system example m=2

src/sections/02_coefficients_via_system_of_equations/coefficients_via_system_of_equations.tex

@@ -30,4 +30,5 @@
         &+ \coeffA{m}{8} \left[ \frac{1}{218790} (-1551693 n + 2042040 n^3 - 516868 n^5 + 26520 n^7 + n^{17}) \right] + \cdots
     \end{split}\label{eq:equation17}
 \end{equation}
-\input{coefficients_via_system_of_equations_example1}
+\input{coefficients_via_system_of_equations_example1}
+\input{coefficients_via_system_of_equations_example2}

src/sections/02_coefficients_via_system_of_equations/coefficients_via_system_of_equations_example2.tex

@@ -1,36 +1,36 @@
-For $m=2$ we have an identity
-
+\begin{examp}
+    Let be fixed $m=2$ so that we have the following relation defined by (eqref)
+    \begin{equation*}
+        \coeffA{m}{0} n
+        + \coeffA{m}{1} \left[ \frac{1}{6} (-n + n^3) \right]
+        + \coeffA{m}{2} \left[ \frac{1}{30} (-n + n^5) \right] - n^5 = 0
+    \end{equation*}
+\end{examp}
+Multiplying by $30$ both, right hand side and left hand side, we get
 \begin{equation*}
-    \coeffA{m}{0} n
-    + \coeffA{m}{1} \left[ \frac{1}{6} (-n + n^3) \right]
-    + \coeffA{m}{2} \left[ \frac{1}{30} (-n + n^5) \right] - n^5 = 0
+    30 \coeffA{2}{0} n + 5 \coeffA{2}{1} (-n + n^3) + \coeffA{2}{2} (-n + n^5) - 30n^5 = 0
 \end{equation*}
-Multiplying by 30 both parts we get
+Opening brackets and rearranging the terms gives
 \begin{equation*}
-    30 \coeffA{m}{0} n
-    + 5 \coeffA{m}{1}  (-n + n^3)
-    + \coeffA{m}{2} (-n + n^5) - 30 n^5 = 0
+    30 \coeffA{2}{0} - 5 \coeffA{2}{1} n + 5 \coeffA{2}{1} n^3 - \coeffA{2}{2} n + \coeffA{2}{2} n^5 - 30n^5 = 0
 \end{equation*}
-
-Opening brackets gives
+Combining the terms yields
 \begin{equation*}
-    30 \coeffA{m}{0} n -5 \coeffA{m}{1} n + 5 \coeffA{m}{1} n^3 -\coeffA{m}{2} n + n^5 \coeffA{m}{2} n^5 - 30 n^5 = 0
+    n (30 \coeffA{2}{0} - 5 \coeffA{2}{1} - \coeffA{2}{2}) + 5 \coeffA{2}{1} n^3 + n^5 (\coeffA{2}{2} - 30) = 0
 \end{equation*}
-
-Arranging terms we get
-
+Therefore, the system of linear equations follows
 \begin{equation*}
-    n (30 \coeffA{m}{0} n - 5 \coeffA{m}{1} -\coeffA{m}{2}) + 5 \coeffA{m}{1} n^3 + n^5 (\coeffA{m}{2} n^5 - 30) = 0, \quad n \geq 1
+    \begin{cases}
+        30 \coeffA{2}{0} - 5 \coeffA{2}{1} - \coeffA{2}{2} = 0 \\
+        \coeffA{2}{1} = 0 \\
+        \coeffA{2}{2} - 30 = 0
+    \end{cases}
 \end{equation*}
-
-Hence
-
+Solving it we get
 \begin{equation*}
     \begin{cases}
-        \coeffA{m}{1} = 0 \\
-        \coeffA{m}{2} n^5 - 30 = 0 \\
-        30 \coeffA{m}{0} n - 5 \coeffA{m}{1} -\coeffA{m}{2} = 0
+        \coeffA{2}{2} = 30 \\
+        \coeffA{2}{1} = 0 \\
+        \coeffA{2}{0} = 1
     \end{cases}
-\end{equation*}
-
-So that $\coeffA{m}{1}=6 \quad \coeffA{m}{1} = 1$
+\end{equation*}

src/sections/02_coefficients_via_system_of_equations/coefficients_via_system_of_equations_example3.tex

@@ -0,0 +1,37 @@
+\begin{examp}
+    Let be fixed $m=3$ so that we have the following relation defined by (eqref)
+    \begin{equation*}
+        \coeffA{m}{0} n
+        + \coeffA{m}{1} \left[ \frac{1}{6} (-n + n^3) \right]
+        + \coeffA{m}{2} \left[ \frac{1}{30} (-n + n^5) \right]
+        + \coeffA{m}{3} \left[ \frac{1}{420} (-10 n + 7 n^3 + 3 n^7) \right] - n^7 = 0
+    \end{equation*}
+\end{examp}
+Multiplying by $30$ both, right hand side and left hand side, we get
+\begin{equation*}
+    30 \coeffA{2}{0} n + 5 \coeffA{2}{1} (-n + n^3) + \coeffA{2}{2} (-n + n^5) - 30n^5 = 0
+\end{equation*}
+Opening brackets and rearranging the terms gives
+\begin{equation*}
+    30 \coeffA{2}{0} - 5 \coeffA{2}{1} n + 5 \coeffA{2}{1} n^3 - \coeffA{2}{2} n + \coeffA{2}{2} n^5 - 30n^5 = 0
+\end{equation*}
+Combining the terms yields
+\begin{equation*}
+    n (30 \coeffA{2}{0} - 5 \coeffA{2}{1} - \coeffA{2}{2}) + 5 \coeffA{2}{1} n^3 + n^5 (\coeffA{2}{2} - 30) = 0
+\end{equation*}
+Therefore, the system of linear equations follows
+\begin{equation*}
+    \begin{cases}
+        30 \coeffA{2}{0} - 5 \coeffA{2}{1} - \coeffA{2}{2} = 0 \\
+        \coeffA{2}{1} = 0 \\
+        \coeffA{2}{2} - 30 = 0
+    \end{cases}
+\end{equation*}
+Solving it we get
+\begin{equation*}
+    \begin{cases}
+        \coeffA{2}{2} = 30 \\
+        \coeffA{2}{1} = 0 \\
+        \coeffA{2}{0} = 1
+    \end{cases}
+\end{equation*}