Paradox: increasing sequence that goes to $0$? It is $10$ o'clock, and I have a box.

Inside the box is a ball marked $1$.

At $10$:$30$, I will remove the ball marked $1$, and add two balls, labeled $2$ and $3$.
At $10$:$45$, I will remove the balls labeled $2$ and $3$, and add $4$ balls, marked $4$, $5$, $6$, and $7$.
$7.5$ minutes before $11$, I will remove the balls labeled $4$, $5$, and $6$, and add $8$ balls, labeled $8$, $9$, $10$, $11$, $12$, $13$, $14$, and $15$.
This pattern continues.

Each time I reach the halfway point between my previous action and $11$ o'clock, I add some balls, and remove some other balls. Each time I remove one more ball than I removed last time, but add twice as many balls as I added last time.

The result is that as it gets closer and closer to $11$, the number of balls in the box continues to increase. Yet every ball that I put in was eventually removed.

So just how many balls will be in the box when the clock strikes $11$?
$0$, or infinitely many?
What's going on here?

[Edit.] I am editing my answer to try to give some more insight, prompted by comments on my original response. I hope to develop a deeper idea into what's going on here.

The short answer is that what is important is not the size of the set of balls at any particular time, but rather how the set of balls changes; and ultimately what the limit of those sets are. The key is to determine what that limit is, and then determine how many balls are in that limiting set. The answer is that the limit is the empty set, which has size 0. The rest of this answer is devoted to describing this in some detail.

Part of what I have added to my answer is to point out that although there are multiple ways of measuring convergence — in terms of various norms on characteristic functions — only one of these actually defines the limit of the sequence, and in this case the limit is well-defined.

What matters is not the number of balls, but the set of balls

In this problem, we have more than just a number of balls which changes with time. What's different is that each of these balls has a unique identity.

This might not seem like it should matter, but it means that the state of the "system" is not a quantity of balls but a set of balls. That set has a certain size, but the size is a derived feature of the system; it follows from which particular set of balls is present. So it is important to determine what the limit of the sequence of sets is.

Description of the problem in terms of sets

Let's consider how the set of balls in the box change with time in the game you present. At step $n$, you add $2^n$ balls, and remove the $n$ lowest-numbered balls. The "state of the system" is given by the following sets:

$S(0) = \{1\}$
$S(1) = \{2,3\}$
$S(2) = \{4,5,6,7\}$
$S(3) = \{7,8,9,10,11,12,13,14,15\}$
$S(4) = \{11,12,\dots,31\}$
etc.

Note that after step 1, the first ball is removed, never to be added again; so it is not an element of the final set. Similarly, at step 2, the second and third balls are removed, never to be added again; so they aren't elements of the final set. And so on. So... none of the balls are in the final set. So then it must be empty! It doesn't matter that the number of balls in the sequence are increasing; what matters is that the number of balls which will never again be in the box is also increasing, and in the limit includes all of the balls.

We can make this more striking by considering, for each step $n$, the set of balls $I(n)$ which is in the set $S(t)$ for all $t \geq n$: that is, $I(n) = S(n) \cap S(n+1) \cap S(n+2) \cap \dots $.
Because each ball is eventually removed, never to be added again; this means that

$I(0) = I(1) = I(2) = \dots = \varnothing$

So while the original description makes it look like, moment-to-moment, the final state of the box should be to hold infinitely many balls, a more "forward looking" approach shows that it's clear that the final state of the box is to be empty.

An analysis in terms of characteristic functions

We can contrast the "intuitive" answer of infinitely many balls in the box, and the more precise answer that there ultimately are no balls in the box, using characteristic functions: that is, we replace each set $S(n)$ by a function $f_n : \mathbb{N} \to \{ 0, 1\}$  which is $1$ for those integers belonging to $S(n)$, and $0$ otherwise.

Consider the various $p$-norms on these functions. The cardinality of each set $S(n)$ is precisely equal to the $1$-norm of the function $f_n$ , which grows without bound. The fact that the 1-norm grows without bound — and in particular, the distance between the function $f_n$ and the zero function $\mathbf{0}$ in the $1$-norm grows without bound — is essentially the source of most people's intuition about this problem, and exactly the reason why they find it counterintuitive that the final set should be empty.

But just as the size of a set is a derived quantity, the norm of a function — or of a sequence of functions — is also a derived quantity; and the norm of the limit of a sequence of functions is not necessarily the limit of the norms. In fact, the functions $f_n$ don't converge to anything, in any of the p-norms; it simply diverges to nothing in particular.

But there is at least one notion of convergence which applies to the functions $f_n$, and that is point-wise convergence — the form of convergence which is the broadest, in the sense that it applies to the most cases (and with which all other notions of convergence must agree, if they show that a sequence of functions converges at all). We may simply show that for each $x$, we have $f_n (x) = 0$ for sufficiently large n. It then follows that the sequence $f_n$ converges to $\mathbf{0}$.

The fact that the sequence $f_n$ doesn't converge to $\mathbf{0}$ under any of the p-norms doesn't matter; ultimately what matters is that the sequence converges point-wise, because what we're interested in is the cardinality of the limit itself, which is defined in terms of point-wise convergence. At worst, from a certain aesthetic point of view, one might say that it doesn't converge particularly gracefully (informally speaking) to $\mathbf{0}$; but it does indeed converge, and that is all that matters.

So: using characteristic functions, which is ultimately equivalent to the sets described in the first place, one can show that the sequence of sets does converge, and what they converge to is the empty set. But take comfort that your intuition that they should not reflects a certain awareness of the concept of $p$-norms. :-)

$2^n$ increases faster than $(n+1)$, therefore

$\sum\limits_{n=0}^{\infty} [ 2^n - (n+1) ] = \infty$


Value of derivative at a particular point We were recently asked in class whether $f:X \to \mathbb{R}$ with $f'\equiv 0$ implies $f$ is constant.

Of course, if we take $X=(0,1)\cup (2,3)$ with $$f(x)=1\text{ when } x\in (0,1)$$
$$f(x)=2\text { when } x \in (2,3)$$

I could work out through this. Then I thought whether taking $X=[0,1]\cup [2,3]$ would yield the same result. I am confused whether $f'(1)$ would be $0$ or not, since of course right hand limit does not exist.

It's not usually (ever?) useful to define derivatives on points that aren't in the interior of an open set. In fact it's discourageable to do so via one-sided limits, because a function could then be differentiable at $a$ over $[a,b]$, but not on $[a-\varepsilon,b]$ (for instance the absolute value function on $[0,1]$). See this question and its answers for some discussion.

With this in mind, the answer to your question is that $f'(1)$ could be $0$ if you want it to be, but this definition of derivative at a point isn't meaningful so there's no reason to consider the point $x = 1$ anyways (and besides you've already answered the question of whether $f' \equiv 0\implies f\equiv \mathrm{constant}$ with the example on open intervals).


Taylor Series centered at $\alpha$. Is $\alpha$ arbitrary? Let's say I have a function $f \in C^{\infty}$ represented by a Taylor Series

$$f(x) = \sum_{n=0}^{\infty}\frac{f^{(n)}(\alpha)}{n!}(x-\alpha)^n$$

What relevance does $\alpha$ have to $f$? Can we just choose any $\alpha$ to be a value where we know how to evaluate $f^{(n)}(\alpha)$?

i.e. Do we just choose a $\alpha$ such that we know the value of the function $f$ and all it's derivatives at that point $\alpha$? If so is $\alpha$ just purely arbitrary?

First we need to note we have to be careful about the hypotheses, because $f\in C^\infty(\Bbb R)$ does not imply that $f$ is represented by its Taylor series. Yes, $\alpha$ is arbitrary, in the sense that if there exists an $\alpha$ that works for every $t$ then any other $\alpha$ also works:

    Theorem. Suppose that $f(t)=\sum\frac{f^{(k)}(\alpha)}{k!}(t-\alpha)^k$ for every $t\in\Bbb R$. Then $f(t)=\sum\frac{f^{(k)}(\beta)}{k!}(t-\beta)^k$ for  every $t,\beta\in\Bbb R$.

No, It's not arbitray. Because if you choose an alpha and in that alpha, f(alpha) or f'(alpha) or ... Become infinite it won't have meaning. For better understanding try expand f(x)=sqrt(x) and choose alpha=0, you will understand why we can't use any alpha.


Prove that if $\|A\|<1$, then $(I+A)^{-1}=I-A+A^2-A^3+\cdots$. Prove that if $\|A\|<1$, then $(I+A)^{-1}=I-A+A^2-A^3+\cdots$.

I'm not sure how to do this. I know the result for $(I-A)^{-1}$, but that won't help me.

If you know the result for $(I-A)^{-1}$ you can employ it: if $\|A\|<1$, then $\|-A\|=\|A\|<1$; hence $I+A = I -(-A)$ is invertible and $$(I+A)^{-1} = (I-(-A))^{-1} =\sum_{n=0}^\infty (-A)^n = \sum_{n=0}^\infty (-1)^nA^n.$$


True or false Calculus sequences Let $\{s_n\}$ be a sequence in $\mathbb{R}$ and assume $c \in \mathbb{R}$. If $|s_n - c| < \frac{3}{n}$ for all $n \in \mathbb{N}$, then $lim_{n \rightarrow \infty} s_n = c$.

Is this false since $|s_n - c|$ has to be less than $\varepsilon$ for every $\varepsilon > 0$? In this case, its only less than $\frac{3}{n}$.

No, actually. While the definition say "for each $\epsilon > 0 \ldots$ it doesn't have to explicity be epsilon. You can choose it to be any positive number. So this statement is actually true.


What is the smallest Euclidean space in which one can embed a given curved space? Given a $d$-dimensional curved space, how many dimensions are required to embed it? As an example think of a sphere's surface, which is a two-dimensional curved space that can be expressed in Euclidean 3D, while a flat plane requires only two dimensions.

Like you say:

The plane $\Bbb R^2$ can be embedded in $\Bbb R^2$ but not $\Bbb R$.
The sphere $\Bbb S^2$ can be embedded in $\Bbb R^3$ but not $\Bbb R^2$.

In particular, the answer depends not only on the dimension of the given curved space (which here I'll take to mean manifold), but the space itself.

We also have:

Real projective space $\Bbb {RP}^2$ and the Klein bottle $K$ can both be embedded in $\Bbb R^4$ but not $\Bbb R^3$.

In fact, $4$ dimensions is sufficient for all surfaces ($2$-manifolds); see below.

To be more precise the answer depends on whether one means a smooth embedding or a Riemannian embedding, that is, whether one has a metric on the manifold one wants to line up with the Euclidean metric of the ambient space. (The results above refer to smooth embeddings.)

Smooth manifolds

The (Strong) Whitney Embedding Theorem gives a universal upper bound: A smooth $m$-manifold can be embedded in $\Bbb R^{2m}$.

Under some common conditions, we can improve this bound some. Wall ($m = 3$) and Haefliger-Hirsch ($m > 3$) showed that if a smooth $m$-manifold is compact and $m$ is not a power of $2$, then it can be embedded in $\Bbb R^{2m - 1}$. (We can even improve on this, but there is no known closed-form formula for the maximum dimension required as a function of $m$.) This restriction re powers of $2$ is necessary, as $\Bbb {RP}^{2^k}$ cannot be embedded in $\Bbb R^{2 \cdot 2^k - 1} = \Bbb R^{2^{k + 1} - 1}$.

The Weak Whitney Embedding Theorem says that we can do slightly better if we allow our manifold to be immersed, rather than embedded (roughly speaking, this means we allow self-intersections that aren't pathological): A (topological, so not necessarily smooth) $m$-manifold can be embedded in $\Bbb R^{2 m - 1}$. (In particular, there is a smooth immersion of $\Bbb {RP}^2$ into $\Bbb R^3$, despite that there is no such embedding.

Riemannian manifolds

If one wants to embed a (smooth) Riemannian manifold $(M, g)$ in Euclidean space $(\Bbb R^N, \bar{g})$, so that the metric $g$ is the metric determined on $M$ by the embedding, one generally requires higher-dimensional Euclidean spaces than Whitney's Embedding Theorem prescribes: The Nash Embedding Theorem (well, one theorem by this name) says that a Riemannian $m$-manifold can be embedded in $\Bbb R^{m (3m + 11) / 2}$ if the manifold is compact and $\Bbb R^{m (m + 1) (3m + 1) / 2}$ if not.

NB that the dimension required can depend on $g$ and not just on $M$. For example, the torus $\Bbb T^2$ can be embedded into $\Bbb R^3$, and any such embedding determines a metric on it. No such metric is flat, however, so the flat torus cannot be embedded into $\Bbb R^3$; it can, though, be embedded in $\Bbb R^4$.

@weineld Yes. You can also embed N dimensional curved space into N dimensional euclidean space. This is what was done by Ed Witten when he applied a select range of landscapes of string theory to twistor space (and we can all appreciate the effect that had on Feynman diagrams). This is only true if your curved space falls within specific landscapes or vacua and assuming your flat space is euclidean in its geometry, thus limiting the total number of dimensions N this can be done with. Further; in a dynamic carteesian coordinate system k', one can treat time in a similiar fashion by introducing imaginary time (Ti) like a y axis to our usual time, thus giving us the notion of "euclidean time". This is a notion Hawking has been exploring in his more recent publications.


How do you find distance from a point to a plane? Compute the distance from the point A = (1,1,-1) to the plane which contains the point B = (3,6,-1) and the line,  \begin{align}
    \begin{bmatrix}
           x \\
           y \\
           z \\
         \end{bmatrix} &= \begin{bmatrix}
           2 \\
           5 \\
           0 \\
         \end{bmatrix} t
  \end{align}

NOT ALLOWED to use cross product to determine the answer!

The answer is 0, because point A is on the plane. But I don't know how to get the plane equation using the point B which is on the plane and the line which is also on the plane, to then derive the distance.

Edit: thanks for all the answers, I decided to go talk to my prof and she explained that we were supposed to first select two points on the line(lets call them G and H), so we would have three points on the plane, including point B. So we could then create two directional vectors, example: BG and BH. Using these two directional vectors, we would find a vector N, which is orthogonal to both of these vectors. this would give us a vector which is orthogonal to the plane. This would be done by have BG∙N=0 and BH∙N=0 and solving the system to give us N. After we get N, we will use N and a point on the plane, B to compute the distance from A to the plane.

One way to write a general equation of a plane in $x,y,z$-space is
$$ ax + by + cz = d $$
where $a, b, c$, and $d$ are constants.
The equation is not unique, because you can multiply all the constants
by the same constant factor (as long as it is not zero) and get a different
equation of the same plane.

Three non-collinear points determine a plane in three-dimensional space.
From the problem description, you have at least these three points
on the plane:
\begin{align}
&(0,0,0) && \text{on the given line at $t=0$} \\
&(2,5,0) && \text{on the given line at $t=1$} \\
&(3,6,-1) && \text{the given point $B$}
\end{align}

Plugging each of these points into the general equation $ax + by + cz = d$
will give you a system of equations in which the only unknowns are
$a, b, c$, and $d$.
If the three points are not collinear, every solution of the system of equations will give you a set of constants $a, b, c$, and $d$
such that $ax + by + cz = d$ is a correct equation of the desired plane.
At this point you can just choose one of the solutions;
which one does not matter (at least not much)
as long as $a, b, c$, and $d$ are not all zero.

Note that this means the vector
$\begin{bmatrix} a & b & c \end{bmatrix}^T$
is perpendicular to the desired plane, and therefore it
is some scalar multiple of the cross product of any two independent
vectors parallel to the plane, which is the other way you could
have found an equation of the plane if you were allowed to use
a cross product.

In this particular case,
since the point $A$ actually is exactly in the plane (distance $0$),
just plugging the coordinates of $A$ into the equation will
answer the question.
But in general, for any arbitrary point $(x,y,z)$,
the distance from $(x,y,z)$ to the plane is
$$
\frac{ax + by + cz - d}{\sqrt{a^2 + b^2 + c^2}}.
$$

hint: Choose $2$ points on the plane: $t = 0 \implies A = (0,0,0)$, $t = 1\implies C = (2,5,0)$,and the given point $B = (3,6,-1)$, then find the normal line to the plane defined as $\vec{N} = \vec{AB} \times \vec{AC}= (m,n,p)$. You have to solve for $m,n,p$ as you do the cross product of the above two vectors. Then the formula for the plane is: $\vec{XB}\cdot \vec{N} = 0$, with $X = (x,y,z)$


Limit of $\frac{\sin(x+y)}{x+y}$ as $(x,y) \to (0,0)$ I want to prove that $\lim_{(x,y)\to (0,0)} \frac{\sin(x+y)}{x+y} = 1$

Is it sufficient to say that if $u = x+y$ then as $(x,y) \rightarrow (0,0)$ then $u \rightarrow 0$ and then evaluate $\lim_{u\to 0} \frac{\sin u}{u}$ which is 1 by L'Hôpital's rule?

I feel that this is somehow "cheating" and I would really like to prove it in a more rigorous way, i.e., by using an $\epsilon$ - $\delta$ argument.

I have tried for a long time to use the one variable version and prove that if $0 \lt \sqrt{x^2 + y^2} \lt \delta$ then somehow $0 \lt |x+y| \lt \delta$ and it would follow from the one variable case that $| \frac{\sin(x+y)}{x+y} - 1| \lt \epsilon$. However, because $|x+y| = 0$ when $y = -x$ I am not getting anywhere with this.

Any input on this would be helpful :)

As @julien points out, the function $\frac{\sin(x+y)}{x+y}$ is not defined in a neighbourhood of the origin, so the limit does not exist. But there is a natural extension of the function which is defined on a neighbourhood of the origin, and for which the limit exists and equals 1. (There are lots of extensions, but this one seems most natural to me.) Define $f:\mathbb{R}^2\to\mathbb{R}$ by $$f(x,y)=\left\{
\begin{array}{cl}
\frac{\sin(x+y)}{x+y} & x+y\neq0; \\
1 & x+y=0.\end{array}\right.$$
Then $\lim_{(x,y)\to(0,0)}f(x,y)=1$. An $\epsilon-\delta$ proof relies on the one variable result $$\lim_{u\to 0}\frac{\sin u}{u}=1.$$
That is, given $\epsilon>0$, there exists $\delta>0$ such that $$0<|u|<\delta \Rightarrow \left|\frac{\sin u}{u}-1\right|<\epsilon.$$
It is helpful to notice that this can be rephrased as follows: given $\epsilon>0$, there exists $\delta>0$ such that $$0<|u|<2\delta \Rightarrow \left|\frac{\sin u}{u}-1\right|<\epsilon. \qquad (*)$$

We want to prove that given $\epsilon>0$, there exists a positive $\delta$ such that $$0<\sqrt{x^2+y^2}<\delta \Rightarrow |f(x,y)-1|<\epsilon.$$

So let $\epsilon>0$. Choose $\delta$ so that $(*)$ holds true, and let $0<\sqrt{x^2+y^2}<\delta$. Then $$|x+y|\leq|x|+|y|\leq\sqrt{|x|^2+|y|^2}+\sqrt{|x|^2+|y|^2}=2\sqrt{x^2+y^2}<2\delta.$$ Then by (*) (for $x+y\neq0$) and by the definition of $f$ (for $x+y=0$), we have $|f(x,y)-1|<\epsilon$ as required.

The same argument works if the function is defined as $g:\mathbb{R}^2\setminus\{(x,y):x+y=0\}\to\mathbb{R}$ with $g(x,y)=\frac{\sin(x+y)}{x+y}$. With a slight amendment, the same result also holds for $$f(x,y)=\left\{
\begin{array}{cl}
\frac{\sin(x+y)}{x+y} & x+y\neq0; \\
h(x) & x+y=0,\end{array}\right.$$
where $h$ is any function defined in a neighbourhood of $x=0$ which has $\lim_{x\to0}h(x)=1$.


Is $\frac{1}{x}$ integrable? I am taking a course on theory of integrals and measure, where I have learned a very fundamental definition of integral. One of the things I've learned is that $f$ is integrable iff $\int|f|<\infty$. Then, according to this definition, $f(x)=\frac{1}{x}$ wouldn't be integrable, since the integral of $|f(x)|=|\frac{1}{x}|$ diverges. However, I can calculate $\int_a^b\frac{1}{x}dx$ for any $a, b \in (-\infty, +\infty)$ and this isn't problematic at all. How do I deal with this?

I think your real difficulty is that we don't say "$f$ is integrable," but rather "$f$ is integrable over $D$" where $D$ is some domain.
Your definition works fine if the domain is an interval that doesn't contain $0$.

Hint:
$$\int_a^b\frac{dx}x=\log\frac ba$$
(with no absolute value) is correct. This implies that $a,b$ must have the same sign and be nonzero. Otherwise the definite integral is undefined.


What is the intuition behind the reduced row echelon form of a matrix? When we convert a matrix into reduced row echelon form , the linearly independent vectors in the pivot columns form a unit vectors in the corresponding columns ? what is really happening here if I visualize it? we are performing row operations i.e operations on row vectors, so we are scaling down or up the row vectors by the same amount. But why this doesn't change the column vectors ?

Have a look at the Wikipedia entry on 'Guassian elimination'.  The example on three simultaneous equations is the starting point.  Then the idea of elementary row operations and what they do to the determinant of a matrix is next, in understanding how echelon form is useful in calculating the determinant of a matrix - but you have to be aware/happy that the determinant of an echelon form matrix is simply the product of its main diagaonal elements.  Next is that reduced row echelon form is a way of getting an inverse matrix.  There are other uses, but these three are reason enough for using the technique.


Prove or disprove that $\int_a^b |f(x)|\ \mathrm{d}x\geq \big | \int_a^b f(x)\ \mathrm{d}x\big| $ Let $f$ be a continuous and integrable function over $[a,b]$. Prove or disprove that
$$\int_a^b |f(x)|\ \mathrm{d}x\geq \left | \int_a^b f(x)\ \mathrm{d}x\right|
$$

If $f$ is real Riemann-integrable function, this inequality is true. (and if $f$ is complex Riemann-integrable, this inequality is true.)

By properties of absolute value, we get
$$-|f(x)|\le f(x)\le |f(x)|.$$
Since $f$ is continous, so $|f|$ is also continous, hence $|f|$ is Riemann integrable. We can integrate each side of this inequality and we get
$$-\int_a^b |f(x)|dx\le \int_a^bf(x)dx\le \int_a^b|f(x)|dx$$
and we get
$$\left| \int_a^b f(x)dx\right|\le \int_a^b|f(x)|dx.$$

Think about what your inequality is saying:

Imagine coordinates $y$ and $x$.
Imagine some arbitrary curve on that plane $f(x)$.
Imagine integrating $f(x)$ by taking the area under the $x$-axis away from the area above the $x$-axis.
Imagine repeating the integration for $\vert f(x) \vert$.

And you can see that your inequality is true for integrable functions.


Showing that a regular curve is a pregeodesic I'm trying to prove the next:

To show that a regular curve $\alpha$ with $\alpha^{'}$ and $\alpha^{''}$ collinear is a pregeodesic, write $\alpha^{''}(s)=f(s)\alpha^{'}(s)$ and prove that

a) $\beta=\alpha\circ h$ is a geodesic if and only if $h''+ (f\circ h) (h')^2 = 0$.

b) If  $\langle \alpha',\alpha'\rangle $ is never zero, then any constant speed reparametrization of $\alpha$ is a geodesic.

Suposse $\beta$ has unit speed, so $\langle \beta''(s), \beta'(s)\rangle = 0$ for all $s.$ From here $h'(s) ( h''(s) + f( h(s)) (h'(s))^2) \langle \alpha'(s),\alpha'(s)\rangle = 0,$ because $\beta^{''}$ is as a) and $\beta^{'}=(\alpha^{'}\circ h)h^{'}.$

c)$\langle \alpha',\alpha'\rangle$ is always zero or never zero.

To see this,  $\langle \alpha'(s),\alpha'(s)\rangle' = 2f(s)\langle \alpha'(s),\alpha'(s)\rangle$, then $\langle \alpha'(s),\alpha'(s)\rangle = Ce^{2\int f(s)\,{\rm d}s}$ for some integration constant $C.$

d) If $\langle \alpha',\alpha'\rangle$ is always zero, then $\alpha$ is pre-geodesic.

I've proved a), b) and c). Such points follow by some computations with $\beta^{'}$ and $\beta^{''},$ the hypotesis that $\alpha$ is regular and the first at the proposition: $\alpha^{''}(s)=f(s)\alpha^{'}(s).$

My first doubt is: How is possible write  $\alpha^{''}(s)=f(s)\alpha^{'}(s)?$ I don't get how to prove this. It is part of the hypotesis?

Also I'm stuck prove d).

Second: Why the proof of the behind ensures $\alpha$ is pregeodesic?

I think the next result ensures that, if the previous holds, then $\alpha$ is pregeodesic:

Let $\gamma:I\rightarrow M$ be a nonconstant geodesic. A reparametrization $\gamma\circ h:J\rightarrow M$ is a geodesic if and only if $h$ has the form
$h(t)=at+b.$ If a curve has a reparametrization as a geodesic we call it pregeodesic.

Any kind of help is thanked in advanced.

(1) If $c=\alpha \circ h(s)$ is a geodesic, then $ c''=0$ implies
$$ \alpha '' (h')^2+\alpha' h'' =0 $$

That is, there is $f$ s.t. $\alpha '' =f\alpha'$

(2) If $\alpha '' =f\alpha '$, then $c:=\alpha \circ h$ has unit speed.
Then $ c'=\alpha'h' $ and $c'\perp c''$. That is, $c''\perp \alpha'$
so that $$ c''=\alpha '' (h')^2+\alpha' h'' = \alpha '\{ f (h')^2+
h''\} =0 $$

Hence $c$ is a geodesic.

In the exercise, the curve $\alpha$ is regular, that is $\alpha'(x)\ne0$ for every $x$. Thus, in the point (d), if $(M,g)$ is semi-Riemannian manifold, we have $\langle\alpha'|\alpha'\rangle=0$ (light-like curve), and, $\alpha'(x)\ne0$ for every $x$ (it's regular curve by hypothesis). I'm Italian, sorry for my English...


Biological processes described by the exponential function What are biological or biochemical processes that are described (stochastically or deterministically) by exponential function $f(x) = e^{ax}$?

Edit: What about examples that are not related to the population growth? I'm especially interested in biochemical examples.

whenever a process evolves at a speed proportionnel to its own advancement, it satisfies the differential equation $y'=a y$ which leads to exponential solutions.

this is a very basic model for all kind of population growth (as you'd say that the bigger the population is, the more baby they make per time unit).

it is universal in the sense that if $y$ satisfies a more complicated evolution equation $y'=f(y)$ near an equilibrium point (for example, if you have a very small initial population, or a starting process) then linearizing this equation you obtain something like $y' \approx f'(y_0) y$ where $y_0$ is the equilibrium point.

Addendum : radioactive decay follows a decreasing exponential law because of the time-independance nature of the probability for a particle to split.

Basic equations in kinetic chemistry are exponential laws (though they get more complicated for complex reactions), I suspect mainly because of the universal property I mentionned : it is "the most simple evolution to an equilibrium state" in a sense.

an example of a BIOLOGICAL exponential decay process is the concentration of a drug in the bloodstream, after it has benn intravenously injected into the body :)


Does $y=(ax^{2}+bx+c)/(dx+e)$ have any lines of symmetry? Does $y = \dfrac{ax^2 + bx + c}{dx + e}$ have any lines of symmetry. If it does, what are they, and if not, how would one prove that it doesn't have any lines of symmetry?

(Please consider the general case. And please ignore trivial cases where the expression reduces to a quadratic expression in $x$, linear expression in $x$, or a constant.)

I have written this solution independently of the solution of @Steven Gregory, with which it shares many common points . The differences are that

1) I have tried to concentrate on the main ideas, escaping computational details.

2) I am guided by the common general shape of the curves, an example of which I give below

3) I consider the exceptional case where the function is in fact affine.

Assuming $a \neq 0$ and $c \neq 0$, Euclidean division of the numerator by the denominator gives :

$$ax^2+bx+c=(Ax+B)(dx+e)+C \ \ \text{for certain real numbers} \ \ A,B,C.$$

Thus

$$f(x)=y=Ax+B+\dfrac{C}{dx+e} \ \ \text{with} \ \ A:=\frac{a}{d} \neq 0 \ \ (1)$$

Two cases:

Either: $C=0$ (which happens when $dx+e$ exactly divides $ax^2+bx+c$) in this case, the graphical representation of $f$ is a straight line which is its own symmetry axis and has an infinite number of axes of symmetry in the perpendicular direction...
Or $C\neq0$: we recognize here the equation of a hyperbola having a vertical asymptote with equation  $x=-e/d$ and oblique (slant) asymptote with equation $y=Ax+B$.
See graphics below. Thus the graphical representation has two axis of symmetry.

Explanation about the fact that the graphical representation of (1) is a hyperbola : let us write (1) under the form $(y-Ax-B)(cx+d)=C$ and make the affine change of variables $X=y-Ax-B, Y=cx+d$ (this does not modify the nature of the conical section) thus we obtain $XY=C$, which is one of the canonical forms for the equation of a hyperbola.

The example of $y=0.5x-1+\epsilon(x)$ with $\epsilon(x)=\frac{2}{x-1}\rightarrow 0 \ $ when $x \rightarrow \infty$, proving that $y=0.5x-1$ is the equation of the oblique asymptote (vertical asymptote at $x=1$):

There does exist a line of symmetry when $a =1, b = c = d = 0, e = 1$, and the line is $x = 0$.


Solve for these 2 ODE equations How do I solve these ODE's? Do I need to find the value of my y variable or do I need to find solutions for the function in the derivative? Would like to know how it is done. Many thanks :). P.S. new to all these higher level mathematics.

The first ODE

Let's write it in a more convenient notation:
$$yy'=0,\,\, y(0)=1$$

What does this ODE says? a function times it's derivative is the constant zero, so at least one of the functions should be the the zero function. One trivial solution is $y=0$ since it satisfies the ODE, but it doesn't satisfy the initial value! So we must have $y'=0$ so $y=c$, from the initial value we conclude the solution is $y=1$.

The second ODE

You can solve this ODE using the Laplace transform. let me know in the comments if you don't know how.

SKETCH FOR SOLUTION USING THE LAPLACE TRANSFORM

We usually solve ODEs by applying the Laplace transform for linear differential equations (not necessarily homogenous!), most popularly of order 2 or less.

The most useful fact about  the Laplace transform, the one that makes it useful to solve linear ODEs is the following relation between the Laplace transform and the derivative:
$$\mathcal{L}(f')=s\mathcal{L}(f)-f(0)$$

Do note we also need and an initial value of $f$ evaluated at $0$.
Another useful fact about the Laplace transform- since it's defined by the integral:

$$\mathcal{L}(f)=\int_{0}^\infty f(t)e^{-st}dt$$

It is a linear transformation! (This can be seen by using the linearity of the integral).

These properties of the Laplace transform actually reduce solving an IVP from a differential problem to an algebraic one (It is is a really rich field of research called Operational Calculus!). After applying the transform to both sides of the equation we re-arrange the equation to that we can use the last important property of the Laplace transformation: It is invertible! so after re-arranging we can apply $\mathcal{L}^{-1}$ to both sides and thus solve for $f$ .

Table of the laplace transform for common functions

For your first equation we get
$$y(x)=0$$ or $$y'(x)=0$$
The second one has the solutions
$$y(x)=c_2 \sin (2 x)+c_1 \cos (2 x)+\dfrac{1}{3}\sin x$$


Analytic solutions to a nonlinear second order PDE I am trying to solve the following 2nd order PDE analytically, but haven't succeeded so far. I have tried to separate the variables, but it doesn't work here. I would be very grateful for any suggestions that could lead me in the right direction in solving it and for all suggestions about where I could look further. The equation is the following:
$$\frac{\partial u(x,t)}{\partial x}=- \frac{b(x)}{2} (1-u(x,t))^2+\frac{b(x)}{c} \frac{ \partial^2 u(x,t)} {\partial x \partial t},$$
where
$u(x,t)$ is the unknown function,
$b(x)$ is a known function of $x$ only, and
$c$ is a constant.

Any help would be much appreciated. Thanks!

$$\frac{\partial u(x,t)}{\partial x}=- \frac{b(x)}{2} (1-u(x,t))^2+\frac{b(x)}{c} \frac{ \partial^2 u(x,t)} {\partial x \partial t},$$
Let $u(x,t)=v(x,t)-1$
$$\frac{\partial v(x,t)}{\partial x}=- \frac{b(x)}{2} v(x,t)^2+\frac{b(x)}{c} \frac{ \partial^2 v(x,t)} {\partial x \partial t},$$
Change of variable : $X(x)=\int{b(x)}{dx}$

$b(x)dx=dX$

Let $V(X(x),t)=v(x,t)$
$$\frac{\partial V(X,t)}{\partial X}=- \frac{1}{2} V(X,t)^2+\frac{1}{c} \frac{ \partial^2 V(X,t)} {\partial X \partial t},$$
Change of variable : $T(t)=ct$

Let $W(X,T(t))=V(X,t)$
$$\frac{\partial W(X,T)}{\partial X}=- \frac{1}{2} W(X,T)^2+ \frac{ \partial^2 W(X,T)} {\partial X \partial T},$$
Now, there is no parameter (constant nor variable) in the PDE, which is a major simplification.

$1$ is the solution of the partial differential equation.


Why can't two vectors span $\Bbb R^3$? I came across a question in my linear algebra textbook and it said: "Given $x_1 = (1, 1, 1)^T$ and $x_2 = (3, -1, 4)^T$: Do $x_1$ and $x_2$ span $\Bbb R^3$? Explain."

I'm pretty sure that the answer is no (I thought you needed n vectors to span $\Bbb R^n$) but I'm not sure how to show that this is the case, if anyone could help it would be greatly appreciated.

A perfectly good proof in any specific case such as this is to take (almost) any vector and show that it is not a linear combination of your vectors.

For the more general proof that you need three vectors, you have to point out that a set of three equations in two variables generally does not have a solution.

Given a basis of a vector space, the dimension is defined to be exactly the number of vectors in the basis. Since we know that $(1, 0, 0)$, $(0, 1, 0)$, and $(0, 0, 1)$ span $\mathbb{R}^3$, hence the dimension is 3. So it can't be spanned by two vectors, otherwise the dimension would be 2.


Comparison of Adams-Bashforth and Runge-Kutta methods of order 4 I have a system of ODE, that must to solve with numerical methods. I solve it by Adams_Bashforth with order4 and Runge-Kutta with order4 methods. Do you know with same length step which methods answers exactly?

Runge-Kutta is a fixed step solver but Adams is a variable step solver. You can't compare them with the same step size because in Adams method you define max. and min. step size with relative tolerance. You don't define a single step size like 0.01 as in Runge Kutta method. Obviously, variable step solvers give smoother results but the computation time also increases.

You can find a comparison of different methods here: Matlab Alghorithms

Edit: This answer is about variable step Adams-Bashforth-Moulton solver in Matlab. I don't know its fixed step version.


Proving that the sequence $n^{(-1)^{n}}$ is divergent I would like to prove that the sequence $n^{(-1)^{n}}$ is divergent.

My thoughts: I know $(-1)^n$ is divergent, so $n$ to the power of a divergent sequence is still divergent? I am not sure how to give a proper proof, pls help!

The argument that "n to the power of a divergent sequence is divergent" does not make sense (consider $n^{-n}$ for example.). Regarding your sequence: if this sequence were convergent against some limit value $a$, then every subsequence would have to converge against the same value $a$. Now look at the subsequences for $n$ even and odd, respectively.

If $n$ even then $-1^n = 1 $
$$n^1 = n\Rightarrow \lim\limits_{n \rightarrow \infty}{({n})} = \infty$$

If $n$ odd then $-1^n = -1 $
$$n^{-1} = \frac{1}{n}\Rightarrow \lim\limits_{n \rightarrow \infty}{(\frac{1}{n})} = 0$$


What is difference between stochastic process and a sequence of random variables? Suppose that I have a sequence of random variables $X=\langle X_1,...,X_n\rangle$ and I do not have any assumption about these continuous random variables, $X_i$'s, (they can be dependent/independent, identically distributed or not). Again suppose that $Y=\langle Y_1,...,Y_n\rangle$ is a stochastic process. Is there any difference between X and Y? can I say that X is also a stochastic process?

I would be grateful if you could help me.

Ahhh... A silent downvote, nearly four years after this was posted. So cute...

According to the comments, there is a problem of comprehension at the heart of this question, which might be the following.

Consider a real valued random variable $Z$. This is a measurable function $Z:\Omega\to\mathbb R$ between a probability space $(\Omega,\mathcal F,P)$ and the measurable space $(\mathbb R,\mathcal B(\mathbb R))$. The distribution of $Z$ (aka the law of $Z$) is definitely not the probability measure $P$, but the image of $P$ by $Z$, often denoted $P_Z$, that is, the probability measure $\mu$ on $\mathcal B(\mathbb R)$ defined by the identity $\mu(B)=P(Z^{-1}(B))$ for every $B$ in $\mathcal B(\mathbb R)$.

Thus, to declare that two real valued random variables $X$ and $Y$ are identically distributed means that $X:\Omega\to\mathbb R$ and $Y:\Phi\to\mathbb R$ for some probability spaces $(\Omega,\mathcal F,P)$ and $(\Phi,\mathcal G,Q)$, and that the probability measures $P_X$ and $Q_Y$ on $\mathcal B(\mathbb R)$ coincide. The fact that $P=Q$ or that $P\ne Q$ (both can happen) is quite unrelated.

To declare that two real valued random variables $X$ and $Y$ are such that $X:\Omega\to\mathbb R$ and $Y:\Omega\to\mathbb R$ for some common probability space $(\Omega,\mathcal F,P)$, as you seem to be intent on saying, one usually simply mentions that $X$ and $Y$ are defined on the same probability space.

If $X$ is a function of $X_i$ and if the $X_i$'s can take random values (like price, height, quantity, etc...), and $X$ is sorted by time, then $X$ is a stochastic process.


How can I find a number $a$ such that this limit is 1 I want to find the number $a$ and $b$ such that $\lim_{x\to 0} \frac{\sqrt{ax+b}-2}{x}=1$.

First of all, I know that $b$ has to be 4, because the limit of the numerator has to be zero because the denominator is zero when we take its limit.

My problem is with the number
$a$. I need help.

Thanks a lot!

Multiplying by the conjugate gives:
$$
\frac{\sqrt{ax+b}-2}{x}=\frac{ax+b-4}{x(\sqrt{ax+b}+2)}.
$$

As you already observed, you need $b=4$, for otherwise the limit is infinite.
Then you can simplify by $x$.

Now the limit is:
$$
\frac{a}{\sqrt{4}+2}=\frac{a}{4}.
$$

So you want $a=4$.

multiply by the conjugate on both sides of the equal sign and you get

$$\frac{ax+b-4}x = \sqrt{ax+b}+2$$
multiply by x and plug 4 into b

$$ax+4-4 = x(\sqrt{ax+4}+2)$$
cross out the x's

$$a = \sqrt{ax+4}+2$$
set $x=0$ because $\lim_{x\to 0}$

$$ a = \sqrt{4}+2$$
    $$a=4$$


Is there a bijective map from $(0,1)$ to $\mathbb{R}$? I couldn't find a bijective map from $(0,1)$ to $\mathbb{R}$. Is there any example?

Here is a nice one ${}{}{}{}{}{}{}{}{}$, can you find the equation?

By virtue of http://natureofmathematics.wordpress.com/lecture-notes/cantor/, here's another picture. The interval at the bottom is $\mathbb{R}$.


Show that the equation $\tan x=x$ has an infinity of solution Given the function $f$ defined as :

for every $x \in (0,1]$ : $f(x)=x\sin(\frac{\pi}{x})$
$f(0)=0$

Show using roll's theorem that the equation $\tan x=x$ has an infinity of solutions

What I have tried

$f$ is differentiable in $(0,1]$

In $(0,1]$ we have $f(1)=\sin(\pi)=f(0)=0$

By Roll's : there exists $c \in(0,1)$ such that $f'(c)=0$

$$f'(x)=\sin(\pi/x)-x\cos(\pi/x)$$

In $(0,1)$ we have $f'(c)=\sin(\pi/c)-x\cos(\pi/c)=0$

so $\sin(\pi/c)=x\cos(\pi/c)$

Dividing by $\cos(\pi/c)$

we obtain $\tan(\pi/x)=x$

I am blocked here
No problem if you give me any other information about an other way .

$f(x)=x\sin \dfrac{\pi}{x}$ is differentiable on $(0,1].$ Moreover, we have that $f(1/n)=\dfrac{1}{n}\sin (n\pi)=0.$ So, we can apply Rolle's theorem on $\left[\dfrac{1}{n+1},\dfrac{1}{n}\right].$ Thus there exists $c_n\in\left(\dfrac{1}{n+1},\dfrac{1}{n}\right)$ such that $f'(c_n)=0.$ Or, equivalently, $\tan c_n=c_n.$ This shows that $\tan x=x$ has infinitely many positive solutions.

Over arbitrary interval of definition tangent goes from minus infinity to infinity. That is, it goes from less than x to more than x (monotonically) Thus by intermediate value theorem it is the same as x precisely once on each interval of definition. There are infinitely many such intervals.


Find the number of ways to form 15 teams out of 15 men and 15 women. In how many ways can 15 teams be formed, each consisting of a man and a woman, from 15 men and 15 women.

This looks like the same problem as finding the number of bijective functions from a set $A$ to $B$, both containing 15 elements each.

So the answer should be $15!$ which is not correct.

Edit: This is not a legitimate question though, could somebody try thinking about what constraints need to be added, so that the answer is close to the options. Yes, I mean 'correct the question'; it uses very rough language and is probably not what the options imply.

Edit: Let's try breaking the problem. What would be the number of ways to form 3 teams, each consisting of a man and a woman?

The possible options given seem far, far too small. I too believe the answer is $15!$.

There are many ways of thinking about it, but perhaps the easiest is to remove the identities of the men and line them up as a series of $15$ bins (hm, maybe 'rooms' instead here), in to which we place the women. Since we assume we only care which woman a man is with, we only care which room a woman is assigned to, the order and labelling of the rooms doesn't matter. There are then, as usual, $15$ options then $14$ options and so on, giving $15!$.

I am struggling to see how any interpretation of the problem could reduce the answer to something as small as roughly $2000$. It seems to me this would require further labels to be removed, for example treating a load of the men as being called Dave, and saying they're the same (no disrespect to Daves).

There are many ways to approach this problem. One is that- the first man will have 15 choices. Then the 2$^{nd}$ one will have 14 and like this till 1. So total choices for men-$15!$
Now in the same way total choices for women will be $15!$.
So total ways= $15!.15!=(15!)^2$

One more way to approach this problem is- For the first team there will be $(15*15)$ choices. For  the second one- $(14*14)$ , till $(1*1)$
So total ways$=(15*15).(14*14)...(1*1)=(15!)^2$


Fourier series for $\sin^2(x)$ I was asked to compute the Fourier series for $\sin^2(x)$ on $[0,\pi]$. Now this is what I did and I'd like to know if I'm right.  $\sin^2(x)=\frac12-\frac12\cos(2x)$ . I got the right hand side using trig identities. I'm wondering If I can do this without using the formulas. Thanks.

One small point...

The way the question is stated, there may be a slight ambiguity.  One way (and almost certainly the intended way) to read the question is:  given the (periodic) function $\sin^2(x)$, find its Fourier series on the interval $[0, \pi]$.  In this case, $(1 - \cos(2x))/2$ is correct.

However, we could also read it as follows:  given the function $\sin^2(x)$ defined on the interval $[0, \pi]$, find its Fourier series.  In this case, we must first decide how to extend the function to be periodic.  Of course the natural choice is to extend it to equal $\sin^2(x)$ for all $x$.  Again, the cosine series is correct.

But... we do have the freedom to extend $\sin^2(x)$ to an odd function on $[-\pi, \pi]$ instead, in which case the Fourier series will contain only sine functions (with the coefficients computed in the usual way). The point being that there is in fact another series, featuring only sines, that converges to $\sin^2(x)$ on the interval $[0, \pi]$.  Of course, the convergence isn't as fast ;-).

honestly, i don't think that $b_n = 0$ people. unless somebody can convince because the integral of $sin(x)^2 * sin(nx) dx$ is  not zero on $(0, \pi ) $. it actually has a value and only zero when n = 2. just saying. the post mentioned this interval, i just wondered why
most of you assumed $(-\pi, \pi)$  if this is still nonsense to you, you might as well delete this post too, mr.modulator.


Evaluate $\int_{0}^{\infty}\frac{\alpha \sin x}{\alpha^2+x^2} \mathrm{dx},\space \alpha>0$ Evaluate
$$\int_{0}^{\infty}\frac{\alpha \sin x}{\alpha^2+x^2} \mathrm{dx},\space \alpha>0$$
I thought of using Feyman way, but it doesn't seem to help that much.
Some hints, suggestions?
Thanks.

One can show that the function $$ f(\alpha) = \int_0^\infty\frac{\alpha \sin(x)}{\alpha^2+x^2}dx = \int_0^\infty \frac{\sin(\alpha x)}{1+x^2}dx $$ satisfies the differential equation $$f''(\alpha) = f(\alpha) -\frac{1}{\alpha}$$ which together with $f(0)=0$ and $\lim_{\alpha\to\infty}f(\alpha)=0$ leads to $$f(\alpha) = \frac{e^{-\alpha}\operatorname{Ei}(\alpha)-e^{\alpha}\operatorname{Ei}(-\alpha)}{2}.$$

Assuming $\alpha$ is finite
$$\int_{0}^{\infty}\frac{\alpha \sin x}{\alpha^2+x^2} \mathrm{dx},\space \alpha>0$$

Let, $tan\theta=\frac{x}{\alpha}$

$$\int_{0}^{\frac{\pi}{2}}\sin (\alpha\tan\theta) \mathrm{d\theta},\space \alpha>0$$


Condition for Implicit Function Theorem Given $f(x,y)$ continuously differentiable in a domain of the plane, $f(a,b) = 0$ and $f_x(a,b)\ne 0$. Can you show $f(x,y) = 0$ in a finite neighborhood of $(a,b)$?

Examples

Sphere tangent to the plane.
$f(a,b )= 0$ at one point but $f_x(a,b) = 0$ and $f_y(a,b) = 0$ and there is no neighborhood of $(a,b)$ in which $f(x,y)=0$.

Sphere intersecting the plane.
There are points $(a,b)$ (on the circle of intersection) where $f(a,b) =0 $ , $f_x(a,b)\ne 0$ , and $f(x,y)=0$ in a neighborhood of $(a,b)$, (circular arc through $(a,b)$ )

But how do you show this in general without geometry?
Application: Implicit Function Theorem.

Utterly false:
$$f(x,y) = x,\qquad f(0,0) = 0,\qquad f_x(0,0) = 1\ne 0.$$

Inverse Function Theorem

Given E a subset of Rn and f:E->Rn

If y=f(x), if Difj=dfj/dxi i,j=1,2,3, exist and are continuous on E, and if det (Difj) =’0, at x=a, and if b=f(a)
Then y=f(x) is invertible in a finite neighborhood of y=b.

Illustration of terminology:

y1=f1(x1,x2,x3), b1=f1(a1,a2,a3)
y2=f2(x1,x2,x3), b2=f2(a1,a2,a3)
y3=f3(x1,x2,x3), b3=f3(a1,a2,a3)

x1=g1(y1,y2,y3)
x2=g2(y1,y2,y3)
x3=g3(y1,y2,y3)

Finite Neighborhood: not just in the limit as (y1,y2,y3)->(b1,b2,b3)

Implicit Function Theorem

Let z=f(x,y)
z1=f1(x1,x2,y)
z2=f2(x1,x2,y)

Let Difj  (D3f1=df1/dy, D3f2=df2/dy) i=1,2,3 j=1,2 exist and be continuous.
Let det (D1f1,D1f2; D2f1,D2f2)=’0 at x,y=a,b where f(a,b)=0.

Consider the map F(z,w) where z=f(x,y), w=y
F1=f1(x1,x2)
F2=f2(x1,x2)
F3=y

Then Difj i,j=1,2,3 exists and is continuous. Furthermore det (DiFj)=’0. By the Inverse Function Theorem it follows that F is invertible in a neighborhood of (f(a,b),b)=(0,b):

x1=g1(F1,F2,F3)
x2=g2(F1,F2,F3)
y=y

So far so good. If F1=0 and F2=0 in a subset of the neighborhood of (f(a,b)=0,b), then

x1=g1(0,0,y)
x2=g2(0,0,y)

in a neighborhood of y=b, ie, x=g(y).
But you can’t assume this. You have to prove F1=0 and F2=0 in a subset of  a neighborhood of (0,b) under given conditions, ie, f(x,y)=0 in a subset of a neighborhood of x,y=a,b. I haven’t seen this proof anywhere.

Conclusion: The Implicit Function theorem is unproven, ie, you can’t start off a discussion of implicit functions by writing;
f1(x1,x2,x3,y1,y2)=0
f2(x1,x2,x3,y1,y2)=0
f3(x1,x2,x3,y1,y2)=0
because you don’t know that condition exists, unless you solve it in which case the Implicit Function Theorem is academic. Assuming it exists is tantamount to assuming it is invertible (under usual conditions).

Ref: Baby Rudin, Inverse and Implicit function theorem. Also 3rd Ed.


How do you prove that $[c,d]$ is not open in $\Bbb R$? We know that $(a,b)$ are open by definition. How do you prove that some arbitrary union of $(a,b)$ cannot give you $[c,d]$ ?

Because for every union of open intervals, let's call it $U$, we have the property:

  $$\forall x\in U \exists\epsilon>0: (x-\epsilon, x+\epsilon) \subseteq U$$

Or, in english,

  Each element $x$ of $U$ has some neighborhood $(x-\epsilon, x+\epsilon)$ that is completely included in $U$.

This property does not hold for $c$ and $d$ in the set $[c,d]$.

The statement above can easily be proven. Let $U$ be some union of intervals,

$$U=\bigcup_{i\in I} (a_i, b_i)$$

(note, $I$ may not be finite or even countable for this proof to work!).

Then, let $x\in U$. By definition of union, there exists some $i\in I$ such that $x\in (a_i, b_i)$. Then, set $\epsilon = \min\{\frac{x-a_i}{2}, \frac{b_i-x}{2}\}$.

We can now prove that $(x-\epsilon, x+\epsilon)\subseteq (a_i, b_i)\subseteq U$:

Let $y\in (x-\epsilon, x+\epsilon)$. Then, we know:

$$y>x-\epsilon$$
$$x-\epsilon \geq x-\frac{x-a_i}{2} =\frac{x}{2} + \frac{a_i}{2} > \frac{a}{2} + \frac{a}{2} = a $$

So we also know that $y>a_i$.

Similarly, we can show that $y<b_i$:

$$y< x+\epsilon$$
$$x+\epsilon \leq x+\frac{b_i-x}{2} = \frac{x+b_i}{2}<\frac{b_i+b_i}{2}=b_i$$

so $y<b_i$.

Together, this means that $y\in (a_i, b_i)$ and, consequently, that $(x-\epsilon, x+\epsilon)\subseteq(a_i, b_i)$.

In $\Bbb R$ both open and closed sets are $\Bbb R$ and $\phi$. Now see the complement of $[c,d]$ is $(-\infty ,c) \cup (d, + \infty)$ which is open. Hence $[c,d]$ is closed.


Why normal approximation to binomial distribution uses np> 5 as a condition I was reading about normal approximation to binomial distribution and I dunno how it works for cases when you say for example p is equal to 0.3 where p is probability of success.

On most websites it is written that normal approximation to binomial distribution works well if average is greater than 5. I.e. np> 5
But I am unable to find where did this empirical formula came from?

If n is quite large and probability of success is equal to .5 then i agree that normal approximation to binomial distribution is going to be quite accurate. But what about other cases? How can one say np> 5 is the condition for doing normal approximation?

The mean $\mu$ of a binomial = np. The standard deviation of a binomial = $\sqrt{np(1-p)}$

For a normal distribution, $\mu$ should be 3 standard deviations away from 0 and n.

Therefore:

$\mu$ - $3\sqrt{np(1-p)} > 0 \hspace{2cm}$ and $\hspace{2cm}\mu$ + $3\sqrt{np(1-p)}<n$

From that starting point, algebraically you can get to the inequalities:

$np>9(1-p)\hspace{2cm}$ and $\hspace{2cm}n(1-p)>9p$

To satisfy these inequalities, as n gets larger, p has a wider range. Or you could also say the closer p is to 0.5, the smaller n you can use.

Using n=10 (for example):

$0.474<p<0.526$

As n gets larger, p does not have to be so close to 0.5. For n = 100,

$0.0826<p<0.9174$

Remarkably, even with a p = 0.9, if n >100 then the mean will be 3 standard deviations away from 0 and n.

This relates to calculating np and n(1-p), as if both are greater than 5, usually these inequalities are satisfied. However something like n=15, p=0.65 does not work, so some textbooks say np>9.

This condition does not guarantee that the binomial will fit a normal dist. but just that the mean will not be skewed too far towards 0 or n.

For $np$ and $nq$ to increase $n$ must increase. $n$ is the number of independent trials, so it should be clear that the more independent trials made, the more accurate your approximation is.  The probability histogram approximates a normal curve pretty accurately when $np$ and $nq$ are greater(or equal to) $5$.  However bigger is better!  If $np$ and $nq$ were greater than $10$ the probability histogram would approximate the normal curve even more.


Number of invertible matrices modulo 26 The number of invertible matrices modulo $26$ can be computed by the Chinese Remainder Theorem. i.e. a matrix is invertible modulo 26 if it is invertible modulo $13$ and modulo $2$ which are given by the order of the groups $GL(n, \mathbb{Z}_{2})$ and $GL(n,\mathbb{Z}_{13})$: $$2^{n^2}(1-1/2)(1-1/2^2)\ldots(1-1/2^n)$$$$13^{n^2}(1-1/13)(1-1/13^2)\ldots(1-1/13^n)$$ and so the number of invertible $n\times n$ matrices is their product, i.e. $$26^{n^2}(1-1/2)(1-1/2^2)\ldots(1-1/2^n)(1-1/13)(1-1/13^2)\ldots(1-1/13^n)$$

My questions are, how does this use Chinese Remainder Theorem and how do the product of these quantities give us the number of matrices? What if a matrix is invertible modulo $13$ but not modulo $2$? Is it still invertible modulo $26$?

The Chinese Remainder Theorem tells us that $\mathbb{Z}_{26}\cong\mathbb{Z}_2\times\mathbb{Z}_{13}$, which in turn is what tells us that $GL(n,\mathbb{Z}_{26})\cong GL(n,\mathbb{Z}_2\times\mathbb{Z}_{13})\cong GL(n,\mathbb{Z}_2)\times GL(n,\mathbb{Z}_{13})$. Then the result follows by the proof you have in your answer.

It is definitely not the case that a matrix which is invertible modulo $13$ but not invertible modulo $2$ will be invertible modulo $26$. For example, take the matrix with all $2$'s on its diagonal and zeros everywhere else. Then this matrix is in $GL(n,\mathbb{Z}_{13})$, is not in $GL(n,\mathbb{Z}_2)$, and is not in $GL(n,\mathbb{Z}_{26})$, as $2$ does not have a multiplicative inverse in $\mathbb{Z}_{26}$.


Existence of inverse Laplace transform of $F(1/s)$ I have two questions about inverse Laplace transform.

Given a function $F(s)$, does its inverse Laplace transform always exists?

If it's not, assume $F(s)$ has an inverse Laplace transform, does the inverse Laplace transform of $Q(s) = F(1/s)$ exists?

You have to have F(s)-->0 as s-->Inf because of how the Laplace transform is defined (integrate f(t) with exp(-st) over all positive t's). As s-->Inf, the exponential factor goes to zero as s --> Inf for all positive t's, so you have to have for any Laplace-transformable f(t) the corresponding transform F(s) --> 0 as s-->Inf. This is a necessary condition, but I'm not sure if sufficient.

This precludes the existence of inverse Laplace transforms of functions such as F(s) = s, sqrt(s) etc.


Generating correlated random numbers: Why does Cholesky decomposition work? Let's say I want to generate correlated random variables. I understand that I can use Cholesky decomposition of the correlation matrix to obtain the correlated values. If $C$ is the correlation matrix, then we can do the cholesky decomposition:

$LL^{T}=C$

Then I can easily generate correlated random variables:

$LX=Y$,

where $X$ are uncorrelated values and $Y$ are correlated values. If I want two correlated random variables then $L$ is:

$L = \left[ {\begin{array}{*{20}c}
   1 & 0  \\
   \rho  & {\sqrt {1 - \rho ^2 } }  \\
\end{array}} \right]
$

I understand that this works, but I don't really understand why... My question is: Why does this work?

The co-variance matrix of any random vector $Y$ is given as $\mathbb{E} \left(YY^T \right)$, where $Y$ is a random column vector of size $n \times 1$. Now take a random vector, $X$, consisting of uncorrelated random variables with each random variable, $X_i$, having zero mean and unit variance $1$. Since $X_i$'s are uncorrelated random variables with zero mean and unit variance, we have $\mathbb{E} \left(X_iX_j\right) = \delta_{ij}$. Hence, $$\mathbb{E} \left( X X^T \right) = I$$ To generate a random vector with a given covariance matrix $Q$, look at the Cholesky decomposition of $Q$ i.e. $Q = LL^T$. Note that it is possible to obtain a Cholesky decomposition of $Q$ since by definition the co-variance matrix $Q$ is symmetric and positive definite.

Now look at the random vector $Z = LX$. We have $$\mathbb{E} \left(ZZ^T\right) = \mathbb{E} \left((LX)(LX)^T \right) = \underbrace{\mathbb{E} \left(LX X^T L^T\right) = L \mathbb{E} \left(XX^T \right) L^T}_{\text{ Since expectation is a linear operator}} = LIL^T = LL^T = Q$$ Hence, the random vector $Z$ has the desired co-variance matrix, $Q$.

The specified correlation, C, can have negative element, i.e. Negative correlation. But, C, must Be positive definite to be a proper corr. matrix. E.g. [ 1 -0.9 0.5
  -0.9 1 -0.1
  0.5 -0.1 1 ]
Is ok, but 0.9,-0.5,0.1 (in all places) produces a negative definite matrix, so not ok.
Actually, for all values positive will have the effect that the sign of the second random variable changes.


How to prove that a function is continuous? Could you give me some hint how to solve this question:

Suppose $f$ is a differentiable function for all $0<x<1$,$f(0)=1,f'(x)>0$ in the given interval.
It is obvious that $f$ is continuous for all $0<x<1$, but is it continuous at $x=0$ ?

Thanks.

No and the example is very simple:
$$
f(x)=\begin{cases}
1 & \text{if $x=0$}\\
x & \text{if $0<x<1$}
\end{cases}
$$
The function is differentiable in the open interval $(0,1)$ and its derivative is $f'(x)=1>0$ for all $x\in(0,1)$. However,
$$
\lim_{x\to0}f(x)=0\ne1=f(0)
$$
so $f$ is not continuous at $0$.

Differentiability at every point of $(0,1)$ can't imply anything about the behavior of the function at $0$, because for each $x>0$ there is $\delta>0$ such that $(x-\delta,x+\delta)\subset(0,1)$ and the differentiability of $f$ at $x$ only involves the values of $f$ in $(x-\delta,x+\delta)$, which $0$ doesn't belong to.

First, knowing the definition of a continuous function helps.

Quoting wikipedia here,

A function f(x) is continuous at point c if the limit of f(x) as x approaches c is f(c).

You have some facts stated earlier:

f(x) is differentiable in 0 < x < 1
f(0) = 1
f'(x)>0 in (0,1)

We should check the limit of f(x) as x approaches 0. The answer is going to be a number, whose derivative is >0. However, Since f'(0) = 0, we can automatically assume that

lim as f(x) approaches 0 isn't equal to f(0) because their derivatives are different.

Hence f(x) is not continuous at 0.

Hope this makes sense.


Players and tournaments 16 players take part in a tennis tournament.
The order of the matches is chosen at random.
 If there is a player better than the other one then the better one wins find the :

a)probability that all the four best place reach the semifinals

b)probability that sixth best reaches the semifinals

My attempt:

I did realise that we need to select 8 players out of 12 less better players so (12C8*8C2*6C2*4C2*2*2*2) in the numerator /16C2*8C2 in the denominator.
But I can't understand why this comes out to be >1. This means I am doing a mistake

Please help me out.

I am assuming the way the tournament works is that there is a first round with $8$ games, then the $8$ winners of the first round play a second round of $4$ games, then the winners of the second round enter the semi-final.  In order for the top $4$ players to all play in the semi-final, no two of them can play against each other in either of the first two rounds.

For the sake of exposition, let's say the top four are the players numbered $1,2,3$ and $4$.  So what is the probability that player $1$ does not play another of the top four in his first game?  There are $15$ other players, of which $3$ are also in the top $4$.  So the probability that player $1$ does not play another of the top $4$ is $12/15$.

Assuming player $1$ does not play another player in the top $4$, what is the probability that player $2$ does not play another in the top $4$?  Excluding player $1$ and his opponent, there are $13$ players left, of which player $2$ must avoid $2$.  So the probability that player $2$ does not play another in the top $4$, given that player $1$ doesn't, is $11/13$.

Proceeding in this way, we see that the probability that all top $4$ players make it through the first round is
$$\frac{12}{15} \cdot \frac{11}{13} \cdot \frac{10}{11} \cdot \frac{9}{9}$$
and the probability that all top $4$ players make it through the second round, given that they made it through the first round, is
$$\frac{4}{7} \cdot \frac{3}{5} \cdot \frac{2}{3} \cdot \frac{1}{1}$$
So the probability that the top $4$ all make it to the semi-final is
$$\frac{12}{15} \cdot \frac{11}{13} \cdot \frac{10}{11} \cdot \frac{9}{9} \cdot \frac{4}{7} \cdot \frac{3}{5} \cdot \frac{2}{3} \cdot \frac{1}{1}$$

* EDIT *

The second part of the problem asks for the probability that player $6$ survives to the semi-finals.  Let's say his portion of the tournament bracket looks like this:

6   A   B   C
|   |   |   |
+-6-+   +-?-+
  |       |
  +---6---+

Player $6$ survives to the semi-final exactly when all the other three players in his bracket of four are not in the top $6$, i.e, are in the set $\{7, 8, 9, \dots ,16\}$.  There are $\binom{15}{3}$ ways to select the other three players in the bracket.  Of these, $\binom{10}{3}$ consist entirely of players not in the top $6$.  So the probability that player $6$ survives to the semi-final is
$$\frac{\binom{10}{3}}{\binom{15}{3}} = \frac{10 \cdot 9 \cdot 8}{15 \cdot 14 \cdot 13}$$

The possible ways to organise the matches of the tournament is in bijection with permutations on $\{1, \dots, 16 \}$. Here we denote $1$ as the best player and $16$ as the worst player. Given a permutation $\sigma$, the rounds are as follows. In round one player $\sigma(2i-1)$ plays $\sigma(2i)$ for each $i \in \{1, \dots, 8 \}$ and then winner is $W_1(i) = \min(\sigma(2i-1), \sigma(2i))$. Then in round two, player $W_1(2i-1)$ plays $W(2i)$ for each $i \in \{1,2,3,4 \}$ and the winner is denoted $W_2(i)$ and so on for the remaining rounds.

In order for the four best players to reach the semi-finals, it is equivalent to saying that the four best players never play against each other before the semi-finals. These are permutations for which $\sigma^{-1}(i_1) \in \{1, \dots, 4 \}, \sigma^{-1}(i_2) \in \{5, \dots, 8 \}, \sigma^{-1}(i_3) \in \{9, \dots, 12 \}, \sigma^{-1}(i_4) \in \{1, \dots, 16 \}$ where $\{i_1, i_2, i_3, i_4\} = \{1,2,3,4 \}$.

To count such permutations, let's begin by assuming $i_1 = 1, i_2 = 2, i_3 = 3, i_4 = 4$. There are four possible options for $\sigma^{-1}(1)$, similarly for the others. Then there are $(n-4)!$ choices for the remaining entries of $\sigma^{-1}$. We could have permuted $i_1, \dots, i_4$ in one of $4!$ ways so the total number of permutations is $4^4 (n-4)! 4!$.

For the second part, you can figure our the considions you need for your permutation to satisfy in order for the sixth best player to reach the semi finals. Hint: it is the same as saying that player six never plays against players one to five in the first or second round.


Selfish Trolley Car Debate So, I'm in disagreement with my boyfriend over the following scenario: given the trolley car problem (1 person on one track and 5 people on a second track, an out of control trolley car will kill the 5 unless you pull the lever to kill the 1), he knows that one of the people is me and the other 5 are strangers. Since I'm amazing, he selfishly wants to maximize the chance that I survive.

His solution: he says he would pull the lever to save the five people. Since he doesn't know which one I am, he should save the most people to maximize the chance of saving me.

My solution: I say I wish he would put more thought into saving me. Instead of assuming a uniform distribution on my position, he should assume a uniform distribution on all distributions. So, he should pull the lever with 5/6 probability and let it kill the 5 with 1/6 probability.

So, what's the right answer? Is there even a right answer?

Update

People keep downvoting the correct answer, so I'll put it here. The correct answer is neither one of the suggestions.

The assumption of no information about my placement means there is no prior information from which to draw a probability of survival. That means there is only one strategy where this probability is even defined: to pull the lever with probability $1/2$. This trivially maximizes the survival probability ($1/2$), since it is the only probability.

This answer is meant to summarise and clarify my viewpoint as it developed in the discussion under the OP’s answer, a discussion I found fruitful; I honestly believe OP is smart, but has gotten on the wrong track with this and sees no way out; I’m trying to switch the lever here.

Firstly, unless we set up a mathematical model and give mathematical definitions to the concepts we are asking for, one cannot make any mathematical statement. OP claimed for a while that "symbols in math are just shortcuts for words", but that’s not exactly true: We are in a situation akin to the Bertrand paradox, where somebody keeps asking "but what is the probability for a randomly chosen chord", and a mathematician (OP among them, I presume) says "well that depends on what you mean by random". In our situation, the words which need a mathematical definition are "survival probability under a given strategy", and, once these are defined, what does it mean for one of them to be "optimal". OP has finally given in to at least proposing a mathematical formulation of the thought experiment, but has not given a mathematical meaning to either of these concepts, which are the ones they ask for.

Both models OP finally proposed (Aug 16) are equivalent to the one I proposed (Aug 3):

  Model. Call $p \in [0,1]$ the "prior" probability that you are positioned on the first track; call a "strategy" a $q\in [0,1]$ such that you have a method which lets the trolley go on the second track with probability $q$.

Now everyone seems to have noticed that for most strategies $q$ we cannot compute a survival probability without knowing the "prior" probability of your position, and OP has a right to insist we do not know this prior.

But what does that mean mathematically? There is a perfectly well-defined function giving out the survival probability, but it’s a function in two variables $p$ and $q$, and which OP also has finally put down in some special cases and different formulations:

  $$P_{surv}(p,q) = pq + (1-p)(1-q) = 1-p-q+2pq  \qquad (*)$$

This is the master formula one should look at here, and from which we can derive everything of interest about the model. Indeed, note that joriki’s answer just amounts to computing $P(1/6, 1/6) < P(1/6, 0)$.

Now OP complained that that answer assumed $p=1/6$, and poses the question for an "optimal" $q$ "without assuming anything about $p$". But this question is as meaningful as asking "what $x$ maximises $f(x,y) = (9-x^2)y-(3-x)y^2$"? For a function in two variables, there is no (at least I’m not aware of, I’m happy to learn otherwise) standard meaning for "optimal value of first (or second) variable".

 NB (OP does not doubt this, but I think it’s worth writing down to show the power of the model and the function $P_{surv}(p,q)$): For fixed $p=p_0$, there is a clear meaning of "optimal" $q$: Find, if it exists, a $q$ such that the function $f(q) := P(p_0,q)$ has a maximum at $q$. Just noting that this function is linear in $q$ with slope $2p_0-1$ shows, in vast generalisation of joriki’s answer:

 If we assume $p_0 < 1/2$, then $q=0$ is the unique optimal $q$ in the sense that $P_{surv}(p_0,0) > P_{surv}(p_0,q)$ for all $q \neq 0$.

If we assume $p_0 > 1/2$, then $q=1$ is the unique optimal $q$, in the sense that $P_{surv}(p_0,1) > P_{surv}(p_0,q)$ for all $q \neq 1$.

If we assume $p_0=1/2$, all $q$ are "optimal" in the sense that $P_{surv}(1/2,q)=1/2$ for all $q\in [0,1]$.

But OP wants an "optimal $q$" without further qualifications, i.e. not for a fixed $p_0$. As said, I am not aware of a standard mathematical meaning these words might have. A function in two variables can have maxima at specific points $(p_0,q_0)$, not "at a line" $q=q_0$. Of course, if need be, I can think of some ways to define such "optimal lines":

O1. First possible meaning of "optimal $q$":

  $q_0$ is called "optimal" if $P_{surv}(p,q_0) \ge P_{surv}(p,q)$ for all $(p,q) \in [0,1]^2$.

Basically meaning that there is an entire line $q=q_0$ which consists of points which all are maxima in the classical sense.

Exercise: With this definition, no optimal $q_0$ exists.

O2. Second possible meaning of "optimal $q$":

Idea: Somehow project the two-variable function $P_{surv}(p,q)$ to a one-variable function $P_{surv}(q)$, and then use the standard definition for maxima of this. Since we have no information about $p$, we could average over them. Since nothing else is given, the most natural way to do this would be to weigh all $p \in [0,1]$ equally, i.e. set $P_{surv}^{avg}(q) := \int_0^1 P_{surv}(p,q) dp$ and say that

  $q_0$ is called "optimal" if  $P^{avg}_{surv}(q_0) \ge P_{surv}^{avg}(q)$ for all $q \in [0,1]$.

Exercise: With this definition, all $q \in [0,1]$ are optimal.

O3. OP’s apparent meaning of "optimal $q$":

  $q_0$ is called "optimal" if $P_{surv}(p,q_0)$ is maximal among those $q$ for which $P_{surv}(p,q_0)$ does not depend on $p$; a set which turns out to consist of the singleton $q=1/2$.

No exercise needed: With this definition, $q_0 =1/2$ clearly is the unique optimal strategy.

The idea seems to be vaguely related to the one behind O2. OP keeps talking about a "survival probability under a given strategy", and a strategy which optimises this survival probability, so they must find a way to give meaning to these words. But these words suggest a function depending only on $q$, thus they also want to extract from the model a function $P’_{surv}(q)$ in one variable $q$, so that there would be a clear meaning of "optimal $q$". OP sees that in a very narrow subset of the model, such a thing could be defined; so they restrict to that very narrow subset. In other words: Instead of projecting the information of the model as in O2, OP wants to slice out the one line $q=1/2$ because restricted to this line, the function $P_{surv}(p,q)$ is independent of $p$ (i.e. constant). After narrowing down like this, there is nothing left to do, OP’s intended answer just pops out.

I claim, meta-mathematically, that O3 is not a reasonable use of the word "optimal". Whether it is, is a philosophical question not to be solved on this site, but let me give my metamathematical arguments:

O1 and O2 allow for some non-trivial math, whereas O3 solves the question by a definition.

Both O1 and O2, via said non-trivial math, come to a conclusion that agrees with a heuristic idea: that without any knowledge about the priors, no quantitative comparison of any strategies is possible (O1), respectively any strategy is as good as any other (O2). Whereas O3, via a very narrow definition, comes to the very unintutive idea that one specific strategy has a better outcome than others.

O1 and O2 make use of information from the entire model. O3 throws out almost the entire model in a first step. In a very vague analogy, O3 to me as an algebraist feels like looking at a subset of representatives when one should be looking at a quotient; modelling parity via $\lbrace 1,2 \rbrace \subset \mathbb Z$ instead of $\mathbb Z/2$. Or like looking at the subset of symmetric tensors instead of passing to the quotient symmetric algebra.

Related to this "slicing out" (O3) instead of "projecting" (O2), or just leaving the model as is and accepting that the question was ill-posed (O1): OP seems to insist for whatever non-mathematical reason that "not assuming a prior" must not be modelled by letting $p$ be a variable, but it must mean that no formula is allowed to contain a $p$, and declares that we are only allowed to look at $P_{surv}(p,1/2) \equiv 1/2$ (albeit they only get this by already using the two-variable function $P_{surv}(p,q)$ evaluated at $q=1/2$), and ignore all $P_{surv}(p,q)$ for $q \neq 1/2$. But what is a good model for something which contains both known and unknown quantities: something where you call the unknown quantity x (or $p$) and write down equations relating the known quantities and x, and then see what you can derive? Or: only allow formulas which do not refer to the unknown quantity at all? I think most mathematicians and scientists would go for the first option.

Added in response to comments: In a comment (Aug 16) to their own answer, OP wrote

  $p(S)=p(L)p(D)+p(R)p(P)$ is the value we seek.

This is the same as my formula $(*)$ with $p :=p(L)$ and $q:=p(D)$. (OP then notes that $p(R) = 1-p(L)$, and I assume they can also notice $p(D) =1-p(P)$, hence be able to write $pq+(1-p)(1-q)$ too.) The plain math is now that this expression is e.g.

$=p$ for $q=1$, or

$=1-p$ for $q=0$, or

$=3/4-p/2$ for $q=1/4$, and no-one doubts it is

$=1/2$ for $q=1/2$.

So e.g. my answer to "what's the survival probability if you pull the lever with certainty ($q=0$)" is "Well, if we agree on that model, it is $1-p$, and since I don't know what $p$ is, I don't know what that is." However, OP seems to know some ontological property of $p$ called "existence". This property is related to other big words, and somehow has the crucial effect that, for the case $q=1/2$, OP's derivation of their survival probability $1/2$ as $=\frac12 p + \frac12 (1-p)$ (comment to their answer, Aug 3) is allowed, but my derivation, e.g. for the case $q=0$, of $1-p$ as $=0p+1(1-p)$ makes no sense. Where naive me is mathematically tempted to say that the formula $(*)$ contains all information about the model, and simplifies to something independent of $p$ for $q=1/2$, but depends on $p$ for other values of $q$, OP writes a lot about "destruction of information" which under "maximal entropic strategies" can "force [a probability] to exist", as "the information" gets "replaced" and "resolves [the formula] into a real number" so that their

  $1/2=1/2p + 1/2(1-p) \qquad (**)$

is a profound insight resp. something totally obvious which I supposedly don't get, but my (e.g.)

  $1-p =0p+1(1-p)$

amounts to "calculating a probability with information you do not have" and gives too much credit to some "auxiliary function" which "exists" only "if $p$ exists", which it does not. (But apparently, OP's equation $(**)$ "exists" even though $p$ "does not exist".) Since I do not know enough about that ontology, have never learned to tell existing from non-existing variables, have no clue which algebraic methods are and are not allowed in the case of non-existing variables, and my metaphysics are quite rusty in general these days, I cannot argue with that any further.

Dear user144527: It’s as if you ask "what $x$ maximises $f(x,y) = (9-x^2)y-(3-x)y^2$", and while I insist that question is ill-posed, but try to come up with interpretations of it which at least involve information about the entire graph of $f$, you insist that the true answer is $x=3$, "because that’s the only $x$ for which the function does not depend on $y$, hence the only one for which my question makes sense, so that’s the solution".

Final conciliatory remark: We’re not that far apart actually. We both agree that with the strategy $q=1/2$, you survive with a probability of 50%. For any other strategy $q$, I say you survive with a probability of $pq +(1-p)(1-q)$, which is unknown because $p$ is unknown; and I cannot say whether that’s better or worse. You say, for those strategies no probability “exists”; and say that dying with a probability of 50% is better than dying with a non-existing probability. If that makes you happy, fine.

Update: In further discussion, faced with my challenge "wherever you get your $1/2$ from, that’s where I get my $pq+(1-p)(1-q)$ from", OP has found a stunning way out. Maybe granting that optimality definition O3 is ridiculously narrow, as it throws out most of the information of the model, OP has decided to -- scrap the model and instead propose the

  New Model. A "survival probability depending on a strategy" is defined as follows: It is $1/2$ for the strategy "flipping a coin", and undefined for every other strategy.

Others could argue that this amounts to optimality definition O3 with an extra grain of chutzpah, and that all my meta-mathematical arguments against O3, and more, hold against this entire model, but OP stresses they are happy I finally got their point, and indeed I cannot argue mathematically against the fact that within this model, their answer has a good chance ($\ge 50\%$ I bet) to be correct.

With no prior information, there's no probability, with the singular exception of the coin flip strategy, because this strategy sets the probability of my survival to $1/2$ independently of your prior information. Since it is the only value of the probability, it trivially maximizes the chance of survival.


Symbol for Euclidean norm (Euclidean distance) Which symbol is more commonly used to denote the Euclidean norm: $ \left \| \textbf a \right \|  $ or $ \left | \textbf b \right |$?

As mentioned above, I don't know what is most common (statistically). However, ff you have a vector $V$ space over say the real numbers $\mathbb{R}$, then you can have a norm $\|\cdot\|$ on the vector field (so you get a normed space). One thing that you would like is:
$$
\| \alpha v\| = \lvert\alpha\lvert\cdot\| v\|.
$$
for $\alpha \in \mathbb{R}$, and $v\in V$. Here the single vertical lines is the norm on the real numbers and the double lines is the norm on the vector space.

If you consider for example the real numbers as a vector space over itself, then you can use the absolute value as a norm.

If you have the vector space $V = \mathbb{R}^n$ as a vector space over the real numbers, then I do believe that the standard notation is the doube lines $\|\cdot \|$. Again, this is because you want to have the single lines for the real numbers. Note that even though the absolute value and the norm seem like the same thing, they are different because the absolute value is evaluated at real numbers, the norm of the vectors. Indeed the Euclidean norm is defined from the absolute value. So for $v = (v_1, \dots, v_n)$, you have
$$
\| v \|^2 = \lvert v_1\lvert^2 + \dots + \lvert v_n\lvert^2.
$$
Note what happens when $n = 1$, then you have $\|\cdot \| = \lvert\cdot\lvert$.

$ \| \mathbf{v} - \mathbf{u} \| $

or

$ \| \mathbf{v} - \mathbf{u} \|_2 $


Find all $n$ such that $3^{2n+1}+2^{n+2}$ is divisible by $7$ Find all $n$ such that

$3^{2n+1}+2^{n+2}$ is divisible by $7$

Prove that your answer is correct

So I am not allowed to use mods,  as is a calculus question, I have tried by induction but can't get to prove that it works for $k+1$, by multiplying the equation by powers of $2$ and $3$.

Thank you for your help

Setting $n=1$, we get $$3^{2\cdot 1+1}+2^{1+2}=35$$ above number $35$ is divisible by $7$ hence it holds for $n=1$
Assume the number $3^{2n+1}+2^{n+2}$ is divisible by $7$ for $n=k$ then $$3^{2k+1}+2^{k+2}=7\lambda \tag 1$$
Setting $n=k+1$, we get $$3^{2k+2+1}+2^{k+1+2}$$
$$=9\cdot 3^{2k+1}+2\cdot 2^{k+1}$$
$$=9\cdot 3^{2k+1}+9\cdot 2^{k+1}-7\cdot 2^{k+1}$$
$$=9(3^{2k+1}+2^{k+1})-7\cdot 2^{k+1}$$
setting the value from (1),
$$=9(7\lambda)-7\cdot 2^{k+1}$$
$$=7(9\lambda-2^{k+1})$$

since, $(9\lambda-2^{k+1})$ is some integer hence the number $7(9\lambda-2^{k+1})$ is divisible by $7$

hence the number $(3^{2n+1}+2^{n+1})$ is divisible by $7$ for all integers $n\ge 1$


Determine where a point lies in relation to a circle, is my answer right? The circle $C$ has equation
$$x^2-8x+y^2+6y=24$$
I completed the square:
$$(x-4)^2+(y+3)^2=49$$
Therefore the radius is 7. I want to know if the point $(9,2)$ is inside the circle. Substituting into the distance formula:
$$\sqrt{(9-4)^2 + (2+3)^2}=\sqrt{50}=7.07\dots$$
So the point lies outside the circle, or have I gone wrong?

Perfectly fine.

However, do note you didn't actually need to compute the square root. Since for all positive real numbers $x>y$ if and only if $x^2>y^2$, it's enough to observe that $50>49$, and that therefore $\sqrt{50}>\sqrt{49}$.

Too complicated.

Simply re-write the given equation in the form  $f(x, y) = 0$

That is,  $x^2-8x+y^2+6y-24 = 0$

$d = \sqrt {f(x_o, y_o)}$ is the "length" of tangent from $P(x_o, y_o)$ to $C:f(x, y) = 0$.

Revised:-

If  ${f(x_o, y_o)} \gt 0$, that tangent is a real one and P is outside of $C$.

If ${f(x_o, y_o)} \lt 0$, that tangent is an imaginary one and P is inside $C$.

If ${f(x_o, y_o)} = 0$; guess what.......


Induction question about triangle pattern Let n be a positive integer. An equilateral triangle is divided into $4^n$ equilateral congruent triangles and the top small triangle is cut off . Show that it can covered trapezoids formed by 3 of the small triangles.

I would start by proving the base case of $n=1$, which is clearly doable, assume $n=k$ is doable, and then prove that $n=k+1$ is also doable by showing that a triangle pattern of a smaller size can fit and the cut-off rearranged to form the trapezoid, showing that tiling it is possible. It doesn't seem that you can move the triangles and make the same case.


Are the unitary homomorphisms $\mathbb{R} \rightarrow \mathbb{R}$ and $\mathbb{C} \rightarrow \mathbb{C}$ the identity? I have some trouble with an algebraic excercise about unitary mappings.

First part

If $f:  \mathbb{R} \rightarrow \mathbb{R} $ is unitary, how to proof that $f$ can only be the identity?

My attempts so far

A unitary mapping $f:\mathbb{Z} \rightarrow \mathbb{Z}$ is the identity, because
$$
f(n) = f(1+1+ \cdots +1) = f(1)+f(1)+ \cdots +f(1)=1+1+\cdots+1=n
$$
This means that a unitary mapping $f: \mathbb{Q} \rightarrow \mathbb{Q}$ is the identity, as shown below.
$$f(q)=f(ab^{-1})=f(a)f(b)^{-1}=ab^{-1}=q$$
I didn't know how to deal with $\mathbb{R}$, but I thought:
$$x = x-\lfloor x \cdot 10^n \rfloor \cdot 10^{-n} + \lfloor x \cdot 10^n \rfloor \cdot 10^{-n}$$
The rightmost term is rational, so we can write
$$f(x)=f(x-\lfloor x \cdot 10^n \rfloor \cdot 10^{-n}) + \lfloor x \cdot 10^n \rfloor \cdot 10^{-n} $$
We can act the same way by ceiling x.
$$x = \lceil x \cdot 10^n \rceil \cdot 10^{-n} - (\lceil x \cdot 10^n \rceil \cdot 10^{-n}-x)$$

The leftmost term is rational, so $f(x) = \lceil x \cdot 10^n \rceil \cdot 10^{-n} - f(\lceil x \cdot 10^n \rceil \cdot 10^{-n}-x)$.

For large $n$ the expressions $\lceil x \cdot 10^n \rceil \cdot 10^{-n}-x$ and $x-\lfloor x \cdot 10^n \rfloor \cdot 10^{-n}$ will be small, but I don't know if $f(\lceil x \cdot 10^n \rceil \cdot 10^{-n}-x)$ and $f(x-\lfloor x \cdot 10^n \rfloor \cdot 10^{-n})$ behave the same way. Is this approach a good one or is there an easier one?

Second part

When this matter is proved, I can ask my second question: Is the only unitary mapping $f:\mathbb{C} \rightarrow \mathbb{C}$ the identity?

This is how far I came. We know that $f(a+bi) =f(a)+f(b)f(i)= a+bf(i)$, so if $f(i)=i$, we are done. But is this true? Intuitively I say yes, because we the rotating and stretching up the complex plane are not unitary, and a projection on the real axis is not a multiplacitive mapping, and I see no other options. How should  this be done?

I am looking forward to your answers guys.

As T. Bongers explained (+1), a ring homomorphism from reals to itself is the identity, because we can use squares to prove that it is order preserving, and hence continuous.

The purpose of my answer is to point out that in the case of complex numbers there are uncountably many isomorphisms! In fact the cardinality of the set of automorphisms of $\mathbb{C}$ is at least $2^{2^{\aleph_0}}$.

The catch is that if $f: \mathbb{C}\to\mathbb{C}$ is an automorphism, there is no need for $f$ to map reals to reals. A different approach is needed. Unfortunately it requires a bit more advanced machinery.

Let us pick a transcendence basis ${\cal B}$ for the extension $\mathbb{C}/\mathbb{Q}$. A cardinality argument shows that ${\cal B}$ must have continuum cardinality. Then obviously any permutation $\pi$ of ${\cal B}$ is an automorphism
of $\mathbb{Q}({\cal B})$. Basically this is because the elements of ${\cal B}$ are algebraically independent variables, and we can permute them in an unconstrained manner and view $\pi$ as a field automorphism.

The next step is to observe that $\mathbb{C}$ is algebraic over $\mathbb{Q}({\cal B})$. As $\mathbb{C}$ is algebraically closed, any automorphism $\pi$ of $\mathbb{Q}({\cal B})$ can be extended to an automorphism of $\mathbb{C}$ (actually in infinitely many ways, but let's ignore that). Therefore
$$
|Aut(\mathbb{C})|\ge |Sym({\cal B})|\ge 2^{2^{\aleph_0}}.
$$
See the ever useful Basic Algebra I, by N. Jacobson for the missing details.

let $f:\mathbb{R}\to \mathbb{R}$ be a unitary ring homomorphism, then $f(1)=1$. But this fixes $\mathbb{Q}$ as you have seen, now you can use a density theorem and you have done it.
use $f(i)=-i$, this is still unitary but not the identy. This is the complex conjugation.


What is the intuition behind the uniqueness constraint for morphisms determining products? I am following Bartosz Milewski's book and weblectures and have a question about the definition of the categorical product.

He uses names that are a bit different from the ones I can find on Google, so here is the definition. An object $c$ is the product of objects $a$ and $b$ (just denoted $c$, not $a \times b$) if it has the projections $p : c \rightarrow a$ and $q : c \rightarrow b$ such that for any object $c'$ together with projections $p' : c' \rightarrow a$ and $q' : c' \rightarrow b$ there exists an unique morphism $m : c' \rightarrow c$ that make the triangle commute, i.e. such that $p \cdot m = p'$ and $q \cdot m = q'$, i.e. "there exists only one of them that makes the triangles commute".

Now, what is the intuition behind this uniqueness constraint? I'm not looking for a counterexample, but for the "reason" behind it. What happens when we drop the uniqueness constraint? I know that if we do we satisfy all requirements for $a = a, b = b, c = a \times a \times b, c' = a \times b$, which we should not be able to. When we drop the uniqueness constraint, we allow for the situation that the morphism $m$ "makes up" something by selecting some random value for the unused $a$.

Can we state something about this in terms of bounds or entropy? I.e. something like 'a product is the "lowest-information situation" possible while still having all information available to be general enough to always satisfy the laws' or something like that? Like some greatest lower bound? An by enforcing $m$ to be unique we ensure that it cannot "add entropy/information" in any way?

Edit: I guess what I'm asking is: what is the consequence of one object having an unique mapping to another? If a morphism $a \rightarrow b$ is unique, what does this tell us about the information content of $a$ and $b$? Surely $a$ is "bigger"/contains more information than $b$, right? Doesn't $b$ even have to be a singleton in this sitution?

Edit2: OK let me reformulate my question. When we look at the object that forms the product, we are looking at the /smallest possible object/ that still satisfies all laws. Any bigger object we can "shave down" to this most basic object, and still retrieve the two components. Dually, when defining a sum, we want the most general object possible, any instantiation we can generalize. We define these greatest lower bounds and least upper bounds trough saying there must be an unique morphism between them. My question: how does the uniqueness of a morphism say anything about the size of the objects at its beginning and end?

When you define the product (not a product) of $X$ and $Y$, you are in fact defining 4 things at the same time:

$X \times Y$
$\pi_1 : X \times Y \; \to \; X$
$\pi_2 : X \times Y \; \to \; Y$
$\langle f, g \rangle$, for any $Z$, $f : Z \to X$ and $g : Z \to Y$

Those things must be unique or else this wouldn't be a definition for the product.

The unicity for the second, third and fourth things is expressed with equality. While the unicity for the first thing is expressed with isomorphism, because in category theory we are only interested in objects up to isomorphism.

I thought about it some more, and have been able to come up with an explination such that it makes sense for me.

When we relate morphisms to set theory, if we have an unique $f :: a \to b$, it means $f$ is surjective: it uses the entire codomain. As functions cannot map one element in the domain to multiple elements in the codomain, it means that $a$ is at least as big as $b$. Now, if we require no further conditions, we eventually morph into the smallest set satisfying no conditions, i.e. the empty set. However, if we set some requirements, we eventually morph into the smallest set still satisfying the conditions. This is how a product is defined: the smallest set such that you can construct both parts again, i.e. their union without any extra information. Even more specific: all elements of $a$ and $b$ must be in $a \times b$ as otherwise the conditions will not hold. Furthermore, no additional elements can be inside $a \times b$ because otherwise there would be an unique mapping from it to the set that does not contain this additional element, i.e. it would not be the smallest.

The dual applies for the sum: here we are looking for the biggest set such that it still satisfies the conditions.


If $x_0=1$ and $x_n=\frac {1}{1+x_{(n-1)}}$, find: $\lim_{x\to\infty} x_n$ If $x_0=1$ and $x_n=\dfrac {1}{1+x_{(n-1)}}$, find $\displaystyle\lim_{x\to\infty} x_n$.

My attempt:
$x_1=1+\dfrac 1 2=\dfrac 3 2$
$x_2=1+\dfrac {1}{1+\frac 3 2}=\dfrac2 5$
Which gives following series:
$$1, \frac32, \frac35,\frac 58, \frac 8 {13}, \dots$$
Since the denominator is greater than the numerator, the limit of $x_n=0$ when $x\to\infty$. I do not think this result is correct. I think this problem must have a solution using continued fractions. Any idea?

You miscalculated the two first initial terms of the sequence.
$x_{1}=\frac{1}{1+x_{0}}=\frac{1}{1+1}=\frac{1}{2}$ and $x_{2}=\frac{1}{1+x_{1}}=\frac{1}{1+\frac{1}{2}}=\frac{2}{3}$.
Let's try to apply the Fixed Point Theorem. Consider $I=[\frac{1}{2},1]$ and $f(x)=\frac{1}{1+x}$ defined in $I$.

$f$ is continuous in $I$
$f(I)\subset I$ once $f'<0$
$f$ is Lipschitz continuous because as $f$ is a function of class $C^1$ you have that $|f'(x)|=\frac{1}{(1+x)^2}\leq \frac{1}{1+\frac{1}{2}}=\frac{2}{3}$. Then for any $x,y \in I$ $\Rightarrow$ $|f(x)-f(y)|\leq\frac{2}{3}|x-y|$.

Then $x_{n}$ converges for the only fixed point of $f$ in $I$, i.e., $l=\frac{1}{1+l}$.

It is very simple Fibonacci squence
$\lim_{n\to+\infty} x_n=\lim_{n\to+\infty}\frac{F_{n-1}}{F_n}=\frac{a}{a}=1$
$a^2+a=1$;
$ F_1=1$ ;
$F_{n+1}=F_n+F_{n-1}$


Alternative to epsilon-delta definition Back in Multivariable Calculus, I remember my professor, when explaining why he decided to skip a rigorous definition of limit (the reason was that those of us that would continue in math would go over it in more depth in real analysis, and those that wouldn't continue didn't need to know it, anyway), said that modern mathematicians don't use the epsilon-delta definition that was in the book, anyway.  He alluded to some kind of set definition.  I wasn't particularly interested at the time, but now I'm wondering what he was referring to.A cursory (very cursory, in fact) search of the internet did not yield said definition.  So I thought I'd just ask.  Do you guys know what definition of limits he was talking about?

He's likely referring to the notion of a continuous map of topological spaces. A topological space is a set equipped with a set of "open sets" satisfying certain properties; familiar examples are metric spaces, for instance.

A map $f: X \rightarrow Y$ between topological spaces is said to be continuous if, for every open set $U \subset Y$, $f^{-1}(U)$ is open in $X$. To motivate this, you might want to prove that this definition is equivalent to the $\epsilon-\delta$ one for metric spaces.

In a topological space, one can say that a sequence $(x_n)$ has limit $x$ if, for every open set $U$ that contains $x$, $x_n \in U$ for $n > N$ (with $N$ sufficiently large). (Check that for metric spaces, this is equivalent to the standard definition!) This definition can be turned into one of 'continuity' as above. In general, the notion of limit of a sequence isn't well-behaved in a topological space; one needs to work more abstractly with the notion of net, which is baggage you shouldn't worry about right now. In a Hausdorff space, they behave (mostly) as they should: a sequence has at most one limit.

If you have this book available to you, I believe it uses the definition you're looking for.  It's also a great foundational book if you're thinking about getting waaay deep in math.
http://www.amazon.com/Calculus-4th-Michael-Spivak/dp/0914098918


Reduction Formula for $I_n=\int \frac{dx}{(a+b \cos x)^n}$ Reduction Formula for $$I_n=\int \frac{dx}{(a+b \cos x)^n}$$

I considered $$I_{n-1}=\int \frac{(a+b \cos x)dx}{(a+b \cos x)^n}=aI_n+b\int \frac{\cos x\:dx}{(a+b\cos x)^n}$$

Let $$J_n=\int \frac{\cos x\:dx}{(a+b\cos x)^n}$$ using parts for $J_n$ we get

$$J_n=\frac{\sin x}{(a+b \cos x)^n}-nb \int \frac{\sin^2 x\:dx}{{(a+b \cos x)^{n+1}}}$$

Can we proceed here?

If I've got my coefficients right, $$\sin^2 x =1-\cos^2 x=-\frac{1}{b^2}(a+b\cos x)^2+\frac{2a}{b^2}(a+b\cos x)+1-\frac{a^2}{b^2}.$$
Thus $$\frac{\sin x (a+b\cos x)^{-n}-J_n}{nb}=\int\frac{1-\cos^2 x}{(a+b\cos x)^{n+1}}dx=(1-\frac{a^2}{b^2})I_{n+1}+\frac{2a}{b^2}I_n-\frac{1}{b^2}I_{n-1}.$$You'll want to double-check what comes next, but I get:$$(1-n)I_{n-1}=(1-2n)aI_n+\frac{b\sin x}{ (a+b\cos x)^n}-nb^2(1-\frac{a^2}{b^2})I_{n+1}.$$

This is not a full answer the the original question but instead evidence that any such answer will not lead to the underlying problem at hand:  calculating the given integral.

Mathematica yields:

$$\frac{\csc (x) \sqrt{\frac{b (\cos (x)+1)}{b-a}} \sqrt{\frac{b-b \cos (x)}{a+b}} (a+b
   \cos (x))^{1-n} F_1\left(1-n;\frac{1}{2},\frac{1}{2};2-n;\frac{a+b \cos
   (x)}{a-b},\frac{a+b \cos (x)}{a+b}\right)}{b (n-1)}$$

which strongly suggests you won't find a simple such reduction.  So in principle there may be a way to proceed as you ask, but it seems virtually certain that such an approach will lead to a dead end.


Prove linear dependence Let $S =\{v_1,...,v_k\}$ be a subset of $T=\{v_1,...,v_m\}$. show if $S$ is linearly dependent then $T$ is linearly dependent.

I know to show that something is linearly dependent you should be able to express at least one vector as a linear combination of the others.
I don't know how to prove this though.
My thought are if I prove $S$ is linearly dependent then I can say that since each vector in $S$ is also in $T$ means that $T$ is also linearly dependent.

If $S$ is a linearly dependent set, and $S$ is a subset of $T$, then $T$ is linearly dependent. If a vector in $S$ can be represented by a linear combination of other vectors in $S$, then that same vector can be represented by the same linear combination in $T$.

Let's call $s_n$ an ordinary vector of $S$. If $s_n$ can be written as a linear combination of, for example, $s_k$ and $s_p$ (also vectors of $S$), then there is linear dependence in $S$, as stated. Note now that $s_n,s_k$ and $s_p$ also belong to $T$, since $S$ is a subset of $T$. Therefore, there is linear dependence in $T$.


Which sum shows up most often when you roll 10 dice? Which sum shows up most often when you roll 10 dice?

The sum that shows up most often when you roll 10 dice is 35.

Are you also interested to see a reason why this is so?

For each individual die, its 'expected value' is 7/2. This is because it can be either 1, 2, 3, 4, 5 or 6, each with a probability of 1/6 (assuming it's a fair die). If you want to know what value will come up most often for 1 die, this is the sum of each of it's possible values times the probability of that value. Which means 1(1/6) + 2(1/6) + 3(1/6) + 4(1/6) + 5(1/6) + 6(1/6) = 1/6(1 + 2 + 3 + 4 + 5 + 6) = 7/2. Obviously, 3.5 will never show up on a die, and this makes sense because if each side has equal probability, you shouldn't expect to see one side more than the others.

To find the expected value of multiple die, you can simply add up each individual expected value. For all 10 dice, this value is 7/2, so you have 7/2 * 10 = 35

Thus, the expected value of rolling 10 dice is 35.

Wikipedia has more information on probability and expected values, if it interests you:

http://en.wikipedia.org/wiki/Expected_value


Probability of Poisson event occuring at least 3 times in given interval At a parking garage, automobiles enter at a rate of $1$ car every $2$ minutes.
I need to find the probability that the number of automobiles entering the garage during any $2$ minute period exceeds $3$.

I know $\Pr(x \gt 3)=  1-\Pr(x\le 3)$.

So, $1-\left(e^{-1} + e^{-1} + \frac{1}{2} e^{-1} + \frac{1}{6} e^{-1}\right) \dots$.

About the statement

The Probability that the event exceeds 3 is not the same that the event happen at least 3 times.

The Probability that the event exceeds 3 is the same as the probability that the event happen at least 4 times.

Then we can perform

P(x>3) = 1 - P(x<=3)

Best Regards


What is the number of combinations of the solutions to $a+b+c=7$ in $\mathbb{N}$? My professor gave me this problem:

  Find the number of combinations of the integer solutions to the equation $a+b+c=7$ using combinatorics.

Thank you.

UPDATE

Positive solutions

If N means natural numbers then you need to be clear whether you mean positive or non-negative integers.
In either case, you can use a stars and bars calculation to get ${7-1 \choose 3-1}$ or ${7+3-1 \choose 3-1}$ respectively
It is easier to explain the non-negative case: you have seven stars and two bars (separators) which you can place in any order to get a solution such as:
$$****|**|*$$
so with $9$ elements which can vary.
For the positive case, you must start with a star and the separators are a bar followed by a star such as:
$$(*)***\underline{|*}*\underline{|*}$$
so with $6$ elements which can vary.

Is this a hoax?

Perhaps I should put that differently. What institution are you studying at?
Was that the whole question? Or was there a part two asking the same thing but =702 or some other rather bigger number?

I ask these questions because the question as posed is absolutely trivial. The only tricky point is deciding what "integers" means. If he/she really wrote just "integers" then your professor is either dumb or careless, because the answer is obviously infinite. If he didn't mean that, then it is ambiguous, as Henry pointed out.

Second, assuming he meant natural numbers, he is a bad setter of questions. Who wants to "use combinatorics" when they can just list the solutions, as Dotanooblet did! Well, if you have it at your fingertips, combinatorics is slightly faster, but in an exam I would list (if "use combinatorics" was not in the question), because it requires less thought allows one to mull over the other questions at the same time.

Third, the approach set out by Henry above is absolutely standard bookwork. There is nothing remotely tricky or hard about it.