expect to calculate a eigenvector in terms of variable test10

close3 are decimal value

> NewMatrix3 := Matrix([[test10, close3(t), close3(t+1), close3(t+2), close3(t+3), close3(t+4)], [close3(t), close3(t+1), close3(t+2), close3(t+3), close3(t+4), close3(t+5)], [close3(t+1), close3(t+2), close3(t+3), close3(t+4), close3(t+5), 0], [close3(t+2), close3(t+3), close3(t+4), close3(t+5), 0, 0], [close3(t+3), close3(t+4), close3(t+5), 0, 0, 0], [close3(t+4), close3(t+5), 0, 0, 0, 0], [close3(t+5), 0, 0, 0, 0, 0]]); New_Asso_eigenvector := Eigenvectors(MatrixMatrixMultiply(Transpose(NewMatrix3), NewMatrix3));

Error, (in LA_Main:-Eigenvectors) cannot determine if this expression is true or false: abs(149.8198+5.59*Re(test10))+27.38*abs(Im(test10))+abs(118.8174+5.74*Re(test10))+abs(90.3603+5.49*Re(test10))+abs(61.9327+5.19*Re(test10))+abs(31.0804+5.37*Re(test10)) < (1/10)*abs(149.8198+5.59*Re(test10))+2.738000000*abs(Im(test10))+(1/10)*abs(118.8174+5.74*Re(test10))+(1/10)*abs(90.3603+5.49*Re(test10))+(1/10)*abs(61.9327+5.19*Re(test10))+(1/10)*abs(31.0804+5.37*Re(test10))

Hello, 

I have a trigonometric equation.

I would like to isolate gamma[1](t) and to determine gamma[1](t) in fonction of alpha(t), beta(t) and z(t). The others variables in the equations are fixed parameters.

I have tried to use isolate function. But it doesn't work.

Of course, my expressions should be complex but that is not a problem if i manage to expresse gamma[1](t) in fonction of alpha(t), beta(t) and z(t).

Here my program

constraints_2.mw

Thank you for you help

Hello, could you give me ideas with such challenge? I have created my model in MapleSim and want to check correctness of the scheme. I need to get values of block variable. by default maplesim displays results as a graph and i do not see what is real value. How I can get these values? 

In his article “Subscripts as Partial Differentiation Operatuuors” , rlopez 1228 showed us a way to denote partial derivatives by repeat subscripts. For example, the sixth derivative of u(x,y) with respect to x will be denoted by u_{x,x,x,x,x,x}.

Is there a way to make the notation u_{x,x,x,x,x,x} even shorter by u_{6x}?

In the same way mixed derivatives u_{x,x,x,y,y,y,y} will be denoted as u_{3x,4y}, etc.

Thank you very much!

superposition said that a vector is a linear combination of other vectors

but even if i calculated the coefficient, i do not know which vector is which other vectors's linear combination

how to prove?

InputMatrix3 := Matrix([[close3(t), close3(t+1) , close3(t+2) , close3(t+3) , close3(t+4) , close3(t+5)],
[close3(t+1) , close3(t+2) , close3(t+3) , close3(t+4) , close3(t+5) , close3(t+6)],
[close3(t+2) , close3(t+3) , close3(t+4) , close3(t+5) , close3(t+6) , 0],
[close3(t+3) , close3(t+4) , close3(t+5) , close3(t+6) , 0 , 0],
[close3(t+4) , close3(t+5) , close3(t+6) , 0 , 0 , 0],
[close3(t+5) , close3(t+6) , 0 , 0 , 0, 0],
[close3(t+6) , 0 , 0 , 0, 0, 0]]):
EigenValue1 := Eigenvalues(MatrixMatrixMultiply(Transpose(InputMatrix3), InputMatrix3)):
Asso_eigenvector := Eigenvectors(MatrixMatrixMultiply(Transpose(InputMatrix3), InputMatrix3)):
AEigenVector[tt+1] := Asso_eigenvector;

Matrix(6, 6, {(1, 1) = .514973850028629+0.*I, (1, 2) = .510603608194333+0.*I, (1, 3) = .469094659512372+0.*I, (1, 4) = .389872713818831+0.*I, (1, 5) = .279479324327359+0.*I, (1, 6) = -.154682461176604+0.*I, (2, 1) = .493994413154560+0.*I, (2, 2) = .306651336822139+0.*I, (2, 3) = -0.583656699197969e-1+0.*I, (2, 4) = -.417550308930506+0.*I, (2, 5) = -.566122865008542+0.*I, (2, 6) = .404579494288380+0.*I, (3, 1) = .449581541124671+0.*I, (3, 2) = -0.266751368453398e-1+0.*I, (3, 3) = -.529663398913996+0.*I, (3, 4) = -.359719616523673+0.*I, (3, 5) = .313717798014566+0.*I, (3, 6) = -.537405340038665+0.*I, (4, 1) = .386952162293470+0.*I, (4, 2) = -.351332186748244+0.*I, (4, 3) = -.390816901794187+0.*I, (4, 4) = .470032416161955+0.*I, (4, 5) = .231969182174424+0.*I, (4, 6) = .547134073332474+0.*I, (5, 1) = .306149178348317+0.*I, (5, 2) = -.530611390076568+0.*I, (5, 3) = .192717713961280+0.*I, (5, 4) = .291213691618787+0.*I, (5, 5) = -.562991429686901+0.*I, (5, 6) = -.431067688369314+0.*I, (6, 1) = .212576094920847+0.*I, (6, 2) = -.489443150196337+0.*I, (6, 3) = .553283259136031+0.*I, (6, 4) = -.488381938231088+0.*I, (6, 5) = .363604594054259+0.*I, (6, 6) = .195982711855368+0.*I})

Matrix(6, 6, {(1, 1) = .515428842592397+0.*I, (1, 2) = .515531996615269+0.*I, (1, 3) = .468108280940919+0.*I, (1, 4) = -.392394120975052+0.*I, (1, 5) = -.280467124908196+0.*I, (1, 6) = -.129613084502380+0.*I, (2, 1) = .494563493180197+0.*I, (2, 2) = .301273494494509+0.*I, (2, 3) = -0.622136916501293e-1+0.*I, (2, 4) = .438383262732459+0.*I, (2, 5) = .571041594120088+0.*I, (2, 6) = .377494770878435+0.*I, (3, 1) = .450886315308369+0.*I, (3, 2) = -0.323387895921418e-1+0.*I, (3, 3) = -.527636820417566+0.*I, (3, 4) = .332744872607714+0.*I, (3, 5) = -.322934536375586+0.*I, (3, 6) = -.549772001891837+0.*I, (4, 1) = .385916641681991+0.*I, (4, 2) = -.352066020655722+0.*I, (4, 3) = -.389655495441319+0.*I, (4, 4) = -.450049711766943+0.*I, (4, 5) = -.221529986447276+0.*I, (4, 6) = .568916672007495+0.*I, (5, 1) = .305485655770791+0.*I, (5, 2) = -.528766119966973+0.*I, (5, 3) = .201065789602278+0.*I, (5, 4) = -.310329356773806+0.*I, (5, 5) = .555973984740943+0.*I, (5, 6) = -.425730045170186+0.*I, (6, 1) = .210210489500614+0.*I, (6, 2) = -.488744465076970+0.*I, (6, 3) = .553484076328700+0.*I, (6, 4) = .494245653290329+0.*I, (6, 5) = -.364390406353340+0.*I, (6, 6) = .183130120876843+0.*I})
mm1 := 1;
solve(
[AEigenVector[mm1][2][1][6] = m1*AEigenVector[mm1][2][1][1]+m2*AEigenVector[mm1][2][1][2]+m3*AEigenVector[mm1][2][1][3]+m4*AEigenVector[mm1][2][1][4]+m5*AEigenVector[mm1][2][1][5],
AEigenVector[mm1][2][2][6] = m1*AEigenVector[mm1][2][2][1]+m2*AEigenVector[mm1][2][2][2]+m3*AEigenVector[mm1][2][2][3]+m4*AEigenVector[mm1][2][2][4]+m5*AEigenVector[mm1][2][2][5],
AEigenVector[mm1][2][3][6] = m1*AEigenVector[mm1][2][3][1]+m2*AEigenVector[mm1][2][3][2]+m3*AEigenVector[mm1][2][3][3]+m4*AEigenVector[mm1][2][3][4]+m5*AEigenVector[mm1][2][3][5],
AEigenVector[mm1][2][4][6] = m1*AEigenVector[mm1][2][4][1]+m2*AEigenVector[mm1][2][4][2]+m3*AEigenVector[mm1][2][4][3]+m4*AEigenVector[mm1][2][4][4]+m5*AEigenVector[mm1][2][4][5],
m1^2 + m2^2 + m3^2 + m4^2 + m5^2 = 1], [m1, m2, m3, m4, m5]);

[m1 = .4027576723+.5022235499*I, m2 = -.5922841426-1.043213223*I, m3 = -.1130969773+.9150300317*I, m4 = .9867039883-.5082455178*I, m5 = -1.400123192+.1536850673*I], [m1 = .4027576723-.5022235499*I, m2 = -.5922841426+1.043213223*I, m3 = -.1130969773-.9150300317*I, m4 = .9867039883+.5082455178*I, m5 = -1.400123192-.1536850673*I]

mm1 := 2;
solve(
[AEigenVector[mm1][2][1][6] = m1*AEigenVector[mm1][2][1][1]+m2*AEigenVector[mm1][2][1][2]+m3*AEigenVector[mm1][2][1][3]+m4*AEigenVector[mm1][2][1][4]+m5*AEigenVector[mm1][2][1][5],
AEigenVector[mm1][2][2][6] = m1*AEigenVector[mm1][2][2][1]+m2*AEigenVector[mm1][2][2][2]+m3*AEigenVector[mm1][2][2][3]+m4*AEigenVector[mm1][2][2][4]+m5*AEigenVector[mm1][2][2][5],
AEigenVector[mm1][2][3][6] = m1*AEigenVector[mm1][2][3][1]+m2*AEigenVector[mm1][2][3][2]+m3*AEigenVector[mm1][2][3][3]+m4*AEigenVector[mm1][2][3][4]+m5*AEigenVector[mm1][2][3][5],
AEigenVector[mm1][2][4][6] = m1*AEigenVector[mm1][2][4][1]+m2*AEigenVector[mm1][2][4][2]+m3*AEigenVector[mm1][2][4][3]+m4*AEigenVector[mm1][2][4][4]+m5*AEigenVector[mm1][2][4][5],
m1^2 + m2^2 + m3^2 + m4^2 + m5^2 = 1], [m1, m2, m3, m4, m5]);

[m1 = .4262845394-.5114193433*I, m2 = -.6313720018+1.072185334*I, m3 = -0.7337582213e-1-.9580760394*I, m4 = -1.036525681-.5400714113*I, m5 = 1.412710014+.1874839516*I], [m1 = .4262845394+.5114193433*I, m2 = -.6313720018-1.072185334*I, m3 = -0.7337582213e-1+.9580760394*I, m4 = -1.036525681+.5400714113*I, m5 = 1.412710014-.1874839516*I]

any user of the community has material ppt or pdf on presentation of clickable math popup and maple in computaconal applied to mathematics.

any help on the origin of math clickable popup;? place the link if they were so friendly!

If A is a matrix 2*2 then how can decompose A as spectral decomposition.

I want to write maple code of the following algorithm with

the following parameters and initial values please help me.

T₀ = 5.5556 × 10⁷ cells, I₀ = 1.1111 × 10⁷ cells, V₀ = 6.3096 × 10⁹ copies/ml,

A_1=A2=1,

c = 0.67, h = 1, d = 3.7877 × 10⁻³, δ = 3.259d,

λ = 2/3× 10⁸d, R₀ = 1.33,

p = (cV₀δR₀₎/λ(R0−1)

and β = dδcR₀/λp .

Algorithm
step 1 :
T(0) = T₀, I(0) = I₀, V (0) = V₀ λi(100 ) = 0 (i=1, ..., 3), u₁(0) = 0 =
u₂(0).

step 2 :
for i=1, ..., n-1, do _:
T_i+1=(T_i + hλ)/(1 + h[d + (1 − u₁ⁱ)βV_i]),

I_i+1 =(I_i + h(1 − u₁ⁱ)βV_iT_i+1)/(1 + hδ),

V_i+1 =(V_i + h(1 − u₂ⁱ)pI_i+1)/(1 + hc),

λ_{1ⁿ⁻ⁱ⁻¹} =(λ₁ⁿ⁻ⁱ + h[1 + (1 − u₁ⁱ)βVi+1])/(1 + h[d + (1 − u₁ⁱ)βV_i+1]),

λ₂ⁿ⁻ⁱ⁻¹ =(λ₂ⁿ⁻ⁱ+ hλ₃ⁿ⁻ⁱ (1 − u₂ⁱ)p)/(1 + hδ),

λ₃ⁿ⁻ⁱ⁻¹ =(λ₃ⁿ⁻ⁱ + h(λ₂ⁿ⁻ⁱ⁻¹− λ₁ⁿ⁻ⁱ⁻¹ )(1 − u₁ⁱ)βT_i+1)/(1 + hc),

R1ⁱ⁺¹ =(1/A1)(λ_{1ⁿ⁻ⁱ⁻¹}−λ_{2ⁿ⁻ⁱ⁻¹} )βV_i+1T_i+1,

R2ⁱ⁺¹ =−(1/A2)λ_{3ⁿ⁻ⁱ⁻¹} pI_i+1,

u_1ⁱ⁺¹ = min(1, max(R1ⁱ⁺¹ , 0)),

u_2ⁱ⁺¹ = min(1, max(R2ⁱ⁺¹ , 0)),

end for

step 3 :
for i=1, ..., n-1, write
T^∗(t_i) = T_i, I^∗(t_i) = I_i, V ^∗(t_i) = V_i,

u_1^∗(t_i) = u_1ⁱ, u2^∗(t_i) = u_2ⁱ.

end for

hello. before I used Mapple 15. But then I`ve run Mapple 16 and now I`ve a problem. I can`t use this program. I open the program, everthing is in the rule, but if I want to write any mathemathical function, or a letter, such as- x or x+2, the program does`t give any reaction. program only gives reaction the numbers.

Please, help me. (my english isn`t very good, and I don`t know I`ve explained my opinion).

Say I have a polynomial x^5 + 4xy^4 + 2y^3 +  x*y^2 + x^2 + y + 3

Can I truncate it up to total degree 3 (for example), so 2y^3 +  x*y^2 + x^2 + y + 3

Greetings!

I only recently started to work with Maple17 and tried to test the Explore command. In Classic worksheet I tried a very simple function with a parameter, but after setting the inteval of the parameter in the Java pop-up window I have no result. In return I take: "Error, (in Explore) invalid input: rtable_dims uses a 1st argument, A (of type rtable), which is missing"

Has anyone encounter the same issue before?

Thanx

a = c1*b + c2*c;
b = c3*a + c4*c;

where a, b, c are vector, a is linear combination of b and c, b is linear combination of a and c, and c1^2 + c2^2 = 1, c3^2 + c4^2 = 1

assume equation for a = c1*b + c2*c; is
Matrix([[y1],[y2]]) = Matrix([[x1,x2],[x3,x4]])*Matrix([[c1],[c2]]);

how to find c1 and c2 when c1^2 + c2^2 = 1

nothing return after solve({y1 = c1*x1 + c2*x2, y2 = c1*x3 + c2*x4, c1^2 + c2^2 = 1}, {c1,c2});

https://drive.google.com/file/d/0B2D69u2pweEvMV92SGhtRGZONFk/edit?usp=sharing

a error and code in this attachment mw

i can pdsolve it, but numeric pdsolve it get error