How to do Taylor expansions generally for multivariate functions?
Ok so I have a function like this: $$f(t + 1, z_t + x_t)$$ where $x_t$ equals $z_{t+1} - z_t$.
I want to Taylor-expand this function so that I can get a first-order approximation. How do I do this? I want to expand around $x_t = 0$.
$z_t$ and $x_t$ are random variables, known at time $t$ and $t+1$ respectively, but just treat them as regular variables.
taylor-expansion
asked Nov 24 at 11:42
JohnDoe
1
add a comment |
Ok so I have a function like this: $$f(t + 1, z_t + x_t)$$ where $x_t$ equals $z_{t+1} - z_t$.
I want to Taylor-expand this function so that I can get a first-order approximation. How do I do this? I want to expand around $x_t = 0$.
$z_t$ and $x_t$ are random variables, known at time $t$ and $t+1$ respectively, but just treat them as regular variables.
taylor-expansion
asked Nov 24 at 11:42
JohnDoe
1
So it's more likely that I get an answer if I make my question less clear and to-the-point by cluttering it with my own messy and incorrect attempts? Yeah, that makes a lot of sense.
– JohnDoe
Nov 24 at 12:11
$$ f(t + 1, z_t) + left.frac{partial f}{partial x_t}right|_{x_t = 0} x_t + frac{1}{2}left.frac{partial^2 f}{partial x_t^2}right|_{x_t = 0} x_t^2 + cdots $$
– caverac
Nov 24 at 12:33
check this
– Masacroso
Nov 24 at 13:04
add a comment |
Ok so I have a function like this: $$f(t + 1, z_t + x_t)$$ where $x_t$ equals $z_{t+1} - z_t$.
I want to Taylor-expand this function so that I can get a first-order approximation. How do I do this? I want to expand around $x_t = 0$.
$z_t$ and $x_t$ are random variables, known at time $t$ and $t+1$ respectively, but just treat them as regular variables.
taylor-expansion
asked Nov 24 at 11:42
JohnDoe
1
Ok so I have a function like this: $$f(t + 1, z_t + x_t)$$ where $x_t$ equals $z_{t+1} - z_t$.
I want to Taylor-expand this function so that I can get a first-order approximation. How do I do this? I want to expand around $x_t = 0$.
$z_t$ and $x_t$ are random variables, known at time $t$ and $t+1$ respectively, but just treat them as regular variables.
taylor-expansion
taylor-expansion
asked Nov 24 at 11:42
JohnDoe
1
asked Nov 24 at 11:42
JohnDoe
1
asked Nov 24 at 11:42
JohnDoe
1
asked Nov 24 at 11:42
JohnDoe
1
asked Nov 24 at 11:42
JohnDoe
1
1
So it's more likely that I get an answer if I make my question less clear and to-the-point by cluttering it with my own messy and incorrect attempts? Yeah, that makes a lot of sense.
– JohnDoe
Nov 24 at 12:11
$$ f(t + 1, z_t) + left.frac{partial f}{partial x_t}right|_{x_t = 0} x_t + frac{1}{2}left.frac{partial^2 f}{partial x_t^2}right|_{x_t = 0} x_t^2 + cdots $$
– caverac
Nov 24 at 12:33
check this
– Masacroso
Nov 24 at 13:04
add a comment |
So it's more likely that I get an answer if I make my question less clear and to-the-point by cluttering it with my own messy and incorrect attempts? Yeah, that makes a lot of sense.
– JohnDoe
Nov 24 at 12:11
$$ f(t + 1, z_t) + left.frac{partial f}{partial x_t}right|_{x_t = 0} x_t + frac{1}{2}left.frac{partial^2 f}{partial x_t^2}right|_{x_t = 0} x_t^2 + cdots $$
– caverac
Nov 24 at 12:33
check this
– Masacroso
Nov 24 at 13:04
So it's more likely that I get an answer if I make my question less clear and to-the-point by cluttering it with my own messy and incorrect attempts? Yeah, that makes a lot of sense.
– JohnDoe
Nov 24 at 12:11
So it's more likely that I get an answer if I make my question less clear and to-the-point by cluttering it with my own messy and incorrect attempts? Yeah, that makes a lot of sense.
– JohnDoe
Nov 24 at 12:11
$$ f(t + 1, z_t) + left.frac{partial f}{partial x_t}right|_{x_t = 0} x_t + frac{1}{2}left.frac{partial^2 f}{partial x_t^2}right|_{x_t = 0} x_t^2 + cdots $$
– caverac
Nov 24 at 12:33
$$ f(t + 1, z_t) + left.frac{partial f}{partial x_t}right|_{x_t = 0} x_t + frac{1}{2}left.frac{partial^2 f}{partial x_t^2}right|_{x_t = 0} x_t^2 + cdots $$
– caverac
Nov 24 at 12:33
check this
– Masacroso
Nov 24 at 13:04
check this
– Masacroso
Nov 24 at 13:04
add a comment |
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So it's more likely that I get an answer if I make my question less clear and to-the-point by cluttering it with my own messy and incorrect attempts? Yeah, that makes a lot of sense.
– JohnDoe
Nov 24 at 12:11
$$ f(t + 1, z_t) + left.frac{partial f}{partial x_t}right|_{x_t = 0} x_t + frac{1}{2}left.frac{partial^2 f}{partial x_t^2}right|_{x_t = 0} x_t^2 + cdots $$
– caverac
Nov 24 at 12:33
check this
– Masacroso
Nov 24 at 13:04