Calculate the Derivative of a polynomial in C. (with dynamic allocation)

I feel difficult to make a question of a list of the university. It is requested that a function that calculates the derivative of a polynomial be implemented. and in the case that it is an error, since it is incomplete. (The libraries are already included).

void drift (double * poly, int degree, double * out) {

int i;
(i = grade, i> = grade, i = i-1) {
cop[i] = (poly[i]* Grade);
}
/ **
(i = 0, i <degree, i ++) {
October[i] = poly[i];
} * /

}

int main () {} ()

setlocale (LC_ALL, "Portuguese");
int i, coefficient, degree;

printf ("DIGEST THE DEGREE OF POLYNOMIUM:"); // How many coefficients does (ax ^ 4 + bx ^ 3 + cx ^ + ... + nx ^ n).
scanf ("% d", & degree);

double = poly = (double *) malloc (grade * sizeof (double)); // Assigning the "poly" vector to store the coefficients.
if (poli == NULL) {
printf ("Assignment failed!");
of exit (1);
}
printf ("DIGIT COEFFICIENTS OF POLYNOMIUM:");
(i = 0; i) < grau; i++){
            scanf("%d", &poli[i]);
        }
double* out = (double*) malloc(grau* sizeof(double));
    if(out == NULL){
        printf("Erro de Alocação2!");
        exit(1);
    }
deriva(poli, grau, out);

for(i = grau; i >= 0; i = 1).
printf ("% d", poly);
}
free (poly);
free (out);

return 0;
}

Calculation and analysis – Using rule with derivative.

You can give Derivative a UpValues for a:

a /: Derivative[n_, 1][a]    : = Derived[n, 0][b]

So:

re[a[t, r], r]+ D[RE[D[RE[D[a[t, r], r], r]+ D[RE[D[RE[D[a[t, r], t], r]// TeXForm

$ b ^ {(0,1)} (t, r) + b ^ ((1,0)} (t, r) + b (t, r) $

Another possibility is:

Clear[a]
re[a[t, r], r]+ D[RE[D[RE[D[a[t, r], r], r]+ D[RE[D[RE[D[a[t, r], t], r]/. a-> Derivative[0,-1][b]    // TeXForm

$ b ^ {(0,1)} (t, r) + b ^ ((1,0)} (t, r) + b (t, r) $

Differential equations – pde coupled with second order spatial derivative.

I am trying to solve a PDE system using NDsolve but the following error accents.

NDSolveValue :: femcmsd: the spatial derived order of the PDE can not
exceed two.

Equations
one
two

Qm and Q are constants.

                pde1 = {
re[qr[r, x, t], t]+ tq * D[RE[D[RE[D[qr[r, x, t], t], t]==
alpha * D[(1/r)*(D[r*qr[r, x, t], r]+ D[r*qx[r, x, t], X]), r]+
alpha * ta *
re[RE[D[RE[D[(1/r)*(D[r*qr[r, x, t], r]+ D[r*qx[r, x, t], X]), r], t]+
alpha * wrc * D[Te[r, x, t], r]+ alpha * wrc * ta * D[RE[D[RE[D[Te[r, x, t], t], r]};

pde2 = {
re[qx[r, x, t], t]+ tq * D[RE[D[RE[D[qx[r, x, t], t], t]==
alpha * D[(1/r)*(D[r*qr[r, x, t], r]+ D[r*qx[r, x, t], X]), X]+
alpha * ta *
re[RE[D[RE[D[(1/r)*(D[r*qr[r, x, t], r]+ D[r*qx[r, x, t], X]), X], t]+
alpha * wrc * D[Te[r, x, t], X]+ alpha * wrc * ta * D[RE[D[RE[D[Te[r, x, t], t], X]};

pde3 = {
ro * c * D[Te[r, x, t],
t]== - (1 / r) * (D[r*qr[r, x, t], r]+ D[r*qx[r, x, t], X]) +
wrc * (Tb - Te[r, x, t]) + Qm
};

sol = NDSolveValue[{pde1, pde2, pde3, BC1, BC2, BC3, BC4, BC5, IC1,
IC2, IC3, IC4, IC5, IC6},
Tea[r, x, t], {t, 0, 5}, {r, x} [Element] [CapitalOmega]];

Is there any way to handle this error?

Functional analysis. How can we make the derivative for this equation w.r.t.to time t> 0

Leave $ x en[0,L]$ and consider the following equation,
$$ varepsilon left (t right) = frac {1} {2} int_ {0} ^ {L} {({{ rho} _ {1}} {{ left | {{vari} } _ {t}} right |} ^ {2}} + {{ rho} _ {2} {{ left | {{ varphi} _ {t}} right |} ^ {2}} + {{ rho} _ {1}} {{ left | {{ omega} _ {t}} right |} ^ {2}}} + b {{ left | {{ psi} _ { x}} right |} ^ {2}} + k {{ left | {{ varphi} _ {x}} + psi + lw right |} ^ {2}} + {{k} _ { 0}} {{ left | {{ omega} _ {x}} – l varphi right |} ^ {2}} + theta _ {1} ^ {2} + theta {2} ^ {2}) dx $$

where $ varphi $, $ psi $ Y $ omega $ they are functions whereas θ1 and θ2 are constants. Further, $ k_0 $,$ k $,$ b $ Y $ {{ rho} _ {1}}, {{ rho} _ {2}} $ with $ l = 1 / R $ are all positive constants where $ R $ is the radius of curvature

In this task we are interested in finding the derivative of $ varepsilon (t) $ w.r.t on time $ t> 0 $ .
First I think I use Leibtiz's rule to make this derivative that is presented in this Leibniz integral, this leads me to something wrong and it is not similar to the result I have in this document. Page 2

$$ frac {d varepsilon} {dt} = – {{ int_ {0} ^ {L} {({{ gamma} _ {1}} {{ left | {{left ({{ varphi} _ {x} + psi + l omega right)} _ {t}} right |} ^ {2}} + {{ gamma} _ {2}} {{ left | {{ psi} _ {xt}} right |} ^ {2}} + {{ gamma} _ {0}} {{ left | {{ omega} _ {xt}} – l {{ varphi} _ {t}} right |} ^ {2}} + left | {{ theta} _ {1}} _ {x} right |}} ^ {2}} + {{ left | {{ theta} 2x}} right |} ^ {2}}) dx $$
where everyone $ gamma $They are the viscosity coefficients. How did you derive it to obtain the previous result?

partial derivative: particle that moves along the unit circle centered on the origin of the xy plane

A particle moves along the unit circle centered on the origin of the xy plane. Find the address of $ nabla times mathbf v $.

My intent:

I found that $ displaystyle frac { partial v_x} { partial and}> 0 $ Y $ displaystyle frac { partial v_y} { partial x} <0 $. As $ v_z equiv 0 $, its partial derivatives with respect to $ x $ Y $ and $ They are also zero. But how do I find the signs of $ displaystyle frac { partial v_x} { partial z} $ Y $ displaystyle frac { partial v_y} { partial z} $?

Functions – Which version of DiracDelta became the derivative of HeavisideTheta of the UnitStep?

The mathematical version of my co-workers is 5.0, they use Unit step To derive the formula, however, I also want to use the parallel So used features of the latest version 11.3. Then I found Unit step In both versions it is different, the Unit stepThe derivation in 5.0 is DiracDelta, but not in 11.3. In version 11.3, DiracDelta becomes HeavisideThetaderivation of. I want to know, what is the first version that made the difference? I can replace Unit step to HeavisdeTeta?
Thank you!

In version 5.0

enter the description of the image here

In version 11.3

enter the description of the image here

statistics – Management of negative variations in the derivative of Gaussian processes

The variance of the derivative of a Gaussian process, $ f $, is given by (9.1):

$$ Var ( frac { partial f} { partial x}) = frac { partial ^ 2 k (x, x)} { partial x ^ 2}, $$

where $ k (·, ·) $ it is both a definite-positive quantity and the covariance function of $ f $. But when evaluating the error corresponding to $ frac { partial f} { partial x} $, we observe that it is not necessarily positive everywhere. Therefore, it is the previous definition for $ Var ( frac { partial f} { partial x}) $ really correct? Is it valid to simply take the absolute value of this quantity when calculating the error or should this variation be handled differently?

As a simple case, if we consider a modified square exponential core centered on $ x_a $ Y $ x_b $, so $ k (x_i, x_j) = exp (- (x_i-x_a) ^ 2 – (x_j-x_b) ^ 2) $. This is positive defined. But $ frac { partial k (x_i, x_j)} { partial x_i} = -2 (x_i – x_a) k (x_i, x_j) $ Y $ frac { partial k (x_i, x_j)} { partial x_i partial x_j} = 4 (x_i – x_a) (x_j – x_b) k (x_i, x_j) $, which can be negative. Therefore, is (9.1) itself a valid covariance function? To obtain the variance, is it appropriate to take the absolute value of the values ​​along the diagonal of the covariance matrix?

partial derivative – Ge the canonical form of the equation

Dice $$ x ^ 2u_ {xx} + 2xyu_ {xy} – 3y ^ 2u_ {yy} – 2xu_x + 4yu_y + 16x ^ 4u = 0 $$
I have to find the canonical form. The determinant is the following $$ D = 16x ^ 2y ^ 2 $$
I considered cases when $ x $ Y $ and $ can be $ 0 $. When calculating for the remaining case (where both $ x $ Y $ and $ It is not $ 0 $ and the type is hyperbolic), I'm not getting its canonical form. I'm receiving this
$$ (- 3y ^ 2 – 6x ^ 3y + 9x ^ 6) u _ { xi xi} + (-3y ^ 2 + frac {2y} {x} + frac {1} {x ^ 2} ) u _ {eta} + (-6y ^ 2 – 6x ^ 2 + frac {2y (1-3x ^ 4)} x) u _ { eta xi} + ( frac {-4} {x} + 4y) u_ eta + 4yu_ xi + 16x ^ 4u = 0 $$
But the point is that this does not meet the requirements for the canonical form of hyperbolic type.
Any help would be appreciated.

Find the closing and the derivative set of $ A $ in a certain topology.

Leave $ X =[0,1] subset mathbb {R} $, leave $ mathscr {B} $ Be a base for the topology. $ mathscr {T} 1 in $ X $ given by $$ mathscr {B} = { {0 }, {1 } } cup {(0,2 ^ {- k}) ,: , k in mathbb {Z} _ { ge0} }, $$and let $ mathscr {T} 2 $ be a topology in $ X $ given by $$ mathscr {T} _ {2} = left {G subset X ,: , frac {1} {2} not in G right } cup {G subset X ,: , (0,1) subset G }. $$Yes $ A = left { left ( frac {1} {4}, frac {1} {2} right) right } $ it is a subset of the product space $ X ^ { ast} = (X, mathscr {T} _ {1}) times (X, mathscr {T} 2) $find the closure $ overline {A} $, and the derived set $ A & # 39; $ in $ X ^ { ast} $.

Find $ A & # 39; $, I looked at $ A = left { frac {1} {4} right } times { frac {1} {2} } $ and using the formula for $ A & # 39; $ in the product space as $ A & # 39; = left ( left { frac {1} {4} right } times { frac {1} {2} } right) & # 39; = left (, overline { left { frac {1} {4} right }} times { frac {1} {2} } & # 39; right) cup Left ( left { frac {1} {4} right } & # 39; times overline { { frac {1} {2} }} , right) = left ( left[frac{1}{4}1right)times{frac{1}{2}}right)cupleft(left(frac{1}{4}1right)times{frac{1}{2}}right)=left[frac{1}{4}1right)times{frac{1}{2}}$[frac{1}{4}1right)times{frac{1}{2}}right)cupleft(left(frac{1}{4}1right)times{frac{1}{2}}right)=left[frac{1}{4}1right)times{frac{1}{2}}$[frac{1}{4}1right)times{frac{1}{2}}right)cupleft(left(frac{1}{4}1derecha)times{frac{1}{2}}derecha)=izquierda[frac{1}{4}1right)times{frac{1}{2}}$[frac{1}{4}1right)times{frac{1}{2}}right)cupleft(left(frac{1}{4}1right)times{frac{1}{2}}right)=left[frac{1}{4}1right)times{frac{1}{2}}$

But, the solution is $ A & # 39; = left ( frac {1} {4}, 1 right) times { frac {1} {2} } $.

Where am I wrong? Or can not you use that formula for a subset of a point in the product space?

The derivative in inverse matrix.

I wonder how to calculate the following derivative w.r.t to the array: $ frac {d (x ^ TW ^ {- T} W ^ {- 1} x)} {dW} $, where $ W $ is a $ mathbb {R} ^ {d times d} $ matrix and $ x $ is a $ mathbb {R} ^ d $ vector. The result must be a $ mathbb {R} ^ {d times d} $ matrix.