My textbook says that any function which can be written as $y(x,t)=f(ax±b)$ represents a wave. It says that that the only condition is that it should be finite everywhere and at all times.   With this definition, I cannot understand why $(x-vt)^2$ does not represent a wave. It is finite for all real $x$ and $t$.  What am I missing?
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A function which represents a wave must satisfy the following differential equation: $$\frac{\partial^2 y}{\partial t^2} = k\frac{\partial^2 y}{\partial x^2}$$ Any function that satisfies the wave differential equation represents a wave provided that it is finite everywhere at all times.  What does &quot;it is finite everywhere at all times&quot; mean?  Question:Which of the following functions represent a wave? a) $(x - vt)^2$ b) $\ln(x + vt)$ c) $e^{-(x - vt)^2}$ d) $(x + vt)^{-1}$  Only option (c) is given as the answer though all 4 satisfy the differential equation. I believe I did not understand the significance &quot;function should be finite everywhere at all times&quot; which is why I am unable to answer the aforementioned question.
103505
I was wondering if there are functions for which $$f'(x) &gt; f(x)$$ for all $x$. Only examples I could think of were $e^x - c$ and simply $- c$ in which $c &gt; 0$. Also, is there any significance in a function that is always less than its derivative?    Edit: Thank you very much for all the replies. It seems almost all functions that apply are exponential by nature...  Are there more examples like - 1/x?  Again are there any applications/physical manifestations of these functions? [for example an object with a velocity that is always greater than its position/acceleration  is always greater than its velocity]
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Let $f:[0,\alpha]\to\mathbb{R}$ be a solution of the Cauchy problem:  $\begin{cases} f'(t)=(f(t))^2+t \\ f(0)=0 \end{cases} $  The question is: prove that $\alpha&lt;3$.  It is clear that the problem admits a unique solution locally on some $[-\epsilon,\epsilon]$, since $(f)^2+t$ is $C^1$ in $(f,t)$, and we have $f(t)&gt;0$ for all $t&gt;0$, looking at the sign of $f'$. But now I don't know how to proceed. Thank you all!
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I've seen $d(xy) = y(dx) + x(dy)$, but I don't understand the principle behind it and memorizing it is lame.  Can anyone explain what is going on here?  For example from physics, $$F = {{dP} \over {dt}}$$ $$F = {{d(mv)} \over {dt}}$$ $$F = {{v(dm) + m(dv)} \over {dt}}$$ Since $$dm = 0$$  $$F = {{m(dv)} \over {dt}}$$ Since $$a = {{dv} \over {dt}}$$ $$F = ma$$
73379
I'm studying Quantum Mechanics and I came across this which I don't quite understand: For a vector space of functions $f(x)$ to be square integrable (i.e $\int{|f(x)|^2dx &lt; \infty)}$, the necessary condition is $\lim_{|x| \to \infty}f(x) \to 0$. Can someone help me understand why that must be the case? I sort of understand that it must not blow up at the infinities but can a (continuous or piece-wise continuous) function have a non-zero finite value at the infinities and still be square integrable?
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Having just finished physics 2, I've been (slightly) exposed to showing that light is a wave with speed $1/\sqrt{\mu _0 \epsilon _0 }$ using the differential forms of  though this is the only derivation I've come across. Can you show the same thing using the integral forms? My first thought is that it may be more difficult since the wave equation is often given as a differential equation.   Note: I have not taken vector calculus (or even multivariable), and do not have sufficient mathematical (or even physical) background to explicitly do the derivation. I'm merely asking for a hopefully understandable solution or a source to the solution.
111724
The Problem:   Let $f : \mathbb R \to \mathbb R$ be a real-valued function defined on the set of real numbers that satisfies $$f(x + y) \leq yf(x) + f(f(x))$$ for all real numbers $x$ and $y$.  Prove that $f(x) = 0$ for all $x ≤ 0$.  (IMO $2011$ , Pr: $3$)  Here is the solution:  Before we prove that $f (x) = 0$ for all $x ≤ 0$, we will prove that $f (0) = 0.$ $\color{black}{\large\text{Case} \thinspace  1:}$ $f(0)&gt;0$ Let, $\varepsilon &gt;\dfrac{f(f(0))}{f(0)}$ and $x&lt;- \dfrac{\varepsilon+f(f(0))}{f(0)}$  we have, from $f(x)\leq xf(0)+f(f(0))$,  $f(x)&lt;-\varepsilon.$ Then from, $f(f(x))\leq f(x)f(0)+f(f(0))$ we have $f(f(x))&lt;0$. Finally,  since $0&lt;f(0)\leq f(f(0))$ ,we get $f(0)\leq-xf(x)+f(f(x))&lt;0$ which imply $f(0)&lt;0$, which gives a contradiction. $\color{black}{\large\text{Case} \thinspace  2:}$ $-\alpha \leq f(0)&lt;0$, where $\alpha\in\mathbb{R^+}.$ From $f(x + y) \leq yf(x) + f(f(x))$ we have: $\begin{cases} f(x)\leq f(f(x)) \\ f(x) \leq xf(0)+ f(f(0)) \end{cases} \Longrightarrow f(x)\leq f(x)f(0)+f(f(0)) \Longrightarrow f(x)(1-f(0))\leq f(f(0))$. Then applying $x=f(0)$, we get $f(f(0))\leq 0$, which imply $f(x)\leq 0$, which gives $f(f(x))\leq 0$. In this case, we have $f(x)&lt;0.$ Because, if $f(x)=0$ at least for some $x$, from $f(x)\leq f(f(x))$, we get $f(0)\geq 0$, which gives a contradiction. Then let, $x&lt;-\sqrt {\alpha}$,  from  $f(x + y) \leq yf(x) + f(f(x))$  we have: $$\begin{align} f(z)\leq(z-x)f(x)+f(f(x)) \Longrightarrow f(f(x)) \leq (f(x)-x)f(x)+f(f(x))\Longrightarrow f(x)(f(x)-x)\geq 0 \Longrightarrow f(x) \leq x &lt;-\sqrt {\alpha} \Longrightarrow f(x)&lt; -\sqrt{\alpha}.\end{align}$$ Finally, from $f(0)\leq-xf(x)+f(f(x))$ we get $f(0)&lt;-\alpha$, which gives a contradiction. So, we deduce that $f(0)=0$. Then, applying $f(0)=0$, from  $f(0)\leq-xf(x)+f(f(x))$ and $f(x) \leq xf(0)+ f(f(0))$ we have, $$\begin{cases} xf(x)-f(f(x)) \leq 0 \\ f(x) \leq 0 \\ f(f(x)) \leq 0 \end{cases} \Longrightarrow  \begin{cases} xf(x)\leq0 \\ f(x)\leq 0 \end{cases} \Longrightarrow \begin{cases} xf(x)\leq0, x\in\mathbb {R} \\ xf(x)\geq 0, x\leq 0 \end{cases} \Longrightarrow xf(x)=0, x\leq 0 \Longrightarrow \color{blue} {\boxed{ f(x)=0, \text{for all} \thinspace x\leq 0.}}$$ End of the Proof.  Can you verify this solution? Are there any mistakes, gaps, etc.? Thank you very much!
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Why $f(x) = \sqrt{x}$ is a function (as I found in my textbook) since for example the square root of $25$ has two different outputs ($-5,5$) and a function is defined as "A function from A to B is a rule of corre- spondence that assigns to each element in set A exactly one element in B.", so $f(x) = \sqrt{x}$ is not a function?
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Wave equations take the form:  $$\frac{ \partial^2 f}  {\partial t^2} = c^2 \nabla ^2f$$  But the Schroedinger equation takes the form:  $$i \hbar \frac{ \partial f}  {\partial t} = - \frac{\hbar ^2}{2m}\nabla ^2f + U(x) f$$  The partials with respect to time are not the same order. How can Schroedinger's equation be regarded as a wave equation?  And why are interference patterns (e.g in the double-slit experiment) so similar for water waves and quantum wavefunctions?
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why: $$\int_0^1\int_{u(t)}^{u(t)+w(t)} f(t,v(t)) dv dt = \int_0^1 f(t,u(t)+\theta w(t))w(t) dt; ~~\theta\in[0,1] $$  how to get this ?  Please help me   Thank you.
62941
Find all functions $f$ that assign a real number $f(x)$ to every real number $x$ such that $$(x+y)f(x)+f(y^2)=(x+y)f(y)+f(x^2)$$  I've tried subbing in heaps of values but I keep getting things like $f(0)=f(0)$ and other such useless results.  Any help would be hugely appreciated.
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I know that it is defined such that there is no more than one y-value output for any given x-value input, but I"m wondering WHY it is defined that way? Why can't we apply everything we know about functions to equations that have more than one y-value output for any given x-value input?   In short, what is the benefit or reason to define a function this way?  At the same time, a little historical context would also be appreciated (i.e., when the concept originated and why).  Thanks,  Moshe
20398
The solution of the wave equation $$u_{tt}-c^2u_{xx}=0$$ is given as $$u(x,t)=f(x+ct)+g(x-ct)$$ where $f,g$ are arbitrary functions of one variable.  One way to prove this is the following:  $$(\partial_{tt}-c^2 \partial_{xx})u=0$$ $$(\partial_t-c \partial_x)(\partial_t+c \partial_x)u=0$$  So solutions must satisfy  $$ (\partial_t-c \partial_x)u=0 \;\;\cup\;\;  (\partial_t+c \partial_x)u=0 $$  $\displaystyle{(\partial_t-c \partial_x)u=0 \Rightarrow u=f(x+ct)}$  $\displaystyle{(\partial_t+c \partial_x)u=0 \Rightarrow u=g(x-ct)}$    Could you explain me why the last two equations stand??
81155
Suppose a function $f : \mathbb{R} \to \mathbb{R} $ satisfies the relation $$f(x^2) = xf(x) \ \ \forall x$$  Does this imply $f$ must be a straight line, $f(x) = mx$? If so, why?  If not, are there other such functions?
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Plotting the function $f(x)=x^{1/3}$ defined for any real number $x$ gives us:   Since $f$ is a function, for any given $x$ value it maps to a single y value (and not more than one $y$ value, because that would mean it's not a function as it fails the vertical line test). This function also has a vertical tangent at $x=0$.   My question is: how can we have a function that also has a vertical tangent? To get a vertical tangent we need 2 vertical points, which means that we are not working with a "proper" function as it has multiple y values mapping to a single $x$. How is it possible for a "proper" function to have a vertical tangent?  As I understand, in the graph I pasted we cannot take the derivative of x=0 because the slope is vertical, hence we cannot see the instantaneous rate of change of x to y as the y value is not a value (or many values, which ever way you want to look at it). How is it possible to have a perfectly vertical slope on a function? In this case I can imagine a very steep curve at 0.... but vertical?!? I can't wrap my mind around it. How can we get a vertical slope on a non vertical function?
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Assuming I have the graph of a function $f(x)$ is there function $f_1(f(x))$ that will give me a rotated version of the graph of that function?  For example if I plot $\sin(x)$ I will get a sine wave which straddles the $x$-axis, can I apply a function to $\sin(x)$ to yield a wave that straddles the line that would result from $y = 2x$?
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I have problems proving the following result:     Each $f: \mathbb{R}^2 \rightarrow \mathbb{R}$ such that $\forall a,b \in \mathbb{R} \ : \ f_a(y) := f(a,y), \ f_b(x) := f(x,b) $ are polynomials is a polynomial with two variables.   If I consider $f$ as a function of $x$, then its derivative  $f'(x) = \frac{\partial f}{\partial b}(x)$. Similarly if we treat $f$ as a function of $y$.   I assume that $f_a(y) = \frac{\partial f}{\partial a} (y) $ and $f_b(x)=\frac{\partial f}{\partial b}(x)$.  But I am not sure if we can assume that, because  the degree of the derivative should be smaller than the degree of the original function (and it isn't).   Actually, I'm not even sure if what I'm trying to prove is true, because in the original formulation of the problems there is written $f_a(y) := (a,y), \ f_b(x):=(x,b)$. But that didn't make sense.  Could you help me here?  Thank you.