Hitting time $h_i(k)\geqslant h_i(j)\cdot h_j(k)$ Let $(X_n)_{n\in\mathbb{N}_0}$ be a Markov chain with state space $E$. The hitting time of a set $A\subseteq E$ is a RV
    $$
H(A)\colon\Omega\to\mathbb{N}_0\cup\left\{\infty\right\},~~\omega\mapsto\inf\left\{n\in\mathbb{N}_0 | X_n(\omega)\in A\right\}.
$$
    The probability starting from $i\in E$ that $(X_n)_{n\in\mathbb{N}_0}$ ever hits $A$  is
    $$
h_i(A):=\mathbb{P}_i(H(A)<\infty).
$$
    For $i,j\in E$ we write $h_i(j):=h_i(\left\{j\right\})$.
    Prove, for all $i,j,k\in E$ that
    $$
h_i(k)\geqslant h_i(j)\cdot h_j(k).
$$

I tried to prove that using
$$
\begin{cases}h_i(A)=1, & i\in A\\h_i(A)=\sum_{j\in E}p_{ij}h_j(A), & i\notin A\end{cases}
$$

Although I do not know if I can use that here and how to continue I at least have
$$
h_i(k)=\sum_{s\in E}p_{is}h_s(k)
$$
and
$$
h_i(j)\cdot h_j(k)=\sum_{s\in E}\sum_{t\in E}p_{is}h_s(j)p_{jt}h_t(k)
$$

This follows from the (strong) Markov property of the Markov chain at the first hitting time of $j$.

Note that $h_i(k)\geqslant P_i(B)$, where $B$ denotes the event that the Markov chain hits $j$ and that, after it hits $j$, it hits $k$. Conditionally on the path of the Markov chain up to the hitting time $\tau$ of $j$, the probability to hit $k$ after $\tau$ is the probability that the Markov chain hits $k$ starting from $j$, that is, $h_j(k)$. Thus, $P_i(B)=h_i(j)\cdot h_j(k)$.

Edit 2 And again a new try...

Let $\tau:=H(j)<\infty$. Otherwise the equation is fullfilled anyway, because then $h_i(k)\geq h_i(j)h_j(k))=0$.

It is
$$
P_i(B)=P(\exists s>t>0, X_s=k,X_t=j)
$$
Furthermore
$$
h_i(j)=P_i(\exists s>0, X_s=j)=\frac{P(\exists s>0, X_s=j,X_0=i)}{P(X_0=i)}
$$
and (using the strong Markov property)
$$
h_j(k)=P_j(\exists u>0, X_u=k)\\=P(\exists u>0,X_u=k|X_0=j)\\=P(\exists u>0, X_u)k|X_{\tau}=j)\\
=\frac{P(\exists u>0, X_u=k,X_{\tau}=j,\ldots,X_0=i)}{P(X_{\tau}=j,\ldots,X_0=i)}.
$$
Thus
$$
h_i(j)h_j(k)=\frac{P(\exists s>0, X_s=j,X_0=i)}{P(X_0=i)}\cdot\frac{P(\exists u>0, X_u=k,X_{\tau}=j,\ldots,X_0=i)}{P(X_{\tau}=j,\ldots,X_0=i)}.
$$
Because
$$
P(\exists s>0, X_s=j,X_0=i)=P(X_{\tau}=j,\ldots,X_0=i)
$$
it remains
$$
h_i(j)h_j(k)=\frac{P(\exists u>0, X_u=k,X_{\tau}=j,\ldots,X_0=i)}{P(X_0=i)}\\
=P_i(\exists u>0,X_u=k,X_{\tau}=j,\ldots,X_1=a_1)=P_i(B).
$$

On the other hand it is
$$h_i(k)\geqslant P_i(B),$$
so
$$
h_i(k)\geqslant h_i(j)h_j(k).
$$


Is there any way to test the existence of left or right inverse matrix? I know that the inverse matrix of a square matrix exists iff its determinant isn't 0.

What about a non-square matrix?
Is there any theorem about the existence of a left- or right-inverse matrix of a non-square matrix?
It seems there does not exist a determinant of a non-square matrix as Determinant of a non-square matrix

If $A$ is a $m \times n$ and $rg(A) = n$, $A$ has a left inverse matrix $B$ which is a $n \times m$ matrix, thus $BA = I_n$.

If $rg(A) = m$, then $A$ has a right inverse matrix $B$, a $n \times m$ matrix. Thus $AB = I_m$.

https://math.berkeley.edu/~shinms/SP14-54/left-right-invertible-matrices.pdf
Let m, n be two positive integers.
(1) Suppose m < n (more columns than rows).
(i) There are no left invertible m × n matrices.
(ii) An m × n matrix is right invertible if and only if its REF has pivots
in every row.3
(2) Suppose m > n (more rows than columns).
(i) There are no right invertible m × n matrices.
(ii) An m × n matrix is left invertible if and only if its REF has pivots in
every column.4

Hope this helps


Four students are in a class. Compute the probability of the event Four students are in a class. Compute the probability of the event that:

1) at least one of them is born on a Saturday

2) at least 2 of them are born on the same day of the week(e.g on a Sunday)

3) all four of them are born during the weekend(i.e on Saturday or Sunday)

These are my solutions:

1) 1/7 + 1/7+ 1/7 + 1/7 = 57%

2) 1/7 * 1/7 = 2.04%

3) 2/7 * 2/7 * 2/7 * 2/7 = 0.66%

Are these correct?

Thank you

Some hints:

What is the negation of the first statement?  (If the statement is "At least one was born on a Saturday," what is the opposite of that?)
Let's call one statement $Q$ and its negation $\neg Q$.  Then, $P(Q) = 1 - P(\neg Q)$.
What is the negation of the second statement?

I was thinking of this for the first one:

53 saturdays in a year -> 4 people in a class

1/53 + 1/53 + 1/53 + 1/53 =7.5%


Maximizing growth rate in betting on multiple events Suppose we have $n$ independent events.
We know their probabilities $p_i,\dotsc,p_n$.
We are given betting odds $c_1,\dotsc,c_n$.

We can make bets to any of the events, and also any combination of events.
For every bet (be it a single or combined-event), the bookkeepers take a certain percentage $t$ from the winnings amount.

It is obvious that we can make $N=\sum_{i=1}^{n}\binom{n}{i}$ different bets.

We bet a fraction $b_i$ of our "bankroll" for the $i$-th of the N possible bets. If $b_i=0$, it means we don't make the $i$-th bet at all.

Is there a way of determining the optimal set of $b_1,\dotsc,b_N$ so that, in the long run, the "bankroll" increases maximally? (Something like the Kelly criterion)

Using the strategy described in the Wikipedia article on the Kelly Criterion, there absolutely are situations where you should place different bets on multiple different outcomes, including outcomes with a less-than-maximum Expected Value.  There are even situations where you should bet on outcomes with a negative Expected Value.

Strategy:

To summarize the strategy described in the article (for multiple, exclusive, "Many horses" event situations):

1) Order all of the possible events from most to least profitable (highest to lowest Expected Value).

2) For each event, see if the Expected Value for that event exceeds the "Reserve Rate" for your existing set of bets. (Your "Reserve Rate" is initially "1" when your set of planned bets is empty.)  If the Expected Value is higher, then add that event to your set of bets.

3) Once you have discovered your optimal set of outcomes to bet on, compute the optimal bet size for each outcome using the "Fraction to Bet" equation below (where "Fraction to Bet" is a fraction of your current bankroll.)

Reserve Rate = (1 – (sum of each probability bet on)) / (1 – (sum of each 1/payoff))

Fraction to Bet = Probability – Reserve Rate/Payoff

Example:

Here is a gambling article which gives a horse-racing example in which your optimal betting set would include a horse with a negative Expected Value.  In this example situation, betting on sub-optimal horses allow you to safely wager a much larger percentage of your overall bankroll, which enables you to maximize your long-term expected profit as defined by the Kelly Criterion.

"Independent" vs. "Exclusive" Events:

Your question deals with Independent events, and the Kelly article deals with Exclusive events. But you can still use the strategy by considering all of the Exclusive possibilities.  First compute your Reserve Rate for not betting. Then compute your Reserve Rate for the highest EV bet. Then compute the Reserve Rate for the top two highest EV bets where you separately compute the EV for each exclusive outcome (A and B, A and Not B, B and not A, etc...)

Yes, there is a way.

Like in a single bet (Kelly) where the
new bankroll $B_n$ will be (the bankroll just before this new one) $B_{n-1}$ multiplied with gain:
$$B_n=B_{n-1}   (1-u)^{1-s} (1+u(k-1))^s,$$
where $k$ is decimal odds, $u$ is invest (% of bankroll) and $s$ is the probability for win.

Thus, for two bets we have:
$$B_n=B_{n-1}  (1-u_1-u_2)^{(1-s_1)(1-s_2)} (1+u_1(k_1-1)-u_2)^{s_1(1-s_2)} \\·(1-u_1+u_2(k_2-1))^{(1-s_1)s_2} (1+u_1(k_1-1)+u_2(k_2-1))^{s_1 s_2}.$$

I hope you get the picture of what comes with three and more bets.

So, if we can find maximums - we can find optimum bet sizes. It is hard to do that kind of calculus without a computer, so I have made an algorithm that does that and have put it in a link, but the original post has unfortunately been deleted (it was .exe).

If administrator sees it appropriate - maybe he will return the link back on. Hope this helps :)
Cheers!

I have made "multiKelly" calculator so you can find your maximums.. :)

Calculator for "multiple Kelly"

Cheers!


Computing geodesics (or shortest paths) I'm interested in computing geodesic curves (or locally shortest paths) on surfaces in 3D. We can assume that the surface is given in parametric form $(u,v) \mapsto \mathbf{S}(u,v)$. If it helps, we can even assume that the parameterization is given by polynomials of fairly low degree (say 3, 4, or 5). Further, we are given two points $\mathbf{P}_0 = \mathbf{S}(u_0,v_0)$ and $\mathbf{P}_1 = \mathbf{S}(u_1,v_1)$ lying on the surface.

Two questions:

(1) I want to compute a curve of minimal length between $\mathbf{P}_0$ and $\mathbf{P}_1$, and lying in the surface.

(2) I want to compute a curve between $\mathbf{P}_0$ and $\mathbf{P}_1$, lying in the surface, on which the geodesic curvature is zero at every point.

I believe questions (1) and (2) have a close relationship, though I'm not exactly sure what it is. The relationship is not particularly important to me; I mentioned it only because it might allow us to solve both problems with one method.

The surfaces I'm dealing with are "real world" ones -- think about the roof of a car, for example, or the upper surface of the wing of an aircraft. The surfaces are very smooth, and are usually convex, if that helps. There are important applications in manufacturing, which I can explain if anyone is interested.

Edit: After some more searching, I found this paper, which seems to have what I need. The ideas are quite similar to those in the accepted answer below.

Ok, let me see if I get it right. You are interested in a computational method that finds the geodesic connecting two points. More precisely, you are looking for a reasonable discretization method, you are not necessarily looking for the differential-geometric explanation.

There is several different but equivalent (i.e. leading to the same result) characterizations of geodesics on a two dimensional surface embedded (immersed) in three space. So what I'm gonna say is by no means exhaustive of the subject.

In what follows I will use $ u = (u_1,u_2)$ instead of $(u,v)$ and a point in three space will be $x = (x_1,x_2,x_3)$. So you have your surface $S : (u_1,u_2) \mapsto (x_1,x_2,x_3)$. We mostly work in $(u_1,u_2)$ coordinates and then, when we have some solution, we map it in 3D via $S$.

First, compute the metric tensor $g_{ij}(u)$ which is a 2 by 2 matrix of function $G(u) = \big( g_{ij}(u)\big)$ with components: $$g_{ij}(u) = \left(\frac{\partial S}{\partial u_i} \cdot  \frac{\partial S}{\partial u_j}\right)$$
where $( \, \cdot \, )$ is the dot product in three space. On a surface in three space, as well as on any smooth manifold, the length of a smooth curve $u : [a,b] \to \mathbb{R}^2$ (curve given with respect to $u = (u_1,u_2)$ coordinates and parametrized as $u=u(t)$) with tangent vector (first derivatives with respect to parameter $t$) $\dot{u} = (\dot{u}_1, \dot{u}_2)^T$ is

$$\mathcal{L}[u] = \int_{a}^{b} \sqrt{\dot{u}(t)^TG\big( u(t)\big)\dot{u}(t)} \,\, dt = \int_{a}^{b} \sqrt{\dot{u}^TG\big( u\big)\dot{u}} \,\, dt$$

The distance between the points $u(a)$ and $u(b)$ is the minimum of $L[u]$ over all smooth curves connecting $u(a)$ to $u(b)$. Geodesics are curves that (locally) minimize $\mathcal{L}[u]$. Now, alongside the length functional, define the (kinetic) energy functional

$$E[u] =\frac{1}{2} \int_{a}^{b} \dot{u}(t)^TG\big( u(t)\big)\dot{u}(t) \,\, dt = \frac{1}{2}\int_{a}^{b} \dot{u}^TG\big( u\big)\dot{u} \,\, dt$$

Due to the Cauchy-Schwarz inequality, $(\mathcal{L}[u])^2 \leq 2(b-a) E[u]$ for any curve $u(t)$ with equality if and only if $\dot{u}^TG\big( u\big)\dot{u} \equiv const$ which happens when both functionals are optimized, which means that $\mathcal{L}[u]$ is optimized exactly when $E[u]$ is optimized and both are optimized by the same curve (geodesic). Bottom line, the optimizers of $E[u]$ are the geodesics so we can simply replace $\mathcal{L}[u]$ by $E[u]$.

If one denotes by $L(u,\dot{u}) =\frac{1}{2} \dot{u}^TG\big( u\big)\dot{u}$, aka the Lagriangian or the Lagrange function,
then the energy functional becomes
$$E[u] = \frac{1}{2} \int_{a}^{b}  \dot{u}^TG\big( u\big)\dot{u}  \,\, dt = \int_{a}^{b} L(u, \dot{u}) \, dt = \int_{a}^{b} L\Big(u(t), \dot{u}(t)\Big) \, dt .$$

Then, according to the calculus of variations, the optimizing curves of the functional $E$ (  which are the geodesics ) are the solutions of the Euler-Lagrange equations $$\frac{d}{dt}\Big(\, \frac{\partial L}{\partial \dot{u}}(u,\dot{u}) \, \Big) = \frac{\partial L}{\partial u}(u,\dot{u}).$$ After rearranging the terms of the latter equations, one gets the equations

\begin{align}
&\ddot{u}_1 + \sum_{ij} \Gamma^1_{ij}(u) \, \dot{u}_i\dot{u}_j =0\\
&\ddot{u}_2 + \sum_{ij} \Gamma^2_{ij}(u) \, \dot{u}_i\dot{u}_j =0
\end{align}

which are exactly the geodesic equation written in terms of the Christoffel symbols $\Gamma^k_{ij}(u)$ of the Levi-Civita connection of the Riemannian metric $G(u)$. Geometrically, they express the fact, written in $(u_1,u_2)$-coordinates, that the curvature vector of the curve $S\big(u(t)\big)$ in three space has zero projection on the tangent plane to the surface $S$ at every point of the curve $S\big(u(t)\big)$ (i.e. the normal vector of the geodesic is aligned with the normal vector of the surface). The latter projection, of the normal vector of a curve onto the tangent plane, is in fact the geodesic curvature vector, so these equations say the geodesic curvature is zero.

Now, let's go back to the functional $E[u]$ because I suggest a straight-forward discretization of the method I outlined above as follows: replace the derivative $\dot{u}$ by a difference $(u^1 - u)/\varepsilon$ (or something even better if you wish) and consider the discrete Lagrangian $$L_{\varepsilon}(u,u^1) := L\left(\, u, \, \frac{u^1-u}{\varepsilon} \, \right) = \frac{1}{2\varepsilon^2} (u^1 - u)^T G\big( u\big)(u^1 - u).$$ By analogy with the continuous case, where the action is $E[u] = \int_{a}^{b} L(u,\dot{u})dt$, in the discrete case the energy is $$E_{\varepsilon}[u] = \sum_{k=0}^{N} L_{\varepsilon}\big(u^k,u^{k+1} \big)$$  where $u^0$ is your initial point and $u^{N+1}$ is your final point. Then, the critical discrete geodesic should simply be the solution to the algebraic equations $\nabla E_{\varepsilon}[u] = 0$ which componentwise leads basically to the discrete version of the Euler-Lagrange equations $$\frac{\partial L_{\varepsilon}}{\partial u^k}\big(u^{k-1}, u^k\big) + \frac{\partial L_{\varepsilon}}{\partial u^k} \big(u^{k}, u^{k+1}\big) = 0 \,\,\, \text{ for } \,\,\,  k=1,...,N$$
$$u^0 = \text{ initial point, } \,\, u^{N+1} = \text{ finial point. }$$
The latter is a system of algebraic equations, something like $2 N$ equations and variables. The solution is a sequence of points $\,\, u^0, \, u^1, \, u^2, \, ... \, , \, u^{N+1} \,$ which should approximate the geodesic between $u^0$ and $u^{N+1}$. When you map them onto the surface in 3D via the map $S$ you get a sequence of points $\,\, S\big(u^0\big), \, S\big(u^1\big), \, S\big(u^2\big), \, ... \, , \, S\big(u^{N+1}\big) \,$
which lie on your surface and should be a good approximation to the smooth geodesic. You can interpolate between consecutive points (either the $u$'s or the $S(u)$'s) if you wish to get an actual curve.

Now, you have two parameters you need to adjust. These are $\varepsilon$ and $N$. Maybe a link between them makes sense, something like $\varepsilon = \frac{1}{N}$ or something even more sophisticated. I haven't thought about it too much. The smaller $\varepsilon$ and the larger the $N$, the better the approximation (hopefully :) ).

Now, to find discrete geodesics with a stating point and a direction vector, one can look locally at the discrete Euler-Lagrange equations and write them down using a simplified superscript notation  $$\frac{\partial L_{\varepsilon}}{\partial u}\big(u^{(-1)}, u\big) + \frac{\partial L_{\varepsilon}}{\partial u}\big(u, u^{1}\big) = 0$$ Introduce the variable $p = \frac{\partial L_{\varepsilon}}{\partial u}\big(u, u^{1}\big)$. Then the discrete Euler-Lagrange equations become $$\frac{\partial L_{\varepsilon}}{\partial u}\big(u^{(-1)}, u\big) +p= 0$$ which shifted by one subscript turn into $\frac{\partial L_{\varepsilon}}{\partial u^1}\big(u, u^1\big) + p^1= 0$. Thus we obtain the equations $$p = \frac{\partial L_{\varepsilon}}{\partial u}\big(u, u^{1}\big)$$
$$p^1 = -\frac{\partial L_{\varepsilon}}{\partial u^1}\big(u, u^1\big).$$ If one can express $u^1$ as a function of $(u,p)$ from the first equation, then the second equation also gives us $p^1$ as a function of $(u,p)$. Thus we have obtained a map $\Phi : (u,p) \mapsto (u^1,p^1)$. Observe that this is a map $\Phi : T^*\mathbb{R}^2 \to T^*\mathbb{R}^2$, which turns out to be symplectic, because the Lagrangian $L_{\varepsilon}$ is in fact a generating function of the symplectomorphism $\Phi$. Then, given an initial point $u^0$ and a direction vector $v_0$, take let's say $u^1 =u_0 + \varepsilon \, v^0 $ and obtain $p^0$. Then starting with $u = u^0$ and $p=p^0$, start iterating the map $\Phi$ getting $\big(u^{k+1}, p^{k+1}\big) = \Phi\big(u^{k}, p^{k}\big)$ obtaining again a sequence $\,\, u^0, \, u^1, \, u^2, \, ... \, , \, u^{k}, \,\, .... \,$ which approximates your geodesic (here you have another sequence $\,\, p^0, \, p^1, \, p^2, \, ... \, , \, p^{k}, \,\, .... \,$ which is a sequence somehow related to the dual vectors to the tangent vectors to your geodesic, but maybe that's not important for you).

Comment: As explained above, the discrete Euler-Lagrange equations are $\nabla E[u] = \nabla E( u^1, u^2, ..., u^k, ... u^n) = 0$, i.e. the multi-point $( u^1, u^2, ..., u^k, ... u^n)$ for which the gradient of $E( u^1, u^2, ..., u^k, ... u^n)$ is zero. Therefore, in the case of given initial and final point for the geodesic, a first line computational approach could be a gradient descent method (or a version of Newton's method). In the case of an initial point and a direction vector, simply iterate the map $\Phi$. Even if the map is not explicit, the problem is still slightly simpler because one goes step by step, every time solving only the system involving the variables $u, u^1, p, p^1$.

Comment: To make the gradient descent method a bit more efficient, one can choose a smart initial guess for the discrete geodesic $u^0,\, u^1, \, ..., \, u^{N+1}$. We have $u^0$ and $u^{N+1}$ fixed. These are the initial point and the end point. Denote by $\bar{u}(m) = \big(u^0, \, u^1(m),\, u^2(m), \, ...,\, u^{N}(m), \, u^{N+1} \big)$ a sequence of $N+2$ points at iteration $m$. Then say we use some sort of gradient descent $$\bar{u}(m+1) = \bar{u}(m) + \alpha_m \, \nabla E \big[\bar{u}(m)\big]$$ with starting point $\bar{u}(0) = \big(u^0, \, u^1(0),\, u^2(0), \, ...,\, u^{N}(0), \, u^{N+1}\big)$. If $\bar{u}(0)$ is picked carefully, then the gradient descent could make less iterations before arriving at a very good approximate solution to the problem $\nabla \, E[u] = 0$.

Suggestion 1: For instance (especially if your metric is not too curved or your points are not too far apart) take simply the segment between $u^0$ and $u^{N+1}$ and choose $N$ equally spaced points in between as a starting point $\bar{u}(0)$.

Suggestion 2: If you have some extra information, you can smartly use $u^0$ as a starting point and a second point $u^1$, then run the initial value iteration to create a discrete geodesic you can use as an initial guess for your gradient descent algorithm (or Newton's method or whatever numerical scheme you use to solve the discrete variation problem). That's why I have also discussed the initial value problem because one can use it as an auxiliary tool.

For instance, take $S(u^0)$ and $S(u^{N+1})$, the images of $u^0$ and $u^{N+1}$ on the surface $S$ in $\mathbb{R}^3$, and form the vector $\overrightarrow{u^{0}u^{N+1}}$, then project it onto the tangent plane at to the surface $S$ at the point $S(u^0)$. Write it as $$\text{proj}\Big( \overrightarrow{u^{0}u^{N+1}} \Big) = v_1 \, \frac{\partial S}{\partial u_1}(u^0) +  v_2 \, \frac{\partial S}{\partial u_2}(u^0).$$ Form the vector $v=(v_1, v_2)^T$ (a tangent vector with respect to $(u_1, u_2)$-coordinates), rescale it to become small enough, and then use it as an intial value problem to generate a discrete geodesic that comes close to $u^{N+1}$. You can even get to a reasonable choice for $N$. Then use this geodesic as a starting point for your gradient descent method in order to get a much more accurate approximation of the geodesic connecting $u^0$ to $u^{N+1}$.

HINTS:

Both are parts of the same problem. Read topic of zero geodesic curvature in differential geometry. An ODE has to be solved. There is a unique geodesic of minimum length. Experimenting with cone geodesics is convenient for theory and visualization.

To practically trace such on an airplane wing for example use a wide 3M cello tape and start sticking along an approximate direction without frills on edges. Nudge/change its start orientation, come back until there is a hit at the other point by trial and error. Like shooting a point on an hill opposite  with a gun along a curved trajectory where elevation angle is adjustable.

Numerical BVP shoot-through techniques are done in a similar way.


geodesics equation in Riemannian manifold Let $(M,g)$ be a Riemannian metric and $U \subset M$. We know that in local coordinates on $U \subset M$ the equation for a curve to be a geodesic is:

$$
0=γ ̈
=(d^2 γ^k)/(dt^2 ) ∂_k+(dγ^i)/dt   (dγ^j)/dt Γ_{ij}^k ∂_k
$$

Where $Γ_{ij}^k$ are the Christoffel symbols. This is a second-order system of differential equations.

Now are the geodesics always smooth in a Riemannian metric?
 Thanks in advance.

Note: The Einstein summation convention is in force throughout this answer.End of Note.

Our OP walaa's question,

Now are the geodesics always smooth in a Riemannian metric?

appears to be somewhat of a misstatement, insofar as smoothness is technically defined with respect to local coordinate patches of a given atlas; both the smoothness of the metric tensor $g_{ij}$ and of the Christoffel symbols $\Gamma_{ij}^k$ are then affirmed if these functions are smooth in such coordinates.  A more carefully stated question along these lines might read:

Are the geodesics of a smooth Riemannian metric smooth?

This is indeed the case, as may be seen by observing that the coefficients $\Gamma_{ij}^k$ occurring in the geodesic equation

$\ddot \gamma = \dfrac{d^2 \gamma^k}{dt^2} \partial_k + \Gamma_{ij}^k \dfrac{d\gamma^i}{dt} \dfrac{d\gamma^j}{dt} \partial_k = 0 \tag 1$

are themselves smooth functions on $M$, being given in terms of the $g_{ij}$ by

$\Gamma_{ij}^k = \dfrac{1}{2}g^{km}(g_{mi, j} + g_{mj, i} - g_{ij, m}), \tag 2$

where

$[g^{ij}] = [g_{ij}]^{-1}; \tag{2.5}$

in these equations we understand that

$\partial_k = \dfrac{\partial}{\partial x_k}, \tag 3$

and

$g_{ij, m} = \dfrac{\partial g_{ij}}{\partial x_m}, \tag 4$

and so forth.

We may cast (1) in the familiar first order form by setting

$\dfrac{d\gamma^i}{dt} = \beta^i, \; 1 \le i \le \dim M; \tag 5$

then

$\dfrac{d^2\gamma^i}{dt^2} = \dfrac{d\beta^i}{dt}; \tag 6$

(1) may then be written

$\dfrac{d\beta^k}{dt}\partial_k + \Gamma_{ij}^k \beta^i \beta^j \partial_k = 0, \tag 7$

which corresponds to the collection of $\dim M$ first order, non-linear ordinary differential equations

$\dfrac{d\beta^k}{dt} + \Gamma_{ij}^k \beta^i \beta^j \ = 0, \; 1 \le k \le \dim M, \tag 8$

that is,

$\dfrac{d\beta^k}{dt} = -\Gamma_{ij}^k \beta^i \beta^j \ = 0, \; 1 \le k \le \dim M; \tag 9$

since the $\Gamma_{ij}^k$ are functions of the position coordinates $x = (x_1, x_2, \ldots, x_{\dim M})$ in $M$, along $\gamma(t)$ we have

$\Gamma_{ij}^k(x) = \Gamma_{ij}^k(\gamma(t)) = \Gamma_{ij}^k(\gamma^1(t), \gamma^2(t), \ldots, \gamma^{\dim M}(t));\tag{10}$

(9) then becomes

$\dfrac{d\beta^k}{dt} = - \Gamma_{ij}^k(\gamma^1(t), \gamma^2(t), \ldots, \gamma^{\dim M}(t))\beta^i \beta^j = 0, \; 1 \le k \le \dim M; \tag{11}$

these together with (5) form a system of $2\dim M$ ordinary, non-linear differential equations for the $\gamma^i(t)$, $\beta^i(t)$, $1 \le i \le \dim M$.

We observe that if the functions $\Gamma_{ij}^k(\gamma^1, \gamma^2, \ldots, \gamma^{\dim M})\beta^i \beta^j$ occurring on the right-hand side of (11) are in fact $C^m$ in the $\gamma^i$, $\beta^i$, then
(5), (11) forms a $C^m$ system of non-linear, ordinary differential equations for the $\gamma^i$, $\beta^i$.  It then follows from the standard theorems that the solution functions $\gamma^i(t)$, $\beta^i(t)$, $1 \le i \le \dim M$ are themselves at least $C^m$;  since the $\Gamma_{ij}^k$ in fact depend on the first derivatives of the $g_{ij}$ (cf. (2)), we may infer that for $C^m$-smooth $g_{ij}$ the $\Gamma_{ij}^k$ are $C^{m - 1}$-smooth, and hence so the $\gamma_i(t)$, $\beta_i(t)$ are at least $C^{m - 1}$.  Since these assertions bind for any $m \ge 1$ it follows that if the metric $g_{ij}$ is $C^\infty$, so are the functions $\gamma_i(t)$, $\beta_i(t)$, $1 \le i \le \dim M$.  Thus the geodesics of a smooth Riemannian metric are themselves smooth.

The geodesic equation says that, locally, geodesics are at least $C^2$. Since the geodesic equation is invariant under coordinates changes, we find out that a geodesic is at least $C^2$ in all of its range. There is no smoothness $C^\infty$ requirement here.


How to rearrange this matrix equation to find x The equation is:
$$ A=x^{T}Bx $$
How to get matrix $x$ knowing $A$ and $B$? Should I somehow divide both sides by $B^{-1}?$

A and B as well as x are square matrices. A and B are symetric with positive diagonal values.

This doesn't have a unique solution. For example if $A=B=I$ then we just need
$$x^Tx=I$$
so $x$ can be any orthogonal matrix.

Since we know $A$ and $B$, one straightforward way is to just matrix multiply using $x$ as a matrix of variables, then use linear algebra to solve the equation.

Depending on whether some matrices are diagonal or orthonormal, we can do some additional tricks or even perform numerical computations, but without knowing the kind of the matrices, it's hard to say which way to go.

Note that matrix multiplication is not commutative, so you can't simply multiply $B^{-1}$ to eliminate $B$ in the middle.


Ratio test when checking the convergence of series Suppose we have the series $\sum a_n$.  Define,

$$
L=\lim_{n\to\infty}\frac{a_{n+1}}{a_n}
$$

Then,

if $L<1$ the series is absolutely convergent (and hence convergent).
if $L>1$ the series is divergent.
if $L=1$ the series may be divergent, conditionally convergent, or absolutely convergent.

What if $\lim_{n\to\infty} \frac{a_{n+1}}{a_n}$ doesn't exist, does it mean that series $\sum a_n$ diverges?
I think that if sometimes our ratio test doesn't work, then by checking other tests we might find that series converges. Am I right?

If I am not right, can you show me the proof. Thanks

You can't conclude anything from the fact $\lim_{n \to \infty} a_{n+1}/a_n$ does not exist.

The series may converge in that case: for example, let
$$a_n = \begin{cases}
\frac{1}{2}^n & \text{if $n$ is odd} \\
\frac{1}{3}^n & \text{if $n$ is even} \\
\end{cases}$$
In this case, $\liminf a_{n+1}/a_n = 2/3$ and $\limsup a_{n+1}/a_n = 3/2$, so $\lim a_{n+1}/a_n$ does not exist. But the series converges because $0 \leq a_n \leq 1/2^n$ for all $n$.

On the other hand, it's also possible for the series to diverge: for example, let
$$a_n = \begin{cases}
\frac{1}{2}^n & \text{if $n$ is odd} \\
3^n & \text{if $n$ is even} \\
\end{cases}$$

Clearly if $a_n=0$ for every $n\in\mathbb{N}$, the limit is undefined but the series still converges.


What exactly are the meaning of the followings in the definition of a category? In Awodey's Category Theory a category is defined as follows.

  A category consists of the following data,

  Objects: $A, B, C,\ldots$
  Arrows: $f,g,h,\ldots$
  For each arrow $f$ there are given objects, $$\operatorname{dom}(f),\ \ \operatorname{cod}(f)$$
  called the domain and codomain of $f$. We write $f : A → B$
  to indicate that $A = \operatorname{dom}(f)$ and $B = \operatorname{cod}(f)$.
  Given arrows $f : A → B$ and $g : B → C$, that is, with
  $\operatorname{cod}(f) = \operatorname{dom}(g)$ there is given an arrow
  $g \circ f : A → C$
  called the composite of $f$ and $g$.
  For each object $A$, there is given an arrow $1_A : A → A$
  called the identity arrow of A.

  These data are equired to satisfy the following laws,

  Associativity:
  $$h \circ (g \circ f)=(h \circ g) \circ f$$
  for all $f : A → B, g : B → C, h : C → D$.
  Unit: $f \circ 1_A = f = 1_B \circ f$ for all $f : A → B$.

My questions are,

What do we mean when we say that a category "consists of" something? Do the functions and objects "belong to" the category in some more general sense of $\in$ as in $\sf{ZFC}$?
Presumably, here Awodey talks of "collection of objects" and "collection of arrows" but what precisely is a collection here? Will it be sets? classes? or, something else?
What is(are) the difference(s) between data and laws?
As has been hinted in Q the Platypus's answer below, I think we may define a category to be a triple of obljects, arrows and composition. But then we need to know whether we are taking the notion of triple as a primitive notion. For if not, then naturally the question is, what is the definition of a triple?

This question is not really about category theory itself (though category theory is the first subject in which the issue you are running into cannot be easily swept under the rug). 1. and 2. could be equally well asked of set theory and basic algebra

"In what way does the collection of all sets consist of sets?"
"What are collections of sets actually?"

Slightly more subtle is 3. but it can be asked in algebra

"What is the difference between declaring a group to be a set equipped with a binary multiplication operation so that every element has an inverse, vs. declaring a group to be a set equipped with a binary multiplication operation and a unary inversion operation (both structures with the appropriate axioms)?"

The answers to 1. and 2. is that you have to set up some theory that allows you to talk about collections. The standard theory do this with is first-order logic, in which every formula with free variables is a description of a collection (aka a class), and these collections may also be described by auxillary functions and relations.

Set theory is then a particular collection with an auxillary relation called "belonging" ($\in$) that satisfies certain axioms (e.g. ZFC). We call the objects that constitute this collection sets.

Peano arithmetic is another collection, called the natural numbers, equipped with a designated object $0$ and special function called "successor" subject to certain axioms. This is different (but related) to the natural numbers considered as a set, because any set determines a collection of its elements with properties mirroring the properties of the set.

Categories in general are (or rather, can be presented as) a pair of collections in the above sense (collections described by formulas with free variables in first-order logic) known as the collections of objects and collections of morphisms, equipped with auxillary functions and relations between them (identity, domain, codomain, partially-defined composition), satisfying certain axioms (identity, associativity).

Things get complicated because if you also have a fixed set theory (e.g. ZFC), then you can build categories as the collections of elements of pairs of sets equipped with set-functions and relations that satisfy the appropriate axioms. These would be so-called small categories. A fundamental category is then the category of sets because traditional mathematics is built on top of it, so you have to study the interactions of set theory (i.e. the category of sets) with all the other categories, paying attention to which ones are small and which ones are large.

A further complication is that there is no class of all class functions, for the same reason there is no set of all sets. But there is a class of all class functions whose domain is a set, i.e. a small class. So, because categories can be large, when constructing categories and generally thinking about them you have to pay attention to size issues, to whether they are small (given by classes of elements of some set or not).

People find this to be annoying to deal with by hand, so instead they enlarge ZFC with an "axiom of universes", which asserts there is a set all of whose elements form a collection satisfying the axioms of ZFC. Such sets are then called small, the others called large, and you can then go through category theory without having to learn about first-order logic and only using this extension of ZFC in a naive fashion. Doing this, you no longer have a category of ALL sets, but you work instead with the category of small sets.

Some thoughts from a definitely non expert:

When doing mathematics, you have to first choose a logical system within which to work. The first axiomatic system introduced to mathematicians is practically always ZFC, but by no means is it the only choice.

Category theory is interested in the consequences of composition of events, and so it only defines itself in terms of what it needs:

You have some logical things called objects that you find interesting
Things happen to these objects in a compositional way that is even $more$ interesting
One possible thing that can happen to your objects is nothing

There is then, in a sense, a different category theory for each definition of "logical thing", but not all logical systems are created equal with respect to discussing categorical questions!

As an example, we certainly have logical things called sets, functions act on sets compositionally, and the identity map certainly doesn't effect a set. So we $should$ be able to talk about the category of sets, but this is a pain to do in ZFC, because, as you have experienced first hand, when you try to talk about all of sets as an object, ZFC screams in your ear, "That's not a set, that's not a set, be suspicious of this thing's existence!"

The solution is simple: if you want to ask categorical questions about ZFC(or the theory of groups, rings, modules, and many other subjects) don't use ZFC as your logical system. You could instead use a logical system that axiomatizes things larger than sets(logical things $a$ for which $a\in b$ is not defined), and then SETS(and many other interesting things!) will be within your grasp.

An example of such a system that is similar to ZFC is NBG(von Neumann Bernays Gödel) theory, wherein one begins with the notion of class as a starting point, but there are of course others.

Alternative, you could ask yourself if you even $want$ set theory when doing category theory. Set theory is, in a word, an axiomatic system used to describe the use of the symbol $\in$ in a way that hopefully doesn't, but possibly might lead to paradoxes. Since category theory tends to focus on $actions$(which compose) instead of $things$(which belong), this is rarely useful. So a simple, and seemingly often suggested course of action is: take what you've learned from set theory, and just stop asking $\in$ questions.

Either way, your worries are now no longer a concern: SETS is a reasonable object to consider, as you are either using a larger theory to study it, or you are not asking questions which presuppose the nonexistence of SETS.


What Can We Say About the Continuity of $y=\frac{x}{x}$ at $x=0$? If we can't divide by $0$, should $\frac{x}{x}$ be discontinuous and undefined at $x=0$ or is it continuous with value $1$? Most online graph calculators plot a continuous curve.

If it's continuous at $x=0$ with $y=1$, then we should be able to say that $\frac{(a-b)}{(a-b)} = 1$ at $a=b$. Or $q^2*\frac{p}{q}$ is $0$ at $q = 0$. And that whole proof of $2=1$ would hold true. The graphs online for say $\frac{x^2-4}{x-2}$ at $x=2$ are puzzling me.

Haha. That's what we call a removable discontinuity.

Now $f(x)=1$ and $g(x)=\displaystyle \frac{x}{x}$ are very similar functions, with one difference you just pointed out. (Discontinuity at $x=0$)

A similar thing for $f(x) = x-2$ and $g(x) = \displaystyle \frac{x^2-4}{x+2}$. (Discontinuity at $x=-2$.)

We notice that even at $x=-1.9999$ the values of the function are the same. We say that the limit of $g(x)$ as $x$ approaches $-2$ exists, even if, as we can see, $g(-2)$ does not exist.

A nonremovable discontinuity would be something like, $\displaystyle \frac{1}{x}$ as $x$ approaches $0$.

I hope that helps you understand. It is still a discontinuity, but since it's removable, we can sometimes patch it up.

Consider the piecewise function $h(x)$.

$h(x)=\displaystyle \begin{cases}
      \frac{x}{x} & x \neq 0 \\
      1 & x=0
   \end{cases}$

We were only able to "patch up" the discontinuity because it was removable. It was a infinitesimally small hole, not an entire asymptote like in the case of $f(x) = \displaystyle \frac{1}{x}$.

You can't divide by $0$, so the value at $0$ is undefined. The screens of graphing calculators have a finite resolution and although a hole should be in the graph, the hole might be "in between" two actual pixels and may not be visible. With high resolution graphs, even if it's displayed it may be too small to see. There are many reasons we don't rely on visuals to make deductions and this is one of them.


What is the natural domain of this function? I was wondering what the natural domain of $$(1+x)^{\frac{1}{x}}$$
was. I know it's defined on interval $x\in(-1,0)\cup (0,\infty)$. However, I'm not exactly sure what the rule is for the rest of the negative reals. Since $(1+x)$ would be negative, I know that I would need to avoid taking a square root, fourth root, etc., which would depend on the decimal that I end up choosing. Is there a rigorous way to define the domain?

I guess the question is talking about $\mathbb R$ to $\mathbb R$ function. Your main concern should be why $x > -1$, the rest is obvious.

For real number, one natural definition of exponential function could be from the exponent of a sequence of rational numbers which converges to the real number, i.e.

$$b^x := \lim_{r \in \mathbb Q \rightarrow x} b^r \, (b \in \mathbb R^+, x \in \mathbb R)$$

But the limit above only exists for positive $b$ (and is unique when $b >0$), that's why we have $x > -1$.

Another definition could be via logarithm:

$$b^x:=e^{x \cdot \ln(b)}; \, e^x:=\sum_{k=0}^\infty\frac{x^k}{k!}$$

This is more straightforward why we need $(1+x)>0$.

If $b<0$, things are more complicated: we might get complex number as results, and also we get multiple values (you could define a principal value, but identity function $(b^r)^s=b^{r\cdot s}$ will fail).

Since $(1+x)^\frac{1}{x}$ undefined for $x=0$ and also the given function  $(1+x)^\frac{1}{x}$ gives only positive value so the domain of this function is
$x\in(-1,0)\cup (0,\infty)$.
You can also verify you domain using graph


Is this homomorphism identically zero? This question was motivated by trying to generalise the derivative of $\ln x$ to a homomorphism.

Let $f: \mathbb R^+ \to \mathbb R$ be such that $f(xy)=f(x)+f(y)$ for all positive real numbers $x$ and $y$.

We can show that $f(1)=0$ by considering $f(1\cdot y)=f(1)+f(y)$.

Assume that, for example, $f(2)=0$. We can show that $f(2^q)=0$ for all rational numbers $q$.

My first instinct was that $f(x)=0$ for all $x>0$, but I'm probably wrong.

QUESTION: If $f$ is continuous for all $x>0$ then does it follow that $f(x) \equiv 0$? If not, then how about $f$ being differentiable?

EDIT: I know about the logarithm. This question was motivated by trying to find the derivative of a function that satisfies the "laws of logs". Finding the derivative of a log function requires that it be one-to-one. So I asked myself: what if it is not one-to-one? We must have $f(1)=0$, so I assumed that $f(2)=0$ and it implied that $f(2^q)=0$ for all $q \in \mathbb Q$. I then thought that perhaps, if $f$ is continuous then $f(x) \equiv 0$.

Proof that $f(2^q)=0$ for all $q \in \mathbb Q$.

We can show that $f(2^{1/k})=0$ for all positive integers $k$:

$$0 \ \ = \ \ f(2^1) \ \ = \ \ f(\underbrace{2^{1/k}\cdots 2^{1/k}}_{k \ \mathrm{times}}) \ \ = \  f(2^{1/k})+\cdots+f(2^{1/k}) \ \ = \ \ k\cdot f(2^{1/k})$$

It follows that $f(2^{-1/k})=0$ for all positive integers $k$:
$$0 \ \ = \ \ f(1) \ \ = \ \ f\left(\frac{1}{2^{1/k}} \cdot 2^{1/k}\right)\ \ = \ \ f\left(\frac{1}{2^{1/k}}\right)+f\left(2^{1/k}\right) \ \ = \ \ f\left(\frac{1}{2^{1/k}}\right)+0$$

We can show that $f(2^q)=0$ for all positive rational numbers $q=s/t$:

$$f(2^{s/t}) \ \ = \ \ f([2^{1/t}]^s) \ \ = \ \ f(\underbrace{2^{1/t} \cdots 2^{1/t}}_{s \ \mathrm{times}}) \ \ = \ \ s \cdot f(2^{1/t}) \ \ = \ \ s \cdot 0 \ \ = \ \ 0$$

It follows that $f(2^{-q})=0$ for all positive rational numbers $q=s/t$:
$$0 \ \ = \ \ f(1) \ \ = \ \ f\left(\frac{1}{2^{s/t}} \cdot 2^{s/t}\right)\ \ = \ \ f\left(\frac{1}{2^{s/t}}\right)+f\left(2^{s/t}\right) \ \ = \ \ f\left(\frac{1}{2^{s/t}}\right)+0$$

Hence $f(2^q)=0$ for all rational numbers $q$.

The endomorphisms of $\mathbb R$ are vast, they are already a lot if you consider the linear functions $\mathbb R \rightarrow \mathbb R$ when looking at $\mathbb R $ as a vector space over $\mathbb Q$, and each of these is a group homomorphism.

The homomorfisms from $\mathbb R^+ $ to $\mathbb R$ are also a lot, since if $\varphi$ is an isomorphicm on $\mathbb R$ you always have the morfism obtained via the following composition:

$\mathbb R^+ \overbrace{\rightarrow}^{\ln} \mathbb R
\overbrace{\rightarrow}^\varphi \mathbb R $

Suppose $f$ is continuous and $f(2)=0$. We can compose $f$ with the function $g(x)=e^x$. This gives us a homomorphims $f\circ g :\mathbb R \rightarrow \mathbb R$ that vanishes at some point $\alpha\neq 0$.  Using the fact that $x$ is a homomorphism we can prove that $f(\alpha/n)=0$ for all $n$ and then prove that $f(m\alpha/n)=0$ also. Since this set is dense in $\mathbb R$ and the function is continuous we get that $f(x)=0$ for all $x$. Since $g$ is surjective it follows that $f$ is $0$ at every point.

In fact it suffices for $f$ to be measurable and vanish at some point, proving this is slightly harder and uses the Steinhaus theorem in measure theory.

The multiplicative group $(\Bbb{R}^{+}, \times)$ is isomorphic to the additive group $(\Bbb{R}, +)$ via the mapping $x \mapsto \ln(x)$. So your problem is equivalent to one about additive homomorphisms: functions $h : \Bbb{R} \to \Bbb{R}$ such that $h(x + y) = h(x) + h(y).$  If such an $h$ is continuous, then it must be given by $h(x) = \lambda x$ for some $\lambda \in \Bbb{R}$ (which you can see by thinking about the values $h(m/n)$ for integer $m$ and $n$ using the homomorphism property and then using continuity). This means that if your $f$ is continuous and has $f(2) = 0$ then $f(x) = 0$ for all $x$. If you don't assume continuity, then information about $f(2)$ will not determine $f(x)$ for all values of $x$.


Determine $x$ such that $\lim\limits_{n\to\infty} \sqrt{1+\sqrt{x+\sqrt{x^2…+\sqrt{x^n}}}} = 2$ Find the value of $x$ such that $\lim\limits_{n\to\infty} \sqrt{1+\sqrt{x+\sqrt{x^2…+\sqrt{x^n}}}} = 2$

I tried getting rid of square roots and got $(...((9-x)^2-x^2)^2-...)^2-x^n = 0$ which I don't think helped. Please point me in the right direction.


How can you say it is true or false? Can we compare an equation like I did. Given that $p(x)$ has real roots. If it true, Would it also be true when roots are imaginary?
Is it true. Please help

It's not quite right. Saying $p(x)$ has two roots $\alpha$ and $\beta$ does not mean that $p(x)=(x-\alpha)(x-\beta)$. You are forgetting that any constant times $(x-\alpha)(x-\beta)$ will still have roots $\alpha$ and $\beta$. So, what you can conclude is that
$$
p(x)=c(x-\alpha)(x-\beta),
$$
where $c$ is a constant. From the expansion, it follows, in fact, that $c=a$.

To prove your statement wrong, it suffices to show a single counter-example.

$$p(x):=2x^2-6x+4$$ and $$q(x):=(x-1)(x-2)$$ both have the roots $1$ and $2$, yet $a\ne1$.


Geometrical Interpretation of Cauchy Riemann equations? Differentiation has an obvious geometric interpretation, and the Cauchy Riemann equations are closely linked with differentiation.

Do the Cauchy Riemann equations have a geometric interpretation?

Suppose $f: \mathbf{C} \rightarrow \mathbf{C}$ is analytic near $a \in \mathbf{C}$. Geometrically, this can be interpreted as saying that the effect of $f$ on a small vector $h$ emanating from $a$ is just multiplication by the complex number $f'(a)$: $$ f(a+h) \approx f(a) +f'(a)h,$$ where the error tends to zero sublinearly as $|h| \rightarrow 0$. That is, the local effect of an analytic function is a rotation plus a scaling. (There's an equivalent interpretation of the real derivative.)

In general, if we multiply $z = x + iy$ by a fixed complex number $a + ib$, we have

$$(x + iy) \mapsto (a +ib)(x+iy) =(ax-by) + i(bx+ay).$$

Thinking of $\mathbf{C}$ as $\mathbf{R}^2$, this transformation corresponds to the matrix multiplication
$$ \begin{bmatrix}a&-b\\b&a\end{bmatrix} \begin{bmatrix} x\\y\end{bmatrix} = \begin{bmatrix} ax - by\\bx+ay\end{bmatrix}.$$

Notice the form of this matrix - it's the composition of a dilation and a rotation about the origin.

Back in $\mathbf{C}$, if $u$ and $v$ are continuously differentiable mappings from $\mathbf{R}^2$ into $\mathbf{R}$, then the Jacobian $J$ of $f = u + iv$ is

$$J = \begin{bmatrix}\frac{\partial u}{\partial x}&\frac{\partial u}{\partial y}\\\frac{\partial v}{\partial x}&\frac{\partial v}{\partial y}\end{bmatrix} .$$

$J$ describes the local effect of $f$. Now, for $f$ to be analytic, the Cauchy-Riemann equations require that $u$ and $v$ satisfy $ \frac{\partial u}{\partial x} = \frac{\partial v}{\partial y}$  and $\frac{\partial u}{\partial y} = -\frac{\partial v}{\partial x} $. But if that's true, then $J$ must be a dilative rotation, like the matrix above. So the Cauchy-Riemann equations express the fact that the local effect of an analytic mapping is a dilative rotation, that is, multiplication by a complex number.

Example of action of a derivative of complex functions. If f(z)=z^2, f'(z)=2z.
For example, at z=1+i, the derivative is f'(1+i)=2(1+i). So if you apply this to something like 0.01, a differential along the horizontal axis in the complex plane, you get
2(i+1)(0.01)=0.02(1+i)=0.02+0.02i. The "2" in 2(i+1) stretches 0.01 and the 1+i turns it 45 degrees, since 0.02/0.02=1 and arctan(1)=45 degrees.  So you can approximate (1.01+i)^2 as
(1+i)^2+2(i+1)(0.01) as a first step. This has the form f(z)+f'(z)dz roughly speaking. Or in words, value of function+amplitwist * differential. (Amplitwist is from Visual Complex Analysis by Needham)


Find first five non-zero terms of power series I have the function f(x) = $\dfrac{2x}{\left(x-5\right)^2}$, and I'm supposed to "find the first five non-zero terms of power series representation centered at x = 0."

Using $\frac{f^{(n)}(0)}{n!}$, the first non-zero terms I get are:

C1 = 2/25

C2 = 4/125

C3 = 6/625

C4 = 8/3125

C5 = 2/3125

However, WebWork marks these as incorrect. Any alternative methods or corrections to the work above are gladly appreciated.

Hint:

You don't have to compute the successive derivatives. Instead rewrite the fraction as
$$\frac{2x}{\left(x-5\right)^2}=\frac{2x}{25}\frac 1{\Bigl(1-\cfrac x5\Bigr)^2}=\frac{2x}5\left(\frac 1{1-\smash[b]{\cfrac x5}}\right)'$$
and use the power series expansion $\;\dfrac1{1-u}=1+u+u^2+\dotsm$


Prove that either the average of the numbers $a_{1}, a_{2},..., a_{n}$ Let $a_{1}, a_{2},..., a_{2n}$ be positive real numbers such that $a_{j}a_{n+j}=1$ for the values $j=1,2,...,n$
($a$)  Prove that either the average of the numbers $a_{1}, a_{2},..., a_{n}$ is at least 1 or the average of the numbers $a_{n+1}, a_{n+2},..., a_{2n}$ is at least $1$
($b$)  Assuming that $n\geq 2$ prove that there exist two distinct numbers $j,k$ in the set $\left \{ 1,2,..,2n \right \}$ such that $$\left | a_{j}-a_{k} \right |< \frac{1}{n-1}$$

I tried to write part $(a)$ like this $$\frac{\sum_{i=1}^{n}a_{i}}{n}+\frac{\sum_{i=1}^{n}a_{n+i}}{n}$$
but I don't know if it's true,
the part $(b)$m I think there will be exactly $n$ values of $i$ for which $0< a_{i}\leq 1$
and by the way, this is Olympiad question, any help will thankful.

Hints:

a) It's enough to show that the average of all the numbers is at least $1$. You can do this with a famous inequality...

b) At least $n$ of the numbers must lie in $(0,1]$. Use the pigeonhole principle, with intervals of length $\frac{1}{n-1}$ as your pigeonholes.

Ok, I will complete with your solution,

Part $(a)$ $$\frac{\sum_{i=1}^{n}a_{i}}{n}+\frac{\sum_{i=1}^{n}a_{n+i}}{n}$$
$$\Rightarrow  \frac{1}{n}\sum_{i-1}^{n}\left ( a_{i}+\frac{1}{a_{i}} \right )\geq 2$$

Therefore either $$\frac{\sum_{i=1}^{n}a_{i}}{n}\geq 1$$ or $$\frac{\sum_{i=1}^{n}a_{n+i}}{n}\geq 1$$

DONE

(b) there will be exactly $n$ values of $i$ for which $0< a_{i}\leq 1$. Partition $(0,1]$ into $n-1$ intervals $(\frac{a}{n-1}, \frac{a+1}{n-1}]$
for $a=0,1,...,n-2$.

Therefore, by the Dirichlet principle, there are distinct $j,k$ such that $a_{j}, a_{k}$ are in the same half-open interval of length $1/(n-1)$ and $$\left | a_{j}-a_{k} \right |< \frac{1}{n-1}$$

AND WE'RE DONE


Why isn't $\mathbb{Z}$ an open set of the set of real numbers in the standard topology? In class my professor said that the interior of $\mathbb{Z}$ is the empty set. That is because the interior is the union of all open sets contained in $A$. But why is it not an open set?
Can someone explain (without using metric definitions because we have not covered that)?

If you take an open interval centered around an integer $n$, it looks like $(n-\varepsilon, n + \varepsilon)$ for some real number $\varepsilon$. But $(n-\varepsilon, n + \varepsilon)$ will never be contained in $\mathbf{Z}$ for any $\varepsilon$; there will always be a non-integer real number in $(n-\varepsilon, n + \varepsilon)$. So $n$, which was arbitrarily chosen, cannot be in the interior of $\mathbf{Z}$.

One point of view is that it's closed (complement is open), and if $\mathbb  Z $ were clopen, then $\Bbb R$ wouldn't be connected.


How is $x^2$ prime in this ring? Let $S$ denote the set of polynomials in $x$ with integer coefficients that have no linear term.  In other words, $S = \{ a_nx^n + a_{n-1}x^{n-1} + \cdots + a_2x^2 + a_0 \,|\, n \in \mathbb{Z}^{\geqslant 0}, a_i \in \mathbb{Z} \}.$

How is $x^2$ prime in this ring? It seems like $x^2 \mid x^6$ by $x^2 \cdot x^4 = x^6$. But $x^6 = x^3 \cdot x^3$ and $x^2$ does not divide $x^3$ in this ring. Help?

You're correct: $x^2$ is not a prime element of the ring $S$.  Indeed, this is a standard example of an element of a domain that is irreducible but not prime.

A frequent definition of primality in rings is the following:  a principal ideal $(w)$ is prime  $\iff$ $\forall u,v$ with $u.v \in (w) \implies u \in (w)$ or $v \in (w)$. Note that the the polynomials of the ideal $(x^2)$ have a zero constant term .Can you write two polynomials that $\notin (x^2)$  such that their product $\in (x^2)$?


Is there a "positive" definition for irrational numbers? An irrational number is a real number that cannot be expressed as a ratio of integers.

Is it possible to formulate a "positive" definition for irrational numbers?

A few examples:

An irrational number is a real number that can be expressed as...
An irrational number is a real number that can be expressed only as...

I cannot find any such definition which does not include the term irrational number to begin with.

Since not all irrational numbers are computable, I believe that the answer is No.

Any idea how to prove it formally?

Thanks

An irrational number is a real number that can be expressed as an infinite
 simple continued fraction.

This question is discussed in some detail on a different answer of mine: Are there real numbers that are neither rational nor irrational? See also What's an example of a number that is neither rational nor irrational?.

A small quote, with the actual definition:

  Thus there is a measurable gap between the irrationality measures of rational and irrational numbers, and this yields an alternative "constructive" definition of irrational: let $x\in\Bbb I$, read "$x$ is irrational", if $|x-p/q|<q^{-2}$ has infinitely many coprime solutions.

Let me add, to forestall confusion about the word "infinite" in the above: I mean that the set of solutions is denumerable, which is to say that there is a function which takes each solution $(p,q)$ to another solution $(p',q')$ such that $(p,q)<(p',q')$ under some suitable total order (such as lexicographic order).

The definition is equivalent to Eric Schmidt's, but the extra comments about application to intuitionistic logic are worth a read if that is your interest.


Differential equation: $y'-\tan x \cdot y=e^{2x}$ I am having difficulty solving the differential equation $y'-\tan x \cdot y=e^{2x}$.

Here is my attempt

Any help is appreciated.

Trivial.

Multiple both sides $\cos(x)$:

$\cos(x)y'-\sin(x)y = e^{2x}\cos(x)$

Notice $d\cos(x)/dx=-\sin(x)$ so we have:

$y'\cos(x)+y\cos'(x)=e^{2x}\cos(x)$

But:

$\dfrac{d(f \cdot g)}{dx} = g \cdot \dfrac{df}{dx} + f \cdot \dfrac{dg}{dx}$  is the left side of our equation

$d(y\cos(x))/dx=e^{2x}\cos(x)$

Now we finally can integrate:

$y\cos(x) = C + 1/5*e^{2x}*(\sin(x)+2\cos(x))$

P.S. Integration by parts of $e^{2x}*cos(x) dx$

First time:

$U = e^{2x}$

$dU = 2e^{2x}dx$

$V = sin(x)$

$dV = cos(x)dx$

(1)  $\int e^{2x}*cos(x) dx = e^{2x}*sin(x)-2*\int e^{2x}*sin(x) dx = $

Second time:

$U = e^{2x}$

$dU = 2e^{2x}dx$

$V = cos(x)$

$dV = -sin(x)dx$

(1) $=e^{2x}*sin(x)+2(e^{2x}*cos(x)-2*\int e^{2x}*cos(x) dx)$

Finally:

$5*\int e^{2x}*cos(x) dx = e^{2x}(\sin(x)+2\cos(x))$

$\int e^{2x}*cos(x) dx = 1/5*e^{2x}*(\sin(x)+2\cos(x)) + C$


How do I get the logarithm of a matrix using another matrix as a base? $$
        \begin{bmatrix}
        0.8 & 0.2 & 0 \\
        0 & 0.5 & 0.5 \\
        0 & 0 & 1 \\
        \end{bmatrix} = T
$$

$$
        \begin{bmatrix}
        0 & 0 & 1 \\
        0 & 0 & 1 \\
        0 & 0 & 1 \\
        \end{bmatrix} = F
$$

This is a Markov Chain for a virus (well, sick, immune) and I'm trying to find out the exact number of cycles (and thus the exact amount of time) that the virus will last.

$T^x = F$ becomes $x = \log_{T} F$. From there... I just have no clue.

The eigenvalues of your $T$ matrix are $0.8,0.5,1$
The 1 eigenvalue corresponds to the $[1,1,1]^T$ vector which is found last column in your matrix. So far so good.
When multiplying, what happens is that the eigenvalues multiply on themselves. So the eigenvalues of $T^k$ will be $0.8^k,0.5^k,1^k$ and as you know the power function will not totally zero any non-zero base, but it will get arbitrarily close as $k$ increases.

So a better question is something like "after how long time can we say with at least X% certainty that we are in this distribution?"

You can calculate the expected probability of being in last state after k hops :

 [1/3,1/3,1/3]*T^k*[0;0;1]

or
$$\frac{1}{3}\begin{bmatrix}1&1&1\end{bmatrix}\begin{bmatrix}
    0.8 & 0.2 & 0 \\
    0 & 0.5 & 0.5 \\
    0 & 0 & 1 \\
    \end{bmatrix}^{\,k}\begin{bmatrix}0\\0\\1\end{bmatrix}$$

Where 1/3 chance of starting in any position. Set k=1 and you should get 50% which you can check by hand.

What you may want to check is then to solve for when this expressions gets "close enough" to 1.

Here is a plot of the above expression for $k\in[1,32]$: As you can see, gets closer and closer but never quite hits $100\%$.


How to prove these using natural deduction I'd like to prove the following logical equivalence by using natural deduction:
$$(\exists x)(p(x) \implies q) \dashv\vdash (\forall x)(p(x)) \implies q.$$

As far as I'm concerned to show that two statements are equivalent we should show proofs for both of the sides. Can anyone help. Thanks in advance

For the first direction, $(\exists x)(p(x)\implies q) \vdash (\forall x)(p(x))\implies q$, we could reason as follows (in a natural deduction system):

$(\exists x)(p(x)\implies q)\quad {\sf P}\quad$ 1
$p(a)\implies q \quad {\sf EI(1)}\quad$ 1
$(\forall x)(p(x))\quad{\sf P}\quad$ 3
$p(a)\quad {\sf UI(3)}\quad$ 3
$q \quad {\sf MP(2,4)}\quad$ 1,3
$(\forall x)(p(x))\implies q\quad {\sf CondI(3,5)}\quad$ 1

For the converse (which you state to have proven already -- this is just for completeness' sake), we could do:

$(\forall x)(p(x)) \lor \neg (\forall x)(p(x))\quad {\sf LEM}\quad$
$(\forall x)(p(x)) \implies q\quad{\sf P}\quad$ 2
$(\forall x)(p(x))\quad{\sf P}\quad$ 3
$q\quad\sf MP(2,3)\quad$ 2,3
$p(a)\quad\sf P\quad$ 5
$p(a)\implies q\quad\sf CondI(5,4)\quad$ 2,3
$(\exists x)(p(x)\implies q)\quad\sf EG(6)\quad$ 2,3
$(\forall x)(p(x)) \implies (\exists x)(p(x)\implies q)\quad \sf CondI(3,7)\quad$ 2
$\neg(\forall x)(p(x))\quad\sf P\quad$ 9
$(\exists x)(\neg p(x))\quad\sf TI(9)\quad$ 9
$\neg p(a)\quad\sf EI(10)\quad$ 9
$p(a)\quad\sf P\quad$ 12
$\bot\quad\sf BI(11,12)\quad$ 9,12
$q\quad\sf BE(13)\quad$ 9,12
$p(a)\implies q\quad\sf CondI(12,14)\quad$ 9
$(\exists x)(p(x)\implies q)\quad\sf EG(15)\quad$ 9
$\neg(\forall x)(p(x))\implies (\exists x)(p(x)\implies q)\quad\sf CondI(9,16)$
$(\exists x)(p(x)\implies q)\quad \sf DisjE(1,8,17)\quad$ 2

Where at line 10, the (somewhat tedious to prove) equivalence between $\neg(\forall x)(p(x))$ and $(\exists x)(\neg p(x))$ was used to keep the proof reasonably short.

The abbreviations used for the respective rules should be easily readable to anyone familiar with natural deduction. If someone would appreciate them being explained explicitly, please let me know in the comments.

The lines in above derivations follow the scheme:

  Line number. Formula$\quad$Rule$\quad$Undischarged premises

Added: It is worthwhile to note that the invocation of the Law of Excluded Middle is used essentially in the second proof. The two propositions are inequivalent in intuitionistic logic (the logic which accepts everything from classical logic except LEM (and its sister DNE, $\neg\neg p \vdash p$)).

We also have proven the drinker paradox as a special case (with $q = (\forall x)(p(x))$). It's fun to read about.

Here is a similar proof using a Fitch-style proof checker:

On line 9, rather than taking the negation of the goal, I tool the negation of the material implication equivalence of the goal.  It was (perhaps) easier to reach a contradiction using that technique. Regardless, from that result on line 14, I still had to derive the desired goal.

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


Is there a quick proof as to why the vector space of $\mathbb{R}$ over $\mathbb{Q}$ is infinite-dimensional? It would seem that one way of proving this would be to show the existence of non-algebraic numbers. Is there a simpler way to show this?

The cardinality argument mentioned by Arturo is probably the simplest. Here is an alternative: an explicit example of an infinite $\, \mathbb Q$-independent set of reals. Consider the set consisting of the logs of all primes $\, p_i.\,$ If $ \, c_1 \log p_1 +\,\cdots\, + c_n\log p_n =\, 0,\  c_i\in\mathbb Q,\,$ multiplying by a common denominator we can assume that all $\ c_i \in \mathbb Z\,$ so, exponentiating, we obtain $\, p_1^{\large c_1}\cdots p_n^{\large c_n}\! = 1\,\Rightarrow\ c_i = 0\,$ for all $\,i,\,$ by the uniqueness of prime factorizations.

As $\pi$ is trascendent over $\mathbb{Q}$. Then the set $\{1, \pi, \pi^{2},\cdots\}$ is linearly independent.


Can the derivative of an absolutely continuous real function have a simple discontinuity? If $f'$ exists everywhere, then we know that it cannot have any simple discontinuities. But in this case we only know that $f'$ exists a.e. (since $f$ is absolutely continuous).

More specifically, I've also been assuming that $f'$ is increasing (on the set where it is defined). In this case, $f'(x+)$ exists everywhere, and I've been asking if there can be a point $x$ such that $f'(x)$ exists but $f'(x)<f'(x+)$. (We know this cannot happen if $f$ is differentiable on $[x, x+\epsilon)$.)

No, $f'(x)$ can  not have simple discontinuities.

Theorem

Let $f(x)$ be differentiable on some interval $(a,b)$, and continuous on $[a,b]$.

If $\lim_{x \to a^+}f'(x)$ exists, then it is equal to $f'_+(a)$ (the right derivative), same for the left derivative. (I will prove this bellow)

Consequence:

Because a simple discontinuity at $c\in(a,b)$ implies existence of sideways limits $\lim_{x \to c^+}f'(x)$ and $\lim_{x \to c^-}f'(x)$, and because $f$ is differentiable at $c \ $ $\left(f'_+(c) = f'_-(c) = f'(c)\right)$, it follows from the theorem above that

$$\lim_{x \to c^+}f'(x) = f'_+(c) = f'(c)  = f'_-(c) = \lim_{x \to c^-}f'(x)$$

And thus $f'$ cannot have a simple discontinuity.

Proof of theorem stated above:
$$
f'_+(a) = \lim_{h \to 0^+}\frac{f(a + h) - f(a)}{h},
$$

Now if you apply the MVT on the interval $[a,a+h]$,

\begin{equation}
\frac{f(a + h) - f(a)}{h} = f'(a +h\theta)
\end{equation}

where $\theta \in (0,1)$

Because $\lim_{x \to a^+}f'(x) = L$ exists,

$$
\lim_{h \to 0^+}f'(a +h\theta) = L = \lim_{h \to 0^+}\frac{f(a + h) - f(a)}{h} = f'_+(a)
$$

I believe that $f(x) = |x|$ is a counterexample. Derivative exists except at $x = 0$, but is discontinuous there.


A Question on certain Hilbert space of continuous functions, and a characteristic of convergence in it Define $T^k(\Omega)$, $\Omega$ an open subset of $\mathbb{R}^m$ (with a smooth boundary), as a space of function equivalance classes, with the norm defined as $$ \|f\|_{T^k(\Omega)}^2 =     \|f\|_{L^2(\Omega)}^2 + \|(\sum\limits_{i=1}^m(\frac{\partial^{k}f}{\partial x_i^{k}})^2)^{\frac{1}{2}}\|_{L^2(\Omega)}^2 $$

It can be easily noted that $T^k(\Omega)$ is a Hilbert space. Also note that this norm is not a Sobolev norm, as we don't consider cross derivatives.

Consider the set $$M = C^0(\bar{\Omega}) \cap T^k(\Omega)  $$

Prove that :

If $k > \frac{m}{2}$, every sequence $\{f_n\},f_n \in M$, that converges in the norm $\|.\|_{T^k(\Omega)}$, also converges in the norm $\|.\|_{C^0(\bar{\Omega})}$ (and to a limit $f \in M$)

Proof :

Consider a sequence $f_n \in M$ and let $f_n\to f \in M$ in the norm $\|.\|_{T^k(\Omega)}$ Idea is to add a small perturbation in the form of a shrinking bump, to produce a simple discontinuity in the limit function $f$. Lets add a small bump function $\psi(n\boldsymbol{x})$ to $f_n(\boldsymbol{x})$ to form a new sequence $$\phi_n(\boldsymbol{x}) = \psi(n\boldsymbol{x}) + f_n(\boldsymbol{x}) $$ Now we show that, in doing so, we blow up the norm and spoil the convergence of the sequence.
For simplicity, assume $\psi_n(\boldsymbol{x}) = \psi(n\boldsymbol{x})$ is radially symmetric. With a change of variable $\boldsymbol{t} = n\boldsymbol{x}$  we can easily see that  $$\|f_n(\boldsymbol{x}) + \psi(n\boldsymbol{x})\|_{L^2(\Omega)} \to \|f\|_{L^2(\Omega)}$$ But when we consider the other term of the norm, again with a change of variable $\boldsymbol{t} = n\boldsymbol{x}$, we can see that $$\begin{align} \int_{\Omega} |\frac{\partial^k{\phi_n}}{\partial{x_i^k}}|^2 \mathop{}\!\mathrm{d}^m\boldsymbol{x} & = \int_{\Omega}|\frac{\partial^k{f_n}}{\partial{x_i^k}}|^2 \mathop{}\!\mathrm{d}^m\boldsymbol{x} + 2\int_{\Omega}\frac{\partial^k{f_n}}{\partial{x_i^k}} \frac{\partial^k{\psi_n}}{\partial{x_i^k}}\mathop{}\!\mathrm{d}^m\boldsymbol{x} +  \int_{\Omega}|\frac{\partial^k{\psi_n}}{\partial{x_i^k}}|^2 \mathop{}\!\mathrm{d}^m\boldsymbol{x} \\\\ & = \|\frac{\partial^k{f_n}}{\partial{t_i^k}}\|_{L^2}^2 + O(n^{(k-m)}\|\frac{\partial^k{f_n}}{\partial{t_i^k}}\|_{L^2} \|\frac{\partial^k{\psi}}{\partial{t_i^k}}\|_{L^2}) +  n^{(2k-m)}\|\frac{\partial^k{\psi}}{\partial{t_i^k}}\|_{L^2}^2\end{align} $$

The last term blows up, when $k > \frac{m}{2}$. So one cannot produce a discontinuity by way of adding a shrinking bump. Hence all sequences in $M$ that converge in the norm $\|.\|_{T^k(\Omega)}$ also converge in the norm $\|.\|_{C^0(\bar{\Omega})}$

Other cases :

For a jump discontinuity, we let the bump have a flatter region and we shrink only the transition region. Same logic applies here.
For a blow up situation, consider $\phi_n(\boldsymbol{x}) = g(n)\psi(n\boldsymbol{x})$, where $g(n) = \omega(1)$ (Bachmann–Landau notations), which means $g(n)$ grows faster than 1, or $\lim\limits_{n\to\infty}g(n) = \infty$. In this case, one can see that the last term in the RHS of the last equation in my proof, blows up when $k\ge\frac{m}{2}$. Hence blow up discontinuity is also ruled out in case of $k>\frac{m}{2}$.
Case of Oscillatory discontinuity remains: For oscillatory case consider $\phi_n(\boldsymbol{x}) = \sin(n\boldsymbol{x})\psi_(n\boldsymbol{x})$

Question : Is the above proof complete?(if at all it makes sense) Is there any simpler proof with a direct application of a well known theorem? Similar references appreciated. For $\Omega = [0,1)^m$, with periodic boundary conditions, this norm is equivalent to Sobolev norm, hence by general Sobolev inequality, the result follows. But I am seeking a direct proof as in this approach, without invoking Sobolev inequality.

I don't think the proof is complete. You have a Banach space $X$ with a norm
and you have a sequence $\{x_{n}\}_{n}$ which converges to some $x$. In
particular, the sequence is bounded so $\Vert x_{n}\Vert\leq M$ for all $n$.
You are considering a sequence $\{y_{n}\}_{n}$ such that $\Vert y_{n}
\Vert\rightarrow\infty$. Then by the triangle inequality
$$
\Vert x_{n}+y_{n}\Vert\geq\Vert y_{n}\Vert-\Vert x_{n}\Vert\geq\Vert
y_{n}\Vert-M
$$
and so you obtain that
$$
\lim_{n\rightarrow\infty}\Vert x_{n}+y_{n}\Vert\geq\lim_{n\rightarrow\infty
}\Vert y_{n}\Vert-M=\infty.
$$
So you have shown that $\Vert x_{n}+y_{n}\Vert\rightarrow\infty$ but that does
not say anything about the original sequence.

The fact that in your case your norm is the sum of two $L^{2}$ norms is not
important. As you computed, if you denote $\psi_{n}(x)=\psi(nx)$, you have
$$
\Vert\psi_{n}\Vert_{T^{k}(\Omega)}\rightarrow\infty,
$$
and so by what I wrote above, $\Vert f_{n}+\psi_{n}\Vert_{T^{k}(\Omega
)}\rightarrow\infty$, but you cannot conclude anything about the original sequence.

Concerning your approach, while for functions of one variable there are simple ways to classify discontinuities (left limit and right limit exist but they are different, left limit and right limit exist and equal but different from the value of the function at the point, one of the left or right limit does not exist), for functions of several variables the situation is much more difficult since you can approach a point in a lot of different ways. In two dimensions you can approach a point along any curve passing through it. So when you say that in the oscillatory case, which I guess means
$$\liminf_{\bf{x}\to\bf{x}_0}f(\bf{x})<\limsup_{\bf{x}\to\bf{x}_0}f(\bf{x}),$$
all you know is that there are two sequences approaching $\bf{x}_0$ along which the limits are different. So your perturbation would need to be tailored on those two sequences but you have little or no information of where they are. So your "oscillation" $\sin (n \bf{x})$ (which I don't know what it means in the vectorial case) would have to follow the two sequences and so it cannot be that simple.

Also, since $f$ is an equivalence class of functions and not a function, the discontinuities change with the representative you choose. Hence, given an equivalence class of functions $[f]$, you have no way to classify the discontinuities, since given a representative you can modify it on a countable dense set and change all its discontinuity points.

As for the original problem, it seems that you would need to use the Morrey embedding theorem for the super critical case, but that would imply that you would need to prove that your space coincides with $H^k(\Omega)$. To be honest I would not know how to start to prove it and even if it is true or not (take a look at this article mixed derivatives). The closest thing I can think of for $k=2$ is an identity that relates the Laplacian to the full Hessian. See the reply
Laplacian-Hessian
but it only applies in $\mathbb{R}^m$ and you need density of smooth functions. There are version of this identity for bounded convex domains (it's in Grisvard book), but you need the function to have trace zero on the boundary and in your space there is no easy way to define a trace. So I don't really know. Sorry I cannot be of more help.

EDIT: If you are in the entire space, say $\mathbb{R}^{2}$, then using Fourier
transforms you have that
\begin{align*}
\int_{\mathbb{R}^{2}}\left\vert \frac{\partial^{2}f}{\partial x^{2}%
}\right\vert ^{2}dxdy  & =\int_{\mathbb{R}^{2}}\xi_{1}^{2}|\hat{f}(\xi_{1}%
,\xi_{2})|^{2}d\xi_{1}d\xi_{2},\\
\int_{\mathbb{R}^{2}}\left\vert \frac{\partial^{2}f}{\partial y^{2}%
}\right\vert ^{2}dxdy  & =\int_{\mathbb{R}^{2}}\xi_{2}^{2}|\hat{f}(\xi_{1}%
,\xi_{2})|^{2}d\xi_{1}d\xi_{2},
\end{align*}
and so
\begin{align*}
\int_{\mathbb{R}^{2}}\left\vert \frac{\partial^{2}f}{\partial x\partial
y}\right\vert ^{2}dxdy  & =\int_{\mathbb{R}^{2}}|\xi_{1}||\xi_{2}||\hat{f}%
(\xi_{1},\xi_{2})|^{2}d\xi_{1}d\xi_{2}\\
& \leq\frac{1}{2}\int_{\mathbb{R}^{2}}(|\xi_{1}|^{2}+|\xi_{2}|^{2})|\hat
{f}(\xi_{1},\xi_{2})|^{2}d\xi_{1}d\xi_{2}\\
& =\frac{1}{2}\int_{\mathbb{R}^{2}}\left\vert \frac{\partial^{2}f}{\partial
x^{2}}\right\vert ^{2}dxdy+\frac{1}{2}\int_{\mathbb{R}^{2}}\left\vert
\frac{\partial^{2}f}{\partial y^{2}}\right\vert ^{2}dxdy.
\end{align*}
But in a bounded domain this is no longer true. Take $f(x,y)=xy$. Then
$\frac{\partial^{2}f}{\partial x^{2}}=\frac{\partial^{2}f}{\partial x^{2}}=0$.


Ito integral's zero mean My Sto Cal prof gave a long proof for the fact that $E[\int_{0}^{t} f_s dW_s] = 0$ where W is Brownian and f is Borel x $\mathscr{F}$-measurable, adapted and satisfies some integrability condition. Apparently it follows from the fact that is is true for simple functions (the proof was actually for simple functions) and then we take the limit for general functions or something like that.

However, it was pointed out to me on another question that the zero-mean property actually follows from the fact that

if we define $I(t) \equiv \int_{0}^{t} f_s dW_s$,

then $I(0) = \int_{0}^{0} f_s dW_s = 0$

and $I(t) = \int_{0}^{t} f_s dW_s$ is a martingale.

So, I guess it then follows that $E[I(t)] = E[I(0)] = E[0] = 0$?

I suspect that my prof may not have done this because proof of the martingale property was not discussed and might come out in our exam, but anyway does the fact that $I(0)=0$ and that $I(t)$ is a martingale really imply that $E[I(t)]=0$ ?

The answer is certainly: yes. By definition of a martingale, $\mathsf E M_t = \mathsf E M_0$ (just using the tower property) and since $M_0 = 0$ in your case, all follows. Note that Ito integral may not be a martingale unless some integrability conditions are satisfied, in general it may be only a 'local martingale' - so you need to see which integrability conditions do you have.


wrongly asked question about precalculus? Im seeing the following question in a precalc textbook: Suppose $f$ is a function whose domain is $[-5,5]$ and $f(x) = \frac{x}{x+3}$ for every $x$ in $[0,5]$. Suppose $f$ is an odd function. Evaluate $f(-3)$.

Isnt this a bad formulated problem?? I mean, $f(-3)$ is not even defined, but since $f$ is odd, then $f(-3) = - f(3) = - \frac{1}{2}$.

Or, am I just misinterpreting the question?

It is not poorly formulated but may seem a bit tricky. This essentially is a piecewise function as

$f(x)=\frac{x}{x+3}: \quad x\in [0,5]$

$f(x)=-\frac{|x|}{|x|+3}: \quad x\in [-5,0)\quad\!\!$ because the function is odd.

Thus, $f(-3)=-\frac{1}{2}$ as you have mentioned.

To give a direct answer using the properties that you have likely already covered. We are given that $f$ is an odd function, meaning that it satisfies the equation

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

Hence to find $f(-3)$ we let $x=3$ and apply the above formula, yielding

$f(-3)=-f(3)$

I assume you know how to find $f(3)$ using the given formula, which defines $f$ for $x$ in $[0,5]$. (Note the negative sign on $f(3)$ in the above formula, and be sure to include it in your final answer!)


Is memorization a good skill to learn to master mathematics? I sometimes spend inordinate amounts of time memorizing math articles or theorems/proofs or formulas. My question is "am I wasting time?" and will 'active thinking' or 'working out problems' be faster way to master mathematics?

I am absolutely a beginner. So at an apprentice stage sometimes I find that best way to grasp a subject is through verbatim scribing. Also, memorization seem to be my forte.

Mathematics is a language and just like when trying to learn the basics one has to memorize grammar, does the same theory apply in this field?

I used to browse MO, this website, wikipedia but since "mathematics is not a spectator sport" I imagine more fruitful way would to be to isolate small problems and work on it?

I am sorry if the question is very general.

As you say, mathematics is a language. In any language, memorizing vocabulary very quickly becomes useless, if it is not paired with active use. Certainly, memorizing an entire Spanish dictionary before attempting to formulate one's first Spanish sentence is not the right way to learn Spanish. You should learn a few basic words, and practice using them in all sorts of combinations until you are comfortable with their properties; perhaps you must take a few things on faith from a fluent speaker. Then learn a few more words, plus maybe a grammar rule, and practice them intensively as well, using them in sentences together with the words you learned earlier, looking at correct and incorrect instances of that grammar rule. Each time, you expand more and more, bringing in more intricate aspects of the language, learning more vocabulary as a natural part of expanding your general facility with the language.

The most important part is that you practice using the language, expressing your own thoughts with it. Memorizing a dictionary will never teach you the language, nor will memorizing books or articles that use it. Reading with understanding is much better. Reading with understanding, together with using the language in conversation and writing as often as you can, is ideal. It is the best way to practice the vocabulary and grammar you've learned from a teacher or book and checking whether you really understand it. After all, consciously trying to memorize anything will only go so far; if you want to reach the point where you can use the language fluently, you have to practice it, so that each time you learn something new, it becomes ingrained, second-nature. This unconscious memorization is ultimately much more important.

In mathematics, we do not use our mouths, but our minds. Think of exercises from a book as conversation prompts. You should answer in the language of mathematics, as best you can, using the vocabulary and grammar you've learned so far. Note that knowing lots of vocabulary and grammar do  not, in and of themselves, let you participate in even the most rudimentary conversations - you have to have something to say first, and you have to practice expressing it in the language. Those are not things one can practice passively. So, without a doubt, learning mathematics requires an immense amount of active thinking.

You might find this post helpful: How Do You Go About Learning Mathematics?

Here are some relevant quotes:

"What you have been obliged to discover by yourself leaves a path in your mind which you can use again when the need arises." - G. C. Lichtenberg
"The only way to learn mathematics is to do mathematics." - Paul Halmos
"Keep in mind that there are millions of theorems but only thousands of proofs, hundreds of proof blocks, and dozens of ideas. Unfortunately, no one has figured out how to transfer the ideas directly yet, so you have to extract them from complicated arguments by yourself." - Fedja Nazarov
"Don't just read it; fight it! Ask your own questions, look for your own examples, discover your own proofs. Is the hypothesis necessary? Is the converse true? What happens in the classical special case? What about the degenerate cases? Where does the proof use the hypothesis?" - Paul Halmos

Yes to some extend you need to memories.

Example: You must remember what a prime number and a factor is before you can find prime factors of numbers. to remember that you need to memories and it obviously differs from person to person.


Proving the identity $\sum_{k=1}^n {k^3} = \big(\sum_{k=1}^n k\big)^2$ without induction I recently proved that

$$\sum_{k=1}^n k^3 = \left(\sum_{k=1}^n k \right)^2$$

using mathematical induction.  I'm interested if there's an intuitive explanation, or even a combinatorial interpretation of this property. I would also like to see any other proofs.

Stare at the following image, taken from this MO answer, long enough:

To the collection of the geometric proofs:


Showing the rationals are not complete We consider the set
$$A=\{x:x \in \mathbb{Q} \text{ and } x^2<2\} $$
We wish to show that this set has no largest member since this is all the rationals less than $\sqrt{2}$. I have been given the hint consider
$$r^2<2, r>0 \qquad 0<\delta<1, \delta<\frac{2-r^2}{2r+1} $$
We wish to show that $(r+\delta)^2<2$.

I have tried expanding this in several ways and haven't been successful

$$(r+\delta)^2=r^2+2r\delta+\delta^2$$

We might factor in the two following ways

$$r(r+2\delta)+\delta^2 \qquad \text{ or } \qquad r^2+\delta(2r+\delta)$$

Any suggestions?

Use the second factorization:
$$\begin{align}
(r+\delta)^2&=r^2+\delta(2r+\delta)\\\\
&<r^2+\delta(2r+1)\tag{$\delta<1$}\\\\
&<r^2+2-r^2\tag{$\delta<\tfrac{2-r^2}{2r+1}$}\\\\
&=2
\end{align}$$

Assume there is $y\in \mathbb{Q}$ such that $y=sup\{x\in \mathbb{Q} : x^2<2\}$. then $y^2\leq 2$. Since the square root of $2$ is not rational $y<2$, i.e. $y\in \{x\in \mathbb{Q} : x^2<2\}$. To get a contradiction we now use the tip. Let $0< \delta \in \mathbb{Q}$ be such that $\delta <\frac{2-y^2}{2y+1}$. Then certainly $y+\delta \in \mathbb{Q}$ and furthermore $(y+\delta)^2<(y+\frac{2-y^2}{2y+1})^2=(\frac{y^2+y+2}{2y+1})^2 \leq 2$. Contradiction.

Edit: The last inequality is equivalent to $(y^2-2) (y^2+2 y-1)\leq 0$ which is true because $y^2<2$ and $y\geq 1$.

Edit 2: We want to prove that $(y^2+y+2)^2 \leq 2(2y+1)^2$ expanding both sides and cancelling yields $y^4+2y^3+2\leq 3y^2+4y$. Subtracting both sides by $3y^2+4y$ and factoring one sees that this is equivalent to $(y^2-2) (y^2+2 y-1)\leq 0$.


A total of 28 percent of American males smoke cigarettes, 7 % smoke cigars, .05 smoke cigars and cigarettes. How many smoke cigars but not cigarettes? I was able to find out how many smoke neither cigars nor cigarettes as:

$P(E \cup F)$ is the event that someone smokes a cigar OR a cigarette

$P(E \cup F)^c$ is the event someone smokes neither

$(P \cup F)^c = 1-( P(E) +P(E) - P(EF))= .70$

The probability someone smokes cigars but not cigarettes could be $P(E \cap F^C)$ if I let E denote the event that someone smokes cigar and let F denote the event that a person smokes a cigarette, thus:

$P(E \cap F^C) =....$ I have no clue, this chapter two material over mutually inclusive-exclusive events and their formula and sample spaces with equally likely outcomes and the probability axioms 1-3. So if someone could explain the answer in those terms it would be great. So no Bayes formula although I think it could apply here, I would mind a side note or second part using Bayes method.

Draw a Venn diagram. On the left, you have 28% that smoke cigarettes. On the right, you have the 7% that smoke cigars. The intersection is 5%. Simple arithmetic will tell you the portions of the circles that smoke one but not the other.

Let $A$ be the set of Americans, $B$ the set of cigar smokers and $C$ the set of cigarette smokers. Then the question translates to:

  Given finite sets $A,B,C$ with $\frac{|C|}{|A|} = .28, \frac{|B|}{|A|} = .07, \frac{|B\cap C|}{|A|} = .0005$, determine $\frac{|B\setminus C|}{|A|}$.

There is absolutely no need to work with probabilities here.

Since for all disjoint finite sets $X,Y$ we have $|X\uplus Y| = |X|+|Y|$, it follows that

$$|B\setminus C| = |B| - |B\cap C| = |A|\cdot(.07-.0005)$$

Thus, $$\frac{|B\setminus C|}{|A|} = .07-.0005 = 0.0695 = 6.95\%$$

Note that the information that 28 percent smoke cigatettes is not required here. Neither are probabilities.


Find all polynomials such that $f(x)$ divides $f(2x^3+x)$ I came across an interesting problem on one of the wiki pages on brilliant.org related to polynomials and the remainder theorem. The problem statement is as follows:

Find all polynomials $f(x)$ which satisfy the following conditions:
1. $f(x)$ is a monic polynomial
2. $f(x)$ has a degree 1000
3. $f(x)$ has integer coefficients
4. $f(x)$ divides $f(2x^3+x)$

If $f(x)$ has the property that $f(a*b)=f(a)*f(b)$ for all $a,\ b$, then condition 4 is satisfied as $2x^3+x$ can be factored as $x*(2x^2+1)$. So I tried to find all polynomials with the above property.

Then, any such polynomial must also satisfy $f(cx)=f(c)f(x)$ where $c$ is a constant..
Setting $c=0$, we get $f(0)=0$ and thus there is no constant term in $f(x)$, but then this leads us to just one possibility for $f(x)$ namely, $f(x)=x^{1000}$.
In this way, we get just one solution to the given problem.

But the solution to the problem is 501, meaning there are 501 such polynomials which satisfy the stated conditions. Where do the other solutions come from??

Link to the web page(site: brilliant.org) : https://brilliant.org/problems/it-isnt-easy-for-fx-to-divide-f2x3x/
Thanks for any answers!!

$f(x)$ can't be multiplicative, polynomials aren't multiplicative though the information given is not strong enough to find $f(x)$, but since $f(x)$ divides $f(2x^3+x)$, and we're determind to find $f(x)$

Then it's easy to assume that $f(x)$ has no constant term because $2x^3+x$ factors $x\cdot(2x^2+1)$, so that $f(x)$ can be divisible

this is a list of shapes $f(x)$ can take and there are $1000$ of them

$f(x) = x^{1000}$

$f(x) = x^{1000}+ax^{999}$

$f(x) = x^{1000}+ax^{999}+bx^{998}$

$f(x) = x^{1000}+ax^{999}+bx^{998}+cx^{997}$

$f(x) = x^{1000}+ax^{999}+bx^{998}+cx^{997}+dx^{996}$

$f(x) = x^{1000}+ax^{999}+bx^{998}+cx^{997}+dx^{996}+e^{995}$

$f(x) = x^{1000}+ax^{999}+bx^{998}+cx^{997}+dx^{996}+ex^{995}+...........+x$

Where $a,b,c,d,e \in Z$,


Prove or refute $f(n) * g(n) \ge f(n) + g(n)$ I received some homework (calculus), which I can't prove:
$f(n) * g(n) \ge f(n) + g(n)$ for some $n \geq 1$ is always true $(f(n),  g(n) \ge 1)$

I think that this is true, so I need to prove that there is some $n \geq 1$ ($n$ is a natural number) for which this inequality will be true always, thanks in advance for any idea

If $f$ and $g$ are both constant functions, then it's easy to see that the claim isn't true in general. Let's assume that $g$ is not constant, and since this problem came from a calculus class, let's assume that $f$ and $g$ are differentiable. Compare the derivative of $f(x) \cdot g(x)$ with the derivative of $f(x) + g(x)$. What can you say about the relative size of the derivatives? What does that tell you about the functions?


Recurrence Relation for and $n$-gon I know there is a solution here: Combinatorics question involving recurrence relations and n-gon

I am having trouble understanding it though.

  Find a recurrence relation for the number of ways to divide an $n$-gon into triangles with non-crossing diagonals.

I know that when $n=4$, there are $2$ ways to do this.

When $n=5$, I found there are $8$ ways of doing this.

So far for $n=6$, I have $10$ ways of doing this.

I am stuck on finding the recurrence relation.

I am going to try to expand upon and clarify the linked solution for you. I want to give credit where credit is due to @metamorphy. If you give this an upvote, make sure to give them one too.

We are going to divide an $n$-gon into triangles. In any such triangulation there is always at least one triangle that has two vertices that are adjacent on the $n$-gon. This is not obvious, but I will prove it at the bottom. Let the two adjacent vertices be $A$ and $B$. In the triangulation, we'll call the third vertex of the triangle $C$. This triangle could divide the $n$-gon into two others polygons (as in the left figure below), or if $C$ is adjacent to either $A$ or $B$, the triangle $ABC$ will not divide the polygon (as in the right figure below).

$\hskip2in$

We know that $ABC$ is a triangle in our triangulation so now we just need to triangulate the remaining (one or two) polygon(s). If our choice for $C$ did $\textit{not}$ divide the polygon (like the right figure above), then the number of triangulations is just $a_{n-1}$, as the remaining polygon is just an $(n-1)$-gon. This situation occurs two different ways: if $C$ is adjacent to $B$ (as in the figure) or if $C$ is adjacent to $A$. So the number of triangulations is $$
2(a_{n-1})
$$
If our choice of $C$ $\textit{did}$ divide the polygon into two remaining polygons, then we must triangulate the remaining polygons. The polygon's sides will add up to $(n+1)$, but depending on our choice of $C$ this could divide the $n$-gon into an $(n-2)$-gon and a $3$-gon, an $(n-3)$-gon and a $4$-gon, ... , or a $3$-gon and a $(n-2)$-gon. These are all distinct possible triangulations, so the number of triangulations is $$a_{n-2}a_3+a_{n-3}a_4+...+a_4a_{n-3}+a_3a_{n-2}$$
Adding these two possibilities  we get $$
a_n=2(a_{n-1})+(a_{n-2}a_3+a_{n-3}a_4+...+a_4a_{n-3}+a_3a_{n-2})
$$
If we define $a_2=1$ and $a_1=0$ (it makes no sense to triangulate a shape with two or less sides since these shapes don't exists), then we can rearrange the sum into$$
a_n=a_{n-1}a_2+a_{n-2}a_3+...+a_3a_{n-2}+a_2a_{n-1}=\sum^{n-1}_{i=2}a_ia_{n-i+1}
$$
These are a well studied group of numbers called the Catalan numbers $C_n$. Here,
$$
a_n=C_{n-2}
$$

If we  choose one vertex against one side, reminders are $n-i-1$ and $i-1$ polygons. From this, we get asked recurrence.


A question about surjective functions. I am looking for a sample of surjective function $f:X \to Y$ and a set $A \subseteq X$ such that $f^{-1}(f(A))\neq A$.

Is the  sample $f(x)=x^2, f^{-1}(x)=\sqrt{x}, X=\mathbb{R}, Y=[0, +\infty), A=[-1,1]$  a correct one?

By definition, $\mathrm{f}^{-1}$ is the function for which $(\mathrm{f}^{-1}\! \circ \mathrm{f})(x) = x$ for all $x$ in the domain of $\mathrm{f}$. As a result, $(\mathrm{f}^{-1} \! \circ \mathrm{f})(A) = A$ for all $A \subseteq X$.

For $\mathrm{f}^{-1}$ to exist, you need $\mathrm{f} : X \to \mathrm{f}(X)$ to be injective. If it were many-to-one, e.g. $\mathrm{f}(x_1) = \mathrm{f}(x_2)=y$ with $x_1 \neq x_2$, then how would you define $\mathrm{f}^{-1}(y)$? Is it $x_1$ or is it $x_2$? It can't be both because a function cannot be one-to-many. (A mapping can be, but not a function.)

In your example $\mathrm{f} : [-1,1] \to [0,1]$ is no injective and so there is no well-defined inverse.

The inverse of $\mathrm{f} : x \mapsto x^2$ is given by $x \mapsto \sqrt{x}$ only when the domain of $\mathrm{f}$ is restricted to a subset of the non-negative real numbers. The inverse of $\mathrm{f} : x \mapsto x^2$ is given by $x \mapsto -\sqrt{x}$ only when the domain of $\mathrm{f}$ is restricted to a subset of the non-positive real numbers.


How to use Borel-Cantelli specifically to show that the probability of a simple random walk returning to the origin in finite time is 1? Suppose we have that $X_i$ are iid random variables with $P(X_i =1) = P(X_i = -1) = 1/2$ and that $X_0 = 0$. Then, we define the simple symmetric random walk to be $S_n = \sum_{i=1}^n X_i$. We define the hitting time as $\tau_{i,j} = min\{n \geq 1 : S_n = j \ |\ S_0 = i\}$. I would like to show that $P(\tau_{0,0} < \infty) = 1$.

Specifically, I would like to use the Borel-Cantelli lemma to show that if I can define some independent sequence of events here, and show that the sum of the events is infinite, then moves of some finite length happen infinitely often, and so I can show that it must return to $0$. However, I am not sure how to define such an indexed event. Could someone give me a hint? Thank you

My reputation is under 50, hence I cannot add comments to contact the user jwg. So this is the only way to warn that, IMHO, the above solution of jwg is probably wrong. He states that

  So the events that it reaches $n$ for the first time, and then reaches $0$
  before $n+1$, are disjoint and independent.

Yes, they are disjoint, which implies that $\sum_{n=1}^\infty P(A_n) = P(\cup_{n=1}^\infty A_n) \leq 1$, which contradicts

  The sum of their probabilities is infinite...

Another problem is the independence:

Independence of, say, $A_2$ and $A_6$ would mean that $P(A_2 \cap A_6) = P(A_2) P(A_6)$. However, the left-hand side is zero (since $A_2$ and $A_6$ are disjoint), while the right-hand side is positive (since  $P(A_n)>0$ for all $n$).

It will be exremely enlightening, if jwg provides a more detailed reasoning. The possibility of proving the posted problem just by Borel-Cantelli makes me very curious, since I have never seen that kind of solution.

As pointed out in comments, there could be good ways of doing this using not Borel-Cantelli, but the properties of either Markov Chains or (I would recommend) martingales, since your random walk is both a Markov chain and a martingale.

However, there is a neat argument which uses the Borel-Cantelli lemma. Suppose that the random walk is at $n$. The probability of reaching $0$ before reaching $n+1$ is $\frac{1}{n+1}$.

For the random walk to go back to $0$, it must reach some maximum $k$, then go back $0$ before ever reaching $k+1$. (Or the same argument by symmetry if it goes to $-1$ before $1$.) So the events that it reaches $n$ for the first time, and then reaches $0$ before $n+1$, are disjoint and independent. The sum of their probabilities is infinite, hence at least one of them must occur with probability $1$.

To see that the probability, when at $n$, of reaching $0$ before $n+1$ is $\frac{1}{n+1}$, you could use a simple martingale argument. If win \$1 for heads and lose \$1 for tails, and stop gambling when you win \$1 or lose \$n, the probability of winning or losing must be such that your expectation is \$0.

One can also see that to remain in a range for ever without ever reaching either of the endpoints has probability $0$. To see this, just consider that there is a positive probability of getting $m$ heads or $m$ tails in a row, where $m$ is the length of the range. If we have infinite tries, we must, with probability $1$, get this sequence eventually.


How to prove that $9^n - 8n - 1$ is divisible by $64$ for $n\ge 0$? My textbook provided the following proof:

  Base case: When $n=0, 9^n-8n-1=0=64\cdot0$, so $64\mid\left(9^n-8n-1\right)$.

  Induction step: Suppose that $n\in\mathbb N$ and $64\mid\left(9^n-8n-1\right)$. Then there is some integer $k$ such that $9^n-8n-1=0=64k$. Therefore:

  $$\begin{align}
9^{n+1}-8(n+1)-1&=9^{n+1}-8n-9\\
&=9^{n+1}-72n-9+64n\\
&=9\left(9^n-8n-1\right)+64n\\
&=9(64k)+64n\\
&=64(9k+n)
\end{align}$$

  so $64\mid\left(9^{n+1}-8(n+1)-1\right)$.

What I don't understand is how the equation goes from

$$= 9^{n+1} - 8n - 9$$

to

$$= 9^{n+1} - 72n - 9 + 64n$$

because $-72n+64n=-8n$.Hence the equation.


Binomial distribution central moment calculation If for a binomial distribution the mean is $4$ and variance is $3$, find th $3^{\text{rd}}$ central moment.

I understand that the first and second central moments are mean and variance respectively, but what is the $3^{\text{rd}}$? A Google search only reveals formulas with $n$ in them, which is information that is, as per the question, unavailable. The answer to the  question is $3/2$, but I can't figure out where it comes from.

Recall that if $X\sim\mathrm{Bin}(n,p)$, then $\mathbb E[X] = np$ and $\mathrm{Var}(X)=np(1-p)$. Given $\mathbb E[X] = 4$ and $\mathrm{Var}(X) = 3$, we have $np = 4$ and $np(1-p)=3$. Hence $n=16$, $p=\frac14$. So the distribution of $X$ is given by
$$\mathbb P(X=k) = \binom {16}k \left(\frac14\right)^k\left(\frac34\right)^{16-k}, k=0,1,\ldots,16.$$
The second moment of $X$ is
$$\mathbb E[X^2] = \mathrm{Var}(X) + \mathbb E[X]^2 = 3 + 4^2 = 19.$$
The generating function of $X$ is
$$\begin{align*}
\mathbb E[s^X] &= \sum_{k=0}^\infty\mathbb P(X=k)s^k.
\end{align*}$$
Since $X$ is the sum of $16$ i.i.d. $\mathrm{Ber}(p)$ random variables, each with generating function $1-p+ps$, we have
$$\mathbb E[s^X] := P(s) = (1-p+ps)^n = \left(\frac34 + \frac14s\right)^{16}.$$
The third central moment of $X$ is
$$\begin{align*}
\mathbb E[(X-\mathbb E[X])^3] &= \mathbb E[(X-4)^3]\\
&= \mathbb E[X^3 - 12X^2 + 48X - 64]\\
&= \mathbb E[X(X-1)(X-2) - 9X^2 + 46X - 64]\\
&= \mathbb E[X(X-1)(X-2)] - 9\mathbb E[X^2] + 46\mathbb E[X] - 64.
\end{align*}$$
Now,
$$\begin{align*}\mathbb E[X(X-1)(X-2)] &= \lim_{s\uparrow1}P^{(3)}(s)\\
&= \lim_{s\uparrow1}\left(\frac14\right)^3 16\cdot15\cdot14\left(\frac34+\frac14s\right)^{13}\\
&= \frac{105}2. \end{align*}$$
Hence
$$\begin{align*}
\mathbb E[(X-\mathbb E[X])^3] &= \mathbb E[X(X-1)(X-2)] - 9\mathbb E[X^2] + 46\mathbb E[X] - 64\\
&= \frac{105}2 - 9\cdot19 + 46\cdot 4 - 64\\
&= \frac32.
\end{align*}$$

In this question
Np = 4 mean (1)
Npq = 3 variance (2)
Solving from (1) and (2)

$P = 1/4$

$Q = 3/4$

$N = 16$

This bionomail 1 moments = 0
2 moments = npq ie 3
3 moment = npq(q-p) = 3 X 0.5 = 3/2
4 moment = 3 ( npq)^2 + npq (1-6pq ) = 3(3) ^2 + 3 ( 1- 6 ( 0.1875) =
= 26.625


Find all additive real valued functions such that $f(x^{2019})=f(x)^{2019}$ The following is the final problem from this page:

  Find all the functions $f:\mathbb{R}\rightarrow \mathbb{R}$ such that $$f(x+y)=f(x)+f(y) \; \; \; \forall \,x,y\in \mathbb{R}$$ and also (this is the important part) $$f(x^{2019})=f(x)^{2019}\tag{$*$}$$

My idea is to prove that $f(x)=x \; \; \forall x \in \mathbb{R}$,  $f(x)=-x \; \; \forall x \in \mathbb{R}$ or $f\equiv 0$.

If we change $2019$ for an even number this is easy because it implies that the image of a positive number is positive and from there $f$ is linear and hence the identity or zero.

If we change $2019$ by $3$ then this is related (although I don't know how to deal with the case $f(1)=0$ or $f(1)=-1$)

But in this case I don't know how to prove any type of regularity from $(*)$ to conclude that $f$ must be linear.

Let $f$ satisfy the premises. Then $f(ax)=af(x)$ for any $x\in\mathbb{R}$ and $a\in\mathbb{Q}$. Now $$f\big((a+x)^{2019}\big)=f(a+x)^{2019}$$ (with both sides expanded using the binomial formula and the above), being a polynomial identity in $a\in\mathbb{Q}$, implies $$f(x^k)=f(1)^{2019-k}f(x)^k\qquad(0\leqslant k\leqslant 2019).$$ Taking $k=2$, we get $f(x^2)=f(1)f(x)^2$. This reduces to the case you have worked out (after replacing $f$ by $-f$ if needed).

We can prove that $f$ is continuous at $\mathbb{R}$. Furthermore $f$ has the first derivative on $(0, 0)$. Using these properties leads to show that $f(x) = x$ for every $x \in \mathbb{R}$ or $f \equiv 0$.


Find the integral $\frac{d }{d x}\int_{\sin x}^{\cos x}\cos ( \pi t^2)\, d t $ Find the integral $$\frac{d}{dx}\int_{\sin x}^{\cos x}\cos ( \pi t^2) \, dt. $$

Am I right that it is $$\frac{d}{dx}\cos x \cdot  \cos (\pi \cdot  \cos^2x) - \frac{d}{dx} \sin x \cdot  \cos (\pi \cdot  \sin^2x))?$$

$$\int_a^b \cos ( \pi t^2))\,\mathrm d t = \frac{1}{\sqrt{2}}\left[C(\sqrt{2} b)-C(\sqrt{2}a)\right].$$  This is a Fresnel integral.

$$\int_{\sin x}^{\cos x}\cos ( \pi t^2))\,\mathrm d t  =\frac{1}{\sqrt{2}}\left[C(\sqrt{2} \cos x)-C(\sqrt{2}\sin x)\right].$$

$$\frac{\mathrm{d} }{\mathrm{d} x}\int_{\sin x}^{\cos x}\cos ( \pi t^2))\,\mathrm d t =-\sin x \, C^\prime(\sqrt{2} \cos x)- \cos x \, C^\prime(\sqrt{2}\sin x). $$

$$= -\sin x \cdot  \cos (\pi \, \cos^2x) -  \cos x \cdot  \cos (\pi \, \sin^2x))$$

or just use the fundamental theorem of calculus, as you did.


Examples of arguments from connectedness Suppose $X$ is a connected topological space. A typical way that we prove a property $P(x)$ holds for all $x \in X$ is to show that $P$ is an open and a closed condition, and that $P(x)$ for some $x \in X$.

Most recently I saw this show up in the proof that a connected, locally path connected space is path connected. (Pick a point $x \in X$ and let $P(y)$ be "There exists a path from $y$ to $x$".)

This seems to also be the idea behind some proofs that $[a,b]$ is compact. See here.

I wanted to accrue some more examples of where we use this technique (for teaching purposes, perhaps). Does anyone have some more?

Uniqueness of lifting: Given a covering space $p \colon X' \to X$, a pointed function $f \colon (Y,y_0) \to (X,x_0)$ where $Y$ is connected and a value $x' \in p^{-1}(x_0)$, there exists at most one lift $f': Y \to X'$ such that $f = pf'$ and $f'(y_0)=x'$.

The proof uses the property $P(x)$: $f'_1(x)=f'_2(x)$; it is open, closed and true for $y_0$.

A nice one I saw recently in Warner's Foundations of Differentiable Manifolds and Lie Groups, in defining orientation on a manifold:

Suppose $M$ is connected. Consider $\bigwedge^{\text{top}}TM \setminus Z$, where $Z$ is the zero section of top-degree forms. Each fiber of $\bigwedge^{\text{top}}TM \setminus Z$ has exactly two components, so a connectedness argument (and local triviality of the bundle) shows that $\bigwedge^{\text{top}}TM \setminus Z$ must have at most two components. If it has one, the manifold is non-orientable; if it has two, $M$ is orientable.