Power series for integral (1/x) dx

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Homework Help Overview

The discussion revolves around finding the power series representation for the integral of (1/x) dx, which relates to the natural logarithm function. Participants are exploring the connection between this integral and the known power series for ln(1+x).

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants are attempting to relate the integral of (1/x) to the power series for ln(1+x) and questioning the correctness of their shifts in the series representation. There is discussion about whether ln(x) can be derived from the series and how to adjust the indices and terms accordingly.

Discussion Status

The discussion is ongoing, with participants sharing their reasoning and questioning their assumptions about the power series representation. Some have proposed adjustments to the series terms, but there is no consensus on the correct form yet.

Contextual Notes

Participants are working under the assumption that x > 0 and are considering the implications of shifting the series for ln(1+x) to represent ln(x). There is a focus on ensuring the correct application of series terms and indices.

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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