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janhardo

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@acer ThanksYes, there is leading m...

janhardo 695
December 23 2021
0 0

@acer 

Thanks

Yes, there is leading minus sign ( as you mean A= - ....) , but forget a second one after the first one  A = -.. = - ...= 

A = -(int(1/t, t = 1 .. t))/(2*bi) and -(int(1/t, t = 1 .. t))/(2*bi) = log[10](t)/(2*bi) and log[10](t)/(2*bi) = log[10]((b*i+z)/(b*i-z))/(2*bi)

@vv ThanksThis example is from Joha...

janhardo 695
December 23 2021
0 0

@vv 

Thanks

This example is from Johann Bernoulli [1702] who started with  A = .....
Interesting to know when the  "complex path integral". is invented ?

Do you think that Johann Bernoulli [1702] has this also in mind what you explained ?

@janhardo Don't get the right i...

janhardo 695
December 23 2021
0 0

@janhardo 
Don't get the right integral see , why not 

> 

restart;
> 

J:= Int(exp(t*x), x);
#J1:= subs(t= a+b*I, J);# heeft Carl gebruikt
J1:= subs(t= alpha+beta*I, J);
evalc(J1);
value(%);
simplify(convert(diff(%, x), exp), {a+b*I= t});
                            exp(x t)

Int(exp(t*x), x)

  

Int(exp((alpha+I*beta)*x), x)

  

Int(exp(x*alpha)*cos(x*beta), x)+I*(Int(exp(x*alpha)*sin(x*beta), x))

  

alpha*exp(x*alpha)*cos(x*beta)/(alpha^2+beta^2)+beta*exp(x*alpha)*sin(x*beta)/(alpha^2+beta^2)+I*(-beta*exp(x*alpha)*cos(x*beta)/(alpha^2+beta^2)+alpha*exp(x*alpha)*sin(x*beta)/(alpha^2+beta^2))

  

exp((alpha+I*beta)*x)

  

Error, missing operator or `;`

  
> 

 

It is theorem about complex exponentials|, in this case with exp(1)as base

If f(x)=exp(t*x)for all real x and a fixed complex t , then f '(x) =  texp(1)"^(t x) "

 

If you take the integral : int(e^(t*x), x) = e^(t*x)/t(1)

When t ≠ 0, if we let t = α + i*betaand t = α +β and equate the real and imaginairy parts of equation (1)  , we obtain the integration formulas

 

"∫e^(alpha x)cos beta x ⅆx = " exp(alpha*x)*(alpha*cos(beta*x)+beta*sin(beta*x))/(alpha^2+beta^2)

int(e^(alpha*x)*sin*beta*x, x) = 2*exp(alpha*x)*(alpha*sin(beta*x)+beta*cos(beta*x))/(alpha^2+beta^2)

wich are valid if α and β are not both zero

 

 

Consider the differential equation :
`y"`+a*(diff(y(x), x))+b*y(x) = 0
The real and imaginairy parts of the function f(x)=exp(t*x)are solutions of this DE
Now as exercise i can solve this DE to check this
This exercise was for me how to use/find commands for complex variables

> 

#alpha*exp(x*alpha)*cos(x*beta)/(alpha^2 + beta^2) + beta*exp(x*alpha)*sin(x*beta)/#(alpha^2 + beta^2);
> 

 

> 

#normal(%);
> 

#simplify(%);
> 

#a:=-beta*exp(x*alpha)*cos(x*beta)/(alpha^2 + beta^2) + alpha*exp(x*alpha)*sin(x*#beta)/(alpha^2 + beta^2);
> 

#simplify(a);
> 

restart;
> 

Int(exp(x*alpha)*cos(x*beta), x)= alpha*exp(x*alpha)*cos(x*beta)/(alpha^2 + beta^2) + beta*exp(x*alpha)*sin(x*beta)/(alpha^2 + beta^2);

Int(exp(x*alpha)*cos(x*beta), x) = alpha*exp(x*alpha)*cos(x*beta)/(alpha^2+beta^2)+beta*exp(x*alpha)*sin(x*beta)/(alpha^2+beta^2)

 (1)
> 

simplify(%);

Int(exp(x*alpha)*cos(x*beta), x) = exp(x*alpha)*(cos(x*beta)*alpha+sin(x*beta)*beta)/(alpha^2+beta^2)

 (2)
> 

Int(exp(x*alpha)*sin(x*beta), x)= -beta*exp(x*alpha)*cos(x*beta)/(alpha^2 + beta^2) + alpha*exp(x*alpha)*sin(x*beta)/(alpha^2 + beta^2);

Int(exp(x*alpha)*sin(x*beta), x) = -beta*exp(x*alpha)*cos(x*beta)/(alpha^2+beta^2)+alpha*exp(x*alpha)*sin(x*beta)/(alpha^2+beta^2)

 (3)
> 

simplify(%);

Int(exp(x*alpha)*sin(x*beta), x) = -exp(x*alpha)*(cos(x*beta)*beta-sin(x*beta)*alpha)/(alpha^2+beta^2)

 (4)
> 

Int(exp(x*alpha)*sin(x*beta), x)*I= (-beta*exp(x*alpha)*cos(x*beta)/(alpha^2 + beta^2) + alpha*exp(x*alpha)*sin(x*beta)/(alpha^2 + beta^2))*I;

(Int(exp(x*alpha)*sin(x*beta), x))*I = (-beta*exp(x*alpha)*cos(x*beta)/(alpha^2+beta^2)+alpha*exp(x*alpha)*sin(x*beta)/(alpha^2+beta^2))*I

 (5)
> 

 

Download complexe_getallen_post_3_integral_(1).mw

@Carl Love Thanks  ......

janhardo 695
December 23 2021
0 0

@Carl Love 

Thanks
Yes, when i evaluate J1 integral i do get two integrals with their values
With some rearrangement i should get them in the wanted form?

> 

restart;
> 

J:= Int(exp(t*x), x);
J1:= subs(t= a+b*I, J);
evalc(J1);
value(%);
simplify(convert(diff(%, x), exp), {a+b*I= t});
                            exp(x t)

Int(exp(t*x), x)

  

Int(exp((a+I*b)*x), x)

  

Int(exp(x*a)*cos(x*b), x)+I*(Int(exp(x*a)*sin(x*b), x))

  

a*exp(x*a)*cos(x*b)/(a^2+b^2)+b*exp(x*a)*sin(x*b)/(a^2+b^2)+I*(-b*exp(x*a)*cos(x*b)/(a^2+b^2)+a*exp(x*a)*sin(x*b)/(a^2+b^2))

  

exp(t*x)

  

Error, missing operator or `;`

  
> 

 

It is theorem about complex exponentials|, in this case with exp(1)as base

If f(x)=exp(t*x)for all real x and a fixed complex t , then f '(x) =  texp(1)"^(t x) "

 

If you take the integral : int(e^(t*x), x) = e^(t*x)/t(1)

When t ≠ 0, if we let t = α + i*betaand t = α +β and equate the real and imaginairy parts of equation (1)  , we obtain the integration formulas

 

"∫e^(alpha x)cos beta x ⅆx = " exp(alpha*x)*(sin(x*beta)*beta+cos(x*beta)*alpha)/(alpha^2+beta^2)

"∫e^(alpha x)sin beta x ⅆx = "??

wich are valid if α and β are not both zero

 

 

Consider the differential equation :
`y"`+a*(diff(y(x), x))+b*y(x) = 0
The real and imaginairy parts of the function f(x)=exp(t*x)defined on whole x -axis, are solutions of this DE
Now as exercise i can solve this DE to check this
This exercise was for me how to use/find commands for complex variables

Download complexe_getallen_post_3_integral.mw

@acer ThanksYes, i know already ear...

janhardo 695
December 22 2021
0 0

@acer 
Thanks

Yes, i know already earlier (realized)  that it make no sense the diff(%) command here.
That was the next step for me to figure out .( there was a x in it and other variables were symbolic )
I got the same result as in textbook with maple for this integral.
Unfortanely its still difficult to see the relation with the start integral and this integral ? 

@janhardo Complicated those complex...

janhardo 695
December 22 2021
0 0

@janhardo 

O i see the post is crossing from @Carl now ..

The expression (3) is the same as found in the textbook  and alpha and beta unknowns are real numbers..now only take the derative of this expression ? 

> 

 

found this

 

> 

exp((alpha + beta*I)*x)/(alpha + beta*I);

exp((alpha+I*beta)*x)/(alpha+I*beta)

 (1)
> 

evalc(exp((alpha + beta*I)*x)/(alpha + beta*I));

exp(x*alpha)*cos(x*beta)*alpha/(alpha^2+beta^2)+exp(x*alpha)*sin(x*beta)*beta/(alpha^2+beta^2)+I*(exp(x*alpha)*sin(x*beta)*alpha/(alpha^2+beta^2)-exp(x*alpha)*cos(x*beta)*beta/(alpha^2+beta^2))

 (2)
> 

simplify(exp(x*alpha)*cos(x*beta)*alpha/(alpha^2 + beta^2) + exp(x*alpha)*sin(x*beta)*beta/(alpha^2 + beta^2));

exp(x*alpha)*(sin(x*beta)*beta+cos(x*beta)*alpha)/(alpha^2+beta^2)

 (3)
> 

diff(%);

Error, invalid input: diff expects 2 or more arguments, but received 1

  
> 

 

Now i must take the derative of this expression ..how ?  in order to say
"∫f ' ⅆx = f to get the integral "

 

Download complexe_som_tom_apostel_-integratie_deel_2.mw

@janhardo I think i must rewrite (4...

janhardo 695
December 22 2021
0 0

@janhardo 

I think i must rewrite (4) in a exponential e form in order to split the formula in a real and imaginair part in x and y 
f(z)=f(x+i y) =u(x,y)+v(x,y)

Fill in (10):  x+i y .. i can solve the integral too?

@Carl Love ThanksIts only limit tha...

janhardo 695
December 22 2021
0 0

@Carl Love 
Thanks

Its only limit that has a complex option
By using I in the calculation for calculus is enough ?

We can also get Maple to show the steps of the solution by using the same template of commands as before.

> 

restart;

 

 

> 

x^2-4*x+5=0;
eq1 := %:
if rhs(eq1)<>0 then eq2 := lhs(eq1)-rhs(eq1)=0 else eq2 := eq1 end if:
eq3 := student[completesquare](eq2,x):
if patmatch(lhs(eq3),_a::algebraic*(x+_p::algebraic)^2+_q::algebraic,'la') then
   pp := subs(la,_p): aa := subs(la,_a): qq := subs(la,_q):
   bb := simplify(2*aa*pp): cc := simplify(qq+bb^2/(4*aa)):
   eq4 := x^2+bb/aa*x+cc/aa=0:
   if eq4<>eq2 and eq4<>eq1 then print(eq4) end if;
   eq5 := x^2+bb/aa*x=-cc/aa:
   if eq5<>eq1 then print(eq5) end if;
   rr := simplify((bb^2-4*aa*cc)/(4*aa^2));
   print(x^2+bb/aa*x+pp^2=rr);
   print((x+pp)^2=rr);
   ss := simplify(bb^2-4*aa*cc);
   print(x+pp=sqrt(ss)/(2*aa),x+pp=-sqrt(ss)/(2*aa));
   print(x=-pp+sqrt(ss)/(2*aa),x=-pp-sqrt(ss)/(2*aa));
end if:

x^2-4*x+5 = 0

  

x^2-4*x = -5

  

x^2-4*x+4 = -1

  

(x-2)^2 = -1

  

x-2 = I, x-2 = -I

  

x = 2+I, x = 2-I

 (1)
> 

;
> 

solve(x^2-4*x+5=0, x);# Complex numbers are the default numbersystem

2+I, 2-I

 (2)
> 

fsolve(x^2-4*x+5=0, x,complex);

2.000000000-1.000000000*I, 2.+1.*I

 (3)
> 

 

> 

 

Did not found not yet other commands that you mentioned for complex calculus, that makes it easier for doing math with complex numbers

 

> 

 

Note : the step by step calculation made by Peter Stone makes no distinction in  x1  and x2  as roots , so that's not clear

 

> 

 

int(exp(t*x), x)NULL

> 

 

> 

Int(exp(t*x),x)=int(exp(t*x),x);# page 370 ,calculus volume 1, tom.m.apostal

Int(exp(t*x), x) = exp(t*x)/t

 (4)
> 

 

When t ≠ 0, .If we let  t = α+ Iβ , so t becomes now a complex number and equate the real and imaginary parts of equation (4) we obtain some integration formulas
How to do this ? , z= a + bI , it is only by using I that maple is using the complex number system.

> 

t:=alpha + beta*I;

alpha+I*beta

 (5)
> 

 

> 

 

Gauss (1777 - 1855) first used the notation a+bi.
"a" is called the real part and
"b" is called the imaginary part.
Also Maple knows this:

> 

Re(2+4*I);

2

 (6)
> 

Im(2+4*I);

4

 (7)

 

> 

 

> 

Re(alpha + beta*I);

Re(alpha)-Im(beta)

 (8)
> 

Im(alpha + beta*I);

Im(alpha)+Re(beta)

 (9)
> 

Int(exp((alpha + beta*I)*x), x)=int(exp((alpha + beta*I)*x), x);

Int(exp((alpha+I*beta)*x), x) = exp((alpha+I*beta)*x)/(alpha+I*beta)

 (10)
> 

 

Try to find the two integrationformulas ?,   but the idea using I  maple comes in the complex mode

Download complexe_getallen_post.mw

@janhardo  I will look at all comm...

janhardo 695
December 20 2021
0 0

@janhardo 

I will look at all commands , but using them is another story 
Got here book with some examples for signal processing, but never did something with it

@vv  Thanks Must look at the dffe...

janhardo 695
December 20 2021
0 0

@vv 

Thanks

Must look at the dfference betwween  of complexplot
create a 2-D complex plot
and conformal plot of a complex function.

@janhardo  Only problem is to inst...

janhardo 695
December 20 2021
0 0

@janhardo 

Only problem is to install the complex.m package for  the worksheets.
Note:  and get the same issue with a package another lessonserie 

@Thomas Richard  Maybe fractals an...

janhardo 695
December 20 2021
0 0

@Thomas Richard 

Maybe fractals and i will look in the packages you mentioned

@acer Thanks i could look in t...

janhardo 695
December 20 2021
0 0

@acer 

Thanks 

i could look in those worksheets 

@Thomas Richard ThanksYes, the work...

janhardo 695
December 20 2021
0 0

@Thomas Richard 

Thanks

Yes, the worksheet is made in 2002 by a person who studied at the same time i did for secondairy schools teacher math. 
Only he studied falso or mathematical engineering too at the same time, so trained with Maple
It was possible to follow two studies at the same time, but i was too late and it just finished the second study then when i ask for. ( in fact years too late , i did not knew it)

I think i will be most complex functions , so it will be generally then.
Don't know yet not else what more to do ?

tomleslie ThanksI am wondering if a...

janhardo 695
December 20 2021
0 0

tomleslie 

Thanks

I am wondering if all code in the book is rubbish , that 's why i came up with this example. (checktri) from the book from wich i think this could be obselote code?
If i am able to recognise bad code, then chance it in good code easily, that was the idea.
How to recognise this obselote code then?

So i don't want to learn how to code badly, no of course not  
I will study your made example with a subprocedure in it and  downloaded the programming guide.

Your code example is not the same a s intentend as in the book 
Its about knowing what type of triangle is concerned ( 4 types) and a plot with a normal vector what acts as a help for visualizing ( read ) the plot in the right position 

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