Mixing
Explanation of Error in Euler's Method (first order differential equations)
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Can anyone provide a simple explanation why halving the step size tends to decrease the numerical error in Euler's method by one-half?
I've looked at some online sources but they do provide very complex explanations.
calculus ordinary-differential-equations numerical-methods
asked Jan 27 at 0:11
Meghan CMeghan C
21828
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add a comment |
$begingroup$
Can anyone provide a simple explanation why halving the step size tends to decrease the numerical error in Euler's method by one-half?
I've looked at some online sources but they do provide very complex explanations.
calculus ordinary-differential-equations numerical-methods
asked Jan 27 at 0:11
Meghan CMeghan C
21828
$endgroup$
$begingroup$
See also Euler Method - Global Error: How to calculate $C_1$ in $error=C_1h$ and similar
$endgroup$
– LutzL
Jan 27 at 9:24
add a comment |
$begingroup$
Can anyone provide a simple explanation why halving the step size tends to decrease the numerical error in Euler's method by one-half?
I've looked at some online sources but they do provide very complex explanations.
calculus ordinary-differential-equations numerical-methods
asked Jan 27 at 0:11
Meghan CMeghan C
21828
$endgroup$
Can anyone provide a simple explanation why halving the step size tends to decrease the numerical error in Euler's method by one-half?
I've looked at some online sources but they do provide very complex explanations.
calculus ordinary-differential-equations numerical-methods
calculus ordinary-differential-equations numerical-methods
asked Jan 27 at 0:11
Meghan CMeghan C
21828
asked Jan 27 at 0:11
Meghan CMeghan C
21828
asked Jan 27 at 0:11
Meghan CMeghan C
21828
asked Jan 27 at 0:11
Meghan CMeghan C
21828
asked Jan 27 at 0:11
Meghan CMeghan C
21828
21828
$begingroup$
See also Euler Method - Global Error: How to calculate $C_1$ in $error=C_1h$ and similar
$endgroup$
– LutzL
Jan 27 at 9:24
add a comment |
$begingroup$
See also Euler Method - Global Error: How to calculate $C_1$ in $error=C_1h$ and similar
$endgroup$
– LutzL
Jan 27 at 9:24
$begingroup$
See also Euler Method - Global Error: How to calculate $C_1$ in $error=C_1h$ and similar
$endgroup$
– LutzL
Jan 27 at 9:24
$begingroup$
See also Euler Method - Global Error: How to calculate $C_1$ in $error=C_1h$ and similar
$endgroup$
– LutzL
Jan 27 at 9:24
add a comment |
1 Answer
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oldest
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Here's an AoPS thread with some fairly simple explanations.
In short: the error added in one step of length $h$ is $h^2frac{y''(c)}{2}$, where $c$ is some point in the interval we stepped over. To cover a certain fixed distance $A$, we need about $frac{A}{h}$ steps, so the total error looks like a constant times $h$.
There's also compounded error - in later steps, our $y$ values will be in error, and that will affect our estimated $y'$ values. With some calculation, we can show that this basically only makes the proportionality constant bigger; the error will still be proportional to $h$.
answered Jan 27 at 0:51
jmerryjmerry
15.8k1632
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add a comment |
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$begingroup$
Here's an AoPS thread with some fairly simple explanations.
In short: the error added in one step of length $h$ is $h^2frac{y''(c)}{2}$, where $c$ is some point in the interval we stepped over. To cover a certain fixed distance $A$, we need about $frac{A}{h}$ steps, so the total error looks like a constant times $h$.
There's also compounded error - in later steps, our $y$ values will be in error, and that will affect our estimated $y'$ values. With some calculation, we can show that this basically only makes the proportionality constant bigger; the error will still be proportional to $h$.
answered Jan 27 at 0:51
jmerryjmerry
15.8k1632
$endgroup$
add a comment |
$begingroup$
Here's an AoPS thread with some fairly simple explanations.
In short: the error added in one step of length $h$ is $h^2frac{y''(c)}{2}$, where $c$ is some point in the interval we stepped over. To cover a certain fixed distance $A$, we need about $frac{A}{h}$ steps, so the total error looks like a constant times $h$.
There's also compounded error - in later steps, our $y$ values will be in error, and that will affect our estimated $y'$ values. With some calculation, we can show that this basically only makes the proportionality constant bigger; the error will still be proportional to $h$.
answered Jan 27 at 0:51
jmerryjmerry
15.8k1632
$endgroup$
add a comment |
$begingroup$
Here's an AoPS thread with some fairly simple explanations.
In short: the error added in one step of length $h$ is $h^2frac{y''(c)}{2}$, where $c$ is some point in the interval we stepped over. To cover a certain fixed distance $A$, we need about $frac{A}{h}$ steps, so the total error looks like a constant times $h$.
There's also compounded error - in later steps, our $y$ values will be in error, and that will affect our estimated $y'$ values. With some calculation, we can show that this basically only makes the proportionality constant bigger; the error will still be proportional to $h$.
answered Jan 27 at 0:51
jmerryjmerry
15.8k1632
$endgroup$
Here's an AoPS thread with some fairly simple explanations.
In short: the error added in one step of length $h$ is $h^2frac{y''(c)}{2}$, where $c$ is some point in the interval we stepped over. To cover a certain fixed distance $A$, we need about $frac{A}{h}$ steps, so the total error looks like a constant times $h$.
There's also compounded error - in later steps, our $y$ values will be in error, and that will affect our estimated $y'$ values. With some calculation, we can show that this basically only makes the proportionality constant bigger; the error will still be proportional to $h$.
answered Jan 27 at 0:51
jmerryjmerry
15.8k1632
answered Jan 27 at 0:51
jmerryjmerry
15.8k1632
answered Jan 27 at 0:51
jmerryjmerry
15.8k1632
answered Jan 27 at 0:51
jmerryjmerry
15.8k1632
15.8k1632
add a comment |
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$begingroup$
See also Euler Method - Global Error: How to calculate $C_1$ in $error=C_1h$ and similar
$endgroup$
– LutzL
Jan 27 at 9:24