janhardo

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11 years, 47 days

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These are replies submitted by janhardo

@Carl Love Thanks
I checked it and it "works"
I will give attention to plotcompare command.
Seems to me that it easier to study both plots side by side 

The plotcompare command can be used for one function to start with and see branchpoints/cuts?

@janhardo 

2d input

This 2d input is converted to maple to 1d input

=I/(2*b)*ln((b*I + z)/(b*I - z));

Seems to be impossible in Maple to get this 2d input as answer, via 1d maple input like in my textbook ( the simplification done by @acer is correct (good job), but it is not in the wanted form)

Ok you can say why bother further about this, although its right?
Its so satifsying to see in Maple exact the same as in a textbook. 

@Carl Love 

Thanks

Probably  i was tired after trying another code getting working and doing figuring out Maple math for some hours. I took it over from a example.
Yes, this what  meant, it is the plotttingrange from x=-1..1 for all 6 plots

The idea was only to show two plots besides eachother for a complex plot with a real and imaginair part. 

@janhardo 

Something like this

restart;
with(plots):
a:=Array(1..2,1..3);

a[1,1]:=plot(sin(Pi*x), x=1..1, color=black):
a[1,2]:=plot(sin(2*Pi*x), x=1..1, color=black):
a[1,3]:=plot(sin(3*Pi*x), x=1..1, color=black):

a[2,1]:=plot(cos(Pi*x), x=1..1, color=black):
a[2,2]:=plot(cos(2*Pi*x), x=1..1, color=black):
a[2,3]:=plot(cos(3*Pi*x), x=1..1, color=black):
display (a, tickmarks= [[],[]],labels=["",""]);

Its only not working yet.

@Ronan Thanks 
I can use it 
But i do need the plots placed in a table structure (  like in Word text program) 
Don't know if there is another way too  

@Carl Love

 Thanks, looks great those plots

The distiction between a conformal plot and the spacial complex plot(  u-v form :)  must be yet understood by me ?
z = r*exp(I*theta) = r*cos(theta) + I*r*sin(theta)

This equation is the complex number written in two of his three forms and the (formula of Euler)  can be derived from this by cancelling out the r.  

Yes, this plotting representation can be used to show two to or more plots side by side 
Its only that there is no direct relation by formula is between the graphs in my case 
Its here for 2D , but now for 3D.

I do want use it for for compare plots

with(plots)c# verder uitzoeken
plotcompare(ln(x), x - x^3/6, x = -3 .. 3); Can this be used?

For the FunctionAdvisor ( ln(z) give lots of interesting information , but for plot it gives only a plot. 
I like to see the plotting code underneath the plot as i saw with a other plot example.

@vv 

Thanks

Unfortenaly  must first study this subject of branche(s) in order to get a idea about it 

@acer 

Thanks for searching to more easier commands.

Looking at both commands then is the combine command a much easier to use then the nested simplify command

But there was  third command used lost in the delete post what uses a e power  in his nested simplify command => no it is done by @acer

So three  rather simple ways to for transforming the expression: expr

> expr := I*( ln(-z*I + b) - ln(z*I + b) );

I*(ln(-I*z+b)-ln(I*z+b))

combine(expr, symbolic);

 

> I*log(simplify(expand(exp(expr/I))));

a third ? ..no ln(exp(-I*expr))*I is done = expr

@acer
Thanks 

The most direct approach by you is something difficult to master for me, or with more explenation how it excactly all works.
For commercial use it cannot be short enough , because time is money
But for me it is only educational  use ,so when i get the same result with more easier command then i am also satisfied. 

Indeed i made mistake by over take the expression, it shows again how to be accurate is important.

@janhardo 

# You might examine these results, for your interest:
   restart;
expr := ln(1 + z*I/b) - ln(1 - z*I/b);
expr2 := normal(expr);
ans1 := combine(expr2, symbolic);
ans2 := evalindets(ans1, specfunc(ln),
                   u->ln(expand(sign(numer(op(u)))*numer(op(u)))
                         /expand(sign(numer(op(u)))*denom(op(u)))));
sort(ans2, order=plex(b,z));
restart;
foo := -(b + z*I)/(-b + z*I);
numer(foo);
sign(numer(foo));
sign(numer(foo))*numer(foo);
sign(numer(foo))*denom(foo);
expand(sign(numer(foo))*numer(foo))/expand(sign(numer(foo))*denom(foo));

ln(1+I*z/b)-ln(1-I*z/b)

 

ln((I*z+b)/b)-ln(-(I*z-b)/b)

 

ln(-(I*z+b)/(I*z-b))

 

ln((I*z+b)/(-I*z+b))

 

ln((b+I*z)/(b-I*z))

 

-(b+I*z)/(-b+I*z)

 

-I*z-b

 

-1

 

b+I*z

 

b-I*z

 

(b+I*z)/(b-I*z)

(1)

 

the expr used here above seems to be   the same expression i used ? ,

I*ln(-z*I + b) - ln(z*I + b);# my expression

I*ln(b-I*z)-ln(b+I*z)

(2)

restart;

expr := I*ln(-z*I + b) - ln(z*I + b);
expr2 := normal(expr);
ans1 := combine(expr2, symbolic);
ans2 := evalindets(ans1, specfunc(ln),
                   u->ln(expand(sign(numer(op(u)))*numer(op(u)))
                         /expand(sign(numer(op(u)))*denom(op(u)))));
sort(ans2, order=plex(b,z));
restart;
foo := -(b + z*I)/(-b + z*I);
numer(foo);
sign(numer(foo));
sign(numer(foo))*numer(foo);
sign(numer(foo))*denom(foo);
expand(sign(numer(foo))*numer(foo))/expand(sign(numer(foo))*denom(foo));

I*ln(-I*z+b)-ln(I*z+b)

 

I*ln(-I*z+b)-ln(I*z+b)

 

I*ln(-I*z+b)+ln(1/(I*z+b))

 

I*ln(-I*z+b)+ln(1/(I*z+b))

 

I*ln(b-I*z)+ln(1/(b+I*z))

 

-(b+I*z)/(-b+I*z)

 

-I*z-b

 

-1

 

b+I*z

 

b-I*z

 

(b+I*z)/(b-I*z)

(3)

 

 

But there was simpler approach by forummeber @nm 6139 : but his post is removed

 

 

 

 

Download post_acer_extra_na_verwijdering_nm_en_kitonum.mw

@janhardo 

Ok, i posted a double post ..it slipped in and the post of @nm 6131 and @kitonum are removed

The question was there was how to rearrange this  complex ln expression 

I*ln(-z*I + b) - ln(z*I + b);

As you can see i did it by hand too, but don't know if it is right? :

I*ln(-z*I + b) - ln(z*I + b)= I*ln ((-z*I + b)/(z*I + b));      gives (16) see above 

If i remember the answer from nm 6131 it was like this :
expr : = I*ln(-z*I + b) - ln(z*I + b); 
I*simplify(log(exp(1)*(expr)) ;   ?        (in general : ln x^a = a*ln x)  or  take ln^a             

Note: should be nice  to  have the answer of @nm 6131 and @kitonum back again

@Kitonum 

Thanks

Your calculation brings  me back to the direction of the start formula 

@nm 

Thanks

Seems that mathematica is more advanced then Maple? 
Well, i stay  with Maple, because learning again a new language for Mathematica is too much. 

Surprize, the manual derivation of the ln expression seems to be correct what i did by hand. (it is the same as your Maple calculation)
But your  code example shows me that i must practice with this ..

Not capable of not using advanced commands makes it neccesary to know the basic rules 
Its here : a power of a logarithm with base e : 


  ln x^n =  n.ln x 

The answer of @acer is almost the same as in the picture 
Now i tried it doing also, but with easier commands... how far do i come to the wanted answer? 

restart;with(plots): # made by @acer

A := Int(1/(b^2 + zz^2), zz = 0 .. z);

Int(1/(b^2+zz^2), zz = 0 .. z)

(1)

value(A);

arctan(z/b)/b

(2)

convert(value(A),ln);

((1/2)*I)*(ln(1-I*z/b)-ln(1+I*z/b))/b

(3)

combine(expand(I*convert(value(A),ln)), symbolic)/I;

-((1/2)*I)*ln(-(I*z+b)/(I*z-b))/b

(4)

 

A little manipulation can grind the above into a form with the target leading sign, etc.

 

evalindets(%, specfunc(ln),
           u->ln(expand(sign(numer(op(u)))*numer(op(u)))
                 /expand(sign(numer(op(u)))*denom(op(u)))));

-((1/2)*I)*ln((I*z+b)/(-I*z+b))/b

(5)

ans := sort(%, order=plex(b,z));

-((1/2)*I)*ln((b+I*z)/(b-I*z))/b

(6)

simplify( combine( convert( value(A) - ans, ln ), symbolic) );

0

(7)

 

 

 

plot3d([Re,Im](arctan(z/b)/b-ans),b=-4..4,z=-4..4,
       color=[red,blue],grid=[100,100],view=-1e-14..1e-14):

 

 

================================ made by  @janhardo

 

I started also trying to get  on the wanted formula

restart;

A := Int(1/(b^2 + z^2), z = 0 .. z);

Int(1/(b^2+z^2), z = 0 .. z)

(8)

value(A);

arctan(z/b)/b

(9)

A1:=convert(value(A),ln);

((1/2)*I)*(ln(1-I*z/b)-ln(1+I*z/b))/b

(10)

A2:=expand(A1):%=combine(%);

((1/2)*I)*ln(1-I*z/b)/b-((1/2)*I)*ln(1+I*z/b)/b = ((1/2)*I)*(ln(1-I*z/b)-ln(1+I*z/b))/b

(11)

A3:=simplify(A2,symbolic);

((1/2)*I)*(ln(-I*z+b)-ln(I*z+b))/b

(12)

A4:=denom(A3);

2*b

(13)

A5:=numer(A3);

I*(ln(-I*z+b)-ln(I*z+b))

(14)

 

 THIS  IS DID BY HAND : (not knowing if possible, this is forum question ) ..i am trying to get a wanted  expression

 

I*ln(-z*I + b) - ln(z*I + b)= I*ln ((-z*I + b)/(z*I + b));# see forum question

I*ln(-I*z+b)-ln(I*z+b) = I*ln((-I*z+b)/(I*z+b))

(15)

 

 

And so on.. If some  of the high level commands are advanced, then a step back

 


 

Download handmatig_betrekking_logaritmisch_en_goniometrisch_forum_nm_6131.mw

@Carl Love 

Thanks

It is clear that after reading your step by step explanation i ever never could deduct this by meself, the working of combination of commands.
For instance could not give meaning to [Re,Im](J1) and the nested simplify command.
Very helpful to try understand the whole picture better with all details involved.

@Carl Love Thanks

It seems that the base-10 logarithm is not the logarithm be found in higher level mathematics..
A ln logarithm use is widespread,why is that, i can't give the answer .
The number e is popping up everwhere in math.

The derative value of e^x is e^x value for x
Also the e^x function can never be 0 
Its a remarkable  number e

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