TITLE: Perturbation expansion with path integrals
QUESTION [1 upvotes]: This is from Hugh Osborn's 'Advanced Quantum Field Theory' notes, Lent 2013, page 15.
I want to evaluate the expression
$$ Z = \exp\Big(\frac{1}{2} \frac{\partial}{\partial \underline{x}} . A^{-1} \frac{\partial}{\partial \underline{x}} \Big) \exp\Big(-V(x) + \underline{b}. \underline{x}\Big) \bigg\vert_{\underline{x} = \underline{0}}$$
assuming that $\underline{b} = \underline{0}$.
We use the notation
$$ V_{i_{1} i_{2} \dots i_{k}} =  \frac{\partial}{\partial x_{i_{1}}} \frac{\partial}{\partial x_{i_{2}}} \dots \frac{\partial}{\partial x_{i_{k}}} V(\underline{x})\Big\vert_{\underline{x} = \underline{0}}$$
Where
$\frac{\partial}{\partial \underline{x}} \equiv \Big( \frac{\partial}{\partial x_{1}}, \dots, \frac{\partial}{\partial x_{n}}\Big)$
We also assume that $V(\underline{0}) = V_{i}(\underline{0}) = 0$.
And $A^{-1}$ is an $n \times n$ matrix. I expanded the exponential with the derivative to get:
$$ Z = \Bigg(1 + \frac{1}{2} \frac{\partial}{\partial x_{i}} A^{-1}_{ij} \frac{\partial}{\partial x_{j}} + \frac{1}{2} \frac{1}{2}\frac{\partial}{\partial x_{i}} A^{-1}_{ij} \frac{\partial}{\partial x_{j}} \frac{1}{2}\frac{\partial}{\partial x_{k}} A^{-1}_{kl} \frac{\partial}{\partial x_{l}} + \frac{1}{6}\frac{1}{2} \frac{\partial}{\partial x_{i}} A^{-1}_{ij} \frac{\partial}{\partial x_{j}}\frac{1}{2} \frac{\partial}{\partial x_{k}} A^{-1}_{kl} \frac{\partial}{\partial x_{l}}\frac{1}{2} \frac{\partial}{\partial x_{m}} A^{-1}_{mn} \frac{\partial}{\partial x_{n}} + \dots\Bigg) \exp(-V(x))\biggr\vert_{\underline{x} = \underline{0}}$$
Then this series acts on $\exp(-V(x))$, where summation notation is implied. Am I proceeding correctly?
The answer is

REPLY [0 votes]: The corrected expansion does indeed give you the right answer. For example,
$$ \frac{1}{2} \frac{\partial}{\partial x_{i}} A^{-1}_{ij} \frac{\partial}{\partial x_{j}} \exp(-V(x)) \big\vert_{x = 0} = \frac{1}{2} \frac{\partial}{\partial x_{i}} A^{-1}_{ij} (-\frac{\partial}{\partial x_{j}}V(x)) \exp(-V(x)) \big\vert_{x = 0} = -\frac{1}{2} A^{-1}_{ij} \Big[\frac{\partial}{\partial x_{i}} \frac{\partial}{\partial x_{j}} V(x)  - \frac{\partial}{\partial x_{i}} V(x)\Big] \exp(-V(x))\Big|_{x = 0} = -\frac{1}{2}A_{ij}^{-1}V_{ij}$$
And so on and so forth for higher derivatives.


TITLE: Inequality involving Sobolev spaces
QUESTION [1 upvotes]: Let us define
$$
\mathbb{H}^{1} =  H^{1}(-L,0) \times H^{1}(0,L) \ \ \text{and} \ \ \mathbb{L}^{2} = L^{2}(-L,0)\times L^{2}(0,L),
$$
where $H^{1}(I) = \big\lbrace u \in L^{2}(I) \ \text{and} \ u_{x} \in L^{2}(I); I = (a,b) \big\rbrace$.
Besides these,
$$
\mathbb{M} = \big\lbrace (u,v) \in \mathbb{H}^{1}; u(-L) = v(L) = 0 \ \text{and} \ u(0) = v(0) \big\rbrace .
$$
Under the above conditions, we have that the phase space is given by
$$
\mathcal{H} = \mathbb{M} \times \mathbb{L}^{2}.
$$
Note that this space equipped with the inner product
$$
\langle (u_{1},v_{1},w_{1},z_{1}), (u_{2},v_{2},w_{2},z_{2}) \rangle = \int_{-L}^{0}u_{1_{x}}\overline{u}_{2_{x}} + \int_{0}^{L}v_{1_{x}}\overline{v}_{2_{x}} + \int_{-L}^{0}w_{1}\overline{w}_{2} + \int_{0}^{L}z_{1}\overline{z}_{2}
$$
is a Hilbert space.

REPLY [2 votes]: Just take a derivative, and you get
$$ i \lambda u_x = f_x - w_x $$
So taking the $L^2(-L,0)$ norms on both sides, you get
$$ \lambda^2 \int |u_x|^2 \leq \int |f_x - w_x|^2 $$
The RHS can be expanded and estimated using AM-GM to be
$$ \lambda^2 \int |u_x|^2 \leq 2 \int |f_x|^2 + |w_x|^2 $$
The first term is bounded by $\|F\|^2$, and the second, by your assumption on the $L^2$ bound of $w_x$, is bounded by $\|U\| \|F\|$.


TITLE: Give all solutions of the system of congruences:
QUESTION [1 upvotes]: Give all solutions of the system of congruences:
$$x \equiv 3 \pmod 5 $$
$$x \equiv 2 \pmod 4$$
$$x \equiv 5 \pmod 7$$
Then give the least strictly positive solution.

REPLY [1 votes]: Another Hint:
As $x\equiv 3 $ mod 5, $x=5k+3$ for some integral $k$. Using $x\equiv 2$ mod 4, we have $k+1\equiv 0$ mod 4 i.e. $k$ is of the form $4j+3$. Proceed in a similar fashion to finish this off!


TITLE: Integration similar to consequences of Riemann lebesgue lemma
QUESTION [0 upvotes]: Let $f\in C^1[-\pi,\pi]$. Define , for $n\in N$ b_n=$\int_{-\pi}^\pi f(x)sinntdx$.
Which of the following statement are true?
a) $b_n\to 0$, as $n\to \infty$
b)$nb_n\to 0$, as $n\to \infty$
Please give me hint for option b)
I tried substitution method, as it value of f(-π)=f(π) were given then I could able to done it easily. But how to think without that condition?

REPLY [0 votes]: b) is false without periodicity. Just take $f(x)=x$ and compute the integral using integration by parts. $(nb_n)$ does not converge in this case.


TITLE: Convergence of $\int_0 ^\infty \frac {dx}{\sqrt {1+x^3}}$
QUESTION [0 upvotes]: Convergence of $\int_0 ^\infty \dfrac {dx}{\sqrt {1+x^3}}$
Attempt: $\lim_{x \rightarrow \infty} \dfrac {x^{\frac{3}{2}}}{\sqrt {1+x^3}} =1$
Hence, $\dfrac {1}{x^{\frac{3}{2}}}$ and $\dfrac {1}{\sqrt {1+x^3}}$ converge/diverge together.
$\int_0 ^\infty \dfrac {dx}{x^{\frac{3}{2}}}$ does not converge since, it's not defined at $0$.
However, my textbook says this integral is convergent. Why is it so?
Thank you for reading through!

REPLY [2 votes]: You got some good answers but I want to highlight some of the reasoning errors you made.  It's important to derive your assertions carefully.

Hence, $\dfrac {1}{x^{\frac{3}{2}}}$ and $\dfrac {1}{\sqrt {1+x^3}}$ converge/diverge together.

Remember that convergence is a property of the improper integral, not the function being integrated.  So a statement like “$\frac{1}{x^{3/2}}$ converges” is ambiguous.  What you mean is

Hence, $\color{red}{\int_1^\infty}\frac {1}{x^{\frac{3}{2}}}\color{red}{\,dx}$ and $\color{red}{\int_1^\infty}\frac {1}{\sqrt {1+x^3}}\color{red}{\,dx}$ converge/diverge together.

This is true.  Now for this:

$\int_0 ^\infty \frac {dx}{x^{\frac{3}{2}}}$ does not converge since, it's not defined at $0$.

A couple of problems here.  One is that you have two improper integrals in one.  Both $\int_0^1 \frac{dx}{x^{3/2}}$ and $\int_1^\infty \frac{dx}{x^{3/2}}$ are improper.  But only the second one is relevant for this problem.  That's because $\int_0^1 \frac{dx}{\sqrt{1+x^3}}$ is not improper, so you are only worried about whether $\int_1^\infty \frac{dx}{\sqrt{1+x^3}}$ converges.
As for $\int_0^1 \frac{dx}{x^{3/2}}$, you are correct that it does not converge, but it does not follow from your assertion that the integrand is not defined at $0$.  After all, $\int_0^1 \frac{dx}{x^{1/2}}$ does converge, but the integrand is not defined at $0$ either.  The divergence of $\int_0^1 \frac{dx}{x^{3/2}}$ follows from the $p$-test: if $p > 1$ then $\int_0^1 \frac{dx}{x^p}$ diverges and $\int_1^\infty \frac{dx}{x^p}$ converges.
To piece this all together: You know $\int_1^\infty \frac{dx}{x^{3/2}}$ converges by the $p$-test.  Therefore, $\int_1^\infty\frac{dx}{\sqrt{1+x^3}}$ converges.  Finally,
$$
    \int_0^\infty \frac{dx}{\sqrt{1+x^3}} = \int_0^1 \frac{dx}{\sqrt{1+x^3}}  + \int_1^\infty\frac{dx}{\sqrt{1+x^3}}
$$
so the given integral converges.


TITLE: Equivalent of random variable sequences in distribution?
QUESTION [0 upvotes]: Suppose that $X_n, Y_n$ are sequences of random variable on probability space $\Omega$. If $Xn,Yn$ converges to $X$ ( some random variable ) in distribution, then is $X_n=Y_n$ almost everywhere (a.s)?
I supposed it is true, and I tried to prove $P\{|X_n-Y_n|>\epsilon\}=0$. I started from adding and subtracting $X$.
Am I in the right path ?

REPLY [1 votes]: The assertion is not true. A counterexample: Pick a random variable $Y$ and define $Y_n:=Y$ and $X_n:=Y+1/n$. Then $X_n$ converges in distribution to $Y$, and obviously $Y_n$ converges in distribution to $Y$. But $X_n$ and $Y_n$ are equal nowhere.