Taylor Series Expansion on error propagation.
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I am reading through Stoer and Bulirsch's Introduction to Numerical Analysis. In their section on error propagation they are describing a derivation of the Jacobian as it related to a problems condition number.
We suppose the that $tilde{x}$ is an approximation for the some $xinmathbb{R}^n$ and that $phi:mathbb{R}^n rightarrow mathbb{R}$ has continuous first derivatives everywhere. The textbook then expands $phi$ by its Taylor series and ignores the higher order terms to write that
$$phi(tilde{x})- phi(x) =sum_{j= 1}^n (tilde{x_j} - x_j)frac{partial phi(x)}{partial x_J} $$
I don't understand how we would be able to only evaluate $phi$ at $x$ and not also at $tilde{x}$. I would think that the first order approximation should appear as
$$phi(tilde{x})- phi(x) =sum_{j= 1}^n tilde{x_j}frac{partial phi(tilde{x})}{partial x_J} - x_jfrac{partial phi(x)}{partial x_J} $$
Am I missing a minor detail somewhere? The only other informationis that $tilde{x}$ is the closest machine number to $x$, but there must be something more to this statement.
Thanks!
taylor-expansion approximation-theory error-propagation
asked Jan 2 at 6:53
Andrew ShedlockAndrew Shedlock
1688
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add a comment |
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I am reading through Stoer and Bulirsch's Introduction to Numerical Analysis. In their section on error propagation they are describing a derivation of the Jacobian as it related to a problems condition number.
We suppose the that $tilde{x}$ is an approximation for the some $xinmathbb{R}^n$ and that $phi:mathbb{R}^n rightarrow mathbb{R}$ has continuous first derivatives everywhere. The textbook then expands $phi$ by its Taylor series and ignores the higher order terms to write that
$$phi(tilde{x})- phi(x) =sum_{j= 1}^n (tilde{x_j} - x_j)frac{partial phi(x)}{partial x_J} $$
I don't understand how we would be able to only evaluate $phi$ at $x$ and not also at $tilde{x}$. I would think that the first order approximation should appear as
$$phi(tilde{x})- phi(x) =sum_{j= 1}^n tilde{x_j}frac{partial phi(tilde{x})}{partial x_J} - x_jfrac{partial phi(x)}{partial x_J} $$
Am I missing a minor detail somewhere? The only other informationis that $tilde{x}$ is the closest machine number to $x$, but there must be something more to this statement.
Thanks!
taylor-expansion approximation-theory error-propagation
asked Jan 2 at 6:53
Andrew ShedlockAndrew Shedlock
1688
$endgroup$
add a comment |
$begingroup$
I am reading through Stoer and Bulirsch's Introduction to Numerical Analysis. In their section on error propagation they are describing a derivation of the Jacobian as it related to a problems condition number.
We suppose the that $tilde{x}$ is an approximation for the some $xinmathbb{R}^n$ and that $phi:mathbb{R}^n rightarrow mathbb{R}$ has continuous first derivatives everywhere. The textbook then expands $phi$ by its Taylor series and ignores the higher order terms to write that
$$phi(tilde{x})- phi(x) =sum_{j= 1}^n (tilde{x_j} - x_j)frac{partial phi(x)}{partial x_J} $$
I don't understand how we would be able to only evaluate $phi$ at $x$ and not also at $tilde{x}$. I would think that the first order approximation should appear as
$$phi(tilde{x})- phi(x) =sum_{j= 1}^n tilde{x_j}frac{partial phi(tilde{x})}{partial x_J} - x_jfrac{partial phi(x)}{partial x_J} $$
Am I missing a minor detail somewhere? The only other informationis that $tilde{x}$ is the closest machine number to $x$, but there must be something more to this statement.
Thanks!
taylor-expansion approximation-theory error-propagation
asked Jan 2 at 6:53
Andrew ShedlockAndrew Shedlock
1688
$endgroup$
I am reading through Stoer and Bulirsch's Introduction to Numerical Analysis. In their section on error propagation they are describing a derivation of the Jacobian as it related to a problems condition number.
We suppose the that $tilde{x}$ is an approximation for the some $xinmathbb{R}^n$ and that $phi:mathbb{R}^n rightarrow mathbb{R}$ has continuous first derivatives everywhere. The textbook then expands $phi$ by its Taylor series and ignores the higher order terms to write that
$$phi(tilde{x})- phi(x) =sum_{j= 1}^n (tilde{x_j} - x_j)frac{partial phi(x)}{partial x_J} $$
I don't understand how we would be able to only evaluate $phi$ at $x$ and not also at $tilde{x}$. I would think that the first order approximation should appear as
$$phi(tilde{x})- phi(x) =sum_{j= 1}^n tilde{x_j}frac{partial phi(tilde{x})}{partial x_J} - x_jfrac{partial phi(x)}{partial x_J} $$
Am I missing a minor detail somewhere? The only other informationis that $tilde{x}$ is the closest machine number to $x$, but there must be something more to this statement.
Thanks!
taylor-expansion approximation-theory error-propagation
taylor-expansion approximation-theory error-propagation
asked Jan 2 at 6:53
Andrew ShedlockAndrew Shedlock
1688
asked Jan 2 at 6:53
Andrew ShedlockAndrew Shedlock
1688
asked Jan 2 at 6:53
Andrew ShedlockAndrew Shedlock
1688
asked Jan 2 at 6:53
Andrew ShedlockAndrew Shedlock
1688
asked Jan 2 at 6:53
Andrew ShedlockAndrew Shedlock
1688
1688
add a comment |
add a comment |
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