Hello,

I am trying to get Maple to recognize and reverse the product rule in more than one dimension. In one dimension, this works:

Int((Diff(f(x), x))*g(x)+(Diff(g(x), x))*f(x), x) = int((diff(f(x), x))*g(x)+f(x)*(diff(g(x), x)), x);

.

But in two dimensions, it no longer evaluates:

Int((Diff(f(x, y), x))*g(x, y)+(Diff(g(x, y), x))*f(x, y), x) = int((diff(f(x, y), x))*g(x, y)+f(x, y)*(diff(g(x, y), x)), x)

As far as I can tell, mathematically these should be identical (except for the antiderivatives being defined up to a constant in the first case and a function of y in the second). Is there a way to get Maple to reverse the product rule to integrate in more than one dimension? Or am I missing something mathematically that makes this incorrect?

Thanks for your help,

Johnathan

Problem:

Suppose you have a bunch of 2D data points which:

1. May include points with the same x-value but different y-values; and
2. May be unevenly-spaced with respect to the x-values.

How do you clean up the data so that, for instance, you are free to construct a connected data plot, or perform a Discrete Fourier Transform? Please note that Curve Fitting and the Lomb–Scargle Method, respectively, are more effective techniques for these particular applications. Let's start with a simple example for illustration. Consider this Matrix:

A := < 2, 5; 5, 8; 2, 1; 7, 8; 10, 10; 5, 7 >;

Consolidate:

First, sort the rows of the Matrix by the first column, and extract the sorted columns separately:

P := sort( A[..,1], output=permutation ); # permutation to sort rows by the values in the first column
U := A[P,1]; # sorted column 1
V := A[P,2]; # sorted column 2

We can regard the sorted bunches of distinct values in U as a step in a stair case, and the goal is replace each step with the average of the y-values in V located on each step.

Second, determine the indices for the first occurrences of values in U, by selecting the indices which give a jump in x-value:

m := numelems( U );
K := [ 1, op( select( i -> ( U[i] > U[i-1] ), [ seq( j, j=2..m ) ] ) ), m+1 ];
n := numelems( K );

The element m+1 is appended for later convenience. Here, we can quickly define the first column of the consolidated Matrix:

X1 := U[K[1..-2]];

Finally, to define the second column of the consolidated Matrix, we take the average of the values in each step, using the indices in K to tell us the ranges of values to consider:

Y1 := Vector[column]( n-1, i -> add( V[ K[i]..K[i+1]-1 ] ) / ( K[i+1] - K[i] ) );

Thus, the consolidated Matrix is given by:

B := < X1 | Y1 >;

Spread Evenly:

To spread-out the x-values, we can use a sequence with fixed step size:

X2 := evalf( Vector[column]( [ seq( X1[1]..X1[-1], (X1[-1]-X1[1])/(m-1) ) ] ) );

For the y-values, we will interpolate:

Y2 := CurveFitting:-ArrayInterpolation( X1, Y1, X2, method=linear );

This gives us a new Matrix, which has both evenly-spaced x-values and consolidated y-values:

C := < X2 | Y2 >;

Plot:

plots:-display( Array( [
        plots:-pointplot( A, view=[0..10,0..10], color=green, symbol=solidcircle, symbolsize=15, title="Original Data", font=[Verdana,15] ),
        plots:-pointplot( B, view=[0..10,0..10], color=red, symbol=solidcircle, symbolsize=15, title="Consolidated Data", font=[Verdana,15] ),
        plots:-pointplot( C, view=[0..10,0..10], color=blue, symbol=solidcircle, symbolsize=15, title="Spread-Out Data", font=[Verdana,15] )
] ) );

Sample Data with Noise:

For another example, let’s take data points from a logistic curve, and add some noise:

# Noise generators
f := 0.5 * rand( -1..1 ):
g := ( 100 - rand( -15..15 ) ) / 100:

# Actual x-values
X := [ seq( i/2, i=1..20 ) ];

# Actual y-values
Y := evalf( map( x -> 4 / ( 1 + 3 * exp(-x) ), X ) );

# Matrix of points with noise
A := Matrix( zip( (x,y) -> [x,y], map( x -> x + f(), X ), map( y -> g() * y, Y ) ) );

Using the method outlined above, and the general procedures defined below, define:

B := ConsolidatedMatrix( A );
C := EquallySpaced( B, 21, method=linear );

Visually:

plots:-display( Array( [
    plots:-pointplot( A, view=[0..10,0..5], symbol=solidcircle, symbolsize=15, color=green, title="Original Data", font=[Verdana,15] ),
    plots:-pointplot( B, view=[0..10,0..5], symbol=solidcircle, symbolsize=15, color=red, title="Consolidated Data", font=[Verdana,15]  ),
    plots:-pointplot( C, view=[0..10,0..5], symbol=solidcircle, symbolsize=15, color=blue, title="Spread-Out Data", font=[Verdana,15] )
] ) );

Generalization:

Below are more generalized custom procedures, which are used in the above example. These also account for special cases.

# Takes a matrix with two columns, and returns a new matrix where the new x-values are unique and sorted,
# and each new y-value is the average of the old y-values corresponding to the x-value.
ConsolidatedMatrix := proc( A :: 'Matrix'(..,2), $ )

        local i, j, K, m, n, P, r, U, V, X, Y:
  
        # The number of rows in the original matrix.
        r := LinearAlgebra:-RowDimension( A ):

        # Return the original matrix should it only have one row.
        if r = 1 then
               return A:
        end if:

        # Permutation to sort first column of A.
        P := sort( A[..,1], ':-output'=permutation ):       

        # Sorted first column of A.
        U := A[P,1]:

        # Corresponding new second column of A.
        V := A[P,2]:

        # Return the sorted matrix should all the x-values be distinct.
        if numelems( convert( U, ':-set' ) ) = r then
               return < U | V >:
        end if:

        # Indices of first occurrences for values in U. The element m+1 is appended for convenience.
        m := numelems( U ):
        K := [ 1, op( select( i -> ( U[i] > U[i-1] ), [ seq( j, j=2..m ) ] ) ), m+1 ]:
        n := numelems( K ):

        # Consolidated first column.
        X := U[K[1..-2]]:

        # Determine the consolidated second column, using the average y-value.
        Y := Vector[':-column']( n-1, i -> add( V[ K[i]..K[i+1]-1 ] ) / ( K[i+1] - K[i] ) ):

        return < X | Y >:

end proc:

# Procedure which takes a matrix with two columns, and returns a new matrix of specified number of rows
# with equally-spaced x-values, and interpolated y-values.
# It accepts options that can be passed to ArrayInterpolation().
EquallySpaced := proc( M :: 'Matrix'(..,2), m :: posint )

        local A, i, r, U, V, X, Y:

        # Consolidated matrix, the corresponding number of rows, and the columns.
        A := ConsolidatedMatrix( M ):
        r := LinearAlgebra:-RowDimension( A ):
        U, V := evalf( A[..,1] ), evalf( A[..,2] ):

        # If the consolidated matrix has only one row, return it.
        if r = 1 then
               return A:
        end if:

        # If m = 1, i.e. only one equally-spaced point is requested, then return a matrix of the averages.
        if m = 1 then
               return 1/r * Matrix( [ [ add( U ), add( V ) ] ] ):
        end if:

        # Equally-spaced x-values.
        X := Vector[':-column']( [ seq( U[1]..U[-1], (U[-1]-U[1])/(m-1), i=1..m ) ] ):

        # Interpolated y-values.
        Y := CurveFitting:-ArrayInterpolation( U, V, X, _rest ):    

        return < X | Y >:

end proc:

Worth Checking Out:

• The CurveFitting:-ArrayInterpolation() command
• The dataplot() command
• The plots:-pointplot() command
• The Statistics:-ScatterPlot() command
• The Interpolation package (new in Maple 2018)

Download question.mw

I am working on solving a set of equations for a number of parameters, and everything looks great except that I am getting a "1." in front of the solution for a couple of variables. What does this mean? In the example below, in the solution for both d__1 and d__2, there is a "1." (3 of them in fact).

s1Mean := -1/(2*(-rho^2+1))*(-2*d__1/c__1)-2*rho*d__2/((2*(-rho^2+1))*c__1^.5*c__2^.5) = -2*(-(-X*beta+y)*kappa/(2*sigma)+xi__2*Z__2*gamma__1/(2*tau__2)-xi__1*Z__1/(2*tau__1))

s2Mean := -1/(2*(-rho^2+1))*(-2*d__2/c__2)-2*rho*d__1/((2*(-rho^2+1))*c__1^.5*c__2^.5) = -2*(-(-X*beta+y)*gamma__2/(2*sigma)-xi__2*Z__2/(2*tau__2))

meanSol := solve({s1Mean, s2Mean}, {d__1, d__2})

with maple 2015

how can type log[2](3) into

Can you help me?

Thank you very much.

log.mw

 Any one can help me to solve the differential equations using maple to get the velocities u ,v and pressure p for the problem mentioned below

I tried the example in BodePlot help.

restart;
with(DynamicSystems):
sys := TransferFunction( 1/(s-10) ):
BodePlot(sys);

That works OK. But, if I invoke Syrup, the example no longer works.

restart;
with(Syrup);
with(DynamicSystems):
ckt := [V, Rsrc(50), C1(15e-9), L1(15e-6), C2(22e-9), L2(15e-6), C3(22e-9), L3(15e-6), C4(15e-9), 1, Rload(50)];

TF := subs(other, V=1, v[Rload]);
sys := TransferFunction(TF);

BodePlot(sys);
I get a message "not a valid plot structure".  OK, try the example, again.

sys := TransferFunction( 1/(s-10) ):
BodePlot(sys);
I also get the "not a valid plot structure" message.

What am I doing wrong?

I have asked this before but am still confused. I have a half-dozen procedures I want to save. Don't want them in a module/package. I am using windows 10.

I just can't get the syntax correct on this.

Obviously after saving restart and load to test.

libname;
       "C:\Program Files\Maple 2018\lib", 

         "C:\Users\Ronan\maple\toolbox\CodeBuilder\lib", 

         "C:\Users\Ronan\maple\toolbox\OEIS\lib", 

         "C:\Users\Ronan\maple\toolbox\personal\lib", 

         "C:\Users\Ronan\maple\toolbox\UTF8\lib"
libdir := "C:/Users/Ronan/maple/toolbox/personal/lib";
     libdir := "C:/Users/Ronan/maple/toolbox/personal/lib"
NULL;

LibraryTools:-Save(Pedal, cat(kernelopts(homedir), "/maple/toolbox/personal/lib/Pedal.mpl"));
Error, (in LibraryTools:-Save) could not open `C:\Users\Ronan/maple/toolbox/personal/lib/Pedal.mpl\Pedal.m` for writing

Please tell me where I go wrong

And I have not even tried lamda in nanometers

Since theata is small I repalce sin(theta) by theta because, with units, sin did not compute

Anyone know a good reference (book/website) explaing Maple units?

Many thanks

TelescopeUnits.mw

Hi

I hope everyone is fine.

Here is a nice question :

I have an inequality ( please see maple code) if I assume a special condition (on |f(u,s)| used in the code )

how can I get an upper bound of the function |x(t)| and is the upper bound converges to zero when t goes to infinity.

Maybe this is can be done using maple because by hand up to know I can't find an upper bound which converges to zero as t goes to infinity.

Maybe, there is  a good, nice and appropriate answer using maple.

Below, please find the upper_bound.mw code.

Many thinks

Upper_bound.mw

Download Upper_bound.mw

I am curious how one can fine the whattype of operation in expression when two variables are actually doing difference or division operation

like

whattype(a-b);
whattype(a/b);

is there any better method to find out actually for all five operations exactly.

1. a-b;  should tell the operation is difference(substraction(`-`))
2. a+b; should tell the operation is addition(`+`)
3. a*b;  should tell the operation is multiplication(`*`)
4. a/b;  should tell the operation is division(`/`)
5. a^b;  should tell the operation is power(`^`)

I understood that maple is reading a-b as a+(-b) and same a/b as a*(1/b). but, for programing it is very much comfortable to know exactly the operation. Especially when solving partial differentiation it is easy if I know exactly.

would be very grateful if I can get this thing solved. Any suggestions and coments are welcomed and thanks in advance
 

Hi,

      I need to compute something involved with the pseudo differential operators.

https://en.wikipedia.org/wiki/Pseudo-differential_operator

Specifically, I need to calculate the inverse of a pseudo differential operator, the multiplication of two pseudo differential operators, and the n-th root of a pseudo differential operator. 

I don't know whether Maple could handle these. 

Thanks.

Respected sir,

I have attached a file that contain integration . Sir I am unable to solve please give me some hint.

PLease note followings 

1. I want to get final expression , want to get the value of constant from boundary condition.

2.I want to get the final expression (.......+c1)

3.I will put boundary condition to determine the value of c1

Dear all;

I need a help to get a simple code about the null hypothesis test.

A drug is administrad to a population X of size 50 while a  placebo is given to a population Y of size 25.

Observed results of good bad and no effects are given in the following vectors for both population.

X=[ 20,11,19];

Y=[4,4,17];

test the null hypothesis H0: population independent of treatment versus the one tailed alternative that they are dependent by computing the theoretical contingency table with entries T[i,j] where i=1,2

for the two rows and j=1,2,3 for the three columns. At what p-value can we reject H0.

Many thanks

There is a simulation of a T handle in space done in matlab https://rotations.berkeley.edu/a-tumbling-t-handle-in-space/

Would like to see someone do this in Maple. More specifically how the Matlab code would look like translated to Maple