Dear all;

Thanks in advance for helping me to plot the solution of this second order ode.

with(plots):
ode := diff(y(x), x, x) = x*y(x)+x^(17/12);
ics := y(1000) = 0, y(1001) = 1;
 dsolve({ics,ode}):

How can I plot the solution obtained in the range (1000, 1001).

Thanks

Dear all,

I have the function

y:=x->-9.8455282400*10^9142*exp(-(2/3)*x^(3/2))/(x^(1/4)*sqrt(Pi))+3.3889331940*10^(-9169)*exp((2/3)*x^(3/2))/(x^(1/4)*sqrt(Pi))+(16/153)*x^(7/6)*sqrt(Pi)*exp((2/3)*x^(3/2))+Pi*((1/2)*exp(-(2/3)*x^(3/2))*(-1+exp((2/3)*x^(2/3)))/(x^(1/4)*Pi)-(16/153)*x^(7/6)*exp((2/3)*x^(3/2))/sqrt(Pi));

how can I plot y versus x with x in the interval (1000, 1001).

First, it's simple to verify that y(1000)=0; y(1001)=1;  So (1000, 0) and (1001,1)  belong to our graph.

I tried plot( y, x=1000..1001); but there is no curves.

Thank you in advance to help me to plot the graph of this function.

with brest regards,

Dear all,

I need you help to finish some steps of this idea to approximate the roots of a given equation (polynom). Thanks in advance for your help. 

I have a sturm sequence, I would like to use Bisection method to approximation the roots using Sturm decomposition of my polynom. For example, my polynom is  P=x^6-4*x^3+x-2

s := sturmseq(x^6-4*x^3+x-2,x);

sturm(s,x,-2,2); # The number of roots in the interval (-2,2)

Here, i would like to find the roots in (-M,M) :

Bounding all roots in [-M,M] where M = max{1, sum^(n-1) |ai|/an}.

f0 = f, f1 = f', then use -remainder,

I know that  sturm(s,x,-M,M); gives the number of roots in (-M,M)  but is it possible to use the variation of sign like :

      gives a Sturm sequence for f.

      variation of sign, varsign(a0,a1,...,ar).

      Thm: (Sturm) varsign(f0(alpha),...,fr(alpha)) - varsign(f0(beta),..., fr(beta))

      is the number of distinct roots of f in [alpha,beta].

then i would like Isolating roots of rational polynomials

Method: reduce, remove rational roots, divide and conquer in [-M,M],

      then use bisection  in disjoint closed intervals ctg one root each

 Bisection method :

Bisection

      Setup: f(a) < 0, f(b) > 0 (or conversely).

      Repeated subdivision of [a,b] guaranteed to get close to a root.

      Error analysis: for error eps, solve (b-a)/ 2^(n+1)  < tol for n. where tol is the tolerance

Thanks

Dear all;

I need you help for solving this problem, and thanks in advantage for your help.

I have a polynom like  P =x^6-4*x^3+x-2;  and i would like to find an approximate value of the roots in some interval [a,b] =[-2,2] using sturm sequence.

The method is based on:

1) first construct the sturm sequence:

For given polynom P =x^6-4*x^3+x-2;

Let S0=P;

S1=diff(p,x);

let   s:=quo(S0,S1,x);
       S2:=-rem(S0,S1,x);

.... S[k+1-rem(S[k-1],S[k]);

S[k] is the sturm sequence.

2) let f(a)= number of change of sign in the sturm sequence and f(b) the same . so f(b)-f(a) give the number of roots in the interval [a,b].

3) If f(b)-f(a) =0 so there are no roots

and if f(a)-f(b)=1 one can find the root

4) if f(a) -f(b) >2  :

given toterance tol=0.001; for example

if the abs(a-b)<2*epsilon we display a message that there are k roots at (b+a)/2

with our error tolerance

5) otherwise if c=(b+a)/2 is not a root of P_k(x)  for any k, ( where p_k is an element of the sturm sequence ) 

we divide the interval into equal halves [a,c] and [x,b] and we run step 2 on each interval

else if c is a root of one of these p_k(x) add any time account to c so that c lies close the middle of [a,b] and not a root

6) Give all the roots ( approximate the rrots with small error epsilon).

I kindly  appreciate your help

Dear all;

Thank you very much for helping me to understand this problem.

I need your help for this question, it's seem for correct but when I run the code there is no dispaly of the solution with this command  dsolve({ode,ics}) ;

ode:=x*diff(y(x),x)+y(x)=x*exp(-x^2);

ics:=y(0)=1;

dsolve(ode);

dsolve({ode,ics}) ;