My intuition tells me that the first case (tangent lines all passing through a fixed point) corresponds to a segment of a line, and the second one corresponds to a circle, but I'm having trouble actually proving it. I thought about assuming (without loss of generality) that the fixed point is the origin, but I didn't get anywhere. For the second one, it would be enough to prove that it has constant curvature... but I'm not seeing where to start.
Let $\alpha$ be a regular curve in $\mathbb{R}^2$ and let all of its tangent lines pass through the origin. Also, let $\beta$ be a regular curve in $\mathbb{R}^2$ and let all of its normal lines pass through the origin.   How can I show that $\alpha$ is contained in a straight line through the origin and that $\beta$ is contained in a circle around the origin?  I know that the tangent line $\alpha(t)$ is the line that points in the direction of the tangent vector $T(t)$ and that the normal line of $\beta(t)$ is the line that points in the direction of $N(t)$, but how can I put this all together to complete the proof?
I mean are we certain that they will be able to provide us a huge improvements (in some tasks) compared to clasical computers? How do I draw the graph of inequalities in the coordinate plane? I want to graph the following inequalities (done in desmos) in latex. I need the code. Find the equation of the plane passing through a point and a vector orthogonal I have come across this question that I need a tip for.     Find the equation (general form) of the plane passing through the point $P(3,1,6)$ that is orthogonal to the vector $v=(1,7,-2)$.   I would be able to do this if it said "parallel to the vector"  I would set the equation up as $(x,y,x) = (3,1,6) + t(1,7,-2)$ and go from there.  I don't get where I can get an orthogonal vector. Normally when I am finding an orthogonal vector I have two other vectors and do the cross product to find it.  I am thinking somehow I have to get three points on the plane, but I'm not sure how to go about doing that.  Any pointers?  thanks in advance. How do you get \mathbb{1} to work (characteristic function of a set)? If $f:\mathbb R \to \mathbb R$ satisfies $f'(x) <L$ for some $L<1$, then $f$ has a unique fixed point Assume that $f: \mathbb{R} \rightarrow \mathbb{R}$  is differentiable.  If there exists an $L&lt;1$ such that for each $x \in \mathbb{R}$, $f'(x)&lt;L$, prove that there exists a unique point $x$ such that $f(x)=x$.  So far I have that where I fixed $a$ to be zero and came up with three cases to show existence for  $f(0) = 0$, $f(0) &lt; 0$, and $f(0) &gt; 0$  $f(0)=0$ was fairly trivial following from the mean value theorem but the other two cases I was only able to eliminate half of the possible fixed points by contradiction and I need help showing it explicitly.  Edit as to avoid confusion here is the work I have garnished so far:  From the mean value theorem it follows for some arbitrary $a,b \in \mathbb{R}$ and $a\neq b$ that there exists a point $c$ such that $a&lt;c&lt;b$ or $b&lt;c&lt;a$ then implies that $\frac{f(b)-f(a)}{b-a}=f'(c)$ and this is because it is differentiable on $\mathbb{R}$ and, by convention, is also continuous on $\mathbb{R}$ as well.  Then the result of $\dfrac{f(b)-f(a)}{b-a}=f'(c)$ then follows for $L&lt;1$ such that $\forall x \in \mathbb{R}$ means $f'(x)&lt;1$ from our hypothesis. Which then follows  (*) $\dfrac{f(b)-f(a)}{b-a}=f'(c)&lt;L&lt;1 \Rightarrow f(b)-f(a)&lt;L(b-a)&lt;b-a$  Now without loss of generality choose $a=0$ and let $b$ be some arbitrary value in $\mathbb{R}$ then there are three cases that need to be considered for the value of $f(a)$:  (i) $f(0) = 0$, (ii) $f(0) &lt; 0$, (iii) $f(0) &gt; 0$  (i) For $f(0)=0$   Then $0$ is a fixed point of point of $f(x)$ and it follows from (*) that if $f(0)=a=0$ then $f(b)-f(0)=f(b)&lt;Lb&lt;b$ which shows that $f(b)&lt;b$ thus shows that for any arbitrary value $b\in \mathbb{R}$ that no $f(b)=b$ thus shows that $0$ is the only fixed point if $f(0)=0$  (ii) For $f(0)&lt;0$  Suppose there exists a fixed point $d&gt;0$ then:  $\frac{f(d)-f(0)}{d}&lt;L&lt;1 \Rightarrow d-f(0)&lt; dL&lt;d \Rightarrow -f(0)&lt;d(L-1)&lt;0\Rightarrow 0&lt;-d(L-1)&lt;f(0)$ but this contradicts that $f(0)&lt;0$  thus no other fixed point exists if $d&gt;0$ when $f(0)&lt;0$  (iii) For $f(0)&gt;0$ Wine or virtual machine (VirtualBox) to run apps (eg. Photoshop) I am wondering if it will be better (stability, convenience and speed) to run apps like Photoshop with Wine or a virtual machine like VirtualBox? How can I set up dual monitors with nVidia? Order of finite fields is $p^n$ Let $F$ be a finite field. How do I prove that the order of $F$ is always of order $p^n$ where $p$ is prime? Well, the original task was to figure out what the following expression evaluates to for any $n$.  $$\begin{pmatrix} 1 &amp; 1 \\ 0 &amp; 1\end{pmatrix}^{\large n}$$  By trying out different values of $n$, I found the pattern: $$\begin{pmatrix} 1 &amp; 1 \\ 0 &amp; 1\end{pmatrix}^{\large n} = \begin{pmatrix} 1 &amp; n \\ 0 &amp; 1\end{pmatrix}$$  But I have yet to figure out how to prove it algebraically.   Suggestions? How do I make a floor plan with specific dimensions? This is my very first time of using blender. I've tried several tutorials but can't find anything that tells me how to do the very basic thing i want to do.  I simply want to draw a floor plan for a room which is 13 feet by 11 feet and make it 3d so that ultimately i can play with where things go in that room.  Can anyone take me through the steps needed to do this?   I will be so grateful.  Thank you in advance Prove that for all non-negative integers $m,n$, $\frac{(2m)!(2n)!}{m!n!(m + n)!}$ is an integer. Prove that for all non-negative integers $m,n$, $\frac{(2m)!(2n)!}{m!n!(m + n)!}$ is an integer.   I'm not familiar to factorial and I don't have much idea, can someone show me how to prove this? Thank you. Repeat copies of array elements: Run-length decoding in MATLAB What is the difference between:   The company would rather each employee be provided with ID card.  The company would rather each employee were provided with ID card. I have a Python regular expression that contains a group which can occur zero or many times - but when I retrieve the list of groups afterwards, only the last one is present. Example:  re.search("(\w)*", "abcdefg").groups()  this returns the list ('g',)  I need it to return ('a','b','c','d','e','f','g',)  Is that possible? How can I do it? Are non-citizens of the Capitol permitted to permanently immigrate? Reading  got me thinking about permitted citizenship.  Are citizens of other districts allowed to permanently move to, or become citizens of the Capitol?   My gut says, "No." The citizens of the other districts are being punished, so it makes little sense to allow them a true, legal escape from their punishments. Even victors are only given new homes in their own districts, keeping them there. I've found strange behavior when closing this  - I definitely see that there are 3 closing votes and 2 of them are at "Off-topic" section:   -but there are no explanations on section itself:   -while, normally, it's expected that it will appear. In custom comments section there are also no closing-votes displayed.   Perhaps it's just my browser (which is FF 26 x64 under Win) - so marking this as a "possible bug". Thank you for your attention. I'll be generous and say it might be reasonable to assume that nature would tend to minimize, or maybe even maximize, the integral over time of $T-V$.  Okay, fine.  You write down the action functional, require that it be a minimum (or maximum), and arrive at the Euler-Lagrange equations.  Great.  But now you want these Euler-Lagrange equations to not just be derivable from the Principle of Least Action, but you want it to be equivalent to the Principle of Least Action.  After thinking about it for awhile, you realize that this implies that the Principle of Least Action isn't really the Principle of Least Action at all:  it's the "Principle of Stationary Action".  Maybe this is just me, but as generous as I may be, I will not grant you that it is "natural" to assume that nature tends to choose the path that is stationary point of the action functional.  Not to mention, it isn't even obvious that there is such a path, or if there is one, that it is unique.  But the problems don't stop there.  Even if you grant the "Principle of Stationary Action" as fundamentally and universally true, you realize that not all the equations of motions that you would like to have are derivable from this if you restrict yourself to a Lagrangian of the form $T-V$.  As far as I can tell, from here it's a matter of playing around until you get a Lagrangian that produces the equations of motion you want.  From my (perhaps naive point of view), there is nothing at all particularly natural (although I will admit, it is quite useful) about the formulation of classical mechanics this way.  Of course, this wouldn't be such a big deal if these classical ideas stayed with the classical physics, but these ideas are absolutely fundamental to how we think about things as modern as quantum field theory.  Could someone please convince me that there is something natural about the choice of the Lagrangian formulation of classical mechanics (I don't mean in comparison with the Hamiltonian formulation; I mean period), and in fact, that it is so natural that we would not even dare abandon these ideas? Proving the quotient of a principal ideal domain by a prime ideal is again a principal ideal domain Please help me prove that      the quotient of a principal ideal domain by a prime ideal is again a principal ideal domain.   This was from Abstract Algebra Serializing anonymous delegates in C# Search for a string in all tables, rows and columns of a DB