janhardo

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These are replies submitted by janhardo

@Ronan 

Thanks

Yes, it is exactly what i want and overlooked your maple worksheet 
Its only that i don't see the logic of your code in relation with a list.

This k = [1, 2, 4, 6] in the plot3d command where is this documented in Maple?

The prime notation has become a standard in Maple

Strange i don't get the prime function standard in Maple not working yet ?

Also the function definition in the lessons hereunder ?

Maple for Differential Equations (yfei.page)

  • In Maple (version >=10), the prime derivative notation also works. For example, given a function f(x):= sin(x^2), the second derivative of can be calculated using the command f''(x). When applying the prime derivative notation to a function or an expression, the result is in the same type. The default differentiation variable for the prime notation is x. To change it the another variable, say tt, you may use the following command

    Typesetting:-Settings(prime = t) 
    # Change the default differentiation variable to t.

    Here, the symbol # is the comment symbol in Maple.

When using diff or the prime derivative notation to an expression, the result is an expression. To evaluate it, you may use the subs command. Another way is to apply unapply to the result to convert it into a function (operator).

restart;

 

some examples for function of two variables as expression

z:=3*x^3*exp(2*y)+x^2*y^2;

3*x^3*exp(2*y)+x^2*y^2

(1)

# differentiate two times partially to x

Diff(z,x,x)=diff(z,x,x);

Diff(3*x^3*exp(2*y)+x^2*y^2, x, x) = 18*x*exp(2*y)+2*y^2

(2)

(* $ is handy to use for higher partials , three  times z partial to x *)

Diff(z,x$3)=diff(z,x$3);

Diff(3*x^3*exp(2*y)+x^2*y^2, x, x, x) = 18*exp(2*y)

(3)

# as a function

f:=(x,y)->3*x^3*exp(2*y)+x^2*y^2;

proc (x, y) options operator, arrow; 3*x^3*exp(2*y)+x^2*y^2 end proc

(4)

fxx:=D[1,1],(f); # ?? wrong in book.. no , !

D[1, 1], f

(5)

?D

fxx:=D[1,1](f);

proc (x, y) options operator, arrow; 18*x*exp(2*y)+2*y^2 end proc

(6)

fx:=D[1](f);

proc (x, y) options operator, arrow; 9*x^2*exp(2*y)+2*x*y^2 end proc

(7)

fxxx:=D[1,1,1](f); # correct

proc (x, y) options operator, arrow; 18*exp(2*y) end proc

(8)

fx3:= D[1$3](f);

proc (x, y) options operator, arrow; 18*exp(2*y) end proc

(9)

fyxx:=D[2,1,1](f);

proc (x, y) options operator, arrow; 36*x*exp(2*y)+4*y end proc

(10)

fxyyx:=D[1,2,2,1](f);

proc (x, y) options operator, arrow; 72*x*exp(2*y)+4 end proc

(11)

fxy2x:=D[1,2$2,1](f);

proc (x, y) options operator, arrow; 72*x*exp(2*y)+4 end proc

(12)

 

Some examples here and using another noatation input with a subscript  for diff in (2) ..can it be simpler?
Using the D operator makes it much easier, so there is no need for a simpler notation, but after the reading the post of dr Lopez

  • subscript as partial differentation operator-1
  • subscript as partial differentation operator 2

What to do?

 

Download prime_en_functie_index.mw

@ecterrab 

Thanks

Make a input with a prime notattion and a input with a indexed function notatation ( like by hand) has now my interest for elementary calculus 

This old post from dr Lopez is just what i am looking for 

Here is a example where @acer asked for:

Its like this by doing it by handwriting  
example : differential equation 

y" -2y' -3y=6 

Or for partial deratives in this notation fxx   

I go further with this in Maple 

Note: i know my English is not that good, but i  like to write manual and not with deepl translation online

@Mariusz Iwaniuk

 Thanks

There is no doubt about it, that it has to do with the functional equation from Zeta(z)  , because i can always alter (convert) from maple input to 2d input 
But seeing the functional equation in Maple input is no problem either in your code.

By substituing : z=2+3*I (x>0, so in the rightplane) in the rightside of the functional equation you can calculate the leftside of the functional equation
(-z is in the left plane of the complex plane)
Now you extend the whole complex plane ,except for a pole z= 1) 

I start this thread about how to calculate the two function values from Zeta (z) for the two complexnumber  in the functional equation.
How precise for a decimal number here?

@Carl Love 
It was too much expected from me that the sum ( zeta function) could produce the Eulerproduct formula
The Zeta function as we now is not a average function
I try to understand more the importance for number theory of the zeta function.

@Mariusz Iwaniuk
 Thanks

It revolves between the connection  -z  and  z+1  
Its not directly clear not yet for me where your example stands for? 

@vv 

Thanks

You can say that the sum and eulerproduct( derived from the sum ) are equal after their proves. (only for real numbers then)

In general form both are only a notation then.
 

 

@vv 
Thanks

With this correct sum now Maple knows perhaps the Euler prorduct formula as closed formula?

@tomleslie 

Thanks

Your definition of the sum and product seems to be correct: they are equal , if look to the comparison
I am unsure if i defined this correct in my worksheet too
Is there a way in Maple doing this direct without filling in first some values in the sum and product?

@vv 

Thanks 
informative the wiki 

At least then : sum(1/n^p, n = 1 .. infinity)

Proving for all numbers in Maple is impossible , so how i can check  this symbolically ?
In order to check these sum and product if they are equal then i must fill in some values then. 
note: i think i must chance p in the sum into s  ( is real number >1 , or the same as  Re s>1) after looked in a book and seen in the @tom leslie code

@tomleslie 
Thanks

It is not that important for examing the graph of the zeta function i think?

@Carl Love 

Thanks

Those definitions can  be tricky

f = u +i v form   f = u(x,y) + i(x,y)  form see example 2.1

Question : use of `  hereunder ?

`f(x,y) ` = f(x,y); ` `;

About the one-argument and two argument form of arctan : for real arctan versus complex arctan ?

restart;

I got this information from this course and thought it can be used for the function definition via the FunctionAdvisor of arctan

FunctionAdvisor(arctan);

There are two approaches to defining a complex function in Maple.

Method 1. Make  f(x, y)  a function of two real variables  x, y .

Method 2. Make  f(z)  a function of the complex variable  z .

 

Example 2.1.  Write  f(z) = z^4  in the  f = i*v+u  form.  

 

Method 1. Make  f(x, y)  a function of two real variables  x, y .

 

f:='f': x:='x': y:='y': z:='z':
f := proc(x,y)
  local z,w;
  z := x + I*y;
  w := expand(z^4);
end:
`f(z) ` = z^4;
`f(x,y) ` = f(x,y); ` `;
`At  z = 1 + 2i: `;
`f(1,2) ` = f(1,2);

`f(z) ` = z^4

`f(x,y) ` = x^4+(4*I)*x^3*y-6*x^2*y^2-(4*I)*x*y^3+y^4

` `

`At  z = 1 + 2i: `

`f(1,2) ` = -7-24*I

 

Method 2. Make  f(z)  a function of  z .

 

F:='F': x:='x': y:='y': z:='z':
F := proc(z)
  local w;
  w := expand(z^4);
end:
`F(z) ` = F(z);
`F(x + I y) ` = F(x + I*y); ` `;
`At  z = 1 + 2i: `;
`F(1 + I 2) ` = F(1 + I*2);

`F(z) ` = z^4

`F(x + I y) ` = x^4+(4*I)*x^3*y-6*x^2*y^2-(4*I)*x*y^3+y^4

` `

`At  z = 1 + 2i: `

`F(1 + I 2) ` = -7-24*I

restart;

# arctan(x + y*I) = arctan(y, x); ?

arctan(y, x)# also defined (see functionadvisor) as complex function in u(x,y),v(x,y) form, but defined in Maple as # arctan(x,y) its a a real function

arctan(2. + 5*I);# complex arctan function

1.499847799+.1732867951*I

(1)

arctan(5., 2); # real arctan function

1.190289950

(2)

# arctan(Pi);

arctan(Pi)

(3)

arctan(-sqrt(3)/3);

-(1/6)*Pi

(4)

arctan(-1);

-(1/4)*Pi

(5)

arctan(y, x) = -I*ln((x + y*I)/sqrt(x^2 + y^2));

 

Download reaction_functionadvisor_arctan.mw

@The function 

I don't think it's a shame that the book does not come up with such answers you got here on the Maple Primes forum.
Here some people on the MaplePrimes  forum are really specialists in the Maple language.

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