I am a math student and I am taking my first physics course of Mechanics. But I don't know the difference between the $w=mg$ force and $F=G\cfrac{m_1 m_2}{r^2}$ (why we don't use the first one when talking about the interaction between our planet and me).   For me, the two forces are related since $g$ is the gravitational acceleration. I know that $F=G\cfrac{m_1 m_2}{r^2}$  is the force with which the earth attracts us (because we have mass).
Newton's Law of Universal Gravitation tells us that the potential energy of object in a gravitational field is $$U ~=~ -\frac{GMm}{r}.\tag{1}$$  The experimentally verified near-Earth gravitational potential is  $$U ~=~ mgh.\tag{2}$$  The near-Earth potential should be an approximation for the general potential energy when $r\approx r_{\text{Earth}}$, but the problem I'm having is that they scale differently with distance.  $(1)$ scales as $\frac 1r$.  So the greater the distance from the Earth, the less potential energy an object should have.  But $(2)$ scales proportionally to distance.  So the greater the distance from the Earth, the more potential energy an object should have.  How is this reconcilable?
I am looking for a closed form (ideally expressed as elementary functions) of the function $\exp_n(x) = \sum_{k=0}^n x^k / k!$.  I am already aware of expressing it in terms of the gamma function. Background / Motivation When counting combinations of objects with generating functions, it is useful to be able to express the partial sum $1 + x + \cdots + x^n$ as $\frac{1-x^{n+1}}{1-x}$. For example, to count the number of ways to pick 5 marbles from a bag of blue, red, and green marbles where we pick at most 3 blue marbles and at most 2 red marbles, we can consider the generating function $f(x) = (1+x+x^2+x^3)(1+x+x^2)(1+x+x^2+\cdots)$. By using the partial sum identity, we can express it as $f(x) = \left(\frac{1-x^4}{1-x}\right)\left(\frac{1-x^3}{1-x}\right)\left(\frac{1}{1-x}\right)$.  Simplify, express as simpler product of series, and find the coefficient of the $x^5$ term. I want to be able to do the same for a generating function in the form $g(x) = \exp_{n_1}(x)^{p_1} \exp_{n_2}(x)^{p_2} \cdots \exp_{n_j}(x)^{p_j}$ The easiest way to extract the coefficient of a given term $x^p / p!$ would be to use a similar closed form expression for $\exp_n(x)$ and a similar technique to $f$. Attempted Solutions Differential equation Recall that the way to prove the identity $1+x+x^2+\cdots+x^n = \frac{1-x^{n+1}}{1-x}$ is to define $S = 1 + x + x^2 + \cdots + x^n$ and notice that: $S - Sx = 1 - x^{n+1}$. Likewise, notice that $y(x) = \exp_n(x)$ satisfies $y - y' = x^n/n!$.  Via SAGE, the solution is $y(x) = \frac{c+\Gamma(n+1,x)}{n!}e^x$.  Our initial condition $y(0) = 0$ so $c=0$.  By , $\Gamma(n+1,x) = n! e^{-x} \exp_n(x)$ so we have $y(x) = \exp_n(x)$. Recurrence Relation Notice that $\exp_n(x) = \exp_{n-1}(x) + x^n/n!$. Using the unilateral Z-Transform and related properties, we find that $\mathcal{Z}[\exp_n(x)] = (z e^{x/z})/(z-1)$. Therefore, $\exp_n(x) = \mathcal{Z}^{-1}\left[(z e^{x/z})/(z-1)\right] = \frac{1}{2 \pi i} \oint_C z^n e^{x/z}/(z-1)\;dz$. $(z^n e^{x/z})/(z-1)$ has two singularities: $z = 1$ and $z = 0$.  The point $z = 1$ is a pole of order one with residue $e^x$.  To find the residue at $z = 0$ consider the product $z^n e^{x/z} (-1/(1-z)) = -z^n \left( \sum_{m=0}^\infty x^m z^{-m} / m! \right) \left( \sum_{j=0}^\infty x^j \right)$.  The coefficient of the $z^{-1}$ term is given when $n - m + j = -1$. The residue of the point $z=0$ is then $-\sum_{m,j} x^m / m! = -\sum_{m=n+1}^\infty x^m / m!$. Let $C$ by the positively oriented unit circle centered at the origin.  By Cauchy's Residue Theorem, $\frac{1}{2 \pi i} \oint_C z^n e^{x/z}/(z-1)\;dz = \frac{1}{2 \pi i} 2 \pi i \left(e^x - \sum_{m=n+1}^\infty x^m / m!\right) = \exp_n(x)$. Finite Calculus I've tried to evaluate the sum using finite calculus, but can't seem to make much progress. Is 'that is' optional? When I write something like this:     I am a man that is doing these things.   Is it correct to shorten it like this?     I am a man doing these things.   And does it require commas? Opening the menu and selecting an item from the "Favorite Sites" menu allows us to see a feed for a specific site. In the top right corner of this page is a drop-down which allows us to select the "sort order" for the posts. The default is "Active" with other options like "Newest", "Hot", etc.  The problem is that when I select "Newest" or any other option. This preference is not remembered when I change between a different "Favorite Site".  So I am requesting that the selection is persisted when changing between site feeds. I have no preference whether this is implemented per site (to allow a different default for each site) or just one single preferred default that works on all sites (last option selected). I just want it to remember that I have selected "Newest" when I am flicking though the sites. Suppose $Y_1, \ldots, Y_n$ is a random sample from a normal distribution with mean $\mu$ and variance $\sigma^2$. Let $S^2$ be the sample variance, which is unbiased for $\sigma^2$.  GOAL: Show that $S = \sqrt{S^2}$ is a biased estimator of $\sigma$.  I realize that $(n-1)S^2/ \sigma ^2$ is chi-square with $(n-1)$ degrees of freedom.  I am given that $$E(S) = E\left[\frac{\sigma}{\sqrt{n-1}} \left[\frac{(n-1)S^2}{\sigma^2}\right]^{1/2}\right] = \frac{\sigma}{\sqrt{n-1}} \int_0^\infty v^{1/2} \frac{1}{\Gamma[(n-1)/2] 2^{(n-1)/2}} v^{(n-1)/2} e^{-v/2} \, dv$$  Can anyone help me understand why they have $v^{1/2}$ and $v^{(n-1)/2}$ (combining them $v^{n/2}$ as I think it should be $v^{1/2}$ and $v^{\frac{(n-1)}{2}-1}$ (combining both terms give $v^{(n-2)/2}$ since the pdf for chi-square distribution is $$ f(v) = \frac{v^{(x/2)-1} e^{-v/2}}{2^{x/2}\Gamma(x/2)},$$ where $x$ is degree of freedom?  Could anyone please tell me which one is the correct answer with explanation?  Thank you linq: order by random I want to prove that every real polynomial of odd degree has at least one real root, using  the intermediate value theorem.  Let $P(x) = x^{2n+1} + a_n x^{2n} + . . . + a_0$ for each $a_i \in \mathbb{R}$ and $n \in \mathbb{N}$.  By the fundamental theorem of algebra I know that $P(x)$ has exactly $2n+1$ complex roots, so  $P(x) = (x+r_1)(x+r_2) . . . (x+r_{2n+1})$ for each $r_i \in \mathbb{C}$  I do not know how to complete this but I do know that, at some point, I probably have to show that each root with imaginary part non zero has to come in conjugate pairs, and since $2n+1$ is odd there is at least $1$ root that is imaginary part $0$ and thus real. I have a time series that shows a nonstationary seasonal autoregressive component as well as known heteroshedasticity. In order to model the series, I have fit a seasonal ARIMA model for the mean with the auto.arima model in the forecast package in R and a GARCH model on the residuals of the ARIMA model.   Is the procedure of sequentially estimating ARIMA and GARCH model correct or would it have been better to jointly model the mean and the variance of the series? If this were correct, is there a (possibly R) function to do it? "An operator is hermitian". Implications? Alastair Rae states that there are 4 postulates of Quantum Mechanics in his text on the subject matter.  The first part of his second postulate can be stated as:      Every dynamical variable may be represented by a Hermitian operator whose eigenvalues represent the result of carrying out a measurement of the value of the dynamical variable...   My question is: just how much can we deduce from saying that the some of the operators involved in Quantum Mechanics are Hermitian?  When I use the word Hermitian, I am referring to the property that $A=A^+$ where $A$ is an operator and $A^+$ is the adjoint of $A$.  Does an operator being Hermitian automatically imply that there will exist eigenfunctions and eigenvalues corresponding to this operator?  Does being Hermitian imply that, should there exist eigenfunctions corresponding to this operator, they will form a complete and orthonormal set?  That any wavefunction can be expanded in terms of these eigenfunction?  Just how much is hidden in the word "Hermitian"? Tiling problem: 100 by 100 grid and 1 by 8 pieces If I utilize the two columns of multicols for my Kurdish sequence of numbers, it arranges the items in column like the following figure:   But I want to arrange the items in rows splitted by some columns (two columns in this case) alike the following figure:   And for the case of "three columns", the output would be alike the following figure:   How may I achieve the functionality? And I don't want to use the sequence of the following commands:  \begin{multicols}{2}‎ \begin{enumerate}[$1)$,noitemsep] \item ‎ Yek \item ‎ Du \end{enumerate} \end{multicols}‎   Instead, I want to specify the split count of the columns and then have the items arranged in the desired arrangement automatically. For the record, I've check  post, but it's too generalized. Residence has changed - Must I re-apply for ESTA? I have an old, but still valid ESTA, with the country of residence showing my home country. However, I have recently moved for education in the UK. Do I have to apply for a new one?  Nothing else has changed - country of citizenship, passport, names etc Previously, someone posted his/her proof of "every neighborhood is open" . I am focusing on the same statement, but it seems that my approach is different from his/hers. Could someone comment on my own proof? Here we go:  Let $X$ be a metric space and $p\in X$. Fix $r&gt;0$. Define a neighborhood around $p$ as $N_r (p)=\{q\in X| d(p, q)&lt;r\}$. Now, take $w = \frac{r}{2}$. Then, $N_w (p)\subset N_r(p)$. In other words, every point of $N_r(p)$ is interior. Therefore, $N_r (p)$ is open. Since $r$ is arbitrary, this proves every neighborhood is open.  Update:  Under my definition of "neighborhood," can I prove in this way instead: For all $m\in N_r(p)$, take $w=\frac{r−d(m,p)}{2}$. Then, $N_w(m)\subset N_r(p)$.   Update2  From the hints given in comments, I solve it:  Assume $m\in N_r(p)$ and $q\in N_w(m)$. Take $w=r-d(m, p)$. Then, $d(p, q)\leq d(p, m) + d(m, q)&lt; d(p, m) + r-d(m,p) =r$.   Thanks to @JoséCarlosSantos and @fleablood ! In chapter 1, of Landau-Lifshitz Mechanics' book, Landau through isotropy and homogeneity of space and homogeneity of time proves that the Lagrangian must depend of magnitude of velocity of the particle.  This seemed fine. But he didn't give a reason as to why it must depend of $|v|^2$ and not on some $|v|^n$ where $n\neq2$. Using the itemize environment inside a caption Proving by induction that $1^3 + 2^3 + 3^3 + \ldots + n^3 = \left[\frac{n(n+1)}{2}\right]^2$ Need guidance on this proof by mathematical induction. I am new to this type of math and don't know how exactly to get started.   $$ 1^3  +  2^3  +  3^3  +  \ldots  +  n^3  =  \left[\frac{n(n+1)}{2}\right]^2 $$ I need help with this exercise. What I've done so far is prove the exercise when $n=1$. So:  $$n=1$$ $$1!\ge2^{(1-1)}$$ $$1\ge2^0$$ $$1\ge1$$ Which is true  Therefore, now that I assume that the assumption is correct, I want to prove that with $n+1$, it will also be true. So, what I've done now is:  $$P(n) \implies P(n+1) $$ $$n!+n+1\ge 2^{(n-1)}...$$ And my  problem is that I do not know how to add the $n+1$ in the right side of the equation, therefore I'm not been able to finish the exercise.  Thank you so much for your help. If something's not very clear, please let me know. I'll try to be clearer next time :) I am wondering how to show that $\mathbb{Z}[\sqrt {10}]$ is not a UFD.    My only idea is to show that there are two factorizations of $10$, say, $ab, uv$ such that $a$ is not a unit times $u$ or $v$. In this ring $10=2\cdot5=\sqrt {10}\cdot \sqrt {10}$, so it suffices to show $2$ is not a unit times $\sqrt {10}$. Suppose $2=\sqrt {10}(a+b\sqrt{10})=a\sqrt{10}+10b$. Then $a=0$ since $\sqrt{10}$ is not rational. So $10b=2$, which has no integer solutions. So $\mathbb{Z}[\sqrt {10}]$ is not a UFD.  Is my reasoning correct? What are the flaws? Combining multiple textures with a white background I'd like to sort of "overlay" these shaders, but the only options are mix and add. Allow me to explain.  I made two images, both white background, one with red scribbles and one with blue scribbles, both 1920x1080. Image textures plugged into diffuse was my node setup. I want both scribbles to be added to the plane while the background stays the same color.  If I mix them I get the following:  The red and blue scribbles are now 50% dimmer than they would be on their own, and the white color hasn't changed in intensity.  If I add them I get the following:  The red and blue scribbles are still 50% dimmer than they normally would be (since the white background is being added over each). But the white background is now 2x brighter than normal.  Can I get a result where the white background is normal brightness and both the blue and red scribbles are at normal brightness? prove that a connected graph with $n$ vertices has at least $n-1$ edges Show that every connected graph with $n$ vertices has at least $n − 1$ edges.  How can I prove this?  Conceptually, I understand that the following graph has 3 vertices, and two edges:  a-----b-----c  with $a$, $b$ and $c$ being vertices, and $\{a,b\}$, $\{b,c\}$ being edges.  Is there some way to prove this logically?  --UPDATE--  Does this look correct?  Any advice on how to improve this proof would be appreciated.  Thank you. If a subring of a ring R has identity, does R also have the identity? I know it does not make sense that if a subring of a ring $R$ is commutative, then $R$ is also commutative. (For example, the set consisting of the matrices whose all entries except (1,1)-entry are zero, is a subring of ring of 2x2 real matrices.)  I also considered the case of a ring containing a subring with identity, but I had no ideas. Maybe it seems that it does not make sense, either.  Who give me some examples supporting my guess?