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Prove that the natural logarithm of (1 + x) for |x| < 1 is: x3 In(1 + x) = x - 3 n! Show that the accuracy of your series solution is higher ifn is larger. So repeat the solution for 2 different values of n, one of which is larger than 4 and the other smaller than 4. You must use a recursive loop within a function to evaluate the right hand side of the equation above.

Prove that the natural logarithm of (1 + x) for |x| < 1 is: x3 In(1 + x) = x - 3 n! Show that the accuracy of your series solution is higher ifn is larger. So repeat the solution for 2 different values of n, one of which is larger than 4 and the other smaller than 4. You must use a recursive loop within a function to evaluate the right hand side of the equation above.

Database System Concepts
Database System Concepts
7th Edition
ISBN:9780078022159
Author:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan
Publisher:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan
Chapter1: Introduction
Section: Chapter Questions
Problem 1PE
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I am doing some practice in basic Python and a bit hung up on this piece. Could someone review the question and my progress thus far?

Prove that the natural logarithm of (1 + x) for |x| < 1 is: x² In(1 + x) = x - 3 4 n! Show that the accuracy of your series solution is higher ifn is larger. So repeat the solution for 2 different values of n, one of which is larger than 4 and the other smaller than 4. You must use a recursive loop within a function to evaluate the right hand side of the equation above.
Transcribed Image Text:Prove that the natural logarithm of (1 + x) for |x| < 1 is: x² In(1 + x) = x - 3 4 n! Show that the accuracy of your series solution is higher ifn is larger. So repeat the solution for 2 different values of n, one of which is larger than 4 and the other smaller than 4. You must use a recursive loop within a function to evaluate the right hand side of the equation above.
In [37]: H import math x = 0.75 n1 = 5 n2 = 3 In [38]: M LHS = (np.log(1 + x)) In [40]: N Sum_RHS_n1 = 1 Factorial_i = 1 In [42]: for i in range (1, n1 + 1): x = (math.fabs (x)) Sum_RHS_n1 = Sum_RHS_n1 - (x**i) * ((1)**i) print((x**i) * ((-1)**i)) -0.75 8.5625 -0.421875 0.31640625 -0.2373046875 In [43]: H print('The value of the expression on the RHS is', Sum_RHS_n1) print('The value of the expression on the LHS is', LHS) The value of the expression on the RHS is -3.576171875 The value of the expression on the LHS is 0.5596157879354227 In [44]: Sum_RHS_n2 = 1 In [35]: N for i in range (1, n2 + 1): Sum_RHS_n2 = Sum_RHS_n2 + (x**i) * ((-1)**i)
Transcribed Image Text:In [37]: H import math x = 0.75 n1 = 5 n2 = 3 In [38]: M LHS = (np.log(1 + x)) In [40]: N Sum_RHS_n1 = 1 Factorial_i = 1 In [42]: for i in range (1, n1 + 1): x = (math.fabs (x)) Sum_RHS_n1 = Sum_RHS_n1 - (x**i) * ((1)**i) print((x**i) * ((-1)**i)) -0.75 8.5625 -0.421875 0.31640625 -0.2373046875 In [43]: H print('The value of the expression on the RHS is', Sum_RHS_n1) print('The value of the expression on the LHS is', LHS) The value of the expression on the RHS is -3.576171875 The value of the expression on the LHS is 0.5596157879354227 In [44]: Sum_RHS_n2 = 1 In [35]: N for i in range (1, n2 + 1): Sum_RHS_n2 = Sum_RHS_n2 + (x**i) * ((-1)**i)
Expert Solution
Step 1

Term  Xn / n!   in RHS seems irrelevant as Taylor or Maclaurin series for ln(1+x) does not have that term.

As question description ask to use recursive approach, solution uses recursive method.

 

recursive function take following parameters,

x  - value of x

n - value of n

current_n  - value of n in current recursive call. Increase by 1 for every recursive call and starts with 1

result - total result up to current recursion call. Act as the accumulator for RHS and it starts with 0.

 

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