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hi everyone,
I wanna to solve this equation using maple 12 but I receive this warning message

cos(x)*cosh(x)=1

cos(x)*cosh(x)=-1

tan(x)=tanh(x)

thanks in advance

(a) Show that if {an} ∞ n=1 is Cauchy then {a 2 n} ∞ n=1 is also Cauchy. (b) Give an example of a Cauchy sequence {a 2 n} ∞ n=1 such that {an} ∞ n=1 is not Cauchy

Show that 2^3 + x ^2 − 3x + 2 is O(x ^3 ).


"the set Q of rational numbers does not have the least-upper-bound property under the usual order. "

Proof:

"Consider the part A = {x in `&Qopf;`; 1<x^(2)<2<}, this part is not empty as [4/(3)]  in A; It is bounded by 2 as if x^(2)<4 then x<2. THe set of greatest elements of A, belonging to `&Qopf;`, is not empty."

Lemma:

"If [p/(q)]  in A with q  in `&Nopf;`^(*), then p>q for p>0 and p^(2)-2*q^(2)<0; or p^(2)-2*q^(2) in `&Zopf;`, thus"

p^2-2*q^2 <= -1

p^2-2*q^2 <= -1

(1)

"for all r  in `&Nopf;`^(*), put y=(r*p+1)/(r.p), we have :"

y > p/q

Now:

"y^(2)-2=supA/(r^(2)*q^(2)), with supA= r^(2)*p^(2)+2 r*p+1-2 r^(2)q^(2),"

otherwise; -1; s = r(p^2-2*q^2)+2*p+1

s = r(p^2-2*q^2)+2*p+1

(2)

"a good choice for r, for instance r=2*p+1, we get from (1)"

(2*p+1)(p^2-2*q^2) <= (2*p+1)(-1)

2*p(p^2-2*q^2) <= 2*p(-1)

(3)

thus:

(2*p+1)(p^2-2*q^2)+2*p <= -1

2*p(p^2-2*q^2)+2*p <= -2

(4)

(2*p+1)*[(2*p+1)(p^2-2*q^2)+2*p] <= -2*p-1

(2*p+1)*[2*p(p^2-2*q^2)+1+2*p] <= -2*p-1

(5)

(2*p+1)*[(2*p+1)(p^2-2*q^2)+2*p]+1 <= -2*p

(2*p+1)*[2*p(p^2-2*q^2)+1+2*p]+1 <= -2*p

(6)

"finally, supA<0"

`and`(thus*y^2 < 2, `in`(y, A*with*y^2) and A*with*y^2 > 1)

"so, for any x  in p/(q) in A, there exists y  in A such that y>x: in conclusion A does'nt admit a greatest element."

Now, given*m = p/q, `and`(not `in`(Typesetting:-delayDotProduct(a*greatest*element*of*A, Then)*m, A), we*get*thus):

`and`(p > q, p^2-2*q^2 >= 0)

as*the*equation*p^2-2*q^2 = (0*has)*no*solutions and `in`((0*has)*no*solutions, nonnegint*nonnegint), we*get:

2*p^2-2*q^2 >= 1:

"for any r  in `&Nopf;`^(*)let's put m'=(r*p)/((r*q+1)); we have:"

(diff(m(x), x))^2-2 = supA/(r*q+1)^2:

With*supA = r^2*p^2-2*(r*q+1)^2 and r^2*p^2-2*(r*q+1)^2 = r[r[p^2-2*q^2]-4*q]-2:

"for a good choice of r, for example r=4*q+1, we get (back to (2)):"

supA > (0*thus)*(diff(m(x), x))^2 and (0*thus)*(diff(m(x), x))^2 > 2:

"So, for any greatest element m of A, it exists m', greatest element of A such as:"

diff(m(x), x) < m(x):

"A does not admit a least upper bound."

``


Download rational_numbers.mw

"the set Q of rational numbers does not have the least-upper-bound property under the usual order. "

Proof:

"Consider the part A = {x in `&Qopf;`; 1<x^(2)<2<}, this part is not empty as [4/(3)]  in A; It is bounded by 2 as if x^(2)<4 then x<2. THe set of greatest elements of A, belonging to `&Qopf;`, is not empty."

Lemma:

"If [p/(q)]  in A with q  in `&Nopf;`^(*), then p>q for p>0 and p^(2)-2*q^(2)<0; or p^(2)-2*q^(2) in `&Zopf;`, thus"

p^2-2*q^2 <= -1

p^2-2*q^2 <= -1

(1)

"for all r  in `&Nopf;`^(*), put y=(r*p+1)/(r.p), we have :"

y > p/q

Now:

"y^(2)-2=supA/(r^(2)*q^(2)), with supA= r^(2)*p^(2)+2 r*p+1-2 r^(2)q^(2),"

otherwise; -1; s = r(p^2-2*q^2)+2*p+1

s = r(p^2-2*q^2)+2*p+1

(2)

"a good choice for r, for instance r=2*p+1, we get from (1)"

(2*p+1)(p^2-2*q^2) <= (2*p+1)(-1)

2*p(p^2-2*q^2) <= 2*p(-1)

(3)

thus:

(2*p+1)(p^2-2*q^2)+2*p <= -1

2*p(p^2-2*q^2)+2*p <= -2

(4)

(2*p+1)*[(2*p+1)(p^2-2*q^2)+2*p] <= -2*p-1

(2*p+1)*[2*p(p^2-2*q^2)+1+2*p] <= -2*p-1

(5)

(2*p+1)*[(2*p+1)(p^2-2*q^2)+2*p]+1 <= -2*p

(2*p+1)*[2*p(p^2-2*q^2)+1+2*p]+1 <= -2*p

(6)

"finally, supA<0"

`and`(thus*y^2 < 2, `in`(y, A*with*y^2) and A*with*y^2 > 1)

"so, for any x  in p/(q) in A, there exists y  in A such that y>x: in conclusion A does'nt admit a greatest element."

Now, given*m = p/q, `and`(not `in`(Typesetting:-delayDotProduct(a*greatest*element*of*A, Then)*m, A), we*get*thus):

`and`(p > q, p^2-2*q^2 >= 0)

as*the*equation*p^2-2*q^2 = (0*has)*no*solutions and `in`((0*has)*no*solutions, nonnegint*nonnegint), we*get:

2*p^2-2*q^2 >= 1:

"for any r  in `&Nopf;`^(*)let's put m'=(r*p)/((r*q+1)); we have:"

(diff(m(x), x))^2-2 = supA/(r*q+1)^2:

With*supA = r^2*p^2-2*(r*q+1)^2 and r^2*p^2-2*(r*q+1)^2 = r[r[p^2-2*q^2]-4*q]-2:

"for a good choice of r, for example r=4*q+1, we get (back to (2)):"

supA > (0*thus)*(diff(m(x), x))^2 and (0*thus)*(diff(m(x), x))^2 > 2:

"So, for any greatest element m of A, it exists m', greatest element of A such as:"

diff(m(x), x) < m(x):

"A does not admit a least upper bound."

``


Download rational_numbers.mw

If a dosage Q units of a certain drug is administrated to an individual, then the amount remaining in the bloodstream at the end of t minutes is given by Q*exp^-ct, where c>0. Suppose this same dosage is given at successive T-minute intervals.

 

a) Show that the amount A(k) of the drug is given by A(k) = ∑n=0k-1 Q*exp(^-ncT).

b) Find an upper bound for the amount of the drug in the bloodsteam after any number of doses.

c) Find the smallest time between doses that will ensure that A(k) does not exceed a certain level M for M>Q.

worksheet/expressions/copypasteMaple

Gerschgorin := proc (A::Matrix) local Delta, m, n, AA, R, C, i, c, eig, P, Plt; Delta := proc (i, j) if i = j then 0 else 1 end if end proc; m, n := LinearAlgebra[Dimension](A); AA := Matrix(m, n, proc (i, j) options operator, arrow; Delta(i, j)*abs(A[i, j]) end proc); R := evalm(`&*`(AA, Vector(m, 1))); C := {seq(('plottools[circle]')([Re(A[i, i]), Im(A[i, i])], R[i], color = violet), i = 1 .. m)}; c := {seq(('plottools[point]')([Re(A[i, i]), Im(A[i, i])], color = blue, symbol = diamond), i = 1 .. m)}; eig := evalf(LinearAlgebra[Eigenvalues](A)); P := {seq(('plottools[point]')([Re(eig[i]), Im(eig[i])], color = red, symbol = box), i = 1 .. m)}; Plt := `union`(`union`(C, c), P); plots[display](eval(Plt), scaling = constrained) end proc

 

A := Matrix([[5, 8, 4, -3], [8, -9, 7, 5], [0, 4, 4, 2], [5, -5, 9, -9]]); evalf(LinearAlgebra[Eigenvalues](A), 3); Gerschgorin(A)

worksheet/expressions/pasteMathML

 

F := Matrix([[2, -1/2, -1/3, 0], [0, 6, 1, 0], [1/3, -1/3, 5, 1/3], [-1/2, 1/4, -1/4, 4]]); evalf(LinearAlgebra[Eigenvalues](F)); Gerschgorin(F)

Could you print A & F ?

 

regards

 

 

Hi, as I can't manage to copy and paste on mapleprimes, I would be glad to get a hint ...

Determine wether the sequence below converges or diverges, and if it converges, find the limit.

{n^(1/n)}  

and,

{exp^n/n^4}

Thanks

how can you graph when piecewise function is not continuous?

 

example:

BB := piecewise (-1.57 < c and c < -1.56, h, -0.06< c and  c < -0.05, aa, -0.5< c and c < 0.04, bb);

 

 

how to graph in maple 

for example

 

-2 < x < -3, h

-1 < x < -2, b

 

why do I get the error Error, (in rtable/Sum) invalid arguments

In positive numbers, I get it ok

code:

restart;
convert([92*x/790+18*x*(1-y)/1000=0.125,ln(46.59/x)=1/1.5*(ln(0.553/y)+2.5*ln((1-y)/(1-0.553)))],rational);
fsolve(%);
plots:-implicitplot(%%,x=0..2,y=0..1,numpoints=1000,thickness=2,gridlines,color=[blue,red]);

question:

there is something missing with the red line. I can't find the intersect point on the figure.

Is there something wrong with implicitplot? 

Dear All,

I am solving 6 ODE equations with boundary conditions using Runge kutta Felbergh 45 (Maple 12). then, i got this problem.. any suggestion??

Thank you :)

ISPC3.mw

``

restart; with(plots); M := 3; k = .2; blt := 6; r := 2; l := .1; Pr := 6.8; Ec := 2; N := .5; rho := .5; Tv := .5; Tt := .5; c := 1; cm := .1; cp := .1

Eq1 := diff(f(eta), eta, eta, eta)+f(eta)*(diff(f(eta), eta, eta))-(diff(f(eta), eta))^2-M*(diff(f(eta), eta))+B*H(eta)*(F(eta)-(diff(f(eta), eta))) = 0;

diff(diff(diff(f(eta), eta), eta), eta)+f(eta)*(diff(diff(f(eta), eta), eta))-(diff(f(eta), eta))^2-3*(diff(f(eta), eta))+B*H(eta)*(F(eta)-(diff(f(eta), eta))) = 0

(1)

Eq2 := G(eta)*(diff(F(eta), eta))+F(eta)^2+B*(F(eta)-(diff(f(eta), eta))) = 0;

G(eta)*(diff(F(eta), eta))+F(eta)^2+B*(F(eta)-(diff(f(eta), eta))) = 0

(2)

Eq3 := G(eta)*(diff(G(eta), eta))+B*(f(eta)+G(eta)) = 0;

G(eta)*(diff(G(eta), eta))+B*(f(eta)+G(eta)) = 0

(3)

Eq4 := G(eta)*(diff(H(eta), eta))+H(eta)*(diff(G(eta), eta))+F(eta)*H(eta) = 0;

G(eta)*(diff(H(eta), eta))+H(eta)*(diff(G(eta), eta))+F(eta)*H(eta) = 0

(4)

Eq5 := diff(theta(eta), eta, eta)+Pr*(f(eta)*(diff(theta(eta), eta))-2*(diff(f(eta), eta))*theta(eta))+N*Pr*(theta1(eta)-theta(eta))/(rho*c*Tt)+N*Pr*Ec*(F(eta)-(diff(f(eta), eta)))^2/(rho*Tv) = 0;

diff(diff(theta(eta), eta), eta)+6.8*f(eta)*(diff(theta(eta), eta))-13.6*(diff(f(eta), eta))*theta(eta)+13.60000000*theta1(eta)-13.60000000*theta(eta)+27.20000000*(F(eta)-(diff(f(eta), eta)))^2 = 0

(5)

Eq6 := 2*F(eta)*theta1(eta)+G(eta)*(diff(theta1(eta), eta))+cp*(theta1(eta)-theta(eta))/(c*cm*Tt) = 0;

2*F(eta)*theta1(eta)+G(eta)*(diff(theta1(eta), eta))+2.000000000*theta1(eta)-2.000000000*theta(eta) = 0

(6)

bcs1 := f(0) = r, (D(f))(0) = -1, (D(f))(blt) = 0, F(blt) = 0, G(blt) = -f(blt), H(blt) = k, theta(0) = 1, theta(blt) = 0, theta1(blt) = 0;

f(0) = 2, (D(f))(0) = -1, (D(f))(6) = 0, F(6) = 0, G(6) = -f(6), H(6) = k, theta(0) = 1, theta(6) = 0, theta1(6) = 0

(7)

L := [0.1e-2];

[0.1e-2]

(8)

for k to 1 do R := dsolve(eval({Eq1, Eq2, Eq3, Eq4, Eq5, Eq6, bcs1}, B = L[k]), [f(eta), F(eta), G(eta), H(eta), theta(eta), theta1(eta)], numeric, output = listprocedure); Y || k := rhs(R[2]); YP || k := rhs(R[3]); YR || k := rhs(R[4]); YQ || k := rhs(R[5]) end do

Error, (in dsolve/numeric/bvp) initial Newton iteration is not converging

 

R

R

(9)

print([(YP || (1 .. 1))(0)]);

[YP1(0)]

(10)

``

P1 := plot([YP || (1 .. 1)], 0 .. 14, labels = [eta, (D(f))(eta)]):

Warning, unable to evaluate the function to numeric values in the region; see the plotting command's help page to ensure the calling sequence is correct

 

plots:-display([P1]);

 

``

``


Download ISPC3.mw

I'd like to plot the following inequalities:

sqrt(x)<=1/sqrt(2)

1/sqrt(2)<sqrt(x)<=1/sqrt(2)

 

Hello,

Im solving 4 ODE equations with BC. im trying to shoot the initial value but im having this error:

""Error, (in isolate) cannot isolate for a function when it appears with different arguments""

anyone could help me???

shooting92.mw

``

restart

Shootlib := "E:\\shooting/":

libname := Shootlib, libname:

with(Shoot):

with(plots):

n := 2:

FNS := {F(eta), H(eta), f(eta), g(eta), u(eta), v(eta)}:

ODE := {g(eta)*(diff(g(eta), eta))+B*(f(eta)+g(eta)) = 0, g(eta)*(diff(F(eta), eta))+F(eta)^2+B*(F(eta)-u(eta)) = 0, g(eta)*(diff(H(eta), eta))+H(eta)*(diff(g(eta), eta))+F(eta)*H(eta) = 0, diff(v(eta), eta)+f(eta)*v(eta)-u(eta)^2+B*H(eta)*(F(eta)-u(eta))-M*u(eta) = 0, diff(f(eta), eta) = u(eta), diff(u(eta), eta) = v(eta)};

{g(eta)*(diff(H(eta), eta))+H(eta)*(diff(g(eta), eta))+F(eta)*H(eta) = 0, g(eta)*(diff(g(eta), eta))+0.2e-1*f(eta)+0.2e-1*g(eta) = 0, g(eta)*(diff(F(eta), eta))+F(eta)^2+0.2e-1*F(eta)-0.2e-1*u(eta) = 0, diff(v(eta), eta)+f(eta)*v(eta)-u(eta)^2+0.2e-1*H(eta)*(F(eta)-u(eta))-3*u(eta) = 0, diff(f(eta), eta) = u(eta), diff(u(eta), eta) = v(eta)}

(1)

IC := {F(0) = gamma, H(0) = Q, f(0) = 0, g(0) = z, u(0) = 1, v(0) = alpha};

{F(0) = gamma, H(0) = Q, f(0) = 0, g(0) = z, u(0) = 1, v(0) = alpha}

(2)

BC := {F(L) = 0, H(L) = n, g(L) = -f(L), u(L) = 0};

{F(6) = 0, H(6) = 2, g(6) = -f(6), u(6) = 0}

(3)

infolevel[shoot] := 1:

S := shoot(ODE, IC, BC, FNS, [alpha = 0, gamma = 0, z = -.2, Q = 0])

Error, (in isolate) cannot isolate for a function when it appears with different arguments

 

``

``


Download shooting92.mw

hello dear freinds

im new comer in maple.

i want to find  particular solution of an ode by following code:

ode := diff(u[1](t), t, t)+u[1](t) = -(1/4)*a^3*cos(3*beta[0]+3*t)-(3/4)*a^3*cos(beta[0]+t)

m := combine(convert(particularsol(ode), trig))

but maple solution is : m := u[1](t) = (81/32)*a^3*cos(-3*beta[0]+t)-(81/16)*a^3*cos(3*beta[0]+t)-(3/8)*a^3*t*sin(beta[0]+t)+(3/16)*a^3*cos(-beta[0]+t)-(27/16)*a^3*cos(beta[0]+t)+(1/32)*a^3*cos(3*beta[0]+3*t)

but  particular solution is :

u[1](t) = -(3/8)*a^3*t*sin(beta[0]+t)+(1/32)*a^3*cos(3*beta[0]+3*t)

is there any idear for finding the solution?

thanks in advance

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