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How to substitute in argument of the exp...

sunit 105 Maple 2018
substitution exp subsindets
August 15 2023
1 3

Hi,

It might be really trivial, but I am struggling in the algebraic manipulation of the argument of the exponential function. As an example, I want to substitute 

I*T[0]*(omega1-2*omega2) = I*omega2*T[0]-I*si*T[2]

in the expression of

exp(-I*T[0]*(omega1-2*omega2)).

However, I am only able to do so by subs command and also by exactly copying the argument in the following manner.

subs(-I*T[0]*(omega1-2*omega2) = I*omega2*T[0]-I*si*T[2], exp(-I*T[0]*(omega1-2*omega2)))

The issue is I have expressions like this all over in the main problem, and I have to copy-paste such expressions for the substitution. So I am wondering if there is a more efficient way to tackle this problem. 

Thanks in Advance,

Regards

Combining algebraic expression...

sunit 105 Maple 16
simplification
October 06 2020
2 1

Hi,

It might be a really silly question, but I am wondering is it possible to simplify expression like this

a^2+b^2+2*a*b+c^2. Just by looking it we know that we can write it in the form of (a+b)^2+c^2. This is the basic exmaple I come up with. I have very lengthy expressions in maple, which can be factorize like this, but factor command will not work as it will try to factorize entire expression. So I am wondering is it doable in maple or I have to do it manually by collecting terms and check whether they can be factorize or not.

Thanks in Advance.

With Regards

sunit

How to get rid of 1.0 in numerical and s...

sunit 105 Maple 16
coefficients floating-point
August 07 2020
2 4

Hi,

This might be a very silly question but it is troubling me a little bit and that's why I need to post it. During the combination of symbolic and numerical comutation this '1.0' is appearing as a coefficient for the variables whose coeffecient is just '1'. It's quite annoying as sometimes if I have to collect coefficient of a variable for an example exp(I*omega*t) then I have to write exp(1.I*omega*t), so chances of making mistake is higher. Please find the attached worksheet for this. In eq(4), you can easily see that for x[3](t) and x[1](t) this '1.0' appears in front of the variables.

I really appreciate  if someone can help me out of this.

With Regards

Sunitquestion.mw
 

> 

restart
> 

with(LinearAlgebra):
> 

par := {a = 2.5, alpha = 2, k_r = .5, k_rc = .2, k_rq = .2, kappa = 0.1e-2, mu_k = .35, mu_s = .44, omega = .766620580157922, sigma_0 = 110, sigma_1 = 1.37, sigma_2 = 0.823e-1, x_s3 = -1.04003626422324936017819852700633621040584050594846801927800, zeta = 0.904504977553123318334601762827181333680096702957228781770315e-1}:
> 

f := proc (v_r) options operator, arrow; mu_k+(mu_s-mu_k)*exp(-a*v_r) end proc:
> 

NULL
> 

g_exp1 := taylor(1/f(v_rv+x21), x21 = 0, 4):
> 

for k to 4 do g_coeff[k] := taylor(subs(v_r = v_rv, coeff(g_exp1, x21, k-1)), epsilon = 0, 3) end do:
> 

g0 := eval(subs(par, coeff(g_coeff[1], epsilon, 0))):
> 

g1 := eval(subs(par, coeff(g_coeff[2], epsilon, 0))):
> 

g2 := eval(subs(par, coeff(g_coeff[3], epsilon, 0))):
> 

g3 := eval(subs(par, coeff(g_coeff[4], epsilon, 0))):
> 

NULL
> 

``
> 

eq[1] := subs(par, diff(x[1](t), t)-x[2](t)):
> 

eq[2] := subs(par, diff(x[2](t), t)+2*zeta*x[2](t)+x[1](t)+k_r*(x[1](t)-x[3](t))+2*kappa*(x[2](t)-x[4](t))+k_rq*(x[1](t)-x[3](t))^2+k_rc*(x[1](t)-x[3](t))^3)

diff(x[2](t), t)+.1829009955*x[2](t)+1.5*x[1](t)-.5*x[3](t)-0.2e-2*x[4](t)+.2*(x[1](t)-x[3](t))^2+.2*(x[1](t)-x[3](t))^3

 (1)
> 

eq[3] := subs(par, diff(x[3](t), t)-x[4](t))

diff(x[3](t), t)-x[4](t)

 (2)
> 

eq[4] := subs(par, diff(x[4](t), t)+2*kappa*alpha*(x[4](t)-x[2](t))+k_r*alpha*(x[3](t)-x[1](t))-k_rq*alpha*(x[3](t)-x[1](t))^2+k_rc*alpha*(x[3](t)-x[1](t))^3+alpha*(sigma_0*x[5](t)+sigma_1*v_r*(1-sigma_0*x[5](t)*(g_0+g_1*x[4](t)+g_2*x[4](t)^2+g_3*x[4](t)^3))+sigma_2*v_r))

diff(x[4](t), t)+0.4e-2*x[4](t)-0.4e-2*x[2](t)+1.0*x[3](t)-1.0*x[1](t)-.4*(x[3](t)-x[1](t))^2+.4*(x[3](t)-x[1](t))^3+220*x[5](t)+2.74*v_r*(1-110*x[5](t)*(g_0+g_1*x[4](t)+g_2*x[4](t)^2+g_3*x[4](t)^3))+.1646*v_r

 (3)
> 

eq[5] := subs(par, diff(x[5](t), t)-v_r*(1-sigma_0*x[5](t)*(g_0+g_1*x[4](t)+g_2*x[4](t)^2+g_3*x[4](t)^3)))

diff(x[5](t), t)-v_r*(1-110*x[5](t)*(g_0+g_1*x[4](t)+g_2*x[4](t)^2+g_3*x[4](t)^3))

 (4)
> 

``


 

Download question.mw

 

Incorrect definite integration of hyperb...

sunit 105 Maple 16
integration numeric hyperbolic
May 06 2020
2 2

Hi Guys,

I am trying to intgrate a function involving hyperbolic functions in a range of 0 to 1 and it is giving me very large value of 10^94. However, on doing integration terms terms i can see that some large terms involving 10^191 cancel out with each other and I can have a fnite value of this integration. It would be really helpful if someone can help me out why it is happening with int function and how can I solve this case involving hyperbolic function. For reference maple file is attached. Thanks in advance and much appreciated.

With Regards

Sunit

question.mw

How to simplify multiple definite integr...

sunit 105 Maple 16
integral simplification combine
May 02 2020
1 1

Hi Guys,

I need help regarding simplifying definite integrals. I have following expression

2*s^2*(int(Phi[1, 1](x, T[2])*(diff(Y1(x), x, x)), x = 0 .. x1))+int(Phi[1, 1](x, T[2])*(diff(Y1(x), x, x, x, x)), x = 0 .. x1)-omega_y^2*(int(Y1(x)*Phi[1, 1](x, T[2]), x = 0 .. x1))

As we can notice the limit for each integral is same, so I should be able to take these integral limit outside. I am trying to use collect and simplify, but both of these command are not working. It would be really helpful for me if I can get the solution for it.

Please find the attached example sheet also.example.mw

With Regards

Sunit

 

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