Maple Questions and Posts

These are Posts and Questions associated with the product, Maple

The video shows the curvilinear components of acceleration in polar coordinates, radial and tangential scalar components. Applied to a structure; in a time interval; to finally be graphed and interpreted. For engineering students.

Speed_and_Acceleration_with_Cylindrical_Components.mw

Lenin Araujo Castillo

Ambassador of Maple

I have a polynomial in the variables ya[i] and yd[i] where i are integers. I want to divide each of the coefficients by the 'shortest' coeficient. What i mean by that is the coefficient that is going to cause me the least trouble when i later do things with groebner bases of on the coefficients - I expect a good proxy for that is the one that has the smallest number of terms.

For example, for the polynomial:

2*yd[0]*k[a1]*k[d1]*ya[1]+(alpha*C[T]*k[a1]*k[m]-alpha*R[b]*k[a1]*k[d1]-alpha*R[m]*k[a1]*k[d1]-alpha*k[d1]*k[m])*ya[1]-2*k[a1]*k[d1]*yd[1]*yd[0]+(-alpha*C[T]*k[a1]*k[m]+alpha*R[b]*k[a1]*k[d1]+alpha*R[m]*k[a1]*k[d1]+alpha*k[d1]*k[m])*yd[1]

2*k[a1]*k[d1] is the shortest monomial coefficient

 

I have a program that produces lists of polynomails in multiple variables; I want to remove any polynomials that have the variable x[i] where i is a number.

An example list would be:
[
y[a0]-y[d0],
k[d1]*y[a1]-k[d2]*y[d2],
k[d1]*y[a1]*x[1]-k[d2]*y[d2]*x[2],
]
 

int(int(x,y^4..16),y=0..2);


yields the output $\int_0^2\int_{y^4}^{16} x(x) dx dy$.

 

any one can help me to get the output of the plot in 3 rows and 2 columns here is my codes. thanks in advance.

 restart;
  h:=z->1-(delta2/2)*(1 + cos(2*(Pi/L1)*(z - d1 - L1))):
  K1:=((4/h(z)^4)-(sin(alpha)/Gamma2)-h(z)^2+Nb*h(z)^4):
  lambda:=unapply(Int(K1,z=0..1), Gamma2):
  L1:=0.2:
  d1:=0.2:
  alpha:=Pi/6:
  with(plots):
  display
  ( Vector[row]
    ( [ seq
        ( plot
          ( [ seq
              ( eval(lambda(Gamma2), Nb=j),
                j in [0.1,0.2,0.3]
              )
            ],
            delta2=0.02..0.1,
            legend=[Nb=0.1,Nb=0.2,Nb=0.3],
            labels=[typeset(`δ1`), typeset(conjugate(`Δp`))],
            title=typeset("Effect of ", ''alpha'', " when ", Gamma,"2=", Gamma2)
          ),
          Gamma2 in [10,20,30,40,50,-10]
        )
      ]
    )
  );
 

After run a batch to cmaple to run a prettyprint=0 and screenwidth=500

it use lprint in window

i set prettyprint=2 

still can not set back to original print for matrix

Excel tool import can not be used in cmaple.exe

Plot the first 20 Fibonacci numbers.

I have this so far..

restart;

nums := [seq(i, i = 1 .. 20)]*with(combinat, fibonacci);

fibnums;

 

  here my loop; after  8 iteration maple couldnt solve the equations and give me this error .
Is there any method to garentee that fsolve could work intire the 1000 iteration 
 

 
 
 

Download exp_new_for_alpha_more_than_22.mw
 

with(LinearAlgebra):

f[1] := VectorCalculus:-`+`(VectorCalculus:-`+`(VectorCalculus:-`*`(n, 1/R), sum(x[i], i = 1 .. n)), VectorCalculus:-`-`(sum(VectorCalculus:-`*`(VectorCalculus:-`*`(VectorCalculus:-`*`(VectorCalculus:-`+`(2, a[i]), x[i]), exp(VectorCalculus:-`*`(R, x[i]))), 1/VectorCalculus:-`+`(VectorCalculus:-`+`(exp(VectorCalculus:-`*`(R, x[i])), -1), Q)), i = 1 .. n))):

f[2] := VectorCalculus:-`+`(VectorCalculus:-`+`(VectorCalculus:-`*`(m, 1/S), sum(y[j], j = 1 .. m)), VectorCalculus:-`-`(sum(VectorCalculus:-`*`(VectorCalculus:-`*`(VectorCalculus:-`*`(VectorCalculus:-`+`(2, b[j]), y[j]), exp(VectorCalculus:-`*`(y[j], S))), 1/VectorCalculus:-`+`(VectorCalculus:-`+`(exp(VectorCalculus:-`*`(y[j], S)), -1), Q)), j = 1 .. m))):

f[3] := VectorCalculus:-`+`(VectorCalculus:-`+`(VectorCalculus:-`*`(VectorCalculus:-`+`(VectorCalculus:-`+`(VectorCalculus:-`+`(n, m), sum(a[i], i = 1 .. n)), sum(b[j], j = 1 .. m)), 1/Q), VectorCalculus:-`-`(sum(VectorCalculus:-`*`(VectorCalculus:-`+`(2, a[i]), 1/VectorCalculus:-`+`(VectorCalculus:-`+`(exp(VectorCalculus:-`*`(R, x[i])), -1), Q)), i = 1 .. n))), VectorCalculus:-`-`(sum(VectorCalculus:-`*`(VectorCalculus:-`+`(2, b[j]), 1/VectorCalculus:-`+`(VectorCalculus:-`+`(exp(VectorCalculus:-`*`(y[j], S)), -1), Q)), j = 1 .. m))):

NULL

E1[1] := 0.5e-1:

E2[1] := 0.5e-1:

E3[1] := 0.5e-1:

n := 45:

n := 45:

a := [seq(0, i = 1 .. 21), 2, 2, 1, seq(0, i = 1 .. 21)]:

NULL

K := 1000:

for so from 0 to K do W := GenerateUniform(n, 0, 1); for iii to n do vv[iii] := W[iii]^(1/(iii+sum(a[jjj], jjj = n-iii+1 .. n))) end do; for sss to n do uu[sss] := 1-product(vv[n-jjj+1], jjj = 1 .. sss); x[sss] := fsolve(1-3/(exp(.3*t)-(1-3)) = uu[sss], t = 0 .. infinity) end do; U := GenerateUniform(m, 0, 1); for ii to m do v[ii] := U[ii]^(1/(ii+sum(b[jj], jj = m-ii+1 .. m))) end do; for ss to m do u[ss] := 1-product(v[m-jj+1], jj = 1 .. ss); y[ss] := fsolve(1-3/(exp(.1*t)-(1-3)) = u[ss], t = 0 .. infinity) end do; c := describe[quartile[1]]([seq(x[i], i = 1 .. n)]); cc := describe[quartile[3]]([seq(x[i], i = 1 .. n)]); L := describe[quartile[1]]([seq(y[i], i = 1 .. m)]); LL := describe[quartile[3]]([seq(y[i], i = 1 .. m)]); R[1] := fsolve(9*exp(R*c)-exp(R*cc) = 8, R = 0 .. infinity); S[1] := fsolve(9*exp(S*L)-exp(S*LL) = 8, S = 0 .. infinity); Q[1] := 3*(exp(R[1]*c)-1+(exp(S[1]*L)-1))*(1/2); for h to 40 while `and`(`and`(`and`(`and`(`and`(abs(E1[h]) > 0.5e-3, abs(E2[h]) > 0.5e-3), abs(E3[h]) > 0.5e-3), Q[h] > 2), S[h] > 0), R[h] > 0) do Q[h+1] := fsolve(eval(f[3], {R = R[h], S = S[h]}) = 0, Q = 2 .. infinity); R[h+1] := fsolve(eval(f[1], Q = Q[h+1]) = 0, R = 0 .. infinity); S[h+1] := fsolve(eval(f[2], Q = Q[h+1]) = 0, S = 0 .. infinity); KK := Matrix([[R[h]], [S[h]], [Q[h]]]); E1[h+1] := abs(R[h+1]-R[h]); E2[h+1] := abs(S[h+1]-S[h]); E3[h+1] := abs(Q[h+1]-Q[h]) end do; A[so] := Determinant(KK[1]); B[so] := Determinant(KK[2]); C[so] := Determinant(KK[3]); P[so] := simplify(int(A[so]*C[so]^2*exp(A[so]*x)/((exp(A[so]*x)-1+C[so])^2*(exp(B[so]*x)-1+C[so])), x = 0 .. infinity, numeric)) end do

W := Vector(4, {(1) = ` 1 .. 45 `*Array, (2) = `Data Type: `*float[8], (3) = `Storage: `*rectangular, (4) = `Order: `*C_order})

 

U := Vector(4, {(1) = ` 1 .. 45 `*Array, (2) = `Data Type: `*float[8], (3) = `Storage: `*rectangular, (4) = `Order: `*C_order})

 

c := 2.251942600

 

cc := 6.413093396

 

L := 8.631577783

 

LL := 25.39584518

 

R[1] := .4287243564

 

S[1] := .1043333848

 

Q[1] := 4.630481096

 

A[0] := .4247642181

 

B[0] := .1149899971

 

C[0] := 6.627593396

 

P[0] := .8815279215

 

W := Vector(4, {(1) = ` 1 .. 45 `*Array, (2) = `Data Type: `*float[8], (3) = `Storage: `*rectangular, (4) = `Order: `*C_order})

 

U := Vector(4, {(1) = ` 1 .. 45 `*Array, (2) = `Data Type: `*float[8], (3) = `Storage: `*rectangular, (4) = `Order: `*C_order})

 

c := 2.918917328

 

cc := 5.547621812

 

L := 3.857225847

 

LL := 21.10240063

 

R[1] := .8018219086

 

S[1] := 0.5213484487e-1

 

Q[1] := 14.41300577

 

A[1] := .3666457947

 

B[1] := .1191082759

 

C[1] := 3.847329446

 

P[1] := .8226338823

 

W := Vector(4, {(1) = ` 1 .. 45 `*Array, (2) = `Data Type: `*float[8], (3) = `Storage: `*rectangular, (4) = `Order: `*C_order})

 

U := Vector(4, {(1) = ` 1 .. 45 `*Array, (2) = `Data Type: `*float[8], (3) = `Storage: `*rectangular, (4) = `Order: `*C_order})

 

c := 2.442365249

 

cc := 7.394009487

 

L := 4.713824874

 

LL := 24.79260797

 

R[1] := .3468711931

 

S[1] := 0.4792653690e-1

 

Q[1] := 2.379817029

 

A[2] := .2337020019

 

B[2] := 0.7824619488e-1

 

C[2] := 2.252708122

 

P[2] := .7880876611

 

W := Vector(4, {(1) = ` 1 .. 45 `*Array, (2) = `Data Type: `*float[8], (3) = `Storage: `*rectangular, (4) = `Order: `*C_order})

 

U := Vector(4, {(1) = ` 1 .. 45 `*Array, (2) = `Data Type: `*float[8], (3) = `Storage: `*rectangular, (4) = `Order: `*C_order})

 

c := 1.950620121

 

cc := 7.490968154

 

L := 7.989340649

 

LL := 22.40840798

 

R[1] := .2570696142

 

S[1] := .1248268277

 

Q[1] := 3.543022180

 

A[3] := .3254617069

 

B[3] := .1177911768

 

C[3] := 4.708933240

 

P[3] := .8124474245

 

W := Vector(4, {(1) = ` 1 .. 45 `*Array, (2) = `Data Type: `*float[8], (3) = `Storage: `*rectangular, (4) = `Order: `*C_order})

 

U := Vector(4, {(1) = ` 1 .. 45 `*Array, (2) = `Data Type: `*float[8], (3) = `Storage: `*rectangular, (4) = `Order: `*C_order})

 

c := 1.973241013

 

cc := 7.005418646

 

L := 6.495611086

 

LL := 22.94839275

 

R[1] := .3034319432

 

S[1] := 0.9318350072e-1

 

Q[1] := 2.477406387

 

A[4] := .3446953632

 

B[4] := .1065224704

 

C[4] := 4.241185270

 

P[4] := .8370643415

 

W := Vector(4, {(1) = ` 1 .. 45 `*Array, (2) = `Data Type: `*float[8], (3) = `Storage: `*rectangular, (4) = `Order: `*C_order})

 

U := Vector(4, {(1) = ` 1 .. 45 `*Array, (2) = `Data Type: `*float[8], (3) = `Storage: `*rectangular, (4) = `Order: `*C_order})

 

c := 1.839457984

 

cc := 7.186959772

 

L := 6.911480924

 

LL := 24.15316459

 

R[1] := .2619901249

 

S[1] := 0.8971740871e-1

 

Q[1] := 2.217385534

 

A[5] := .2889717346

 

B[5] := 0.8838227764e-1

 

C[5] := 2.965514897

 

P[5] := .8196572233

 

W := Vector(4, {(1) = ` 1 .. 45 `*Array, (2) = `Data Type: `*float[8], (3) = `Storage: `*rectangular, (4) = `Order: `*C_order})

 

U := Vector(4, {(1) = ` 1 .. 45 `*Array, (2) = `Data Type: `*float[8], (3) = `Storage: `*rectangular, (4) = `Order: `*C_order})

 

c := 2.075842156

 

cc := 6.451509594

 

L := 7.355551514

 

LL := 22.57154486

 

R[1] := .3857437868

 

S[1] := .1118795995

 

Q[1] := 3.756557664

 

A[6] := .3672067772

 

B[6] := .1169243938

 

C[6] := 4.269406267

 

P[6] := .8320277948

 

W := Vector(4, {(1) = ` 1 .. 45 `*Array, (2) = `Data Type: `*float[8], (3) = `Storage: `*rectangular, (4) = `Order: `*C_order})

 

U := Vector(4, {(1) = ` 1 .. 45 `*Array, (2) = `Data Type: `*float[8], (3) = `Storage: `*rectangular, (4) = `Order: `*C_order})

 

c := 2.052084434

 

cc := 5.954773664

 

L := 7.287571569

 

LL := 19.13650694

 

R[1] := .4519986760

 

S[1] := .1574582015

 

Q[1] := 5.517824640

 

A[7] := .3834726657

 

B[7] := .1300598692

 

C[7] := 4.443377611

 

P[7] := .8222904412

 

W := Vector(4, {(1) = ` 1 .. 45 `*Array, (2) = `Data Type: `*float[8], (3) = `Storage: `*rectangular, (4) = `Order: `*C_order})

 

U := Vector(4, {(1) = ` 1 .. 45 `*Array, (2) = `Data Type: `*float[8], (3) = `Storage: `*rectangular, (4) = `Order: `*C_order})

 

1.288903514

 

6.337860209

 

6.623173031

 

20.57453160

 

.2114575385

 

.1210220497

 

2.313450170

 

Error, (in fsolve) Q is in the equation, and is not solved for

 

a := [seq(A[i], i = 1 .. 1000)]:

for i to 1000 do aa_[i] := `if`(0 < P[i] and P[i] < 1, a[i], 0); bb_[i] := `if`(0 < P[i] and P[i] < 1, b[i], 0); cc_[i] := `if`(0 < P[i] and P[i] < 1, c[i], 0); gg_[i] := `if`(0 < P[i] and P[i] < 1, p[i], 0) end do:

NULL

Tau := [seq(aa_[i], i = 1 .. 1000)]:

rr := [seq(`if`(Tau[i] = 0, NULL, i), i = 1 .. 1000)]:

r := Tau[rr]:

1000

 

1000

 

1000

 

1000

(1)

lambda[1] := Mean([seq(r[i], i = 1 .. nops(r))]); lambda[2] := Mean([seq(s[i], i = 1 .. nops(s))]); alpha := Mean([seq(q[i], i = 1 .. nops(q))]); Pro := Mean([seq(w[i], i = 1 .. nops(w))]); Bi_ := .647737-Pro; ME_ := Bi_^2

Error, (in Statistics:-Mean) unable to evaluate `if`(0 < P[8] and P[8] < 1, a[i], 0) to floating-point

 

Error, (in Statistics:-Mean) unable to evaluate `if`(0 < P[8] and P[8] < 1, b[i], 0) to floating-point

 

Error, (in Statistics:-Mean) unable to evaluate `if`(0 < P[8] and P[8] < 1, c[i], 0) to floating-point

 

Error, (in Statistics:-Mean) unable to evaluate `if`(0 < P[8] and P[8] < 1, p[i], 0) to floating-point

 

.647737-Pro

 

(.647737-Pro)^2

(2)

NULL


 

Download exp_new_for_alpha_more_than_22.mw

 

I am calling a function (GTS2) multiple times with varying inputs, using the curry function, and i want to record how long/how much RAM the function takes with each input, and put those in seperate matrices that i can plot later
 

Sols3 := proc (H::algebraic, F::(list(algebraic)), i::posint, j::posint) options operator, arrow; GTS2(H, F, i, j) end proc;
n, m := 5, 4;
M=Matrix(n, m, curry(Sols3, H, F))


You can find all the functions required in this worksheet. The curried call to this function is in section 4.

MHD_cchf_2.mw
 

NULL

NULL

NULL

NULL

w := .572433:

NULL

for j to nops(N, m) do sol1 := dsolve([diff(diff(diff(f(eta), eta), eta), eta)+w*x(f(eta)*(diff(diff(f(eta), eta), eta))-(m[j]*m[j])*(diff(f(eta), eta))-(diff(f(eta), eta))^2) = 0, y*(diff(diff(theta(eta), eta), eta))/(pr*z)-b*f(eta)*(diff(f(eta), eta))*(diff(theta(eta), eta))-b*f(eta)^2*(diff(diff(theta(eta), eta), eta))+f(eta)*(diff(theta(eta), eta)) = 0, f(0) = N[j], (D(f))(0) = 1, (D(f))(20) = 0, theta(0) = 1, theta(20) = 0], numeric, method = bvp); plots[odeplot](sol1, [eta, ((D@@2)(f))(eta)], color = red); plots[odeplot](sol1, color = red); plots[odeplot](sol1, [eta, theta(eta)], color = K[j], linestyle = L[j]); fplt[j] := plots[odeplot](sol1, [eta, f(eta)], color = K[j], axes = boxed, linestyle = L[j]); tplt[j] := plots[odeplot](sol1, [[eta, theta(eta)]], color = K[j], axes = box, linestyle = L[j]) end do; plots:-display([seq(fplt[j], j = 1 .. nops(N, m))]); plots:-display([seq(tplt[j], j = 1 .. nops(N, m))])

Error, invalid input: nops expects 1 argument, but received 2

 

Error, invalid input: nops expects 1 argument, but received 2

 

Error, invalid input: nops expects 1 argument, but received 2

 

``

``


 

Download MHD_cchf_2.mw

 

Respected sir, I try to plot graphs using two parameters once. But it showing the error as

Error, invalid input: nops expects 1 argument, but received 2
Error, invalid input: nops expects 1 argument, but received 2
Error, invalid input: nops expects 1 argument, but received 2

can anybody do help in this regard?

I have a problem writing a program for the numerical solution of nonlinear volterra integral equation using the method of reproducing kernel space. I have my algorithm as well as the program I tried to write, though they are full of error messages. Please could anyone give me a clue on how to go about my challenges. The algorithm is as follows:

Step 1. Fix π‘Ž ≤ π‘₯ and 𝑑 ≤ 𝑏.
If 𝑑 ≤ π‘₯, set 𝑅π‘₯(𝑑) = 1 − π‘Ž + 𝑑.
Else set 𝑅π‘₯(𝑑) = 1 − π‘Ž + π‘₯.
Step 2. For 𝑖 = 1, 2, . . . , π‘š set π‘₯i = (𝑖 − 1)/(π‘š − 1).

Set πœ“i(π‘₯) = 𝐿t𝑅π‘₯(𝑑)|𝑑=π‘₯i .
Step 3. Set 𝑒0(π‘₯1) = 𝑒(π‘₯1).
Step 4. For 𝑖 = 1, 2, . . . , π‘š set 𝛾ij = [πœ“-1]ij.
Step 5. 𝑛 = 1.
Step 6. Set Sn = Σ𝑛
π‘˜=1 𝛾nk𝑒k-1(π‘₯k).
Step 7. Set 𝑒n(π‘₯) = Σ𝑛
𝑖=1 Siπœ“i(π‘₯).
Step 8. If 𝑛 < π‘šthen set 𝑛 = 𝑛 + 1 and go to step 6.
Else stop.

how can i plot outside of an sphere? for example x^2+y^2+z^2>1 ? tnx for help

So I have this system of equations with which I am not sure if the result is the same or not using "series" and "limit" or what is going on here.

I hope it is clear what I mean.


 

restart; with(MathematicalFunctions); Assume(k__2H2O > 0, `k__HA+OH` > 0, `k__A+H2O` > 0, `k__H3O+OH` > 0, `k__HA+H2O` > 0, `k__H3O+A` > 0, HA__0 > 0, H2O > 0); sys := k__2H2O*H2O^2+`k__A+H2O`*H2O*(HA__0-HA)-(H3O*`k__H3O+OH`+HA*`k__HA+OH`)*OH = 0, k__2H2O*H2O^2+`k__HA+H2O`*H2O*HA-(`k__H3O+A`*(HA__0-HA)+`k__H3O+OH`*OH)*H3O = 0, (H2O*`k__HA+H2O`+OH*`k__HA+OH`)*HA-(H2O*`k__A+H2O`+H3O*`k__H3O+A`)*(HA__0-HA) = 0; sys := `~`[simplify]([eval(eval(sys, HA = HA__0+OH-H3O), HA__0 = x__HA0*H2O)]); sol := solve(sys, [OH, H3O]); sol := sol[1]; OH__sol := simplify(rhs(sol[1])); H3O__sol := simplify(rhs(sol[2])); simplify(OH__sol*H3O__sol); OHH3O := simplify(limit(%, `k__HA+OH` = 0)); series(OHH3O, x__HA0 = 0, 2); collect(convert(%, polynom), x__HA0, simplify, factor); r1 := limit(%, x__HA0 = 0); r2 := radnormal(limit(OHH3O, x__HA0 = 0)); simplify(r1-r2)

[`&Intersect`, `&Minus`, `&Union`, Assume, Coulditbe, Evalf, Get, Is, SearchFunction, Sequences, Series]

 

{H2O::(RealRange(Open(0), infinity))}, {HA__0::(RealRange(Open(0), infinity))}, {k__2H2O::(RealRange(Open(0), infinity))}, {`k__A+H2O`::(RealRange(Open(0), infinity))}, {`k__H3O+A`::(RealRange(Open(0), infinity))}, {`k__H3O+OH`::(RealRange(Open(0), infinity))}, {`k__HA+H2O`::(RealRange(Open(0), infinity))}, {`k__HA+OH`::(RealRange(Open(0), infinity))}

 

k__2H2O*H2O^2+`k__A+H2O`*H2O*(HA__0-HA)-(H3O*`k__H3O+OH`+HA*`k__HA+OH`)*OH = 0, k__2H2O*H2O^2+`k__HA+H2O`*H2O*HA-(`k__H3O+A`*(HA__0-HA)+`k__H3O+OH`*OH)*H3O = 0, (H2O*`k__HA+H2O`+OH*`k__HA+OH`)*HA-(H2O*`k__A+H2O`+H3O*`k__H3O+A`)*(HA__0-HA) = 0

 

[-OH^2*`k__HA+OH`+((-x__HA0*`k__HA+OH`-`k__A+H2O`)*H2O+H3O*(`k__HA+OH`-`k__H3O+OH`))*OH+k__2H2O*H2O^2+`k__A+H2O`*H2O*H3O = 0, (x__HA0*`k__HA+H2O`+k__2H2O)*H2O^2+`k__HA+H2O`*(OH-H3O)*H2O+(-`k__H3O+A`*H3O+OH*(`k__H3O+A`-`k__H3O+OH`))*H3O = 0, H2O^2*x__HA0*`k__HA+H2O`+((x__HA0*`k__HA+OH`+`k__A+H2O`+`k__HA+H2O`)*OH-H3O*(`k__A+H2O`+`k__HA+H2O`))*H2O+(OH-H3O)*(H3O*`k__H3O+A`+OH*`k__HA+OH`) = 0]

 

-RootOf(-x__HA0*`k__A+H2O`^2*`k__HA+H2O`+k__2H2O^2*`k__H3O+A`-k__2H2O*`k__A+H2O`^2-k__2H2O*`k__A+H2O`*`k__HA+H2O`+(2*x__HA0*`k__A+H2O`*`k__H3O+OH`*`k__HA+H2O`-k__2H2O*`k__A+H2O`*`k__H3O+A`+k__2H2O*`k__A+H2O`*`k__H3O+OH`+k__2H2O*`k__H3O+OH`*`k__HA+H2O`)*_Z+(-x__HA0*`k__H3O+OH`^2*`k__HA+H2O`-k__2H2O*`k__H3O+A`*`k__H3O+OH`+`k__A+H2O`^2*`k__H3O+OH`+`k__A+H2O`*`k__H3O+OH`*`k__HA+H2O`)*_Z^2+(`k__A+H2O`*`k__H3O+A`*`k__H3O+OH`-`k__A+H2O`*`k__H3O+OH`^2-`k__H3O+OH`^2*`k__HA+H2O`)*_Z^3)*H2O^2*(-`k__A+H2O`*RootOf(-x__HA0*`k__A+H2O`^2*`k__HA+H2O`+k__2H2O^2*`k__H3O+A`-k__2H2O*`k__A+H2O`^2-k__2H2O*`k__A+H2O`*`k__HA+H2O`+(2*x__HA0*`k__A+H2O`*`k__H3O+OH`*`k__HA+H2O`-k__2H2O*`k__A+H2O`*`k__H3O+A`+k__2H2O*`k__A+H2O`*`k__H3O+OH`+k__2H2O*`k__H3O+OH`*`k__HA+H2O`)*_Z+(-x__HA0*`k__H3O+OH`^2*`k__HA+H2O`-k__2H2O*`k__H3O+A`*`k__H3O+OH`+`k__A+H2O`^2*`k__H3O+OH`+`k__A+H2O`*`k__H3O+OH`*`k__HA+H2O`)*_Z^2+(`k__A+H2O`*`k__H3O+A`*`k__H3O+OH`-`k__A+H2O`*`k__H3O+OH`^2-`k__H3O+OH`^2*`k__HA+H2O`)*_Z^3)+k__2H2O)/(-`k__H3O+OH`*RootOf(-x__HA0*`k__A+H2O`^2*`k__HA+H2O`+k__2H2O^2*`k__H3O+A`-k__2H2O*`k__A+H2O`^2-k__2H2O*`k__A+H2O`*`k__HA+H2O`+(2*x__HA0*`k__A+H2O`*`k__H3O+OH`*`k__HA+H2O`-k__2H2O*`k__A+H2O`*`k__H3O+A`+k__2H2O*`k__A+H2O`*`k__H3O+OH`+k__2H2O*`k__H3O+OH`*`k__HA+H2O`)*_Z+(-x__HA0*`k__H3O+OH`^2*`k__HA+H2O`-k__2H2O*`k__H3O+A`*`k__H3O+OH`+`k__A+H2O`^2*`k__H3O+OH`+`k__A+H2O`*`k__H3O+OH`*`k__HA+H2O`)*_Z^2+(`k__A+H2O`*`k__H3O+A`*`k__H3O+OH`-`k__A+H2O`*`k__H3O+OH`^2-`k__H3O+OH`^2*`k__HA+H2O`)*_Z^3)+`k__A+H2O`)

 

-(k__2H2O*`k__H3O+A`^2-2*`k__A+H2O`^2*`k__H3O+A`+`k__A+H2O`^2*`k__H3O+OH`-2*`k__A+H2O`*`k__H3O+A`*`k__HA+H2O`+2*`k__A+H2O`*`k__H3O+OH`*`k__HA+H2O`+`k__H3O+OH`*`k__HA+H2O`^2)*`k__A+H2O`*`k__HA+H2O`*H2O^2*x__HA0/((`k__A+H2O`*`k__H3O+A`-`k__A+H2O`*`k__H3O+OH`-`k__H3O+OH`*`k__HA+H2O`)*(k__2H2O*`k__H3O+A`^2-`k__A+H2O`^2*`k__H3O+OH`-2*`k__A+H2O`*`k__H3O+OH`*`k__HA+H2O`-`k__H3O+OH`*`k__HA+H2O`^2))-(k__2H2O*`k__H3O+A`-`k__A+H2O`^2-`k__A+H2O`*`k__HA+H2O`)*H2O^2*(`k__A+H2O`+`k__HA+H2O`)/(`k__H3O+A`*(`k__A+H2O`*`k__H3O+A`-`k__A+H2O`*`k__H3O+OH`-`k__H3O+OH`*`k__HA+H2O`))

 

-(k__2H2O*`k__H3O+A`-`k__A+H2O`^2-`k__A+H2O`*`k__HA+H2O`)*H2O^2*(`k__A+H2O`+`k__HA+H2O`)/(`k__H3O+A`*(`k__A+H2O`*`k__H3O+A`-`k__A+H2O`*`k__H3O+OH`-`k__H3O+OH`*`k__HA+H2O`))

 

k__2H2O*H2O^2/`k__H3O+OH`

 

-`k__A+H2O`*(-(`k__A+H2O`+`k__HA+H2O`)^2*`k__H3O+OH`+k__2H2O*`k__H3O+A`^2)*H2O^2/(`k__H3O+OH`*`k__H3O+A`*((-`k__A+H2O`-`k__HA+H2O`)*`k__H3O+OH`+`k__A+H2O`*`k__H3O+A`))

(1)

``


 

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