Find the Maclaurin series for the following function

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SUMMARY

The Maclaurin series for the function f(x) = ln(1 - x^3) / x^2 is derived using the series expansion for ln(1 + x). The correct series representation is [(-1)^(2n-1) (x)^(3n-2)] / n, as confirmed by the professor. The initial attempt yielded an incorrect series of [(-1)^(2n-1) (x)^(n)] / n. The discrepancy arises from the need to properly account for the transformation of the variable in the logarithmic function.

PREREQUISITES
  • Understanding of Maclaurin series expansion
  • Familiarity with logarithmic functions and their properties
  • Knowledge of series notation and summation
  • Basic calculus concepts related to Taylor series
NEXT STEPS
  • Study the derivation of the Maclaurin series for ln(1 - x)
  • Explore the application of series transformations in calculus
  • Learn about convergence criteria for power series
  • Investigate the relationship between Maclaurin and Taylor series
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Students studying calculus, particularly those focusing on series expansions, and educators seeking to clarify the derivation of Maclaurin series for logarithmic functions.

jorgegalvan93
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Homework Statement



f(x) =ln (1-x^3) / (x^2)

Homework Equations



Using the maclaurin series ln (1 +x) = Ʃ (-1)^(n-1) (x^n)/(n)

The Attempt at a Solution



the maclaurin series for the function i get is [(-1)^(2n-1) (x)^(n)] / (n)
however, the answer according to my prof is [(-1)^(2n-1) (x)^(3n-2)] / (n)
How?
 

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jorgegalvan93 said:

Homework Statement



f(x) =ln (1-x^3) / (x^2)

Homework Equations



Using the maclaurin series ln (1 +x) = Ʃ (-1)^(n-1) (x^n)/(n)

The Attempt at a Solution



the maclaurin series for the function i get is [(-1)^(2n-1) (x)^(n)] / (n)
however, the answer according to my prof is [(-1)^(2n-1) (x)^(3n-2)] / (n)
How?
How about showing your steps ?

What is the Maclaurin series for ln(1-x) ?

Then what is the Maclaurin series for ln(1-x3) ?
 

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