Ramakrishnan

Ramakrishnan Vaidyanathan

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6 years, 158 days

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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.

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These are answers submitted by Ramakrishnan

Your code does not make the plot disappear.

The solution shows at (2, 2) or (0, 0), the answer is 100.

The plot shows two horizontal lines at 100. This may be the solution for other values therein.

Check for (x , y) = (2, y), or (0, y), the answer should be 100. Maple shows this answer.
 

``

restart; PDEtools

f := proc (x, y) options operator, arrow; piecewise((x, y) = (.8, .5) or (x, y) = (1.6, 1.5), -250, 0) end proc

proc (x, y) options operator, arrow; piecewise((x, y) = (.8, .5) or (x, y) = (1.6, 1.5), -250, 0) end proc

(1)

pde := -(diff(u(x, y), `$`(x, 2)))-(diff(u(x, y), `$`(y, 2))) = f

-(diff(diff(u(x, y), x), x))-(diff(diff(u(x, y), y), y)) = f

(2)

Bcs := u(0, y) = 100, u(2, y) = 100, (D[2](u))(x, 0) = 0, (D[2](u))(x, 2) = 0

u(0, y) = 100, u(2, y) = 100, (D[2](u))(x, 0) = 0, (D[2](u))(x, 2) = 0

(3)

sol := pdsolve([pde, Bcs])

u(x, y) = 100+(1/2)*(-x^2+2*x)*f

(4)

plot3d(rhs(sol), x = 0 .. 2, y = 0 .. 2)

 

``

``


 

Download plot_3d_curve.mw

Dear Sir,

3d plot is a three dimensional plot. If there are more than three variables, then for each value of the fourth variable, the graph has to be plotted.

However, the problem you have chosen is very simple and there is no need at all for a fourth variable.

Consider the variables, x, y and z. These are only added in the function. So considering the maximum limits of x, y and z, the addition is going to be limited to xmax + y max + z max

For,  x = 0 .. 10, y = 0 .. 10, x = y varies 

from 0 .. 20. Similarly, for z = 0..10, x+y+z varies from 0..30

A variable p = x + y  + z  with variation from zero to maximum value (i.e. xmax + y max + z max)

Hence a plot3d(tanh(p+w*t), t = 0..10, p=0..30) will give you the required plot.

Also note that the plot indicates a value varying from 0 to 1 only for all cases. The result is 1 for all values of p beyond a limit that depends upom w.

Hope the clarifications are in order for understanding. Thanks.

Ramakrishnan V

plot3d(x+y, x = 0 .. 10, y = 0 .. 10)

 

plot3d(z+x, z = 0 .. 10, x = 0 .. 20)

``

w := 1.0; plot3d(tanh(p+2*t), t = 0 .. 1, p = 0 .. 30)

 

w := 1.5; plot3d(tanh(p+2*t), t = 0 .. 1, p = 0 .. 30)

 

``

 

``


 

Download tanhplot.mw

 

Dear Sir,

I have made the plot of the function and arrow separately and then displayed them together. As far as the arrow is concerned, we need to practice or know the terms thoroughly to get the correct plot with correct arrows.

I have made a single plot and two plots as well in separately for a better uderstanding.

Arrow commands are individually made to ensure the shape coicides with curves. Once experienced, you can directly enter the correct arrow command to make it inline with curve.

Commands are:

b1 := plot(x^2, x = 0.. 5)
b2 := arrow(<5, 5^2>, <0.1, 1>, width = 0.05, color = red)

display(b1, b2, scaling = unconstrained)

Another Plots (two graphs)

a1 := plot([x^2, (x + 1)^2], x = 0 .. 10);

b2 := arrow(<10, 10^2>, <1, 16>, width = 0.1, color = red);

b3 := arrow(<10, 11^2>, <1, 18>, width = 0.1, color = blue);

display(a1, b2, b3, scaling = unconstrained);

Hope this contributes at least to a little extent.

Cheers.

Ramakrishnan V

 

 

We can use two 2D plots and then display the two together

with(plots)

b1 := plot(x^2, x = 0 .. 5)b1 := plot(x^2, x = 0 .. 5)

 

b2 := arrow(`<,>`(5, 5^2), `<,>`(.1, 1), width = 0.5e-1, color = red)

 

display(b1, b2, scaling = unconstrained)

 

``

 

a1 := plot([x^2, (x+1)^2], x = 0 .. 10)

 

``

b2 := arrow(`<,>`(10, 10^2), `<,>`(1, 16), width = .1, color = red)

 

b3 := arrow(`<,>`(10, 11^2), `<,>`(1, 18), width = .1, color = blue)

 

``

``

````

display(a1, b2, b3, scaling = unconstrained)

 

``


 

Download PlotwithArrow.mw

@Angelo Melino 

In my opinion, it does not eveluate if r value is not specified. It evaluates if r value is specified. I enclose my doc. May please be referred. Cheers
 

restart

g0 := exp(r*cos(theta))/Pi

exp(r*cos(theta))/Pi

(1)

G0 := `assuming`([int(g0, theta = 0 .. Pi)], [r > 0])

BesselI(0, r)

(2)

g1 := exp(r*cos(theta))*cos(theta)/Pi

exp(r*cos(theta))*cos(theta)/Pi

(3)

G1 := `assuming`([int(g1, theta = 0 .. Pi)], [r > 0])

BesselI(1, r)

(4)

g := g0+g1

exp(r*cos(theta))/Pi+exp(r*cos(theta))*cos(theta)/Pi

(5)

G := `assuming`([int(g, theta = 0 .. Pi)], [r > 0])

int(exp(r*cos(theta))/Pi+exp(r*cos(theta))*cos(theta)/Pi, theta = 0 .. Pi)

(6)

G := G0+G1

BesselI(0, r)+BesselI(1, r)

(7)

"(=)"

BesselI(0, r)+BesselI(1, r)

(8)

g0 := exp(r*cos(theta))/Pi

exp(r*cos(theta))/Pi

(9)

G0 := `assuming`([int(g0, theta = 0 .. Pi)], [r = 1])

BesselI(0, 1)

(10)

g1 := exp(r*cos(theta))*cos(theta)/Pi

exp(r*cos(theta))*cos(theta)/Pi

(11)

G1 := `assuming`([int(g1, theta = 0 .. Pi)], [r = 1])

BesselI(1, 1)

(12)

g := g0+g1

exp(r*cos(theta))/Pi+exp(r*cos(theta))*cos(theta)/Pi

(13)

G := evalf(`assuming`([int(g, theta = 0 .. Pi)], [r = 1]))

1.831224982

(14)

G0v := evalf(G0)

1.266065878

(15)

``

G1v := evalf(G1)

.5651591040

(16)

G := evalf(G0+G1)

1.831224982

(17)

``


 

Download Maplesumint_works_for_specific_rvalues_VRK.mw

 

The plot in your pdf shows a plot of a function f(t) vs t and not g2 vs g1

with(plots);
logplot([x, tan(x), x = 1 .. 10], color = green);

logplot([cos, sin, -Pi .. Pi], color = yellow);

"Find*values*of*g1 and g2*first in your*program*as*function*of*t"
                          "Then plot."

Here;
x and tan x are obtained.
In your program g1 and g2 are not obtained.

 

Examples logplot.mw
 

with(plots); logplot([x, tan(x), x = 1 .. 10], color = green)

 

logplot([cos, sin, -Pi .. Pi], color = yellow)

 

`and`(Find*values*of*g1, `in`(g2*first, your*program*as*function*of*t))

"Then plot."

``

Herex and tan x are obtained.

In your program g1 and g2 are not obtained.


 

Download logplot.mw

.
 

@tomleslie 

I am not the right person to contribute an answer here. However I feel that the following suggestion may help in getting a clear solution for all types of problems covered in the question raised. 

Let us take simple examples of non-differential equations and then get maple to solve it aand analyse the answer.

There are different sigularities that may arise. A simple case is f(z) = z2/z. This gives f(z) = z for all values of z except at z = 0.

Such problems maple should be able to answer.

Consider the equation y" + P(x)y'+Q(x)y = 0

Non-essential singularity occurs when P(x) or Q(x) diverges at x = x0, but (x-x0)P(x) and (x-x0)2 Q(x) are finite as x tends to x0.

Essential singularity occurs when (x-x0)P(x) diverges or  (x-x0)2 Q(x) diverges as x tends to x0.

A pole of order m  is a point  z of f(z) such that the Laurent series of f(z) has  an=0 for n< m and

a != 0

Essential singularities and poles of infinite order: A pole of order n is a singularity z0 of  f(z) for which the function (z-z0)n f(z) is nonsingular and for which (z-z0)k f(z) is singular for k=0, 1, ..., n-1.

Consider the equation

    x2 y"(x)-2x y'(x)+2y(x) = 0

 The solutions are y1(x) = x and y2(x) = x2 both are well behaved at singular point x=0.

Consider the equation

    x2 y"(x)-2x y'(x) -12y(x) = 0

 The solutions are y1(x) = x3 behaved at singular point x=0 and the other one and y2(x) = x-4 diverges at x = 0!

Consider the function f(z) cosec (1/z). It has poles at z = 1 / [2 (pi) n] where n is integer,  and non-essential singularity at z = 0.

The question raise may have to go through individual problems, its nature and expected singularity and then analyse the answer from maple by its answer.

hope some clarity will arise in solving the problems using maple now. 

Cheers.

Ramki.

 I am surprised Maple is giving you the answer. Normally it will keep quiet if problem is wrong!!

So your problem is correct and the answer as well.

You have given a parameter t without specifying its value or values for a sequential answer.

For odd values of t, the summation is zero. For even values, it is not so. I have not gone deep into this, but maple can do if required.

The answers as required are below.

I hope I am somewhat clear if not completely.

Cheers. Ramki.solution_to_SummationSum.mw
 

restart; t := 1; MySum := evalf(sum(sin((1/2)*Pi*n)*sin((1/2)*n*Pi*(x+1))*cos((1/2)*n*Pi*t), n = 1 .. infinity), 2)

0.

(1)

restart; t := 3; evalf(sum(sin((1/2)*Pi*n)*sin((1/2)*n*Pi*(x+1))*cos((1/2)*Pi*n), n = 1 .. infinity), 2)

0.

(2)

restart; t := 4; evalf(sum(sin((1/2)*Pi*n)*sin((1/2)*n*Pi*(x+1))*cos(4*Pi*n*(1/2)), n = 1 .. infinity), 2)

sum(sin((1/2)*Pi*n)*sin((1/2)*n*Pi*(x+1))*cos(2*Pi*n), n = 1 .. infinity)

(3)

restart; t := 2; evalf(sum(sin((1/2)*Pi*n)*sin((1/2)*n*Pi*(x+1))*cos((1/2)*n*Pi*t), n = 1 .. infinity), 2)

sum(sin((1/2)*Pi*n)*sin((1/2)*n*Pi*(x+1))*cos(Pi*n), n = 1 .. infinity)

(4)

NULL


 

Download solution_to_SummationSum.mw

 

Dear Sir, Please find attached document from help page.

If the final contour plot shown at the beginning is not clear, then read from the bottom to know in depth data and commands to get what you want.

Once the system equation is known, plots are easy from the known points by substitution.

If the answer seems irrelevant to the question, please ignore this. Thanks
 

PLOT3D(POLYGONS([[0, 0, 0], [1, 0, 0], [1, 1, 0], [0, 1, 0]], [[0, 0, 0], [0, 1, 0], [0, 1, 1], [0, 0, 1]], [[1, 0, 0], [1, 1, 0], [1, 1, 1], [1, 0, 1]], [[0, 0, 0], [1, 0, 0], [1, 0, 1], [0, 0, 1]], [[0, 1, 0], [1, 1, 0], [1, 1, 1], [0, 1, 1]], [[0, 0, 1], [1, 0, 1], [1, 1, 1], [0, 1, 1]]), LIGHT(0, 0, 0., .7, 0.), LIGHT(100, 45, .7, 0., 0.), LIGHT(100, `-`(45), 0., 0., .7), AMBIENTLIGHT(.4, .4, .4), TITLE(CUBE), STYLE(PATCH), COLOR(ZHUE))

 

PLOT(POINTS([0, 0], SYMBOL(DIAMOND)), TEXT([0, 0], 'Origin', ALIGNBELOW, ALIGNRIGHT, FONT(HELVETICA, OBLIQUE, 10)), CURVES([[`-`(3), .5], [3, .5]], THICKNESS(3), LINESTYLE(DOT)), TEXT([0, .5], 'Dotted', ALIGNBELOW), TEXT([3.1415, 0], 'p', FONT(SYMBOL, 12)), TEXT([`-`(3.1415), 0], 'P', FONT(SYMBOL, 12)), POLYGONS([[`-`(2), `-`(.25)], [`-`(2), `-`(.5)], [2, `-`(.5)], [2, `-`(.25)]], COLOR(HUE, .5)), TEXT([0, `-`(.37)], 'Red', COLOR(RGB, 1, 0, 0)), AXESSTYLE(FRAME), VIEW(`-`(4) .. 4, `-`(1) .. 1))

 

PLOT(POINTS([0, 0], SYMBOL(DIAMOND)), TEXT([0, 0], 'Origin', ALIGNBELOW, ALIGNRIGHT, FONT(HELVETICA, OBLIQUE, 10)), CURVES([[`-`(3), .5], [3, .5]], THICKNESS(3), LINESTYLE(DOT)), TEXT([0, .5], 'Dotted', ALIGNBELOW), TEXT([3.1415, 0], 'p', FONT(SYMBOL, 12)), TEXT([`-`(3.1415), 0], 'P', FONT(SYMBOL, 12)), POLYGONS([[`-`(2), `-`(.25)], [`-`(2), `-`(.5)], [2, `-`(.5)], [2, `-`(.25)]], COLOR(HUE, .5)), TEXT([0, `-`(.37)], 'Red', COLOR(RGB, 1, 0, 0)), AXESSTYLE(FRAME))

 

PLOT(POINTS([0, 0], SYMBOL(DIAMOND)), TEXT([0, 0], 'Origin', ALIGNBELOW, ALIGNRIGHT, FONT(HELVETICA, OBLIQUE, 10)), CURVES([[`-`(3), .5], [3, .5]], THICKNESS(3), LINESTYLE(DOT)), TEXT([0, .5], 'Dotted', ALIGNBELOW), TEXT([3.1415, 0], 'p', FONT(SYMBOL, 12)), TEXT([`-`(3.1415), 0], 'P', FONT(SYMBOL, 12)), POLYGONS([[`-`(2), `-`(.25)], [`-`(2), `-`(.5)], [2, `-`(.5)], [2, `-`(.25)]], COLOR(HUE, .5)), TEXT([0, `-`(.37)], 'Red', COLOR(RGB, 1, 0, 0)))

 

PLOT(POINTS([0, 0], SYMBOL(DIAMOND)), TEXT([0, 0], 'Origin', ALIGNBELOW, ALIGNRIGHT, FONT(HELVETICA, OBLIQUE, 10)), CURVES([[`-`(3), .5], [3, .5]], THICKNESS(3), LINESTYLE(DOT)), TEXT([0, .5], 'Dotted', ALIGNBELOW), TEXT([3.1415, 0], 'p', FONT(SYMBOL, 12)), TEXT([`-`(3.1415), 0], 'P', FONT(SYMBOL, 12)), POLYGONS([[`-`(2), `-`(.25)], [`-`(2), `-`(.5)], [2, `-`(.5)], [2, `-`(.25)]], COLOR(HUE, .5)), TEXT([0, `-`(.37)], 'Red', COLOR(RGB, 1, 0, 0)))

 

PLOT(POINTS([0, 0], SYMBOL(DIAMOND)), TEXT([0, 0], 'Origin', ALIGNBELOW, ALIGNRIGHT, FONT(HELVETICA, OBLIQUE, 10)), CURVES([[`-`(3), .5], [3, .5]], THICKNESS(3), LINESTYLE(DOT)), TEXT([0, .5], 'Dotted', ALIGNBELOW), TEXT([3.1415, 0], 'p', FONT(SYMBOL, 12)), TEXT([`-`(3.1415), 0], 'P', FONT(SYMBOL, 12)))

 

PLOT(POINTS([0, 0], SYMBOL(DIAMOND)), TEXT([0, 0], 'Origin', ALIGNBELOW, ALIGNRIGHT, FONT(HELVETICA, OBLIQUE, 10)), CURVES([[`-`(3), .5], [3, .5]], THICKNESS(3), LINESTYLE(DOT)), TEXT([0, .5], 'Dotted', ALIGNBELOW), TEXT([3.1415, 0], 'p', FONT(SYMBOL, 12)))

 

PLOT(POINTS([0, 0], SYMBOL(DIAMOND)), TEXT([0, 0], 'Origin', ALIGNBELOW, ALIGNRIGHT, FONT(HELVETICA, OBLIQUE, 10)), CURVES([[`-`(3), .5], [3, .5]], THICKNESS(3), LINESTYLE(DOT)), TEXT([0, .5], 'Dotted', ALIGNBELOW))

 

PLOT(POINTS([0, 0], SYMBOL(DIAMOND)), TEXT([0, 0], 'Origin', ALIGNBELOW, ALIGNRIGHT, FONT(HELVETICA, OBLIQUE, 10)), CURVES([[`-`(3), .5], [3, .5]], THICKNESS(3), LINESTYLE(DOT)))

 

PLOT(POINTS([0, 0], SYMBOL(DIAMOND, 15)), TEXT([0, 0], 'Origin', ALIGNBELOW, ALIGNRIGHT, FONT(HELVETICA, OBLIQUE, 10)), CURVES([[`-`(1), .5], [1, .5]]))

 

PLOT(POINTS([0, 0], SYMBOL(DIAMOND, 100)), TEXT([0, 0], 'Origin', ALIGNBELOW, ALIGNRIGHT, FONT(HELVETICA, OBLIQUE, 15)))

 

PLOT(POINTS([0, 0], SYMBOL(DIAMOND, 125)), TEXT([0, 0], 'Origin', ALIGNBELOW))

 

PLOT(POINTS([5, 7]), SYMBOL(DIAMOND, 122))

 

Explore(plots:-display(plottools:-polygon([[0, 1], [a, b], [a, c], [0, 1]], color = "Orange"), view = [0 .. 5, 0 .. 5]), a = 1.0 .. 4.0, b = 1.0 .. 4.0, c = 1.0 .. 4.0, initialvalues = [a = 3, b = 2], markers = [[a, b, symbol = circle, symbolsize = 35, color = "Red"], [0, c, symbol = solidbox, symbolsize = 20, color = "Green"]], tableborders = false, componentborders = false)

``


 

Download READ_FROM_BOTTOM_PLEASE.mw

Cheers.

Ramki.

Select Text from the context bar avove. Type anything subscript and superscript.
Select subscript. Select superscript using left mouse click drag. Select Format. Hover over popout menu first item Character. Hover over second popup menu fourth item Superscript and left click to select it. You are done.
Again go to format Character and deselect the clicked item and begin to get normal character.
Hope this clarifies to an extent getting superscript and subscript. Cntrl plus u gets u underline the word. This also can be done by the above method. Many such character formats can be done like this.

Cheers.

Ramki
 

``

Select Yext from the context bar avove. Type anything subscript and superscript.

Select subscript. Select superscript using left mouse click drag. Select Format. Hover over popout menu first item Character. Hover over second popup menu fourth item Superscript and left click to select it. You are done.

Again go to format Character and deselect the clicked item and begin to get normal character.


 

Download subscriptlearning.mw

I just use copy and insert rows in the table. Once you make atleast one table, then you can use it as a template and add columns to the left or right by copy paste and cancel and insert selections in that order.

Select, copy, paste, cancel delete dialogue, insert dialogue appears, select insert and you are done.
 

``

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

``

``


 

Download textTable.mw

Cheers.

Ramki

Hope the enclosed doc covers all possible problems with answers.


 

restart

f := proc (x) options operator, arrow; piecewise(x <= 0, -3, 0 < x and x <= 1, x^2, 1 < x, -x) end proc

proc (x) options operator, arrow; piecewise(x <= 0, -3, 0 < x and x <= 1, x^2, 1 < x, -x) end proc

(1)

f(-1) = -3NULL

f(1/2) = 1/4NULL

f(4) = -4NULL

Inverse relation required by the OP

 

a

x

x; f(x) > a

a < -3

a < -3

(2)

``

x; x = RealRange(Open(-infinity), 0)

x = RealRange(-infinity, 0)

(3)

-3

"<=0" or 3

x; 0 < x and x < 3

a < 0, a <> -3

x = abs(a)

x; x < abs(a)

0 <= a and a <= 1

sqrt(x)

x; sqrt(a) < x and x < 1

1 > a, a <> -3

a

x*is*undefined

 

 

Procedure defined here

xValue := proc (a) local x, x1; if a < -3 then x := "x:x=(-&infin;,0] " elif a = -3 then x := "x:0<x<3" elif -3 < a and a < 0 then x1 := abs(a); x := ["x1=", x1, "x:x<x1"] elif 0 <= a and a < 1 then x1 := sqrt(a); x := ["x1=", x1, " x: x1<x&le;1"] elif 1 <= a then x := "undefined" end if end proc

xValue(4)

"undefined"

(4)

NULL

xValue(-3)

"x:0<x<3"

(5)

xValue(0.9e-1)

["x1=", .3000000000, " x: x1<x&le;1"]

(6)

x1

x1

(7)

``


 

Download piecewiseFunctionSolution.mw

Cheers.

Ramki

The given problem has domain {R}, but Range is (-inf, 1].

having known this, the problem can be stated R: f(x)>a, a: (-inf, 1] and solved.

for i from -1 by 0.5 to 2 do
 if   f(i)-a>0 then answer:=i;
end do

A crude program to convey my idea is enclosed.

Cheers.

Ramki
 

restart

f := proc (x) options operator, arrow; piecewise(x <= 0, -3, 0 < x and x <= 1, x^2, 1 < x, -x) end proc

proc (x) options operator, arrow; piecewise(x <= 0, -3, 0 < x and x <= 1, x^2, 1 < x, -x) end proc

(1)

f(-1) = -3NULL

f(1/2) = 1/4NULL

f(4) = -4NULL

To find R: f(x)>3

f(x)-3 should be >0

f(0)-3

-6

(2)

f(7)-3

-10

(3)

f(.2)

0.4e-1

(4)

``

for i from -1 by .5 to 2 do f(i) end do

-2.0

(5)

The domain for the problem is {R} and the Range is (-∞,1]

for i by -0.5e-1 to -1 do f(i)-1 end do


NULL 


 

Download piecewiseFunction.mw

 I am not an expert in squarefree ploynomial. However, I thought the doc I have made for nth root of a polynomial and its derivatives may be of use to you. First few lines of doc are of not importance here. Hope the rest of the algorithm will be of use. Cheers.

Ramki
 

eval(root(x^3, 3))

(x^3)^(1/3)

(1)

(x^3)^(1/3)

(x^3)^(1/3)

(2)

(-8.0)^(1/3); root(-8.0, 3); surd(-8.0, 3);

1.000000000+1.732050807*I

 

1.000000000+1.732050807*I

 

-2.000000000

(3)

(-8)^(1/3)

(-8)^(1/3)

(4)

(1+1.732*I)^3

-7.999472+0.304832e-3*I

(5)

root(-8, 3)

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

(6)

surd(-8, 3)

-2

(7)

surd(x^3, 3)

surd(x^3, 3)

(8)

root(x^3, 3)

(x^3)^(1/3)

(9)

y := proc (x) options operator, arrow; x^2 end proc

proc (x) options operator, arrow; x^2 end proc

(10)

y(2)

4

(11)

y := x, proc (lambda) options operator, arrow; x^6+(2*lambda+1)*x^3+2 end proc

x, proc (lambda) options operator, arrow; x^6+(2*lambda+1)*x^3+2 end proc

(12)

root[3](x^6+(2*lambda+1)*x^3+2, symbolic)``

(x^6+2*lambda*x^3+x^3+2)^(1/3)

(13)

diff(x^6+(2*lambda+1)*x^3+2, x)

6*x^5+3*(2*lambda+1)*x^2

(14)

root[3](6*x^5+(3*(2*lambda+1))*x^2, symbolic)

(6*x^5+6*lambda*x^2+3*x^2)^(1/3)

(15)

diff(6*x^5+6*lambda*x^2+3*x^2, x)

30*x^4+12*lambda*x+6*x

(16)

root[3](30*x^4+12*lambda*x+6*x, symbolic)

(30*x^4+12*lambda*x+6*x)^(1/3)

(17)

diff(30*x^4+12*lambda*x+6*x, x)

120*x^3+12*lambda+6

(18)

root[3](120*x^3+12*lambda+6, symbolic)

(120*x^3+12*lambda+6)^(1/3)

(19)

diff(120*x^3+12*lambda, x)

360*x^2

(20)

root[3](360*x^2, symbolic)

2*45^(1/3)*(x^2)^(1/3)

(21)

diff(360*x^2, x)

720*x

(22)

root[3](720*x, symbolic)

2*90^(1/3)*x^(1/3)

(23)

diff(720*x, x)

720

(24)

``

root[3](720)

2*90^(1/3)

(25)

``


 

Download RootsOfPolynomial.mw

Dear member,

A maple student version, if availbale, will be more useful for a student to learn right from basics to advanced studies in a PhD programme. All this with an understanding of what you do and what maths does and how it looks to your eyse as a visualization of the happenings in maths?

A simple example of domain and range in a function can be visualised as a vending machine letting you icecream bars for the coins from your purse.

The transformation of a simple parent function moving in space and stretching its size by simple parameters involved in a function. A great place to learn maths in maple in my opinion.

Ramki

Dear friend, The expression 

((√1 + y′ ² )/dy')

cannot be solved. Only equations can be solved. Softwares normally solve an expression by equating the expression to zero. If that is the way you want, yes maple can solve it.

((√1 + y′ ² )/dy') = 0 -> √1 + y′ ²  = 0  -> 1 + y′ ²  = 0 -> y′ ²= -1 ->  y′ =  i -> dy = i dx -> y = i x²/2

Cheers. 

Ramakrishnan V

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