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AHSAN

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@acer basically I want to calculate...

AHSAN 165
May 22 2020
0 0

@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&nb...

AHSAN 165
May 21 2020
0 0

@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...

AHSAN 165
May 20 2020
0 0

@Carl Love 

ok pls solve my issue

ODE solution...

AHSAN 165
May 20 2020
0 0

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 

ODE solution...

AHSAN 165
May 20 2020
0 0

@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

ODE solution...

AHSAN 165
May 20 2020
0 0

@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.

ODE solution...

AHSAN 165
May 03 2020
0 0

@Carl Love 

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

ODE solution...

AHSAN 165
May 01 2020
0 0

@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. 

ODE solution...

AHSAN 165
April 30 2020
0 0

@Carl Love 

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

ODE solution...

AHSAN 165
April 28 2020
0 0

@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

ODE solution...

AHSAN 165
April 28 2020
0 0

@tomleslie 

sir thanks allot for your a lot help and cooperation

ODE solution...

AHSAN 165
April 27 2020
0 0

@tomleslie 

it's really good for me and thanks a lot sir . can you pls told me the name of this approach means regular false method, newton method, etc

ODE solution...

AHSAN 165
April 27 2020
0 0

many regard sir for your help and cooperation.and the problem you fixed with the student(precalculus)is it possible to solve same problem by using a false position method? @tomleslie 

try.mw   can you please have l...

AHSAN 165
April 26 2020
0 0

try.mw   can you please have look on this file, which I have loaded. I am trying to solve it but could not get the result. and is it possible,you'll solve this problem with(Student[NumericalAnalysis]) false position method @tomleslie 

ODE solution...

AHSAN 165
April 26 2020
0 0

i am trying to solve this equation but could not get the result. need help

> 

restart
> 

with(Student[Precalculus])
> 

k := .1

.1

 (1)
> 

F := (-(288*(lambda+2*k*(1/3)-2/3))*sqrt(2)*lambda^2*arctan((1/2)*sqrt(4*lambda/(1-k)-2)*sqrt(2))/(1-k)^2-(144*(lambda+2*k*(1/3)-2/3))*Pi*lambda^2*sqrt(2)/(1-k)^2-(36*((lambda+2*k*(1/3)-2/3)*(4*lambda/(1-k)-2)+10*lambda*(1/3)+4*k*(1/3)-4/3))*sqrt(4*lambda/(1-k)-2))*(1-k)^2/(256*lambda^2)

0.3164062500e-2*(-355.5555555*(lambda-.6000000000)*2^(1/2)*lambda^2*arctan((1/2)*(4.444444444*lambda-2)^(1/2)*2^(1/2))-558.5053607*(lambda-.6000000000)*lambda^2*2^(1/2)-36*((lambda-.6000000000)*(4.444444444*lambda-2)+(10/3)*lambda-1.200000000)*(4.444444444*lambda-2)^(1/2))/lambda^2

 (2)
> 

`&lambda;__1` := 0

0

 (3)
> 

`&lambda;__2` := 1

1

 (4)
> 

tol := 0.10e-3

0.10e-3

 (5)
> 

for j do p := evalF([Line([`&lambda;__1`, F(`&lambda;__1`)], [`&lambda;__2`, F(`&lambda;__2`)])][5]); if abs(F(p)) < tol then break elif F(p) > 0 then `&lambda;__2` := p else `&lambda;__1` := p end if end do; root = p, residual = F(p)

Error, invalid subscript selector

  

root = p, residual = 0.3164062500e-2*(-355.5555555*(lambda(p)-.6000000000)*2^(1/2)*lambda(p)^2*(arctan((1/2)*(4.444444444*lambda-2)^(1/2)*2^(1/2)))(p)-558.5053607*(lambda(p)-.6000000000)*lambda(p)^2*2^(1/2)-36*((lambda(p)-.6000000000)*(4.444444444*lambda(p)-2)+(10/3)*lambda(p)-1.200000000)*(4.444444444*lambda(p)-2)^(1/2))/lambda(p)^2

 (6)
> 

``
> 

``


 

Download try.mw

 

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