Maple 17 Questions and Posts

These are Posts and Questions associated with the product, Maple 17

Hello,

 

I tried to plot the problem presented below:

restart; with(plots); C := setcolors(); with(LinearAlgebra);

formula1 := 2.6*BodyWeight*abs(sin(4*Pi*t));
2.6 BodyWeight |sin(4 Pi t)|
BodyWeight := 80*9.81;
plot(formula1, t = 0 .. 2);


eq2 := formula1-SpringConstant*y(t)-DampConstant*(diff(y(t), t)) = Mass*(diff(y(t), `$`(t, 2)));
2040.480 |sin(4 Pi t)| - SpringConstant y(t)

/ d \ / d / d \\
- DampConstant |--- y(t)| = Mass |--- |--- y(t)||
\ dt / \ dt \ dt //
DampConstant := 50;
50
Mass := .200;
Springt := 200;
200
SpringConstant := Youngsmodulus*Surface/DeltaLength;
DeltaLength := 0.2e-1-y(t);
Surface := .15;
Youngsmodulus := 6.5*10^6/(t+1)+6.5*10^6;
plot(Youngsmodulus, t = 0 .. 10000);

eq2;
2040.480 |sin(4 Pi t)|

/ 6 \
|6.5000000 10 6|
0.15 |------------- + 6.5000000 10 | y(t)
\ t + 1 / / d \
- ----------------------------------------- - 50 |--- y(t)| =
0.02 - y(t) \ dt /

/ d / d \\
0.200 |--- |--- y(t)||
\ dt \ dt //

incs := y(0) = 0, (D(y))(0) = 0;
eq4 := dsolve({eq2, incs}, y(t), type = numeric, method = lsode[backfull], maxfun = 0);
proc(x_lsode) ... end;

plots:-odeplot(eq4, [t, y(t)], 0 .. 5);

 When I try to plot it beyond t=5, Maple gives the following error:

Warning, could not obtain numerical solution at all points, plot may be incomplete

Does anyone know how to plot it even further?

 

 

vz := 2*(-eta^2+1);

D_im := .22;

r0 := 1;

pde := diff(vz*Y(eta, z), z)-D_im*((diff(eta*(diff(Y(eta, z), eta)), eta))/eta+diff(Y(eta, z), `$`(z, 2)))/r0 = 0;

pde := expand(%);

ibc := [Y(1, z) = 0, (D[1](Y))(0, z) = 0, Y(eta, 0) = 1, (D[2](Y))(eta, 0) = 0];

sol := pdsolve(pde, ibc, numeric, time = z, range = 0 .. 1);

pds := sol:-value(z = 0, output = listprocedure);

sol:-plot(z = 0.1e-3, numpoints = 50, color = blue, view = 0 .. 1)

So I was trying to solve this conservation equation for the radial coordinate eta and the z coordinate being treated as time. The flow is in z direction. Now unfortunately it is diverging. Not sure why though. What am I doing wrong?

Im trying to draw a shpere but it always saying: 

Error, (in plot3d) unexpected option: z = -2 .. 2


this is the equation: x^2+y^2+z^2-4=0

i'm writing this way

plot3d(x^2+y^2+z^2-2^2, x = -2 .. 2, y = -2 .. 2, z = -2 .. 2)


what should I do? this is my first time with this software

 

best from Brazil,
Nina

Hi all,

Let $A$ be a 0-1 square matrix of order $n$. I want to obtain a matrix $D$ from powers $A$, $A^2$, $\dots$, $A^{n-1}$, where the ($i,j$)-element of $D$ is the smallest $k$ for which the ($i,j$)-element of $A^k$ is nonzero. I only consider the non-diagonal elements of $D$.For example, if the matrix $A$ is Matrix( [[0, 1, 0, 0, 0], [1, 0, 1, 0, 0], [0, 1, 0, 1, 0], [0, 0, 1, 0, 1], [0, 0, 0, 1, 0]]), then $D$ shoule be Matrix([[0, 1, 2, 3, 4], [1, 0, 1, 2, 3], [2, 1, 0, 1, 2], [3, 2, 1, 0, 1], [4, 3, 2, 1, 0]]). However, I cannot obtain this result.

The code is as follows.

p := proc ()

   local A, B, D, m, n, k, r;

   A := Matrix([[0, 1, 0, 0, 0], [1, 0, 1, 0, 0], [0, 1, 0, 1, 0], [0, 0, 1, 0, 1], [0, 0, 0, 1, 0]]);

   r := LinearAlgebra:-RowDimension(A);

   D := A;

   for k from 2 by 1 to r-1 do

       B := A^k;

       for m from 1 by 1 to r do

           for n from 1 by 1 to r do

               if m <> n and B[m, n] <> 0 and D[m, n] = 0 then

                    D[m, n] := k

                end if;

           end do;

        end do;

   end do;

   D;

end proc;

By executing this procedure, I obtain D=Matrix([[0,1,2,3,3],[1,0,1,2,3],[2,1,0,1,2],[3,2,1,0,1],[3,3,2,1,0]]), which is not I want.

Thanks.

When the Physics package is loaded, Maple returns the following error in connection with the simple statement 2*D[1]:

Error, (in TypeTools/ac_var_local) unable to determine the anticommutative character of D[1]

Why that? Although to no avail, for D[1] there is no problem. Note that the above error results even without having set up any anticommutative prefixes or the likes in Physics Setup.

PS: I am using Maple 17.

This is the coding till i do dhe decryption process. 

Do(plaintext=GetProperty("message",value));
Do(plaintext=convert(GetProperty("message",value),name));
Do(plaintextInt = convert(plaintext, bytes));
Do(plaintextBin = `~`[convert](plaintextInt, binary));
Do(plaintextBin2 = map2(nprintf, "%07d", plaintextBin));
Do(n0 = plaintextBin2[]);
Do(length1 = length(n0));
Do(plaintextCode = cat("", plaintextBin2[]));
Do(length2 = length(plaintextCode));
Do(z = convert(plaintextCode, decimal, binary));
Do(z1 = z+1);
Do(z2 = z1+%sk1);
Do(z3 = convert(z2, base, 2));
Do(b = cat("", z3[]));
Do(z4 = length(b));
Do(z5 = [Bits:-GetBits(-z2, -1 .. 0, bits = z4)]);
Do(z6 = cat("", z5[]));
Do(z7 = convert(z6, decimal, binary));
Do(%C = `mod`(Power(z7, %txte), %txtN));
Do(%C1 = `mod`(Power(%sk1, %txte), %txtN));

Do(%m = `mod`(Power(%C, %d), %N));

Do(%sk2=`mod`(Power(%C1,%d),%N));

Then nw i need to decrypt back to the original message with the coding:

Do(z8 = [Bits:-GetBits(-%m,-1 .. 0, bits = z4)]);
Do(c = cat("", z8[]));
Do(z9 = convert(c, decimal, binary));
Do(z10 = z9-sk2);
Do(z11 = z10-1);
Do(z12 = [Bits:-GetBits(z11, -1 .. 0, bits = length2)]);
Do(d = cat("", z12[]));
Do(plaintextBin2 = [StringTools:-LengthSplit(d, length1)]);
Do(plaintextInt2 = `~`[convert](plaintextBin2, decimal, binary));
Do(%message1 = convert(plaintextInt, bytes));

when i execute the program it shows the error

so how should I solve this as although i think that it should be problem of parsing the number z4 in the sentence that i highlighed, but whenever i correct it it still can't work.Thus anyone who know please help.Thanks.

 

restart:with(plots):
eq:=(diff(f(eta),eta$2))-a*f(eta)+b*(1+diff(f(eta),eta)^2)^(-1/2)=0;
bc:=f(1)=0,D(f)(0)=0;
ans := dsolve(eq);

I am trying to find the root of an equation. The problem is, I keep getting the error

"Error, (in fsolve) Can't handle expressions with typed procedures"

and 

Warning, solutions may have been lost


whenever I try to solve it. Anyone have any ideas? My worksheet is here:  1.mw

Coding in key generation of RSA Cryptosystem

message :=123456;

u := nextprime(RandomTools[Generate](integer(range = 10^100 .. 10^(110-1))));

v := nextprime(RandomTools[Generate](integer(range = 10^100 .. 10^(110-1))));

N := u*v;

phi := (u-1)*(v-1);

In RSA cryptosystem, the encryption or decryption of message only can be existed between range 1<=message<=N. This means that if the value of message bigger than N value then we can't get back the original value when performing decryption.Thus, wanna to ask that how can we create a coding so that the system will recognise the random integer u and v whether lie between range 1<=message<=N, means wanna crete a coding that if  N>=message then it will continue to the phi step whereas if message>=then the system need to regenerate the u and v until it satisfy the condition  1<=message<=N .Thus please help as i am a beginner in Maple.Thanks.

>M1:=146996733613391:

>M2:=1348471408813:

>teks:=numtheory[cfrac](M1/M2,'quotients');

tekssifer:=[109,101,115,115,97,103,101]

 

Hi, how should i rewrite my command to decrease the execution time?

Thank you~=]]

 

ANALYSIS AND DESIGN OF MACHINE FOUNDATION

 


restart

Loading Optimization

 

Loading LinearAlgebra  

 

Loading plots  

with(ScientificConstants):

Loading DynamicSystems  

with(Units:-Standard)

with(Units:-Natural)

with(StringTools)

FormatTime("%m-%d-%Y, %H:%M")

FormatTime("%m-%d-%Y, %H:%M")

(1)

NULL

Introduction

 

This document deals with vibration analysis and design of machine foundations subjected to dynamic load.

NULL

NULL

NULL

NULL

GetConstant(g);

standard_acceleration_of_gravity, symbol = g, value = 9.80665, uncertainty = 0, units = m/s^2

(2)

g__SI := evalf(Constant(g, system = SI, units))

9.80665*Units:-Unit(('m')/('s')^2)

(3)

NULL

Richart and Lysmer's Model

 

Richart et al. (1970) idealised the foundation as a lumped mass supported on soil which is idealised as frequency independent springs which he described in term of soil parameter

dynamic shear modulus or shear wave velocity of the soil for circular footing when footings having equivalent circular radius. The Tables below shows the different values of spring and damping vlaues as per Richart and Lysmer.

NULL

In which, G = dynamic shar modulus of he soil and is given G = `&rho;__s`*V__s^2 ; ν = Piosson's ratio of the soil; ρs = mass density of the soil; Vs = shear wave velocity of the soil obtained

from soil testing; g = acceleration due to gravity; m = mass of the machine and foundation; J = mass moment of inertia of the machine and foundation about the appropriate axes; K = equivalent spring stiffness of the soil; C = damping value of the soil; B = interia factor contributing to the damping factor; D = damping ratio of the soil; r = equivalent radius of a circular foundation; L = length of foundation, and B = width of the foundation.

NULL

NULL

NULL

Sketch

 

NULL

NULL

NULL

NULL

nu := .25

Table : Values of soil springs as per Richart and Lysmer (1970) model

 

NULL

NULL

SI No.

Direction

Spring value

Equivalent radius

Remarks

1

Vertical

K__z = 4*G*r__z/(1-nu)"(->)"

r__z = sqrt(L*B/Pi)"(->)"

This is in vertical Z direction

2

Horizontal

K__x = (32*(1-nu))*G*r__x/(7-8*nu)
"(->)"

r__x = sqrt(L*B/Pi)"(->)"

This induce sliding in horizontal X

2.1

Horizontal

K__y = (32*(1-nu))*G*r__y/(7-8*nu)
"(->)"

r__y = sqrt(L*B/Pi)"(->)"

This induce sliding in horizontal Y

3

Rocking

`K__&phi;x` = 8*G*`r__&phi;x`^3/(3*(1-nu))"(->)"

`r__&phi;x` = (L*B^3/(3*Pi))^(1/4)"(->)"

This produces roxking about Y axis

3.1

Rocking

`K__&phi;y` = 8*G*`r__&phi;y`^3/(3*(1-nu))"(->)"

`r__&phi;y` = (L^3*B/(3*Pi))^(1/4)"(->)"

This produces roxking about X axis

4

Twisting

`K__&psi;` = 16*G*`r__&psi;`^3*(1/3)"(->)"

`r__&psi;` = ((B^3*L+B*L^3)/(6*Pi))^(1/4)
"(->)"

This produces twisting about vertical Z axis

 

NULL

NULL

NULL

NULL

NULL

Table : Values of soil damping as per Richart and Lysmer (1970) model

 

NULL

SI No.

Direction

Mass ratio (B)

Damping ratio and Damping values

Remarks

1

Vertical

B__z = .25*m__U*(1-nu)*g__SI/(`&rho;__s`*r__z^3)
"(->)"

`&zeta;__z` = .425/sqrt(B__z)"(->)"C__z = 2*`&zeta;__z`*sqrt(K__z*m__U)"(->)"

This damping value is in vertical Z direction

2

Horizontal

B__x = (7-8*nu)*m__U*g__SI/((32*(1-nu))*`&rho;__s`*r__x^3)
"(->)"

`&zeta;__x` = .288/sqrt(B__x)"(->)"

C__x = 2*`&zeta;__x`*sqrt(K__x*m__U)"(->)"

This damping value is in lateral X direction

2.1

Horizontal

B__y = (7-8*nu)*m__U*g__SI/((32*(1-nu))*`&rho;__s`*r__y^3)
"(->)"

`&zeta;__y` = .288/sqrt(B__y)"(->)"

`&zeta;__y` = .288/((2.145204688-2.451662500*nu)*m__U*Units:-Unit(('m')/('s')^2)/((1-nu)*`&rho;__s`*(L*B/Pi)^(3/2)))^(1/2)

(5.1)

NULLError, invalid left hand side in assignmentError, invalid left hand side in assignment

`&zeta;__&psi;` = .5/(1+117.6798000*`J__&psi;`*Units:-Unit(('m')/('s')^2)*6^(1/4)/(`&rho;__s`*((B^3*L+B*L^3)/Pi)^(5/4)))

(5.2)

C__y = 2*`&zeta;__y`*sqrt(K__y*m__U)"(->)"

This damping value is in lateral Y direction

3

Rocking

`B__&phi;x` = (.375*(1-nu))*`J__&phi;x`*g__SI/(`&rho;__s`*`r__&phi;x`^5)
"(->)"

`&zeta;__&phi;x` = .15/((1+`B__&phi;x`)*sqrt(`B__&phi;x`))
"(->)"

`&zeta;__&phi;x` = .15/((1+11.03248125*(1-nu)*`J__&phi;x`*Units:-Unit(('m')/('s')^2)*3^(1/4)/(`&rho;__s`*(L*B^3/Pi)^(5/4)))*((11.03248125-11.03248125*nu)*`J__&phi;x`*Units:-Unit(('m')/('s')^2)*3^(1/4)/(`&rho;__s`*(L*B^3/Pi)^(5/4)))^(1/2))

(5.3)

Error, invalid left hand side in assignmentError, invalid left hand side in assignmentNULLError, invalid left hand side in assignment

.15/((1+11.03248125*(1-nu)*`J__&phi;x`*Units:-Unit(('m')/('s')^2)*3^(1/4)/(`&rho;__s`*(L*B^3/Pi)^(5/4)))*((11.03248125-11.03248125*nu)*`J__&phi;x`*Units:-Unit(('m')/('s')^2)*3^(1/4)/(`&rho;__s`*(L*B^3/Pi)^(5/4)))^(1/2)) = .15/((1+11.03248125*(1-nu)*`J__&phi;x`*Units:-Unit(('m')/('s')^2)*3^(1/4)/(`&rho;__s`*(L*B^3/Pi)^(5/4)))*((11.03248125-11.03248125*nu)*`J__&phi;x`*Units:-Unit(('m')/('s')^2)*3^(1/4)/(`&rho;__s`*(L*B^3/Pi)^(5/4)))^(1/2))

(5.4)

`C__&phi;x` = 2*`&zeta;__&phi;x`*sqrt(`K__&phi;x`*`J__&phi;x`)"(->)"

NULL

This damping value is for rocking about Y direction

3.1

Rocking

`B__&phi;y` = (.375*(1-nu))*`J__&phi;y`*g__SI/(`&rho;__s`*`r__&phi;y`^5)
"(->)"

NULL

`&zeta;__&phi;y` = .15/((1+`B__&phi;y`)*sqrt(`B__&phi;y`))
"(->)"

.15/((1+`B__&phi;y`)*`B__&phi;y`^(1/2)) = .15/((1+`B__&phi;y`)*`B__&phi;y`^(1/2))

(5.5)

`C__&phi;y` = 2*`&zeta;__&phi;y`*sqrt(`K__&phi;y`*`J__&phi;y`)"(->)"

NULL

This damping value is for rocking about X direction

4

Twisting

`B__&psi;` = `J__&psi;`*g__SI/(`&rho;__s`*`r__&psi;`^5)"(->)"

`&zeta;__&psi;` = .5/(1+2*`B__&psi;`)"(->)"NULLError, invalid left hand side in assignmentError, invalid left hand side in assignmentNULLError, invalid left hand side in assignmentNULLError, invalid left hand side in assignment

.5/(1+117.6798000*`J__&psi;`*Units:-Unit(('m')/('s')^2)*6^(1/4)/(`&rho;__s`*((B^3*L+B*L^3)/Pi)^(5/4))) = .5/(1+117.6798000*`J__&psi;`*Units:-Unit(('m')/('s')^2)*6^(1/4)/(`&rho;__s`*((B^3*L+B*L^3)/Pi)^(5/4)))

(5.6)

`C__&psi;` = 2*`&zeta;__&psi;`*sqrt(`K__&psi;`*`J__&psi;`)"(->)"NULL

``

NULL

This damping value is valid for twisting about vertical Z axis

 

NULL

NULL

NULLNULL

(4)

``

NULL

Vertical Motion Considering damping of the Soil

 

For vertical direction the equation becomes that of a lumped mass having single degree of freedom when

deq := m__U*(diff(z(t), t, t))+'C__z'*(diff(z(t), t))+'K__z'*`#mi("z")` = P__0*sin(`&omega;__m`*t)

m__U*(diff(diff(z(t), t), t))+C__z*(diff(z(t), t))+K__z*`#mi("z")` = P__0*sin(`&omega;__m`*t)

(6.1)

NULL

t1 := subs(P__0*sin(`&omega;__m`*t)/m__U = F, expand(deq/m__U));

diff(diff(z(t), t), t)+1.085721853*(G*(L*B/Pi)^(1/2)*m__U/(1-nu))^(1/2)*(diff(z(t), t))/(m__U*(m__U*Units:-Unit(('m')/('s')^2)/(`&rho;__s`*(L*B/Pi)^(3/2))-m__U*Units:-Unit(('m')/('s')^2)*nu/(`&rho;__s`*(L*B/Pi)^(3/2)))^(1/2))+4*G*(L*B/Pi)^(1/2)*`#mi("z")`/(m__U*(1-nu)) = F

(6.2)

NULL

(6.3)

By algebraically manipulating the expression, the form traditionally used by engineers is derived:

t2 := algsubs('C__z'/m__U = 2*zeta*omega, t1)

diff(diff(z(t), t), t)-(-1.085721853*(-G*(L*B/Pi)^(1/2)*m__U/(nu-1))^(1/2)*(diff(z(t), t))*nu+1.085721853*(-G*(L*B/Pi)^(1/2)*m__U/(nu-1))^(1/2)*(diff(z(t), t))+4.*G*(L*B/Pi)^(1/2)*`#mi("z")`*(-Units:-Unit(('m')/('s')^2)*(nu-1)*m__U/(`&rho;__s`*(L*B/Pi)^(3/2)))^(1/2))/((nu-1.)*(-Units:-Unit(('m')/('s')^2)*(nu-1)*m__U/(`&rho;__s`*(L*B/Pi)^(3/2)))^(1/2)*m__U) = F

(6.4)

NULL

t3 := algsubs('K__z'/m__U = omega^2, t2)

diff(diff(z(t), t), t)-(-1.085721853*(-G*(L*B/Pi)^(1/2)*m__U/(nu-1))^(1/2)*(diff(z(t), t))*nu+1.085721853*(-G*(L*B/Pi)^(1/2)*m__U/(nu-1))^(1/2)*(diff(z(t), t))+4.*G*(L*B/Pi)^(1/2)*`#mi("z")`*(-Units:-Unit(('m')/('s')^2)*(nu-1)*m__U/(`&rho;__s`*(L*B/Pi)^(3/2)))^(1/2))/((nu-1.)*(-Units:-Unit(('m')/('s')^2)*(nu-1)*m__U/(`&rho;__s`*(L*B/Pi)^(3/2)))^(1/2)*m__U) = F

(6.5)

This form includes the damping ratio , the natural frequency , and the external forcing term .  Consider only free vibration by setting

gen3 := subs(F = 0, t3)

diff(diff(z(t), t), t)-(-1.085721853*(-G*(L*B/Pi)^(1/2)*m__U/(nu-1))^(1/2)*(diff(z(t), t))*nu+1.085721853*(-G*(L*B/Pi)^(1/2)*m__U/(nu-1))^(1/2)*(diff(z(t), t))+4.*G*(L*B/Pi)^(1/2)*`#mi("z")`*(-Units:-Unit(('m')/('s')^2)*(nu-1)*m__U/(`&rho;__s`*(L*B/Pi)^(3/2)))^(1/2))/((nu-1.)*(-Units:-Unit(('m')/('s')^2)*(nu-1)*m__U/(`&rho;__s`*(L*B/Pi)^(3/2)))^(1/2)*m__U) = 0

(6.6)

NULL

sol1 := dsolve({gen3, z(0) = P, (D(z))(0) = V}, z(t))

z(t) = -(1000000000/1178791942081753609)*Pi*m__U*exp(-(1085721853/1000000000)*Units:-Unit(('s')/('m')^(1/2))*t/(Pi*((nu-1)^2/(`&rho;__s`*L^2*B^2*G))^(1/2)*m__U))*(1085721853*V*Units:-Unit(('s')/('m')^(1/2))*((nu-1)^2/(`&rho;__s`*L^2*B^2*G))^(1/2)*`&rho;__s`*L^2*B^2-4000000000*(L*B/Pi)^(1/2)*Pi*nu*`#mi("z")`+4000000000*`#mi("z")`*Pi*(L*B/Pi)^(1/2))/(Units:-Unit(('s')/('m')^(1/2))^2*B^2*L^2*`&rho;__s`)+(4000000000/1085721853)*`#mi("z")`*G*Pi*(L*B/Pi)^(1/2)*((nu-1)^2/(`&rho;__s`*L^2*B^2*G))^(1/2)*t/((nu-1)*Units:-Unit(('s')/('m')^(1/2)))+(1/1178791942081753609)*(1085721853000000000*V*Units:-Unit(('s')/('m')^(1/2))*Pi*((nu-1)^2/(`&rho;__s`*L^2*B^2*G))^(1/2)*m__U*`&rho;__s`*L^2*B^2+1178791942081753609*Units:-Unit(('s')/('m')^(1/2))^2*P*`&rho;__s`*L^2*B^2-4000000000000000000*(L*B/Pi)^(1/2)*Pi^2*m__U*nu*`#mi("z")`+4000000000000000000*`#mi("z")`*Pi^2*(L*B/Pi)^(1/2)*m__U)/(B^2*L^2*`&rho;__s`*Units:-Unit(('s')/('m')^(1/2))^2)

(6.7)

NULL

NULL

 

Download Analysis_and_Design_of_Machine_Foundations_1.mw

Good Morning Mapleprime Community,

Would anybody please help in the attached worksheet. I'm trying to use the new function in Maple that is the clicable method, but I was having problem in some of my output such as zeta_y and zeta_phi as this two equations are generating an error message.

 

Regards,

Moses

H, how i need to write my command to get n times of the sentences i insert.

For example, 

>teksbiasa:= `Hello Bob`;

Hello Bob

So, if i want 5 times of `Hello Bob` like 

Hello Bob Hello Bob Hello Bob Hello Bob Hello Bob 

What should i write?

Thank you ~=]]

How i need to write 

>teksbiasa:=`Hello! Bob`;

in button Action When Clicked at simple graphical interface instead of

Do(teksbiasa=%txtTeksBiasa);

because when i enter Hello! Bob in %txtTeksBiasa, system pop up ERROR

 

Error

Error in Component button with caption "Botton":

(in unknown) incorrect syntax in parse: missing operator or

`;`(near 7th character of parsed string)

 

Thank you~=]]

>message:=`67A`;

67A

>P:=convert(message, bytes);

[54, 55, 65]

>with(Bits):

>bitP1:=Split(P1);

[0, 1, 1, 0, 1, 1]

>bitP2:=Split(P2);

[1, 1, 1, 0, 1, 1]

>bitP3:=Split(P3);

[1, 0, 0, 0, 0, 0, 1]

>with(Statistics):

>b1:=Count(bitP1);

6

>b2:=Count(bitP2);

6

>b3:=Count(bitP3);

7

>totalBits=b1+b2+b3;

19

 

Hi, how i need to modify my command so when i write any message with any lenght, i can get the totalBits directly..

Thank you~=]]

How I can solve it ? If I want a solution dependent of a. With fsolve? But how?

-x3+ax2-lnx=0
 

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