# Verify that ## L(1, \chi)\neq 0 ##

• Math100
In summary: Thank you for clarifying!In summary, the table of values for nonprincipal Dirichlet characters modulo ## 16 ## shows that ## L(1, \chi) ## is calculated for three specific characters: ## \chi_{2}, \chi_{5}, \chi_{6} ##. The sums for each character are found to be nonzero, indicating that ## L(1, \chi) ## is also nonzero for these characters. This demonstrates that ## L(1, \chi) \neq 0 ## for each real-valued nonprincipal character modulo ## 16 ##. It is worth noting that ## L(1, \chi) \in \mathbb{C} ##.
Math100
Homework Statement
For each real-valued nonprincipal character ## \chi ## mod ## 16 ##, verify that ## L(1, \chi)\neq 0 ##.
Relevant Equations
## L(1, \chi)=\sum_{n=1}^{\infty}\frac{\chi(n)}{n} ##
Note that ## L(1, \chi)\neq 0 ## when ## \chi ## is a nonprincipal character.
From a table of values for all the nonprincipal Dirichlet characters modulo ## 16 ##, we have that
\begin{array}{|c|c|c|c|c|c|c|c|c|}
\hline n & 1 & 3 & 5 & 7 & 9 & 11 & 13 & 15 \\
\hline \chi_{1}(n) & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 \\
\hline \chi_{2}(n) & 1 & -1 & 1 & -1 & 1 & -1 & 1 & -1 \\
\hline \chi_{3}(n) & 1 & i & i & 1 & -1 & -i & -i & -1 \\
\hline \chi_{4}(n) & 1 & -i & i & -1 & -1 & i & -i & 1 \\
\hline \chi_{5}(n) & 1 & 1 & -1 & -1 & 1 & 1 & -1 & -1 \\
\hline \chi_{6}(n) & 1 & -1 & -1 & 1 & 1 & -1 & -1 & 1 \\
\hline \chi_{7}(n) & 1 & i & -i & -1 & -1 & -i & i & 1 \\
\hline \chi_{8}(n) & 1 & -i & -i & 1 & -1 & i & i & -1 \\
\hline
\end{array}

So ## L(1, \chi)=\sum_{n=1}^{\infty}\frac{\chi_{k}(n)}{n} ## with ## k\in\left \{ 2, 5, 6 \right \} ##.
Observe that
\begin{align*}
&\sum_{n=1}^{\infty}\frac{\chi_{2}(n)}{n}=\frac{\chi_{2}(1)}{1}+\frac{\chi_{2}(3)}{3}+\frac{\chi_{2}(5)}{5}+\frac{\chi_{2}(7)}{7}+\frac{\chi_{2}(9)}{9}+\frac{\chi_{2}(11)}{11}+\frac{\chi_{2}(13)}{13}+\frac{\chi_{2}(15)}{15}\\
&1+(-\frac{1}{3})+\frac{1}{5}+(-\frac{1}{7})+\frac{1}{9}+(-\frac{1}{11})+\frac{1}{13}+(-\frac{1}{15})=\frac{33976}{45045}\neq 0\\
&\sum_{n=1}^{\infty}\frac{\chi_{5}(n)}{n}=\frac{\chi_{5}(1)}{1}+\frac{\chi_{5}(3)}{3}+\frac{\chi_{5}(5)}{5}+\frac{\chi_{5}(7)}{7}+\frac{\chi_{5}(9)}{9}+\frac{\chi_{5}(11)}{11}+\frac{\chi_{5}(13)}{13}+\frac{\chi_{5}(15)}{15}\\
&1+\frac{1}{3}+(-\frac{1}{5})+(-\frac{1}{7})+\frac{1}{9}+\frac{1}{11}+(-\frac{1}{13})+(-\frac{1}{15})=\frac{47248}{45045}\neq 0\\
&\sum_{n=1}^{\infty}\frac{\chi_{6}(n)}{n}=\frac{\chi_{6}(1)}{1}+\frac{\chi_{6}(3)}{3}+\frac{\chi_{6}(5)}{5}+\frac{\chi_{6}(7)}{7}+\frac{\chi_{6}(9)}{9}+\frac{\chi_{6}(11)}{11}+\frac{\chi_{6}(13)}{13}+\frac{\chi_{6}(15)}{15}\\
&1+(-\frac{1}{3})+(-\frac{1}{5})+\frac{1}{7}+\frac{1}{9}+(-\frac{1}{11})+(-\frac{1}{13})+\frac{1}{15}=\frac{27904}{45045}\neq 0.\\
\end{align*}
Thus
\begin{align*}
&L(1, \chi)=\sum_{n=1}^{\infty}\frac{\chi_{2}(n)}{n}+\sum_{n=1}^{\infty}\frac{\chi_{5}(n)}{n}+\sum_{n=1}^{\infty}\frac{\chi_{6}(n)}{n}\\
&=\frac{33976}{45045}+\frac{47248}{45045}+\frac{27904}{45045}\\
&=\frac{36376}{15015}\\
&\neq 0.\\
\end{align*}
Therefore, ## L(1, \chi)\neq 0 ## for each real-valued nonprincipal character ## \chi ## mod ## 16 ##.

You have ##\chi\in \{\chi_2,\chi_5,\chi_6\}.## ##L(1,\chi)## is not summed over all non-principal characters so everything from "Thus" on must be deleted. You have to calculate ##L(1,\chi_2),L(1,\chi_5),L(1,\chi_6)## which you did.

The rest is ok. Maybe it's worth mention that ##L(1,\chi)\in \mathbb{C}.##

Math100
fresh_42 said:
You have ##\chi\in \{\chi_2,\chi_5,\chi_6\}.## ##L(1,\chi)## is not summed over all non-principal characters so everything from "Thus" on must be deleted. You have to calculate ##L(1,\chi_2),L(1,\chi_5),L(1,\chi_6)## which you did.

The rest is ok. Maybe it's worth mention that ##L(1,\chi)\in \mathbb{C}.##
So it's asking for the summation for each of the ## \chi_{2}, \chi_{5}, \chi_{6} ## then?

Math100 said:
So it's asking for the summation for each of the ## \chi_{2}, \chi_{5}, \chi_{6} ## then?
No. One sum per character, makes three sums. Everything was fine until you started with "Thus".

\begin{align*}
L(1,\chi_2)&=\sum_{n=1}^\infty \dfrac{\chi_2(n)}{n}=\ldots \neq 0\\
L(1,\chi_5)&=\sum_{n=1}^\infty \dfrac{\chi_5(n)}{n}=\ldots \neq 0\\
L(1,\chi_6)&=\sum_{n=1}^\infty \dfrac{\chi_6(n)}{n}=\ldots \neq 0
\end{align*}

Math100
fresh_42 said:
No. One sum per character, makes three sums. Everything was fine until you started with "Thus".

\begin{align*}
L(1,\chi_2)&=\sum_{n=1}^\infty \dfrac{\chi_2(n)}{n}=\ldots \neq 0\\
L(1,\chi_5)&=\sum_{n=1}^\infty \dfrac{\chi_5(n)}{n}=\ldots \neq 0\\
L(1,\chi_6)&=\sum_{n=1}^\infty \dfrac{\chi_6(n)}{n}=\ldots \neq 0
\end{align*}
I see now.

## 1. What does the notation ## L(1, \chi) ## mean?

The notation ## L(1, \chi) ## refers to the Dirichlet L-function evaluated at the point 1 with a specific character or Dirichlet character, denoted by the symbol ## \chi ##.

## 2. Why is it important to verify that ## L(1, \chi)\neq 0 ##?

It is important to verify that ## L(1, \chi)\neq 0 ## because it is a necessary condition for the associated Dirichlet character to be primitive. This means that the character cannot be expressed as a product of other characters and is essential in various number theory applications.

## 3. How can one verify that ## L(1, \chi)\neq 0 ##?

There are various methods for verifying that ## L(1, \chi)\neq 0 ##, depending on the specific character and the available tools. One common approach is to use the functional equation of the Dirichlet L-function, which relates the values at different points and can be used to show that ## L(1, \chi) ## cannot be equal to 0. Other methods may involve analyzing the properties of the character or using other known results in number theory.

## 4. What happens if ## L(1, \chi) = 0 ##?

If ## L(1, \chi) = 0 ##, then the associated Dirichlet character is not primitive and can be factored into simpler characters. This can limit its usefulness in certain number theory applications and may also indicate that further analysis is needed to fully understand the properties of the character.

## 5. Are there any exceptions to the condition ## L(1, \chi)\neq 0 ##?

Yes, there are some exceptions to this condition, known as exceptional characters. These are special cases where the character can still be primitive even though ## L(1, \chi) = 0 ##. However, these cases are rare and typically require additional conditions to be satisfied.

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