Here is the problem: Find all continuous $f:\mathbb{R}\rightarrow\mathbb{R}$ which satisfy
$$f(x+y)+f(x-y)=2[f(x)+f(y)]\;\;\;(1).$$
Here is my attempt:
Fix $\delta>0$ and let $C=\int_{0}^{\delta}2f(y)dy.$
Then
$$\begin{align*}
2\delta f(x)+C&=\int_{0}^{\delta}2[f(x)+f(y)]dy\\
&=\int_{0}^{\...
The problem arose in the context of kinetic energy, where it can be proven from symmetry principles that $E(2v)=4E(v)$ without assuming $E=mv^2/2$ (see e.g. physics stackexchange).
While one may do further physics from this point to prove the desired result (that $E$ is quadratic in $v$) -- cons...
Hi everyone. I was reading an answer on Physics SE where it is said: "$E(2v)=4E(v)$ [...] implies that $E$ is quadratic. Now, it doesn't seem obvious to me. How can we prove that $f(2x)=4f(x)$ doesn't have a non-quadratic solution?
@SineoftheTime For physical reasons. $x=y$ seems to be just a particular case of this general functional equation. There is a long discussion before this message on this chat and in "The h Bar".
@BenSteffan No, but it doesn't seem an assumption in the posts I mentioned. Surely $g(x)$, being the kinetic energy function must always be nonzero, so $g(x) \geq 0$, so $a > 0$.
@BenSteffan I don't know if this can be achieved mathematically.
i would also say that signs r just conventions in physics. if KE+PE is conserved, so is -KE-PE
so there is nothing in nature which says that KE must be positive
there is nothing that truly fixes the factor of 1/2 either. there r some arguments from relativity, but still, if E is conserved, so is cE for any constant c
whenever u get these fixed constants, it is because of some arbitrariness in the definitions
unless it's some unitless quantity or stuff like that
@RyderRude Yes, I agree, but wherever you find it written that $E_{kinetic} \geq 0$. Somewhere I had read that this condition was important for some reason.
@RyderRude Yes, the term $1/2$ is a convention that makes the energy laws easier. It follows from the assumption of Galilei-invariance with some arguments from special relativity.
For elastic collisions of n particles, we know that momentum in the three orthogonal directions are independently conserved:$$ \frac{d}{dt}\sum\limits_i^n m_iv_{ij} =0,\quad j=1,2,3$$
From this, it follows there's also a corresponding scalar quantity conserved:$$\frac{d}{dt}\sum\limits_i^n m_i(v...
As usual, the top-rated answer is illuminating, and I noticed that the answerer is the same as the most voted answer in the relationship kinetic energy-velocity square. I find all this stuff spot on, I have no idea where these arguments came from, where he read them or how he got there. Never seen them in all the Physics books I have read in my life.
@AlessandroCodenotti I don't really know what your question was, but my relation to pi-base is that I sometimes contribute to it, and I've been main contributor of realcompactness to pi-base
The problem arose in the context of kinetic energy, where it can be proven from symmetry principles that $E(2v)=4E(v)$ without assuming $E=mv^2/2$ (see e.g. physics stackexchange).
While one may do further physics from this point to prove the desired result (that $E$ is quadratic in $v$) -- cons...
For some reason I thought that being hereditarily normal and being able to put disjoint open sets around separated sets were equivalent only for regular spaces, but they are always equivalent
@RyderRude The OP of the question I linked asked about the same argument I asked yesterday. Why did he assume the function increasing over positive $v$?
@SineoftheTime Instead for $2x \sin x$ I found that $\alpha + i\beta$ is a solution of $ay^2+by+c=0$ then $x(A+Bx) e^x \cdot \cos(3x) + x(C + Dx) e^x \sin(3x)$
Hi all. Unfortunately, I am still stuck on the proof I mentioned yesterday. I see that the homothety of ratio $\det(\mathcal U)$ is injective, i.e. the endomorphism $M\to M$ given by $m\mapsto\det(\mathcal U)m$ is injective. But I don't see how it follows that $u$ is injective.
I don't want to confuse you, but in general if you have $y''+ay'+by=k_1f(x)+k_2g(x)$, and if you denote by $y_{p_1}$ the particular solution of $y''+ay'+by=f(x)$ and by $y_{p_2}$ the particular solution of $y''+ay'+by=g(x)$ , then $y_p=k_1y_{p_1}+k_2y_{p_2}$ is a particular solution of $k_1f(x)+k_2g(x)$ @Pizza
Consider the property $\int cf=c\int f$ for nonnegative measurable functions, $c\in[0,\infty)$. I know a proof of this that invokes the MCT, but in Folland's text this comes before the MCT. We have the definition $$\int f\,d\mu=\sup\left\{\int\phi\, d\mu:0\leq\phi\leq f,\phi\text{ simple}\right\}.$$
If I replace $f$ with $cf$ here, we get $\sup\left\{\int\phi\, d\mu:0\leq\phi\leq cf,\phi\text{ simple}\right\}$ and $c\int f$ is the same as $\sup\left\{\int c\phi\, d\mu:0\leq\phi\leq f,\phi\text{ simple}\right\}$. I still do not quite see how we get equality...
that's what I feel like I did in claiming \begin{align}c\int f&=c\cdot\sup\left\{\int \phi\, d\mu:0\leq\phi\leq f,\phi\text{ simple}\right\} \\ &=\sup\left\{c\int \phi\, d\mu:0\leq\phi\leq f,\phi\text{ simple}\right\}\\ &=\sup\left\{\int c\phi\, d\mu:0\leq\phi\leq f,\phi\text{ simple}\right\},\end{align}where the last equality is due to scaling simple functions.
Hello everyone, I have a question related to differential inequality. $$ \dot{d}(t) \geq -a d(t), \quad a\in\mathbb{R}^+. $$ I believe the solution is this $$ d(t) \geq d(0) e^{-a t} $$ is this correct? I haven't taken any course in differential inenqaulity. Any suggestions?
This bound is always true but not every function satisfying this will satisfy the differential inequality. For example for $a = 1$, $d(0) = 0$ we can take $d(t) = 1$ for $0 \leq t \leq 2$, $d(t) = e^{-4-2t}$ for $2 \leq t \leq 3$, $e^{1-t}$ for $t \geq 3$. The inequality fails on $(2,3)$. To make the function smooth, needs some small adjustment at the non-smooth points $2$ and $3$,
Make substitution $u = de^{-at}$, then it's $\dot{u} \geq 0$. So the solution is $d(t) = m(t)e^{-at}$ where $m$ is nondecreasing.
In mathematics, Grönwall's inequality (also called Grönwall's lemma or the Grönwall–Bellman inequality) allows one to bound a function that is known to satisfy a certain differential or integral inequality by the solution of the corresponding differential or integral equation. There are two forms of the lemma, a differential form and an integral form. For the latter there are several variants.
Grönwall's inequality is an important tool to obtain various estimates in the theory of ordinary and stochastic differential equations. In particular, it provides a comparison theorem that can be used to...
Basic question. Suppose $f$ is a nonnegative function. What is the negation of $f=0$ almost everywhere? Does this mean $f>0$ almost everywhere? The reason I'm asking is because I'm trying to deduce that the integral of a strictly positive function is strictly positive from the equivalence $\int f=0\iff f=0$ a.e. (which is Proposition 2.16 in Folland, though he doesn't state what I want as a corollary). I'm not sure it follows from this Proposition, but I think it's a useful result.
If $F:X\times B\rightarrow Y\times B$ is a map between nice enough metric spaces (not necessarily compact) such that $F(x,b)=(F_b(x),b)$ for all $(x,b)$. If $F$ is a local homeomorphism, is $f_b$ for each $b$, a local homeomorphism?
@monoidaltransform If there exists $(x, b)\in U\times B_0\subseteq X\times B$ such that $F\restriction_{U\times B_0}:U\times B_0\to F(U\times B_0)$ is a homeomorphism, then $F\restriction_{U\times \{b\}}$ is clearly also a homeomorphism onto its image. This is the map $(x, c)\mapsto (f_b(x), c)$ i.e. its essentially $f_b$ restricted to $U$
@BenSteffan so $f$ could still equal $0$ if $f=0$ not almost everywhere? So the converse of $ f=0$ a.e. $\implies \int f=0$ would be simply that if $f$ is not zero almost everywhere (i.e. possibly $0$ somewhere but not everywhere), then the integral is strictly positive. In particular, if $f$ is strictly positive, the integral is strictly positive :)
@Jakobian for openness so if $U$ open in $X$, then $F(U\times b)=F(U\times B)\cap (Y\times b)$. Since $F(U\times B)$ is open, $F(U\times b)$ is open in $Y\times b$. Why does this imply $f_b(U)$ is open in $Y$?
yes, it follows that, albeit not completely directly, that if $f$ is strictly positive and measurable, and measure of the whole space is positive, then $\int f > 0$
@Jakobian as a very silly sanity check, the above argument holds (to show opennness) if I restrict to any open set of $X\times B$, of form $V=V_1\times V_2$. So if $F$ is apriori defined on $V\subseteq X\times B$, and$F:V\rightarrow Y\times B$ open then $f_b:V_1\rightarrow Y$ is open, right?
say I have a transformation $T \in \mathscr{L}(V,W)$ and a basis for $V$:=$\{v_1,v_2 \cdots v_n\}$ given. It needn't generally be that I can always "split" this basis into a part that spans the kernel and the rest whose image spans the range right?
@SineoftheTime with gauss green I set $Q = 0$ so as to have an integral with constant $x$ then $\int \frac{x^3y}{\sqrt{x^2+y^2}} dy = x^3\sqrt{x^2+y^2}$
@nickbros123 given T, the possibility of choosing such a basis is a big part of elementary linear algebra, but this is certainly not guaranteed for a randomly given basis and T
So I started by calculating the integral on the segment with vertices (1,0) (2,0), I did well to write $\gamma(t) = (t,0)$, $\gamma'(t) = (1,0)$ and $t \in [1,2]$. So the integral should be 0, but I'm not sure what I did
@nickbros123 you can even visualise it, consider R^3 and the transformation being projection on the xy-plane. Now if you choose the standard basis, you can split, but if you slightly rotate the standard basis to get a new one, you may not now split.
@BenSteffan @Jakobian I'm trying to create a render in C + + as fast as possible. Is it possible to approximate a closed form expression for a quadratic recurrence relation?
@BenSteffan My understanding is quadratic recurrence relations don't usually have closed form solutions but the Sylevester sequence for example does when you use an irrational number in the solution so it can be approximated
yeah, as much as people want to short circuit the process of learning limits, the better kind of textbook will not encourage students to think/speak in those terms
its up there with "plugging in infinity" in terms of pedagogical soundness
but there's history behind it so it's not like made up
usually, you don't need it. But I think inderminate forms are taught because first you consider limits of continuous functions in a continuous point and student think you can just plug in the value. But when dealing with "indeterminate forms" you can't do that
@leslietownes yeah,. I was working a problem in which we are given a basis for $W$, the codomain of $T$, and we are tasked to find a basis for $V$ so that the matrix of $T$ wrt the found basis for $V$ and the given basis for $W$ is such that the first column is either all 0, or has a 1 in its (1,1) place.
it helps to center the actual concept (the limit) and not, like, approach the subject as a kind of extended arithmetic with infinity which is either possible or not possible depending on the situation
@SineoftheTime you say that but they literally do this and then operate based on those indeterminate forms, a lot of the time
so is it better to teach students you can't always plug things in the first place, or to plug them and then mention "oh yeah but this symbol that we just wrote doesn't make sense" where the student will willfully ignore you at that point
because I think its better to not even give them the incentive in the first place
I hate when my professors used to write $\frac{\infty}{\infty}$ or $\frac00$
I agree with you
it's not really that relevant, but one side effect of defining $0^0=1$ would be the confusion it generates among students because if the have an indeterminate form 0^0 they may think the limit is 1
Here is the problem: Find all continuous $f:\mathbb{R}\rightarrow\mathbb{R}$ which satisfy
$$f(x+y)+f(x-y)=2[f(x)+f(y)]\;\;\;(1).$$
Here is my attempt:
Fix $\delta>0$ and let $C=\int_{0}^{\delta}2f(y)dy.$
Then
$$\begin{align*}
2\delta f(x)+C&=\int_{0}^{\delta}2[f(x)+f(y)]dy\\
&=\int_{0}^{\...
Find all the functions $f\colon\mathbb{Q} \to \mathbb{Q}$ such that
$f(x+y) + f(x-y) = 2f(x) + 2f(y)$, for all rationals $x,y$.
