## Functional analysis of fa – Isoperimetric inequality for analytical functions in a ring

Leave $$f$$ be an anylytic function on the disk drive $$| z | < 1$$. It is well known that
$$left ( int_0 ^ {2 pi} f (e ^ {i theta}) d theta right) ^ 2 geq 4 pi iint_ {| z | <1} | f (re ^ {i theta}) | ^ 2nd dr d theta.$$

I wonder if the constant $$4 pi$$ can be improved in a ring $$a . More precisely, does the following inequality

$$left ( int_0 ^ {2 pi} f (e ^ {i theta}) d theta + int_0 ^ {2 pi} f (ae ^ {i theta}) d theta right ) ^ 2 geq C iint_ {a <| z | <1} | f (re ^ {i theta}) | ^ 2r dr d theta$$

wait for some constant $$C (a)> 4 pi$$ independent of $$f$$? They can $$C (a)$$ be calculated in terms of $$to$$? This seems to be a classic problem, but I couldn't find a reference and couldn't test it after trying it for a couple of days.

## mathematical optimization – Iterate Minimize in a list of functions

I have a function $$f$$ defined in $$(- 1,1)$$.
For a minimal example, it is sufficient to define

``````f(z_):=z^2 - 1
``````

I need to find a list of points so that $$f (z_0) = f (z_1)$$, points whose image "is at the same height".
I continued to find the minimum through

``````  min = First@Minimize(f(z), {z})
``````

what happens for $$z$$ equal to

``````argmin = Values@Last@Minimize(f(z), {z})
``````

Also I created a list with

``````  rang = Subdivide(a, 0,10)
``````

spanning the range from minimum to predefined value.

Now I would like to find, for each item on this list, points such that $$f (z_0) = f (z_1) = rang_j$$, for each item in the list.

I couldn't find a better plan than defining a list of features $$fun_j = (f (z) + rang_j) ^ 2$$. By changing the original function and squaring, I am sure that the functions $$fun_j$$, one for each item in the list $$rang$$ they are positive everywhere except the roots.

So I wanted to iterate over the list of functions a restricted minimization through the commands (the argument $$f_j$$ just to clarify my question, I understand that the syntax will be different, that's exactly what the question is about):

``````   Minimize({f_j, z > argmin}, {z})
Minimize({f_j, z > argmin}, {z})
``````

that is, run two minimizations one on the left and one on the right of the $${arg , min}$$. I know for mathematical reasons that there are two unique solutions.

I create my list of functions as

`````` f1(z_,c_):=f(z)+c
``````

and then using

`````` f1(z,rang)
``````

but I find it hard to iterate Minimize, any suggestion would be helpful.

Annoying

``````  Minimize({f1( z, rang), z > b}, z)
``````

produces an error message, since the Minimize function argument is expected to be a scalar function.
I'd also like to hear about better methods, in general and in reference to Mathematica.
Health

## functions: new data framework for different groups

I have a data frame with a & # 39; Country & # 39; column. Each country has multiple records. I want to write a function in Python using pandas so that for each country I can return a separate data frame.

Have a list of countries like, countries = [& # 39; USA, & # 39; Spain & # 39;, & # 39; France & # 39;],

Individually I can do for

df_us = df [df.country == & # 39; US & # 39;]
df_spain = df [df.country == & # 39; Spain & # 39;],

But how can it be done using a function! Thanks in advance

## set-return functions in postgres

I've been trying to use Postgres's generate_series () function to get a table like this:

I still haven't come up with a solution, so any ideas would be appreciated.

## Complex variables cv.com: switch arrays of complex functions

Leave $$z_0 in Bbb {R}$$ be arbitrary matrices $$A_0: = A (z_0)$$ Y $$B_0: = B (z_0)$$ are normal; in view of $$A_0 ^ * = A ^ # ( bar {z_0}) = A ^ # (z_0)$$ Y $$B_0 ^ * = B ^ # ( bar {z_0}) = B ^ # (z_0)$$ they travel with their Hermitian deputies. They also travel with each other. So $$A_0$$ Y $$B_0$$ it could be simultaneously diagonalized by a unitary matrix. That matrix is ​​also diagonalized $$A_0 ^ *$$ Y $$B_0 ^ *$$. So the four matrices $$A_0 = A (z_0), B_0 = B (z_0), A_0 ^ * = A ^ # (z_0), B_0 ^ * = B ^ # (z_0)$$ it could be diagonalized simultaneously, implying that they all travel with each other. So $$B (AA ^ #) = (AA ^ #) B$$ at any point on the actual line. Since the inputs are complete functions, they coincide throughout the complex plane.

## calculus and analysis: limit on infinity of arbitrary functions

Here is the code that takes the limit of an expression.

``````Limit((-I E^(I x) f1(y))/(g2^(Prime)(Prime))(y), x -> (Infinity) )
``````

and the output returned is `INDETERMINATE` while the desired output is $$infty$$. Or if I had to do this instead

``````Limit((g2^(Prime)(Prime))(y)/(-I E^(I x) f1(y)), x -> (Infinity) )
``````

I would like to get 0 and not `INDETERMINATE`.

How would you tell Mathematica that the $$f$$ Y $$g$$ What functions are irrelevant when evaluating the limit?

Thanks for any help.

## functions: how does following two (same) calculations give two different results?

I have the following two pieces of code that give me two different results,

``````  N((-Kp tp + Lc - tp Lc)/(Kp tp)) /. {Lc -> 6, tc -> 0.8, tp -> 0.2, Kp -> 1/3}
``````

``````(-Kp tp + Lc - tp Lc)/( Kp tp) /. {Lc -> 6, tc -> 0.8, tp -> 0.2, Kp -> 1/3}
``````

Which gives the answer as 3 (which is obviously wrong).

Could someone explain how / why this happens and how to avoid this (possible) error?

Below is the image of the calculation on my machine:

Can anyone explain the following behavior in Mathematica 12.0?

``````EllipticK[N[1/2, 100]]
``````

spit `ComplexInfinity`. Nevertheless

``````EllipticK[1/2] // N[#, 100] &
``````

It seems to give the correct result. In my current code, I only know the argument of the function `EllipticK` numerically then the second option is not exactly what i want.

## CHAIN ​​REVERSION in PYTHON (without using built-in functions)

What is wrong with the following code?
It is giving an error: ROPE OUT OF REACH

``````def reverse(s):
i=0
l=()

while i``````
``` jQuery(document).ready(function() {jQuery.ajax({type: "POST",url : rating_form_script.ajaxurl,data : { action : "display_rating_form_ajax", args : {"id":2,"post_id":201862,"comment_id":0,"custom_id":"0","user_id":0,"term_id":0,"title":false,"score":true,"total":true,"stats":true,"user_stats":false,"tooltip":true,"result":true,"rich_snippet":true,"is_widget":false,"state":"","before_content":"","after_content":"","rates":"rating,ratings","txt_score":"%1\$s\/%2\$s"} }, success : function(data) { jQuery("body").find("[data-id=\"RFR2P201862\"]").html(data); }});}); ```
``` Posted on April 1, 2020Author Proxies123Tags builtin, chain, functions, Python, REVERSION ```
``` parametric functions – Problem: solving the system of equations I am trying to solve a "simple" system of equations for a formal theoretical model that I am developing. Although I think I had correctly specified the system, I can't find a solution so far. Is it a problem with so many parameters? Can someone help me with this? Solve({A1*A2*y - A3*y - A4 - A5*((1/A6)^(1-A7))*((x*A8 + A9)^(A10)) + A11*((A6)^(-A12))*((y*A13)^(A14)) == 0, A15*A16*x - A17*x -A19 - A18*A6+ A5*((1/A6)^(1-A7))*((x*A8 + A9)^(A10)) - A11*((A6)^(-A12))*((y*A13)^(A14)) == 0, 00, A2>0,A3>0,A4>0,A5>0,A6>0,A7>0,A8>0,A9>0,A10>0,A11>0,A12>0,A13>0,A14>0,A15>0,A16>0,A17>0, A18>0, A19>0}, {x,y}) Thank you! jQuery(document).ready(function() {jQuery.ajax({type: "POST",url : rating_form_script.ajaxurl,data : { action : "display_rating_form_ajax", args : {"id":2,"post_id":201567,"comment_id":0,"custom_id":"0","user_id":0,"term_id":0,"title":false,"score":true,"total":true,"stats":true,"user_stats":false,"tooltip":true,"result":true,"rich_snippet":true,"is_widget":false,"state":"","before_content":"","after_content":"","rates":"rating,ratings","txt_score":"%1\$s\/%2\$s"} }, success : function(data) { jQuery("body").find("[data-id=\"RFR2P201567\"]").html(data); }});}); Posted on March 31, 2020Author Proxies123Tags equations, functions, parametric, Problem, solving, system Posts navigation Page 1 Page 2 … Page 82 Next page ```