Tag: function

fourier analysis – Definite integral with Dirac delta and Heaviside function

In relation with the question I posed on MathSE here, I want to ask how can Mathematica give an answer to my problem.

The context

I am trying to get rid of the integral over $y$ in
$$begin{align}
int_0^{infty}dy ; psileft(frac{y}{q}right)phi(y) int_{-infty}^{infty}frac{dp}{sqrt{2pi}} p^{gamma} e^{-ip(y-x)};,
end{align}$$
and to obtain an analytical expression in terms of $phi(x)$, $psi(x/q)$ and their derivatives. I assume that $gamma=1-delta$, with $0<delta<1$. I tried to break down the problem (though I don’t know if there is a better way) so that I am left with
$$ begin{align}
int_0^{infty} dy ; psileft(frac{y}{q}right)phi(y) times left(frac{delta(y-x)}{(y-x)^{1-delta}} – (1-delta)frac{H(y-x)}{(y-x)^{2-delta}}right);,
end{align}$$
where $delta$ is the Dirac delta and $H$ the Heaviside function.

The problem

Since the parameter $delta$ (not to be confused with the Dirac) is constrained as $0<delta<1$, I expect troubles when evaluating the first term in the brackets. However, when I define $psi$ and $phi$, MMA returns an expression for this integral, which leaves me confused. Basically, my code is

  (Psi)(x_) := 9*6^{-1/2} x^{3/2} Exp(-3 x/2);
  (Phi)(x_) := x; (* Or x^2, or E^(x)...*)
  fourpfrac(x_, y_, (Delta)_) := (2*Pi)^(1/2)/Gamma((Delta))*HeavisideTheta(y - x)*(y - x)^((Delta) - 1); (* Fourier Transform of p^(1-(Delta)) *)
  x1frac(x_, (Nu)_, (Delta)_, (Alpha)_, q_) := Integrate(Derivative(0, (Nu), 0)(fourpfrac)(x, y, (Delta))*(Psi)(y/q)*(Phi)(y)*y^(-(Alpha) - 1), {y, 0, Infinity}); (* The whole integral *)
  x1frac(x, 1, (Delta), -1, q)

which returns in traditional form
$$
left{text{ConditionalExpression}left(frac{45 sqrt{3} x ((pi delta ) sin ) Gamma left(-delta -frac{3}{2}right) (-x)^{delta +frac{1}{2}} , _1F_1left(frac{7}{2};delta +frac{5}{2};-frac{3 x}{2 q}right)}{8 q^{3/2}}+frac{sqrt{pi } 2^{delta +frac{3}{2}} 3^{-delta } Gamma left(delta +frac{3}{2}right) q^{delta } , _1F_1left(2-delta ;-delta -frac{1}{2};-frac{3 x}{2 q}right)}{Gamma (delta -1)},x<0right)right}
$$

and in input form

{ConditionalExpression((2^(3/2 + (Delta)) 3^-(Delta) Sqrt((Pi)) q^(Delta) Gamma(3/2 + (Delta)) Hypergeometric1F1(2 - (Delta), -(1/2) - (Delta), -((3 x)/(2 q))))/Gamma(-1 + (Delta))
+ (45 Sqrt(3) (-x)^(1/2 + (Delta))x Gamma(-(3/2) - (Delta)) Hypergeometric1F1(7/2, 5/2 + (Delta), -((3 x)/(2 q))) Sin((Pi) (Delta)))/(8 q^(3/2)),  x < 0)}

The thing is that in no way the form I have given to $phi$ and $psi$ could have canceled the singularity.

  1. Is the result I have obtained with MMA correct?
  2. What did MMA do to evaluate the integral?
  3. Is there a better way to resolve the problem?

Many thanks!

NDSolve::dvlen: The function [Theta][t] does not have the same number of arguments as independent variables

please help me to see what wrong with my code?

s = NDSolve({(m*g*
   Sin((Theta)(t))) - (m*(r''(t) - r (((Theta)'(t))^2))) == 
k (r(t) - 14),  g*Cos((Theta)(t)) == 
r(t) (Theta)''(t) + 2*r'(t) (Theta)'(t), (Theta)'(0) == 0, 
 r'(0) == 0}, {r, (Theta)}, {t, 0, 60}, {k, 0, 2000})

i want to solve this two equation to find equation of r(t). I’ve already set up m and g

plotting – How to calculate Fourier coefficients in the transmittance function?

Good morning, could someone help me to propose this exercise to be able to solve it

Calculate the Fourier coefficients c_s of the transmittance function t(x)= t(x+P_x) given in|x|<P_x/2,characterizing the following 1-D diffraction gratings of period P_x:
The square-wave amplitude grating composed of evenly spaced parallel slits, givingt(x) =rect(x/w), where w < P_x is the slit width.

sql server – Why changing the column in the ORDER BY section of window function “MAX() OVER()” affects the final result?

I have a table with below structure and it’s data :

create table test_table
(
Item_index   int,
Item_name    varchar(50)
)

insert into test_table (Item_index,Item_name) values (0,'A')
insert into test_table (Item_index,Item_name) values (1,'B')
insert into test_table (Item_index,Item_name) values (0,'C')
insert into test_table (Item_index,Item_name) values (1,'D')
insert into test_table (Item_index,Item_name) values (0,'E')

I want to know why changing the column in order by section of the query , changes the result? In QUERY-1 , I used item_index and in the QUERY-2 I used item_name column in the order by section. I thought that both queries must generate the same result because I used item_index in both queries for partitioning! I’m completely confused now ! why the order by column should affect the final result?

QUERY-1:

select t.*,
       max(t.Item_name)over(partition by t.item_index order by item_index) new_column
from test_table t;

RESULT:

Item_index  Item_name     new_column
----------- --------------------------
0           A                E
0           C                E
0           E                E
1           D                D
1           B                D

QUERY-2:

select t.*,
       max(t.Item_name)over(partition by t.item_index order by item_name) new_column
from test_table t;

RESULT:

Item_index  Item_name  new_column
----------- -----------------------
0           A             A
0           C             C
0           E             E
1           B             B
1           D             D

Can anybody explain how exactly these two queries are being executed and why each of them generates different result?

Thanks in advance

argument patterns – Unusual performace of the Cases function

Recently I was surprised by unusual performance of Cases function. The code

Cases[{{1, a}, {2, b}, {3, c}, {4, e}}, X_ /;X[[1]]==Part[RandomSample[{1, 3}, 1],1]]

may return {1, a}, {3, c}, {{1, a}, {3, c}}, or {}. Try reproduce this by running

Table[Cases[{{1, a}, {2, b}, {3, c}, {4, e}}, X_ /; X[[1]] == Part[RandomSample[{1, 3}, 1],1]], {q,1,200}]

Whereas explicit substitution of pattern like here

Selected = Part[RandomSample[{1, 3}, 1],1];
Cases[{{1, a}, {2, b}, {3, c}, {4, e}}, X_ /; X[[1]] == Selected],

always works properly. As I understand, Mathematica has difficulties with calculation of the pattern in the body of Cases

real analysis – Do we have full control the oscillation of a function by modifying it on a small set?

Definitions and some motivation:

Let $mathcal B$ be the set of bounded measurable functions from $(0, 1)$ to $mathbb R$. Denote by $mathcal N$ the set of measurable subsets of $(0, 1)$ with Lebesgue measure $0$.

Given a function $f in mathcal B$, define the function $mathcal Of$ by

$mathcal Of(x) := inf_{N in mathcal N} lim_{delta to 0} sup_{y, z in B_delta (x) setminus N} |f(y) – f(z)|$.

Thanks to Lusin’s theorem, we know that we can modify $f$ on an arbitrarily small set and get a continuous function, and so we force the oscillation to be $0$ everywhere. But can we force it to be whatever we want?

Question:

Does there exist, for any $f, g in mathcal B$ and $varepsilon > 0$, a function $f’ in mathcal B$ such that the following conditions are satisfied?

i) $f’ = f$ everywhere except for a set of measure at most $varepsilon$.

ii) $mathcal Of’ = mathcal Og$ everywhere.

Note: All functions are genuine functions and not equivalence classes modulo null sets of such.

probability – “First principles” proof of the limit of the expected excess of the uniform renewal function

The closed form of the expected number of samples for $sum_r X_r geqslant t, X_r sim text{U(0,1)}$ is given by:

$$m(t) = sum_{k=0}^{lfloor t rfloor} frac{(k-t)^k}{k!}e^{t-k}$$

From this we can deduce the expected amount by which this sum exceeds $t$, namely:

$$varepsilon(t) = frac{m(t)}{2} – t$$

From knowing that $m(t) to 2t+dfrac{2}{3}$, we can easily see that $varepsilon(t) to dfrac{1}{3}$.

Is there a simple (“low tech”) way of proving that $varepsilon(t) to dfrac{1}{3}$ without first passing through proving $m(t) to 2t+dfrac{2}{3}$ ?

javascript – Do all dynamically typed languages not support function overloading?

The issue is easier to understand if you consider carefully what it really means to overload a function.

Overloaded methods e.g. in Java are really two completely separate entities; they share no byte code, no address, nothing except their name; and their name isn’t really the same either, since in the compiler symbol table, a print() method for ints and a print() method Strings actually have a mangled name that contains both the user-visible identifier (“print”) and additional information encoding the argument type.

Now contrast this with Javascript, where a print() function really is called print and nothing else. The runtime system only knows that it is a function; what arguments it expects and how it deals with them is entirely defined by the code in the function’s body. Therefore, defining a second function “print” simply overwrites the previous one rather than add a second implementation.

The details vary a bit from language to language, but the gist is, if you don’t have an explicit representation of data types in your compile-time/run-time system, you can’t use them to tell elements of the system apart, and that is why overloading on types is largely restricted to systems with a strong presence of types in the language definition.