Mixing
Second order approximation of first derivative gives odd results
$begingroup$
Assume I want to find the second order approximation of the first derivative in 0 of a function $f(x)$ (not defined for $x<0$). I will use the formulation for the forward derivative:
$f'(0)approx frac{-3f(0)+4f(h)-f(2h)}{2h}$
where $h$ is the step in the regular grid I am using. Now assume that the values of the function are:
$f(0)=0$, $f(h)=5$, $f(2h)=25$.
The function is clearly monotonically increasing and the first derivative in 0 is expected to be positive but, from the formula above, I get a negative value.
What am I missing? Does the issue all lie in the error (that I neglected)?
derivatives numerical-methods
asked Jan 28 at 17:08
pugliampugliam
111
$endgroup$
add a comment |
$begingroup$
Assume I want to find the second order approximation of the first derivative in 0 of a function $f(x)$ (not defined for $x<0$). I will use the formulation for the forward derivative:
$f'(0)approx frac{-3f(0)+4f(h)-f(2h)}{2h}$
where $h$ is the step in the regular grid I am using. Now assume that the values of the function are:
$f(0)=0$, $f(h)=5$, $f(2h)=25$.
The function is clearly monotonically increasing and the first derivative in 0 is expected to be positive but, from the formula above, I get a negative value.
What am I missing? Does the issue all lie in the error (that I neglected)?
derivatives numerical-methods
asked Jan 28 at 17:08
pugliampugliam
111
$endgroup$
$begingroup$
When you say "the formulation", I'm not sure that's the right way to think about this. Assuming your calculations are correct, your weird result could simply be a function of the particular way you approximated the first derivative. Just as there are lots of ways to approximate a function, there are lots of ways to approximate its derivative, even in only one direction.
$endgroup$
– Adrian Keister
Jan 28 at 17:15
1
$begingroup$
For a function like $5^x$ a step size of $h=1$ is just too large. The error term $sim ln(5)^35^hh^2$ should be smaller than the result, so $ln(5)h<0.5$ is necessary to start to get good results.
$endgroup$
– LutzL
Jan 28 at 21:49
$begingroup$
@LutzL, I was not meaning $5^x$, my third point could have also been 23, but yes, the derivative is apparently too large to get good results. Thanks!
$endgroup$
– pugliam
Jan 29 at 9:13
$begingroup$
I was just looking for a simple growing function that had the same type of growth in value and first derivative. All other interpolations will have the same type of behavior, either a rapid oscillation so that the negative derivative is correct, or rapid growth in the higher order derivatives so that the error term is large.
$endgroup$
– LutzL
Jan 29 at 9:35
add a comment |
$begingroup$
Assume I want to find the second order approximation of the first derivative in 0 of a function $f(x)$ (not defined for $x<0$). I will use the formulation for the forward derivative:
$f'(0)approx frac{-3f(0)+4f(h)-f(2h)}{2h}$
where $h$ is the step in the regular grid I am using. Now assume that the values of the function are:
$f(0)=0$, $f(h)=5$, $f(2h)=25$.
The function is clearly monotonically increasing and the first derivative in 0 is expected to be positive but, from the formula above, I get a negative value.
What am I missing? Does the issue all lie in the error (that I neglected)?
derivatives numerical-methods
asked Jan 28 at 17:08
pugliampugliam
111
$endgroup$
Assume I want to find the second order approximation of the first derivative in 0 of a function $f(x)$ (not defined for $x<0$). I will use the formulation for the forward derivative:
$f'(0)approx frac{-3f(0)+4f(h)-f(2h)}{2h}$
where $h$ is the step in the regular grid I am using. Now assume that the values of the function are:
$f(0)=0$, $f(h)=5$, $f(2h)=25$.
The function is clearly monotonically increasing and the first derivative in 0 is expected to be positive but, from the formula above, I get a negative value.
What am I missing? Does the issue all lie in the error (that I neglected)?
derivatives numerical-methods
derivatives numerical-methods
asked Jan 28 at 17:08
pugliampugliam
111
asked Jan 28 at 17:08
pugliampugliam
111
asked Jan 28 at 17:08
pugliampugliam
111
asked Jan 28 at 17:08
pugliampugliam
111
asked Jan 28 at 17:08
pugliampugliam
111
111
$begingroup$
When you say "the formulation", I'm not sure that's the right way to think about this. Assuming your calculations are correct, your weird result could simply be a function of the particular way you approximated the first derivative. Just as there are lots of ways to approximate a function, there are lots of ways to approximate its derivative, even in only one direction.
$endgroup$
– Adrian Keister
Jan 28 at 17:15
1
$begingroup$
For a function like $5^x$ a step size of $h=1$ is just too large. The error term $sim ln(5)^35^hh^2$ should be smaller than the result, so $ln(5)h<0.5$ is necessary to start to get good results.
$endgroup$
– LutzL
Jan 28 at 21:49
$begingroup$
@LutzL, I was not meaning $5^x$, my third point could have also been 23, but yes, the derivative is apparently too large to get good results. Thanks!
$endgroup$
– pugliam
Jan 29 at 9:13
$begingroup$
I was just looking for a simple growing function that had the same type of growth in value and first derivative. All other interpolations will have the same type of behavior, either a rapid oscillation so that the negative derivative is correct, or rapid growth in the higher order derivatives so that the error term is large.
$endgroup$
– LutzL
Jan 29 at 9:35
add a comment |
$begingroup$
When you say "the formulation", I'm not sure that's the right way to think about this. Assuming your calculations are correct, your weird result could simply be a function of the particular way you approximated the first derivative. Just as there are lots of ways to approximate a function, there are lots of ways to approximate its derivative, even in only one direction.
$endgroup$
– Adrian Keister
Jan 28 at 17:15
1
$begingroup$
For a function like $5^x$ a step size of $h=1$ is just too large. The error term $sim ln(5)^35^hh^2$ should be smaller than the result, so $ln(5)h<0.5$ is necessary to start to get good results.
$endgroup$
– LutzL
Jan 28 at 21:49
$begingroup$
@LutzL, I was not meaning $5^x$, my third point could have also been 23, but yes, the derivative is apparently too large to get good results. Thanks!
$endgroup$
– pugliam
Jan 29 at 9:13
$begingroup$
I was just looking for a simple growing function that had the same type of growth in value and first derivative. All other interpolations will have the same type of behavior, either a rapid oscillation so that the negative derivative is correct, or rapid growth in the higher order derivatives so that the error term is large.
$endgroup$
– LutzL
Jan 29 at 9:35
$begingroup$
When you say "the formulation", I'm not sure that's the right way to think about this. Assuming your calculations are correct, your weird result could simply be a function of the particular way you approximated the first derivative. Just as there are lots of ways to approximate a function, there are lots of ways to approximate its derivative, even in only one direction.
$endgroup$
– Adrian Keister
Jan 28 at 17:15
$begingroup$
When you say "the formulation", I'm not sure that's the right way to think about this. Assuming your calculations are correct, your weird result could simply be a function of the particular way you approximated the first derivative. Just as there are lots of ways to approximate a function, there are lots of ways to approximate its derivative, even in only one direction.
$endgroup$
– Adrian Keister
Jan 28 at 17:15
1
1
$begingroup$
For a function like $5^x$ a step size of $h=1$ is just too large. The error term $sim ln(5)^35^hh^2$ should be smaller than the result, so $ln(5)h<0.5$ is necessary to start to get good results.
$endgroup$
– LutzL
Jan 28 at 21:49
$begingroup$
For a function like $5^x$ a step size of $h=1$ is just too large. The error term $sim ln(5)^35^hh^2$ should be smaller than the result, so $ln(5)h<0.5$ is necessary to start to get good results.
$endgroup$
– LutzL
Jan 28 at 21:49
$begingroup$
@LutzL, I was not meaning $5^x$, my third point could have also been 23, but yes, the derivative is apparently too large to get good results. Thanks!
$endgroup$
– pugliam
Jan 29 at 9:13
$begingroup$
@LutzL, I was not meaning $5^x$, my third point could have also been 23, but yes, the derivative is apparently too large to get good results. Thanks!
$endgroup$
– pugliam
Jan 29 at 9:13
$begingroup$
I was just looking for a simple growing function that had the same type of growth in value and first derivative. All other interpolations will have the same type of behavior, either a rapid oscillation so that the negative derivative is correct, or rapid growth in the higher order derivatives so that the error term is large.
$endgroup$
– LutzL
Jan 29 at 9:35
$begingroup$
I was just looking for a simple growing function that had the same type of growth in value and first derivative. All other interpolations will have the same type of behavior, either a rapid oscillation so that the negative derivative is correct, or rapid growth in the higher order derivatives so that the error term is large.
$endgroup$
– LutzL
Jan 29 at 9:35
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
A difference quotient with three terms should be exact for quadratics. You can even use this fact to derive difference quotients. So, let's fit a quadratic to the points ${(0,0),(h,5),(2h,25)}$; we find that
$$
f(x) = frac{15 x^2}{2 h^2}-frac{5 x}{2 h}.
$$
It's easy enough to check that this quadratic passes through the desired points. It's also easy to check that
$$
f'(0)=frac{4 f(h)-f(2 h)-3 f(0)}{2 h}=-frac{5}{2 h}
$$
and, in particular, that the derivative is negative when $h>0$. In fact, it grows in absolute value as $hsearrow0$.
answered Jan 29 at 1:12
Mark McClureMark McClure
23.9k34472
$endgroup$
add a comment |
$begingroup$
Using
$f(x+h)
=f(x)+hf'(x)+h^2f''(x)/2+O(h^3)$,
we get
$begin{array}\
dfrac{-3f(0)+4f(h)-f(2h)}{2h}
&=dfrac{-3f(0)+4(f(0)+hf'(0)+h^2f''(0)/2+O(h^3))-(f(0)+2hf'(0)+4h^2f''(0)/2+O(h^3))}{2h}\
&=dfrac{2hf'(0)-3h^2f''(0)/2+O(h^3)}{2h}\
&=f'(0)-3hf''(0)/2+O(h^2)\
end{array}
$
so the approximation goes
in the opposite direction
than $f''(0)$.
answered Jan 29 at 1:57
marty cohenmarty cohen
74.9k549130
$endgroup$
$begingroup$
The second derivative terms cancel, that's why this is a second order derivative approximation. You will need the third derivative term to get the coefficient for the leading error term.
$endgroup$
– LutzL
Jan 29 at 9:30
add a comment |
$begingroup$
Try
$$ f'(0)approx frac{(2-h)f(2h)-(4-4h)f(h) +(2-3h)f(0)}{2h^2}.$$
answered Jan 29 at 1:38
John Z. LiJohn Z. Li
194
$endgroup$
$begingroup$
That's a difference quotient for $f''(h)$.
$endgroup$
– Mark McClure
Jan 29 at 2:31
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
A difference quotient with three terms should be exact for quadratics. You can even use this fact to derive difference quotients. So, let's fit a quadratic to the points ${(0,0),(h,5),(2h,25)}$; we find that
$$
f(x) = frac{15 x^2}{2 h^2}-frac{5 x}{2 h}.
$$
It's easy enough to check that this quadratic passes through the desired points. It's also easy to check that
$$
f'(0)=frac{4 f(h)-f(2 h)-3 f(0)}{2 h}=-frac{5}{2 h}
$$
and, in particular, that the derivative is negative when $h>0$. In fact, it grows in absolute value as $hsearrow0$.
answered Jan 29 at 1:12
Mark McClureMark McClure
23.9k34472
$endgroup$
add a comment |
$begingroup$
A difference quotient with three terms should be exact for quadratics. You can even use this fact to derive difference quotients. So, let's fit a quadratic to the points ${(0,0),(h,5),(2h,25)}$; we find that
$$
f(x) = frac{15 x^2}{2 h^2}-frac{5 x}{2 h}.
$$
It's easy enough to check that this quadratic passes through the desired points. It's also easy to check that
$$
f'(0)=frac{4 f(h)-f(2 h)-3 f(0)}{2 h}=-frac{5}{2 h}
$$
and, in particular, that the derivative is negative when $h>0$. In fact, it grows in absolute value as $hsearrow0$.
answered Jan 29 at 1:12
Mark McClureMark McClure
23.9k34472
$endgroup$
add a comment |
$begingroup$
A difference quotient with three terms should be exact for quadratics. You can even use this fact to derive difference quotients. So, let's fit a quadratic to the points ${(0,0),(h,5),(2h,25)}$; we find that
$$
f(x) = frac{15 x^2}{2 h^2}-frac{5 x}{2 h}.
$$
It's easy enough to check that this quadratic passes through the desired points. It's also easy to check that
$$
f'(0)=frac{4 f(h)-f(2 h)-3 f(0)}{2 h}=-frac{5}{2 h}
$$
and, in particular, that the derivative is negative when $h>0$. In fact, it grows in absolute value as $hsearrow0$.
answered Jan 29 at 1:12
Mark McClureMark McClure
23.9k34472
$endgroup$
A difference quotient with three terms should be exact for quadratics. You can even use this fact to derive difference quotients. So, let's fit a quadratic to the points ${(0,0),(h,5),(2h,25)}$; we find that
$$
f(x) = frac{15 x^2}{2 h^2}-frac{5 x}{2 h}.
$$
It's easy enough to check that this quadratic passes through the desired points. It's also easy to check that
$$
f'(0)=frac{4 f(h)-f(2 h)-3 f(0)}{2 h}=-frac{5}{2 h}
$$
and, in particular, that the derivative is negative when $h>0$. In fact, it grows in absolute value as $hsearrow0$.
answered Jan 29 at 1:12
Mark McClureMark McClure
23.9k34472
answered Jan 29 at 1:12
Mark McClureMark McClure
23.9k34472
answered Jan 29 at 1:12
Mark McClureMark McClure
23.9k34472
answered Jan 29 at 1:12
Mark McClureMark McClure
23.9k34472
23.9k34472
add a comment |
add a comment |
$begingroup$
Using
$f(x+h)
=f(x)+hf'(x)+h^2f''(x)/2+O(h^3)$,
we get
$begin{array}\
dfrac{-3f(0)+4f(h)-f(2h)}{2h}
&=dfrac{-3f(0)+4(f(0)+hf'(0)+h^2f''(0)/2+O(h^3))-(f(0)+2hf'(0)+4h^2f''(0)/2+O(h^3))}{2h}\
&=dfrac{2hf'(0)-3h^2f''(0)/2+O(h^3)}{2h}\
&=f'(0)-3hf''(0)/2+O(h^2)\
end{array}
$
so the approximation goes
in the opposite direction
than $f''(0)$.
answered Jan 29 at 1:57
marty cohenmarty cohen
74.9k549130
$endgroup$
$begingroup$
The second derivative terms cancel, that's why this is a second order derivative approximation. You will need the third derivative term to get the coefficient for the leading error term.
$endgroup$
– LutzL
Jan 29 at 9:30
add a comment |
$begingroup$
Using
$f(x+h)
=f(x)+hf'(x)+h^2f''(x)/2+O(h^3)$,
we get
$begin{array}\
dfrac{-3f(0)+4f(h)-f(2h)}{2h}
&=dfrac{-3f(0)+4(f(0)+hf'(0)+h^2f''(0)/2+O(h^3))-(f(0)+2hf'(0)+4h^2f''(0)/2+O(h^3))}{2h}\
&=dfrac{2hf'(0)-3h^2f''(0)/2+O(h^3)}{2h}\
&=f'(0)-3hf''(0)/2+O(h^2)\
end{array}
$
so the approximation goes
in the opposite direction
than $f''(0)$.
answered Jan 29 at 1:57
marty cohenmarty cohen
74.9k549130
$endgroup$
$begingroup$
The second derivative terms cancel, that's why this is a second order derivative approximation. You will need the third derivative term to get the coefficient for the leading error term.
$endgroup$
– LutzL
Jan 29 at 9:30
add a comment |
$begingroup$
Using
$f(x+h)
=f(x)+hf'(x)+h^2f''(x)/2+O(h^3)$,
we get
$begin{array}\
dfrac{-3f(0)+4f(h)-f(2h)}{2h}
&=dfrac{-3f(0)+4(f(0)+hf'(0)+h^2f''(0)/2+O(h^3))-(f(0)+2hf'(0)+4h^2f''(0)/2+O(h^3))}{2h}\
&=dfrac{2hf'(0)-3h^2f''(0)/2+O(h^3)}{2h}\
&=f'(0)-3hf''(0)/2+O(h^2)\
end{array}
$
so the approximation goes
in the opposite direction
than $f''(0)$.
answered Jan 29 at 1:57
marty cohenmarty cohen
74.9k549130
$endgroup$
Using
$f(x+h)
=f(x)+hf'(x)+h^2f''(x)/2+O(h^3)$,
we get
$begin{array}\
dfrac{-3f(0)+4f(h)-f(2h)}{2h}
&=dfrac{-3f(0)+4(f(0)+hf'(0)+h^2f''(0)/2+O(h^3))-(f(0)+2hf'(0)+4h^2f''(0)/2+O(h^3))}{2h}\
&=dfrac{2hf'(0)-3h^2f''(0)/2+O(h^3)}{2h}\
&=f'(0)-3hf''(0)/2+O(h^2)\
end{array}
$
so the approximation goes
in the opposite direction
than $f''(0)$.
answered Jan 29 at 1:57
marty cohenmarty cohen
74.9k549130
answered Jan 29 at 1:57
marty cohenmarty cohen
74.9k549130
answered Jan 29 at 1:57
marty cohenmarty cohen
74.9k549130
answered Jan 29 at 1:57
marty cohenmarty cohen
74.9k549130
74.9k549130
$begingroup$
The second derivative terms cancel, that's why this is a second order derivative approximation. You will need the third derivative term to get the coefficient for the leading error term.
$endgroup$
– LutzL
Jan 29 at 9:30
add a comment |
$begingroup$
The second derivative terms cancel, that's why this is a second order derivative approximation. You will need the third derivative term to get the coefficient for the leading error term.
$endgroup$
– LutzL
Jan 29 at 9:30
$begingroup$
The second derivative terms cancel, that's why this is a second order derivative approximation. You will need the third derivative term to get the coefficient for the leading error term.
$endgroup$
– LutzL
Jan 29 at 9:30
$begingroup$
The second derivative terms cancel, that's why this is a second order derivative approximation. You will need the third derivative term to get the coefficient for the leading error term.
$endgroup$
– LutzL
Jan 29 at 9:30
add a comment |
$begingroup$
Try
$$ f'(0)approx frac{(2-h)f(2h)-(4-4h)f(h) +(2-3h)f(0)}{2h^2}.$$
answered Jan 29 at 1:38
John Z. LiJohn Z. Li
194
$endgroup$
$begingroup$
That's a difference quotient for $f''(h)$.
$endgroup$
– Mark McClure
Jan 29 at 2:31
add a comment |
$begingroup$
Try
$$ f'(0)approx frac{(2-h)f(2h)-(4-4h)f(h) +(2-3h)f(0)}{2h^2}.$$
answered Jan 29 at 1:38
John Z. LiJohn Z. Li
194
$endgroup$
$begingroup$
That's a difference quotient for $f''(h)$.
$endgroup$
– Mark McClure
Jan 29 at 2:31
add a comment |
$begingroup$
Try
$$ f'(0)approx frac{(2-h)f(2h)-(4-4h)f(h) +(2-3h)f(0)}{2h^2}.$$
answered Jan 29 at 1:38
John Z. LiJohn Z. Li
194
$endgroup$
Try
$$ f'(0)approx frac{(2-h)f(2h)-(4-4h)f(h) +(2-3h)f(0)}{2h^2}.$$
answered Jan 29 at 1:38
John Z. LiJohn Z. Li
194
edited Jan 29 at 3:07
answered Jan 29 at 1:38
John Z. LiJohn Z. Li
194
answered Jan 29 at 1:38
John Z. LiJohn Z. Li
194
answered Jan 29 at 1:38
John Z. LiJohn Z. Li
194
194
$begingroup$
That's a difference quotient for $f''(h)$.
$endgroup$
– Mark McClure
Jan 29 at 2:31
add a comment |
$begingroup$
That's a difference quotient for $f''(h)$.
$endgroup$
– Mark McClure
Jan 29 at 2:31
$begingroup$
That's a difference quotient for $f''(h)$.
$endgroup$
– Mark McClure
Jan 29 at 2:31
$begingroup$
That's a difference quotient for $f''(h)$.
$endgroup$
– Mark McClure
Jan 29 at 2:31
add a comment |
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$begingroup$
When you say "the formulation", I'm not sure that's the right way to think about this. Assuming your calculations are correct, your weird result could simply be a function of the particular way you approximated the first derivative. Just as there are lots of ways to approximate a function, there are lots of ways to approximate its derivative, even in only one direction.
$endgroup$
– Adrian Keister
Jan 28 at 17:15
1
$begingroup$
For a function like $5^x$ a step size of $h=1$ is just too large. The error term $sim ln(5)^35^hh^2$ should be smaller than the result, so $ln(5)h<0.5$ is necessary to start to get good results.
$endgroup$
– LutzL
Jan 28 at 21:49
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@LutzL, I was not meaning $5^x$, my third point could have also been 23, but yes, the derivative is apparently too large to get good results. Thanks!
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– pugliam
Jan 29 at 9:13
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I was just looking for a simple growing function that had the same type of growth in value and first derivative. All other interpolations will have the same type of behavior, either a rapid oscillation so that the negative derivative is correct, or rapid growth in the higher order derivatives so that the error term is large.
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– LutzL
Jan 29 at 9:35