Where does the "i" go

  • #1
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Summary:
When solving a differential equation
Good Morning

I have a second order, linear differential equation. I solve it with complex exponentials. When I am done, I contract terms producing this:

q = exp(some term regarding damping) * ( C+D) cos(wt) + i (C-D) sin(wt) )

Then I define new constants (prior to using the initial conditions)

A = C+D
B = i(C-D) // I forgot the "i" here when I first posted this, but the rest of my post still has the question that I am asking. My dropping of the "i" here was a grave mistake in stating my own question... I am sorry.

And I have

q = exp(some term) * Acos(wt) +Bsin(wt) )

OK; what happened to the "i?"

Now, I know I can leave the original constants and work those through.
I also have a "feeling" that the new numbers COULD involve anything, even imaginary terms.
And I work it out. So all I did was assign B and ignored the "i"

But I always get a funny feeling at this step, and would like (hope) for someone to explain: "why can I justify, at that moment, why I can drop the "i" from consideration of the new definition B. No book I have read, ever addresses this point, and it may, indeed, be silly/Obvious, but I would really appreciate some thoughts by others.

I mean, I get taht I can say "iB is just a constant." But to my stubborn mind, it is a constant from the imaginary world.
 
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Answers and Replies

  • #2
BvU
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OK; what happened to the "i?"
If you first have i(C-D) and then write B as the coefficient in front of the sine, you have simply left it out :smile: !?!?!
(swept it under the rug, forgot it, whatever).

Please write out and post the differential equation and the steps you take in detail.
 
  • #3
243
36
If you first have i(C-D) and then write B as the coefficient in front of the sine, you have simply left it out :smile: !?!?!
(swept it under the rug, forgot it, whatever).

Please write out and post the differential equation and the steps you take in detail.


No

Every book does this. They drop the i, and now I am fairly sure they fail to explain this process. Just look at any book.

Mine is just a second order, linear, diff.eq for spring, mass and damping and no forcing function. Every book gets to this point and absorbs the i.
 
  • #4
DaveE
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Then I would guess they are assuming A and B are complex numbers. Otherwise it doesn't make sense. You are correct, the "i" can't just vanish.

Anyway, it's in the definition you provided:
C+D) cos(wt) + i (C-D) sin(wt) = Acos(wt) +Bsin(wt)
 
  • #5
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Then I would guess they are assuming A and B are complex numbers. Otherwise it doesn't make sense. You are correct, the "i" can't just vanish.

Anyway, it's in the definition you provided:
C+D) cos(wt) + i (C-D) sin(wt) = Acos(wt) +Bsin(wt)


Well, that is interesting... that they assume that A and B could be potentially complex. You could be correct, but I am still uneasy.

Yet, even there, I have never seen it stated in any book.
 
  • #6
DaveE
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The other possibility is that they are assuming that you only care about the real part of the equation, and they are saying it's a generic sinusoid. In that case B ≠ i(C-D). So, because they've shown the answer is a sinusoid they switch to an alternate expression for it. i.e. Re( (C+D)cos(wt) + i(C-D)sin(wt) ) = Acos(wt) + Bsin(wt).
Personally, I find this sloppy and distasteful. If they are going to do that they need to say it explicitly (like I did).

Of course then C & D must be complex and A & B real, or else it's just stupid.
 
  • #7
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The other possibility is that they are assuming that you only care about the real part of the equation, and they are saying it's a generic sinusoid. In that case B ≠ i(C-D). So, because they've shown the answer is a sinusoid they switch to an alternate expression for it. i.e. Re( (C+D)cos(wt) + i(C-D)sin(wt) ) = Acos(wt) + Bsin(wt).
Personally, I find this sloppy and distasteful. If they are going to do that they need to say it explicitly (like I did).

Of course then C & D must be complex and A & B real, or else it's just stupid.


Well, to address your first point, they are setting it up for the forcing functions (initially sine and cosine -- long before they get to Fourier series).

Then, once they get a solution, the books say "If the forcing function was cosine, take the real and drop the imaginary. If the forcing was sine, drop the real part." I am used to that sloppiness.

But I am still uncomfortable with how they just drop the "i" when they are done with the machinery of complex exponentials and wish to return to the world of sinusoids.


Your comment "Personally, I find this sloppy and distasteful."
is reassuring to me, but still, I am hoping for further clarificaiotn.

Maybe it is just that engineering-math books DO get lazy.
 
  • #8
Gaussian97
Homework Helper
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Summary:: When solving a differential equation

Good Morning

I have a second order, linear differential equation. I solve it with complex exponentials. When I am done, I contract terms producing this:

q = exp(some term regarding damping) * ( C+D) cos(wt) + i (C-D) sin(wt) )

Then I define new constants (prior to using the initial conditions)

A = C+D
B = C-D

And I have

q = exp(some term) * Acos(wt) +Bsin(wt) )

OK; what happened to the "i?"
This is wrong, if you define ##A## and ##B## like that, then you will have the term $$A\cos(wt) +iB\sin(wt) )$$. To get rid of the ##i## you must define ##B = i(C-D)##.
I mean, I get taht I can say "iB is just a constant." But to my stubborn mind, it is a constant from the imaginary world.
Well, and ##C## and ##D## may be a constant from the "negative world", or worst, a constant of the "irrational world", whatever these "worlds" mean.
They could even be non-computable constants! What is your problem with that?
 
  • #9
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This is wrong, if you define ##A## and ##B## like that, then you will have the term $$A\cos(wt) +iB\sin(wt) )$$. To get rid of the ##i## you must define ##B = i(C-D)##.

Well, and ##C## and ##D## may be a constant from the "negative world", or worst, a constant of the "irrational world", whatever these "worlds" mean.
They could even be non-computable constants! What is your problem with that?
OK, that was ONE sloppiness in my typing. I have fixed the original post.
I left out the i

Anyway, your last sentence is the one that seems to address the point.

Thank you!
 
  • #10
BvU
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OK, that was ONE sloppiness in my typing. I have fixed the original post.
Even though my post #2 now looks silly: thanks !
Every book does this. They drop the i, and now I am fairly sure they fail to explain this process. Just look at any book.

Mine is just a second order, linear, diff.eq for spring, mass and damping and no forcing function. Every book gets to this point and absorbs the i.
Ok, so I took 'any book' (borrowed James Stewart: Calculus, 5th ed from my son :cool: ). Chapter 17.

It says a second order linear differential equation like you describe in its most general form looks like $$P(x)\,y'' + Q(x)\,y'+ R(x)\,y = G(x)$$ and it's homogeneous if ##G(x)=0## .
For the spring/mass with damping ##P, Q, R## are constants $$ay'' + by'+ c = 0$$ and a trial solution of the form ##\ y = e^{rx}\ ## seems to be a good candidate, leading to $$\left (ar^2 + br + c\right ) e^{rx} = 0$$ and a characteristic equation$$ar^2 + br + c= 0$$which has two (possibly complex) solutions ##r_1## and ##r_2##.
The most general solution of ## ay'' + by'+ c = 0 ## is then$$y = C_1 e^{r_1x} + C_2 e^{r_2x}$$with ##C_1## and ##C_2## (which may well be complex) determined by initial (or other) boundary conditions.

We look at the case ##r_1## and ##r_2## unequal and both complex.

And sure enough we now come to your point of confusion when ##r_1## and ##r_2## are written as $$r_1 = \alpha + i\beta,\quad r_2 = \alpha - i\beta$$leading to $$y = e^{\alpha x} \left ( c_1\cos\beta x + c_2\sin\beta x \right )\qquad\quad (1)$$with $$c_1 = C_1+C_2, \quad c_2 = i(C_1-C_2)\ .$$Nothing has been omitted, left out, swept under the rug or anything. Equation (1) gives all solutions, real or complex. Then:
Stewart said:
The solutions are real when the constants ##c_1## and ##c_2## are real
and this is the case for the spring/mass system.

I also have a "feeling" that the new numbers COULD involve anything, even imaginary terms.
You are absolutely right mathematically spoken. But for us physicists, the fact that what we describe is usually real makes that we forget the most general solution.

(the fun starts when we nevertheless add an imaginary part to the solution, just to drop it again at the very end of a whole lot of processing, claiming that we only need to look at the real part :biggrin: )

##\ ##


##\ ##
 
Last edited:
  • #11
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36
Even though my post #2 now looks silly: thanks !

I am so sorry about that first error.

However, thank you for the rest of your comment. I understand now.
 
  • #12
ergospherical
Gold Member
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It's not necessary to mess around with this algebra; because ##e^{-f(t)} e^{\pm i\omega t}## is a basis of the space of complex solutions, then the real and imaginary parts of the first complex solution ##\{ e^{-f(t)} \cos{\omega t}, e^{-f(t)} \sin{\omega t} \}## constitute a basis of the space of real solutions. That means, any linear combination ##a e^{-f(t)} \cos{\omega t} + be^{-f(t)} \sin{\omega t}##, with ##a## and ##b## real, is a real solution.

[##f(t)## is the unspecified damping term mentioned in the OP.]
 
  • #13
pasmith
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Summary:: When solving a differential equation

Good Morning

I have a second order, linear differential equation. I solve it with complex exponentials. When I am done, I contract terms producing this:

q = exp(some term regarding damping) * ( C+D) cos(wt) + i (C-D) sin(wt) )

Then I define new constants (prior to using the initial conditions)

A = C+D
B = i(C-D) // I forgot the "i" here when I first posted this, but the rest of my post still has the question that I am asking. My dropping of the "i" here was a grave mistake in stating my own question... I am sorry.

And I have

q = exp(some term) * Acos(wt) +Bsin(wt) )

OK; what happened to the "i?"

Now, I know I can leave the original constants and work those through.
I also have a "feeling" that the new numbers COULD involve anything, even imaginary terms.
And I work it out. So all I did was assign B and ignored the "i"

But I always get a funny feeling at this step, and would like (hope) for someone to explain: "why can I justify, at that moment, why I can drop the "i" from consideration of the new definition B. No book I have read, ever addresses this point, and it may, indeed, be silly/Obvious, but I would really appreciate some thoughts by others.

You want [itex]q = Ce^{r_1t} + De^{r_2t}[/itex] to be real. The easiet way to do that is to take [itex]C[/itex] and [itex]D[/itex] to be complex conjugates, because then [itex]q[/itex] is a sum of complex conjugates and therefore real. So: [tex]
\begin{align*}
A = C + D &\mbox{ is real because it's the sum of complex conjugates.} \\
B = i(C - D) &\mbox{ is real because it's $i$ times the difference of complex conjugates.}
\end{align*}
[/tex]
 
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