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Finite-difference of the derivative at an arbitrary location
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The second-order, centered-approximation, finite difference of the derivative of $f(x)$ at $x$ is:
$$
f'(x)=frac-f(x-Delta x)+f(x+Delta x)2Delta x
$$
Question: Suppose we have the value of $f'(x)$ throughout our finite-difference grid with grid spacing $Delta x$. Let $f'(x_0)$ be such a value at $x_0$. Is there a way to find $f'(x)$ at an arbitray location between $(x_0-Delta x,x_0+Delta x)$ based on the value of $f'(x_0)$?
Attempt: I tried linearly interpolating the value of $f'(x_0)$ between $(x_0-Delta x,x_0+Delta x)$. However, the result was incorrect.
numerical-methods finite-differences
edited yesterday
Somos
10.8k1831
asked yesterday
A Slow Learner
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 |Â
show 2 more comments
up vote
0
down vote
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The second-order, centered-approximation, finite difference of the derivative of $f(x)$ at $x$ is:
$$
f'(x)=frac-f(x-Delta x)+f(x+Delta x)2Delta x
$$
Question: Suppose we have the value of $f'(x)$ throughout our finite-difference grid with grid spacing $Delta x$. Let $f'(x_0)$ be such a value at $x_0$. Is there a way to find $f'(x)$ at an arbitray location between $(x_0-Delta x,x_0+Delta x)$ based on the value of $f'(x_0)$?
Attempt: I tried linearly interpolating the value of $f'(x_0)$ between $(x_0-Delta x,x_0+Delta x)$. However, the result was incorrect.
numerical-methods finite-differences
edited yesterday
Somos
10.8k1831
asked yesterday
A Slow Learner
378211
What do you mean "the result was incorrect"? What did you do, and what is the correct answer?
– Andrei
yesterday
Why do you think this is "second order"?
– David G. Stork
yesterday
@DavidG.Stork Because the error is proportional to $(Delta x)^2$. This is taken from my textbook, too.
– A Slow Learner
yesterday
So what is a "first order" method? (And $0$th order?)
– David G. Stork
yesterday
@Andrei The way I performed interpolation is: I let $f'(x)$ at $x_0-Delta x$ to be the maximum value, and $f'(x)$ at $x_0$ to be half the max value. I also assumed that $f'(x)$ is linear between $(x-Delta x_0,x+ Delta x_0)$. I don't know the correct result, but the result that I obtained didn't make sense in a bigger problem that I am working on.
– A Slow Learner
yesterday
 |Â
show 2 more comments
up vote
0
down vote
favorite
up vote
0
down vote
favorite
The second-order, centered-approximation, finite difference of the derivative of $f(x)$ at $x$ is:
$$
f'(x)=frac-f(x-Delta x)+f(x+Delta x)2Delta x
$$
Question: Suppose we have the value of $f'(x)$ throughout our finite-difference grid with grid spacing $Delta x$. Let $f'(x_0)$ be such a value at $x_0$. Is there a way to find $f'(x)$ at an arbitray location between $(x_0-Delta x,x_0+Delta x)$ based on the value of $f'(x_0)$?
Attempt: I tried linearly interpolating the value of $f'(x_0)$ between $(x_0-Delta x,x_0+Delta x)$. However, the result was incorrect.
numerical-methods finite-differences
edited yesterday
Somos
10.8k1831
asked yesterday
A Slow Learner
378211
The second-order, centered-approximation, finite difference of the derivative of $f(x)$ at $x$ is:
$$
f'(x)=frac-f(x-Delta x)+f(x+Delta x)2Delta x
$$
Question: Suppose we have the value of $f'(x)$ throughout our finite-difference grid with grid spacing $Delta x$. Let $f'(x_0)$ be such a value at $x_0$. Is there a way to find $f'(x)$ at an arbitray location between $(x_0-Delta x,x_0+Delta x)$ based on the value of $f'(x_0)$?
Attempt: I tried linearly interpolating the value of $f'(x_0)$ between $(x_0-Delta x,x_0+Delta x)$. However, the result was incorrect.
numerical-methods finite-differences
edited yesterday
Somos
10.8k1831
asked yesterday
A Slow Learner
378211
edited yesterday
Somos
10.8k1831
edited yesterday
Somos
10.8k1831
edited yesterday
Somos
10.8k1831
10.8k1831
asked yesterday
A Slow Learner
378211
asked yesterday
A Slow Learner
378211
asked yesterday
A Slow Learner
378211
378211
What do you mean "the result was incorrect"? What did you do, and what is the correct answer?
– Andrei
yesterday
Why do you think this is "second order"?
– David G. Stork
yesterday
@DavidG.Stork Because the error is proportional to $(Delta x)^2$. This is taken from my textbook, too.
– A Slow Learner
yesterday
So what is a "first order" method? (And $0$th order?)
– David G. Stork
yesterday
@Andrei The way I performed interpolation is: I let $f'(x)$ at $x_0-Delta x$ to be the maximum value, and $f'(x)$ at $x_0$ to be half the max value. I also assumed that $f'(x)$ is linear between $(x-Delta x_0,x+ Delta x_0)$. I don't know the correct result, but the result that I obtained didn't make sense in a bigger problem that I am working on.
– A Slow Learner
yesterday
 |Â
show 2 more comments
What do you mean "the result was incorrect"? What did you do, and what is the correct answer?
– Andrei
yesterday
Why do you think this is "second order"?
– David G. Stork
yesterday
@DavidG.Stork Because the error is proportional to $(Delta x)^2$. This is taken from my textbook, too.
– A Slow Learner
yesterday
So what is a "first order" method? (And $0$th order?)
– David G. Stork
yesterday
@Andrei The way I performed interpolation is: I let $f'(x)$ at $x_0-Delta x$ to be the maximum value, and $f'(x)$ at $x_0$ to be half the max value. I also assumed that $f'(x)$ is linear between $(x-Delta x_0,x+ Delta x_0)$. I don't know the correct result, but the result that I obtained didn't make sense in a bigger problem that I am working on.
– A Slow Learner
yesterday
What do you mean "the result was incorrect"? What did you do, and what is the correct answer?
– Andrei
yesterday
What do you mean "the result was incorrect"? What did you do, and what is the correct answer?
– Andrei
yesterday
Why do you think this is "second order"?
– David G. Stork
yesterday
Why do you think this is "second order"?
– David G. Stork
yesterday
@DavidG.Stork Because the error is proportional to $(Delta x)^2$. This is taken from my textbook, too.
– A Slow Learner
yesterday
@DavidG.Stork Because the error is proportional to $(Delta x)^2$. This is taken from my textbook, too.
– A Slow Learner
yesterday
So what is a "first order" method? (And $0$th order?)
– David G. Stork
yesterday
So what is a "first order" method? (And $0$th order?)
– David G. Stork
yesterday
@Andrei The way I performed interpolation is: I let $f'(x)$ at $x_0-Delta x$ to be the maximum value, and $f'(x)$ at $x_0$ to be half the max value. I also assumed that $f'(x)$ is linear between $(x-Delta x_0,x+ Delta x_0)$. I don't know the correct result, but the result that I obtained didn't make sense in a bigger problem that I am working on.
– A Slow Learner
yesterday
@Andrei The way I performed interpolation is: I let $f'(x)$ at $x_0-Delta x$ to be the maximum value, and $f'(x)$ at $x_0$ to be half the max value. I also assumed that $f'(x)$ is linear between $(x-Delta x_0,x+ Delta x_0)$. I don't know the correct result, but the result that I obtained didn't make sense in a bigger problem that I am working on.
– A Slow Learner
yesterday
 |Â
show 2 more comments
1 Answer
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You asked for an approximation to $ f'(x). $ Using simple algebra the answer is
$$ f'(x) approx frac f(x_0+Delta x) - f(x_0-Delta x) 2 Delta x +
2(x - x_0)frac f(x_0+Delta x) - f(x_0) + f(x_0-Delta x) Delta x^2 $$
where the two fractions are approximations to the first and second derivatives of $ f(x) $ at $ x = x_0. $
You can check that when $ f(x) $ is a quadratic polynomial, then the approximation is exact.
answered yesterday
Somos
10.8k1831
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
You asked for an approximation to $ f'(x). $ Using simple algebra the answer is
$$ f'(x) approx frac f(x_0+Delta x) - f(x_0-Delta x) 2 Delta x +
2(x - x_0)frac f(x_0+Delta x) - f(x_0) + f(x_0-Delta x) Delta x^2 $$
where the two fractions are approximations to the first and second derivatives of $ f(x) $ at $ x = x_0. $
You can check that when $ f(x) $ is a quadratic polynomial, then the approximation is exact.
answered yesterday
Somos
10.8k1831
add a comment |Â
up vote
0
down vote
You asked for an approximation to $ f'(x). $ Using simple algebra the answer is
$$ f'(x) approx frac f(x_0+Delta x) - f(x_0-Delta x) 2 Delta x +
2(x - x_0)frac f(x_0+Delta x) - f(x_0) + f(x_0-Delta x) Delta x^2 $$
where the two fractions are approximations to the first and second derivatives of $ f(x) $ at $ x = x_0. $
You can check that when $ f(x) $ is a quadratic polynomial, then the approximation is exact.
answered yesterday
Somos
10.8k1831
add a comment |Â
up vote
0
down vote
up vote
0
down vote
You asked for an approximation to $ f'(x). $ Using simple algebra the answer is
$$ f'(x) approx frac f(x_0+Delta x) - f(x_0-Delta x) 2 Delta x +
2(x - x_0)frac f(x_0+Delta x) - f(x_0) + f(x_0-Delta x) Delta x^2 $$
where the two fractions are approximations to the first and second derivatives of $ f(x) $ at $ x = x_0. $
You can check that when $ f(x) $ is a quadratic polynomial, then the approximation is exact.
answered yesterday
Somos
10.8k1831
You asked for an approximation to $ f'(x). $ Using simple algebra the answer is
$$ f'(x) approx frac f(x_0+Delta x) - f(x_0-Delta x) 2 Delta x +
2(x - x_0)frac f(x_0+Delta x) - f(x_0) + f(x_0-Delta x) Delta x^2 $$
where the two fractions are approximations to the first and second derivatives of $ f(x) $ at $ x = x_0. $
You can check that when $ f(x) $ is a quadratic polynomial, then the approximation is exact.
answered yesterday
Somos
10.8k1831
edited yesterday
answered yesterday
Somos
10.8k1831
answered yesterday
Somos
10.8k1831
answered yesterday
Somos
10.8k1831
10.8k1831
add a comment |Â
add a comment |Â
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What do you mean "the result was incorrect"? What did you do, and what is the correct answer?
– Andrei
yesterday
Why do you think this is "second order"?
– David G. Stork
yesterday
@DavidG.Stork Because the error is proportional to $(Delta x)^2$. This is taken from my textbook, too.
– A Slow Learner
yesterday
So what is a "first order" method? (And $0$th order?)
– David G. Stork
yesterday
@Andrei The way I performed interpolation is: I let $f'(x)$ at $x_0-Delta x$ to be the maximum value, and $f'(x)$ at $x_0$ to be half the max value. I also assumed that $f'(x)$ is linear between $(x-Delta x_0,x+ Delta x_0)$. I don't know the correct result, but the result that I obtained didn't make sense in a bigger problem that I am working on.
– A Slow Learner
yesterday