succeed, really works

succeed, really works

how about adding a initial condition

pdsolve({(Diff(f(x, y), x, y, x, y))*x+(Diff(f(x, y), x, y))*y = 0, f(0, 0) = 1}, f(x, y))

actually i do not know which intial condition are needed

how about adding a initial condition

pdsolve({(Diff(f(x, y), x, y, x, y))*x+(Diff(f(x, y), x, y))*y = 0, f(0, 0) = 1}, f(x, y))

actually i do not know which intial condition are needed

i has using this differential equation to get a pdf.

however, it seems the input binomial make the graph bad.

if this differential equation solution is not a pdf, then what is f ?

i has using this differential equation to get a pdf.

however, it seems the input binomial make the graph bad.

if this differential equation solution is not a pdf, then what is f ?

@Carl Love 

i guess this equation is crafted from all kinds of distribution, in other words, it may be from inverse direction. 

hope that can be solved in future by other method

@Carl Love 

T := 1/x!*Eval(Diff(g(z)^x, z$(x-1)), z=0) = n*x;
dsolve(T, g(z));
solve(T, g(z));

how to solve after using Eval,

actually right hand side can be any function such as binomial(n,k)*.....etc

can summation in this way?

as i find some pattern of this

L := Summation(Summation(g(m, n)-g(m-1, n)-g(m, n-1), m = 1 .. 2), n = 1 .. 2);
Matrix([[0, g(0,1), g(0,2)],[g(1,0), g(1,1), 0],[g(2,0), 0, g(2,2)]]);

can summation in this way?

as i find some pattern of this

L := Summation(Summation(g(m, n)-g(m-1, n)-g(m, n-1), m = 1 .. 2), n = 1 .. 2);
Matrix([[0, g(0,1), g(0,2)],[g(1,0), g(1,1), 0],[g(2,0), 0, g(2,2)]]);

i understand now, use 'hyporesult'(A,B) 

i understand now, use 'hyporesult'(A,B) 

@Carl Love 

https://skydrive.live.com/redir?resid=E0ED7271C68BE47C!357

this formula come from this picture.

how to rsolve it?

is it possible to define my own notation, for example implication, deduceto,

resultinhypothesis( Conj(A,B) , C ) which in book is |<

is it possible to define my own notation, for example implication, deduceto,

resultinhypothesis( Conj(A,B) , C ) which in book is |<