MHB How do we get the imaginary part?

[evinda]

Hello! (Wave)

We have this: $y''+y=\frac{1}{\cos x} , y'(0)=0, y(\pi)=0$.

Using the Green function I got that $y(x)= x \sin x+ \cos x( -\ln |\cos \pi|+ \ln |\cos x|)=x \sin x+ \cos x (\ln |\cos x|)$.

But according to Wolfram: y'''''''+'y'='1'/'cosx , y''''('0')''='0, y'('pi')''='0 - Wolfram|AlphaHow do we get the imaginary part?

 

[I like Serena]

Hello! (Wave)

We have this: $y''+y=\frac{1}{\cos x} , y'(0)=0, y(\pi)=0$.

Using the Green function I got that $y(x)= x \sin x+ \cos x( -\ln |\cos \pi|+ \ln |\cos x|)=x \sin x+ \cos x (\ln |\cos x|)$.

But according to Wolfram: y'''''''+'y'='1'/'cosx , y''''('0')''='0, y'('pi')''='0 - Wolfram|Alpha

How do we get the imaginary part?

Hey evinda! (Smile)

Wolfram always treats functions as complex functions.
That has in particular impact on the natural logarithm.

Consider that if we're talking about real functions that we actually have:
$$
\int \frac{dx}{x} = \begin{cases}\ln(x) + C_1&\text{if }x>0 \\ \ln(-x) + C_2&\text{if }x<0\end{cases}
$$
That's because the domain is split into 2 disconnected intervals for which we have independent integration constants.
This is usually written as:
$$\int \frac{dx}{x} = \ln|x| + C$$
with the understanding that the integration constant $C$ can be different on each part of the domain. (Nerd)

However, with complex functions there is no such distinction - all of the domain is connected with only an undefined point at $z=0$. That is:
$$
\int \frac{dz}{z} = \ln z + C
$$
Furthermore, $\ln z$ is a so called multivalued function that behaves a bit different than a normal function.
That is, we can't just take the natural logarithm of a complex number. (Sweating)So instead of:
$$
-\ln|\cos\pi| + \ln|\cos x|
$$
we really have something like:
$$
-\ln(\cos\pi) + \ln(\cos x) = \ln\frac{\cos x}{\cos\pi} = \ln(-\cos x)
$$

Now let's define $z = \ln(-\cos x)$, so that we have $e^z = -\cos x$.
Since we're talking about complex numbers, we can write this as:
$$e^{z+2\pi ik} = -\cos x
\quad\Rightarrow\quad e^{z+i\pi+2\pi ik} = \cos x
\quad\Rightarrow\quad z+i\pi+2\pi ik = \ln(\cos x)
\quad\Rightarrow\quad z = \ln(\cos x)-i\pi-2\pi ik
$$
(Whew)

 

[evinda]

Hey evinda! (Smile)

Wolfram always treats functions as complex functions.
That has in particular impact on the natural logarithm.

Consider that if we're talking about real functions that we actually have:
$$
\int \frac{dx}{x} = \begin{cases}\ln(x) + C_1&\text{if }x>0 \\ \ln(-x) + C_2&\text{if }x<0\end{cases}
$$
That's because the domain is split into 2 disconnected intervals for which we have independent integration constants.
This is usually written as:
$$\int \frac{dx}{x} = \ln|x| + C$$
with the understanding that the integration constant $C$ can be different on each part of the domain. (Nerd)

However, with complex functions there is no such distinction - all of the domain is connected with only an undefined point at $z=0$. That is:
$$
\int \frac{dz}{z} = \ln z + C
$$
Furthermore, $\ln z$ is a so called multivalued function that behaves a bit different than a normal function.
That is, we can't just take the natural logarithm of a complex number. (Sweating)So instead of:
$$
-\ln|\cos\pi| + \ln|\cos x|
$$
we really have something like:
$$
-\ln(\cos\pi) + \ln(\cos x) = \ln\frac{\cos x}{\cos\pi} = \ln(-\cos x)
$$

Now let's define $z = \ln(-\cos x)$, so that we have $e^z = -\cos x$.
Since we're talking about complex numbers, we can write this as:
$$e^{z+2\pi ik} = -\cos x
\quad\Rightarrow\quad e^{z+i\pi+2\pi ik} = \cos x
\quad\Rightarrow\quad z+i\pi+2\pi ik = \ln(\cos x)
\quad\Rightarrow\quad z = \ln(\cos x)-i\pi-2\pi ik
$$
Note that Wolfram's answer is actually incomplete, since it leaves out the $2\pi i k$. (Whew)

I see... (Yes)

But if we do not treat functions as complex functions, my result is correct, right?

 

[I like Serena]

I see... (Yes)

But if we do not treat functions as complex functions, my result is correct, right?

Yes.
We can verify (and we should) by filling it in in the original equation, and by also checking the boundary conditions. (Nerd)

 

[evinda]

Yes.
We can verify (and we should) by filling it in in the original equation, and by also checking the boundary conditions. (Nerd)

Ok... Thank you! (Smile)