## 120 Reputation

5 years, 175 days

## Time consuming code...

Hi dear friends

How do I improve my ability to manipulate code?

Thanks

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restart:

Digits := 30: m := 20: p0 := 8.: g := -0.3: nu := 0.3: a := 1:

w := sum(b[n]*r^(1.02*n), n = 1 .. m):

ODE := r^3*(g*r+a)^3*(diff(w, r, r, r, r))+2*(diff(w, r, r, r))*r^2*(4*g*r+a)*(g*r+a)^2+(diff(w, r, r))*r*(3*g^3*nu*r^3+P*a^3*r^2+6*a*g^2*nu*r^2+11*g^3*r^3+3*a^2*g*nu*r+15*a*g^2*r^2+3*a^2*g*r-a^3)+(diff(w, r))*(6*g^3*nu*r^3+P*a^3*r^2+6*a*g^2*nu*r^2-2*g^3*r^3-3*a*g^2*r^2+a^3):

for P from p0 by 0.1e-1 to p0+1 do
W := simplify(subs(solve({seq(evalf(subs(r = j/(m+2), ODE)), j = 2 .. m-1), subs(r = 1, diff(w, r))}, {seq(b[j], j = 1 .. m)}), w/b[m])):
F := (int(((diff(W, `\$`(r, 2))+(diff(W, r))/r)^2-(2*(1-nu))*(diff(W, `\$`(r, 2)))*(diff(W, r))/r)*r*(1+g*r/a)^3, r = 0 .. a))/(int((diff(W, r))^2*r, r = 0 .. a)):
delta[trunc(100*(P-p0))] := abs(F-P) end do:

plot(W, r = 0 .. a);

Min := min(seq(delta[j], j = 0 .. 100)):

for j from 0 to 100 do if delta[j] = Min then print(p0+j/(100.)) else  end if end do;

## Best source for learning maple ...

What is the best source of learning maple for an abecedarian to become a professional?

## difference between plots...

Hi

I can't understand difference between plots!

Thanks.

plot({sqrt(x+2*sqrt(x-1))+sqrt(x-2*sqrt(x-1)), sqrt(x-2*sqrt(x-1)), sqrt(x+2*sqrt(x-1))}, x = -3 .. 3)

## Time-consuming fsolve ...

Maple

I know 1st root of a function locates in a small interval (by drawing plots of function).

but fsolve command uses unreasonable time to find roots.

Also NextZero dosent work.

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restart; s := sqrt(n)*Pi*(sqrt(n)*Pi-tan(sqrt(n)*Pi))/(sqrt(n)*Pi*tan(sqrt(n)*Pi)+2*(1-sec(sqrt(n)*Pi)));

C := (tan(sqrt(n)*Pi)-sqrt(n)*Pi*sec(sqrt(n)*Pi))/(sqrt(n)*Pi-tan(sqrt(n)*Pi));

S := s*(-C^2+1);

Gamma[b] := E[b2]/E[b1];

Gamma[c] := E[c2]/E[c1];

alpha[b] := t[b1]*t[b2]/(t[b1]+t[b2])^2;

alpha[c] := t[c1]*t[c2]/(t[c1]+t[c2])^2;

EIb := b*E[b1]*(t[b1]^4+2*Gamma[b]*alpha[b]*(2-alpha[b])*(t[b1]+t[b2])^4+Gamma[b]^2*t[b2]^4)/(12*(Gamma[b]*t[b2]+t[b1]));

EIc := b*E[c1]*(t[c1]^4+2*Gamma[c]*alpha[c]*(2-alpha[c])*(t[c1]+t[c2])^4+Gamma[c]^2*t[c2]^4)/(12*(Gamma[c]*t[c2]+t[c1]));

b := 1;Lb := 2; Lc1 := 2.5; Lc2 := 3; E[b1] := 180; E[b2] := 200; t[b1] := 0.3e-1; t[b2] := 0.4e-1; E[c2] := 220; t[c1] := 0.5e-1; t[c2] := 0.2e-1;

for k from 0 by 10 to 100 do

E[c1] := 150+k; nce := (1/2)*(Lc1/Lc2)^2*n; nb := (1/2)*(Lb/Lc2)^2*n; q[k] := 1000*fsolve((subs(n = nce, C*s)*EIc/Lc1)^2/(S*EIc/Lc2+subs(n = nce, s)*EIc/Lc1+subs(n = nb, s*(1-C))*EIb/Lb)-subs(n = nce, s)*EIc/Lc1-2*EIb/Lb, n = 1.45 .. 1.56)*evalf(Pi^2)*EIc/(2*Lc2^2); print(q[k], k)

end do;

with(CurveFitting); F := PolynomialInterpolation([seq([10*i, q[10*i]], i = 0 .. 10)], z)

## How does Maple calculate floor limit?...

Maple

I want to calculate

Lim x-->1+ ([X2]-1)/([X]-1)

I let  x=1+c  and  c-->0 then we have [X2]=[c2+2c+1]=[2c]+1

Lim x-->1+ ([X2]-1)/([X]-1) = Lim c-->0+ [2c]/[c]

Maple shows the both limitations equal to zero

I want know how maple claculates it

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