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hi

i have the following ode

.7246228659*(diff(f(x), x, x, x))+f(x)*(diff(f(x), x, x))+.6666666666-.6666666666*(diff(f(x), x))^2

with boundary conditions

f(0) = 0, D(f)(0) = a, (D(f))(b) = 1   (b corresponds to infinity and we can assume it 4 or 10 such that the solution converge)

i know the the solutions exist up to a critical value of "a" that beyonds no solution exist. for example for the following ode the critical value is...

Hello everyone

i forgot to add, subtract, divid and multiple all infinties (∞) together and with any number and i want to know the indetermanite forms.

if anyone know, please try to respond quickly.

thank you all.

Hi

I'm solving a differential equation which has a boundary at infinity, when I add this third condition it returns nothing. Of course I it's easy to solve by hand but it should be done in a loop. can you give me a hint on this?

Thanks

Is there any way for maple to compute the nth degree taylor polynomial?
I tried using taylor(f(x),x=0,n) but that gave me an error. So then I tried taylor(f(x),x=0,infinity) but that just ended up giving me a memory error. Any help?

hi

i have the following function

f[1](x):=_C2 (e)^x-_C1 (e)^(-x)+1/2 x (e)^(-x) h1+1/2 (e)^(-x) h1+_C3

with thease boundary conditions:   f[1](x))=0,diff(f[1](x),x))=1,diff(f[1](x),x),x=infinity)

i want to find cinstants C1,C2,C3 where h1 is arbitrary parametr.

i write the following code but its doesnt work correctly

s1:=simplify(subs(x=0,f[1](x))=0);
s2:=simplify(subs(x=0,diff(f[1](x),x))=1);

hi,

please Help me in the following code :

> restart;
i:=0;
             
u[i](s):=A+B*s-C*exp(-s);
              
lambda(s,x):= -(1/2)*(s-x)^2;
                    

tre(s,x):=lambda(s,x)*(diff(u[i...

Is there an easy way to find the limit of the following equations as Pec_i approches infinity.

 Equation 1

(Q*(R*S+1-S)/(1-R*R3*k*(1-R1)/C/Pec_i*S)*(H_vap+1/(1-exp(Pec_i)))*exp(C*Pec_i/(R*S+1-S)*(1-R*R3*k*(1-R1)/C/Pec_i*S)*S/k)-H_vap)*(1-exp(Pec_i*(1-S)/(R*S+1-S)*(1-R*R3*k*(1-R1)/C/Pec_i*S))) = 1;

 Equation 2

 Theta0/C = (exp(Pec_i/(R*S+1-S)*(1-R*R3*k*(1-R1)/C/Pec_i*S)*(1-S))-1)*(H_liq+1/(1-exp(-C/k*Pec_i/(R*S+1-S)*(1-R*R3*k*(1-R1)/C/Pec_i*S)*S)));

Why does this happen to Maple 15?

    `assuming`([sum(k*p*Beta(k, p+1), k = 1 .. infinity)], [p > 1]); eval(%, p = 2)

Is it possible to show with Maple that for any real p>1 the series converges to p/(p-1), e.g.,

    `assuming`([sum(k*p*Beta(k, p+1), k = 1 .. 1000)], [p > 1]): subs(p = 2, %): evalf(%)

How do I show this symbolically? Thanks.

I want to assign

D1:=-1/4*int((f*exp(-v*s)*(1+v*s)/v^3),0..infinity);

When I do this, it print f(x), v(x), s(x) in the integral and I don't want that...it should stay as f,v,s. Am I doing something wrong?

I am interested in finding the asymptofic constant (Big O(1/n^(2m+2)) for the following expansion

Upon using the preceding command in the maple i get

Error, (in asympt) unable to compute series

Hi,

I would like to define the following exceptions:

0/0 should be equal with 0, but any other x/0 should be infinity, if x>0 or -infinity, if x<0.

In the following example, I show what I achieved, however, I also would like to have equation (8) to be infinity.

BR,

Zoltan Faigl

What can be said about this divergent series:

 sum((cos(m)+I*sin(m))*(cosh(w)+sinh(w)),n=1 ..infinity) where m=Pi*n and w = ln(n)/n?

Hi All,

I'm new here and to Maple and this is my first post.

How do I get maple to evalaute the following expression:

sum(exp(-1/2*(((psi-theta) + 2*pi*n) / (sigma))^2), n=1..infinity);

psi and theta are angles, sigma is a constant - basically psi, theta and sigma are numbers. In fact the summation is where n is a member of Z (set of integers) but I will be happy with a solution for n= 1..infinity.

I have a complicated function (combination of exponential and sin), and I want to find the infinite integral of that from 0 to infinity which I know that the function itself converges to zero at infinity. I used evalf to force maple to do that but it fails to give the numerical answer. How can I resolve the problem?

Thanks,

Hamid

I got the range result as limit

ki:= t -> (sum(alpha1(S[j])*(exp(2*I*k*delta*t)-exp(S[j]*t))/((2*I*k*delta-S[j])*Q1(S[j])),j=1..3));limit(ki(t),t=infinity);
where : k,delta are constant and sum(alpha1([sj]),j=1..3) is number

the output is:

-.1294653014455612741589137389024249282843455188351695736221586639528345838460087707530271384645825938-0.8233085340292186435280681029574142108148445032573172243750416705060582056155292536631305980808392462e-1*I...