longrob

Mr. Robert Long

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15 years, 309 days
Leeds, United Kingdom

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These are replies submitted by longrob

To make Preben's point even clearer, there is no need to even re-type the input in 1D

I copied the 2D output from your worksheet into a Maple 13  classic (1D) worksheet. This comes along with all the hard-to-read 2D formatting thus

T := `#mfrac(mi("RL"),mrow(mi("j"),mo("⋅"),mi("ω",fontstyle = "normal"),mo("⋅"),mi("Cp")),linethickness = "1")`/((RL+1/(j*omega*Cp))*(`#mfrac(mi("R"),mrow(mi("j"),mo("⋅"),mi("ω",fontstyle = "normal"),mo("⋅"),msub(mi("C"),mi("s"))),linethickness = "1")`/(R+1/(j*omega*`#msub(mi("C"),mi("s"))`))+`#mfrac(mi("RL"),mrow(mi("j"),mo("⋅"),mi("ω",fontstyle = "normal"),mo("⋅"),mi("Cp")),linethickness = "1")`/(RL+1/(j*omega*Cp))));

But if you then copy the output of this to the next line, you get this

T := RL/(j*omega*Cp*(RL+1/(j*omega*Cp))*(R/(j*omega*C[s]*(R+1/(j*omega*C[s])))+RL/(j*omega*Cp*(RL+1/(j*omega*Cp)))));

which then simplifies to

(R*j*omega*C[s]+1)*RL/(R*RL*j*omega*Cp+R+RL*R*j*omega*C[s]+RL)

 

I also never use anything else than 1D input, because I find 2D hard to deal with and I learned using 1D, so I also don't know why 2D doesn't work.

What I meant was, can you use the mouse to close/quit Maple after it freezes and does Maple generate any error message ?

Anyway, I think you need to contact Maplesoft technical support. My guess is that something went wrong in the installation.

What I meant was, can you use the mouse to close/quit Maple after it freezes and does Maple generate any error message ?

Anyway, I think you need to contact Maplesoft technical support. My guess is that something went wrong in the installation.

Ahh thanks.... the infamous frontend !  I've always had trouble getting my head around it ! The help pages are a bit terse - thanks for the link - I think I've managed it now. At least it gives me the result I expected. Here's an example of the kind of functional I'm dealing with 

z := exp(-x)*sqrt(y(x)-exp(x)*diff(y(x),x));

z1:=frontend(diff, [z,diff(y(x),x)],[{`*`, `+`, radical}]);

z2:=frontend(diff, [z,y(x)],[{`*`, `+`, radical}]);

z3:=diff(z1,x);

EL:=simplify(z3-z2);

VC:=VariationalCalculus[EulerLagrange](z,x,y(x));

is(VC[1]=EL);

is(VC[1]=-EL);

#...which was the whole point of doing this.

# Hooray !!

For completeness, here's how I did it with subs 

F := z;

F1:=subs(diff(y(x),x)=yp,F);          
F2:=subs(y(x)=y,F1);
F3:=diff(F2,y);
F4:=subs(y=y(x),F3);
F5:=subs(yp=diff(y(x),x),F4);  # This gives the partial wrt to y(x)  

# Then...
FF1:=diff(F1,yp);  
FF2:=subs(yp=diff(y(x),x),FF1); # This gives the partial wrt to diff(y(x),x)

# So now we can formulate the Euler-Lagrange equation:
FF3:=diff(FF2,x);  #This gives the derivative with respect to x of the partial derivative with respect to diff(y(x),x)
ELL:=simplify(FF3-F5);

is(EL=ELL);

# Hooray again !

Are there any other ways of doing this ? 

Ahh thanks.... the infamous frontend !  I've always had trouble getting my head around it ! The help pages are a bit terse - thanks for the link - I think I've managed it now. At least it gives me the result I expected. Here's an example of the kind of functional I'm dealing with 

z := exp(-x)*sqrt(y(x)-exp(x)*diff(y(x),x));

z1:=frontend(diff, [z,diff(y(x),x)],[{`*`, `+`, radical}]);

z2:=frontend(diff, [z,y(x)],[{`*`, `+`, radical}]);

z3:=diff(z1,x);

EL:=simplify(z3-z2);

VC:=VariationalCalculus[EulerLagrange](z,x,y(x));

is(VC[1]=EL);

is(VC[1]=-EL);

#...which was the whole point of doing this.

# Hooray !!

For completeness, here's how I did it with subs 

F := z;

F1:=subs(diff(y(x),x)=yp,F);          
F2:=subs(y(x)=y,F1);
F3:=diff(F2,y);
F4:=subs(y=y(x),F3);
F5:=subs(yp=diff(y(x),x),F4);  # This gives the partial wrt to y(x)  

# Then...
FF1:=diff(F1,yp);  
FF2:=subs(yp=diff(y(x),x),FF1); # This gives the partial wrt to diff(y(x),x)

# So now we can formulate the Euler-Lagrange equation:
FF3:=diff(FF2,x);  #This gives the derivative with respect to x of the partial derivative with respect to diff(y(x),x)
ELL:=simplify(FF3-F5);

is(EL=ELL);

# Hooray again !

Are there any other ways of doing this ? 

Well, I've managed to do this, using subs, to replace the y(x) and diff(y(x),x) with new variables, then doing the necessary differentiations and reversing out the subs, but it seems like a very awkward and contrived way to do it.

 

As a trivial example, suppose we have

F:=y(x)^2;

the I want to differentiate this to give 2y(x) without resorting to using subs to change the variable.

 

Is there a better way ?

Wow. Impressed.
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