- #1
- 1,000
- 6
I have another dilemma with terminology that is puzzling and would appreciate some advice.
Consider the following truncated Taylor Series:
$$\begin{equation*}
f(\vec{z}_{k+1}) \approx f(\vec{z}_k)
+ \frac{\partial f(\vec{z}_k)}{\partial x} \Delta x
+ \frac{\partial f(\vec{z}_k)}{\partial \beta_1} \Delta \beta_1
+ \frac{\partial f(\vec{z}_k)}{\partial \beta_2} \Delta \beta_2 + \dots
+ \frac{\partial f(\vec{z}_k)}{\partial \beta_n} \Delta \beta_n
\end{equation*}$$
where:
$$\begin{align*} f &= f(\vec{z}\,) &
\vec{z} &= (x;\vec{\beta}\,) \\
\vec{z}_k &= (x;\vec{\beta}\,)_k &
\vec{z}_{k+1} &= (x;\vec{\beta}\,)_{k+1} \\
\Delta x &= x_{k+1} - x_k &
\Delta \beta_j &= (\Delta \beta_j)_{k+1} - (\Delta \beta_j)_k \\
\vec{\beta} &= (\beta_1, \beta_2, \dots, \beta_n)
\end{align*}$$
Then form the following function with the truncated Taylor Series:
$$\begin{equation*}
L = \sum_{i=0}^m \left[ f_i
+ \frac{\partial f_i}{\partial x_i} \Delta x_i
+ \frac{\partial f_i}{\partial \beta_1} \Delta \beta_1
+ \frac{\partial f_i}{\partial \beta_2} \Delta \beta_2 + \dots
+ \frac{\partial f_i}{\partial \beta_n} \Delta \beta_n - y_i
\right]^2 \end{equation*}$$
where:
$$\begin{align*}
f_i &= f(z_k) &
z_k &= (x_i; \beta)_k \\
\Delta x_i &= (x_i)_{k+1} - (x_i)_k &
\Delta \beta_j &= (\Delta \beta_j)_{k+1} - (\Delta \beta_j)_k \\
\vec{x} &= (x_1, x_2, \dots, x_m) &
\vec{y} &= (y_1, y_2, \dots, y_m)
\end{align*}$$
The ##\partial x_i## in the second term of the sum matches the ##\Delta x_i## but doesn't seem right because it implies ##f=f(\vec{x};\vec{\beta})## which isn't true, but using ##\partial x## instead doesn't seem right either because it doesn't match ##\Delta x_i##. How to handle?
Consider the following truncated Taylor Series:
$$\begin{equation*}
f(\vec{z}_{k+1}) \approx f(\vec{z}_k)
+ \frac{\partial f(\vec{z}_k)}{\partial x} \Delta x
+ \frac{\partial f(\vec{z}_k)}{\partial \beta_1} \Delta \beta_1
+ \frac{\partial f(\vec{z}_k)}{\partial \beta_2} \Delta \beta_2 + \dots
+ \frac{\partial f(\vec{z}_k)}{\partial \beta_n} \Delta \beta_n
\end{equation*}$$
where:
$$\begin{align*} f &= f(\vec{z}\,) &
\vec{z} &= (x;\vec{\beta}\,) \\
\vec{z}_k &= (x;\vec{\beta}\,)_k &
\vec{z}_{k+1} &= (x;\vec{\beta}\,)_{k+1} \\
\Delta x &= x_{k+1} - x_k &
\Delta \beta_j &= (\Delta \beta_j)_{k+1} - (\Delta \beta_j)_k \\
\vec{\beta} &= (\beta_1, \beta_2, \dots, \beta_n)
\end{align*}$$
Then form the following function with the truncated Taylor Series:
$$\begin{equation*}
L = \sum_{i=0}^m \left[ f_i
+ \frac{\partial f_i}{\partial x_i} \Delta x_i
+ \frac{\partial f_i}{\partial \beta_1} \Delta \beta_1
+ \frac{\partial f_i}{\partial \beta_2} \Delta \beta_2 + \dots
+ \frac{\partial f_i}{\partial \beta_n} \Delta \beta_n - y_i
\right]^2 \end{equation*}$$
where:
$$\begin{align*}
f_i &= f(z_k) &
z_k &= (x_i; \beta)_k \\
\Delta x_i &= (x_i)_{k+1} - (x_i)_k &
\Delta \beta_j &= (\Delta \beta_j)_{k+1} - (\Delta \beta_j)_k \\
\vec{x} &= (x_1, x_2, \dots, x_m) &
\vec{y} &= (y_1, y_2, \dots, y_m)
\end{align*}$$
The ##\partial x_i## in the second term of the sum matches the ##\Delta x_i## but doesn't seem right because it implies ##f=f(\vec{x};\vec{\beta})## which isn't true, but using ##\partial x## instead doesn't seem right either because it doesn't match ##\Delta x_i##. How to handle?