MHB How can we prove that k is an integer?

[mathmari]

Hey!

I want to prove the following implication:

$$k \in \mathbb{Z} \Leftrightarrow ce^x-1 \mid c^ke^{kx}-1$$ For the direction $\Rightarrow$ I tried the following:

• $k >0$:
  $$\sum_{i=0}^{k-1} (ce^x)^i=\frac{(ce^x)^k-1}{ce^x-1} \\ \Rightarrow (ce^x)^k-1=(ce^x-1)\sum_{i=0}^{k-1} (ce^x)^i$$
  So when $k$ is an integer $>0$ we have that $ce^x-1 \mid c^ke^{kx}-1$.
  
• $k <0$:
  $$\sum_{i=0}^{k-1} ((ce^x)^{-1})^i=\frac{\left (\frac{1}{ce^x}\right )^k-1}{\frac{1}{ce^x}-1}=\frac{\frac{1-(ce^x)^k}{(ce^x)^k}}{\frac{1-ce^x}{ce^x}}=\frac{ce^x}{(ce^x)^k} \frac{1-(ce^x)^k}{1-ce^x} =\frac{1}{(ce^x)^{k-1}} \frac{(ce^x)^k-1}{ce^x-1} \\ \Rightarrow (ce^x)^k-1=(ce^x-1)(ce^x)^{k-1}\sum_{i=0}^{k-1} ((ce^x)^{-1})^i$$
  So when $k$ is an integer $<0$ we have that $ce^x-1 \mid c^ke^{kx}-1$.

Is this correct? (Wondering)
Could you give me a hint how we could show the other direction?

We suppose that $ce^x-1 \mid c^ke^{kx}-1$, that means the solutions of $ce^x-1$ are also solutions of $c^ke^{kx}-1$, right?

Does this help? (Wondering)

 

[mathmari]

Since we are working in the ring $\mathbb{C}[e^{\lambda z} \mid \lambda \in \mathbb{C}]$ we can set $c = e^{-\beta}$.

Then
$$ce^x-1=e^{-\beta}e^x-1 =e^{-\beta}(e^x-e^{\beta})$$
$$c^ke^{kx}-1=e^{-\beta k}e^{kx}-1 =e^{-\beta k}(e^{kx}-e^{k\beta})$$

$$ce^x-1 \mid c^ke^{kx}-1 \Rightarrow c^ke^{kx}-1 =A(ce^x-1) \Rightarrow e^{-\beta k}(e^{kx}-e^{k\beta})=Ae^{-\beta}(e^x-e^{\beta}) \\ \Rightarrow e^{kx}-e^{k\beta}=Ae^{\beta (k-1)}(e^x-e^{\beta}) \\ \Rightarrow e^x-e^{\beta} \mid e^{kx}-e^{k\beta}$$ Is this correct? (Wondering) We suppose that $k \notin \mathbb{Z}$.

Since $e^x-e^{\beta} \mid e^{kx}-e^{k \beta}$, we have that the roots of $e^x-e^{\beta}$ are also roots of $e^{kx}-e^{k\beta}$.
Let $\alpha$ be a root of $e^x-e^{\beta}=0$. Then $e^{\alpha}-e^{\beta} \Rightarrow \alpha=\beta+2n\pi i$.

Since $\alpha$ is also a root of $e^{kx}-e^{k\beta}=0$, we have that $e^{k\alpha}-e^{k\beta}=0 \Rightarrow e^{k(\beta+2n\pi i)}-e^{k\beta}=0 \Rightarrow e^{k\beta} e^{2kn\pi i}-e^{k\beta}=0 \Rightarrow e^{k\beta} (e^{2kn\pi i}-1)=0 \Rightarrow e^{2kn\pi i}=1$ Is this correct so far? How do we get a contradiction? (Wondering)

 

[Opalg]

Since we are working in the ring $\mathbb{C}[e^{\lambda z} \mid \lambda \in \mathbb{C}]$ we can set $c = e^{-\beta}$.

Then
$$ce^x-1=e^{-\beta}e^x-1 =e^{-\beta}(e^x-e^{\beta})$$
$$c^ke^{kx}-1=e^{-\beta k}e^{kx}-1 =e^{-\beta k}(e^{kx}-e^{k\beta})$$

$$ce^x-1 \mid c^ke^{kx}-1 \Rightarrow c^ke^{kx}-1 =A(ce^x-1) \Rightarrow e^{-\beta k}(e^{kx}-e^{k\beta})=Ae^{-\beta}(e^x-e^{\beta}) \\ \Rightarrow e^{kx}-e^{k\beta}=Ae^{\beta (k-1)}(e^x-e^{\beta}) \\ \Rightarrow e^x-e^{\beta} \mid e^{kx}-e^{k\beta}$$ Is this correct? (Wondering) We suppose that $k \notin \mathbb{Z}$.

Since $e^x-e^{\beta} \mid e^{kx}-e^{k \beta}$, we have that the roots of $e^x-e^{\beta}$ are also roots of $e^{kx}-e^{k\beta}$.
Let $\alpha$ be a root of $e^x-e^{\beta}=0$. Then $e^{\alpha}-e^{\beta} \Rightarrow \alpha=\beta+2n\pi i$.

Since $\alpha$ is also a root of $e^{kx}-e^{k\beta}=0$, we have that $e^{k\alpha}-e^{k\beta}=0 \Rightarrow e^{k(\beta+2n\pi i)}-e^{k\beta}=0 \Rightarrow e^{k\beta} e^{2kn\pi i}-e^{k\beta}=0 \Rightarrow e^{k\beta} (e^{2kn\pi i}-1)=0 \Rightarrow e^{2kn\pi i}=1$ Is this correct so far? How do we get a contradiction? (Wondering)

Without going through that carefully, it looks to me as though you have correctly shown that $e^{2kn\pi i}=1$, and that this holds for all integers $n$. In particular, it holds for $n=1$, so that $e^{2k\pi i}=1.$ But that implies that $k$ is an integer.

 

[mathmari]

Without going through that carefully, it looks to me as though you have correctly shown that $e^{2kn\pi i}=1$, and that this holds for all integers $n$. In particular, it holds for $n=1$, so that $e^{2k\pi i}=1.$ But that implies that $k$ is an integer.

I see... Thanks a lot! (Mmm) We have that $c=e^{-\beta}$, where $c , \beta \in \mathbb{C}$.

Does at the following part

$$ce^x-1=e^{-\beta}e^x-1 =e^{-\beta}(e^x-e^{\beta})$$
$$c^ke^{kx}-1=e^{-\beta k}e^{kx}-1 =e^{-\beta k}(e^{kx}-e^{k\beta})$$

stand that $c^k=e^{-\beta k}$ ? (Wondering)

This means the following:

$$c^k=(e^{-\beta})^k$$ Is this equal to $e^{-\beta k}$ where $\beta \in \mathbb{C}$ ? (Wondering)

 

[Opalg]

$$c^k=(e^{-\beta})^k$$ Is this equal to $e^{-\beta k}$ where $\beta \in \mathbb{C}$ ? (Wondering)

Yes, the familiar power laws for real numbers apply equally well in the complex case. So $c^zc^w = c^{z+w}$, and $(c^z)^w = c^{zw}$.

 

[mathmari]

Yes, the familiar power laws for real numbers apply equally well in the complex case. So $c^zc^w = c^{z+w}$, and $(c^z)^w = c^{zw}$.

I found in Wikipedia that it doesn't stand in the complex case.

 

[Opalg]

Yes, the familiar power laws for real numbers apply equally well in the complex case. So $c^zc^w = c^{z+w}$, and $(c^z)^w = c^{zw}$.

I found in Wikipedia that it doesn't stand in the complex case.

I should have put that more precisely. The power laws hold for all complex $z$ and $w$, when $c$ is a positive real number (as is the case in your problem).

 

[mathmari]

So the implication $$k \in \mathbb{Z} \Leftrightarrow ce^x-1 \mid c^ke^{kx}-1$$ stands only when $c$ is a positive real number? (Wondering)

 

[mathmari]

Does the following also stand $$k \in \mathbb{Z} \iff ce^x-1 \mid de^{kx}-1$$ ? (Wondering) Or does $d$ have to be of the form $c^k$ ? (Wondering)