- #1

MathewsMD

- 433

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Question:

I was just wondering if there was any error in what I've done in the following steps to find the series representation of ##lnx##. I know ## \frac {1}{x}## is given in the following link by doing having the a function centred at 0, you can let ##f(x) = ∑^∞_{n=0} \frac {f^{(n)}(0)}{n!}x^n## after proving the function ## \frac {1}{x} ## can be represented as a series by doing the remainder test and finding R(x) → 0 as n → 0. Then just integrating.

I was just looking at an alternate method, which I don't find used as commonly but seems a lot simpler. Here are the steps:

Relevant Forumla(s):

##\frac {1}{1-r} = ∑^∞_{n=0} r^n## for a = 1

Attempted Solution:

If we let r = x + 1, then:

##\frac {1}{1- (x+1)} = \frac {1}{x} = ∑^∞_{n=0} (x+1)^n##

Then integrate this function with respect to x to find the representation for lnx

## ∑^∞_{n=0} \frac {(x+1)^{n+1}}{n+1} = lnx ##

For R = 1 on x:(-2,0)

Is my radius and interval of convergence correct? Just wondering, what exactly does the interval and radius of convergence represent for lnx in this case if this is the domain for the series? Since lnx is not defined on (-2,0) I am not exactly understanding what this means for f(x).

I was just wondering if there was any error in what I've done in the following steps to find the series representation of ##lnx##. I know ## \frac {1}{x}## is given in the following link by doing having the a function centred at 0, you can let ##f(x) = ∑^∞_{n=0} \frac {f^{(n)}(0)}{n!}x^n## after proving the function ## \frac {1}{x} ## can be represented as a series by doing the remainder test and finding R(x) → 0 as n → 0. Then just integrating.

I was just looking at an alternate method, which I don't find used as commonly but seems a lot simpler. Here are the steps:

Relevant Forumla(s):

##\frac {1}{1-r} = ∑^∞_{n=0} r^n## for a = 1

Attempted Solution:

If we let r = x + 1, then:

##\frac {1}{1- (x+1)} = \frac {1}{x} = ∑^∞_{n=0} (x+1)^n##

Then integrate this function with respect to x to find the representation for lnx

## ∑^∞_{n=0} \frac {(x+1)^{n+1}}{n+1} = lnx ##

For R = 1 on x:(-2,0)

Is my radius and interval of convergence correct? Just wondering, what exactly does the interval and radius of convergence represent for lnx in this case if this is the domain for the series? Since lnx is not defined on (-2,0) I am not exactly understanding what this means for f(x).

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