diff --git a/Lectures_my/NumMet/Lecture2/mchrzasz.tex b/Lectures_my/NumMet/Lecture2/mchrzasz.tex
index 16a204d..d374721 100644
--- a/Lectures_my/NumMet/Lecture2/mchrzasz.tex
+++ b/Lectures_my/NumMet/Lecture2/mchrzasz.tex
@@ -591,16 +591,115 @@
 \begin{align*}
 a=x_0<x_1<x_2<...<x_n=b
 \end{align*}
+\ARROW The simplest is to use linear functions:
+\begin{align*}
+s(x)=\begin{cases}
+s_0(x)~~~~x \in \left[x_0,x_1 \right)\\
+s_1(x)~~~~x \in \left[x_1,x_2 \right)\\
+...\\
+s_{n-1}(x)~~~~x \in \left[x_{n-1},x_n \right),
+\end{cases}
+\end{align*}
+where
+\begin{align*}
+x_k(x)=a_{k,0}+a_{k,1}(x-x_k)~~x \in \left[ x_k,  x_{k+1} \right)
+\end{align*}
 
+\end{small}
+\end{frame}
+
+
+\begin{frame}\frametitle{First degree splain function}
+\begin{small}
+
+\ARROW The above defined functions $s_k(x)$ need to obey the conditions:
+\begin{align*}
+s_k(x_k ) & = y_k\\
+s_k(x_{k+1}) &  = s_{k+1}(x_{k+1})
+\end{align*}
+\ARROW From the above we can calculate the linear equation coefficients:
+\begin{align*}
+a_{k,0} & =y_k\\
+a_{k,1} & =\frac{y_{k+1} - y_k}{x_{k+1}-x_k}
+\end{align*}
+
+\begin{exampleblock}{Advantage:}
+\ARROWR The biggest advantage of this method is it's simplify. If you have a large number of interpolating points that are ''dents'' packed you can get very good and fast aproximation with this function.
+\end{exampleblock}
+
+\begin{alertblock}{Disadvantage:}
+\ARROW The interpolating function is not smooth on the interpolation points.
+\end{alertblock}
+
+\end{small}
+\end{frame}
+
+\begin{frame}\frametitle{Third degree splain function}
+\begin{small}
+\ARROW As mentioned before the first degree splain function suffer from the continuity problem. \\
+\ARROW A solution would be to use the higher order polynomials to aproximate the function. For example the 3rd order:
+\begin{align*}
+s_k(x)=a_{k,0} + a_{k,1}(x-x_k)+  a_{k,2}(x-x_k)^2 +  a_{k,3}(x-x_k)^3,~~~~x \in \left[x_k,x_{k+1} \right)
+\end{align*}
+\ARROW Now this polynomial needs to obey conditions:
+\begin{align*}
+s_k(x_k)&=y_k\\
+s_k(x_{k+1})& = s_{k+1}(x_{k+1})\\
+s^{\prime}_k(x_{k+1})& = s^{\prime}_{k+1}(x_{k+1})\\
+s^{\prime\prime}_k(x_{k+1})& = s^{\prime\prime}_{k+1}(x_{k+1})
+\end{align*}
+\ARROW In total we will have $4n$ coefficients to find. But we will only have $4n-2$ equations.\\
+\ARROW To solve the system we need to assume somethings about the endpoints (there are many options here).\\
+\ARROW The most popular ones are:
+\begin{align*}
+&s^{\prime}(a)=f^{\prime}(a)~~~~s^{\prime}(b)=f^{\prime}(b)~~~~ {\rm or}\\
+&s^{\prime\prime}(a)=s^{\prime\prime}(b)=0
+\end{align*}
 
 
 \end{small}
 \end{frame}
 
 
+\begin{frame}\frametitle{Third degree splain function}
+\begin{small}
+\ARROW Now to efficiently solve the system we will define a temporary variables:
+\begin{align*}
+h_k&=x_{k+1}-x_k\\
+d_k&=\frac{y_{k+1}-y_k}{h_k}\\
+m_k&=s^{\prime\prime}(x_k)
+\end{align*}
+\ARROW Now the second derivative can be interpolated using the Lagrange polynomial:
+\begin{align*}
+s_k^{\prime\prime}(x)=s^{\prime\prime}(x)\frac{x-x_{k+1}}{x_{k}-x_{k+1}}+s^{\prime\prime}(x)\frac{x-x_{k}}{x_{k+1}-x_{k}}=\frac{m_k}{h_k}(x_{k+1}-x) + \frac{m_{k+}}{h_k}(x-x_k)
+\end{align*}
+\ARROW Now if we integrate the above equation (two times):
+\begin{align*}
+s_k^{\prime}(x)=-\frac{m_k}{2h_k}(x_{k+1}-x)^2 + \frac{m_{k+}}{2h_k}(x-x_k)^2 -p_k + q_k\\
+s_k(x)=\frac{m_k}{6h_k}(x_{k+1}-x)^3 + \frac{m_{k+1}}{3h_k}(x-x_k)^3+p_k(x_{k+1}-x)+q_k(x-x_k)
+\end{align*}
+
+\end{small}
+\end{frame}
 
 
 
+\begin{frame}\frametitle{Third degree splain function}
+\begin{small}
+\ARROW Now we need to calculate the $s_k(x)$ for $x_k$ and $x_{k+1}$ using relations: $s_k(x_k)=y_k$ and $s_k(x_k+1)=y_{k+1}$:
+\begin{align*}
+x_k(x_k) & =y_k  = \frac{m_k}{6h_k}(x_{k+1}-x_k)^3 + p_k(x_{k+1}-x_k)\\
+x_k(x_{k+1}) & =y_{k+1}  = \frac{m_{k+1}}{3h_k}(x_{k+1}-x_k)^3 + q_k (x_{k+1} -x_k)
+\end{align*}
+\ARROW From which we get:
+\begin{align*}
+p_k&=\frac{y_k}{h_k}-\frac{m_k h_k}{6}\\
+p_{k+1}&=\frac{y_{k+1}}{h_{k+1}}-\frac{m_{k+1} h_{k+1}}{6}
+\end{align*}
+
+\end{small}
+\end{frame}
+