pallav

70 Reputation

5 Badges

11 years, 219 days

MaplePrimes Activity


These are questions asked by pallav

Can I solve the given biquadratic equation in terms of sigma. I only need real positive root, if any

6125*_Z^4 + 68644*_Z^3*sigma - 219625*_Z^3 + 255712*_Z^2*sigma - 959250*_Z^2 + 238144*_Z*sigma - 1113500*_Z - 245000

I tried with solve for variable but it does not work. 

I want to make the system of ODE into its dimensionless version:

Dimensional version: 

dN/dT= R*N (1 −N/K)−alpha*N*P/(A + N);

dP/dT= gamma*N*P/( A + N) + C*P/(1 + Q*P) −MP;

N (0) ≡N_0 ≥0 and P (0) ≡P_0 ≥0

R, K alpha, gamma, M, C, Q are all positive constant. 

Using one choice of dimensionless variable x = N/K , y = alpha*P/(R*K), t = R*T, the system of ODE can be reduced to its dimensionless version as follows:

dx/dt = x*(1 −x ) −x*y/(a + x);

dy/dt = b*x*y/(a + x) + c*y/(1 + q*y) −m*y

where the dimensionless parameters are a = A/K , b = gamma/R , c = C/R , q = Q*R*K/alpha, and m = M/R.

How to do this in maple. Please help. 

Suppose I have to solve _Z^2 + (epsilon - 1)*_Z - epsilon + theta=0. If I use the code

solve(_Z^2 + (epsilon - 1)*_Z - epsilon + theta, _Z)

I get the solutions  as -epsilon/2 + 1/2 + sqrt(epsilon^2 + 2*epsilon - 4*theta + 1)/2, -epsilon/2 + 1/2 - sqrt(epsilon^2 + 2*epsilon - 4*theta + 1)/2

But I need them as (1 - epsilon)/2 + sqrt((epsilon + 1)^2 - 4*theta)/2, (1 - epsilon)/2 - sqrt((epsilon + 1)^2 - 4*theta)/2

How to modify the code so that I get the roots in the more simplified form, as I have mentioned above. 

Please help. 

I am unable to simplify a large expression using one given condition.

For example, suppose, I want to simplify the expression 

A=(20*a40*b10^2*b20 - 6*a50*b10^3)*a01^3 + ((-10*a11*a40 + 36*a21*a30)*b10^3 + (7*a11*a30*b20 - 30*a20^2*b21 - 9*a20*a30*b11)*b10^2 + (-12*a11*a20*b20^2 + 39*a20^2*b11*b20)*b10 + 15*b20^2*a11*a20*b11)*a01^2 + (-2*a11^2*a30*b10^3 + (27*a11^2*a20*b20 + 9*a11*a20^2*b11)*b10^2)*a01 + 18*a11^3*a20*b10^3

where it is given that a01*(a20*b20-a30*b10)+a11*a20*b11=0

Is there any way out to simplify A by reducing the number of terms in A? 

I need a proper solution. 

Is there any method to handle the huge expression of A where it is given that sigma+x=a, sigma*x=b

It is huge. I tryed with

B := mtaylor(A, [sigma, x], 8)

Unable to handle. I know that the expression can be written in powers in sigma^ix^j and thereafter sigma+x=a, sigma*x=b could be substituted..

Any help.

A := -sigma*(sigma^2+2*sigma*(alpha-1)+alpha*beta+1-alpha)*(alpha+sigma-1)^2*(sigma*theta+delta)*(gamma1*(alpha+sigma-1)+(1-sigma)*(beta+sigma))*alpha*(1-x)^2*(beta+x)^3*((theta*x+delta-gamma1)*(alpha+x-1)-(1-x)*(beta+x))+alpha*(1-sigma)^2*(beta+sigma)^3*((sigma*theta+delta-gamma1)*(alpha+sigma-1)-(1-sigma)*(beta+sigma))*x*(x^2+2*x*(alpha-1)+alpha*beta+1-alpha)*(alpha+x-1)^2*(theta*x+delta)*(gamma1*(alpha+x-1)+(1-x)*(beta+x))

-sigma*(sigma^2+2*sigma*(alpha-1)+alpha*beta+1-alpha)*(alpha+sigma-1)^2*(sigma*theta+delta)*(gamma1*(alpha+sigma-1)+(1-sigma)*(beta+sigma))*alpha*(1-x)^2*(beta+x)^3*((theta*x+delta-gamma1)*(alpha+x-1)-(1-x)*(beta+x))+alpha*(1-sigma)^2*(beta+sigma)^3*((sigma*theta+delta-gamma1)*(alpha+sigma-1)-(1-sigma)*(beta+sigma))*x*(x^2+2*x*(alpha-1)+alpha*beta+1-alpha)*(alpha+x-1)^2*(theta*x+delta)*(gamma1*(alpha+x-1)+(1-x)*(beta+x))

(1)

NULL``

``

Download 1.mw1.mw

1 2 3 4 Page 1 of 4