Error Bound of composite trapezium rule Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Trapezoidal Rule (Quadrature) Error ApproximationIntegral of $x^2ln(x)$ using Simpson's ruleerror bound of a integrationSimpson's Error Bound EstimationHow to find upper bound on absolute error with composite trapezoid ruleNumerical Integration Error Bound$sum_n=3^infty frac1n(ln n)^4$ what upper bound does it yield for the error S-S30Error of composite trapezium ruleHaving trouble finding error bound due to an undefined termNumerical Integration, error bound setting step size.
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Error Bound of composite trapezium rule
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Trapezoidal Rule (Quadrature) Error ApproximationIntegral of $x^2ln(x)$ using Simpson's ruleerror bound of a integrationSimpson's Error Bound EstimationHow to find upper bound on absolute error with composite trapezoid ruleNumerical Integration Error Bound$sum_n=3^infty frac1n(ln n)^4$ what upper bound does it yield for the error S-S30Error of composite trapezium ruleHaving trouble finding error bound due to an undefined termNumerical Integration, error bound setting step size.
$begingroup$
Given the function: $f(x) = cos(2x) expleft(-x^2right)$
I estimated $int_-2^2 f(x) dx$ using the formula.
I need to calculate the error bound using the formula:
$$
R = −fracb−a12 cdot h^2 cdot fracd^2 fdx^2(xi)
$$
Where $(d^2 y/dx^2)(ξ)$ is the maximum of the second derivative. I can tried calculating the function at both limits getting the same answer of, -0.17075...
Is there a way to measure if this answer is correct or do I just hope for the best, I cant see the pattern as to where the function will be a maximum.
integration approximation approximate-integration
edited Apr 1 at 14:48
gt6989b
36k22557
asked Apr 1 at 14:24
Mitul SuchakMitul Suchak
65
$endgroup$
add a comment |
$begingroup$
Given the function: $f(x) = cos(2x) expleft(-x^2right)$
I estimated $int_-2^2 f(x) dx$ using the formula.
I need to calculate the error bound using the formula:
$$
R = −fracb−a12 cdot h^2 cdot fracd^2 fdx^2(xi)
$$
Where $(d^2 y/dx^2)(ξ)$ is the maximum of the second derivative. I can tried calculating the function at both limits getting the same answer of, -0.17075...
Is there a way to measure if this answer is correct or do I just hope for the best, I cant see the pattern as to where the function will be a maximum.
integration approximation approximate-integration
edited Apr 1 at 14:48
gt6989b
36k22557
asked Apr 1 at 14:24
Mitul SuchakMitul Suchak
65
$endgroup$
$begingroup$
so does that mean the maximum value for the second derivative is 1?
$endgroup$
– Mitul Suchak
Apr 1 at 22:09
add a comment |
$begingroup$
Given the function: $f(x) = cos(2x) expleft(-x^2right)$
I estimated $int_-2^2 f(x) dx$ using the formula.
I need to calculate the error bound using the formula:
$$
R = −fracb−a12 cdot h^2 cdot fracd^2 fdx^2(xi)
$$
Where $(d^2 y/dx^2)(ξ)$ is the maximum of the second derivative. I can tried calculating the function at both limits getting the same answer of, -0.17075...
Is there a way to measure if this answer is correct or do I just hope for the best, I cant see the pattern as to where the function will be a maximum.
integration approximation approximate-integration
edited Apr 1 at 14:48
gt6989b
36k22557
asked Apr 1 at 14:24
Mitul SuchakMitul Suchak
65
$endgroup$
Given the function: $f(x) = cos(2x) expleft(-x^2right)$
I estimated $int_-2^2 f(x) dx$ using the formula.
I need to calculate the error bound using the formula:
$$
R = −fracb−a12 cdot h^2 cdot fracd^2 fdx^2(xi)
$$
Where $(d^2 y/dx^2)(ξ)$ is the maximum of the second derivative. I can tried calculating the function at both limits getting the same answer of, -0.17075...
Is there a way to measure if this answer is correct or do I just hope for the best, I cant see the pattern as to where the function will be a maximum.
integration approximation approximate-integration
integration approximation approximate-integration
edited Apr 1 at 14:48
gt6989b
36k22557
asked Apr 1 at 14:24
Mitul SuchakMitul Suchak
65
edited Apr 1 at 14:48
gt6989b
36k22557
asked Apr 1 at 14:24
Mitul SuchakMitul Suchak
65
edited Apr 1 at 14:48
gt6989b
36k22557
edited Apr 1 at 14:48
gt6989b
36k22557
edited Apr 1 at 14:48
gt6989b
36k22557
36k22557
asked Apr 1 at 14:24
Mitul SuchakMitul Suchak
65
asked Apr 1 at 14:24
Mitul SuchakMitul Suchak
65
asked Apr 1 at 14:24
Mitul SuchakMitul Suchak
65
65
$begingroup$
so does that mean the maximum value for the second derivative is 1?
$endgroup$
– Mitul Suchak
Apr 1 at 22:09
add a comment |
$begingroup$
so does that mean the maximum value for the second derivative is 1?
$endgroup$
– Mitul Suchak
Apr 1 at 22:09
$begingroup$
so does that mean the maximum value for the second derivative is 1?
$endgroup$
– Mitul Suchak
Apr 1 at 22:09
$begingroup$
so does that mean the maximum value for the second derivative is 1?
$endgroup$
– Mitul Suchak
Apr 1 at 22:09
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
It's easy to see that the function $f$ is symmetric around zero, i.e. $f(-x) = f(x)$. Sometimes we say such $f$ is even. One other obvious place to check is at zero, and $f(0)=1$, and it's easy to see that is the global maximum, since both the cosine and the exponent cannot exceed one.
answered Apr 1 at 14:46
gt6989bgt6989b
36k22557
$endgroup$
$begingroup$
so does that mean the maximum value for the second derivative is 1?
$endgroup$
– Mitul Suchak
Apr 2 at 16:33
$begingroup$
@MitulSuchak no, for the function itself. You can compute the second derivative analytically and find the maximum
$endgroup$
– gt6989b
Apr 2 at 17:07
$begingroup$
I mean in the error equation do i sub in 1 for d^2y/dx^2 part?
$endgroup$
– Mitul Suchak
Apr 3 at 21:25
$begingroup$
@MitulSuchak No, given $f$, compute $f''(x)$ and bound similarly
$endgroup$
– gt6989b
Apr 4 at 3:59
add a comment |
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1 Answer
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active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
It's easy to see that the function $f$ is symmetric around zero, i.e. $f(-x) = f(x)$. Sometimes we say such $f$ is even. One other obvious place to check is at zero, and $f(0)=1$, and it's easy to see that is the global maximum, since both the cosine and the exponent cannot exceed one.
answered Apr 1 at 14:46
gt6989bgt6989b
36k22557
$endgroup$
$begingroup$
so does that mean the maximum value for the second derivative is 1?
$endgroup$
– Mitul Suchak
Apr 2 at 16:33
$begingroup$
@MitulSuchak no, for the function itself. You can compute the second derivative analytically and find the maximum
$endgroup$
– gt6989b
Apr 2 at 17:07
$begingroup$
I mean in the error equation do i sub in 1 for d^2y/dx^2 part?
$endgroup$
– Mitul Suchak
Apr 3 at 21:25
$begingroup$
@MitulSuchak No, given $f$, compute $f''(x)$ and bound similarly
$endgroup$
– gt6989b
Apr 4 at 3:59
add a comment |
$begingroup$
It's easy to see that the function $f$ is symmetric around zero, i.e. $f(-x) = f(x)$. Sometimes we say such $f$ is even. One other obvious place to check is at zero, and $f(0)=1$, and it's easy to see that is the global maximum, since both the cosine and the exponent cannot exceed one.
answered Apr 1 at 14:46
gt6989bgt6989b
36k22557
$endgroup$
$begingroup$
so does that mean the maximum value for the second derivative is 1?
$endgroup$
– Mitul Suchak
Apr 2 at 16:33
$begingroup$
@MitulSuchak no, for the function itself. You can compute the second derivative analytically and find the maximum
$endgroup$
– gt6989b
Apr 2 at 17:07
$begingroup$
I mean in the error equation do i sub in 1 for d^2y/dx^2 part?
$endgroup$
– Mitul Suchak
Apr 3 at 21:25
$begingroup$
@MitulSuchak No, given $f$, compute $f''(x)$ and bound similarly
$endgroup$
– gt6989b
Apr 4 at 3:59
add a comment |
$begingroup$
It's easy to see that the function $f$ is symmetric around zero, i.e. $f(-x) = f(x)$. Sometimes we say such $f$ is even. One other obvious place to check is at zero, and $f(0)=1$, and it's easy to see that is the global maximum, since both the cosine and the exponent cannot exceed one.
answered Apr 1 at 14:46
gt6989bgt6989b
36k22557
$endgroup$
It's easy to see that the function $f$ is symmetric around zero, i.e. $f(-x) = f(x)$. Sometimes we say such $f$ is even. One other obvious place to check is at zero, and $f(0)=1$, and it's easy to see that is the global maximum, since both the cosine and the exponent cannot exceed one.
answered Apr 1 at 14:46
gt6989bgt6989b
36k22557
answered Apr 1 at 14:46
gt6989bgt6989b
36k22557
answered Apr 1 at 14:46
gt6989bgt6989b
36k22557
answered Apr 1 at 14:46
gt6989bgt6989b
36k22557
36k22557
$begingroup$
so does that mean the maximum value for the second derivative is 1?
$endgroup$
– Mitul Suchak
Apr 2 at 16:33
$begingroup$
@MitulSuchak no, for the function itself. You can compute the second derivative analytically and find the maximum
$endgroup$
– gt6989b
Apr 2 at 17:07
$begingroup$
I mean in the error equation do i sub in 1 for d^2y/dx^2 part?
$endgroup$
– Mitul Suchak
Apr 3 at 21:25
$begingroup$
@MitulSuchak No, given $f$, compute $f''(x)$ and bound similarly
$endgroup$
– gt6989b
Apr 4 at 3:59
add a comment |
$begingroup$
so does that mean the maximum value for the second derivative is 1?
$endgroup$
– Mitul Suchak
Apr 2 at 16:33
$begingroup$
@MitulSuchak no, for the function itself. You can compute the second derivative analytically and find the maximum
$endgroup$
– gt6989b
Apr 2 at 17:07
$begingroup$
I mean in the error equation do i sub in 1 for d^2y/dx^2 part?
$endgroup$
– Mitul Suchak
Apr 3 at 21:25
$begingroup$
@MitulSuchak No, given $f$, compute $f''(x)$ and bound similarly
$endgroup$
– gt6989b
Apr 4 at 3:59
$begingroup$
so does that mean the maximum value for the second derivative is 1?
$endgroup$
– Mitul Suchak
Apr 2 at 16:33
$begingroup$
so does that mean the maximum value for the second derivative is 1?
$endgroup$
– Mitul Suchak
Apr 2 at 16:33
$begingroup$
@MitulSuchak no, for the function itself. You can compute the second derivative analytically and find the maximum
$endgroup$
– gt6989b
Apr 2 at 17:07
$begingroup$
@MitulSuchak no, for the function itself. You can compute the second derivative analytically and find the maximum
$endgroup$
– gt6989b
Apr 2 at 17:07
$begingroup$
I mean in the error equation do i sub in 1 for d^2y/dx^2 part?
$endgroup$
– Mitul Suchak
Apr 3 at 21:25
$begingroup$
I mean in the error equation do i sub in 1 for d^2y/dx^2 part?
$endgroup$
– Mitul Suchak
Apr 3 at 21:25
$begingroup$
@MitulSuchak No, given $f$, compute $f''(x)$ and bound similarly
$endgroup$
– gt6989b
Apr 4 at 3:59
$begingroup$
@MitulSuchak No, given $f$, compute $f''(x)$ and bound similarly
$endgroup$
– gt6989b
Apr 4 at 3:59
add a comment |
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$begingroup$
so does that mean the maximum value for the second derivative is 1?
$endgroup$
– Mitul Suchak
Apr 1 at 22:09