Using P-Q decomposition method and Newton Raphson method for power flow calculation, which is the faster convergence speed

Using P-Q decomposition method and Newton Raphson method for power flow calculation, which is the faster convergence speed

It depends on the convergence speed you said. If you refer to the number of iterations, the niula method has the upper hand. But for the large matrix calculation formula, it is possible that the calculation time of niula method for five iterations is longer than that of PQ decomposition method for ten iterations. PS: PQ decomposition method reduces a lot of time in the calculation process due to the constancy of Jacobian matrix

Newton's method for solving positive roots of equations x*ln((x^2-1)^0.5+x)-(x^2-1)^0.5-0.5x=0 Iterative solution, VC program code Wait for your expert help! thank you! x*ln((x^2-1)^0.5+x)-(x^2-1)^0.5-0.5*x=0

I use C
The result is 2.1155229
/*
Solution of equations by Newton iterative method
X0 is the initial value of iteration, n is the number of iterations, and Jingdu is the accuracy
Function is the algebraic expression for finding roots, and d2functoin is its derivative
Returns the root that finally meets a certain precision
*/
double newton_ diedai(double x0,int *n,double jingdu)
{
double x,temp;
temp=d2function(x0);
If (Fabs (Temp) > 1e-10) / * prevent divisor from being 0*/
{
x=x0-function(x0)/temp;
printf("n=%d\tx=%.5lf\n",*n,x);
}
else
{
printf("error:div 0:\nPress any key to exit:");
getch();
exit(1);
}
if (++(*n)>MAX_DIEDAI_TIME)
{
printf("diedai time:%d > MAX_DIEDAI_TIME:\nPress any key to exit:",*n);
getch();
exit(1);
}
temp=function(x);
if (fabs(temp)

What is "Newton method" or "Newton iterative method"? Please briefly describe the process and principle, and examples are better

Newton method is a method proposed by Newton in the 17th century to solve the equation f (x) = 0. There is no root formula for most equations, so it is very difficult or even impossible to find the exact root, so it is particularly important to find the approximate root of the equation
Let R be the root of F (x) = 0, choose x0 as the initial approximate value of R, make the tangent l of curve y = f (x) through point (x0, f (x0)), the equation of L is y = f (x0) + F '(x0) (x-x0), calculate the abscissa X1 = x0-f (x0) / F' (x0) of the intersection of L and X axis, call X1 the primary approximate value of R, make the tangent of curve y = f (x) through point (x1, f (x1)), and find the abscissa x2 = x1-f (x1) / F '(x1) of this tangent and X axis, call x2 the secondary approximate value of R, Repeat the above process to obtain the approximate value sequence {xn} of R, where xn + 1 = xn-f (xn) / F '(xn), which is called the N + 1 approximation of R. the above formula is called Newton iterative formula

What does "g-view" mean in physics? Speed, online, etc

We can use the spring scale to weigh the weight of the object. Its principle is to use the two force balance and the relationship between the force and the reaction force. We call the data displayed by the scale as the apparent weight (that is, the pressure on the platform scale or the tension on the spring scale). When the object is in the balance off state, according to the above principle, it can be concluded that the apparent weight is equal to the object

Given the function f (x) = x-lnx, G (x) = LNX / X (1), find the monotone interval of function f (x) 2) Verification: for any m, n belongs to (0, e}, f (m) - G (n) > 1 / 2 (Note: e is about 2.71828... Is the base of natural logarithm)

1）
f(x)=x-lnx (x>0)
f'(x)=1-1/x=(x-1)/x
∴00
The increment interval of F (x) is (1, + ∞),
Decreasing interval is (0,1)
2)
From 1), when x ∈ (0, e], f (x) min = f (1) = 1
g(x)=lnx/x
g'(x)=(1-lnx)/x ²
0

The function is continuous on the closed interval [a, b], differentiable in the open interval (a, b), f (a) = f (b) = 0. It is proved that at least one point x is in (a, b), Make f (x) + X * f '(x) = 0

Let f (T) = TF (T)
Then f '(T) = f (T) + TF' (T)
Because f (a) = f (b) = 0,
So f (a) = AF (a) = 0
F(b)=bf(b)=0
Therefore, according to Rolle's theorem, at least one point x is in (a, b), so that f '(x) = 0, that is, f (x) + X * f' (x) = 0