Ramakrishnan

Ramakrishnan Vaidyanathan

374 Reputation

12 Badges

7 years, 74 days

Social Networks and Content at Maplesoft.com

With twenty years of Industrial experience and twenty years of teaching experience, I am now as retired Professor, using Maple to teach mathematics subject for students studying X to XII standards. Published XII Mathematics books.

MaplePrimes Activity


These are answers submitted by Ramakrishnan


a := 7: 

3*a*4 = 84NULL

Now insert after = sign with Text mode C Text mode in the context tool bar [Or simply go next to equal to sign andyou can see these modes appear.Type now whatever you want.

3*a*4 = a.3.4=84

``

Other way is use substitution in the command itself as below.

restart

4*((a = 7)*3) = 12*a = 84``

``


Download how_do_i_calculate_like_this.mw

You have to type after getting the answer. or use substitution in the command line itself.

(a=7)*3*4 and press (control plus =)

Hope it satisfies you. Cheers. Ramakrishnan v


``

 

``

The given equation with unknowns is

phi := sin((1/4)*Pi*x)*(c__1*cosh((1/4)*Pi*y)+c__3*y*cosh(Pi/(4*y))+c__4*y*sinh((1/4)*Pi*y)) = sin((1/4)*Pi*x)*(c__1*cosh((1/4)*Pi*y)+c__3*y*cosh((1/4)*Pi/y)+c__4*y*sinh((1/4)*Pi*y))````

``

 

NULL

phi

sin((1/4)*Pi*x)*(c__1*cosh((1/4)*Pi*y)+c__3*y*cosh((1/4)*Pi/y)+c__4*y*sinh((1/4)*Pi*y))

(1)

"(->)"

sin((1/4)*Pi*x)*((1/4)*c__1*Pi*sinh((1/4)*Pi*y)+c__3*cosh((1/4)*Pi/y)-(1/4)*c__3*Pi*sinh((1/4)*Pi/y)/y+c__4*sinh((1/4)*Pi*y)+(1/4)*c__4*y*Pi*cosh((1/4)*Pi*y))

(2)

"(->)"

sin((1/4)*Pi*x)*((1/16)*c__1*Pi^2*cosh((1/4)*Pi*y)+(1/16)*c__3*Pi^2*cosh((1/4)*Pi/y)/y^3+(1/2)*c__4*Pi*cosh((1/4)*Pi*y)+(1/16)*c__4*y*Pi^2*sinh((1/4)*Pi*y))

(3)

``

Above Equn(3) is second order differentiation with y

 

 

``

phi

sin((1/4)*Pi*x)*(c__1*cosh((1/4)*Pi*y)+c__3*y*cosh((1/4)*Pi/y)+c__4*y*sinh((1/4)*Pi*y))

(4)

"(->)"

(1/4)*Pi*cos((1/4)*Pi*x)*(c__1*cosh((1/4)*Pi*y)+c__3*y*cosh((1/4)*Pi/y)+c__4*y*sinh((1/4)*Pi*y))

(5)

"(->)"

-(1/16)*Pi^2*sin((1/4)*Pi*x)*(c__1*cosh((1/4)*Pi*y)+c__3*y*cosh((1/4)*Pi/y)+c__4*y*sinh((1/4)*Pi*y))

(6)

Above Equn (6) Second order differentiation with x

 

``

phi

sin((1/4)*Pi*x)*(c__1*cosh((1/4)*Pi*y)+c__3*y*cosh((1/4)*Pi/y)+c__4*y*sinh((1/4)*Pi*y))

(7)

"(->)"

(1/4)*Pi*cos((1/4)*Pi*x)*(c__1*cosh((1/4)*Pi*y)+c__3*y*cosh((1/4)*Pi/y)+c__4*y*sinh((1/4)*Pi*y))

(8)

"(->)"

(1/4)*Pi*cos((1/4)*Pi*x)*((1/4)*c__1*Pi*sinh((1/4)*Pi*y)+c__3*cosh((1/4)*Pi/y)-(1/4)*c__3*Pi*sinh((1/4)*Pi/y)/y+c__4*sinh((1/4)*Pi*y)+(1/4)*c__4*y*Pi*cosh((1/4)*Pi*y))

(9)

Above Equn (9) Second order differentiation with xy

Equn 3, 6 and 9 are 2nd differential wrt to y, y and xy.

``

At x = 0, the condition is

sin((1/4)*Pi*x)*((1/16)*c__1*Pi^2*cosh((1/4)*Pi*y)+(1/16)*c__3*Pi^2*cosh((1/4)*Pi/y)/y^3+(1/2)*c__4*Pi*cosh((1/4)*Pi*y)+(1/16)*c__4*y*Pi^2*sinh((1/4)*Pi*y)) = 0

sin((1/4)*Pi*x)*((1/16)*c__1*Pi^2*cosh((1/4)*Pi*y)+(1/16)*c__3*Pi^2*cosh((1/4)*Pi/y)/y^3+(1/2)*c__4*Pi*cosh((1/4)*Pi*y)+(1/16)*c__4*y*Pi^2*sinh((1/4)*Pi*y)) = 0

(10)

"(->)"

0 = 0

(11)

``

``

At x = 20, the condition is

sin((1/4)*Pi*x)*((1/16)*c__1*Pi^2*cosh((1/4)*Pi*y)+(1/16)*c__3*Pi^2*cosh((1/4)*Pi/y)/y^3+(1/2)*c__4*Pi*cosh((1/4)*Pi*y)+(1/16)*c__4*y*Pi^2*sinh((1/4)*Pi*y)) = 0

sin((1/4)*Pi*x)*((1/16)*c__1*Pi^2*cosh((1/4)*Pi*y)+(1/16)*c__3*Pi^2*cosh((1/4)*Pi/y)/y^3+(1/2)*c__4*Pi*cosh((1/4)*Pi*y)+(1/16)*c__4*y*Pi^2*sinh((1/4)*Pi*y)) = 0

(12)

"(->)"

0 = 0

(13)

``

``

At y = 9, the condition is

-(1/16)*Pi^2*sin((1/4)*Pi*x)*(c__1*cosh((1/4)*Pi*y)+c__3*y*cosh((1/4)*Pi/y)+c__4*y*sinh((1/4)*Pi*y)) = 0

-(1/16)*Pi^2*sin((1/4)*Pi*x)*(c__1*cosh((1/4)*Pi*y)+c__3*y*cosh((1/4)*Pi/y)+c__4*y*sinh((1/4)*Pi*y)) = 0

(14)

"(->)"

-(1/16)*Pi^2*sin((1/4)*Pi*x)*(c__1*cosh((9/4)*Pi)+9*c__3*cosh((1/36)*Pi)+9*c__4*sinh((9/4)*Pi)) = 0

(15)

"(->)"

-.61686*sin(.78540*x)*(587.25*c__1+9.0342*c__3+5285.2*c__4) = 0.

(16)

``

``

At y =-9, the condition is

-(1/16)*Pi^2*sin((1/4)*Pi*x)*(c__1*cosh((1/4)*Pi*y)+c__3*y*cosh((1/4)*Pi/y)+c__4*y*sinh((1/4)*Pi*y)) = arcsin((1/4)*Pi*x)

-(1/16)*Pi^2*sin((1/4)*Pi*x)*(c__1*cosh((1/4)*Pi*y)+c__3*y*cosh((1/4)*Pi/y)+c__4*y*sinh((1/4)*Pi*y)) = arcsin((1/4)*Pi*x)

(17)

"(->)"

-(1/16)*Pi^2*sin((1/4)*Pi*x)*(c__1*cosh((9/4)*Pi)-9*c__3*cosh((1/36)*Pi)+9*c__4*sinh((9/4)*Pi)) = arcsin((1/4)*Pi*x)

(18)

"(->)"

-.61686*sin(.78540*x)*(587.25*c__1-9.0342*c__3+5285.2*c__4) = arcsin(.78540*x)

(19)

``

At y = 9, the condition is

(1/4)*Pi*cos((1/4)*Pi*x)*((1/4)*c__1*Pi*sinh((1/4)*Pi*y)+c__3*cosh((1/4)*Pi/y)-(1/4)*c__3*Pi*sinh((1/4)*Pi/y)/y+c__4*sinh((1/4)*Pi*y)+(1/4)*c__4*y*Pi*cosh((1/4)*Pi*y)) = 0

(1/4)*Pi*cos((1/4)*Pi*x)*((1/4)*c__1*Pi*sinh((1/4)*Pi*y)+c__3*cosh((1/4)*Pi/y)-(1/4)*c__3*Pi*sinh((1/4)*Pi/y)/y+c__4*sinh((1/4)*Pi*y)+(1/4)*c__4*y*Pi*cosh((1/4)*Pi*y)) = 0

(20)

"(->)"

(1/4)*Pi*cos((1/4)*Pi*x)*((1/4)*c__1*Pi*sinh((9/4)*Pi)+c__3*cosh((1/36)*Pi)-(1/36)*c__3*Pi*sinh((1/36)*Pi)+c__4*sinh((9/4)*Pi)+(9/4)*c__4*Pi*cosh((9/4)*Pi)) = 0

(21)

"(->)"

.78540*cos(.78540*x)*(461.22*c__1+.99617*c__3+4738.2*c__4) = 0.

(22)

``

``

At y = -9, the condition is

(1/4)*Pi*cos((1/4)*Pi*x)*((1/4)*c__1*Pi*sinh((1/4)*Pi*y)+c__3*cosh((1/4)*Pi/y)-(1/4)*c__3*Pi*sinh((1/4)*Pi/y)/y+c__4*sinh((1/4)*Pi*y)+(1/4)*c__4*y*Pi*cosh((1/4)*Pi*y)) = 0

(1/4)*Pi*cos((1/4)*Pi*x)*((1/4)*c__1*Pi*sinh((1/4)*Pi*y)+c__3*cosh((1/4)*Pi/y)-(1/4)*c__3*Pi*sinh((1/4)*Pi/y)/y+c__4*sinh((1/4)*Pi*y)+(1/4)*c__4*y*Pi*cosh((1/4)*Pi*y)) = 0

(23)

"(->)"

(1/4)*Pi*cos((1/4)*Pi*x)*(-(1/4)*c__1*Pi*sinh((9/4)*Pi)+c__3*cosh((1/36)*Pi)-(1/36)*c__3*Pi*sinh((1/36)*Pi)-c__4*sinh((9/4)*Pi)-(9/4)*c__4*Pi*cosh((9/4)*Pi)) = 0

(24)

"(->)"

.78540*cos(.78540*x)*(-461.22*c__1+.99617*c__3-4738.2*c__4) = 0.

(25)

-.61686*sin(.78540*x)*(587.25*c__1+9.0342*c__3+5285.2*c__4) = 0.

-.61686*sin(.78540*x)*(587.25*c__1+9.0342*c__3+5285.2*c__4) = 0.

(26)

-.61686*sin(.78540*x)*(587.25*c__1-9.0342*c__3+5285.2*c__4) = arcsin(.78540*x)

-.61686*sin(.78540*x)*(587.25*c__1-9.0342*c__3+5285.2*c__4) = arcsin(.78540*x)

(27)

.78540*cos(.78540*x)*(461.22*c__1+.99617*c__3+4738.2*c__4) = 0.

.78540*cos(.78540*x)*(461.22*c__1+.99617*c__3+4738.2*c__4) = 0.

(28)

 

But solving equations .78540*cos(.78540*x)*(-461.22*c__1+.99617*c__3-4738.2*c__4) = 0., -.61686*sin(.78540*x)*(587.25*c__1+9.0342*c__3+5285.2*c__4) = 0., -.61686*sin(.78540*x)*(587.25*c__1-9.0342*c__3+5285.2*c__4) = arcsin(.78540*x), .78540*cos(.78540*x)*(461.22*c__1+.99617*c__3+4738.2*c__4) = 0., c__1, c__2, c__3, c__4 is difficult for me.

Anyone can comment my work.

``


Download my_attempt_ets_you_6_equations_for_4_unknown.mw

my_attempt_ets_you_6_equations_for_4_unknown.mwMy attempt has brought 4 equations for 4 unknowns. But then solving them is not easy because of arcsin terms. Hope my attempt is useful to you and take you near the answer. Cheers. Ramakrishnan V


``

``

``   

``


In radian text box type the following command

Do(%degree = 3.14*%radian*(1/180)):


In degree text box type the following command

Do(%dradian = 180*%degree/(3.14)):

``


Download conversion_degree_to_radian_or_radian_to_degree.mwconversion_degree_to_radian_or_radian_to_degree.mw


Eq := diff(y(x), x, x) = -(x^2+1)*y(x)+k;

diff(diff(y(x), x), x) = -(x^2+1)*y(x)+k

(1)

``

sol1 := dsolve([diff(diff(y(x),x),x) = -(x^2+1)*y(x)+k, y(-1) = 0, y(1) = 0]);
 # returned NULL

 

sol1 := dsolve([diff(diff(y(x), x), x) = -(x^2+1)*y(x)+2, y(-1) = 0, y(1) = 0]);

``

sol1 := dsolve([diff(diff(y(x), x), x) = -2, y(-1) = 0, y(1) = 0]);

y(x) = -x^2+1

(2)

sol1 := dsolve([diff(diff(y(x), x), x) = -k, y(-1) = 0, y(1) = 0])

y(x) = -(1/2)*k*x^2+(1/2)*k

(3)

sol1 := dsolve([diff(diff(y(x), x), x) = y(x)+k, y(-1) = 0, y(1) = 0])

y(x) = exp(x)*k/(exp(-1)+exp(1))+exp(-x)*k/(exp(-1)+exp(1))-k

(4)

sol1 := dsolve([diff(diff(y(x), x), x) = x^2*y(x)+k, y(-1) = 0, y(1) = 0])

y(x) = x^(1/2)*BesselI(1/4, (1/2)*x^2)*_C2+x^(1/2)*BesselK(1/4, (1/2)*x^2)*(-BesselI(1/4, 1/2)*_C2/BesselK(1/4, 1/2)+(1/4)*k*(GAMMA(3/4)^2*hypergeom([1/2], [11/8, 3/2], 1/16)*2^(1/2)*BesselI(1/4, 1/2)*Pi+2*GAMMA(3/4)^2*hypergeom([1/2], [5/4, 3/2], 1/16)*BesselK(1/4, 1/2)-2*hypergeom([1/4], [3/4, 3/2], 1/16)*BesselI(1/4, 1/2)*Pi^2)/(GAMMA(3/4)*Pi*BesselK(1/4, 1/2)))-(1/2)*x^(3/2)*k*((1/2)*GAMMA(3/4)^2*hypergeom([1/2], [11/8, 3/2], (1/16)*x^4)*2^(1/2)*x*BesselI(1/4, (1/2)*x^2)*Pi+GAMMA(3/4)^2*hypergeom([1/2], [5/4, 3/2], (1/16)*x^4)*x*BesselK(1/4, (1/2)*x^2)-csgn(x)*hypergeom([1/4], [3/4, 3/2], (1/16)*x^4)*BesselI(1/4, (1/2)*x^2)*Pi^2)/(GAMMA(3/4)*Pi)

(5)

"->"

 

ol1 := dsolve([diff(diff(y(x), x), x) = (x^2+1)*y(x)+k, y(-1) = 0, y(1) = 0])

y(x) = exp((1/2)*x^2)*((1/4)*(Int(exp((1/2)*_z1^2)*erf(_z1), _z1 = -1 .. 1))*Pi^(1/2)*k+((1/4)*I)*erf(1)*erf(((1/2)*I)*2^(1/2))*2^(1/2)*Pi*k)+(1/4)*exp((1/2)*x^2)*erf(x)*(Int(exp((1/2)*_z1^2)*erf(_z1), _z1 = -1 .. 1))*Pi^(1/2)*k/erf(1)-(1/4)*exp((1/2)*x^2)*k*(I*Pi*2^(1/2)*erf(((1/2)*I)*2^(1/2)*x)*erf(x)+2*Pi^(1/2)*(Int(exp((1/2)*_z1^2)*erf(_z1), _z1 = -1 .. x)))

(6)

"->"

 

"->"

plots[implicitplot](y(x) = exp((1/2)*x^2)*((1/4)*(Int(exp((1/2)*_z1^2)*erf(_z1), _z1 = -1 .. 1))*Pi^(1/2)*k+((1/4)*I)*erf(1)*erf(((1/2)*I)*2^(1/2))*2^(1/2)*Pi*k)+(1/4)*exp((1/2)*x^2)*erf(x)*(Int(exp((1/2)*_z1^2)*erf(_z1), _z1 = -1 .. 1))*Pi^(1/2)*k/erf(1)-(1/4)*exp((1/2)*x^2)*k*(I*Pi*2^(1/2)*erf(((1/2)*I)*2^(1/2)*x)*erf(x)+2*Pi^(1/2)*(Int(exp((1/2)*_z1^2)*erf(_z1), _z1 = -1 .. x))), k = -5 .. 5, x = -5 .. 5, scaling = constrained)

(7)

``

``

`` 

 

``


Download Detailspossible.mw

I am sending details possible here.


sol1 := dsolve([diff(diff(y(x), x), x) = -(x^2+1)*y(x)+2, y(-1) = 0, y(1) = 0], numeric, method = bvp);
``

 

``


Download ODESolnSent14-11-2015byRVtMaplePrimeoCommunity.mw

Please find attached my answer; Give a value to K. I get the answer.

 


restart

resfresh

resfresh

(1)

With(Units[Standard])

With(Units:-Standard)

(2)

with(Units)

UsingSystem()

SI

(3)

UseSystem('FPS')

k := 2*Unit('poundforce'/'inch'^2)

2*Units:-Unit(('lbf')/('`in`')^2)

(4)

k1 := 2*Unit(Unit('poundal'))

2*Units:-Unit('poundal')

(5)

``

For unit conversion , all you need to do is just right cick the answer (blue color answer) and select unit formating and in the menu that pops up select Unit System as SI. You get the conversion automatically.

``

k := 2*Unit('poundforce'/'inch'^2)

(8896443230521/645160000)*Units:-Unit(Pa)

(6)

k1 := 2*Unit(Unit('poundal'))

(17281869297/62500000000)*Units:-Unit(N)

(7)

For numerical answer conversion , all you need to do is just right cick the answer (blue color answer) and select Numerical Formating and in the menu that pops up select Fixed, select decimal places as 2 and click 'apply'. You get the numerical conversion automatically.

k := 2*Unit('poundforce'/'inch'^2)

(8896443230521/645160000)*Units:-Unit(Pa)

(8)

k1 := 2*Unit(Unit('poundal'))

(17281869297/62500000000)*Units:-Unit(N)

(9)

Please note that lb is mass unit and therefore can not be converted to force unit Newton.

pondal is the force unit in FPS and hence convertible to Newton in SI unit.

 

Weight := Unit('poundal')

Units:-Unit('poundal')

(10)

Weight := Unit('poundal')

(17281869297/125000000000)*Units:-Unit(N)

(11)

One poundal is one pound mass acclerated by 1 ft/s2

Force := Unit('lb')*Unit(Unit('ft')/'s'^2)

Units:-Unit('lb')*Units:-Unit(('ft')/('s')^2)

(12)

Force := Unit('lb')*Unit(Unit('ft')/'s'^2)

(17281869297/125000000000)*Units:-Unit(N)

(13)

``

``

NULL

NULL


Download UnitConversion_from_FPS_to_SI_units_my_notes.mw

I attach herewith my document . Hope this clarifies the usage so simple.

Simply right click and and select the unit you need. It gives without any waste of second!

 


(diff(x^3, x, y))*y^4

0

(1)

diff(x^4, x, x)

12*x^2

(2)

diff(x^3, x, y)+y^4

y^4

(3)

diff(x^4, x, y)+y^3

y^3

(4)

diff(x^4+y^4, x, y)

0

(5)

(diff(x^5, x, y))*y^5

0

 

Explore((diff(x^5, x, y))*y^5, parameters = [[x = 0 .. 10, controller = slider], [y = 0 .. 10, controller = slider]], loop = never, numeric = false, size = NoUserValue)

(6)

diff(x/y, x, y)

-1/y^2

(7)

``


Download partial_differential_equation_(Fractional_differential_equation).mw

Go to calculas pallet. Select the d^2/dxdy f and modify the functionf as u require using the pallet expression.

I shall attach an example. Hope u can view this. The error code shown is because i did not use pallet expression.

The second one is correctly evaluated since i used expression pallet to define the function f as x/y.

Cheers. Ramakrishnan V


``

NULL

diff(a(t), t) = 20+0.3e-1*a(t)

diff(a(t), t) = 20+0.3e-1*a(t)

(1)

a(0) = 1

a(0) = 1

(2)

dsolve([diff(a(t), t) = 20+0.3e-1*a(t), a(0) = 1])

a(t) = -2000/3+(2003/3)*exp((3/100)*t)

(3)

``

``

diff(b(t), t) = 20+2*a(t)^2

diff(b(t), t) = 20+2*a(t)^2

(4)

b(1) = 1

b(1) = 1

(5)

dsolve([diff(a(t), t) = 20+0.3e-1*a(t), a(0) = 1, diff(b(t), t) = 20+2*a(t)^2, b(1) = 1])

{a(t) = -2000/3+(2003/3)*exp((3/100)*t), b(t) = (401200900/27)*(exp((3/100)*t))^2-(1602400000/27)*exp((3/100)*t)+(800000000/27)*ln(exp((3/100)*t))+20*t-(401200900/27)*(exp(3/100))^2+(1602400000/27)*exp(3/100)-8000171/9}

(6)

a := proc (t) options operator, arrow; -2000/3+(2000/3)*exp((3/100)*t) end proc

proc (t) options operator, arrow; -2000/3+(2000/3)*exp((3/100)*t) end proc

(7)

"->"

 

``

 

 

 

 

plot


Download AnswerToMaplePrimeQuestion2015Mar.mw

``

NULL

diff(a(t), t) = 20+0.3e-1*a(t)

diff(a(t), t) = 20+0.3e-1*a(t)

(1)

a(0) = 1

a(0) = 1

(2)

dsolve([diff(a(t), t) = 20+0.3e-1*a(t), a(0) = 1])

a(t) = -2000/3+(2003/3)*exp((3/100)*t)

(3)

``

``

diff(b(t), t) = 20+2*a(t)^2

diff(b(t), t) = 20+2*a(t)^2

(4)

b(1) = 1

b(1) = 1

(5)

dsolve([diff(a(t), t) = 20+0.3e-1*a(t), a(0) = 1, diff(b(t), t) = 20+2*a(t)^2, b(1) = 1])

{a(t) = -2000/3+(2003/3)*exp((3/100)*t), b(t) = (401200900/27)*(exp((3/100)*t))^2-(1602400000/27)*exp((3/100)*t)+(800000000/27)*ln(exp((3/100)*t))+20*t-(401200900/27)*(exp(3/100))^2+(1602400000/27)*exp(3/100)-8000171/9}

(6)

a := proc (t) options operator, arrow; -2000/3+(2000/3)*exp((3/100)*t) end proc

proc (t) options operator, arrow; -2000/3+(2000/3)*exp((3/100)*t) end proc

(7)

"->"

 

``

 

 

 

 

plot


Download AnswerToMaplePrimeQuestion2015Mar.mw

I hv answered first part. Solving two diff. equns. as an example. You can proceed in a similar way for any number of equations!

Pl. see uploaded worksheet.

Graph plotting is possible once a solution is obtained in an equation form. Any equation can be plotted with a simple plot command.

Ramakrishnan V

I ran your program in Maple18 and the error that comes is 

Error, unable to match delimiters

This error appears when there are unmatched delimiters within an expression. In Maple, the bracket delimiters (, ), [, ], {, }, <, and > work in pairs, and an error occurs if a delimiter is used and cannot be matched with its pair.

Hope in any way it helps you.

#return "foo"; 

The above statement needs to be

#return "f()";

it will work.



``

restart; with(plots); with(plottools); printf("Enter the number of players:\n"); numplayers := readline(terminal); printf("Number of players is %s\n", numplayers)

Enter the number of players:

 

Number of players is 6

 

``



Download das1404Answer.mw

solve(2*y-(x-1)^2=2,y) is in text format and hence invalid input.

Ensure that all the characters are in maths format ( maths 2D format)

solve(2*y-(x-1)^2 = 2, y);
print(`output redirected...`); # input placeholder
1 2 3
- x - x + -
2 2
solve(2*y-(x-1)^2-2 = 0, y);
print(`output redirected...`); # input placeholder
1 2 3
- x - x + -
2 2error_in_solving_Maple18.2SolutionByVR.mw

Solve*(2*x^2+6*x-3 = 9, x)

Solve*(2*x^2+6*x-3 = 9, x)

(1)

"(->)"

Error, (in Engine:-Dispatch) badly formed input to solve: not fully algebraic

 

2*x+3 = 0

2*x+3 = 0

(2)

"(->)"

[[x = -3/2]]

(3)

2*x^2+6*x = 0

2*x^2+6*x = 0

(4)

"(->)"

[[x = -3], [x = 0]]

(5)

2*x^2+6*x-3 = 0

2*x^2+6*x-3 = 0

(6)

"(->)"

[[x = -3/2+(1/2)*15^(1/2)], [x = -3/2-(1/2)*15^(1/2)]]

(7)

Solve*(2*x^2+6*x-3 = 0, x)

Solve*(2*x^2+6*x-3 = 0, x)

(8)

"(->)"

Error, (in Engine:-Dispatch) badly formed input to solve: not fully algebraic

 

Capital S not allowed

 

solve(x-2 = 0, x)

2

(9)

solve(2*x^2+6*x = 0, x)

-3, 0

(10)

solve(2*x^2+6*x-3 = 0, x)

-3/2+(1/2)*15^(1/2), -3/2-(1/2)*15^(1/2)

(11)

solve(2*y-(x-1)^2 = 2, y)

(1/2)*x^2-x+3/2

(12)

solve(2*y-(x-1)^2-2 = 0, y)

(1/2)*x^2-x+3/2

(13)

 

NULL


Download error_in_solving_Maple18.2SolutionByVR.mw

for i to 9 do if A(i) = 1 then A(k) := 100 end if end do

will work

Use i inside loop also

Assign not equation

A:= 0 and not A = 0

Hope this clarifies. I am also learner and hope to learn more from mistakes from peer group as well. Cheers.

 

 

 

 


All 3 procedures given below give answer plot.

filter := proc (x) options operator, arrow; if x < 10 then 1-(1/10)*x else 0 end if end proc; plot('filter(x)', x = 0 .. 20)

 

filter := proc (x) options operator, arrow; if x < 10 then 1-(1/10)*x else 0 end if end proc; plot('filter(x)', 'x' = 0 .. 20)

 



filter := proc (x) options operator, arrow; if x < 10 then 1-(1/10)*x else 0 end if end proc

plot('filter(x)');

All the above three works. May be the plot command takes default values for x even if not given as per my third command above.

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