## 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 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:

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

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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 *}
begin {align *} Green green x Green green _ { sqrt {2}} = max { Green x Green _ {2}, sqrt {2} Green x Green _ { infty} } mbox {.} end {align *}
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$$(B _ { ell ^ {2}} ^ {+}) = 1$$But I do not know why $$r _ {x} (B _ { ell ^ {2}} ^ {+}) = 1$$. What element in $$B _ { ell ^ {2}} ^ {+}$$ can take to prove that $$r _ {x} (B _ { ell ^ {2}} ^ {+}) = 1$$?