AHSAN

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1 years, 184 days

MaplePrimes Activity


These are replies submitted by AHSAN

@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="?""

``


 

Download help.mw

@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)

``


 

Download help2.mw

@Carl Love 

ok pls solve my issue

help2.mw

can you please have a look on my this uploaded new maple file. is this write way and pls mention a way how to plot this data and one thing more if we want to adjust tolerance up to 10^-6 then how could we fix this?

@Kitonum 

@Kitonum 

basically I need relation for x in terms of lamda to use in another relation, which is not possible to find from in my case. so I want to generate data points first and then write a polynomial in term of lamda and then use that polynomial to interpolate my original problem.

I appreciate your help but you write in terms of x, but I need a polynomial in terms of lamda, not x means x is a function of lambda. i hope you will understand my issue and solve my problem

@acer 

basically i need relation for x in terms of lamda to use in another relation, which is not possible to find from in my case. so I want to generate data points first and then write a polynomial in term of lamda and then use that polynomial to interpolate my original problem.

@Carl Love 

can you pls upload its maple file because this is not working...

@Carl Love

I am interested in any numerical technique to develop interest e.g false position method, newton raphan method, etc rather than in built-in command like fsolve. 

@Carl Love 

k is a constant and fixed it valve  k=0.1 then solve with respect to lamda as a variable.

@tomleslie 

sir can you pls have a look on it .. I am very beginner to maple so that why face a lot of problems
 

restart

newton := proc (g, X__A, X__B, err) local j, p, X__1, X__2, iters; X__1 := X__A; X__2 := X__B; iters := 0; for j to 50 do iters := iters+1; p := evalf(X__1-(eval(g(X__1)/(diff(g(x), x)), x = X__1))); if abs(g(p)) < err then break else X__1 := p end if end do; return root = p, residual = g(p), numberOfIterations = iters end proc

F := proc (Q) options operator, arrow; 0.2185844281e-2*(428.8960600*sqrt(2)*(0.127600000e-1-(6/5)*Q)*arctan((1/2)*sqrt(2.42704/(2*Q+.99)-2)*sqrt(2))/(2*Q+.99)^3+673.7083558*(0.127600000e-1-(6/5)*Q)*sqrt(2)/(2*Q+.99)^3+60*sqrt(2.42704/(2*Q+.99)-2)*((0.127600000e-1-(6/5)*Q)*(2.42704/(2*Q+.99)-2)^2+(0.68053333e-1-(32/5)*Q)*(2.42704/(2*Q+.99)-2)+1.379488000-8*Q))*(2*Q+.99)^3 end proc

proc (Q) options operator, arrow; 0.2185844281e-2*(428.8960600*sqrt(2)*(0.127600000e-1-(6/5)*Q)*arctan((1/2)*sqrt(2.42704/(2*Q+.99)-2)*sqrt(2))/(2*Q+.99)^3+673.7083558*(0.127600000e-1-(6/5)*Q)*sqrt(2)/(2*Q+.99)^3+60*sqrt(2.42704/(2*Q+.99)-2)*((0.127600000e-1-(6/5)*Q)*(2.42704/(2*Q+.99)-2)^2+(0.68053333e-1-(32/5)*Q)*(2.42704/(2*Q+.99)-2)+1.379488000-8*Q))*(2*Q+.99)^3 end proc

(1)

``

``


 

Download try1.mw

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