Error from central difference seems large
$begingroup$
for a function $f(x)=e^{2x}-cos(2x)$,
at grid points $x in {-0.3,-0.2,-0.1,0}$
I perform a central difference for the derivative at $x=-0.2$
$$frac{df}{dx}=frac{f(-0.1)-f(-0.3)}{2*(-0.1--0.3)}=0.28795$$
The derivative of this function (per mathematica) is $0.561803$
So the error is $$Abs[0.561803-0.28795]=0.273853$$
I have a formula for the upper bound of the error, and that formula is $$frac{M^*Delta x^2}{3!}=frac{8*(-0.1--0.3)}{6}=0.0533333\text{Where }M^*=text{Max}(f''(x)|xin[-0.3,0])=f(0)=8$$
The issue is the error, $0.27$, is greater than the supposid error bound, $0.05$. When I graph this function with its central difference though, it looks correct.
Any idea why my derivatives are higher than the upper bound? The error does seem pretty high.
finite-differences
edited Dec 1 '18 at 0:48
Key Flex
8,18261233
asked Nov 30 '18 at 22:31
FrankFrank
16210
$endgroup$
add a comment |
$begingroup$
for a function $f(x)=e^{2x}-cos(2x)$,
at grid points $x in {-0.3,-0.2,-0.1,0}$
I perform a central difference for the derivative at $x=-0.2$
$$frac{df}{dx}=frac{f(-0.1)-f(-0.3)}{2*(-0.1--0.3)}=0.28795$$
The derivative of this function (per mathematica) is $0.561803$
So the error is $$Abs[0.561803-0.28795]=0.273853$$
I have a formula for the upper bound of the error, and that formula is $$frac{M^*Delta x^2}{3!}=frac{8*(-0.1--0.3)}{6}=0.0533333\text{Where }M^*=text{Max}(f''(x)|xin[-0.3,0])=f(0)=8$$
The issue is the error, $0.27$, is greater than the supposid error bound, $0.05$. When I graph this function with its central difference though, it looks correct.
Any idea why my derivatives are higher than the upper bound? The error does seem pretty high.
finite-differences
edited Dec 1 '18 at 0:48
Key Flex
8,18261233
asked Nov 30 '18 at 22:31
FrankFrank
16210
$endgroup$
add a comment |
$begingroup$
for a function $f(x)=e^{2x}-cos(2x)$,
at grid points $x in {-0.3,-0.2,-0.1,0}$
I perform a central difference for the derivative at $x=-0.2$
$$frac{df}{dx}=frac{f(-0.1)-f(-0.3)}{2*(-0.1--0.3)}=0.28795$$
The derivative of this function (per mathematica) is $0.561803$
So the error is $$Abs[0.561803-0.28795]=0.273853$$
I have a formula for the upper bound of the error, and that formula is $$frac{M^*Delta x^2}{3!}=frac{8*(-0.1--0.3)}{6}=0.0533333\text{Where }M^*=text{Max}(f''(x)|xin[-0.3,0])=f(0)=8$$
The issue is the error, $0.27$, is greater than the supposid error bound, $0.05$. When I graph this function with its central difference though, it looks correct.
Any idea why my derivatives are higher than the upper bound? The error does seem pretty high.
finite-differences
edited Dec 1 '18 at 0:48
Key Flex
8,18261233
asked Nov 30 '18 at 22:31
FrankFrank
16210
$endgroup$
for a function $f(x)=e^{2x}-cos(2x)$,
at grid points $x in {-0.3,-0.2,-0.1,0}$
I perform a central difference for the derivative at $x=-0.2$
$$frac{df}{dx}=frac{f(-0.1)-f(-0.3)}{2*(-0.1--0.3)}=0.28795$$
The derivative of this function (per mathematica) is $0.561803$
So the error is $$Abs[0.561803-0.28795]=0.273853$$
I have a formula for the upper bound of the error, and that formula is $$frac{M^*Delta x^2}{3!}=frac{8*(-0.1--0.3)}{6}=0.0533333\text{Where }M^*=text{Max}(f''(x)|xin[-0.3,0])=f(0)=8$$
The issue is the error, $0.27$, is greater than the supposid error bound, $0.05$. When I graph this function with its central difference though, it looks correct.
Any idea why my derivatives are higher than the upper bound? The error does seem pretty high.
finite-differences
finite-differences
edited Dec 1 '18 at 0:48
Key Flex
8,18261233
asked Nov 30 '18 at 22:31
FrankFrank
16210
edited Dec 1 '18 at 0:48
Key Flex
8,18261233
asked Nov 30 '18 at 22:31
FrankFrank
16210
edited Dec 1 '18 at 0:48
Key Flex
8,18261233
edited Dec 1 '18 at 0:48
Key Flex
8,18261233
edited Dec 1 '18 at 0:48
Key Flex
8,18261233
8,18261233
asked Nov 30 '18 at 22:31
FrankFrank
16210
asked Nov 30 '18 at 22:31
FrankFrank
16210
asked Nov 30 '18 at 22:31
FrankFrank
16210
16210
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The problem lies in your definition of $Delta x$, remember that
$$
f'_{FD}(x) approx frac{f(x + Delta x) - f(x - Delta x)}{2Delta x}
$$
Taking $x = -0.2$ and $Delta x = 0.1$ we get
$$
f'_{FD}(-0.2) approx frac{f(-0.1) - f(-0.3)}{2 cdot 0.1} = 0.575941 tag{1}
$$
So the actual error is
$$
epsilon = |f'_{FD}(-0.2) - f'(-0.2)|= 0.0141374 tag{2}
$$
Now, the prediction for the error is
$$
epsilon_{max} = frac{|f^{(3)}(c)|Delta^2 x}{6} stackrel{c=-0.3}{=} 0.0148461 tag{3}
$$
You can indeed see that
$$
epsilon < epsilon_max
$$
answered Nov 30 '18 at 23:45
caveraccaverac
14.6k31130
$endgroup$
$begingroup$
Thanks so much! But is the formula for the error bound really supposed to be the third derivative of $f$ or the second? I got the formula from someone else's notes! I couldn't find it with a google search hahaha.
$endgroup$
– Frank
Dec 1 '18 at 0:04
$begingroup$
@Frank It is the third derivative
$endgroup$
– caverac
Dec 1 '18 at 0:40
add a comment |
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1 Answer
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active
oldest
votes
1 Answer
1
active
oldest
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active
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active
oldest
votes
$begingroup$
The problem lies in your definition of $Delta x$, remember that
$$
f'_{FD}(x) approx frac{f(x + Delta x) - f(x - Delta x)}{2Delta x}
$$
Taking $x = -0.2$ and $Delta x = 0.1$ we get
$$
f'_{FD}(-0.2) approx frac{f(-0.1) - f(-0.3)}{2 cdot 0.1} = 0.575941 tag{1}
$$
So the actual error is
$$
epsilon = |f'_{FD}(-0.2) - f'(-0.2)|= 0.0141374 tag{2}
$$
Now, the prediction for the error is
$$
epsilon_{max} = frac{|f^{(3)}(c)|Delta^2 x}{6} stackrel{c=-0.3}{=} 0.0148461 tag{3}
$$
You can indeed see that
$$
epsilon < epsilon_max
$$
answered Nov 30 '18 at 23:45
caveraccaverac
14.6k31130
$endgroup$
$begingroup$
Thanks so much! But is the formula for the error bound really supposed to be the third derivative of $f$ or the second? I got the formula from someone else's notes! I couldn't find it with a google search hahaha.
$endgroup$
– Frank
Dec 1 '18 at 0:04
$begingroup$
@Frank It is the third derivative
$endgroup$
– caverac
Dec 1 '18 at 0:40
add a comment |
$begingroup$
The problem lies in your definition of $Delta x$, remember that
$$
f'_{FD}(x) approx frac{f(x + Delta x) - f(x - Delta x)}{2Delta x}
$$
Taking $x = -0.2$ and $Delta x = 0.1$ we get
$$
f'_{FD}(-0.2) approx frac{f(-0.1) - f(-0.3)}{2 cdot 0.1} = 0.575941 tag{1}
$$
So the actual error is
$$
epsilon = |f'_{FD}(-0.2) - f'(-0.2)|= 0.0141374 tag{2}
$$
Now, the prediction for the error is
$$
epsilon_{max} = frac{|f^{(3)}(c)|Delta^2 x}{6} stackrel{c=-0.3}{=} 0.0148461 tag{3}
$$
You can indeed see that
$$
epsilon < epsilon_max
$$
answered Nov 30 '18 at 23:45
caveraccaverac
14.6k31130
$endgroup$
$begingroup$
Thanks so much! But is the formula for the error bound really supposed to be the third derivative of $f$ or the second? I got the formula from someone else's notes! I couldn't find it with a google search hahaha.
$endgroup$
– Frank
Dec 1 '18 at 0:04
$begingroup$
@Frank It is the third derivative
$endgroup$
– caverac
Dec 1 '18 at 0:40
add a comment |
$begingroup$
The problem lies in your definition of $Delta x$, remember that
$$
f'_{FD}(x) approx frac{f(x + Delta x) - f(x - Delta x)}{2Delta x}
$$
Taking $x = -0.2$ and $Delta x = 0.1$ we get
$$
f'_{FD}(-0.2) approx frac{f(-0.1) - f(-0.3)}{2 cdot 0.1} = 0.575941 tag{1}
$$
So the actual error is
$$
epsilon = |f'_{FD}(-0.2) - f'(-0.2)|= 0.0141374 tag{2}
$$
Now, the prediction for the error is
$$
epsilon_{max} = frac{|f^{(3)}(c)|Delta^2 x}{6} stackrel{c=-0.3}{=} 0.0148461 tag{3}
$$
You can indeed see that
$$
epsilon < epsilon_max
$$
answered Nov 30 '18 at 23:45
caveraccaverac
14.6k31130
$endgroup$
The problem lies in your definition of $Delta x$, remember that
$$
f'_{FD}(x) approx frac{f(x + Delta x) - f(x - Delta x)}{2Delta x}
$$
Taking $x = -0.2$ and $Delta x = 0.1$ we get
$$
f'_{FD}(-0.2) approx frac{f(-0.1) - f(-0.3)}{2 cdot 0.1} = 0.575941 tag{1}
$$
So the actual error is
$$
epsilon = |f'_{FD}(-0.2) - f'(-0.2)|= 0.0141374 tag{2}
$$
Now, the prediction for the error is
$$
epsilon_{max} = frac{|f^{(3)}(c)|Delta^2 x}{6} stackrel{c=-0.3}{=} 0.0148461 tag{3}
$$
You can indeed see that
$$
epsilon < epsilon_max
$$
answered Nov 30 '18 at 23:45
caveraccaverac
14.6k31130
answered Nov 30 '18 at 23:45
caveraccaverac
14.6k31130
answered Nov 30 '18 at 23:45
caveraccaverac
14.6k31130
answered Nov 30 '18 at 23:45
caveraccaverac
14.6k31130
14.6k31130
$begingroup$
Thanks so much! But is the formula for the error bound really supposed to be the third derivative of $f$ or the second? I got the formula from someone else's notes! I couldn't find it with a google search hahaha.
$endgroup$
– Frank
Dec 1 '18 at 0:04
$begingroup$
@Frank It is the third derivative
$endgroup$
– caverac
Dec 1 '18 at 0:40
add a comment |
$begingroup$
Thanks so much! But is the formula for the error bound really supposed to be the third derivative of $f$ or the second? I got the formula from someone else's notes! I couldn't find it with a google search hahaha.
$endgroup$
– Frank
Dec 1 '18 at 0:04
$begingroup$
@Frank It is the third derivative
$endgroup$
– caverac
Dec 1 '18 at 0:40
$begingroup$
Thanks so much! But is the formula for the error bound really supposed to be the third derivative of $f$ or the second? I got the formula from someone else's notes! I couldn't find it with a google search hahaha.
$endgroup$
– Frank
Dec 1 '18 at 0:04
$begingroup$
Thanks so much! But is the formula for the error bound really supposed to be the third derivative of $f$ or the second? I got the formula from someone else's notes! I couldn't find it with a google search hahaha.
$endgroup$
– Frank
Dec 1 '18 at 0:04
$begingroup$
@Frank It is the third derivative
$endgroup$
– caverac
Dec 1 '18 at 0:40
$begingroup$
@Frank It is the third derivative
$endgroup$
– caverac
Dec 1 '18 at 0:40
add a comment |
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