Remainder Estimate for Integral test
$begingroup$
I have the following question, it is a fill in the blank type question, however when I submit my answer, the system which verifies it say it is incorrect. I believe I am right, so I was hoping for someone to show me where I am going wrong.
Here is the question.
Accuracy to within 0.0005 means that we have to find the value of $n$ such that $R_n le$ ____.
Since $R_n le int_{n}^infty frac{4dx}{x^3} = lim_{ttoinfty} int_{n}^t frac{4dx}{x^3} = $_____.
We want ____ $le $ _____.
Solving this inequality we get $ n ge$ ____ $approx$ ______.
Here is my solution
Accuracy to within 0.0005 means that we have to find the value of $n$ such that $R_n le$ 0.0005.
Since $R_n le int_{n}^infty frac{4dx}{x^3} = lim_{ttoinfty} int_{n}^t frac{4dx}{x^3} = frac{4}{2n^2}$.
We want $frac{4}{2n^2}le 0.0005$
Solving this inequality we get $ n ge sqrt{frac{4}{2cdot0.0005}} approx$ 63.24555.
Some other notes... I am pretty convinced that all the blanks are correct except the last blank, thus I have tried 64 as another answer (however that was also flagged as incorrect).
calculus sequences-and-series improper-integrals solution-verification
asked Jul 28 '15 at 23:16
darockadarocka
112
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add a comment |
$begingroup$
I have the following question, it is a fill in the blank type question, however when I submit my answer, the system which verifies it say it is incorrect. I believe I am right, so I was hoping for someone to show me where I am going wrong.
Here is the question.
Accuracy to within 0.0005 means that we have to find the value of $n$ such that $R_n le$ ____.
Since $R_n le int_{n}^infty frac{4dx}{x^3} = lim_{ttoinfty} int_{n}^t frac{4dx}{x^3} = $_____.
We want ____ $le $ _____.
Solving this inequality we get $ n ge$ ____ $approx$ ______.
Here is my solution
Accuracy to within 0.0005 means that we have to find the value of $n$ such that $R_n le$ 0.0005.
Since $R_n le int_{n}^infty frac{4dx}{x^3} = lim_{ttoinfty} int_{n}^t frac{4dx}{x^3} = frac{4}{2n^2}$.
We want $frac{4}{2n^2}le 0.0005$
Solving this inequality we get $ n ge sqrt{frac{4}{2cdot0.0005}} approx$ 63.24555.
Some other notes... I am pretty convinced that all the blanks are correct except the last blank, thus I have tried 64 as another answer (however that was also flagged as incorrect).
calculus sequences-and-series improper-integrals solution-verification
asked Jul 28 '15 at 23:16
darockadarocka
112
$endgroup$
$begingroup$
The notation $R_n$ has many uses. Just what does it mean here? And the expression you put in the second blank is not simplified. The computer probably wants $dfrac{2}{n^2}$ or perhaps $2n^{-2}$.
$endgroup$
– Rory Daulton
Jul 29 '15 at 0:35
add a comment |
$begingroup$
I have the following question, it is a fill in the blank type question, however when I submit my answer, the system which verifies it say it is incorrect. I believe I am right, so I was hoping for someone to show me where I am going wrong.
Here is the question.
Accuracy to within 0.0005 means that we have to find the value of $n$ such that $R_n le$ ____.
Since $R_n le int_{n}^infty frac{4dx}{x^3} = lim_{ttoinfty} int_{n}^t frac{4dx}{x^3} = $_____.
We want ____ $le $ _____.
Solving this inequality we get $ n ge$ ____ $approx$ ______.
Here is my solution
Accuracy to within 0.0005 means that we have to find the value of $n$ such that $R_n le$ 0.0005.
Since $R_n le int_{n}^infty frac{4dx}{x^3} = lim_{ttoinfty} int_{n}^t frac{4dx}{x^3} = frac{4}{2n^2}$.
We want $frac{4}{2n^2}le 0.0005$
Solving this inequality we get $ n ge sqrt{frac{4}{2cdot0.0005}} approx$ 63.24555.
Some other notes... I am pretty convinced that all the blanks are correct except the last blank, thus I have tried 64 as another answer (however that was also flagged as incorrect).
calculus sequences-and-series improper-integrals solution-verification
asked Jul 28 '15 at 23:16
darockadarocka
112
$endgroup$
I have the following question, it is a fill in the blank type question, however when I submit my answer, the system which verifies it say it is incorrect. I believe I am right, so I was hoping for someone to show me where I am going wrong.
Here is the question.
Accuracy to within 0.0005 means that we have to find the value of $n$ such that $R_n le$ ____.
Since $R_n le int_{n}^infty frac{4dx}{x^3} = lim_{ttoinfty} int_{n}^t frac{4dx}{x^3} = $_____.
We want ____ $le $ _____.
Solving this inequality we get $ n ge$ ____ $approx$ ______.
Here is my solution
Accuracy to within 0.0005 means that we have to find the value of $n$ such that $R_n le$ 0.0005.
Since $R_n le int_{n}^infty frac{4dx}{x^3} = lim_{ttoinfty} int_{n}^t frac{4dx}{x^3} = frac{4}{2n^2}$.
We want $frac{4}{2n^2}le 0.0005$
Solving this inequality we get $ n ge sqrt{frac{4}{2cdot0.0005}} approx$ 63.24555.
Some other notes... I am pretty convinced that all the blanks are correct except the last blank, thus I have tried 64 as another answer (however that was also flagged as incorrect).
calculus sequences-and-series improper-integrals solution-verification
calculus sequences-and-series improper-integrals solution-verification
asked Jul 28 '15 at 23:16
darockadarocka
112
asked Jul 28 '15 at 23:16
darockadarocka
112
edited Jul 28 '15 at 23:44
user147263
asked Jul 28 '15 at 23:16
darockadarocka
112
asked Jul 28 '15 at 23:16
darockadarocka
112
asked Jul 28 '15 at 23:16
darockadarocka
112
112
$begingroup$
The notation $R_n$ has many uses. Just what does it mean here? And the expression you put in the second blank is not simplified. The computer probably wants $dfrac{2}{n^2}$ or perhaps $2n^{-2}$.
$endgroup$
– Rory Daulton
Jul 29 '15 at 0:35
add a comment |
$begingroup$
The notation $R_n$ has many uses. Just what does it mean here? And the expression you put in the second blank is not simplified. The computer probably wants $dfrac{2}{n^2}$ or perhaps $2n^{-2}$.
$endgroup$
– Rory Daulton
Jul 29 '15 at 0:35
$begingroup$
The notation $R_n$ has many uses. Just what does it mean here? And the expression you put in the second blank is not simplified. The computer probably wants $dfrac{2}{n^2}$ or perhaps $2n^{-2}$.
$endgroup$
– Rory Daulton
Jul 29 '15 at 0:35
$begingroup$
The notation $R_n$ has many uses. Just what does it mean here? And the expression you put in the second blank is not simplified. The computer probably wants $dfrac{2}{n^2}$ or perhaps $2n^{-2}$.
$endgroup$
– Rory Daulton
Jul 29 '15 at 0:35
add a comment |
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$begingroup$
The notation $R_n$ has many uses. Just what does it mean here? And the expression you put in the second blank is not simplified. The computer probably wants $dfrac{2}{n^2}$ or perhaps $2n^{-2}$.
$endgroup$
– Rory Daulton
Jul 29 '15 at 0:35