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janhardo 745
December 24 2021
0 0

@janhardo 

> 

 

> 

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

 (1)
> 

#==========================================
> 

restart;         
J:= Int(exp(t*x), x);
J1:= subs((SUB:= t= alpha+beta*I), J);
J2:= evalc([Re,Im](J1));
J3:= J2 =~ simplify(value(J2));
print~(J3):
 
#reversion to original integrand:
simplify(
    convert(diff(evalc(Complex(rhs~(J3)[])), x), exp),
    {rhs=lhs}(SUB)
);

Int(exp(t*x), x)

  

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

  

[Int(exp(x*alpha)*cos(x*beta), x), Int(exp(x*alpha)*sin(x*beta), x)]

  

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

  

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

  

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

  

exp(t*x)

 (2)
> 

#================================================
> 

 

The difference in code starts with  evalc and the use of  J,J1,J2 and J3 makes it easier to read

Its a difficult command : J2:= evalc([Re,Im](J1)); in order to get rid of the I symbol, and probably there is no need for?
I must study further this to see how it excactly works.

 

Can you explain in your own words how this second code works ?       

Download complex_integration-tom_apostel_example_-bernouille.mw

@acer ThanksIt was tricky, because ...

janhardo 745
December 24 2021
0 0

@acer 

Thanks

Your code is also a difficult one to understand quick , so perhaps you can explain this in your own words too for me to get a idea for me.

It was tricky, because this  expression written (see hereunder)i s not the same in Holland or America/England? 
Log and ln are the same in America ( and in the book example too probably) and not in Holland
Log and ln are different. 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)

So log written in the formula must read as ln then.
Your answer is almost correct , its only the I (imaginairy unit) is still there

In a another example of calculating a complex integral is used to remove the I   ....see hereunder

J:= Int(exp(t*x), x);
J1:= subs((SUB:= t= alpha+beta*I), J);
J2:= evalc([Re,Im](J1));

@Carl Love ThanksThere will be a re...

janhardo 745
December 24 2021
0 0

@Carl Love 

Thanks

There will be a reason why maple uses  log and ln for the same thing ..yes use in English math books ? 
In written math books here in Holland there is a distinction between the two: log and ln and they differ in meaning
Probably not in English math books : log and ln are the same then.

@Carl Love ThanksI noticed the diff...

janhardo 745
December 24 2021
0 0

@Carl Love 

Thanks

I noticed the difference between input and output coloring, so its a mistake 
This second code piece construct the two wanted integrals from 

 J := Int(exp(t*x), x)

Amazing second code piece ,but it needs some clarifcation compared with the first code piece. 

@Carl Love I used 2d input expressi...

janhardo 745
December 24 2021
0 0

@Carl Love 

I used 2d input expression palette and was writing , so it is not executable math then.
Was not further thinking on it , but it is possible to use the log and  ln for the same logaritme with  base e  : 

log[10](100);    can be used also in worksheet .
                             2
The book example use log as notation ( i assume with base 10 ) 

A  2 d input example                         

log[10](100);    
               2
ln(100.0);

             4.605170186
 

@janhardo Complex integration is so...

janhardo 745
December 23 2021
0 0

@janhardo 

Complex integration is somewhere started and ...

-------------------------------------------------------------------------------------------------info------------------------

So far, we’ve seen how to evaluate integrals of simple functions of a complex variable—that were defi ned in terms of a single real parameter we called t. Now it’s time to generalize and consider a more general case, where we just say we’re integrating a function of a complex variable f (z ), where ∈ C. This can be done using a technique called contour integration. The reason integrals of complex functions are done the way they are is that while an integral of a real-valued function is defi ned on an interval of the line, an integral  of a complex-valued function is defi ned on a curve in the complex plane.

@acer ThanksYes, there is leading m...

janhardo 745
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 745
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 745
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 745
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 745
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 745
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 745
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 745
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 745
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

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