Finding error in a an approximation
$begingroup$
We want to see the total error in approximating
$$ f'(x) approx frac{ f(x+h)-f(x) }{h} $$
where $f: R to R$ is differentiable. We can find $theta in [x,x+h]$
by Taylor's to that
$$ f(x+h) = f(x) + f'(x) h + f''( theta ) h^2 /2 $$
If the error in function values is bounded by $epsilon$, prove that
the rounding error is bounded by $2 epsilon /h$ and the truncation
error is bounded by $Mh/2$ where $M$ is a bound for $|f''(t)|$ for $t$
near $x$.
Try
We know truncation error is difference between true result and the result that would be produced by algorithm. By the result given above, we see that
$$ frac{ f(x+h) - f(x) }{h} = f'(x) + f''(theta)h/2 $$
So that
$$ underbrace{ frac{ f(x+h) - f(x) }{h} }_{approx} - underbrace{f'(x)}_{true ; result} = f''(theta)h/2 $$
So that trucantion error $E_T$ is absolute value of the above:
$$ E_T = |f''(theta) | h/2 leq Mh/2 $$
So we have our first result. However, for the rounding error I dont see how it is $2 epsilon / h $. Can someone explain what they really mean? Perhaps am I misunderstanding this part.
numerical-methods rounding-error truncation-error
asked Jan 2 at 7:02
Jimmy SabaterJimmy Sabater
2,579325
$endgroup$
add a comment |
$begingroup$
We want to see the total error in approximating
$$ f'(x) approx frac{ f(x+h)-f(x) }{h} $$
where $f: R to R$ is differentiable. We can find $theta in [x,x+h]$
by Taylor's to that
$$ f(x+h) = f(x) + f'(x) h + f''( theta ) h^2 /2 $$
If the error in function values is bounded by $epsilon$, prove that
the rounding error is bounded by $2 epsilon /h$ and the truncation
error is bounded by $Mh/2$ where $M$ is a bound for $|f''(t)|$ for $t$
near $x$.
Try
We know truncation error is difference between true result and the result that would be produced by algorithm. By the result given above, we see that
$$ frac{ f(x+h) - f(x) }{h} = f'(x) + f''(theta)h/2 $$
So that
$$ underbrace{ frac{ f(x+h) - f(x) }{h} }_{approx} - underbrace{f'(x)}_{true ; result} = f''(theta)h/2 $$
So that trucantion error $E_T$ is absolute value of the above:
$$ E_T = |f''(theta) | h/2 leq Mh/2 $$
So we have our first result. However, for the rounding error I dont see how it is $2 epsilon / h $. Can someone explain what they really mean? Perhaps am I misunderstanding this part.
numerical-methods rounding-error truncation-error
asked Jan 2 at 7:02
Jimmy SabaterJimmy Sabater
2,579325
$endgroup$
add a comment |
$begingroup$
We want to see the total error in approximating
$$ f'(x) approx frac{ f(x+h)-f(x) }{h} $$
where $f: R to R$ is differentiable. We can find $theta in [x,x+h]$
by Taylor's to that
$$ f(x+h) = f(x) + f'(x) h + f''( theta ) h^2 /2 $$
If the error in function values is bounded by $epsilon$, prove that
the rounding error is bounded by $2 epsilon /h$ and the truncation
error is bounded by $Mh/2$ where $M$ is a bound for $|f''(t)|$ for $t$
near $x$.
Try
We know truncation error is difference between true result and the result that would be produced by algorithm. By the result given above, we see that
$$ frac{ f(x+h) - f(x) }{h} = f'(x) + f''(theta)h/2 $$
So that
$$ underbrace{ frac{ f(x+h) - f(x) }{h} }_{approx} - underbrace{f'(x)}_{true ; result} = f''(theta)h/2 $$
So that trucantion error $E_T$ is absolute value of the above:
$$ E_T = |f''(theta) | h/2 leq Mh/2 $$
So we have our first result. However, for the rounding error I dont see how it is $2 epsilon / h $. Can someone explain what they really mean? Perhaps am I misunderstanding this part.
numerical-methods rounding-error truncation-error
asked Jan 2 at 7:02
Jimmy SabaterJimmy Sabater
2,579325
$endgroup$
We want to see the total error in approximating
$$ f'(x) approx frac{ f(x+h)-f(x) }{h} $$
where $f: R to R$ is differentiable. We can find $theta in [x,x+h]$
by Taylor's to that
$$ f(x+h) = f(x) + f'(x) h + f''( theta ) h^2 /2 $$
If the error in function values is bounded by $epsilon$, prove that
the rounding error is bounded by $2 epsilon /h$ and the truncation
error is bounded by $Mh/2$ where $M$ is a bound for $|f''(t)|$ for $t$
near $x$.
Try
We know truncation error is difference between true result and the result that would be produced by algorithm. By the result given above, we see that
$$ frac{ f(x+h) - f(x) }{h} = f'(x) + f''(theta)h/2 $$
So that
$$ underbrace{ frac{ f(x+h) - f(x) }{h} }_{approx} - underbrace{f'(x)}_{true ; result} = f''(theta)h/2 $$
So that trucantion error $E_T$ is absolute value of the above:
$$ E_T = |f''(theta) | h/2 leq Mh/2 $$
So we have our first result. However, for the rounding error I dont see how it is $2 epsilon / h $. Can someone explain what they really mean? Perhaps am I misunderstanding this part.
numerical-methods rounding-error truncation-error
numerical-methods rounding-error truncation-error
asked Jan 2 at 7:02
Jimmy SabaterJimmy Sabater
2,579325
asked Jan 2 at 7:02
Jimmy SabaterJimmy Sabater
2,579325
asked Jan 2 at 7:02
Jimmy SabaterJimmy Sabater
2,579325
asked Jan 2 at 7:02
Jimmy SabaterJimmy Sabater
2,579325
asked Jan 2 at 7:02
Jimmy SabaterJimmy Sabater
2,579325
2,579325
add a comment |
add a comment |
1 Answer
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$begingroup$
The rounding error? Take that upper bound $epsilon$ for the error in function values. We're subtracting two instances of it, which doubles that (at most $epsilon - (-epsilon)$). Then we divide by $h$, for $frac{2epsilon}{h}$. That's how far our calculated value of the difference quotient $frac{f(x+epsilon)-f(x)}{epsilon}$ can be from the true value.
answered Jan 2 at 7:14
jmerryjmerry
16.8k11633
$endgroup$
$begingroup$
what do they mean by function values? Are we trying to estimate f(x) exact with $f(x')$ approx.? That is we are given that $|f(x) - f(x') | leq epsilon $? and we want to estimate $|f'(x) - f'(x')|$?
$endgroup$
– Jimmy Sabater
Jan 2 at 7:28
1
$begingroup$
Say we're trying to calculate some reasonably nice function value - say, $exp(1.234)$. But this is a computer system - we're looking for an explicit number, and our numbers are represented with finite blocks of memory (likely 64 or 128 bits on a modern system). There's only so many numbers that can be represented exactly that way, and $e^{1.234}$ isn't one of them - so, to get our function value, we'll have to round it off. We'll estimate it as best we can, but we can't possibly do better than the closest number that has a floating-point representation. That's roundoff error.
$endgroup$
– jmerry
Jan 2 at 7:55
add a comment |
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1 Answer
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1 Answer
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$begingroup$
The rounding error? Take that upper bound $epsilon$ for the error in function values. We're subtracting two instances of it, which doubles that (at most $epsilon - (-epsilon)$). Then we divide by $h$, for $frac{2epsilon}{h}$. That's how far our calculated value of the difference quotient $frac{f(x+epsilon)-f(x)}{epsilon}$ can be from the true value.
answered Jan 2 at 7:14
jmerryjmerry
16.8k11633
$endgroup$
$begingroup$
what do they mean by function values? Are we trying to estimate f(x) exact with $f(x')$ approx.? That is we are given that $|f(x) - f(x') | leq epsilon $? and we want to estimate $|f'(x) - f'(x')|$?
$endgroup$
– Jimmy Sabater
Jan 2 at 7:28
1
$begingroup$
Say we're trying to calculate some reasonably nice function value - say, $exp(1.234)$. But this is a computer system - we're looking for an explicit number, and our numbers are represented with finite blocks of memory (likely 64 or 128 bits on a modern system). There's only so many numbers that can be represented exactly that way, and $e^{1.234}$ isn't one of them - so, to get our function value, we'll have to round it off. We'll estimate it as best we can, but we can't possibly do better than the closest number that has a floating-point representation. That's roundoff error.
$endgroup$
– jmerry
Jan 2 at 7:55
add a comment |
$begingroup$
The rounding error? Take that upper bound $epsilon$ for the error in function values. We're subtracting two instances of it, which doubles that (at most $epsilon - (-epsilon)$). Then we divide by $h$, for $frac{2epsilon}{h}$. That's how far our calculated value of the difference quotient $frac{f(x+epsilon)-f(x)}{epsilon}$ can be from the true value.
answered Jan 2 at 7:14
jmerryjmerry
16.8k11633
$endgroup$
$begingroup$
what do they mean by function values? Are we trying to estimate f(x) exact with $f(x')$ approx.? That is we are given that $|f(x) - f(x') | leq epsilon $? and we want to estimate $|f'(x) - f'(x')|$?
$endgroup$
– Jimmy Sabater
Jan 2 at 7:28
1
$begingroup$
Say we're trying to calculate some reasonably nice function value - say, $exp(1.234)$. But this is a computer system - we're looking for an explicit number, and our numbers are represented with finite blocks of memory (likely 64 or 128 bits on a modern system). There's only so many numbers that can be represented exactly that way, and $e^{1.234}$ isn't one of them - so, to get our function value, we'll have to round it off. We'll estimate it as best we can, but we can't possibly do better than the closest number that has a floating-point representation. That's roundoff error.
$endgroup$
– jmerry
Jan 2 at 7:55
add a comment |
$begingroup$
The rounding error? Take that upper bound $epsilon$ for the error in function values. We're subtracting two instances of it, which doubles that (at most $epsilon - (-epsilon)$). Then we divide by $h$, for $frac{2epsilon}{h}$. That's how far our calculated value of the difference quotient $frac{f(x+epsilon)-f(x)}{epsilon}$ can be from the true value.
answered Jan 2 at 7:14
jmerryjmerry
16.8k11633
$endgroup$
The rounding error? Take that upper bound $epsilon$ for the error in function values. We're subtracting two instances of it, which doubles that (at most $epsilon - (-epsilon)$). Then we divide by $h$, for $frac{2epsilon}{h}$. That's how far our calculated value of the difference quotient $frac{f(x+epsilon)-f(x)}{epsilon}$ can be from the true value.
answered Jan 2 at 7:14
jmerryjmerry
16.8k11633
answered Jan 2 at 7:14
jmerryjmerry
16.8k11633
answered Jan 2 at 7:14
jmerryjmerry
16.8k11633
answered Jan 2 at 7:14
jmerryjmerry
16.8k11633
16.8k11633
$begingroup$
what do they mean by function values? Are we trying to estimate f(x) exact with $f(x')$ approx.? That is we are given that $|f(x) - f(x') | leq epsilon $? and we want to estimate $|f'(x) - f'(x')|$?
$endgroup$
– Jimmy Sabater
Jan 2 at 7:28
1
$begingroup$
Say we're trying to calculate some reasonably nice function value - say, $exp(1.234)$. But this is a computer system - we're looking for an explicit number, and our numbers are represented with finite blocks of memory (likely 64 or 128 bits on a modern system). There's only so many numbers that can be represented exactly that way, and $e^{1.234}$ isn't one of them - so, to get our function value, we'll have to round it off. We'll estimate it as best we can, but we can't possibly do better than the closest number that has a floating-point representation. That's roundoff error.
$endgroup$
– jmerry
Jan 2 at 7:55
add a comment |
$begingroup$
what do they mean by function values? Are we trying to estimate f(x) exact with $f(x')$ approx.? That is we are given that $|f(x) - f(x') | leq epsilon $? and we want to estimate $|f'(x) - f'(x')|$?
$endgroup$
– Jimmy Sabater
Jan 2 at 7:28
1
$begingroup$
Say we're trying to calculate some reasonably nice function value - say, $exp(1.234)$. But this is a computer system - we're looking for an explicit number, and our numbers are represented with finite blocks of memory (likely 64 or 128 bits on a modern system). There's only so many numbers that can be represented exactly that way, and $e^{1.234}$ isn't one of them - so, to get our function value, we'll have to round it off. We'll estimate it as best we can, but we can't possibly do better than the closest number that has a floating-point representation. That's roundoff error.
$endgroup$
– jmerry
Jan 2 at 7:55
$begingroup$
what do they mean by function values? Are we trying to estimate f(x) exact with $f(x')$ approx.? That is we are given that $|f(x) - f(x') | leq epsilon $? and we want to estimate $|f'(x) - f'(x')|$?
$endgroup$
– Jimmy Sabater
Jan 2 at 7:28
$begingroup$
what do they mean by function values? Are we trying to estimate f(x) exact with $f(x')$ approx.? That is we are given that $|f(x) - f(x') | leq epsilon $? and we want to estimate $|f'(x) - f'(x')|$?
$endgroup$
– Jimmy Sabater
Jan 2 at 7:28
1
1
$begingroup$
Say we're trying to calculate some reasonably nice function value - say, $exp(1.234)$. But this is a computer system - we're looking for an explicit number, and our numbers are represented with finite blocks of memory (likely 64 or 128 bits on a modern system). There's only so many numbers that can be represented exactly that way, and $e^{1.234}$ isn't one of them - so, to get our function value, we'll have to round it off. We'll estimate it as best we can, but we can't possibly do better than the closest number that has a floating-point representation. That's roundoff error.
$endgroup$
– jmerry
Jan 2 at 7:55
$begingroup$
Say we're trying to calculate some reasonably nice function value - say, $exp(1.234)$. But this is a computer system - we're looking for an explicit number, and our numbers are represented with finite blocks of memory (likely 64 or 128 bits on a modern system). There's only so many numbers that can be represented exactly that way, and $e^{1.234}$ isn't one of them - so, to get our function value, we'll have to round it off. We'll estimate it as best we can, but we can't possibly do better than the closest number that has a floating-point representation. That's roundoff error.
$endgroup$
– jmerry
Jan 2 at 7:55
add a comment |
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