shani2775

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4 years, 12 days

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These are replies submitted by shani2775

@rlopez 

Thanx a lot

Its really kind of you.

 

@Kitonum 

> restart; u[0] := (4/3)*c^2*cos((1/4)*x)^2; alpha := 2;

> iteration := 3;
> for k from 0 by 1 while k <= Iteration do
u[s] := eval(u[k], t = xi); u[k+1] := simplify(u[k]-(int(diff(u[s], [`$`(xi, alpha)])+diff(u[s]*u[s], x)+diff(u[s]*u[s], x, x, x), xi = 0 .. t)))
end do;

but i didn't get the answer

@tomleslie 

sorry again to trouble you kindly see the attached file.

 

my_problem.mw

@tomleslie 

i am working on multiplicity of root of an equation f(x)=0
let m is the multiplicity of root of f(x) we use method y:=x-f(x)*f'(x)/((f'(x))^2-f(x)*f''(x)

  1. i want to apply Taylor series 
  2. then do some substitution like seq(((D@@i)(f))(alpha) = 0,i=1..m-1), f(alpha)=0 andx=e[n]+alpha

  3. then get the error term e[n+1]

11.mws

thmx a lot for your time i m very very thankful to u again sory for this trouble

 

@tomleslie 
basically I want to show the convergence of a numerical method so i come up to the above method  it will b oky whn we chose f(x) or D(f)(x) 
but problem arises when we chose a function like 

f(x)*(D(f))(x)/((diff(f(x), x))^2-f(x)*((D@@2)(f))(x))
 it devided by D(f)(alpha) which tend to give error 
"divided by zero "

diff(f(x), x)

@tomleslie 
sorry for trouble
i  want to minimize the terms and commands help me to eliminate the extra terms and commands

a1:= f(x) :

> T1 :=simplify((taylor(a1,x=alpha,N+3))):
> E1:=subs([seq(((D@@i)(f))(alpha) = 0,i=1..m-1),f(alpha)=0,x=e[n]+alpha],T1):
> g1 :=(convert(simplify(series((E1,e[n]=0,N))),polynom));

 

@acer 
 thanx for your sharing 
i solved it and  i get a value in frachtion form 
e[n]*(c[1]+c[2]*e[n]+c[3]*e[n]^2+c[4]*e[n]^3+c[5]*e[n]^4+c[6]*e[n]^5+c[7]*e[n]^6)/c[1]+2*c[2]*e[n]+3*c[3]*e[n]^2+4*c[4]*e[n]^3+5*c[5]*e[n]^4+6*c[6]*e[n]^5+7*c[7]*e[n]^6
but i could not simplify it 
plx help me in it thanx a lot

@Kitonum  thnx a lot 
no it is f(gamma) at the end we take common and use it to find the convergence of the derived formula 
again thanx a lot

 

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