Power series for integral (1/x) dx

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SUMMARY

The discussion focuses on finding the power series representation for the integral of (1/x) dx, which is ln(x) + C for x > 0. Participants reference the known power series for ln(1+x), given by Ʃ (-1)^(n-1) (x^n)/n from n = 1 to ∞. They explore the relationship between ln(1+x) and ln(x), concluding that the power series for ln(x) can be derived by adjusting the series for ln(1+x) to account for the shift in the function. The correct power series representation for ln(x) is Ʃ (-1)^(n-2) (x^(n-1))/(n-1) from n = 1 to ∞.

PREREQUISITES
  • Understanding of power series and their convergence
  • Familiarity with the natural logarithm function and its properties
  • Knowledge of Taylor series expansions
  • Basic calculus concepts, particularly integration techniques
NEXT STEPS
  • Study the derivation of Taylor series for ln(x) and ln(1+x)
  • Learn about the convergence criteria for power series
  • Explore applications of power series in solving integrals
  • Investigate the relationship between power series and other functions, such as exponential functions
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Students studying calculus, particularly those focusing on integration techniques and power series, as well as educators looking for examples of series expansions in mathematical discussions.

Jbreezy
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Homework Statement


I have to find the power series representation for integral (1/x) dx


Homework Equations



ln (1+x)

The Attempt at a Solution


This is very similar to ln(1+x) but I don't know if this helps me.

Is this ln(x) shifted one to the right? So maybe I can use what is already the power series for ln(1+x) = Ʃ (-1)^(n-1) (x^n)/n from n = 1 to ∞

so could I do ln(x) = [(-1)^(n) (x^(n+1)] / (n+1)

NO? Maybe I shifted it wrong?
 
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Jbreezy said:

Homework Statement


I have to find the power series representation for integral (1/x) dx


Homework Equations



ln (1+x)

The Attempt at a Solution


This is very similar to ln(1+x) but I don't know if this helps me.
##\int \frac{dx}{x} = ln(x) + C##, assuming x > 0.

ln(x + 1) is the translation by one unit left, not right, of the graph of y = ln(x).
Jbreezy said:
Is this ln(x) shifted one to the right? So maybe I can use what is already the power series for ln(1+x) = Ʃ (-1)^(n-1) (x^n)/n from n = 1 to ∞

so could I do ln(x) = [(-1)^(n) (x^(n+1)] / (n+1)

NO? Maybe I shifted it wrong?
 
OK so this

Ʃ (-1)^(n-1) (x^n)/n from n = 1 to ∞

Should be Ʃ (-1)^(n-2) (x^(n-1))/(n-1) from n = 1 to ∞

Right?
 
Jbreezy said:
OK so this

Ʃ (-1)^(n-1) (x^n)/n from n = 1 to ∞

Should be Ʃ (-1)^(n-2) (x^(n-1))/(n-1) from n = 1 to ∞

Right?
Instead of writing "answers" show me some mathematics reasoning.
 

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