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Suppose such a function exists with $x\in (a,b)$ such that $f(x)=d$. Pick points $y,z\in(a,b)$ with $y<x$ and $z>x$. We know such points exist because $x$ is an interior point of $(a,b)$. Without loss of generality, suppose $f(y)<f(z)$. We know that both $f(y)$ and $f(z)$ are less than $d$. By the intermediate value theorem, there exists $u\in[y,x]$ ...

5

I suggest that you should think of $\delta$ a bit differently than you do, namely either as a distribution or as a measure. I will discuss the first point of view. The notation with integrals are handy but often lead to confusion when the "integrand" is not a real function. So, what is a distribution? Well, you could say it is something that eats nice ...

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The common definition of the delta function is $$\int_{-\infty}^{\infty} g(x) \delta(x) \, dx = g(0), \tag{1}$$ for any continuous $g$. Now let's look at what $\delta(bt)$ does. We have no formula for this, so let's do that substitution, $t'=bt$, so $dt'=b \, dt$ and the limits remain unchanged because $b>0$. Then we have $$\int_{-\infty}^{\infty} ... 5 You are expanding a function f in a basis for which differentiation \frac{d}{dx} is diagonalized. That's why the Fourier transform is so useful for studying differential equations. That normalized basis is e_{s}(x)=\frac{1}{\sqrt{2\pi}}e^{isx}, and e_s is an eigenfunction of differentiation with eigenvalue is, which gives you a diagonalization:$$ ...

5

The way the limits in red were changed is simply that the piecewise definition of $H$, stated earlier in the question, says that $H(x)=0$ when $x<0$. Thus $$\int_{-\infty}^0 (\text{anything}\times H(x))\, dx = 0.$$ The idea that $\delta(0)=\infty$ should not be taken too literally. Notice that $$\int_{-\infty}^\infty 3.4\delta(x) f(x)\,dx = 3.4f(0), ... 4 Note that the "definition" of \delta(x) that you cite is more of an informal description, to aid with intuition. The actual definition is by the relation$$\int f(x)\delta(x)\,\mathrm dx = f(0)$$Question 1: We have, by definition,$$\int f(x)\delta(x)\,\mathrm dx = f(0)$$We also obviously have$$\int f(0)\delta(x)\,\mathrm dx = f(0)$$but$$\int ...

3

In most common definitions of the Dirac delta (generalized) function, the formula in your post is taken as a definition. But if, for example, you accept that $\delta(x)$ is the Fourier transform of $1$, then it can be proven as follows: $$\int_{-\infty}^{\infty} f(x) \delta(x) dx = \int_{-\infty}^{\infty} f(x) \left( \int_{-\infty}^{\infty} e^{-2\pi i kx} ... 3 It's easy to confuse the image of the curve with the "animation" of a point tracing that curve. Think about that for a second. You can have a perfectly smooth looking image, even a straight line, and still have \gamma'(t)=0 for some t. You just need to imagine a point tracing the line, slowing to a stop, and then picking back up again. On the other ... 3 Very roughly speaking, you can think of a stochastic process as a process that evolves in a random way. The randomness can be involved in when the process evolves, and also how it evolves. A very simple example of a stochastic process is the decay of a radioactive sample (with only one parent and one daughter product). Initially, it has some large number ... 3 A riemannian metric on a manifold provides a way to measure lengths and angles between tangent vectors. This in turns gives ways to measure areas and volumes, as well as distance between points in a manifold. A riemannian manifold is a manifold supporting a given riemannian metric. Nice manifolds, satisfying some topological assumptions, admit such a ... 3 Degree is best thought of as a property of projective varieties, since in \mathbb{C}^n, a linear space is not an intrinsic property. For example, in two variables, x=0 and x=y^2 define indistinguishable varieties, but the degrees of their equations are different. On the other hand, in \mathbb{P}^n varieties are defined as zeroes of homogeneous ... 2 If you're looking for some explanation as to why the two factorial expressions are equal, just break each expression down into prime factors.$$1*2*3*4*5*6*7=7*8*9*101*2*3*(2^2)*5*(2*3)*7=7*(2^3)*(3^2)*(2*5)1*2^4*3^2*5*7=1*2^4*3^2*5*7$$I know this only scratches at the surface of your questions about it, but I hope you still find it of some ... 2 \gamma'(t) \neq 0 can be seen as "The particle never stops.", since the position always changes. 2 As you stated, (P \vee Q) \wedge (P \vee R)  =[EITHER I will eat chocolate OR vanilla ice] AND [EITHER I will eat chocolate or strawberry ice cream] There are also three distinct possibilities that will ensure this statement is true: 1.P: If you eat chocolate ice cream both (P \vee Q) and (P \vee R) are true, and the proposition is true. 2.Q ... 2 Since I don't know how to post drawings of vectors you'll have to do the drawing for me. Draw a vector with any orientation you like and call it \mathbf N. Consider the point at the tail of the vector and call it \mathbf p. Consider all vectors of all lengths whose tails are at p and are perpendicular to \mathbf N. Can you see that this set of ... 2 It's can be true that f(x)\delta(x) = f(0)\delta(x), depending on your initial definition of \delta(x). It is not the case with your definition. Really, \delta(x) is not a function. This seems to be causing you a bit of trouble. It is something termed a "distribution", which morally means it is defined only via how it acts with other functions when ... 2 PRIMER: In This Answer and This Answer, I provided more detailed primers on the Dirac Delta. Herein, we condense the content of those answers. The Dirac is not a function, but rather a Generalized Function also known as a Distribution. The symbol \int_{-\infty}^{\infty}\delta (x)f(x)\,dx is ,in fact, not an integral. It is a ... 2 Manifolds are built from coordinate patches, if you do it the old-fashioned way. This means that in general functions are only defined locally on open sets (there is, for example, no global coordinate system on a sphere, from the hairy ball theorem). The partition of unity is a way to patch compatible definitions of functions on different open subsets ... 2 Details of the steps: \color{red}{(1)} to \color{blue}{(2)}, function substitution f(t)\mapsto f(t+x). \color{blue}{(2)} to \color{#F80}{(3)}, substitute \xi\mapsto\xi-x. \color{#F80}{(3)} to \color{#080}{(4)}, \delta is even; i.e. \delta(-x)=\delta(x). 2 The second formula is the standard expression for the probability density function (PDF) corresponding to the normal (or Gaussian) distribution with mean \mu and standard deviation \sigma. As it is a PDF, it is normalised to 1, i.e., its integral over admissible values of x is 1. The first formula is missing the 1/\sigma factor, thus it is not a ... 2 A gaussian distribution is the same as a normal distribution. The standard gaussian or standard normal distribution is the gaussian distribution with \mu = 0, \sigma = 1. BTW, the first equation above is incorrect, as its integral over the reals is \sigma, and not 1. As the probability of the sample space, by definition, has to be 1, any distribution ... 1 You cannot expect a correct proof of a formula which makes no sense within the formalism at your disposal: There is no function x\mapsto\delta(x) such that for "all" f one has$$\int_{\mathbb R} f(x)\>\delta(x)\>dx=f(0)\ .\tag{1}$$Instead we should view (1) as shorthand for the following mental process: We consider a sequence of functions ... 1 The normal distibution, also called the Gaussian distribution is the probability distribution that assigns to every measurable set A of real numbers the probability$$ \int_A \frac 1 {\sqrt{2\pi}} \exp\left( \frac{-1} 2 \left( \frac{x-\mu} \sigma \right)^2 \right) \, \frac{dx} \sigma. $$In particular, there is the cumulative probability distribution ... 1 The first time I encountered the Dirac delta was when I did my first (and only) course in electrostatics. It was introduced in the following way$$ \delta(x) = \begin{cases} 0 & \space \mathrm{for} \space x \ne 0 \\\infty&\ \mathrm{for} \space x = 0 \end{cases} $$and$$ \int_{-\infty}^{\infty} \delta(x) dx = 1 $$We were told that we should ... 1 The main difference is that or almost uniform convergence, you fix the exceptional set E with \mu(E)<\gamma once and for all (only depending on \gamma) and then you want f_n \to f on uniformly on E^c. Note that E is not allowed to change with n or with the \epsilon distance that we choose when we want to exploit the uniform convergence. ... 1 "I will not eat both strawberry and chocolate ice cream" in the intended sense, i.e. "It is not the case that I will eat both strawberry and chocolate ice cream" is obviously distinct from "I will not eat strawberry ice cream and I will not eat chocolate ice cream". Suppose I eat strawberry but not chocolate. The first then is true and the second false. "I ... 1 The line of intersection of the two planes is the solution of the system:$$ \begin{cases} 3x+2y-z=28\\ x-4y+2z=0 \end{cases} $$solving you find: x=8,y=\frac{t}{2}+2 and z=t \quad \forall t \in \mathbb{R}. This means that the equation of the line (in vector form) is:$$ \begin{bmatrix} x\\y\\z \end{bmatrix}= \begin{bmatrix} 0\\\frac{1}{2}\\1 ...

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Let's write the equations of the planes as $$z=3x+2y-28 \tag 1$$ and $$z=-\frac12 x+2y \tag 2$$ If the planes intersect, then obviously the values of $z$ on the line of intersection must be equal. So, setting the right-hand sides of $(1)$ and $(2)$ equal yields $3x+2y-28=-\frac12 x+2y\implies x=8$ Thus, the two planes intersect at a line on the plane ...

1

Hint: Try working out the total number of ways that you could be given a black card and an ace in the 2 cards that you take from the pack. For example if we assume that one of the aces is the ace of hearts, then there are 26 black cards which could accompany it, meaning there are 26 combinations when we assume we get the ace of hearts. We can continue, ...

1

Firstly, English (or any other natural language) is not designed for logic and it is often counter-productive to attempt solving logic problems using such a poor tool. The English word "and" is not the same as ∧, and "or" is definitely not a good stand-in for ∨ (as it doesn't distinguish between inclusive or exclusive). They are only similar. Moreover, the ...

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