Complex Analysis – Third Simplified Root of (24 * sqrt (3))

I have trouble finding a solution to simplify a given polynomial.

$$ Polynomial: p (x) = x ^ 5 + { sqrt 3} x ^ 4 + 24 { sqrt 3} x ^ 2 + 72x $$

the zeros of this function (polynomial roots? English is not my native language, so I do not know how to express the point or points where the function meets the X axis) are:
$$ x_0 = – { sqrt 3} \
x_1 = 0 \ $$

and the complexes, computed with what "stays" after the polynomial division, pulling the 3rd root, etc .:
$$ x = sqrt[Large 3]{-24 sqrt 3} $$

The first problem comes now. The next step, without explanation, simplifies the above:
$$ x = sqrt[Large 3]{8 * sqrt 3 ^ 3rd ^ { Large_ {i pi}} $$
How is it or rather what is the logic behind all this? Especially the 8 which in some way was transformed from 24.

calculus – Show that $ int ^ {R + delta} _ { delta / 2} sqrt {R ^ 2- (x- delta) ^ 2} , dx- int ^ {R} _ { delta / 2} sqrt {R ^ 2-x ^ 2} , dx leq R delta $, for $ 0

I need to show that if $ 0 < delta <R, $ then

$$ int ^ {R + delta} _ { delta / 2} sqrt {R ^ 2- (x- delta) ^ 2} , dx- int ^ {R} _ { delta / 2 } sqrt {R ^ 2-x ^ 2} , dx leq R delta. $$

The result of the evaluation of the integrals is awful and seems quite difficult to manage. Is there a simpler way to get the result using a binding argument?

All clues are highly appreciated. Thank you for your time.

html – Problem with sqrt () in PHP

I have a problem with the function sqrt () from PHP, the result returns it to me in the following way.

Example:


In this way, he returns the result:

enter the description of the image here

Now the question is: how could it be or is it possible to have it just show the the number and no float ()?

c – How to properly use the exp and sqrt properties

dual purpose
-use sqrt () and exponential function exp ()
-use * to calculate the square
-Do not use pow ()

I'm getting values, they just are not what I thought. I tried to sign them all, but that did not change anything. I have tried to print with 12 decimal places and nothing seems to work. I've linked the math library and defined this one.

double normal (double x, double sigma, double mu)
{
double function = 1.0 / (sigma * sqrt (2.0 * M_PI));
double raise = 1.0 / 2.0 * ((x-mu) / sigma);
double func1 = func * exp (increase);
double comp_func = (func1 * func1);

returns comp_func;
}
int main (void)
{
// create two constant variables for μ and σ
const double sigma, mu;
// create a variable for x - only dynamic variable in the equation
unsigned int x;

// create a variable for N x values ​​to use for the loop
int no_x;

// scaniing value in mu
printf ("Enter mean u:");
scanf ("% lf", & mu);

// digitization of the value in sigma
printf ("Enter standard deviation:");
scanf ("% lf", & sigma);

// if sigma = 0 then leave
if (sigma == 0)
{
printf ("error you entered: 0");
output (0);
}

// store the number of x values ​​in no_x
printf ("Number of values ​​x:");
scanf ("% d", & no_x);

// the for loop where I call the normal function N times
for (int i = 1; i <= no_x; i ++)
{

// i print for the counter in the invited x values
printf ("x value% d:", i);

// scan in x
scanf ("% lf", & x);


x = normal (x, sigma, mu);

printf ("f (x) =:% lf.12", x);

printf (" n");
}

returns 0;
}

C:>. A.exe
Enter average: 3.489
Enter std dev s: 1.203
Number of values ​​x: 3
x value 1: 3.4
f (X) = 0.330716549275
x value 2: -3.4
f (X) = 0.000000025104
x value 3: 4
f (X) = 0.303015189801

But that's what I get

C: Csource> a.exe
Enter average: 3.489
Enter the standard deviation: 1.203
Number of values ​​x: 3
x value 1: 3.4
f (x) =: 15086080.000000
x value 2: -3.4
f (x) =: 15086080.000000
x value 3: 4
f (x) =: 1610612736.000000

calculation – Calculate the circulation of the field $ V (x, y) = frac {3x ^ 3 + xy ^ 2} { sqrt {x ^ 2 + y ^ 2}} i + frac {3y ^ 3 + x ^ 2y} { sqrt {x ^ 2 + y ^ 2}} j $

Calculate the circulation of the field $ V (x, y) = frac {3x ^ 3 + xy ^ 2} { sqrt {x ^ 2 + y ^ 2}} i + frac {3y ^ 3 + x ^ 2y} { sqrt {x ^ 2 + y ^ 2}} j $ on the curve $ r = a (1 + cost), t in [0,frac {pi}2]$.

I know that the formula for the circulation of a field is as follows:

$$ int_ Gamma V dr = int_ Gamma frac {3x ^ 3 + xy ^ 2} { sqrt {x ^ 2 + y ^ 2}} dx- frac {3y ^ 3 + x ^ 2y} { sqrt {x ^ 2 + y ^ 2}} dy $$

But what do they mean when they talk about traffic on this specific curve? How to calculate it and how to set it?

nt.number theory – New formula for the quadratic field class number $ mathbb Q ( sqrt {(- 1) ^ {(p-1) / 2} p}) $?

I have the following conjecture involving a possible new formula for the class number of the quadratic field $ mathbb Q ( sqrt {(1) ^ {(p-1) / 2} p}) $ with $ p $ a strange bonus.

Conjecture. Let $ p $ to be an odd number and leave $ p ^ * = (- 1) ^ {(p-1) / 2} p $. Then the class number $ h (p ^ *) $ of the quadratic field $ mathbb Q ( sqrt {p ^ *}) $ coincides with the number
$$ frac {( frac {-2} p)} {2 ^ {(p-3) / 2} p ^ {(p-5) / 4}}} det left[cotpifrac{jk}pright]_ {1 le, k le (p-1) / 2}, $$
or $ ( frac { cdot} p) $ is the symbol of Legendre.

This is conjecture 5.1 in my preprint arXiv: 1901.04837. I checked it for all the odd firsts $ p <$ 29. Note that $ h (p ^ *) = $ 1 for each odd premium $ p <23 $, and $ h (-23) = $ 3.

Here, I invite some of you to further test this hypothesis. My computer can not check it even for $ p = $ 29.

algebra precalculus – Simplify $ sqrt { frac { sqrt[3]{64} + sqrt[4]{256}} { sqrt {64} + sqrt {256}}} $ in $ frac { sqrt {3}} {3} $

I am on the last question of a chapter of a manual on the radicals and this question seems more difficult, it is perhaps the idea. If you look at the history of my publications, I usually try to simplify the expression to a certain extent, but I am very confused as to the direction to take or the first steps.

I am to simplify:

$$ sqrt { frac { sqrt[3]{64} + sqrt[4]{256}} { sqrt {64} + sqrt {256}}} $$

The solution is $ frac { sqrt {3}} {3} $

The expression gives me the "impression" that there is a rule when dividing radicals having the same radical, but a different index "with a different index". Is it true? In this case, how to divide $ frac { sqrt[3]{64}} { sqrt {64}} $? I know that the 3rd root and sq roots are 4 and 8, which would leave me 1/2. Using a calculator, I can see that the 4th root of 256 is 4, but I think I need to get to the solution without a calculator.

Is there a prescribed approach or order of operation to simplify an expression of this type?

How can I get to $ frac { sqrt {3}} {3} $

How to solve $ int_0 ^ 1 frac { ln frac {x ^ 2 + sqrt 3 x + 1} {x ^ 2 – sqrt 3 x + 1}} x dx $?

I do not know how to solve this integral. Can someone give me a step-by-step solution to explain the exact result, please?

written test – $ sqrt {98} + 4 ^ { frac {1} {3}} $ is irrational

Some of your past responses have not been well received and you may be stuck.

Please pay close attention to the following tips:

  • Please make sure to respond to the question. Provide details and share your research!

But to avoid

  • Ask for help, clarification, or answer other answers.
  • Make statements based on opinions; save them with references or personal experience.

To learn more, read our tips for writing good answers.

functional analysis – $ B _ { ell ^ {2}} ^ {+} $ with the standard $ Green green cdot Green green _ { sqrt {2}} $ n has no structure par

Consider the space $ ell ^ {2} $ with the standard
begin {align *}
Green x Green _ {2} = left ( sum _ {i = 1} ^ { infty} x _ {i} ^ {2} right) ^ {1/2}
end {align *}

and we define the equivalent standard
begin {align *}
Green green x Green green _ { sqrt {2}} = max { Green x Green _ {2}, sqrt {2} Green x Green _ { infty} } mbox {.}
end {align *}

Define the positive part of the unit ball
begin {align *}
B _ { ell ^ {2}} ^ {+} = lbrace x in ell ^ {2}: ; Green x Green _ {2} leqslant 1, ; x _ {i} geqslant 0 rbrace mbox {.}
end {align *}

I want to show that $ B _ { ell ^ {2}} ^ {+} $ with the standard $ Green green cdot Green green _ { sqrt {2}} $ do not have normal structure, and to show that, I should show that diam$ (B _ { ell ^ {2}} ^ {+}) $ = $ r _ {x} (B _ { ell ^ {2}} ^ {+} $. I've shown this diam$ (B _ { ell ^ {2}} ^ {+}) = $ 1But I do not know why $ r _ {x} (B _ { ell ^ {2}} ^ {+}) = $ 1. What element in $ B _ { ell ^ {2}} ^ {+} $ can take to prove that $ r _ {x} (B _ { ell ^ {2}} ^ {+}) = $ 1?