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Sage#4
Project: boip
Path: sage#4.ipynb
Views: 26License: AGPL3
Image: ubuntu2204
Kernel: SageMath 10.0
Hello Dr.
Exercise #1
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Out[3]:
('The sequence an represents 8n! and converges to 0 as n approaches infinity. It is a monotone decreasing sequence.',
'The sequence bn is defined as n + 1 and does not converge. It is not a monotone sequence.',
'The sequence cn is defined as [ln(n)]^4 and does not converge. It is not a monotone sequence.')
Exercise #2
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Out[4]:
Partial Sum S(1) = 1.00000000000000
Partial Sum S(2) = 1.25000000000000
Partial Sum S(3) = 1.36111111111111
Partial Sum S(4) = 1.42361111111111
Partial Sum S(5) = 1.46361111111111
Partial Sum S(6) = 1.49138888888889
Partial Sum S(7) = 1.51179705215420
Partial Sum S(8) = 1.52742205215420
Partial Sum S(9) = 1.53976773116654
Partial Sum S(10) = 1.54976773116654
Partial Sum S(11) = 1.55803219397646
Partial Sum S(12) = 1.56497663842090
Partial Sum S(13) = 1.57089379818422
Partial Sum S(14) = 1.57599583900054
Partial Sum S(15) = 1.58044028344499
Partial Sum S(16) = 1.58434653344499
Partial Sum S(17) = 1.58780674105744
Partial Sum S(18) = 1.59089316081053
Partial Sum S(19) = 1.59366324391302
Partial Sum S(20) = 1.59616324391302
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Out[5]:
Exercise #3
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Out[6]:
['Result of sum_i: The result of ∑∞ 1/i^2 is',
'Divergent (Non-convergent Series)',
'Result of sum_ii:',
'Strategies to calculate this sum may include partial fraction decomposition.',
' The value of ∑∞ 1/((j^2 - 1)) is',
3/4,
'Result of sum_iii:',
'Strategies to calculate this sum may include partial fraction decomposition.',
'The value of ∑∞ 1/(k^2 - 3) is',
-1/6*sqrt(3)*harmonic_number(-sqrt(3)) + 1/6*sqrt(3)*harmonic_number(sqrt(3)),
'Result of sum_iv: The result of ∑∞ 1/n is',
'Divergent (Non-convergent Series)',
'Result of problematic_sum: Symbolic expression for a problematic sum is',
2*sum(log(n + 1)/n^2, n, 1, +Infinity),
'Numerical approximation of problematic_sum is',
'positive infinity',
'Result of integral_result: The integral of ln((x+1)^2)/x^2 from 1 to infinity is',
4*log(2)]
Exercise 4:
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Out[7]:
{'Result Explanation': 'The numerical approximation of s50 for the series ∑(ln((n+1)^2)/n^2) from n=1 to 50 is approximately 3.40619538529727.',
'Error Interval': 'The error interval is approximately [3.20931710539801, 3.60307366519653].'}
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'Hi'
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Out[9]:
{'Result Explanation': 'The numerical approximation of s4 for the series is approximately 4.94215686274510.',
'Integral Explanation': 'The integral result (from 1 to infinity) is approximately 5/4*pi + 1/2*log(2).',
'Error Explanation': 'The error is estimated to be between 0.668592455477884 and 9.21572127001231. The integral essentially represents the exact accumulation (area under the curve) of the function from 1 to infinity. The difference between the numerical approximation and the integral gives an estimate of the error introduced by simplifying the series at a finite point (in this case, at n=4). By adding and subtracting the integral result, we create a range within which we expect the true sum of the series to lie. This range serves as the upper and lower bounds on the error.'}