In my previous questions like 'how do compressions and relaxation carry energy...' and 'how are they not isothermal...' and 'where does the heat come from in the compressed gas...',i've made  enquiries reagarding the phenomenon of sound energy propagation and got answers from experts.So i am now writting my concept:when a vibrating fork compresses the air,it does work on it increasing the K.E. of the layer.When the fork goes opp. side,it works on the compressed layer and relaxes it increasing theP.E.Meanwhile,during expansion,the layer works on the next layer and compresses it using the excess kinetic energy.And the next layer does the same to its next layer by compressing using that K.E. given to it by the fork via the first layer.Ultimately,the last layer works on our ear drum to produce sound using the kinetic energy.The question is am i right in my assumption?Plz help...
Suppose a vibrating fork exerts force to the air particles to compress which leads to the increase of internal energy (heat).This heat or energy given by the fork is given to the next layer of air and in this way, this energy makes sound. But if it was so, it will be isothermal. And Laplace told it was adiabatic. How can it be? I am confused. The heat energy infact is the energy given by the fork. If it is adiabatic, how can the air pass this energy ?
If a sequence $(u_n)$ be such that its every subsequence has a subsequence that converges to $0$, then $\lim u_n= 0$ Parsing DD MM YY HH MM SS columns from TXT file using Python's pandas How to deselect the checkbox when max row selection is set to 1 in LWC? This occurs in the  released recently to the public. When someone posts a code block in their question or answer, the code block is cut off on the right hand side, and I can't scroll to the right to view the rest of it. How can I view these code blocks? How do I make Chromium use Flash from Google Chrome? Now, that Adobe has stopped issuing new releases for Linux desktops, the only option for users is to use Google Chrome, for those who want to use the latest Flash updates. I use Chromium. Is there any way to get Chromium use the Flash from Google Chrome? The reason I as this question is that I have noticed both these browsers run in the background, so technically, the browsers' plugins are available to use. Who is responsible for defining a site logo? I mean, there are moderators of the site that define it, or stackexchange defines it? $n$ Lines in the Plane How am I to "[u]se induction to show that $n$ straight lines in the plane divide the plane into $\frac{n^2+n+2}{2}$ regions"? It is assumed here that no two lines are parallel and that no three lines have a common point. Further, this is not a non-Euclidean problem, but I wouldn't mind a discussion on the non-Euclidean nature of the problem.  I was thinking it might be easier to show using the unit sphere.    A GOOD REFERENCE I FOUND THAT WILL BE USEFUL TO ANYONE COMING ACROSS THIS PROBLEM IS Concrete Mathematics, Graham, Knuth, Patashnik pp. 4–8. How to derive this version of variance formula A probably very easy computational question, but I don't really understand how it's done:  I try to compute the point estimates of a normal distribution as $N \sim (\beta, \sigma)$. Using the method of moments, we get:  $$\frac{1}{n} \sum_{i=1}^{n} X_{i} = \hat{\beta}$$  $$\frac{1}{n} \sum_{i=1}^{n} X_{i}^{2} = \hat{\sigma}^2 + \hat{\beta}^2$$  Solving the two equations for the two parameters, we get:  $$\hat{\beta} = \overline{X} $$   and   $$\hat{\sigma}^2 = \frac{1}{n} \sum_{i=1}^{n} (X_{i} - \overline{X}_n)^2 $$  My question is: How do we get to the term of $\hat{\sigma}^2$? If I try to compute it, I get:  $$\hat{\sigma}^2 = \frac{1}{n} \sum_{i=1}^{n} X_{i}^2 - (\frac{1}{n} \sum_{i=1}^{n} X_{i} )^2 $$  Can someone show me how to simplify the last term to what is stated above and what "rule" or law" is used there? Since it is used so often, I really wanna understand this. I'm reading the proof of the multiplicative property of $$s_k(n)=\sum_{d|(n,k)}f(d)g\bigg( \frac kd\bigg)$$ The book wrote that in order to understand the proof, we need to know if $a,b,k,m$ are integers such that $(a,k)=(b,m)=1$, then $$(ab,km)=(a,m)(b,k).$$  I have written proof, but I don't know is it consistent, and this seems used more tools than it necessarily has to.   My proof: Let $$S_1=\{d: d|(a,m), d&gt;0\}\\S_2=\{d: d|(b,k), d&gt;0\}\\S=\{d:d|(ab,km), d&gt;0\}.$$ I will prove the statement by showing that the function $\phi:S_1\times S_2\to S$ defined as $$\phi(d_1,d_2)=d_1d_2$$ is bijective, and complete the proof by connecting the maximal elements from these partial ordering sets.  Obviously, for $d_1\in S_1, d_2\in S_2$, this tells us $d_1 $ is both divisor of $a$ and $m$, and $d_2$ is both divisor of $b$ and $k$, hence $d_1d_2$ is both divisor of $ab$ and $km$, so we can tell $\phi(S_1\times S_2)\subseteq S$.  To show one-to-one, if we have $c_1,d_1\in S_1$ and $c_2,d_2\in S_2$ such that $$\phi(c_1,c_2)=\phi(d_1,d_2)$$ then $$c_1c_2=d_1d_2$$  In particular, $c_1|d_1d_2$, we need to show $c_1|d_1$. Indeed, if $(c_1,d_2)&gt;1$, this implies $a,k$ have common factor larger than $1$, contradicts the assumption. Therefore $c_1|d_1$. A similar argument implies $d_1|c_1$, therefore $c_1=d_1$. Again a similar argument implies $c_2=d_2$, which shows one-to-one.   To show onto, let $d\in S$, since $d|ab$, we can split $d$ into $d_1d_2$ such that $d_1|a$ and $d_2|b$. Since $(a,k)=1$, we see $(d_1,k)=1$, by property of $S$ we see also $d_1|m$, therefore $d_1\in S_1$, a similar argument implies $d_2\in S_2$. So every number in $S$ has a preimage in $S_1\times S_2$  This completes the proof. Is there any mistake in the proof?     Edit: A proof about the statement: If $d|ab$ then there are $d_1,d_2$ such that $d=d_1d_2$ and $d_1|a, d_2|b$.  Let $ab=p_1^{e_1}\cdots p_r^{e_r}$ for distinct prime factors $p_1,\cdots,p_r$, then we can write $$a=p_1^{x_1}\cdots p_r^{x_r}\\b=p_1^{y_1}\cdots p_r^{y_r}$$ where some of $x_i, y_i$ can be zero, and $x_i+y_i=e_i$ for all $i=1,\cdots,r$.  Since $d|ab$, we can write $$d=p_1^{f_1}\cdots p_r^{f_r}$$ with $f_i\leq e_i$ for all $i=1,\cdots,r$. Then since $$f_i\leq x_i+y_i$$ Split into two cases:   Case 1: $f_i&gt; x_i$, then we can choose $g_i=x_i$ and $h_i=f_i-x_i$.  Case 2: $f_i\leq x_i$, then we can choose $g_i=f_i$ and $h_i=0$.  Then the choice $$d_1=\prod_{i=1}^r p_i^{g_i}, \quad d_2=\prod_{i=1}^r p_i^{h_i}$$ will prove the statement. Bound for the correlation of three random variables There are three random variables, $x,y,z$. The three correlations between the three variables are the same. That is,  $$\rho=\textrm{cor}(x,y)=\textrm{cor}(x,z)=\textrm{cor}(y,z)$$  What is the tightest bound you can give for $\rho$? Is it appropriate to buy a "thank you" gift for a PhD supervisor? I am nearing the end of my PhD and I would quite like to buy a "thank you" gift for my supervisor (although, she's the socially awkward type who I'm sure will be greatly embarrassed to receive one).  I'm just wondering whether a gift is the done thing? And if so, what sort of gift seems appropriate? I suppose a bottle of wine would be a safe choice? Is it possible to write all mathematical formulas in a separate file and add them in main file on-demand? Illustrator: Is there an efficient and accurate way to join two circles by their tangents (and keep it editable)? It's easy to join two circles if they are the same size and horizontal/vertical, into a pill shape:    If the circles are at an add angle then it's still possible but you can no longer rely on smart guides.     You can add lines roughly where the tangents should be and then use the scissors tool on the circles, cut out the inner area and then join the two crescents into one shape.    Is there a better way to do this? Is there a way to do this that is still editable afterwards too?  There are some other questions that cover the same area as this but don't cover editability or joining circles that are not in line with each other (where smart guides can't help). Can I use a variable which has a non-linear relationship to the dependent variable in logistic regression? Let's say I am building a logistic regression model where the dependent variable is binary and can take the values $0$ or $1$. Let the independent variables be $x_1, x_2, ..., x_m$ - there are $m$ independent variables. Let's say for the $k$th independent variable, the bivariate analysis shows a U-shaped trend - i.e., if I group $x_k$ into $20$ bins each containing roughly equal number of observations and calculate the 'bad rate' for each bin - # observations where y = 0 / total observations in each bin - then I get a U shaped curve.  My questions are:   Can I directly use $x_k$ as input while estimating the beta parameters? Are any statistical assumptions violated which might cause significant error in estimating the parameters? Is it necessary to 'linearize' this variable through a transformation (log, square, product with itself, etc.)? Computing second partial derivative with polar coordinates Consider the polar coordinates with $x=r\cos\theta$ and $y=r\sin\theta$.  I can show using the chain rule that $$\frac{\partial}{\partial x} = \frac{x}{r} \frac{\partial}{\partial r} -\frac{y}{r^2} \frac{\partial}{\partial \theta}$$   What is the method to compute $\dfrac{\partial^2}{\partial x^2}$? I don't know how to do it. Requirements for a clan war? Was Mjolnir ever used for building something? In the movie Thor (2011), while he was about to crown Thor as the next king of Asgard, Allfather Odin describes the mighty hammer Mjolnir as the following:      So long entrusted with this mighty hammer, Mjolnir. Forged in the heart of a dying star, from the sacred metal of Uru. Only one may ift it. Only one is worthy. Who wields this hammer commands the lightning and the storm. Its power has no equal -- as a weapon, to destroy, or as a tool, to build. It is a fit companion for a King.   We know Mjolnir as a weapon of destruction but are there references where it is used to build something? Is there a way to copy maintenance plans from one SQL Server 2012 to another? Both servers are not accessible via the same copy of Management Studio at the same time (on different networks). After some searching on stack / google I decided to ask this here, this is the second time I'm facing this issue this time with blender 2.8, previously with 2.7 and without a clue to how I got here.  Some of my objects appear hidden(10/12), and I can't seem to find a way to unhide them.  Solutions to the most common rookie mistakes seems to be   In local-view solved by pressing numpad / hidden by pressing H or Shift + H, unhiding with Alt + H   I'm not in local view, I can toggle visibility for 2/12 objects with Shift + H &amp; Alt + H, but the rest stays gray in the collection.  Last time I solved this by exporting the whole model as STL and importing it back again, I then had to re-group all the pieces but there must be a more simple solution.  Here's the .blend if someone wants to have a go at it: Prove or disprove: if $\sum\limits_{n=0}^∞a_n$ converges and $\lim\limits_{n\to\infty}b_n =0$, then $\sum\limits_{n=0}^{\infty}a_n b_n$ converges.