win without making any guesses.

About first click not being a mine, this is obvious, since the mine positions should be generated after your first click, and I think this is the case in the known minesweeper games.

Although, perfect play is not enough, if the distribution of mines is completely random. For instance, I encountered many times the following configuration in a corner of the board
$\begin{matrix} \square& X & M \\ M & M & M \end{matrix}$, where $\square$ is free square, $M$ is a marked mine, and $X$ is an unknown mine. Imagine this configuration in the upper left corner of the table, and the counter says that there is only $1$ mine left. You would have to guess and have only $50\%$ chances of winning, since there is no clue as to where the mine is.

About implementing an algorithm of solving minesweeper games with perfect play, there are some things you should consider, since some of the mines are not always obvious to find. I have in mind a few steps when I solve minesweeper games:

first mark the obvious mines;
open the safe squares;
look for some patterns learned before (e.g. $1-2-1$ or $1-2-2-1$);

If at one point, none of these steps can be applied, you can only guess the next step. Considering probabilities, is not very conclusive, since the values would be relatively close to $50\%$.

Considering an Expert board is randomly generated and your only "help" in winning is that the first click you are guaranteed not to lose (since the board is set only after your first click), there is no set odds of winning. I just played an Expert game that came down to 4 separate 50/50 guesses, giving me a 6.25% chance of winning after playing 8 minutes of "perfect strategy." I managed to get 1 right and failed on the next guess. Minesweeper does help keep your logical reasoning skills up for a while, but Expert games almost always come down to multiple, unavoidable coin flips.


Lebesgue measurability and a subgroup of $\mathbb{R}$ Let $G$ be a sugroup of $\mathbb{R}$ under addition and $G \neq \mathbb{R}$ and $G$ is Lebesgue measurable.Prove that $m(G)=0$.

One idea to solve this, is that if $m(G)>0$ then $G+G$(or $G-G$) contains an interval.Also because of the fact that $G$ is a group we have that $G+G \subseteq G$.

My idea is that i can construct a non measurable set and derive a contradiction,using the fact that $G \triangleleft \mathbb{R}$ nad use the quotient group $\mathbb{R} /G$.But i cannot achieve countability.le

Maybe i have to use a different approach.

Can someone give me some ideas to solve this(not the whole answer)?

Thank you in advance!

If one can show that $G$ contains an interval, then there is some interval $(a,b)\subseteq G$.  Let $m$ be the midpoint of the interval, since $m\in G$, $-m\in G$, so we can subtract $m$ and remain in $G$ to get that $G$ contains $(a-m,b-m)$ which is an interval centered at the origin, i.e., of the form $(-c,c)$.  Therefore, $(-c,c)\subseteq G$.  Now, if you consider $G+G$ (which is a subset of $G$), you can conclude that $(-2c,2c)\subseteq G$.  By continuing this, one gets that $(-2^nc,2^nc)\subseteq G$ for all $n$, so $G$ is all of $\mathbb{R}$.

The idea the group of $R$ must contain an interval in general it is not true since we can take the rational number is sub group of  $R$ ( filed)  but it doesn't contain any interval.


Is there a mathematical difference between currying and partial application? I know the following example doesn't make what I am saying rigorous, but hopefully it clarifies to some extent what I mean.

  For various computer implementations, dividing by 2 and multiplying by 0.5 require a different number of CPU cycles, even though the two operations are mathematically equivalent. (The first is performing the inverse operation of multiplication with the number 2, which is defined to be multiplication by the multiplicative inverse of 2 which is $\frac{1}{2}$ or 0.5 in decimal.)

Google "practical difference between currying and partial application" and at least the entire first page of results explains some of my skepticism about whether there is a mathematical difference between currying and partial application -- none of the results treat the subject mathematically, i.e. by defining them in terms of Hom functors, and instead discuss how currying and partial application are implemented differently in most functional programming languages.

(In a nod to other websites in the StackExchange network, I will post the results from them:

https://stackoverflow.com/questions/218025/what-is-the-difference-between-currying-and-partial-application

https://softwareengineering.stackexchange.com/questions/290131/what-is-the-difference-between-currying-and-partial-function-application-in-prac)

Notice that in both of the examples given above, they fail to explain any mathematical difference -- instead the difference is explained with examples of lines of code.

While in practice, as in the case when differentiating dividng by 2 and multiplying by 0.5, there is a difference in implementation, it does not seem to amount to a theoretical difference.

Currying and partial application are two different concepts, which impose different requirements on the category $\mathcal{C}$. Currying is a strictly stronger concept: one can express partial application using currying, but partial application is not sufficient for currying.

Currying (in cartesian closed category $\mathcal{C}\,$)

In order to talk about currying, the category has to be cartesian closed. In particular, it has to have exponential objects $Y^X$ for any two objects $X$ and $Y$. Currying is then one direction of the adjunction

$$ \mathcal{C}(X \times Y, Z) \cong \mathcal{C}(X, Z^Y)\qquad,$$

meaning that whenever we are given an $f: X \times Y \to Z$, we can obtain the curried version $\lambda f: X \to Z^Y$.

Partial application (in category with finite products)

For partial application, one needs only a category with all finite products (not necessarily cartesian closed). Since there is a terminal object $\mathbb{T}$ (nullary product), it makes sense to talk about elements of objects $X$: an element of $X$ can be defined as a morphism from $\mathbb{T} \to X$. Given an $f: X \times Y \to Z$, we can partially apply $f$ to $x$ as follows:
  $$ f \circ \langle x \circ \dagger_Y, Id_Y \rangle $$
where $\dagger_Y$ is the terminal morphism $Y \to \mathbb{T}$, the little circle $\circ$ denotes morphism composition, and $\langle - , - \rangle$ denotes the mediating morphism for products. If we now take any element of $Y$, that is, a morphism $y: \mathbb{T} \to Y$, and feed it to the above construction, we obtain:
$$ f \circ \langle x \circ \dagger_Y, Id_Y \rangle \circ y = f \circ \langle x \circ \dagger_Y \circ y, Id_Y \circ y \rangle = f \circ \langle x \circ Id_{\mathbb{T}}, y \rangle = f \circ \langle x, y \rangle \quad,$$
which is exactly what we would expect from a "partially applied function". Notice: we have built the morphism for "f partially applied to x" without using $\lambda$ or exponential objects. We don't need the full-blown cartesian closedness for that.

Expressing partial application through currying

Suppose that $\mathcal{C}$ is cartesian closed, let $f$, $x$, $y$ as above.
We can curry $f$ and apply it to $x$:
   $$ \lambda f \circ x \quad ,$$
obtaining an element of $Z^Y$. In order to be able to postcompose it with $y$, we have to "unpack" the element of $Z^Y$ using the evaluation morphism $\epsilon: Z^Y \times Y \to Z$. Thus, the morphism that represents "f partially applied to x" is:
  $$ \epsilon \circ \langle \lambda f \circ x \circ \dagger_Y, Id_Y\rangle \,.$$
Using the universal property of $\lambda f$ and $\epsilon$ (the commuting triangle that one sees in the definition of exponential objects), we can calculate:
  $$ \epsilon \circ \langle \lambda f \circ x \circ \dagger_Y, Id_Y\rangle =
     \epsilon \circ (\lambda f \times Id_Y) \circ \langle x \circ \dagger_Y, Id_Y \rangle = f \circ \langle x \circ \dagger_Y, Id_Y \rangle \,,$$
thus our notion of "partial application" defined directly coincides with the notion of "partial application" defined through currying. That means, it does not matter which defintion we take in cartesian closed categories.

When does the difference matter ...

...in programming:

Since categories built into functional programming languages (like $\mathrm{Hask}$) are cartesian closed, it does not really matter when creating simple functional programs.

However, as soon as we start to implement other categories inside our functional programming languages, the difference between "currying" and "partial application" begins to influence our design decisions. In general, it should be much easier to provide only finite products (which allow us to implement partial application), without providing currying. If we use those categories to represent certain computations, providing exponential objects and currying makes the framework more expressive, but we usually lose the possibility to inspect the structure of the computation before evaluating it. This is somewhat similar to trade-off between monads and applicatives: monads are more powerful, but applicatives have a more rigid structure, which can be analysed and optimized much better.

...in mathematics:

There are countless examples when one wants to have partial application, but does not want to think about whether the category one is working in is cartesian closed.

For example, in classical multivariate calculus, we look at smooth functions from $\mathbb{R}^n \to \mathbb{R}^m$, and often want to fix some of the variables, and compute partial derivatives. It is crucial that we do this using only the product structure and partial application, but no currying: if we would take a multivariate function $f: \mathbb{R}^2 \to \mathbb{R}$, and then curry it, we would obtain something like a function from $\mathbb{R}$ to the function space $\mathbb{R} \to \mathbb{R}$, and would thereby fall out of the realm of classical multivariate calculus, and land in the realm of functional analysis.

Same with topology: we often want to take some continuous function that depends on multiple variables (some homotopy or something), then fix some of the variables, and argue that the partially applied function is still continuous, and work with that. In basic topology, we do not want to spend lot of time on defining topologies on function spaces, and trying to make the category of topological spaces cartesian closed. We don't have to, because partial application works without currying.

Code examples

The fact that we do not need currying in order to implement partial application can be illustrated in any functional programming language.
The following code snippets show how to implement partialApplication in terms of compose, identity and product, without using lambdas.

Scala (with detailed comments):

// suppose that we are given only identity morphisms,
// composition, terminal morphisms, and products of morphisms
// (mediating morphism for product object)
// [BEGIN: assume that we can treat Scala functions as a "category with products"]
def identity[X]: X => X = x => x
def compose[X,Y,Z](g: Y => Z, f: X => Y): X => Z = x => g(f(x))
def terminal[X]: X => Unit = x => ()
def product[X, Y, Z](fst: X => Y, snd: X => Z): X => (Y, Z) = x => (fst(x), snd(x))
def constant[X](x: X): Unit => X = u => x
// [END: assume]

// We do not need any lambdas in order to express
// partial application!

// non-strict version (x is a morphism that takes unit)
def partialApplication[X, Y, Z](f: ((X, Y)) => Z, x: Unit => X): Y => Z =
  compose(f, product(compose(x, terminal[Y]), identity[Y]))

// strict version, uses const to transform `x` into constant function
def partialApplication[X, Y, Z](f: ((X, Y)) => Z, x: X): Y => Z =
  compose(f, product(compose(constant(x), terminal[Y]), identity[Y]))

// example
val f: ((Int, Double)) => Double = {
  case (n, t) => math.cos(2 * math.Pi * t)
}
val f42 = partialApplication(f, 42)
println(f42(1.0))

The same in Haskell (same comments as in Scala apply):

identity :: a -> a
identity x = x
compose :: (y -> z) -> (x -> y) -> (x -> z)
compose g f = g . f
prod :: (x -> y) -> (x -> z) -> (x -> (y, z))
prod a b = \x -> (a x, b x)
terminal :: a -> ()
terminal x = ()
constant :: x -> (() -> x)
constant x = \u -> x

partialApplication :: ((x, y) -> z) -> (() -> x) -> (y -> z)
partialApplication f x =
  compose f (prod (compose x terminal) identity)

As far as I can tell, the only real difference that is ever noted is that partial application combines two steps into one: (1) first it curries a function $f$ and then (2) evaluates the resulting curried function $\tilde{f}$ with a given value.

But then, at least in my unimportant opinion, that shouldn't be called "partial application", since one would expect such a term to be agnostic with respect to what the given value will be in advance. That should instead be "partial application at a given value". But if we partially apply a function of two arguments, for example, but don't specify in advance at what value the curried function will be evaluated, then isn't that just currying? I.e., if given $f(x,y)$ we return $\tilde{f}(z): f(x,y) \mapsto f(z,y)$, then aren't we returning a function with one argument which returns a function of one argument, rather than a function of one input argument which returns a real number, say like $f(2,y)$?

From this viewpoint it doesn't seem like currying and partial application are different at all, since mathematically if you "fix" a value without specifying what the exact value will be, you are just returning the curried function. In other words, I would describe currying as "fixing one of the  $n$ arguments of $f$ to return a new function $\tilde{f}$ which takes one argument as input and returns a function of $n-1$ arguments as output, and partial application as currying and then evaluating the curried function, e.g. $\tilde{f}(2)=f(2,y)$.

EDIT: According to Wikipedia, this is exactly the only difference.

See: https://en.wikipedia.org/wiki/Currying#Contrast_with_partial_function_application

  Partial application can bee seen as evaluating a curried function at a fixed point

Therefore I am just going to post this as community wiki, so it won't affect my reputation, but will hopefully show up in Google search results for anyone in the future who ends up asking a similar question.


What are some examples of notation that really improved mathematics? I've always felt that the concise, suggestive nature of the written language of mathematics is one of the reasons it can be so powerful.  Off the top of my head I can think of a few notational conventions that simplify problem statements and ideas, (these are all almost ubiquitous today):

$\binom{n}{k}$
$\left \lfloor x \right \rfloor$ and $\left \lceil x \right \rceil$
$\sum f(n)$
$\int f(x) dx$
$[P] = \begin{cases} 1 & \text{if } P \text{ is true;} \\ 0 & \text{otherwise} \end{cases}$

The last one being the Iverson Bracket.  A motivating example for the use of this notation can be found here.

What are some other examples of notation that really improved mathematics over the years?  Maybe also it is appropriate to ask what notational issues exist in mathematics today?

EDIT (11/7/13 4:35 PM): Just thought of this now, but the introduction of the Cartesian Coordinate System for plotting functions was a HUGE improvement!  I don't think this is outside the bounds of my original question and note that I am considering the actual graphical object here and not the use of $(x,y)$ to denote a point in the plane.

Vector notation!
The fact that you can write: $$\vec{a}\cdot\vec{b}$$ Instead of: $$a_1 b_1 + a_2 b_2 + \ldots$$
Or that you can use vector operators such as $\vec\nabla$, really helped develop linear algebra and its use in applied fields. To point out one famous example, Maxwell's original equations take up an entire page!

$\quad\quad\quad\quad\quad$

And here's the same using vector notation for comparison:

$$\boxed{\begin{align}\\\,\\\qquad&\qquad\nabla\cdot\mathbf D=\rho&&\text{(1)} \qquad\qquad\text{Gauss' law}&\\\,\\
\qquad&\qquad\nabla\cdot\mathbf B=0&&\text{(2)}\quad\text{Gauss' law for magnetism}&\\\,\\
\qquad&\,\,\,\nabla\times\mathbf E=-\dfrac{\partial\mathbf B}{\partial t}&&\text{(3)}\qquad\quad\,\text{Faraday's law}&\\\,\\
\qquad&\,\nabla\times\mathbf H=\dfrac{\partial\mathbf D}{\partial t}+\mathbf J&&\text{(4)}\qquad\text{Ampère-Maxwell law}\qquad\\\,\\
\end{align}}\\\,\\\textit{Maxwell's equations in  vector form}$$

(Image from IEEE's history page)

I love the exclamation mark for factorial

x! = 1 * 2 * .. * x

because of the way that bang indicates intensity in natural language. Seems quite fitting to its fast growth.


Prove that if $D^kf=0$ then $f$ is polynomial Let $E\subseteq\mathbb{R}^n$ an open connected set and let $f:E\rightarrow \mathbb{R}$ a $k+1$ times differentiable function such that $D^kf=0$. Show that $f$ is a polynomial in $x_1,...,x_n$ at a degree of at most $k$.

For any $\pi\in\{1,...,n\}^n$ we're given that $\frac{\partial^{k+1}f}{\partial x_{\pi (1)}...\partial x_{\pi (k+1)}}=0$. Thus $\frac{\partial^{k}f}{\partial x_{\pi (1)}...\partial x_{\pi(k)}}=c_k$ and by induction $f=\partial^0 f=x_{\pi(1)}(x_{\pi(2)}(...(x_{\pi(k)}c_k+c_{k-1})...)+c_1)+c_0$ for some $c_i\in\mathbb{R}, \forall 1\leq i \leq n$. I feel that this proof doesn't hold but I don't know where is the mistake and what is the right way. Thanks

Abridged solution. Proceed by induction on $k.$ The case $k = 0$ is very standard; consider a "base point" $a \in E$ and the set $\mathrm{X}$ of points in $x \in \mathrm{E}$ such that $f(x)=f(a).$ Then $\mathrm{X}$ is closed in $\mathrm{E}$ as $f$ is continuous. Now, $\mathrm{X}$ is also open in $\mathrm{E}$ for if $x \in \mathrm{X}$ then there is a small enough ball centred at $x$ contained in $\mathrm{E}$ and for such ball the mean value theorem (applied to the line segment between $x$ and another point in the ball) gives that $f$ is constant in such ball, so such ball is subset of $\mathrm{X}$ and then $\mathrm{X}$ is open as claimed. Being $\mathrm{E}$ connected, $\mathrm{X} = \mathrm{E}$ and the base case $k = 0$ is proved.

For the general case, assume $f$ satisfies $f^{(k+1)}=0$ on $\mathrm{E}.$ The case $k = 0$ shows that $f^{(k)}$ is a constant $k$-linear function on $\mathrm{E},$ call it $M.$ We show that $\varphi:x \mapsto f(x) - \dfrac{1}{k!}M(x, \ldots, x)$ satisfies $\varphi^{(k)}=0$ on $\mathrm{E}$ and hence the result. By standard results about derivatives of multilinear functions and the chain rule, we get at once that $\varphi^{(k)}=f^{(k)}-\dfrac{1}{k!} k!M = 0.$ Q.E.D.

Another alternative proof (which I guess is worng as well because I don't use the fact that $E$ is open and connected):

From Taylor's formula we know that
$$
\\  f(x_1,...,x_n)=\sum_{i_1=1}^\infty ...\sum_{i_n=1}^\infty \frac{(x_1-a_1)^{i_1}\cdot...\cdot(x_n-a_n)^{i_n}}{i_1!\cdot...\cdot i_n!}\cdot\frac{\partial ^{i_1+...+i_n}f}{\partial x_1^{i_1}...\partial x_n^{i_n}}(a_1,...,a_n) \
$$
Let $0\leq i_1,...,i_n\leq k$. If $\sum_{j=1}^n i_j>k$ then $\frac{\partial ^{i_1+...+i_n}f}{\partial x_1^{i_1}...\partial x_n^{i_n}}(a)=0$. Otherwise, $\frac{(x_1-a_1)^{i_1}\cdot...\cdot(x_n-a_n)^{i_n}}{i_1!\cdot...\cdot i_n!}$ is polynom of a degree at most $k$, and $\frac{\partial ^{i_1+...+i_n}f}{\partial x_1^{i_1}...\partial x_n^{i_n}}(a_1,...,a_n)$ is a constant linear tranformation. Thus $f(x_1,...,x_n)$ is a polynom of degree at most $k$.


How do you prove a radical is a circle on the Cartesian plane? I know something like $ \sqrt{x^{2}-1}$ could be a circle because that's just what I've always been told and it relates to the Pythagorean theorem. But, I looked at $ \sqrt{-(x-1)x}.$ I know that certainly looks like a circle, and it probably is, but how do I know for sure it's a circle in the Cartesian plane?

Please note that $y=\sqrt{x^2-1}$ is not a circle. It is a hyperbola because after squaring, you get $x^2-y^2=1$ which is the equation of a hyperbola. Plus, since $y$ has to be non-negative, it can't be all of the hyperbola given by that equation, but only the half of the two branches of the hyperbola that are in the upper half of the plane.

For the second equation, again, note that it can't be a complete circle because we know that $y$ has to be non-negative. By squaring, you get

$$y^2 = -x^2 -x = -(x+\frac{1}{2})^2+\frac{1}{4}$$

Therefore, $$(x+\frac{1}{2})^2+y^2=\frac{1}{4}$$

which is the equation of a circle centered at $(-\frac{1}{2},0)$ with radius $\frac{1}{2}$, but since $y$ has to be positive, it gives you only the upper half of the circle, not the rest.

One other way to see that it can't be a complete circle is to note that your equation defines $y$ as a function of $x$ and that means that its graph must pass the vertical line test. Therefore, it can't be a complete circle.

Please note that a circle is defined as the set of points that are at a fixed distance away from a point called the center. Knowing that distance in the plane for the two points $(x,y)$ and $(a,b)$ is defined as $\sqrt{(x-a)^2+(y-b)^2}$, the equation of a circle centered at $(a,b)$ with radius $R$ is then translated to:

$$\sqrt{(x-a)^2+(y-b)^2}=R$$
$$(x-a)^2+(y-b)^2=R^2$$

That's why $(x+\frac{1}{2})^2+y^2=\frac{1}{4}$ is in the form of the equation of a circle centered at $(-\frac{1}{2},0)$ with radius $\frac{1}{2}$, and the set of our points would've been a circle had we not have any restricting conditions on $y$. But since we do know that $y$ is non-negative, we have to restrict it to where $y$ is non-negative, which is the upper half of the plane.

$-x^2-x=\frac{1}{4}-(\frac{1}{4}+x+x^2)=\frac{1}{4}-(x+\frac{1}{2})^2$. Therefore, your equation is a circle with center $(-\frac{1}{2},0)$ and radius $\frac{1}{2}$


The base in which $1065 = 13 \cdot 54$ is true How can one find the base $r$ in which $1065 = 13 \cdot 54$ is a true statement? My attempt was constructing an equation including $r$ but because the left side has 4 digits, I got a polynomial of degree 3.

$$r^{3}-5r^{2}-13r-7=0$$
I could realize from the rational root theorem that $r=7$ is the solution, but I don't think that was what I expected to do.

Is there an alternative way to figure out $r$? Perhaps without "guessing" anything? Thanks in advance

Since $3\cdot4-5$ ends in a zero, your base divides $12-5=7$.  Therefore, your base is either $1$ or $7$.  Base $1$ is rather silly and usually avoided (see the comments for a discussion for how to interpret base $1$).

To give some details, if $r$ is the base, then, in base $10$, $13_r=1\cdot r+3$ and $54_r=5r+4$.  Therefore, by the distributive law,
$$
13_r\cdot54_r=(r+3)(5r+4)=5r^2+4r+15r+12\equiv12\pmod{r}.
$$
On the other hand, in base $10$,
$$
1065_r=1\cdot r^3+0\cdot r^2+6r+5\equiv 5\pmod{r}.
$$
Therefore, for equality, it must be that $5\equiv 12\pmod{r}.$.


If n balls are thrown into k bins, what is the probability that every bin gets at least one ball? If $n$ balls are thrown into $k$ bins (uniformly at random and independently), what is the probability that every bin gets at least one ball?  i.e. If we write $X$ for the number of empty bins, what is $P(X=0)$?

I was able to calculate the $E(X)$ and thus bound with Markov's inequality $P(X>=1) \le E(X)$ but I don't how to work out an exact answer.

http://www.inference.phy.cam.ac.uk/mackay/itprnn/ps/588.596.pdf

What is the chance  that all $k$ bins are occupied?

For $1\leq i\leq k$, define $A_i$ to be the event that the $i$th bin stays empty.
These are exchangeable events with $P(A_1\cdots A_j)=(1-{j\over k})^n$ and so
by inclusion-exclusion, the probability that there are no empty bins is
$$P(X=0)=\sum_{j=0}^k (-1)^j {k\choose j}\left(1-{j\over k}\right)^n.$$

Stirling numbers of the second kind can be used to give
an alternative solution to the occupancy
problem. We can fill all  $k$ bins as follows: partition the balls
$\{1,2,\dots, n\}$ into $k$ non-empty sets, then assign the bin values $1,2,\dots, k$ to
these sets. There are ${n\brace k}$ partitions, and for each partition $k!$ ways to
assign the bin values. Thus,
$$P(X=0)={{n\brace k}\,k!\over k^n}.$$

Set $X_i = 1$ if there is at least 1 ball in the $\textit{i}$th bin; and $0$ otherwise [$\textit{i}$ goes from 1 to k]. Then this question is asking what is $P[\sum\limits_{i=1}^k X_i = k]$. Given that each $X_i$s are independent of each other, $P[\sum\limits_{i=1}^k X_i=k]) = (P[X_i=1])^k =(1-P[X_i=0])^k =(1-(\frac{k-1}{k})^n)^k$


Simplify $\arccos\left(2\cos x\right)$. Let $x\in[\pi/3,2\pi/3]$. We know that $\arccos (\cos x)=x$ but what we can say about $\arccos\left(2\cos x\right)$?

Are there, for example, any "half-angle formula" also for inverse trigonometric functions?

As mentioned by others already, there is no nice simplification of $\cos^{-1}(2\cos(x))$, but the following similar-looking identities hold:

$\cos(2\cos^{-1}(x)) = 2x^2-1$ for any $x \in [-1,1]$.
$\cos(\frac12\cos^{-1}(x)) = \sqrt{\frac12(x+1)}$ for any $x \in [-1,1]$.

Both can be easily proven by the double-angle formula.

$\arccos(2\cos x)$ can not be defined for $x∈(0,π)$ it can only be defined for $x∈[π/3,2π/3]$. So I don't think it is useful to make calculation for a function with such a domain (Neither useful nor practical)


logic puzzle (truth-tellers / liars) I am trying to solve this puzzle. It is supposed to be solvable.

There are 3 people A,B,C. One of them is always lying, one of them is always telling the truth and one of them is always answering randomly. You can address one person at a time. They understand your language but they will always answer you in their own language with "da" or "ja" which mean yes and no but you don't know which is which. You must find who is who in 3 questions.
They know who is who amongst them, and if the truth teller or false teller doesnt know if he is lying or not (ie if you ask something that he cant possibly know so he can lie or say the truth) he answers randomly.

This is supposed to be solvable as I mentioned earlier. Any help would be appreciated.

This is famously called by George Boolos the "Hardest logic puzzle ever" (1996).

He, that, as far as I know, was the first to present and solve it, states the puzzle as follows:

  Three gods A, B, and C are called, in no particular order, True, False, and Random. True always speaks truly, False always speaks falsely, but whether Random speaks truly or falsely is a completely random matter. Your task is to determine the identities of A, B, and C by asking three yes-no questions; each question must be put to exactly one god. The gods understand English, but will answer all questions in their own language, in which the words for yes and no are da and ja, in some order. You do not know which word means which.

Some hints:

Since either A, B or C is the True, False and Random, we have $3!=6$ possible cases.

Now consider those three questions:

Is A the Random or B is the False? ($Q_1$)
Is B the Random or C is the False? ($Q_2$)
Is C the Random or A is the False? ($Q_3$)

You should ask them to no matter who, in the structure:

  If I asked you Q, would you answer me yes?

Note that the above indirect sentence contains two questions one inside another. The result is that even if the False lies in the innermost question, he will have to lie about lying in the out-most question and this amounts to the cancelling of the negations.

In a normal scenario, that is, where the foreign language problem is not put, then those three questions are enough to find out the identity of A, B, C.

To see this, simply construct a truth table of $2^3=8$ lines, corresponding to the possible combinations of answer to those questions:
$$\begin{array} {|c|}
\hline
Q_1 & Q_2 & Q_3 & Interpretation\\ \hline
\color{red}{1} & \color{red}{1} & \color{red}{1} & --\\ \hline
1 & 1 & 0 & \text{A: Random, B: True, C: False} \\ \hline
1 & 0 & 1 & \text{A: True, B: False, C: Random}\\ \hline
1 & 0 & 0 & \text{A: Random, B: False, C: True}\\ \hline
0 & 1 & 1 & \text{A: False, B: Random, C: True}\\ \hline
0 & 1 & 0  & \text{A: True, B: Random, C: False}\\ \hline
0 & 0 & 1 & \text{A: False, B: True, C: Random}\\ \hline
\color{red}{0} & \color{red}{0} & \color{red}{0} & -- \\ \hline
\end{array}$$

Where:

$$Q_1 \equiv Random(A) \vee False(B)$$
$$Q_2 \equiv Random(B) \vee False(C)$$
$$Q_3 \equiv Random(C) \vee False(A)$$

Indeed, since there is one and only one True, False and Random, it is easy to see that the first and last lines of the table are impossible (show it!). Hence, of of it amounts to, again, 6 possible interpretations, where in every case we can find the True, False and Random's identity.

The above solution assumes we know the language of the True, False and Random.

Now, in its own language variation, we can try to overcome the fact we don't know what ja or da means by considering the following question, as firstly observed by Roberts (2001):

  If I asked you Q, would you answer me ja?

In fact, this answer to this question will come out as ja if the truthful answer to Q is yes, and as da if the truthful answer to Q is no. Now, what about applying this reasoning with the above solution?

The best answer is here:

Rabern & Rabern (2008), A simple solution to the hardest logic puzzle ever, Analysis 68(298).


Prove that $f$ is continuous in $\mathbb{R}$ Let  $f:\mathbb{R} \to \mathbb{R}$ be a monotone function such that $f(V)$ is open for every open set $V\subset \mathbb{R}$. Prove that $f$ is continuous in $\mathbb{R}$.

Any hint for proving this I will appreciate.

Without loss of generality we may assume that $f$ is non decreasing.

First, note that $f$ is injective. Suppose $f(x_1) = f(x_2)$, then if $x_1<x_2$,
the set $(x_1,x_2)$ would be open and the set $f((x_1,x_2)) = \{ f(x_1) \}$ is not open which is a contradiction. Hence $f$ is injective.

Now I claim that $f(\mathbb{R})$ is connected (that is, an interval). Suppose that there is some $y_0 \in (\inf f, \sup f) \setminus f(\mathbb{R})$ (note
that $f(\mathbb{R}) \subset (\inf f, \sup f)$).

Then $I_-=f^{-1}((-\infty, y_0)), I_+f^{-1}((y_0, \infty))$ form a partition of $\mathbb{R}$ into two intervals.

There are two possibilities: (i) $I_- $ has the form $I_- = (-\infty, x_0]$, with
$f(x_0) < y_0$  and $y_0 < f(x) $ for all $x > x_0$. In particular,
$f(x_0) \in f(\mathbb{R})$ but $(f(x_0), y_0) \cap f(\mathbb{R}) = \emptyset$,
and hence $f(\mathbb{R})$ is not open.
(ii) $I_+ $ has the form $I_+ = [x_0, \infty)$, and a similar analysis
leads to the same contradiction.

Hence $f(\mathbb{R}) = (\inf f, \sup f)$,
in particular, it is an interval.

Now pick $x_0$, $\epsilon >0$. Choose $0< \epsilon' \le \epsilon$ such that
$\inf f < f(x_0) - \epsilon'$ and $f(x_0)+ \epsilon' < \sup f$. Now
choose $x_-, x_+ $ such that $f(x_-) = f(x_0) - \epsilon'$ and
$f(x_+) = f(x_0) + \epsilon'$.
Note that $x_- < x_0 < x_+$ and let $\delta = \min (x_0-x_-, x_+-x_0)$.
Then if $|x-x_0| < \delta$, we have $|f(x)-f(x_0)| < \epsilon' \le \epsilon$.

Perhaps the contrapositive:

If $f$ is monotonic and not continuous on $\mathbb{R}$, then $\exists$ an open $V \subset \mathbb{R}$ such that $f(V)$ is not open.

EDIT:

proof:

Suppose $f: \mathbb{R} \to \mathbb{R}$ is monotonic and not continuous at $p \in \mathbb{R}$.  Let $A = (p-\delta, p+\delta)$ for some $\delta \gt 0$. A is open in $\mathbb{R}$.  The domain of $f$ is $\mathbb{R}$. Then $\forall x \in A, f(x) \in f(A)\subset\mathbb{R}$.

Let $s = sup \{f(x)\in f(A)|x \in (p-\delta, p)\}$ and $t = inf\{f(x)\in f(A)|x \in (p, p+\delta)\}$

Without loss of generality, suppose $f$ is non-decreasing.  if $s = t, f(p) \not= s = t$, otherwise $f$ would be continuous at p.  This would imply, for some  $x_1 \lt p \lt x_2$, $f(p) \gt f(x_1) \land f(p) \gt f(x_2)$ or $f(p) \lt f(x_1) \land f(p) \lt f(x_2)$. Then $f$ is not monotonic.  Hence $s \not = t$.

Then $s \lt t$ and $s \le f(p) \lt t \lor s \lt f(p) \le t$.

Assume without loss of generality that $s \lt f(p) \le t$.  Then $s \lt f(p)$, which implies that $\exists y \in \mathbb{R}$, such that $s \lt y \lt f(p)$.  But $\forall x \in (p-\delta,p), f(x) \le s$.  Thus $y \notin f((p-\delta,p])$. Therefore $\forall \alpha$ such that $0 \lt \alpha \le f(p) - y,$ $\exists y^{'} \in \mathbb{R}$ such that $y^{'} \notin f[(p-\alpha,p+\alpha)]\subset f(A)$.

Thus $f(p) \in f(A)$ cannot be an interior point of $f(A)\subset \mathbb{R}$, which implies that $f(A)$ is not open.  Because $p \in A$ and $A$ is open, $A$ is an open subset of $\mathbb{R}$ such that $f(A)$ is not open.


Counterexample to: "If a function is continuous in a point $x_0$ then it is defined in a neighborhood of that point." I'm looking for a counterexample to the statement:

  If a function is continuous in a point $x_0$ then it is defined in a
  neighborhood of that point.

If I take

$f: \mathbb{R} \setminus \mathbb{Q} \rightarrow \mathbb{R}$ defined as $f(x):=x$

Then it is continuous in any point $x_0 \in \mathbb{R}\setminus \mathbb{Q}$ since $lim_{x\to x_0} f(x)=x_0 \forall x_0 \in \mathbb{R}$, nevertheless it is not defined in any neighbourhood of any point $x_0 \in \mathbb{R}$.

Is this counterexample correct?

Edit

Definition of neighborhood of $x_0$: Let $x_0 \in \mathbb{R}$ and $r>0$ a real number. We call a neighborhood of $x_0$ with radius $r$ the open and bounded interval in $\mathbb{R}$ $(x_0-r,x_0+r)$.

Definition of limit: Let $x_0$ be a limit point for a function $f(x)$. We write $lim_{x \to x_0} f(x)=l \iff \forall \epsilon >0 \exists \delta >0 \mid \forall x\in domf \wedge 0<\mid x-x_0 \mid <\delta \implies \mid f(x)-l\mid <\epsilon$

Definition of continuity of $f$ in $x_0$: $f$ is continuous in a point $x_0$ of its domain iff $\forall \epsilon >0 \exists \delta>0 \mid \forall x \in dom(f) \wedge \mid x-x_0 \mid < \delta \implies \mid f(x)-f(x_0)\mid <\epsilon$

Assuming you are using the following definition your counterexample is correct. You could take $f(x)=0$ for all $x\in \mathbb{Q}$ for an even simpler example with a dense domain.

Let $x \in X$. If $f:X \to Y$ then we say that $f$ is continuous at $x$ if for all $\varepsilon>0$ exists $\delta>0$ such that

$$d_Y(f(x),f(y)) < \varepsilon$$

whenever
$$y\in X, ~ d_X(x,y) < \delta$$

Note that this definition is equivalent to yours when $x$ is a limit point of the domain.

No, this is not a valid counterexample because $f$ is not continuous anywhere which I will show as follows.

$$\lim_{x\to x_0}f(x)=f(x) \iff \forall \varepsilon > 0, \exists \delta >0: [x \in V_\delta(x_0) \implies f(x) \in V_\varepsilon(x_0)]$$

(where $V_a(b)$ is the open $a$-neighbourhood of $b$)

For any $\varepsilon,\delta>0$, I can obviously pick for $x$ any rational $q \in V_\delta(x_0)$, in which case $f(x)$ will not be defined and thus cannot be an element of $V_\varepsilon(x_0)$.


Orbits of $GL(m,k)\times GL(n,k)$ on the set $M(m\times n, k)$ I am interested in the orbits of $GL(m,k)\times GL(n,k)$ on the set $M(m\times n, k)$, where $(g_1,g_2)\circ x:=g_1 xg_2^{-1}$.

I know that left-multiplication with a regular matrix is basically a row-transformation and right-multiplication a column-transformation. So in total the rank stays the same under this group operation.
But is it true that for matrices $x\neq y$ of the same rank there always exist column- and row-transformations such that $g_1 xg_2^{-1}=y$?

Here is a direct solution using the Singular Value Decomposition (SVD)

(see for example (https://davetang.org/file/Singular_Value_Decomposition_Tutorial.pdf))

The principle of this decomposition is that every $m \times n$ matrix $M$ can be written in the following way:

$$\tag{1}M = U \Sigma V^{-1} \ \ \ \ \ (\text{=} \ U \Sigma V^{T}) $$

where

$U \in O(m,\mathbb{R}) \subset GL(m,\mathbb{R})$,
$V \in O(n,\mathbb{R}) \subset GL(n,\mathbb{R})$,  and
$\Sigma$ is $m \times n$, with "diagonal elements":

$$\Sigma_{11} \geq \Sigma_{22} \geq \cdots \Sigma_p \geq 0 \ \ \text{where} \ \  p=\min(m,n),$$

all other values $\Sigma_{ij}$ for $i \neq j$ being zero (a kind of generalized diagonal matrix). Here is an example with $m=4$ and $n=3$:

$$\underbrace{\begin{pmatrix}7   &  1  &   1\\
    1  &  -1 &  -7\\
    7 &   -1 &    1\\
    1  &   1  & -7\end{pmatrix}}_{M}=
\underbrace{\begin{pmatrix}   -0.7  &  0.1 &  0.5  & -0.5\\
   -0.1  & -0.7 &  -0.5  & -0.5\\
   -0.7  &  0.1  & -0.5  &  0.5\\
   -0.1  & -0.7  & 0.5  &  0.5\end{pmatrix}}_{U}
\underbrace{\begin{pmatrix} 10 &         0     &    0 \\
         0  & 10 &  0\\
         0  &    0 & 2\\
         0   &    0  &     0\end{pmatrix}}_{\Sigma}\underbrace{\begin{pmatrix}    -1  &   0  &   0\\
     0  &   0   &  1\\
     0  &   1   &  0\end{pmatrix}}_{V^{-1}}$$

Let consider now the particular case $m=3$ and $n=2$.

The orbits can thus be ''represented'' by matrices of the form:

$$\Sigma=\begin{pmatrix}a&0\\0&b\\0&0\end{pmatrix}$$

but in fact, if $a$ and/or  $b$ are nonzero coefficients, they can be "incorporated" into matrix $U$ ; thus, the "representatives" can be taken as:

$$\Sigma_0=\begin{pmatrix}0&0\\0&0\\0&0\end{pmatrix}, \ \ \Sigma_1=\begin{pmatrix}1&0\\0&0\\0&0\end{pmatrix}, \ \
\Sigma_2=\begin{pmatrix}1&0\\0&1\\0&0\end{pmatrix}$$

(resp. null matrix, rank 1 matrices, and rank 2 matrices).

From here, the general case is straightforward.


Convergence of the series Find the convergence and absolute convergence of the series $∑\frac{(-1)^{n+1} n}{1+n^2}$

For Absolute convergence, I found out by comparison tests, it is not absolutely convergent. But I couldnt find for convergence. Please help me

Let $a_n=\frac{(-1)^n n}{1+n^2}$. Then $a_{2n}\geq 0$ and $a_{2n-1}\leq 0$ for all $n$.

$|a_n|=\frac{ n}{1+n^2}$. Need to show $|a_n|\geq |a_{n+1}|$ for all $n$. i. e. need $\frac{ n}{1+n^2}\geq \frac{ n+1}{1+(1+n)^2}$. Here, backward method will work. So starting from $n^2+n\geq 1$, you can show it.

Now, $\lim_{n\rightarrow \infty} a_n=0$. So by alternating series test(Leibnitz test) the series converges.


If a fair die is rolled 3 times, what are the odds of getting an even number on each of the first 2 rolls, and an odd number on the third roll? If a fair die is rolled 3 times, what are the odds of getting an even number on each of the first 2 rolls, and an odd number on the third roll?

I think the permutations formula is needed i.e. $n!/(n-r)!$ because order matters but I'm not sure if n is 3 or 6 and what would r be?

Any help would be much appreciated!

Let's calculate the probability, then convert that to odds.

On a fair die, half the numbers are even and half the numbers are odd. So, the probability for a single roll of getting an even number or an odd number is $\dfrac{1}{2}$.

The probability for a specific roll are unaffected by previous rolls, so we can apply the product principle and multiply probabilities for each roll. Each roll has probability of $\dfrac 1 2$ of obtaining the desired result. So, we have:

$$P(E,E,O) = \dfrac 1 2 \cdot \dfrac 1 2 \cdot \dfrac 1 2 = \dfrac 1 8$$

Now, the probability of that not happening is $$1-\dfrac 1 8 = \dfrac 7 8$$

So, the odds are 7:1 against the desired outcome.

All of the above or below :-} answers are correct that is 50:50 for each throw so your "formula" is

  n
(2 -1):1 against where n is the number of throws


Prove these subrings are ideals Let $R = R_1⊕R_2$ and let $I_1$ and $I_2$ be ideals of rings $R_1$ and $R_2$ respectively. We consider $R_1$ and $R_2$ as subrings of R under canonical embeddings $R_i \rightarrow R$ and thus $I_1$ and $I_2$ are considered as subrings of R.

(a) Prove that $I_1$ and $I_2$ are ideals of $R$.
(b) If each $I_i$ is a maximal ideal of $R_i$, is the ideal $I_1 +I_2$ of $R$ maximal? Prime?

I honestly don't know how to approach either.  I think for a I want to show it just by using the definition, but that seems kind of tedious.  Is there a more efficient way?
For b, I know that an ideal I is maximal if and only if R/I is a field and an ideal P is prime if and only if R/P is an integral domain

The set $I_1\oplus 0$ is an ideal since if $x\oplus y$ is any element of $R$ and $a\oplus 0$ any element of the set, then
$$(x\oplus y)(a\oplus 0)=xa\oplus 0\in I_1\oplus 0$$

However $I_1\oplus I_2$ need not be prime for example let $R_1=R_2=\mathbb{Z}$ and
$$(2\oplus 1)(1\oplus 3)\in (2)\oplus (3)$$ but $2\oplus 1\not \in (2)\oplus 0$ and similarly.

For part (b), the sum of two prime ideals is not necessarily prime. Consider $\mathbb{Z}$ with the ideals $\langle 2 \rangle, \langle 3 \rangle$.

As for maximality, is $I_1 + I_2$ an ideal? $I_1$ and $I_2$ are maximal, so what can you say about $I_1 + I_2$?


Is topology invariant under conformal transformation? Can conformal transformation change the topology of a manifold?
In other words, if two manifolds are conformal, should they have the same topology?

If $(M,g)$ and $(N,h)$ are conformally equivalent we have a diffeomorphism $f:M\rightarrow N$ such that $f^*h$ is conformally equivalent to $g$. Since $f$ is a diffeomorphism the topology on $M$ and $N$ are the same (they have the same open sets).

I don't think they should have the same topology, but if $B_1$ is a topology base of your first manifold $S_1$, and $f$ is a conformal transformation such as $f(S_1) = S_2$ (where $S_2$ is your second manifold), my guess is that $B_2 = f(B_1)$ is a topology base of $B_1$... but I'm not sure of it


Prove that $A^k = 0 $ iff $A^2 = 0$ Let $A$ be a $  2 \times  2 $ matrix and  a positive integer $k \geq 2$. Prove that $A^k = 0 $ iff $A^2 = 0$.

I can make it to do this exercise if I have $ \det (A^k) = (\det A)^k $. But this question comes before this.

Thank you very much for your help!

The solution using the minimal polynomial and Cayley-Hamilton is a bit of an over-kill and somewhat of a magic solution. I prefer the following non-magic solution (for the non-trivial implication, and there is no need to assume the matrix is $2\times 2$).

Think of $A$ as a linear operator $A: V \to V$ with $V$ an $n$-dimensional vector space, and suppose $A^t=0$. We'll show that $A^n=0$. Now, for each $m\ge 1$ consider the space $K_m$, the kernel of $A^m$. It is immediate that $K_{m}\subseteq K_{m+1}$. Since $A^t=0$ it follows that $K_t=V$. It is also immediate that if $K_m=K_{m+1}$, then $K_m=K_{r}$ for all $r>m$.

Thus, the sequence of kernels is an increasing sequence of subspaces, that stabilizes as soon as the next step is equal to the previous one, and it is eventually all of $V$. Now we will use the fact that $V$ is of dimension $n$. Considering the dimensions of the kernels, the above implies that the sequence of dimensions is strictly increasing until it stabilizes. Since it reaches $V$, the dimensions reach $n$ and since the dimension of $K_1$ is not zero, the sequence of dimensions starts at $1$ or more. That means it has to stabilize after no more than $n$ steps, and thus $K_t=V$ for some $t\le n$. But then $K_n=V$, and thus $A^n=0$.

A^k=0 iff A^2=0  k positive interger >2 you can write k=2+n and n>=1. Using induction: A^2+n=0
Then A^2=0. Take A^3=A^2A=0. And  prove  A^2+n+1=0 =>A^2=0  and A^2+nA=0.  Then A^2=0.


Roots of quadratic equation $ax^2+bx+c$ at infinity Our maths teacher taught us that conditions for roots of a quadratic equation $ax^2+bx+c=0$ to lie at infinity are :

A) (for exactly one root at infinity) $a=0$, $b$ = non-zero, & $c$ can be zero or non-zero

B) (for both roots at infinity) $a = 0$, $b= 0$, and $c$ = non-zero

By equal to zero, I mean the coefficients are so small that they are nearly zero, i.e approaching zero. They can’t be exactly equal to zero right? Or it would no longer be a quadratic. Please correct me if I’m wrong.

This is not my question though. What I want to ask is, are these the only two conditions for roots of a quadratic equation to lie at infinity?

For example, if we consider the equations

1) $x^2-4\times 10^{30} = 0$

2) $0.00000001x^2+ 0.0000005x + 6 = 0$

I have attached two screenshots. In the first equation, both roots are at infinity even though coefficient of $x^2$ is non-zero (contrary to what condition (A) says.

In the second equation, none of the roots are lying at infinity even thought coefficients of $x^2$ and $x$ are nearly equal to zero, which again is not consistent with what condition  (B) says.

How do I explain these discrepancies?

(Looks like I can’t attach more than one pic here. So I’m attaching just one, graph of equation (2). It’s clear that both roots of equation (1) are at infinity.)

In a sense you are re-running the arguments that mathematicians had in the time of Newton and Leibniz, about what you can do with infinitely large and infinitely small quantities.

The conclusion was this:

You can’t put infinity as such into your formulae and expect them to make sense. You can’t put division by zero into your formulae and expect them to make sense.

So we have come up with some firm rules and ways of handling the question.

First of all, though, infinity is a specific concept and not a way of saying “very large”. You must get that clear because otherwise you will find yourself using a language different from everyone else. “Does not fit on any reasonable sheet of paper” is not the same as “Infinite”. The roots of your first equation are not infinite. $2×10^{15}$ is not infinite.

Once you have stopped using “infinity” in a different way from everyone else, you have actually pointed out a very interesting fact.

Basically, what you have found, to take your $A=0$ case, is that the closer to $0$ $A$ becomes, the closer to infinity one of the roots of the quadratic becomes. Intuitively, it is being squeezed out of the real world until when $A=0$ exactly, it has completely gone. Algebraically, you can look at the traditional solution of a quadratic and see how, the smaller that $a$ becomes, one solution (the $(-b-\sqrt{b^2-4ac})/2a$ one) runs off towards infinity because the denominator tends to zero while the numerator doesn’t, while in the other solution (the $(-b+\sqrt{b^2-4ac})/2a$ one) the numerator is also tending to be zero, and although $0/0$ could mean anything or everything or nothing, that is actually a hopeful sign. Some subtle algebraic manoeuvring will eventually reveal that that root tends to $-c/b$.

People run the process backwards as well. Given $Ax^2+Bx+C=0$, where $A$ is a very complicated formula but is also very small, one solves $Bx+C=0$ first and then looks into how that solution is perturbed by the presence of a non-zero $A$. Many advanced calculations in theoretical physics work in this way.

So to conclude:

Don’t say “infinite” to mean “very large” or “zero” to mean “very small”.
Do think about what happens when something tends to infinity or tends to zero.
And congratulations on thinking of this question in the first place. One of the best ways to mathematical discovery is to say to yourself “What a beautiful equation! I wonder how I can break it”.

I'm sorry to say that your maths teacher is talking nonsense.

If $a=0$ and $b\ne 0$, the equation $bx+c=0$ has the single root $x=-c/b$.
If $a=b=0$, the "equation" $c=0$ is simply true or false, according as $c$ is zero or non-zero. It doesn't have any "roots".

I can only hope that you are simply misremembering what you were told. The roots of a quadratic equation can never "lie at infinity".

By the way, your comment about "so small that they are nearly zero" is also nonsense. Either they are zero or they are not.


Inequality with positive reals For each $x,y,z>0$, define
$$
f(x,y,z)=\frac{3x^2(y+z)+2xyz}{(x+y)(y+z)(z+x)}.
$$

Fix also $a,b,c,d>0$ with $a\le b$. How can we prove "reasonably" the following inequality?
$$
\frac{f(a,b,d)+f(c,b,d)}{f(b,a,d)+f(c,a,d)} \le \frac{a+c}{b+c}.
$$

The expression
$$\frac{f(a,b,d)+f(c,b,d)}{f(b,a,d)+f(c,a,d)}
 \le \frac{a+c}{b+c}$$ is equivalent to
$$(b+c)(f(a,b,d)+f(c,b,d))\le(a+c)(f(b,a,d)+f(c,a,d))$$ The $LHS$ is equal to
$$\frac{3a^2(b+c)}{(a+b)(a+d)}+\frac{2adb^2+2abcd}{(a+b)(a+d)(b+d)}+\frac{3c^2}{c+d}+\frac{2bcd}{(b+d)(c+d)}$$
The $RHS$ is equal to
$$\frac{3b^2(a+c)}{(a+b)(b+d)}+\frac{2bda^2+2abcd}{(a+b)(a+d)(b+d)}+\frac{3c^2}{c+d}+\frac{2acd}{(a+d)(c+d)}$$ so we have to prove that
$$\frac{3a^2(b+c)}{(a+b)(a+d)}+\frac{2adb^2}{(a+b)(a+d)(b+d)}+\frac{2bcd}{(b+d)(c+d)}$$ is less or equal than
$$\frac{3b^2(a+c)}{(a+b)(b+d)}+\frac{2bda^2}{(a+b)(a+d)(b+d)}+\frac{2acd}{(a+d)(c+d)}$$ Multiplying by $(a+b)(a+d)(b+d)(c+d)$ one has respectively
$$(b+d)(c+d)(b+c)(3a^2)+(c+d)(2adb^2)+(a+b)(a+d)(2bcd)$$ and
$$(a+d)(a+c)(c+d)(3b^2)+(c+d)(2bda^2)+(a+b)(b+d)(2acd)$$

What precedes becomes after simplification $$a^2(b+c)(b+d)\le b^2(a+c)(a+d)$$ where $a,b,c,d\gt 0$ and $a\le b$.

Finally we have the evident inequality
$$\color{red}{0\le ab(b-a)(c+d)+(b^2-a^2)cd}$$


Evaluating $ \lim_{n\to\infty}\sum_{i=0}^n\frac{n}{n^2+i^2} $ A few days ago I confronted with this limit question, and all my attempts to solve it failed somehow.
$$
\lim_{n\to\infty}\sum_{i=0}^n\frac{1}{n^2+i^2}
$$
I feel I should relate it to $\arctan(n)$. I tried integration and derivation to reach to my point, it went well at first but eventually nowhere.
Any advice or idea would be appreciated.

$$
\sum_{i=0}^n\frac{1}{n^2+i^2} = \frac 1 n \sum_{i=0}^n \frac 1 {1 + \left( \frac i n \right)^2} \cdot \frac 1 n = \frac 1 n \sum_{i=0}^n \arctan'\left( \tfrac i n \right)\cdot \tfrac 1 n
$$
That last sum is a Riemann sum that approximates $\displaystyle\int_0^1 \arctan'(x)\,dx = \arctan 1 - \arctan 0 = \frac \pi 4,$ but outside the sum there is a factor of $1/n$ that approaches $0,$ so ultimately the limit is $0.$

$$
\begin{aligned}
\sum_{0\le i\le n}{1\over n^2+i^2}
&={1\over n^2}+\sum_{1\le i\le n}{1\over n^2+i^2} \\
&={1\over n^2}+\int_{1-\varepsilon}^n \,{\mathrm d\lfloor t\rfloor\over n^2+t^2} \\
&={1\over n^2}+\left.{\lfloor t\rfloor\over n^2+t^2}\right|_{1-\varepsilon}^n-\int_1^n[t-\{t\}] \, \mathrm d\left(1\over n^2+t^2\right) \\
&={1\over n^2}+{1\over2n}-\int_1^nt \, \mathrm d\left(1\over n^2+t^2\right)+\int_1^n\{t\} \, \mathrm d\left(1\over n^2+t^2\right) \\
\end{aligned}
$$
wherein the integrals are evaluated as follows
$$
\begin{aligned}
\int_1^nt \, \mathrm d\left(1\over n^2+t^2\right)
&=\left.{t\over n^2+t^2}\right|_1^n-\int_1^n \, {\mathrm dt\over n^2+t^2} \\
&={1\over2n}-{1\over n^2+1}-\left.{1\over n}\arctan\left(\frac tn\right)\right|_1^n \\
&={1\over2n}-{1\over n^2+1}-{\pi\over4n}+\frac1n\arctan\left(\frac1n\right) \\
&=\mathcal O\left(\frac1n\right)
\end{aligned}
$$
$$
\int_1^n\{t\} \, \mathrm d\left(1\over n^2+t^2\right)=\mathcal O\left(\frac1n\right)
$$
As a result, we have
$$
\sum_{0\le i\le n}{1\over n^2+i^2}=\mathcal O \left(\frac1n\right)
$$


That looks correct. It becomes a tiny bit simpler if you expand the fraction with $e^x$ instead of expanding with $e^{-x}$:
$$
\frac{1+e^x}{1+e^{-x}}  = \frac{1+e^x}{1+e^{-x}} \frac{e^x}{e^x} = \frac{1+e^x} {e^x+1} e^x = e^x
$$

This looks correct to me but remember that "$\log$" is typically an abbreviation for "$\log_{10}$" in certain contexts such as physics and engineering. It's pretty clear that you are referring to the natural logarithm here, but if you would like there to be zero ambiguity as what you are referring to, then I suggest using the more widespread notation for the natural log given by "$\ln$."


concurrency of angle bisectors, medians, perpendicular bisectors, altitudes When you think of a triangle the basic constructions you think of are perpendicular bisectors of the sides, angle bisectors, altitudes, and medians. Now if you were someone who just started learning about these concepts it would seem very unusual that all these basic constructions related to the triangle are concurrent. To me that seems really unlikely and I ask, is there some deep lying meaning or reason as to why this is so? Any help would be appreciated.

NOTE : I am not asking for the proofs of the concurrency theorems

The concept of concurrencies of various centers of a triangle can be appreciated from 3 different levels.

At the junior level, the said constructions are mainly done by the help of RULERs and PROTRACTORs. The point is to arouse interest and see if lines can be accurately or not.

At the intermediate level, lines should be drawn using STRAIGHT-EDGE and COMPASSES only. Techniques required are preliminary knowledge of congruency. Thus, the learning tasks are scheduled as studying congruency first and the topic of constructions right after it. The construction exercise can therefore also be considered as a consolidation of the topics on congruency.

At the advanced level, the truth of the concurrencies should be proved formally.


What is the remainder when $6^{99} + 8^{99}$ is divided by $49$? How can we solve this using the Binomial Theorem? I tried it as $(7-1)^{99} + (7+1)^{99}$ divided by $49$. I am getting stuck after this. Please help.

Hint: Expand each of $(7-1)^{99}$ and $(7+1)^{99}$ using the Binomial Theorem. Note which factors are divisible by $49 = 7^2$, and see what's left over.

how about using modular-exponentiation ?
It's very simple and I think it could help you...

int b_power_of_n_mod_m(int b, int n, int m) {
    //convert n to binary and save in A array
    int A[] = new int[20];
    int i = 0;
    // n must be positive
    while (n > 0) {
        A[i++] = n & 1;
        n = n >> 1;
    }
    int x = 1;
    int power = b % m;
    for (int j = 0; j < i; j++) {
        if (A[j] == 1) x = (x * power) % m;
        power = (power * power) % m;
    }
    return x;
}

just use it and you can see that it is very fast :)


Is it easier to calculate a factorial or the inverse of a factorial (1/n!) for extremely large n? I need to calculate extremely large factorials but they grow extremely fast! I was wondering if it might be easier to calculate $\frac{1}{n!}$ rather than n! itself because as n goes to infinity $\frac{1}{n!}$ goes to zero, which I think may be easier to calculate. Is it easier to calculate the factorial or the inverse of it?

The period of the repeating decimal of $\frac{1}{n!}$ can be very large, closer to $n!$ than to the number of digits in $n!$.  For example, $\frac{1}{19!}$ has period $81648$ even though there are only $18$ digits in $19!$ itself.  And we would need to calculate $36$ digits of $\frac{1}{19!}$ to see that it differs from $\frac{1}{19!+1}$.  So this doesn't seem like a very promising strategy.

why do you want to calculate n! in the first place?
most times you will devide it by other factorials and then it is better to wait with calculating till the formula contains as least factors as  possible
for example if you have to calculate n! / ((n-k!) X k!) don't start with calculating n!


Prime numbers of the form $(1\times11\times111\times1111\times...)-(1+11+111+1111+...)$ Let
$$R(1) = 1-1,$$
$$R(2) = (1\times11) - (1+11),$$
$$R(3) = (1\times11\times111) - (1+11+111),$$ and so on...

$$R(4)=1355297\quad\text{(a prime number!)}$$

$R(4)$ is the only prime I found of such form up to $R(200)$. Are there anymore primes of such form?

Probabilistic answer: If $\phi_n\in {\Bbb N}$ is a sequence of 'random' integers going to infinity then the probability of $\phi_n$ being a prime is $\sim 1/\ln \phi_n$. When $\sum_{n\geq N} \frac{1}{\ln \phi_n}$ goes to zero fast with $N$ it is most 'likely' that there are no primes among these numbers for $N$ large. The words 'random' and 'likely' are obviously subject to interpretations.

I wanted to post this as a comment but it was too long.

I don't know if it will help but consider to rewrite your function in this way:

$$R(n) = (1 \times 11 \times 111 \times \ldots) - (1 + 11 + 111 + 1111 + \ldots)$$

$$R(n) = \left((10^0) \times (10^0 + 10^1) \times (10^0 + 10^1 + 10^2) \times \ldots\right) - \left((10^0) + (10^1 + 10^0) + (10^2 + 10^1 + 10^0) + \ldots \right)$$

Now with a bit of maths you can check that

$$\left((10^0) \times (10^0 + 10^1) \times (10^0 + 10^1 + 10^2) \times \ldots\right) = \prod_{k = 1}^n \frac{10^k - 1}{9}$$

For what concerns the second part, it's more amusing. Indeed we have

$$\left((10^0) + (10^1 + 10^0) + (10^2 + 10^1 + 10^0) + \ldots \right)$$

But as we tend to $n$, we can easily check that there will be $n$ terms of $10^0$, $n-1$ terms of $10^1$, $n-2$ terms of $10^2$ and so on, which means

$$\left((10^0) + (10^1 + 10^0) + (10^2 + 10^1 + 10^0) + \ldots \right) = \sum_{k = 0}^{n-1} (n-k)10^k$$

Now with a bit of help (mathematica rules), the productory gives

$$\prod_{k = 1}^n \frac{10^k - 1}{9} = 9^{-k} {Q}_p (10, 10, k)$$

Where ${Q}_p (10, 10, k)$ is the so called q-Pochammer symbol, whose definition is

$${Q}_p (a, q, k) = \prod_{i = 0}^{k-1} (1 - aq^i)$$

Ref https://en.wikipedia.org/wiki/Q-Pochhammer_symbol

Whereas the sum gives

$$\sum_{k = 0}^{n-1} (n-k)10^k = \frac{1}{81}\left(10^{1+n} - 9n - 10\right)$$

Thus your function is

$$R(n) = 9^{-k} {Q}_p (10, 10, k) - \frac{1}{81}\left(10^{1+n} - 9n - 10\right)$$

As I said, I don't know if this helps, but it's a good way to check $R(n)$ quite fast.

Please If I made some mistake, make a comment. I liked this question and I immediately got involved in trying to find a suitable general form for $R(n)$.


Looking for a particular generalization of the concept of a mathematical graph I'm am trying to define a data structure to represent road networks.  The immediately obvious structure is that of a graph - a set of nodes and edges that connect pairs of nodes.  The nodes would represent things like intersections, and the edges would represent lanes.

However, the basic concept of a graph is insufficient to describe what I want.  I also want to be able to describe how certain lanes are "reachable" or "adjacent" to other lanes.  (Imagine parallel lanes with no median between them.  You can just switch lanes.)  This seems to imply a need to have some sort of meta-edge that connects pairs of edges.  Is there any mathematical structure that sounds like this?  Is this reducible to the standard concept of a graph? Thanks.

You could make the lanes themselves nodes as well

It's not clear whether nodes for intersections are necessary.

You need a data structure that can accomodate (state transitions resulting from) the operations continue and switch lanes.   The minimal set of states needed is: those pieces of road where, if you continue driving along them, the set of "switch" options is constant. This is the smallest set of states such that any switch operation is between different states.

You could impose additional nodes thought of as STATIONARY states of motion or FIXED-LOCATION geographic data, to express starting points, destinations, intersections, stopovers or geographic locations, but only the lanes on which a traveller moves are logically necessary.  At a road intersection, one can switch from lane A on one road to lane B on another road, which is expressible as a single state transition, without going through an "inside the intersection" state.  Having an intersection state may be troublesome, as passing through the intersection within different lane-to-lane transition paths will equate situations inside the intersection that are logically different (they support different switch options).


Linear Algebra Problem (Chemistry) $$C_2H_2 + O_2\rightarrow H_2O + CO_2$$

How do I solve this equation problem with linear algebra? How do I take vectors and matrix so what should I do at here? But the answer is: $$2C_2H_6 + 7O_2\rightarrow 6H_2O + 4CO_2$$

Would someone show me this?

The solution is of the form

$$a\,C_2H_2 + b\ O_2\rightarrow c\ H_2O + d\ CO_2$$
 leading to the matter conservation equations

$$\begin{cases}2a-d=0,\\2a-2c=0,\\2b-c-2d=0,\end{cases}$$

which is a system of three linear equations in four unknowns.

As the system is homogeneous, the unknowns are indeterminate to a constant factor, that you can choose to obtain integer values.

Here,

$$a=2,b=5,c=2,d=4.$$

This may help(for combustion reactions):
$$C_xH_y + \big{(}x+\frac{y}{4} \big{)} O_2\rightarrow xCO_2+\frac{y}{2}H_2O$$