Why aren't integration and differentiation inverses of each other? Integration is supposed to be the inverse of differentiation, but the integral of the derivative is not equal to the derivative of the integral:

  $$\dfrac{\mathrm{d}}{\mathrm{d}x}\left(\int f(x)\mathrm{d}x\right) = f(x) \neq \int\left(\dfrac{\mathrm{d}}{\mathrm{d}x}f(x)\right)\mathrm{d}x$$

For instance:
$$\begin{align*} &\dfrac{\mathrm{d}}{\mathrm{d}x}\left(\int 2x+1\;\mathrm{d}x\right) &&= \dfrac{\mathrm{d}}{\mathrm{d}x}\left(x^2+x+C\right) &= 2x+1\\
&\int\left(\dfrac{\mathrm{d}}{\mathrm{d}x}\left(2x+1\right)\right)\mathrm{d}x &&= \int 2\;\mathrm{d}x &= 2x+C\end{align*}$$

Why isn't it defined such that $\dfrac{\mathrm{d}}{\mathrm{d}x}a = \dfrac{\mathrm{d}a}{\mathrm{d}x}$, where $a$ is a constant, and $\int f(x)\;\mathrm{d}x = F(x)$?
Then we would have:
$$\begin{align*} &\dfrac{\mathrm{d}}{\mathrm{d}x}\left(\int 2x+1\;\mathrm{d}x\right) &&= \dfrac{\mathrm{d}}{\mathrm{d}x}\left(x^2+x\right) &= 2x+1\\
&\int\left(\dfrac{\mathrm{d}}{\mathrm{d}x}\left(2x+1\right)\right)\mathrm{d}x &&= \int \left(2+\dfrac{\mathrm{d1}}{\mathrm{d}x}\right)\;\mathrm{d}x &= 2x+1\end{align*}$$

Then we would have:

  $$\dfrac{\mathrm{d}}{\mathrm{d}x}\left(\int f(x)\mathrm{d}x\right) = f(x) = \int\left(\dfrac{\mathrm{d}}{\mathrm{d}x}f(x)\right)\mathrm{d}x$$

So what is wrong with my thinking, and why isn't this the used definition?

The downvotes on this question aren't really fair. Sure, it's a rather simple question to answer trivially, but understanding exactly why derivatives and integrals aren't perfect inverses (and how to deal with that) is hugely important for everything from solving differential equations to constructing topological invariants on manifolds. Furthermore, it's common for students to think invertible operations are always preferable since they preserve all the information, so it's useful to present examples which show that noninvertibility and information loss is not only acceptable, but in many cases desirable. Some of this may go over your head initially (I tried to keep the level accessible to undergraduate students but in the end I had to assume a bit of background), but I hope you'll at least get some idea of what I'm saying and possibly take up studying some of these things in more depth if you're interested in them.

At the most fundamental level, it's easy to see exactly what fails. Suppose we take any two arbitrary constant functions $f_1: x \mapsto c_1$ and $f_2 : x \mapsto c_2$. Then $\frac{d f_1}{dx} = \frac{d f_2}{dx} = 0$. When you define an antiderivative $\int$, you'll have to pick a particular (constant) function $\int 0 dx = c$. Assuming $c_1 \ne c_2$, we can't simultaneously satisfy both $\int \frac{d f_1}{dx} dx = c_1$ and $\int\frac{d f_2}{dx}dx=c_2$, since $c$ can't equal both $c_1$ and $c_2$. This comes directly from the definition of the derivative as $$\displaystyle \frac{df}{dx}|_{x = x_0} = \lim_{x \rightarrow x_0} \frac{f(x) - f(x_0)}{x-x_0},$$ and so there's no helping it so long as we keep this our definition of the derivative and allow arbitrary functions.

Furthermore, this is in some sense the "only" way this can fail. Let's suppose we have two arbitrary smooth functions $g_1$ and $g_2$, such that $\frac{dg_1}{dx} = \frac{dg_2}{dx}$. Then it must hold that $\frac{d}{dx}(g_1-g_2)=0$, and so $g_1(x) - g_2(x) = c$ is some constant function. The only way the derivative of two smooth functions on $\mathbb R$ to be the same is if the difference of the two functions is a constant. Of course, for differentiation to be invertible, we need to have a function which is infinitely differentiable i.e. smooth.

If you've studied linear algebra, you'll realize what we're doing is actually familiar in that context. (If not, while this answer could probably be written without making extensive use of linear algebra, it would necessarily be far longer, and it's probably already too long, so I apologize in advance that you'll probably have a hard time reading some of it.) Rather than thinking of functions as individual objects, it's useful to collect all smooth functions together into a single object. This is called $C^\infty(\mathbb R)$, the space of smooth functions on $\mathbb R$. On this space, we have point-wise addition and scalar multiplication, so it forms a vector space. The derivative can be thought of as a linear operator on this space $D: C^{\infty}(\mathbb R) \rightarrow C^\infty(\mathbb R), f \mapsto \frac{df}{dx}$. In this language, what we've shown is that $\ker D$ (the kernel of $D$, that is, everything which $D$ sends to $0$) is exactly the set of constant functions. A general result from linear algebra is that a linear map is one-to-one if and only if its kernel is $0$.

If we want to make $D$ invertible (so that we could define an anti-derivative $\int$ which is a literal inverse to $D$), we have a few options. I have to warn you that the terminology and notation here is nonstandard (there is no standard for most of it) so you'll need to use caution comparing to other sources.

Option 1: Try to add more "functions" to to $C^\infty (\mathbb R)$ which will make $D$ invertible.

Here's one approach you can use. Define a larger space $\bar C^\infty(\mathbb R) \supset C^\infty(\mathbb R)$ by allowing addition to each function by an arbitrary finite formal sum of the form $a_1 \zeta_1 + a_2 \zeta_2 + a_3 \zeta_3 + \cdots + a_n \zeta_n$, where the $\zeta_i$ are formal parameters. A generic element of $\bar C^\infty(\mathbb R)$ is something like $f(x) + a_1 \zeta_1 + a_2 \zeta_2 + a_3 \zeta_3 + \cdots + a_n \zeta_n$ for some $f \in C^\infty(\mathbb R)$, $n \ge 0$, and $a_1, \ldots, a_n \in \mathbb R$. The $\zeta_i$ will keep track of the information we'd lose normally by differentiating, so that differentiation will be invertible.

We'll define a derivative $\bar D$ on $\bar C^\infty(\mathbb R)$ based on the derivative $D$ we already know for $C^\infty(\mathbb R)$. For any function $f \in C^\infty (\mathbb R)$, let $(\bar D f)(x) = \frac{df}{dx} + f(0) \zeta_1$. The choice of $f(0)$ here is somewhat arbitrary; you could replace $0$ with any other real number, or even do some things which are more complicated. Also define $\bar D(\zeta_i) = \zeta_{i+1}$. This defines $\bar D$ on a set which spans our space, and so we can linearly extend it to the whole space. It's easy to see that $\bar D$ is one-to-one and onto; hence, invertible. The inverse to $\bar D$ is then $$\int (F(x') + a_1 \zeta_1 + \cdots + a_n \zeta_n) dx' = \int_0^x F(x') dx' + a_1 + a_2 \zeta_1 + \cdots+ a_n \zeta_{n-1}.$$ This is a full inverse to $\bar D$ in the sense that ordinary anti-derivatives fail to be.

This is, in fact, almost exactly what the OP was trying to do. He wanted to add formal variables like $\frac{da}{dx}$ so that if $a \ne b$ are constants,  $\frac{da}{dx} \ne \frac{db}{dx}$. In my notation, "$\frac{da}{dx}$" is $a \zeta_1$. That alone doesn't save you, since you don't know how to differentiate $\zeta_1$, so you have to add another parameter $\zeta_2 = \bar D \zeta_1$, and to make $\bar D$ invertible you need an infinite tower of $\zeta_i$. You also need to handle what to do when neither $f_1$ nor $f_2$ is constant, but $f_1 - f_2$ is constant. But you can definitely handle all these things if you really want to.

The question here isn't really whether you can do it, but what it's good for. $\bar D$ isn't really a derivative anymore. It's a derivative plus extra information which remembers exactly what the derivative forgets. But derivatives aren't a purely abstract concept; they were created for applications (e.g. measuring the slope of curves), and in those applications you generally do want to get rid of this information. If you want to think of a function as something you can measure (e.g. in physics), the $\zeta_i$ are absolutely not measurable. Even in purely mathematical contexts, it's hard to see any immediate application of this concept. You could rewrite some of the literature in terms of your new lossless "derivative", but it's hard to see this rewriting adding any deep new insight. You shouldn't let that discourage you if you think it's an interesting thing to study, but for me, I have a hard time calling the object $\bar D$ a derivative in any sense. It isn't really anything like the slope of a curve, which doesn't care if you translate the curve vertically.

There is a useful construction hiding in this, though. Specifically, if you restrict the domain of $\bar D$ to $C^\infty(\mathbb R)$ and the range to its image (which is everything of the form $f + a \zeta_1$ for $f \in C^\infty(\mathbb R)$ and $a \in \mathbb R$), then $\bar D |_{C^{\infty(\mathbb R)}}$ is still an isomorphism, though now between two different vector spaces. Inverting $\bar D$ on this space is just solving an initial value problem:

\begin{align}
F(x) &= \frac{df}{dx} \\
F(0) &= a
\end{align}

The solution is $F(x) = a + \int_0^x f(x) dx$. You can do similar constructions for higher derivatives. Obviously, solving initial value problems is highly important, and in fact the above construction is a natural generalization of solving initial value problems like this which allows for arbitrarily many derivatives to be taken. However, even though $\bar D |_{C^\infty(\mathbb R)}$ is still a linear isomorphism, I have a hard time thinking of this as a true way to invert derivatives in the way the OP wants, since the domain and range are different. Hence, in my view this is philosophically more along the lines of Option 3 below, which is to say that it's a way to work around the noninvertibility of the derivative rather than a way to "fix" the derivative.

Option 2a: Get rid of problematic functions by removing them from $C^\infty (\mathbb R)$

You may say "rather than adding functions, let's get rid of problematic ones which make this not work". Let's just define the antiderivative to be $\int_0^x f(x') dx'$, and restrict to the largest subspace $\hat C^\infty (\mathbb R)$ on which $D$ and $\int$ are inverses. the choice of $0$ for the lower limit is again arbitrary, but changing it doesn't change the story much.

If you do this, you can see that any function $f \in \hat C^\infty (\mathbb R)$ needs to have $f(0) = 0$, since $f(0) = \int_0^0 \frac{df}{dx} dx = 0$. But requiring $\frac{df}{dx} \in \hat C^\infty (\mathbb R)$ as well means that we must also have $\frac{df}{dx}|_{x=0} = 0$, and by induction $\frac{d^nf}{dx^n}|_{x=0} = 0$ for all $n$. One way of saying this is that the Maclaurin series of $f$ must be identically $0$. If you haven't taken a class in real analysis (which you should if you're deeply interested in these things), you may have never encountered such a function before which has all derivatives $0$ at $x=0$ (other than the zero function), but they do exist; one example is

\begin{equation*}
f(x) = \left\{
     \begin{array}{lr}
       e^{-1/x^2} & : x \ne 0\\
       0 & : x = 0.
     \end{array}
   \right.
\end{equation*}

There are many other examples, but I'll leave it at that; you can check the Wikipedia article on smooth, non-analytic functions for more information. On $\hat C^\infty (\mathbb R)$, the differential equation $D f = g$ for fixed $g \in \hat C^\infty (\mathbb R)$ has a unique solution given by $f(x) = \int_0^x g(x) dx$. It's worth pointing out that the only polynomial function in $\hat C^\infty (\mathbb R)$ is the $0$ function.

This space looks bizarre and uninteresting at first glance, but it's actually not that divorced from things you should care a lot about. If we return to $ C^\infty (\mathbb R)$, typically, we'll want to solve differential equations on this space. A typical example might look something like $$a_n D^n f + a_{n-1} D^{n-1} f + \cdots + a_0 f = g$$ for some fixed $g$. As you will learn from studying differential equations (if you haven't already), one way to get a unique solution is to impose initial conditions on $f$ of the form $f(0) = c_0, (Df)(0) = c_1 , \ldots, (D^{n-1} f) (0) = c_{n-1}$. All we've done in creating this space is find a class of functions upon which we can uniformly impose a particular set of initial conditions.

This set of functions also has another nice property which you might not notice at first pass: you can multiply functions. The object we constructed in Option 1 didn't really have an intuitive notion of multiplication, but now, we can multiply point-wise and still end up in the same space. This follows from the general Leibniz rule. In fact, not only can we multiply any two functions in $\hat C^\infty(\mathbb R)$, but we can multiply a function in $\hat C^\infty(\mathbb R)$ by an arbitrary smooth function in $C^\infty(\mathbb R)$. The result will still be in $\hat C^\infty(\mathbb R)$. In the language of algebra, $\hat C^\infty(\mathbb R)$ is an ideal of $C^\infty(\mathbb R)$, and ideals are well-studied and have lots of interesting and nice properties. We can also compose two functions in $\hat C^\infty(\mathbb R)$, and by the chain rule and we'll still end up in $\hat C^\infty(\mathbb R)$.

Unfortunately, in most applications, the space we've constructed isn't sufficient for solving differential equations which arise, despite all its nice properties. The space is missing very many important functions (e.g. $e^x$), and so the right hand side $g$ above will often not satisfy the constraints we impose. So while we can still solve the differential equations, to do it, we need to go back to a space on which $D$ is not one-to-one. However, it's worth pointing out that there are variants of this construction. Rather than putting initial conditions at $x=0$, in many applications, the more natural thing is to put conditions on the asymptotic behavior as $x \rightarrow \infty$. There are interesting parallels to our construction above in this context, but unfortunately, they're a bit too advanced to include here, as they require significantly more than what a typical undergraduate would know. In any case, these constructions are useful, as you would guess, in the study of differential equations, and if you decide to pursue that in detail, eventually you'll see something which looks qualitatively like this.

Option 2b: Get rid of problematic functions by equating functions in $C^\infty (\mathbb R)$

Those who have studied linear algebra will remember that there are two different concepts of a vector space being "smaller" than another vector space. One is subspaces. In 2a we constructed a subspace on which $D$ is one-to-one and onto. But we could just as well construct a quotient space with the same property.

Let's define an equivalence relation on $C^\infty(\mathbb R)$. We'll say $f_1 \sim f_2$ if and only if $f_1(x) - f_2(x)$ is a polynomial function in $x$, i.e. a function such that for some $n$, $D^n (f_1 - f_2) = 0$. Let $[f]$ be the set of all functions $g$ such that $g \sim f$ i.e. the equivalence class of $f$ under $\sim$. We'll call the set of all equivalence classes $\tilde C^\infty(\mathbb R)$. A generic element of $\tilde C^\infty(\mathbb R)$ is a collection $[f]$ of functions in $C^\infty(\mathbb R)$, all of which differ from each other only by polynomial functions.

On this space, we have a derivative operator $\tilde D$ given by $\tilde D [f] = [D f]$. You can check that this is well-defined; that is, that if you pick a different representative $g$ for the class $[f]$, that $[D f] = [D g]$. This is just the fact that the derivative of a polynomial function is again a polynomial function. An important thing to realize is that if $f \in C^\infty(\mathbb R)$ is such that $[f] = 0$, then $f$ is a polynomial function.

In some sense, this is a strictly bigger space than $\hat C^\infty(\mathbb R)$, because it's easy to check that the map $\hat C^\infty(\mathbb R) \rightarrow \tilde C^\infty(\mathbb R)$ given by $f \mapsto [f]$ is one-to-one (since $\hat C^\infty(\mathbb R)$ has no nonzero polynomial functions). But we've now got access to other important functions, like $f(x) = e^x$, so long as we're content to think of them not as actual functions, but as equivalence classes. Of course, if we impose initial conditions on a function and all derivatives at a given point, we can reconstruct any function from its equivalence class, and thus exactly solve differential equations that way.

The expanded set of functions does come at a price. We've lost the ability to multiply functions. To see this, note that $[0] = [1]$ as constant functions, but if we multiply both by $e^x$, $[0] \ne [e^x]$. This flaw is necessary for us, but it does make it harder to study this space in any depth. While $\hat C^\infty(\mathbb R)$ could be studied using algebraic techniques, we're less able to say anything really interesting about $\tilde C^\infty(\mathbb R)$.

The real big reason to use a space like this though (for me at least) is for something like asymptotic analysis. If you only care about the behavior of the function for very large arguments, it turns out often to not matter very much (in a precise way) exactly how you pick integration constants. The space we've constructed here is really only good for being able to solve differential equations of the form $D^n f = g$, but with some work you can extend this to other kinds of differential equations. With some more work (i.e. quotienting out functions which are irrelevant asymptotically), you can get rid of other functions which you don't care about asymptotically.

Such spaces are often used (implicitly and typically without rigorous definition) as a first approximation in theoretical physics and in certain types of research on ordinary differential equations. You can get a very course view of what the solution to a differential equation looks like without caring too much about the details. I believe (and am far from an expert on this) that they also have some application in differential Galois theory, and while that would be an interesting direction to go for this answer, it's probably too advanced for someone learning this for the first time, and certainly outside my expertise.

Option 3: Realize that $D$ not being invertible isn't necessarily a bad thing

This is the direction most good mathematics takes. Just stick with $D$ and $C^\infty(\mathbb R)$, but realize that the non-invertibility of $D$ isn't necessarily a very bad thing. Sure, it means you can't solve differential equations without some initial/boundary conditions, which is to say that you don't have unique anti-derivatives, but let's think about what that buys you. An interesting trend in modern mathematics is realizing that by forgetting information, sometimes you can make other information more manifest and hence easy to work with. So let's see what the non-invertibility of $D$ can actually tell us.

Let's first look at an application, and since I'm a physicist by day, I'll pick physics. Newton's second law is perhaps the most fundamental law in classical mechanics. For a particle travelling on a line, this looks like $\frac{d^2}{dt^2} x(t) = f(\frac{d}{dt}x(t) ,x(t), t)$. I've switched the variables to position for $x$ (the function of interest) and the independent variable as $t$. In general this equation isn't very easy to solve, but we note that it has two derivatives maximum, so we expect that (morally speaking) we'll have to integrate twice, and be left with 2 free parameters.

The two-dimensional space of all such solutions to a system is called "phase space", and as it turns out, this is a very fundamental concept for all of physics. We can take the parameters to be the position and velocity (or momentum) at the initial time. The fact that the full behavior of the system depends only on the initial position and velocity of the system is something that isn't emphasized enough in introductory courses, but is really fundamental to the way that physicists think about physics, to the point that anything involving three derivatives is viewed as confusing and often unphysical. Jerk (i.e. the third derivative of position) isn't expected to play any role at a fundamental level, and when it does show up (e.g. the Abraham–Lorentz force), it is a source of much confusion.

That isn't the end of the story though. We know that the world is better described by quantum mechanics than classical mechanics at short distances. In quantum mechanics, we don't have a two-dimensional phase space. Rather, the space of states for a particle on a line are "wave"functions $\mathbb R \rightarrow \mathbb C$. The magnitude squared describes the probability distribution for measurements of position of the particle, while the complex phase of the function describes the distribution of momenta in a precise way. So rather than having 2 real dimensions, we really only have 1 complex one, and use it to describe the distributions of both position and momenta. The Schrödinger equation (which is the quantum mechanical version of Newton's 2nd law) describes how the wavefunction evolves in time, but it involves only one time derivative. This process of cutting down the two-dimensional phase space to a 1-dimensional space (upon which you look at probability distributions) is a source of ongoing research among mathematicians to understand the full extend and conditions on which it can be performed; this program goes by the name of geometric quantization.

Let's move back to pure mathematics, to see how you can use the non-invertibility of the derivative in that context as well. We already know that the kernel of $D$ is just the set of constant functions. This is a 1-dimensional space; that is, $\dim \ker D = 1$. But what would happen if instead of looking at smooth functions $\mathbb R \rightarrow \mathbb R$, we remove $0$ from the domain? So we're looking at a function $f \in C^\infty((\infty, 0) \cup (0, \infty))$, and we want it to be in the kernel of the derivative operator on this space.

The differential equation $Df=0$ isn't hard to solve. You can see that $f$ needs to be locally constant just by integrating. But you can't integrate past $0$ since the function isn't defined there, so the general form of any such $f$ is

\begin{equation*}
f(x) = \left\{
     \begin{array}{lr}
       c_1 & : x <0 \\
       c_2 & : x > 0.
     \end{array}
   \right.
\end{equation*}

But wait! Now this function depends on 2 parameters; that is, $\dim \ker D = 2$ here. You can convince yourself that if you take the domain of the functions in question to be the union of a collection of $n$ disjoint intervals on $\mathbb R$, then a function solving $Df = 0$ for the derivative on this space is constant on each interval, and hence these form an $n$-dimensional space; i.e. $\dim \ker D = n$.

So, while we have to forget about the constant functions, we're getting information back in the form of the kernel of $D$. The kernel's dimension is exactly the number of connected components of the domain of the functions. This continues to be true even if we allow the domain to be much more complicated than just a disjoint union of lines. For any smooth manifold (i.e. a space which looks locally like Euclidean space but may be connected in ways which a line or a plane isn't, e.g. a circle or a sphere) $M$, if $D$ is the gradient on $C^\infty(M)$, then $\dim \ker D$ is the number of connected components of $M$. You may not yet know multivariable calculus, so this may go over your head a bit, but it's nonetheless true and important.

This may seem like a random coincidence, but it's a rather deep fact with profound generalizations for topology. Unfortunately, there's a limit to how much we can do with one dimension, but I'll push that limit as much as we can.

Let's look at smooth functions on a circle $S^1$. Now, of course, we already know how $D$ fails to be injective (one-to-one) on $C^\infty(S^1)$, and that the kernel is 1-dimensional, since a circle is connected. What I want to look at is the failure of surjectivity; that is, will $D$ be onto? Of course, what exactly we mean by differentiating on $S^1$ isn't obvious. You can think of $Df$ as a gradient vector which points in the direction which $f$ is growing proportional to the rate of growth at any given point. 1-dimensional vectors are (for our purposes, which are purely topological), just numbers, so we get a number at every point. This turns out to not be exactly the right way to think about this for the sake of generalizing it, but it's good enough for now.

As it turns out, when you go all the way around a circle, you need to end up exactly back at the same value of the function you started at. Let's parametrize points on the circle by angle, running from $0$ to $2\pi$. This means that, if we want $f = D g$ for some function $g \in C^\infty(S^1)$, we need $$\int_0^{2 \pi} f(\theta) d \theta = \int_0^{2 \pi} \frac{dg}{d\theta} d\theta = g(2 \pi) - g(0) = 0.$$ So the average value of $f$ needs to be $0$.

This means that $\operatorname{im} D$ isn't all of $C^\infty (S^1)$. It's just those functions with average value $0$. That wasn't the case when we had $\mathbb R$. Now, if you take a function $f$, you can decompose $f(x) = f_0 + f_d(x)$, where $f_0$ is the average of $f$, and $f_d(x)$ is the deviation from the average at the point $x$. $f_d$ is a function that has has $0$ average, and so it's in the image of $D$. So if we look at the quotient space $C^\infty (S^1) / \operatorname{im} D$, it's essentially just the space of constant functions on $S^1$, which is 1-dimensional. That is, $\dim C^\infty (S^1) / \operatorname{im} D = 1$.

At this point, we've actually come upon something pretty surprising. An arbitrary 1-manifold $M$ is necessarily a disjoint union of $n$ intervals and $m$ circles, for some values of $n$ and $m$ (which could be infinite, but I'll ignore this). And we can classify these just by understanding what $D$ does on $M$; specifically, $\dim \ker D = n + m$ is the number of components, and $\dim C^\infty (S^1) / \operatorname{im} D = m$ is the number of circles.

It turns out that one can generalize this to not only measure circular holes in a manifold, but spherical holes of other dimensions. The generalization is neither obvious nor easy, but it does take you to something rather close to modern mathematical research. Specifically, you'd end up at the de Rham cohomology of a manifold. This is a way to compute topological invariants of your manifold $M$ which depend just on the (non)existence to solutions of certain simple classes of differential equations in some unrigorous sense the de Rham cohomology (or more precisely the Betti numbers, which are the dimension of the cohomology) counts the number of $n$-dimensional holes of $M$ for each nonnegative integer $n$. This answer is already too long and the amount of material to construct this is far too much to include here, but you may take a look at these blog posts to get some more information on this for starters: More than Infinitesimal: What is “dx”? and Homology: counting holes in doughnuts and why balls and disks are radically different..

Anyway, modern-day research in algebraic topology has progressed far enough that we no longer particularly need to construct cohomology this way; there are a plethora of other options and a great number of generalizations. But this is still among the most tractible and intuitive ones. There are many further applications of these things, the most incredible (for me at least) of which is probably the Atiyah-Singer index theorem, but that's unfortunately far too advanced to describe here.

So, for me anyway, the question at the end of the day isn't "Why do we define the derivative so that it isn't invertible", so much as "How can we use the fact that the derivative is not invertible to our advantage to do more mathematics?". I've really only given a couple of examples, but hopefully they're enough to convince you that this non-invertibility is actually not entirely a bad thing, and that it can be used to our advantage if we're intelligent about it.

They are inverses up to an additive (arbitrary) constant (given certain conditions). The Fundamental Theorem of Calculus gives the details.

Your definition is a bit confusing. I think there may be brackets missing. But one reason why your definition doesn't work can be seen if you use $2x+2$ instead of $2x+1$. You end up with the same problem.


Find the number of ordered triples (a,b,c) of positive integers such that $30a + 50b + 70c \le 343.$ Find the number of ordered triples $(a,b,c)$ of positive integers such that $30a + 50b + 70c \le 343.$

My confusion is that while solving the question a, b,c can be zero or not

It is the question of combinations every variable can have the values from 1-4 except the variable c bcz 70*4=280 and 50+ 30=80 which exceeds 343.so variable c can have values from 1-3.No we will make combinations for each value of a,b,c and hence the answer will come and it will be 30


Is this an error in my textbook? (Fluid dynamics: Chorin & Marsden) The book uses the notation of a complex flow $F(z)=u-iv$, with a function $W=\phi + i\psi$ such that $F=\frac{dW}{dz}$.

The book gives the following example:

  Choose $$W(z) =\frac{\Gamma}{2\pi i}\log(z-z_0)$$ where $\Gamma$ and $z_0$ are some constants. Then [...] $\psi$ is constant on any circle centered at $z_0$.

Shouldn't it be $\phi$ that's constant? The real part of $W$ is $\phi=\frac{\Gamma}{2\pi i}\log(|z-z_0|)$, which is constant on circles centred at $z_0$. The imaginary part, $\psi$ is a function of $\arg(z)$, which is certainly not constant on any circle.

The book also has the following diagram:

What am I missing here?

Let $z = x+iy$ be a complex number. Define
$$
 \ln z = \frac{1}{2} \ln \color{blue}{\left( x^{2} + y^{2} \right)} + i \text{Arg } (x + y)
$$

When $\color{blue}{\left( x^{2} + y^{2} \right)}$ is constant, the $\color{blue}{real}$ part of the logarithm is constant.

On the circular domains, the real portion of the logarithm is constant.

A simple case is
$$\ln e = 1, \qquad \ln (i e) = 1 + i \frac{\pi}{2}$$.


Number of Arithmetic Operations in Gaussian-elimination/Gauss-Jordan Hybrid Method for Solving Linear Systems I am stucked at this problem from the book Numerical Analysis 8-th Edition (Burden) (Exercise 6.1.16) :

Consider the following Gaussian-elimination/Gauss-Jordan hybrid method for solving linear systems:

First apply the Gaussian-elimination technique to reduce the system to triangular form. Then use the $n$-th equation to eliminate the coefficients of $x_{n}$ in each of the first $n-1$ rows. After this is completed use the $(n-1)$-st equation to eliminate the coefficients of $x_{n-1}$ in the first $n-2$ rows, ans so on. The system will eventually appear as the reduced system we get by applying Gauss-Jordan method to the original system.

Show that this method requires $\frac{n^3}{3}+\frac{3}{2}n^2-\frac{5}{6}n$ multiplication/divisions and $\frac{n^3}{3}+\frac{n^2}{2}-\frac{5}{6}n$ additions/subtractions.

But when I tried to calculate the total number of multiplication/divisions I got $\frac{n^3}{3}+\frac{3}{2}n^2-\frac{11}{6}n$, Here's what I've done (Sorry for being a bit long):

The augmented matrix for the system has the form

$$ \left[
    \begin{array}{cccc|c}
        a_{1,1} & a_{1,2} & \cdots & a_{1,n} & a_{1,n+1} \\
        a_{2,1} & a_{2,2} & \cdots & a_{2,n} & a_{2,n+1} \\
        \vdots  & \vdots  & \ddots & \vdots & \vdots  \\
        a_{n,1} & a_{n,2} & \cdots & a_{n,n} & a_{n,n+1}
    \end{array}
\right] $$

after applying Gaussian-elimination (without backward substitution) to this system we get

$$ \left[
    \begin{array}{cccc|c}
        \hat{a}_{1,1} & \hat{a}_{1,2} & \cdots & \hat{a}_{1,n} & \hat{a}_{1,n+1} \\
        0 & \hat{a}_{2,2} & \cdots & \hat{a}_{2,n} & \hat{a}_{2,n+1} \\
        \vdots  & \ddots  & \ddots & \vdots & \vdots  \\
        0 & \cdots & 0 & \hat{a}_{n,n} & \hat{a}_{n,n+1}
    \end{array}
\right] $$

We know that the total number of multiplications/divisions and additions/subtractions of the Gaussian-elimination technique is $\frac{n^3}{3}+\frac{1}{2}n^2-\frac{5}{6}n$ and $\frac{1}{3}n^3-\frac{1}{3}n$ respectively.

Now we will use the $i$-th row to eliminate the coefficients of $x_i$ in each of the first $i-1$ rows for each $i=n,n-1,...,2$ ($i$ starts at $n$ and goes down to $2$), and we will get:

$$ \left[
    \begin{array}{cccc|c}
        \hat{\hat{a}}_{1,1} & 0 & \cdots & 0 & \hat{\hat{a}}_{1,n+1} \\
        0 & \hat{\hat{a}}_{2,2} & \cdots & 0 & \hat{\hat{a}}_{2,n+1} \\
        \vdots  & \ddots  & \ddots & \vdots & \vdots  \\
        0 & \cdots & 0 & \hat{\hat{a}}_{n,n} & \hat{\hat{a}}_{n,n+1}
    \end{array}
\right] $$

Since for each $i$ we do $i - 1$ divisions (inorder to calculate $\frac{\hat{a}_1,i}{\hat{a}_{i,i}}$ , $\frac{\hat{a}_2,i}{\hat{a}_{i,i}}$ , ... , $\frac{\hat{a}_{i-1},i}{\hat{a}_{i,i}}$), After that we subtract each of the elements in the row $j=1,2,...,i-1$, $\frac{\hat{a}_j,i}{\hat{a}_{i,i}}$ times the $i$-th row, and so additional $i-1$ multiplications are carried out, and we get that the total number of multiplications/divisions in this step is $\Sigma_{i=2}^n2(i-1)=n^2-n$.

And so we get that the total number of multiplications/divisions required to apply the hybrid method is $\frac{n^3}{3}+\frac{1}{2}n^2-\frac{5}{6}n + \frac{1}{2}n^2-\frac{1}{2}n = \frac{1}{3}n^3+\frac{3}{2}n^2-\frac{11}{6}n$ which is different from $\frac{n^3}{3}+\frac{3}{2}n^2-\frac{5}{6}n$.

Thanks for any help.

To provide an answer to the question from Algebraic Pavel's comment:

  You should add the final step: $n$  divisions required to compute the final solution of the diagonal system (or, equivalently, "normalizing" the diagonal matrix to the identity). That will give you the desired term $-\frac56n$.

That is to say, your entries need to be $1$ at the end to actually get the solution.

Gaussian elimination :
(n+1)(n-1)+(n)(n-2)+...+(3)(1)=sigma(k=1 to n-1)(k+2)(k)=n^3/3+n^2/2-5n/6

2.Backward gaussian method :
2(n-1)+2(n-2)+...+2(1)=2*sigma(k=1 to n-1)(k)=n^2-n

3.Find the solution :
n

Sum of multi./div. = sum of 1,2,3 = n^3/3+3n^2/2-5n/6

I tried to upload my picture to explain it in detail but i dont know how to do it...


What is the derivative of $\ln|x|$? I am somewhat confused about finding the derivative of $\ln|x|$. Is it $1/|x|$ or $1/x$?
What if instead of $x$ we have a general function $f(x)$? Most importantly: why?

It's $\,\dfrac 1x$. Indeed, if $x<0$, $\bigl(\ln\lvert x \rvert\bigr)'=\bigl(\ln(-x)\bigr)'=\dfrac1{-x}\times -1$ by the chain rule.

It's $\frac{1}{x}$ (for positive $x$ value, for negative $x$ you would need the absolute value) , the derivative itself can be derived from the fact that derivative of $e^{x}$ is also $e^{x}$ (by letting $y = ln(x)$, so $x = e^{y}$, then you can derive the derivative of $\ln(x)$ with implicit differentiation), you can also do the formal definition of derivative to get it, using logarithm properties and also the limit definition of $e^{x}$ (you will need those two thing if you're using the formal definition of derivative)


The Klein Configuration -- representations In the Klein configuration there are 60 points and 60 planes, with each point lying on 15 planes and each plane passing through 15 points. A $60_{15}$ configuration.

Page 34 of Abstract Configurations in Algebraic Geometry has one description, and Kummer's Quartic Surface by Hudson (which I've ordered) has another.  But I find both of these descriptions murky.

The incidence graph between the planes and points would be a regular graph of 120 points and valence 15, but I can't find any references to such a graph.  For the Kummer $16_6$ configuration, it's easy to find the Kummer graph.

A labeled set of planes and point would also have nice properties in combinatorial designs, but I haven't found it in the Handbook of Combinatorial Designs and other sources.  I can't find it as a pseudoline arrangement.

Does anyone have a handy representation?  Is there a famous graph I'm missing?


Does the same determinant make two matrices equal to each other? Does the same determinant make two matrices equal to each other?

If I have:

Find all values of $x$ that make
$$\begin{pmatrix}2 & -1 &4\\3 & 0 & 5\\4 & 1 & 6\end{pmatrix}=\begin{pmatrix} x & 4\\5 & x\end{pmatrix} $$

Would I calculate and equate the determinants of both matrices to solve this problem?
Edit: Below is the exact question. Do the style of brackets refer to the determinants?

That style of brackets usually refers to the matrix itself, rather than the determinant. The question either uses an unusual style, or is in error.

It is customary to use square brackets $[\ddots]$ to refer to the matrix as a matrix.  It is also customary to use  vertical bars $|\ddots|$ to refer to the determinant of the matrix.

It is not customary to use ordinary brackets $(\ddots)$ for anything.

However, mathematicians are fond of making their own notation.  So, look in your text book and/or your lecture notes to see how they define this.

Matrices are equal to each other only if they are the same size and every member is equal to the member in the same place.  So, in this case there are no solutions.

Determinants are just numbers and can be equal even if the matrices are not.

Since schools like to give exercises which actually have solutions, they are most likely to be talking about determinants, but it can possibly be a trick question.


Probabilistic modelling This question is about how to make sense of a probabilistic model I'm reading about:
You have three random variable $A$, $B$ and $C$ that are all real-valued. You want to model an input-output relationship with these random variables, by assuming the following relationships between them:
It is assumed that $A$ is the constant random variable that only takes the value $a\in \mathbb
{R}$, where $a$ is thought of as input.
Furthermore, $P(B=b | A=a)=\mathcal{N}(b;3\cdot e^a,1)$ and  $P(C=c | B=b)=\mathcal{N}(c;2\cdot b^2,1)$, where $\mathcal{N}(x;\mu,1)$ denotes the probability density of a normal distribution with mean $\mu$ and variance $1$.
My questions are:

Does an equation like $P(B=b | A=a)=\mathcal{N}(b;3\cdot e^a,1)$ even make sense?
If we takes samples, on the left we have a number between $0$ and $1$ but on the right we have a number that can be larger, as the pdf does not need to stay within $[0,1]$, right?

What is the output?
In the (not online available) text I'm reading, the above is all that is mentioned. I would assume that to arrive at the output, one needs to proceed in the following way:
Given $a$, draw a number $b$ from a normally distribution with mean $3\cdot e^a$ and unit variance and then similarly draw $c$, the output, from a normally distribution with mean with mean $2\cdot b^2$ and unit variance. But I'm not sure if that's correct; also, it is not clear to me, in case this interpretation is indeed correct, why conditional probabilities in the model formulation are needed.

Firstly, the following

It is assumed that $A$ is the constant random variable that only takes the value $a\in \mathbb{R}$, where $a$ is thought of as input.

does not really make sense. If something is assumed to be constant, then there is no need to model it as a random variable. Moreover if $A$ is going to model the input, it must vary from time to time. So let us assume that this is rather poorly written - $A$ is a random variable with some (maybe unknown) non-constant distribution. Maybe it should be discrete, or taking only two values? It might make sense.
Nextly, we have

Furthermore, $\mathbb{P}(B=b|A=a)=\mathcal{N}(b;3⋅e^a,1)$ and $\mathbb{P}(C=c|B=b)=\mathcal{N}(c;2⋅b^2,1)$, where $\mathcal{N}(x;\mu,1)$ denotes the probability density of a normal distribution with mean $\mu$ and variance $1$.

This is also weirdly written, and makes little sense. You have written

...on the left we have a number between 0 and 1 but on the right we have a number that can be larger...

You are write, we should not use $\mathbb{P}$ sign when the probability density function is really what we are referring to.
What the author probably wanted to say, was that conditionally on $A$, the distribution of $B$ is normal with given ($A$ dependent) parameters. Later on, after $(A, B$) have been sampled, the distribution of $C$ is also normal with the assigned ($B, A$ dependent) parameters.
The above is only my educated guess, as in the cited text it is written
$$\mathbb{P}(B=b|A=a)=\mathcal{N}(b;3⋅e^a,1),$$
\begin{equation} \mathbb{P}(C=c|B=b)=\mathcal{N}(c;2⋅b^2,1), \tag{1} \end{equation}
and I assume that it should have been
$$\mathbb{P}(B=b|A=a)=\mathcal{N}(b;3⋅e^a,1),$$
\begin{equation}\mathbb{P}(C=c|B=b, A=a)=\mathcal{N}(c;2⋅b^2,1). \tag{2} \end{equation}
My guess is based on the fact, that as You say, we are actually going to model some process. As long as $(1)$ might be valuable piece of information, it does not define a unique model.
Moreover, this should be written in a more clear way though. For example:
$$B|A \sim \mathcal{N}(3\cdot e^A,1),$$
and
$$C|B,A \sim \mathcal{N} (2\cdot B^2,1).$$
Note that this is exactly what You have written in Your interpretation

Given $a$, draw a number $b$ from a normally distribution with mean $3\cdot e^a$ and unit variance and then similarly draw $c$, the output, from a normally distribution with mean with mean $2\cdot b^2$ and unit variance.

I think it is OK. Saying it once again in different words:

we start by sampling $A$,
given $A$, we sample $B$ from $\mathcal{N}(3\cdot e^A,1)$,
given $B, A$, we sample $C$ from $\mathcal{N} (2\cdot B^2,1).$

Such models - known as hierarchical models - are very important  in modern statistics.
Basically, Bayesian Statistics is devoted almost entirely to the study of such models.

I think the question is almost correct except for a few notational issues
The first:
Note $B|A \sim \mathcal{N}(3e^a,1)$. Therefore $\mathbf{P}(B=b|A) = 0$ This follows from a basic definition of continous random variables. What would be true is the density function that is $f_{B|A}(b) = 1/\sqrt{2\pi}e^{\frac{-(b - 3e^a)^2}{2}}$
The second the output is ofcourse C
The output is a probabilistic output that is it's a random variable whose density function conditioned on the input is
$f_{C|A}(c) = \int_\mathbf{R}f_{C|B(x)}f_{B(x)|A}dx$
$ = \int_\mathbf{R}1/\sqrt{2\pi}e^{\frac{-(c - 2x^2)^2}{2}}1/\sqrt{2\pi}e^{\frac{-(x - 3e^a)^2}{2}}dx$
$= 1/2\pi \int_\mathbf{R}e^{\frac{-(c - 2x^2)^2 -(x - 3e^a)^2}{2}}dx$
EDIT Since you mentioned about the credibility, I really dont know what to say. Do we have to show our academic transcript or something?


Let X and Y be random variables with joint pdf $f(x,y)=x+y$ for $0<x<1$ and $0<y<1$. Are X and Y independent? My heart says yes but my math says no. I am using the fact that $X$ and $Y$ are independent if and only if $f_X(x)f_Y(y)=f(x,y)$. So I have

$$f_{X}
(x)=\int_{0}^{1}f(x,y)dy\\=\int_{0}^{1}x+ydy\\=[xy+\frac{y^{2}}{2}]_{y=0}^{y=1}\\=x+\frac{1}{2}$$

and by basically exactly the same math, $f_Y(y)=y+\frac{1}{2}$. Then $$f_X(x)f_y(y)=(x+\frac{1}{2})(y+\frac{1}{2})=xy+\frac{1}{2}(x+y)+\frac{1}{4}\ne f(x,y)$$

And hence they are not independent. But can that be right? Why would the value of X have anything to do with the value of Y? It's not like one is a function of the other.

Or have I made a simple mistake? I looked through a couple of times and I'm pretty sure my math is right...

Note that the conditional density f(x|y) depends on y.  f(x|y)=f(x,y)/f(y).

You showed that f(y)=y+1/2 and f(x,y)=x+y.  So f(x|y)=(x+y)/(y+1/2)

for any 0<=x<=1 and 0<=y<=1.

It clearly is not independent of y. So knowing y does addect the probability that X is in an fixed interval about x.

Your right. You can also check $E[XY]=E[X]E[Y]$ for the independence condition. This may be clearer because expected values are constant.
Also, independence is interpreted as that even if you know the outcome $X=x$, you cannot use that to guess $Y$: $f(Y|X)=f(Y)$.
For $f(x,y)=x+y$, if you know $X=x$, you have a better idea of what $Y$ will be.
For example, if $X=0.1$ or $X=0.5$, the probability of $Y=0.5$ under these two conditions are different.


Show that $\sqrt{1+x}0$ I am a little stuck on this question and would appreciate some help. The question asks me to prove that $\sqrt{1+x}<1+\frac{x}{2}$ for all $x>0$.

I squared both sides of the question to get $1+x<\frac{x^2}{4}+x+1$ for all $x>0$. Then, I multiplied both sides by $4$ to get $4+4x<x^2+4x+4$ for all $x>0$.

I am a little stuck and was wondering what to do after this step and how to actually provide sufficient proof to say that this statement is true.

You did well. Let's finish it:
we have $x^2>0$ thus $\dfrac{x^2}{4}>0$. Adding $x+1$ to both sides, we have
$$\dfrac{x^2}{4}+x+1> x+1$$
or $$(\frac{x}{2}+1)^2>x+1$$
or
$$\frac{x}{2}+1>\sqrt{x+1}$$

The important principle here is that if $a,b$ are nonnegative real numbers, then $a<b$ iff $a^2<b^2.$ This property is used all the time. If we grant this, then all we need to show is that

$$(\sqrt{1+x})^2 < (1+x/2)^2,$$

which is the same as saying $1+x < 1+x + x^2/4.$ That is obviously true, and we're done.


Verifying that $\mathcal{P}(\mathbb{N})\times\mathcal{P}(\mathbb{N})=\text{ the }\sigma\text{- algebra generated by }\mathbb{N}\times\mathbb{N}$ I am asked to show that the product $\sigma$-algebra of $\mathcal{P}(\mathbb{N})$ with $\mathcal{P}(\mathbb{N})$ consists of all subsets of $\mathbb{N}\times\mathbb{N}$, or in other words that $\mathcal{P}(\mathbb{N})\times\mathcal{P}(\mathbb{N})=\mathcal{P}(\mathbb{N}\times\mathbb{N})$ (where $\mathcal{P}(\mathbb{N}\times\mathbb{N})$ denotes the algebra generated by $\mathbb{N}\times\mathbb{N}$). The power set of the natural numbers, $\mathbb{N}$, is here denoted $\mathcal{P}(\mathbb{N})$.

I am unsure how to go about doing this. I feel like
\begin{align}
\mathcal{P}(\mathbb{N})\times\mathcal{P}(\mathbb{N}) &= \{A:A\subset\mathbb{N}\}\times\{B:B\subset\mathbb{N}\} \\ &= \{(A,B):A\subset\mathbb{N},B\subset\mathbb{N}\} \\
&= \{(A,B):A,B\subset\mathbb{N}\}
\end{align}
is exactly all possible subsets of $\mathbb{N}\times\mathbb{N}$, and as such should be just equal to $\mathcal{P}(\mathbb{N}\times\mathbb{N})$. However, I am given the hint that $\mathbb{N}\times\mathbb{N}$ is countable, which leads me to think there is more going on here. Both $\mathcal{P}(\mathbb{N})$ and $\mathcal{P}(\mathbb{N}\times\mathbb{N})$ are uncountable, so I don't know how to put this information to use.

Any hints would be greatly appreciated!

Edit: Attempted to make the notation more clear.

The product $\sigma$-algebra of $\mathcal{P}(\mathbb{N})$ with itself (usually denoted $\mathcal{P}(\mathbb{N}) \otimes \mathcal{P}(\mathbb{N})$) is a $\sigma$-algebra of $\mathbb{N} \times \mathbb{N}$, so it will automatically be a subset of $\mathcal{P}(\mathbb{N} \times \mathbb{N})$. You just need to show the reverse inclusion: any set in $\mathcal{P}(\mathbb{N} \times \mathbb{N})$ will also be in the product $\sigma$-algebra.

Recall that by definition, $\mathcal{P}(\mathbb{N}) \otimes \mathcal{P}(\mathbb{N})$ is the $\sigma$-algebra generated by sets of the form $A \times B\subset \mathbb{N} \times \mathbb{N}$ where $A,B \in \mathcal{P}(\mathbb{N})$. In particular $A$ and $B$ can be one-point sets, which shows that one-point sets $\{(a,b)\}$ are in the product $\sigma$-algebra. Now use the properties of $\sigma$-algebras to show that consequently, any set in $\mathcal{P}(\mathbb{N})$ is also in the product $\sigma$-algebra.

You have two injections:
$\mathcal{P}(\mathbb{N})\subset\mathcal{P}(\mathbb{N})\times\mathcal{P}(\mathbb{N})\subset\mathcal{P}(\mathbb{N}\times\mathbb{N})$.

On the other hand, $\mathbb{N}=\mathbb{N}\times\mathbb{N}$, hence $\mathcal{P}(\mathbb{N})=\mathcal{P}(\mathbb{N}\times\mathbb{N})$.

QED.


Finding a function based on their limits I have points and limits of a function and even the shape of the function and I'm looking for the function, something that very interesting for me how could I control the curve of the function?

(1) $\lim\limits_{x \to inf} f(x) = 1 $

(2) $f(\frac{1}{c}) = 1 $

(3) $0\lt x$

(4) $0\lt c \leq 1$

One such function is $1-(1-c^{2}x^{2})^{+}$ where $y^{+}$ denotes $\max \{y,0 \}$.

Edit: this function is convex on $(0,\frac 1 c)$. To make it concave consider $\min \{1, \frac {1-e^{-cx}} {1-e^{-1}} \}$.

This function satisfies all conditions except one of them, but maybe it'll help you, try $f(x)= -\frac{1}{x+1} +1$


Why is it the case that $\rm\langle Socrates\rangle = Socrates$, while $\rm\{Socrates\}\neq Socrates$? In Nolt's book called Logics (part III, chapter 7, § 7.1 Sets and $n$-tuples) one can read:

  Ordered one-tuples, unlike unit sets, are the same things as the object they contain. Thus , for example, the one-tuple $\rm (Socrates) = Socrates$; and more generally, for any object $x$, $(x)  = x$.

Could you please explain Nolt's (too) short explanation (at least for me)?

Is this explanation a way of saying that a "$1$-tuple" is not a well defined object?

Is there a way to derive the formula $(x) = x$ from the set theoretic definition of an $n$-tuple?

The explanation identifies a $1$-tuple with its object. So while $x$ is generally not the same as $\{x\}$, the $1$-tuple $(x)$ is identified with $x$ itself.

This is perfectly reasonable to do, and so people do it. But does that clash with the set theoretic definition? Well, that depends on your definition of a $1$-tuple to begin with. Since the language of set theory only has $\in$, any definition of $1$-tuples is just a way to interpret this concept of a sequence of length $1$.

Generally, we think about $n$-tuples as functions from $\{0,\ldots,n-1\}$, so a $1$-tuple would be a function from $\{0\}$ into something. But this is not something which is hardcoded into set theory, it's just one way to interpret $n$-tuples. You can do it just as well in a different way, or even explicitly decide this is fine for $n>1$, and for $n=1$, $(x)$ is just $x$ itself (which then raises the question what is a $1$-tuple whose object is a $3$-tuple?).

This is also why we can't require $\{x\}=x$ in general. Since we understand $\{x\}$ as a set whose unique member is $x$. And then $\varnothing\neq\{\varnothing\}$, since exactly one of these has an element. But tuples are looser, since as we see, they are not hardcoded into set theory like singletons.

Your book's author was being careless and accidentally contradicted set theory. There are two ways tuples are defined, and neither of them allow for $\{x\} = x$.

A function $T : \mathbb{N} \to S$ together with a collection of projection functions $\pi_i : \{i\} \to \mathbb{N}$ such that $\pi_i \circ T$ is the ith element of $T$. Here obviously $(x) \neq x$.
Some set theoretic definition involving the Kuratowski pair. And in these definitions, $(a) = \{a\}$ which according to the axiom of regularity cannot be equal to $a$.

So no, your book has made a mistake.


Find the radius of convergence of the series $y=\sum_{n=0}^{\infty}\binom{p}{n} x^{n}$ Let $p\in R$ Find the radius of convergence of the series:

$$y=\sum_{n=0}^{\infty}\binom{p}{n} x^{n}$$

Show that y satisfies the differential equation $(1+x)y'=py$ and initial condition $y(0)=1$

I'm not sure if this question has been asked, but I searched and could not find it. So I know that the radius of convergence is the radius of the largest disk that the series converges in. I'm confused by this problem in general and am unsure of where to even begin. I am looking for a clear solution and explanation to this problem.

It is just the generalization of the usual binomial series:
$$ f(x)=\sum_{n\geq 0}\binom{p}{n} x^n = (x+1)^p.\tag{1} $$
$(1+x)\,f'=p\,f$ is a separable ODE, and $\frac{f'}{f}=\frac{p}{x+1}$ gives that $x=-1$ is a singularity for $f(x)$.

It is the closest singularity to the origin, hence the radius of convergence of $(1)$ is just $\color{red}{1}$.

$$\sum_{n=0}^\infty \binom{p}{n}x^n=\sum_{k=1}^p\binom{p}{k}x^k=(1+x)^p$$
for all $x$. Therefore the radius of convergence is $\mathbb R$.


Why is $\frac{987654321}{123456789} = 8.0000000729?!$ Many years ago,
I noticed that $987654321/123456789 = 8.0000000729\ldots$.

I sent it in to Martin Gardner at Scientific American
and he published it in his column!!!

My life has gone downhill since then:)

My questions are:

Why is this so?
What happens beyond the "$729$"?
What happens in bases other than $10$?

I think @BorisNovikov's answer is the best here, but I had to do some work to understand it; this is my attempt at clarifying his answer, hopefully this can help others too:

In the following, we adopt the notation $123456789 = [ 1, 2, ..., B-1 ]_B$ in base $B=10$.

Boris introduced the following sum:
$$
  S_d(B) = [d, d-1, ..., 1]_B = 1 + 2B + ... + dB^{d-1} = \sum_{k=1}^d kB^{k-1} = \frac{dB^{d+1} - (d+1)B^d + 1}{(B-1)^2}
$$
where $d$ is the number of digits, $B$ the base, and the final formula is obtained by derivation of the usual geometric sum.

From there, it is easy to see that:
$$
  [1, 2, ..., d]_B = d + (d-1)B + ... + 1B^{d-1} = B^{d-1} S_d(1/B)
$$
and putting both results together, we have:
$$
  \frac{ [B-1, B-2, ..., 1]_B }{ [1, 2, ..., B-1]_B } = \frac{ S_{B-1}(B) }{ B^{B-2} S_{B-1}(1/B) } = \frac{1 + (B-2)B^B}{B^B - B(B-1) - 1}
$$

Clearly both the numerator and denominator are dominated by the terms in $B^B$, so this fraction is equivalent to $B-2$ as $B$ tends to infinity, and we can see that the exact value is very close already for $B=10$ (see also @Mark Adler's post for other examples). Beautiful, isn't it? :)


How do you calculate a fee percentage to handle a fee being charged? Problem is that we get charged a 3% fee.  We add this 3% fee to the invoice.  When we get the amount back they charge 3% on the invoice plus on the fee we added.  What formula can I use to figure out what percentage to charge on this so that we cover their 3% on both pieces?

Example 10.00 charge.  We add 0.30 to it to cover the fee, we send in 10.30.  They charge 3% on 10 and on the 0.30 so they send us back 9.99 when we expected 10.00.  How do we cover this?  Also, how do we cover these discrepancies when they charge fees and do rounding per line item?

-Edit-

For clarity we are sending an invoice in where we know they will charge 3% on it, we need to add a fee to our invoice that will negate this 3%.  It has to take into account that they will also charge 3% on the amount we add to negate the 3%.  Need to know what % to add to our invoice or what formula to calculate this.  Basically some sort of break even formula.

If $r$ is the fee (here $r=.03$) and $C$ is the charge you want to recover at the end, the amount to submit is $C/(1-r)$.  So the mark-up rate is $r/(1-r)$, e.g., $.03/.97\approx.0309278$, or approximately $3.1\%$.

If you don't want to switch back and forth between percents and decimals, and prefer to think of the variable here as $r=3$, the pertinent formula for the mark-up rate is $100r/(100-r)$.

The simplest I could make it.
You want to make say $30 and they are going to charge you 3% fee
$30/.97 <—(.03 or 3% less than 1 or 100%)
$30.93 (rounded up)
That’s it


Resolving $T(n) = 9T(n-2)+1$ and $T(n) = T(n-4)$ I'm tasked to give the best possible asymptotic bound for the following recurrences:

a) $T(n) = 9T(n-2)+1$

and

b) $T(n) = T(n-4)$

For a) I used the iteration method:

$$ T(n) = 9 \cdot T(n-2)+1 = 9^k \cdot T(n-2k) + 9 \cdot (k-1) + 1 =... $$

By solving the equation $n-2k = 1$ we get:

$$ ... =  9^{\frac{n-1}{2}} + 9 \cdot \left(\frac{n-1}{2} - 1\right) + 1 \\
 = 3^{n-1} + 4.5n -9.5 $$

Therefore we get: $$ T(n) = \Theta(3^{n-1})$$

For b) I also used the iteration method:

$$T(n) = T(n-4) = ... = T(n-4k) =... $$

and by solving $ n - 4k = 1$ we get:

$$ ...= 1 $$

This proves that $$ T(n) = \Theta(1) $$

Is this correct?

Thanks for any answers.

Applying a) twice, we can deduce that $$T(k+4) = 9T(k+2)+1 = 9(9T(k)+1)+1 = 81T(k)+10$$ for any $k$.  Applying b), we have $$T(k+4) = T(k).$$  Combine these two conclusions we have $$T(k) = \frac18,$$ for any $k$.


Counting number of integer solutions How many integer solutions are there to the equation $a + b + c = 21$ if $a \geq3$,
$b \geq 1$, and $c \geq 2b$?

If $a = 3+A$, $b =1+B$, $c = 2b + d=2B+d+2$, the equation is $A + 3 B + d = 15$ with $A, B, d \ge 0$.  Now for each $B$ from $0$ to $5$,
the number of $(A,d)$ pairs is $16 - 3 B$.  So...

$a$ can vary from $3$ to $18$ $b$ can vary from $1$ to $16$ $c$ can vary from $2$ to $17$ Hence according to me number of solutions should be $16*16*16=4096$


Integral involving bessel function/gaussian/rational function I'd like to solve:

$$\int_0^{\infty}\quad J_1(ak)\,\frac{b+k^2}{(k-\alpha_1)(k-\alpha_2)(k-\alpha_3)(k-\alpha_4)}\,\exp(-ck^2)\,\,dk$$

Is there any specific rule for it?
Thanks!

Perhaps this is the most simple approach:

$\int_0^\infty\dfrac{(b+k^2)\exp(-ck^2)J_1(ak)}{(k-\alpha_1)(k-\alpha_2)(k-\alpha_3)(k-\alpha_4)}dk$

$=\int_0^\infty\sum\limits_{n=0}^\infty\dfrac{(-1)^na^{2n+1}k^{2n+1}(b+k^2)\exp(-ck^2)}{(k-\alpha_1)(k-\alpha_2)(k-\alpha_3)(k-\alpha_4)n!(n+1)!}dk$

Note that $\int_0^\infty\dfrac{e^{-wk^2}}{\alpha k+\beta}dk=\dfrac{1}{2\pi\beta\sqrt{w}}G^{3,2}_{2,3}\left({\dfrac{\beta^2w}{\alpha^2}}\Bigg\vert\,^{\frac{1}{2},1}_{1,\frac{1}{2},\frac{1}{2}}\right)$ , where $G^{3, 2}_{2, 3}$ is the Meijer G-function.

The above integral can be expressed in terms of the infinite series of this result.


'Obvious' theorems that are actually false It's one of my real analysis professor's favourite sayings that "being obvious does not imply that it's true".

Now, I know a fair few examples of things that are obviously true and that can be proved to be true (like the Jordan curve theorem).

But what are some theorems (preferably short ones) which, when put into layman's terms, the average person would claim to be true, but, which, actually, are false
(i.e. counter-intuitively-false theorems)?

The only ones that spring to my mind are the Monty Hall problem and the divergence of $\sum\limits_{n=1}^{\infty}\frac{1}{n}$ (counter-intuitive for me, at least, since $\frac{1}{n} \to 0$
).

I suppose, also, that $$\lim\limits_{n \to \infty}\left(1+\frac{1}{n}\right)^n = e=\sum\limits_{n=0}^{\infty}\frac{1}{n!}$$ is not obvious, since one 'expects' that $\left(1+\frac{1}{n}\right)^n \to (1+0)^n=1$.

I'm looking just for theorems and not their (dis)proof -- I'm happy to research that myself.

Thanks!

Theorem (false):

  One can arbitrarily rearrange the terms in a convergent series without changing its value.

An analytic function with compact support vanishes identically.


Find if lines intersect, only knowing gradient. Is it possible to find out if and where two lines intersect with only knowing the gradient of said lines?

If you don't take the overlap case into consideration, then they intersect if and only if their gradients are different

Yes just look whether they are not paralel or skew , i,e , cross product is not zero.


Why is the value of r in $r=\cos θ$ never negative? The graph of $r=\cos θ$ forms a circle with radius $.5$ and around $(1,0)$. However, in terms of Cartesian plane coordinates, why doesn't the $r$ value go into the second and third quadrants? Wouldn't $r$ be negative when $θ$ is in the second and third quadrants?

You are correct $r$ is negative when $\frac{\pi}{2} \leq \theta < \frac{3\pi}{2}$, but since $r$ is negative, it points in the opposite direction of the $\theta$. Thus, the point $(r,\theta)=(-1,\pi)$ does not map to $(x,y)=(-1,0)$, but to $(x,y) = (1,0)$

I’d like you to first think of the equation $y=x^2$ in the Cartesian coordinate $(x,y)$-plane. That condition is a restriction over the plane that famously identifies those points whose $y$ value (height fron $x$ axis) is exactly the square of their $x$ value.

Now move back to polar coordinate $(r, \theta)$-plane. The equation $r = \cos\theta$ identifies those points whose $r$-value (distance from the origin) is exactly $\cos$ of their $\theta$ value (angle from $x$ axis). But $r$ is always greater than zero
(and $\theta$ is always between $\pm\pi$), so the condition is not meaningless unless $\theta$ makes the cosine positive, that is $\theta$ is between $\pm\pi/2$.

You’d just gotten it the other way around.

ADDENDUM. (Technical) The reason why I require $r>0$ and $-\pi< \theta < \pi$ is because I want polar coordinates to be consistent with the usual definitions of curvilinear coordinates in analysis, or coordinate charts in differential geometry. I want my coordinates to be a homeomorphism between an open subset of the plane (in this case the plane without the negative $x$ axis and the origin) and another open subset of the plane ($(0,+\infty)\times(-\pi,+\pi)$), and in particular a bijection. Otherwise it would be meaningless to even say that the polar coordinates of a point on the plane are such and such, because multiple answers would do. Notice that my approach provides a sensible answer to the OP’s doubts without recurring to weird ideas such as “negative distance from the origin”.


Approximating a double sum by a double integral Related to this question, I'm interested in bounding from above the following sum
$$
S:=\sum_{x=0}^\infty \sum_{y=0}^\infty (x+y)^m e^{-\frac{x^2}{2i} - \frac{y^2}{2j}},
$$
which I hope to do by relating it to the integral
$$
I:=\int_0^\infty \int_0^\infty (x+y)^m e^{-\frac{x^2}{2i} - \frac{y^2}{2j}} dx\,dy.
$$
Answers to the previous questions confirmed my expectation that $I = O\left(\exp\left(m\log\sqrt{(i+j)(m)}-\frac{m}{2}\right)\sqrt{ij}\right)$, the intuition for which is probably that the function behaves approximately like a gaussian around its maximum at $(x_0,y_0) = \left(i \sqrt{\frac{m}{i+j}},j \sqrt{\frac{m}{i+j}} \right)$, where the function takes the value $\exp\left(m\log\sqrt{(i+j)(m)}-\frac{m}{2}\right)$.
However I've been unable to show that the difference $|I-S|$ is significantly smaller than this bound. For simple one dimensional integrals, for example with a unique maximum, it's not too hard to bound this difference in terms of the maximum by considering appropriate telescoping sums. However, a naive analogue of this argument doesn't seem to work in two dimensions, and trying to apply this argument to each `slice' of the integral led to some pretty horrendous calculations. I also looked into using the Euler-Maclaurin formula but it's a bit out of my area of expertise.
I suspect that there should be a relatively standard way to approximate $|I-S|$, and I also wouldn't be surprised if someone more proficient in computing can get a CAS to provide a proof. The former would be more useful, just so that I have a tool for approaching similar questions.
So, very explicitly, I would like to know if
$$
|I-S| = o\left(\exp\left(m\log\sqrt{(i+j)(m)}-\frac{m}{2}\right)\sqrt{ij}\right),
$$
where even big-O would be sufficient for the application I have in mind, and I wouldn't be surprised if the difference is even bounded by a multiple of the maximum of the function. I'm interested in the asymptotics for $i$ and $j$ tending to infinity, $m$ can be fixed or also a function of $i$ and $j$. For the application I have in mind it would probably be sufficient to have such a result for $i = (1+o(1))j$ and $m = o(i)$.

I cannot provide an actual answer , but only some considerations and hint
which might hopefully be helpful.
The function
$$
f(x,y) = \left( {x + y} \right)^{\,m} e^{ - \,{{x^{\,2} } \over {2\,i}}\, - {{y^{\,2} } \over {2\,j}}}
$$
having a (cutted) bell-shape in the first quadrant, means that it is concave around the maximum and convex
further from it.
This makes quite difficult to relate the integral to the  Riemann sum with a $>, <$, because the sign
of the inequality changes in the two areas.
Furthermore, at increasing $i, \, j$, while the position of the max move $\approx \sqrt{i}$, and so approximately does its spread
the bell peak increases $\approx i^{m/2}$.
Since the $\Delta x , \, \Delta y$ of the sum are fixed at $1$, I doubt that the sum might converge to the integral.
Concerning the integral I would try the following approach
$$
\eqalign{
  & I = \int_{y\, = \,0}^{\,\infty } {\int_{x\, = \,0}^{\,\infty }
 {\left( {x + y} \right)^{\,m} e^{ - \,{{x^{\,2} } \over {2\,i}}\, - {{y^{\,2} } \over {2\,j}}} dxdy} }  =   \cr
  &  \Rightarrow \left\{ \matrix{
  s = x + y \hfill \cr
  t = x - y \hfill \cr}  \right.\quad  \Leftrightarrow \quad \left\{ \matrix{
  x = \left( {s + t} \right)/2 \hfill \cr
  y = \left( {s - t} \right)/2 \hfill \cr}  \right.\quad  \Rightarrow   \cr
  &  = \int_{y\, = \,0}^{\,\infty } {\int_{x\, = \,0}^{\,\infty }
 {s^{\,m} e^{ - \,{{\left( {s + t} \right)^{\,2} } \over {2\,i}}\,
 - {{\left( {s - t} \right)^{\,2} } \over {2\,j}}} {1 \over 2}dsdt} }  =   \cr
  &  = \int_{s\, = \,0}^{\,\infty } {\int_{t\, = \, - s}^{\,s}
 {s^{\,m} e^{ - \,{{\left( {s + t} \right)^{\,2} } \over {2\,i}}\,
 - {{\left( {s - t} \right)^{\,2} } \over {2\,j}}} {1 \over 2}dsdt} }  \cr}
$$
then also consider that
$$
\eqalign{
  &  - \,\left( {{{s^{\,2}  + t^{\,2}  + 2st} \over {2\,i}}\,
 + {{s^{\,2}  + t^{\,2}  - 2st} \over {2\,j}}} \right) =   \cr
  &  =  - \,{{\left( {i + j} \right)s^{\,2} } \over {2\,i\,j}}
\left( {\left( {{t \over s}} \right)^{\,2}  - 2{{i - j} \over {i + j}}
\left( {{t \over s}} \right) + 1} \right) =   \cr
  &  =  - \,{{\left( {i + j} \right)s^{\,2} } \over {2i\,j}}\left( {\left( {{t \over s}} \right)^{\,2}
  - 2{{i - j} \over {i + j}}\left( {{t \over s}} \right) + \left( {{{i - j} \over {i + j}}} \right)^{\,2}  + 1 - \left( {{{i - j} \over {i + j}}} \right)^{\,2} } \right) =   \cr
  &  =  - \,{{\left( {i + j} \right)s^{\,2} } \over {2\,i\,j}}
\left( {\left( {\left( {{t \over s}} \right) - {{i - j} \over {i + j}}} \right)^{\,2}
  + 1 - \left( {{{i - j} \over {i + j}}} \right)^{\,2} } \right) \cr}
$$
we can change the variables again
$$
\left\{ \matrix{
  s = s \hfill \cr
  r = t/s \hfill \cr}  \right.\quad J = \left| {\left( {\matrix{
   1 & 0  \cr
   { - t/s^{\,2} } & {1/s}  \cr
 } } \right)} \right| = {1 \over s}
$$
and then proceed with approximation or series expansion of the Error Function.

%94 // FullSimplify

(1/(m (1 + m)))2^(
 m/2) (Sqrt[i j] (i + j)^(m/2) m Sqrt[\[Pi]] Gamma[(3 + m)/2] -
   Gamma[1 + m/
      2] (j^(1 + m/2)
        m Hypergeometric2F1[1/2, 1, (3 + m)/2, -(j/i)] + (
      i^(m/2) (1 + m) (-i^2 +
         i j (-3 + m) + (i + j)^2 Hypergeometric2F1[1, (1 - m)/
           2, -(1/2), -(j/i)]))/(j (-2 + m))))

This shows a professional grade solution. It is meant to be exact and not an approximation of some kind. It sets the requirements ot the question properly first and then integrate for all values that appear in the integrant and integral correctly. To type this in MathML is a really tedious task. So this is Mathematica, Wolfram Language code and can for some extent be verified on Wolfram Alpha. Have a try on the free Maple app too.
The names of the functions appear in the reference question partly and are not reduced to somehow more elementary functions in this calculation. This respects Mathematica great internal convention in favour to some universitary or college simplicity.
This is still a function of {i,j,m} and very lengthy. Only two terms have this $\sqrt{i j}$. Others are dependent on m in the exponent.
This can be calculated to numbers. For example for $j=10=i$ and $m=1,2,3$:
{
{79.2665},
{514.159},
{3963.33},
{34849.6},
{340846.},
{3.6449610^6},
{4.2090510^7},
{5.1989410^8},
{6.8177210^9},
{9.43489*10^10}
}
That looks pretty divergent for bigger ${i,j}$.
For example for $i=10$ and $j$ free and $m=1,2,3,...0$ this looks this way:
$
 5 \sqrt{j} \sqrt{2 \pi} + j \sqrt{5 \pi},$
$20 j + \sqrt{5/2} \sqrt{j} (10 + j) \pi,$
$ \sqrt{j} (100 \sqrt{2} + 30 \sqrt{5} \sqrt{j} + 15 \sqrt{2} j +
 2 \sqrt{5} j^(3/2)) \sqrt{\pi},$
$ (80 j (10 + j) + 3 \sqrt{5/2} \sqrt{j} (10 + j)^2 \pi,$
$ \sqrt{j} (4000 \sqrt{2} + 1500 \sqrt{5} \sqrt{j} + 1000 \sqrt{2} j +
 200 \sqrt{5} j^(3/2) + 75 Sqrt{2} j^2 +
 8 \sqrt{5} j^(5/2)) \sqrt{\pi}, ...$
These are polynomials in j and $\sqrt{j}$.
If there would be another condition for the solution from the Mathematica engine for the input it would show that in the output or on the message cue. But there is none. OK. I am lazy because this is a tedious expression. I added an answer to the reference question too to earn more seriosity for my answer. This is not an expansion of any form and high mathematical intuition to take and offer that solution path for this highly complex problem and the restrictions on it.
$I$ is a function independent on $x$ and $y$. $I(i,j,m)$!
S rises strongly. If the summation is trunkated for fixed $m$ than the function may be plotted. The higher the terms that are taken into the sum the higher the function value. The plot is taken over $l$ and $k$.
I is a type of convolution integral. It acts in information theory and related as s low pass filter.
The binomial polynom is dominating for lower summation parts. The summands to not converge fast enough. This is what concerns one factor: Convolution. This requires that the other function is finite on the reals. That is not the case for $(x+y)^m$ unless it is limited to an finite interval. That is not the case in the question. Therefore the infinite summation fails. $S$ is returned unevaluated by Mathematica and other CAS.
The cognitive dissonance is the the summation is over discretized $x$ and $y$ with out and exhaustion. Integers are not dense in the reals. Approximation of convolution integrals require the finiteness of the convoluted function. This binomial is unlimited.
I selected $m=1$.
p1 = Plot3D[
  With[{m = 1},
   1/(m (1 + m)) 2^(
    m/2) (Sqrt[i j] (i + j)^(m/2) m Sqrt[\[Pi]] Gamma[(3 + m)/2] -
      Gamma[1 + m/
         2] (j^(1 + m/2)
           m Hypergeometric2F1[1/2, 1, (3 + m)/2, -(j/i)] + (
         i^(m/2) (1 + m) (-i^2 +
            i j (-3 + m) + (i + j)^2 Hypergeometric2F1[1, (1 - m)/
              2, -(1/2), -(j/i)]))/(j (-2 + m))))], {i, 1, 20}, {j, 1,
    20}]

p2 = Plot3D[\!\(
\*UnderoverscriptBox[\(\[Sum]\), \(x = 0\), \(100\)]\(
\*UnderoverscriptBox[\(\[Sum]\), \(y = 0\), \(100\)]
\*SuperscriptBox[\((x + y)\), \(1\)] Exp[\(-1\)\ \((
\*SuperscriptBox[\((
\*FractionBox[\(x\), \(k\)])\), \(2\)] +
\*SuperscriptBox[\((
\*FractionBox[\(y\), \(l\)])\), \(2\)])\)]\)\), {k, 1, 20}, {l, 1,
   20}]

Apparently not very much alike. I seems to have something of a Puiseux_series.
1/(m (1 + m)) 2^(
   m/2) (Sqrt[i j] (i + j)^(m/2) m Sqrt[\[Pi]] Gamma[(3 + m)/2] -
     Gamma[1 + m/
        2] (j^(1 + m/2)
          m Hypergeometric2F1[1/2, 1, (3 + m)/2, -(j/i)] + (
        i^(m/2) (1 + m) (-i^2 +
           i j (-3 + m) + (i + j)^2 Hypergeometric2F1[1, (1 - m)/
             2, -(1/2), -(j/i)]))/(j (-2 + m)))) /.
  m -> 1 // FullSimplify

$(\sqrt{i}*j + \sqrt{i*j}*\sqrt{i + j} -
   i*\sqrt{j}*(-1 + \sqrt{\frac{(i + j)}{i}}))*\sqrt{\frac{\pi}{2}}$
On the diagonal $i=j$ or $k=l$ for $m=1$
Plot[{\!\(
\*UnderoverscriptBox[\(\[Sum]\), \(x = 0\), \(100\)]\(
\*SuperscriptBox[\((2  x)\), \(1\)] Exp[\(-1\)\ 2
\*SuperscriptBox[\((
\*FractionBox[\(x\), \(k\)])\), \(2\)]]\)\), (k^(
     3/2) - (-1 + Sqrt[2]) k^(3/2) +
     Sqrt[2] Sqrt[k] Sqrt[k^2]) Sqrt[\[Pi]/2]}, {k, 1, 1000},
 PlotRange -> Full, AxesLabel -> Automatic]

The sum S exits for all values, since exponential is stronger than polynomial and polynomial is kept fix while the argument of the exponential drops to zero with inverse square order. Blue is S and I is goldish. The infinite summation takes endless on my machine. The FullSimplify might get complex and needs extra care.
It is in general very complicated or delicate to approximate convolution integrals. The Gaussian function is problematic in representation and integration. Such profile function play an important roles in all measurements of modern technologies. So they are studied with great care. Since we know this function we just have to handle it with care. Mathematica has extra feature for dealing with the representation of very large and very small numbers. This both are needed unless the solution is symbolic.
There some complexity function around on this community. Mixing polynomials with exponentials is always very hard and complicated. Help might get easier if the wordings would be presented.
There might be impulses for a different approach by interpretation as a Kampe de Feriet Function. The given sum is related to the answer of this question: evaluation-of-sum-x-0-infty-e-x2. Mathematica can do this:
Sum[E^(-r^2), {r, 0, Infinity}]
1/2 (1 + EllipticTheta[3, 0, 1/E])

If we do this:
Sum[D[D[E^(-r^2), r], r], {r, 0, Infinity}]
Sum[-2/E^r^2 + (4*r^2)/E^r^2, {r, 0, Infinity}]

Nice for it is this Jacobi Theta Functions. So S is the second derivative of the EllipticTheta with first argument 3 with respect to the second argument evaluated at 0 for the second argument and 1/e for the third up to $2^3$ as a factor following the formula for the corresponding derivative on EllipticThetaPrime4. Now we have a representation for $i=j=1$ and $m=1$. This is a very simple cases of $S$.
We can plug our $e{-(\frac{x^2}{i^2}+\frac{y^2}{j^2})}$ into this too. That is only representation. But for $m>1$ there is still no help. The $i$ and $j$ make the exponent still elliptical. The $m$ add circularity.
So the next step is hard again. Find relationships between EllipticTheta and Hypergeometric2F1. There are some for derivatives but this needs identities. Then generalize. Candidates are Elliptic hypergeometric series. Perhaps there is need for a revision of the hypothesis.


Showing that $\mathrm{Var}(Y) \leq \mathrm{Var}(X)$ Let $X$ be a random variable with $\mathbb{E}(X^2) <\infty$. Let $Y$ defined as
  $$Y=\begin{cases}X &\text{ if }& \mid X\mid\leq k \\
0 &\text{ if }& \mid X\mid > k
\end{cases}$$
  Show that $\mathrm{Var}(Y) \leq \mathrm{Var}(X)$.

I tried solving it this way:

\begin{align}\mathrm{Var(Y)}&=\int (Y-\mathbb{E}(Y))^2d\mathrm{P}=\int_{\mid x \mid \leq k} (Y-\mathbb{E}(Y))^2d\mathrm{P} +\int_{\mid x \mid > k} (Y-\mathbb{E}(Y))^2d\mathrm{P}\\
&=\mathbb{E}(Y^2)P(({\mid x \mid \leq k})\cup ({\mid x \mid > k}))\\
& \leq \mathbb{E}(Y)^2+\mathbb{E}(X^2)\\
&=\mathbb{E}(X^2)-\mathbb{E}(Y)^2\\
&\leq \mathrm{Var}(X)\end{align}

Is this correct?

I am not sure the statement is true.

Take a discrete variable $X$ such that:

$X=0.8$ with probability $0.5$ and $X=1.0$ with probability $0.5$.

We have $E[X]=0.9$, $E[X^2]=0.82$ and $Var(X)=0.01$.

Now take $k=0.9$. Than:

$Y=0.8$ with probability $0.5$ and $Y=0$ with probability $0.5$.

This leads to $E[Y]=0.4$, $E[Y^2]=0.32$ and $Var(Y)=0.16$.

So the variance has in this case incremented: $Var(Y) \ge Var(X)$. Is the counterexample correct ?

Try to do it using definiton : $$\mathbb{Var}(X) = \int_{\mathbb{R}} (X-\operatorname{E}X)^2 \mu(dx) \le \int_{{[-k,k]}}(X - \operatorname{E}X)^2 = \dots$$


Norm of a Positive definite matrix is the largest eigenvalue Let $A$ be a positive definite symmetric matrix. I need to show that
$$\lambda_n=\max \{\frac{\|Ax\|}{\|x\|}: x\ne 0\}$$ is the largest eigen value of $A$. My try is $\frac{\|Ax\|}{\|x\|}\le \lambda,\forall \lambda$ as a not eigen value of $A$, and the equality occurs when $\lambda$ is an eigen value. So $\lambda$ is maximum? May be I am not even understanding the question.

By the spectral theorem, there is an orthonormal basis of eigenvectors of $A$, say  $x_1,\dots,x_n$. WLOG, assume $0 < \lambda_1 \leq \dots \leq \lambda_n$, where $Ax_i = \lambda_i x_i$. For any non-zero $x$, write $x = c_1x_1 + \cdots + c_nx_n$. Then
$$ \frac{\lVert{Ax\rVert}^2}{\lVert{x\rVert}^2} = \frac{\lVert{c_1\lambda_1x_1 + \cdots + c_n\lambda_nx_n\rVert}^2}{c_1^2 + \cdots + c_n^2} = \frac{c_1^2\lambda_1^2 + \cdots + c_n^2\lambda_n^2}{c_1^2 + \cdots + c_n^2} \leq \lambda_n^2 \frac{c_1^2 + \cdots + c_n^2}{c_1^2 + \cdots + c_n^2} = \lambda_n^2,$$
with equality achieved when $x = x_n$ (whence $\lVert{Ax\rVert}=\lVert{\lambda_n x_n\rVert}=\lambda_n$).
You may wish to read about Rayleigh quotients or see Problem 37 in Brezis.

It is all of the answer
first you can prove that for each eigenvalue of positive definite matrix $M$ you have
$$ \operatorname{eig}(M^2)=\operatorname{eig}(M)^2.$$
In the other words if $s$ is the eigenvalue of $M$ then $s^2$ is the eigenvalue of $M^2$ since $s$ is the eigenvalue of $M$: $|s\cdot I-M|=0$
now by means of determinant properties we can write
|s^2*I-M^2|=|(sI-M)(sI+M)|=|sI-M||sI+M|=0
we proved that s^2 is the eigenvalue of M^2
we now that 2 norm of a matrix M is derived from this equation
||M||=sqrt(max(eig(M'*M)))
you should now that your claim  is only true for symmetric matrix M
so we can write
||M||=sqrt(max(eig(M^2)))
according to our proof we can write
||M||=sqrt(max(s^2)))
since M is positive definite s is positive now we can write
||M||=max(s)=max(eig(M))


Determine the following set is open or closed in $\mathbb{A}^2$ Let $k$ be a field.
Consider the set $$S=A_1\cup \{( 0,0)\}$$ where $A_1=\{(x,y):x\neq 0,y\in k\}$

How do we determine whether this set is open or closed in $\mathbb{A}^2$?

My Attempt: Note $A_1^c=\{(0,y):y\in k\}$, and $A_1^c=V(x,xy)$. Thus $A_1$ is open as the complement is closed. Then note $\{(0,0)\}$ is closed, so $S$ is a union of two disjoint sets where one is open and another is closed. Thus itself cannot be open nor closed.

Note: I'm not sure if my solution is correct or not. Even I believe a disjoint union of closed and open sets cannot possibly be open nor closed (in Hausdorff spaces at least I believe...) but I recall that $\mathbb{A}^2$ is not Hausdorff, so I'm uncertain whether my intuition is correct.

Your solution is not correct: $V(x)\cup D(x)$ expresses $\Bbb A^2$ as a disjoint union of an open set and a closed set, but $\Bbb A^2$ is both open and closed in itself. So your claim about sets of this form being neither open nor closed is not true.

Here's a hint towards a correct solution: recall that any copy of $\Bbb A^1\subset \Bbb A^2$ has the subspace topology. So if $S$ is open (resp. closed), then $S\cap \Bbb A^1$ for any copy of $\Bbb A^1\subset \Bbb A^2$ will be an open (resp. closed) subset of $\Bbb A^1$. Do you see what to do from here?

Full solution (that the OP came to in the comments) under the spoiler:

   Consider the two lines $V(y)$ and $V(x-1)$ inside $\Bbb A^2$. Both have the subspace topology, so if $S$ were to be closed or open, the intersection of $S$ with either of these lines would need to be closed or open respectively. But $S\cap V(y)$ is just the origin inside that line, which is not open, and $S\cap V(x-1)$ is the complement of the origin inside that line, which is not closed. So $S$ is neither closed nor open.

I'm not sure if this is correct or answer your question, but if you think in $\mathbb{R}^2$ ($k=\mathbb{R}$)and take the complement of $S$, $\hat{S}=\left\lbrace (0,y)\in \mathbb{R}^2\ \mid\ y\neq 0\right\rbrace ,$ we can see that $\hat{S}$ is not open (doesn't contain any neighborhood of any point), neither closed (take the sequence $\left\lbrace(0,\frac{1}{n}))\right\rbrace_{n\in\mathbb{N}}$, which does not converge in $\hat{S}$ ). So $S$ is not open neither closed.


When writing in math, do you use a comma or colon preceding an equation? This is a general question about mathematical writing especially for writing research papers and the like.

  Question: Do you precede an equation with a comma or colon?

Example A:

The following equation is the Yosida-Hawking-Penrose-Dantzig function
\begin{equation}
f(x) = \frac{1}{2} D_\alpha(x,y)
\end{equation}
The following equation is the Yosida-Hawking-Penrose-Dantzig function,
\begin{equation}
f(x) = \frac{1}{2} D_\alpha(x,y)
\end{equation}
The following equation is the Yosida-Hawking-Penrose-Dantzig function:
\begin{equation}
f(x) = \frac{1}{2} D_\alpha(x,y)
\end{equation}

Example B:

In fact, we can express the earlier function using a much simpler expression
\begin{equation}
f(x) = \varphi(x,y)
\end{equation}
where $\varphi(x,y)$ is the Demiane functional
In fact, we can express the earlier function using a much simpler expression,
\begin{equation}
f(x) = \varphi(x,y)
\end{equation}
where $\varphi(x,y)$ is the Demiane functional
In fact, we can express the earlier function using a much simpler expression:
\begin{equation}
f(x) = \varphi(x,y)
\end{equation}
where $\varphi(x,y)$ is the Demiane functional

Can anyone comment on which one is the best practice?

Generally, I would treat the equation as if it were any ordinary noun phrase, and use the usual rules for comma, colon, or no punctuation.

A colon is used if the equation is an elaboration, or an item.  So, just as you might write

  Lips are characterized by the following properties: fleshy, paired, red.

you would write

  An ellipse is characterized by the following equation:

  $$ \frac{x^2}{a^2}+\frac{y^2}{b^2} = 1 $$

A comma precedes a non-restrictive clause (one that describes rather than identifies the noun phrase), so by analogy with

  The line can be assigned to a simpler character, Polonius.

we might write

  A line can be described with a simpler equation,

  $$ y = mx+b $$

In comparison, with a restrictive clause, we use no comma, so just as we would write

  From this, the oiler obtained the formula CH$_3$C$_6$H$_4$C$_2$H$_5$.

we would also write

  From this, Euler obtained the formula

  $$ e^{i\pi}+1 = 0 $$

I suspect there aren't any hard and fast rules for this, however.  Whatever you choose to do, be consistent and reasonable.

ETA: You'll notice that I have no periods at the ends of these equations.  The papers I have generally (though not universally) observe this pattern.  However, in other fields, equations may have ending punctuation depending on how they occur within a sentence.  It may be useful for a writer to consult the publication's style guide, if applicable, or at least examine previous articles within the same publication or outlet.

I was told to use correct punctuation as defined by grammatical rules, but to always place the equation or formula on a line of its own, even if this meant starting a line with a punctuation mark. Even though the layout may look a little odd there can be no confusion as to the meaning. While it would be very rare for a comma to be misconstrued imagine the confusion an exclamation mark would cause.


How to prove that the set of maximal elements of a set of prime ideals is finite Let $A$ be a subset of ${\rm Spec}(R)$ with $R$ noetherian

Are there any techniques to prove that ${\rm max}(A)$ (ie the set of maximal elements of $A$) is finite?

I'm looking for equivalent conditions or conditions that imply this finiteness.

Particularly I'm thinking on the case $A:={\rm Ass}_R(M)$

there are different ways:
Matsumura:
from

and

Concludes

Sharp in "Steps in Commutative Algebra" uses minimal primary
decomposition of modules (see 9.25 - 9.33)


Symmetry group of a triangular lattice The question is:

  Let $G$ be the group of symmetries of an equilateral triangular lattice $L$. Find the index in $G$ of the subgroup $T \cap G$. ($T$ is the group of translations)

The solution says that $G$ is $D_6$ (dihedral group) instead of $D_3$, so the answer is the order of $D_6$, which is $12$. Why isn't $G$ $D_3$?

It's $D_6$ instead of $D_3$ because it is a triangular lattice. You can rotate the entire lattice around any single vertex by $60^\circ$, and each lattice point maps to another lattice point. In the triangular lattice below, rotation around the black lattice point by any multiple of $60^\circ$ will permute the blue lattice points among themselves and similarly the green lattice points. The composition of six such rotations, not three, takes all the points back to their original positions.

Similarly you can reflect the lattice around any of the six lines of symmetry that pass through a single vertex. (In the illustration above, three are pink, three red.)

Because a symmetry can make a specific triangle upside-down, which can not be killed by modulo translantion.


Quasilinear equation $u_t = uu_x$ I am not quite sure how to deal with discrete IVP

Find self-similar solution
\begin{equation}
u_t=u u_x\qquad -\infty <x <\infty,\ t>0
\end{equation}

satisfying initial conditions

\begin{equation}
u|_{t=0}=\left \{\begin{aligned}
-1& &x\le 0,\\
1& &x> 0
\end{aligned}\right.
\end{equation}

Here is my attempt

Characteristic equation:
\begin{align}
\frac{dt}{1} &=\ \frac{dx}{-u} = \frac{du}{0}\\
\frac{du}{dt}&=\ 0 \implies u=f(C)\\
\frac{dx}{dt} &=\ -u = -f(C)\\
x &=\ C - tf(C)\\
\end{align}

Impose boundary condition: $t=0$ and $x = C$,
\begin{align}
u &=\ f(C) = u|_{t=0}=\left\{\begin{aligned}
-1& &x\le 0,\\
1& &x> 0
\end{aligned}\right.
\end{align}
\begin{align}
x = C+t \left\{\begin{aligned}
-1& &x\le 0,\\
1& &x> 0
\end{aligned}\right.
\end{align}
Where the characteristic lines are never across each other

This is a Riemann problem for the PDE $u_t - u u_x = 0$, which cannot be solved by applying the method of characteristics alone: weak solutions must be considered. This PDE becomes equal to the inviscid Burgers' equation $v_t + vv_y = 0$ under the transformation $y := -x$. The initial data transforms as $v(y,0) = 1$ if $y < 0$ and $v(y,0) = -1$ if $y \geq 0$. The Lax entropy conditions states that the solution is a shock wave, which speed $s = \frac{1}{2} (1 -1) = 0$ is given by the Rankine-Hugoniot condition. Finally, a stationary shock solution is obtained:
$$
u(x,t) = v(-x,t) = \left\lbrace
\begin{aligned}
&{-1} && \text{if}\quad x\leq 0\\
&1 && \text{if}\quad x> 0
\end{aligned}\right.
$$

Follow the method in http://en.wikipedia.org/wiki/Method_of_characteristics#Example:

$\dfrac{dt}{ds}=1$ , letting $t(0)=0$ , we have $t=s$

$\dfrac{du}{ds}=0$ , letting $u(0)=u_0$ , we have $u=u_0$

$\dfrac{dx}{ds}=-u=-u_0$ , letting $x(0)=f(u_0)$ , we have $x=f(u_0)-u_0s=f(u)-ut$ , i.e. $u=F(x+ut)$

$u(x,0)=\begin{cases}-1&\text{when}~x\leq0\\1&\text{when}~x>0\end{cases}$ :

$\therefore u=\begin{cases}-1&\text{when}~x+ut\leq0\\1&\text{when}~x+ut>0\end{cases}=\begin{cases}-1&\text{when}~x-t\leq0\\1&\text{when}~x+t>0\end{cases}$

Hence $u(x,t)=\begin{cases}-1&\text{when}~x\leq t\\1&\text{when}~x>-t\\c&\text{otherwise}\end{cases}$


find exact value of $\sin(10^\circ)$ How we can find exact value of $\sin(10^\circ)$?

I tried trigonometric ways but I get this equation:
$\ 8y^3-6y+1=0$ and $y = \sin(10^\circ)$ and all the roots are complex.

I saw the pages in site, but I can't find the solution.
Thanks

Using Precise-Rewritten method for Sin 10∘, Exact value may be determined as follows:

+45..............45.....................+2

-22.5............22.5...................-2

-11.25............11.25................+2

-5.625............5.625................+2

+2.8125............8.4375............-2

+1.40625...........9.84375...........+2

+0.703125............10.546875........+2

-0.3515625..........10.1953125......-2

Hint for table: First column is half of earlier (called as Central) with sign towards target angle 10∘; second is cumulative of the first column; and the third is division of current Central with earlier Central.

If we follow above pattern infinitely, the sum will be 10. However, we can see the repeating pattern in third column (-2, +2, +2); hence this is indication for the exact value.

Now write the third column top-to-down approach as:$$+2\overline{-2+2+2} $$.

In the next step just replace 2 by  √(2; the result will be :
$$√(2 \overline{-√(2+√(2+√(2} $$.

Sin 10∘ will be half of this (closing brackets are collapsed by "]").
Therefore, Sin 10∘ = $$√(2 \overline{-√(2+√(2+√(2}]/2 $$

Above method is called as Precise-Rewritten method. You can find exact trigonometric value of all integer angles using this method.

I apologize above bad formatting, discourage to explain the details here.

Source: Breaking Classical Rules in Trigonometry:Precise-Rewritten method, 2016

Request to the scholars: You may assist to copy edit the main document of above method. I have not idea to formatting and use of mathematical tools at all. I cannot format as online forum requires formatting the mathematics. If someone assisted me for copy edit on Precise-Rewritten method (and other new methods), every scholar may know the new idea for new method for exact trigonometric values. My (Breaking Classical Rules in Trigonometry- Mission 2050) un-skill on mathematics formating discouraging me to expose all of those new idea.


Bases of dual spaces I am given that dual to the basis $B=\{v_1,v_2,...,v_n\}$ of the vector space $V$ is the dual basis $\{f_1,f_2,...,f_n\}$ of $V^*$ where $V^*$ is the dual space of $V$.

How do I find the dual basis with respect to another basis, $B'$ of $V$?

I know that any basis of $V$ are composed of  vectors which are linear combinations of $\{v_1,v_2,...,v_n\}$. And then...?

Thanks.

How do you know $\{f_{1},...,f_{n}\}$ ? If you know the relation between $\{v_{1},...,v_{n}\}$ and $\{f_{1},...,f_{n}\}$ then you can find the dual basis with respect to another basis.

I do not think this kind of question appropriate for MathstackExchange. Next time, you should try to find the answer for this kind of question at wikipedia first :
http://en.wikipedia.org/wiki/Dual_basis


Show that if matrix $X$ satisfies the relation $X \cdot Y = Y \cdot X$ it has a specific form Show that: If $X \in \mathbb{R}^{2x2}$ is a matrix which satisfies the relation $X \cdot Y = Y \cdot X$ for all $Y \in \mathbb{R}^{2x2}$, then $X$ has the form $X = \begin{pmatrix}
\lambda & 0\\
0 & \lambda
\end{pmatrix}$ with $\lambda \in \mathbb{R}$.

I don't know how to solve this task? Maybe we are allowed to set specific matrix for $Y$? But I'm not sure because it's saying "for all" so we must stay general?

Else I would say we can satisfy this relation when we set $Y = \begin{pmatrix}
1 & 0\\
0 & 0
\end{pmatrix}$ I think.

But that's not enough to prove it right? : /

Expanding the comment of Ofir:
$$
\begin{pmatrix}
a & 0\\
c & 0
\end{pmatrix} =
\begin{pmatrix}
a & b\\
c & d
\end{pmatrix}
\begin{pmatrix}
1 & 0\\
0 & 0
\end{pmatrix} =
\begin{pmatrix}
1 & 0\\
0 & 0
\end{pmatrix}
\begin{pmatrix}
a & b\\
c & d
\end{pmatrix} =
\begin{pmatrix}
a & b\\
0 & 0
\end{pmatrix}
$$
implies $b = c = 0$.
$$
\begin{pmatrix}
0 & a\\
0 & c
\end{pmatrix} =
\begin{pmatrix}
a & b\\
c & d
\end{pmatrix}
\begin{pmatrix}
0 & 1\\
0 & 0
\end{pmatrix} =
\begin{pmatrix}
0 & 1\\
0 & 0
\end{pmatrix}
\begin{pmatrix}
a & b\\
c & d
\end{pmatrix} =
\begin{pmatrix}
c & d\\
0 & 0
\end{pmatrix}
$$
implies $a = d$.

Since $X = \begin{pmatrix}
\lambda & 0\\
0 & \lambda
\end{pmatrix}$ and $XY = YX$ then define $Y=\begin{pmatrix}
\ y_1 & y_2\\
y_3 & y_4
\end{pmatrix}$

Then we have $$\begin{pmatrix}
\lambda & 0\\
0 & \lambda
\end{pmatrix}\cdot \begin{pmatrix}
\ y_1 & y_2\\
y_3 & y_4
\end{pmatrix} = \begin{pmatrix}
\ y_1 & y_2\\
y_3 & y_4
\end{pmatrix}\cdot \begin{pmatrix}
\lambda & 0\\
0 & \lambda
\end{pmatrix}$$

This gives us the following system of linear equations:

$$\begin{cases}\lambda y_1 = \lambda y_1 \\ \lambda y_2 = \lambda y_2 \\ \lambda y_3 = \lambda y_3 \\ \lambda y_4 = \lambda y_4\end{cases}$$

This system of linear equations tells you that $X$ and $Y$ commute.


Calculus: Integrate Fraction Containing Arcsin Component I'm struggling with the following problem:

I searched online for a solution and came upon below.  But I'm confused by the solution itself:

Specifically, how did the problem go from here:

to here:

I have a hunch it's just this formula below -- this formula that I just need to remember? ... is this true?

Yes the formula used is just
$$
\int \frac{1}{\sqrt{a^2-x^2}}dx =\int \frac{1}{a\sqrt{1-\frac{x^2}{a^2}}}dx= \sin^{-1}\left(\frac{x} {a}\right) +C
$$
that can be easily recovered from the derivative:
$$
\frac{d}{dx} \sin^{-1}(x)=\frac{1}{\sqrt{1-x^2}}
$$

If this is a question on your test in AP Calculus BC, and if you’re going to need to solve questions of this pattern frequently, then I would suggest memorizing all the derivatives for inverse trig functions. That being said, I was able to solve it using trigonometric substitution. Construct a right triangle triangle with sides $a$, $x$, and $\sqrt{a^{2} - x^{2}}$. Now you can solve it as follows:

Let $I = \int \frac{dx}{\sqrt{a^{2} - x^{2}}}$

Construct a right triangle with hypotenuse $a$, and legs, $x$, $\sqrt{a^{2} - x^{2}}$

Let $\theta$ be the angle that is opposite to the side $x$

$\sin{\theta} = \frac{x}{a}$

Differentiating both sides:

$a\cos{\theta}d\theta = dx$

We also know:

$\cos{\theta} = \frac{\sqrt{a^{2} - x^{2}}}{a}$

$a\cos{\theta} = \sqrt{a^{2} - x^{2}}$

Substituting everything into the integral, we get:

$I = \int \frac{a\cos{\theta}}{a\cos{\theta}}d\theta = \int d\theta$

$ = \theta + C$

$ \sin{\theta} = \frac{x}{a}$

$ \arcsin{\frac{x}{a}} = \theta$

$I = \arcsin{\frac{x}{a}} + C$

picture of actual solution

Edit: How does this not resolve the question? I showed you how you get from your integral to your solution. I also said that it's better to memorize the formula if these questions appear frequently.


Why is the cardinality of irrational numbers greater than rational numbers? This was asked by blogegog on a YouTube comment (gasp!):

[Regarding Cantor's diagonal argument:]
Couldn't I just make the same statement about rational numbers and say, 'take the largest number [sic he probably meant the one with the most digits] on the chart and append to the end of it a 1 if I'm hungry, or a 2 if I'm not' to show that his list of rational numbers doesn't contain them all?

Another way of looking at this is via continued fractions.  Note the following facts:

Every rational number finite continued fraction representation of the form $[a;b_1,b_2,\ldots,b_n]$ -- although every rational number has two such representations.  Also, every such finite continued fraction represents a rational number.
Every irrational number has  a unique infinite continued fraction representation of the form $[a;b_1,b_2,\ldots]$, and every such infinite continued fraction represents an irrational number in this interval.

If we restrict our attention to the interval $[0,1)$, we may assume that $a=0$ in all of these representations.

It follows easily that the family of rational numbers in $[0,1)$ has cardinality not greater than the family of all finite sequences of positive integers.  Similarly, the family of all irrational numbers in $[0,1)$ has cardinality exactly equal to the family of all infinite sequences of positive integers.  We have thus reduced our problem to comparing the families of finite and infinite sequences of positive integers.

It is not too difficult to show that the family of all finite sequences of positive integers is equinumerous to the family of all positive integers itself.  From this and the above observation we have that $\mathbb{Q} \cap [0,1)$ is countable.

Exactly using Cantor's diagonalisation technique one can show that the family of infinite sequences of positive integers is uncountable, and therefore $[0,1) \setminus \mathbb{Q}$ is also uncountable.

A sequence of all possible rational numbers though infinite (regardless of their magnitude) has a definite unbreakable order. Between two rational numbers in such sequence, you can't squeeze in more rational numbers.
But that is not the case with irrational numbers. In any given sequence of irrational numbers, you can squeeze in infinite irrational numbers between any two irrational numbers. So irrational numbers are uncountable and thus have greater cardinality beyond the countable rational numbers.


Evaluate $\lim_{x\to 0}\frac{x\tan x}{\sqrt{1-x^2}-1}$ Without Lhopital $$\lim_{x\to 0}\frac{x\tan x}{\sqrt{1-x^2}-1}$$

Tried using $\tan x\approx x$ and $\lim_{x\to 0}\frac{\tan x}{x}=1$

$$
\begin{align} \lim_{x\to 0}\frac{x\tan x}{\sqrt{1-x^2}-1}\approx\lim_{x\to 0}\frac{x^2}{\sqrt{1-x^2}-1}  &=\lim_{x\to 0}-\left(-\frac{x^2}{\sqrt{1-x^2}-1}\right)\\  &=\lim_{x\to 0}-\left(\frac{1-x^2-1}{\sqrt{1-x^2}-1}\right)\end{align}$$

Assumin that you could like more than the limit itself, consider the composition of Taylor series
$$\sqrt{1-x^2}=1-\frac{x^2}{2}-\frac{x^4}{8}+O\left(x^6\right)$$
$$\sqrt{1-x^2}-1=-\frac{x^2}{2}-\frac{x^4}{8}+O\left(x^6\right)$$
$$x\frac{\tan (x)}{\sqrt{1-x^2}-1}=x\frac{x+\frac{x^3}{3}+\frac{2 x^5}{15}+O\left(x^7\right) } {-\frac{x^2}{2}-\frac{x^4}{8}+O\left(x^6\right) }$$ Now, long division to get
$$\frac{x\tan (x)}{\sqrt{1-x^2}-1}=-2-\frac{x^2}{6}+O\left(x^4\right)$$

For the fun of it, try for $x=\frac \pi 6$ whch is quite far away from $0$. The exact value is
$$\frac{\pi }{\sqrt{3} \left(\sqrt{36-\pi ^2}-6\right)}\approx -2.042$$ while the series would give
$$-2-\frac{\pi ^2}{216}\approx -2.046$$

First of all, we have $\lim_{x\to 0}\frac{x\tan x}{\sqrt{1-x^2}-1}=\lim_{x\to 0}\frac{x^2}{\sqrt{1-x^2}-1}\cdot \frac{\tan x}x=\lim_{x\to 0}\frac{x^2}{\sqrt{1-x^2}-1}.$ Changing variables ($y=-x^2$,) we have
\begin{equation}
\begin{split}
\lim_{x\to 0}\frac{x^2}{\sqrt{1-x^2}-1}&=\lim_{y\to 0^+}-\frac{y}{\sqrt{1+y}-1}\\
&=\lim_{y\to 0^+}-\frac{y(\sqrt{1+y}+1)}{y}\\
&=-2
\end{split}
\end{equation}


Is there a way to generate a list of distinct numbers such that no two subsets ever have an equal sum? I'm trying to figure out a way to assign weights to a group of servers (a galera cluster of database servers), and I want to always be able to compute a quorum, meaning no set of weights should ever be allowed to add up to exactly 50% (a quorum in this case means over 50%).

Is there a mathematical formula to generate a set of (probably unique) numbers so that you can never sum any subset of those numbers to equal any subset of the remaining numbers? Additionally, no individual number should be double or more than double of any other number.

For example, with [3, 4, 5], there is no way to take any set of 1, 2, or 3 of those to add up to be equal to any subset of remaining numbers. There will always be an inequality, so a quorum can be computed (or it can be determined that no quorum is available, in the case where too many servers are disconnected from each other).

I understand this is a problem relating to server administration, but it seems to be of a mathematical nature.

What I'd like to be able to do is assign individual weights to a initial pool of servers, but ideally be able to generate another weight if another server gets added to the pool in the future.

The practical application is that all servers know their own weight, and they know the total weight of all servers. If a server suddenly dies, or connectivity fails between a few of them, the servers try to determine if they have a quorum. Each server that can still communicate with another will add up their weights, and if the total of their weights is more than exactly 50% of the initial set's total, then there is a quorum, and those servers will declare themselves to be the new canonical group. If they fail to get over 50%, they don't have a quorum and will declare themselves to be offline or otherwise unable to continue service.

For $n$ servers, consider weights
$$
1 + m, 2 + m, 4 + m, \ldots, 2^n + m
$$
for $m$ large enough to make sure each is less than twice each of the others.

The uniqueness of subset sums for subsets of equal size follows from the uniqueness of binary expansions.

Perhaps you are over-thinking this? What do you lose by taking a quorum as being simply more than 50% of the servers? If you want a way to break ties when the servers are split into two groups of equal size, just name one server as being special, and take the group that contains the special server.

Or am I missing something?