AHSAN

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5 years, 240 days

MaplePrimes Activity


These are replies submitted by AHSAN

@rlopez sir can you ls upload your maple file with the newton method that you did.

@acer basically this problem is related to my research. i want to mention the name of method that we used to compute the valve of lambda. so can you pls mention the name of the method that you used to obtain the valve of lambda

@acer

can you pls explain it little. and why you are taking digits 16? what does it means and is it possible to solve with regula falsi position method 

@tomleslie 

what is the value of C1? for more detail I uploaded the picture of my problem in the above mention comments, please have look at it.

@nm 

ok, I am waiting.  pls, help me to solve my problem. I try this problem for a couple of days.

@Preben Alsholm thank you very much for your support and time.

@nm  this the solution of the problem. the difference is that I want to change boundary condition that is mention mention in my maple file

i am attaching two picture pls have a look, basically, i want to solve this equation mention in equation 22 to 24@nm 

@Preben Alsholm i could understand this, can you please explain little more

sorry second boundary condition is u(y)=j@nm 

@Rouben Rostamian   sorry its u(y)=j

basically I want to find the value of u by using the value of q from first@AHSAN 

ode is ok. the question is defined in the same way we need to calculate any valve q or u first and then use in second but how do I? do not know.@nm 

@acer 

basically I want to calculate the value of lambda from equation (2) as mention in the attached maple file. for this purpose first I strictly bounded to replace x with the help of equation (1). as equation (1) is an implicit function of x and cannot easily find the value of x so that why first I want to generate data points from equation (1) for x=0,0.1,0.2....0.9 and then write a polynomial in term of lambda and then replace that polynomial in equation (2) to find the valve of lambda from equation (2) by using condition given at the end of the maple file. now I may hope you will understand my question and trying to solve my problem. pls have look on maple file
 

restart

lambda := (-1.565845910+2.393779704*x^2+1.564996800*x^4+1.800900000*x^6)/(x^2+2)^4

(-1.565845910+2.393779704*x^2+1.564996800*x^4+1.800900000*x^6)/(x^2+2)^4

(1)

p := (-60*lambda*(1+(1/2)*x^2)^4+54.027*(1+(1/2)*x^2)^3-138.6060480*(1+(1/2)*x^2)^2+78.70712580+66.54222187*x^2)/(20*(1+(1/2)*x^2)^7)

(1/20)*(-60*(-1.565845910+2.393779704*x^2+1.564996800*x^4+1.800900000*x^6)*(1+(1/2)*x^2)^4/(x^2+2)^4+54.027*(1+(1/2)*x^2)^3-138.6060480*(1+(1/2)*x^2)^2+78.70712580+66.54222187*x^2)/(1+(1/2)*x^2)^7

(2)

NULL

NULL

NULL

"#lambda="?""

``


 

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@Kitonum 

can you please have a  look on my maple and guide me on how do I fixed tolerance up to 10^-6 and how do I plot between x and lambda.


 

restart

lambda := proc (x) options operator, arrow; (-1.565845910+2.393779704*x^2+1.564996800*x^4+1.800900000*x^6)/(x^2+2)^4 end proc

proc (x) options operator, arrow; (-1.565845910+2.393779704*x^2+1.564996800*x^4+1.800900000*x^6)/(x^2+2)^4 end proc

(1)

data := [seq([lambda(x), x], x = 0 .. 1, .1)]

[[-0.9786536938e-1, 0], [-0.9445602702e-1, .1], [-0.8473253124e-1, .2], [-0.7004169612e-1, .3], [-0.5215959052e-1, .4], [-0.3283205351e-1, .5], [-0.1345044703e-1, .6], [0.5065808511e-2, .7], [0.2221891338e-1, .8], [0.3780341321e-1, .9], [0.5177568635e-1, 1.0]]

(2)

with(CurveFitting)

x := PolynomialInterpolation(data, lambda)

-0.6298940090e12*lambda^10-0.1443096227e12*lambda^9-7314622270.*lambda^8+489000321.4*lambda^7+45718357.19*lambda^6-73452.0126*lambda^5-68414.72355*lambda^4-508.641296*lambda^3+39.75510893*lambda^2+5.699823870*lambda+.6702161742

(3)

``


 

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