Equivalent expressions for infinite series

[MathewsMD]

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).

 

[BvU]

Could you use the template and proofread what you post ? What relevant equations do you have available ? Do you have a series representation of ln(1+x) at hand ?

 

[micromass]

First of all, integrating ##\frac{1}{x}## doesn't give ##\mathrm{ln}(x)##, but ##\mathrm{ln} |x|##.

 

[MathewsMD]

First of all, integrating ##\frac{1}{x}## doesn't give ##\mathrm{ln}(x)##, but ##\mathrm{ln} |x|##.

Okay. How exactly would the domain of the series be restricted as to only show for (0,∞) for lnx? Is it possible? Just to confirm, would my solution be correct if I was looking at lnlxl = f(x)? Wouldn't the answers for letting x = -1 give me ln1 for example? So if i just let y = -x and sub y in for x, then i would get the interval for (0,2) right? Please correct me where I am going and explain please since I'd like to get all these misconceptions out of my head before they affect my future understanding of this content.

If I'm not mistaken, the interval of convergence for the power series representation is the same interval of f(x) that can be assessed using the power series representation, right?

 

[Dick]

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:

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

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).

You can make that work, but you are being way too sloppy. ##\frac {1}{1- (x+1)}## isn't equal to ##\frac {1}{x}##, it's ##\frac {1}{-x}##. And the integral of that is -log(|x|)+C. It's probably worth checking that the integration constant is zero. Since x is in (-2,0) to make the geometric series converge, you can write that as -log(-x). That's the function you are deriving a series for.

 

[micromass]

Okay. How exactly would the domain of the series be restricted as to only show for (0,∞) for lnx?

You can't find a power series representation for ##\mathrm{ln}(x)## that works on entire ##(0,+\infty)##. However, for any ##L>0##, you can find one that works on ##(0,L)##.

Just to confirm, would my solution be correct if I was looking at lnlxl = f(x)?

No, since you made some arithmetic errors. See Dick's reply. You should obtain

[tex]\sum_{n=0}^{+\infty} \frac{(-1)^{n+1}}{n+1} = -\mathrm{ln}|x|[/tex]

for ##x\in (-2,0)##.

 

[Dick]

You can't find a power series representation for ##\mathrm{ln}(x)## that works on entire ##(0,+\infty)##. However, for any ##L>0##, you can find one that works on ##(0,L)##.



No, since you made some arithmetic errors. See Dick's reply. You should obtain

[tex]\sum_{n=0}^{+\infty} \frac{(-1)^{n+1}}{n+1} = -\mathrm{ln}|x|[/tex]

for ##x\in (-2,0)##.

No, that's not what you should get. There's no x on the left side.

 

[micromass]

No, that's not what you should get. There's no x on the left side.

Obviously. I'm an idiot.

 

[Dick]

Obviously. I'm an idiot.

Yes, but an idiot of extraordinary talent! MatthewsMD should try and do better.

 

[MathewsMD]

You can make that work, but you are being way too sloppy. ##\frac {1}{1- (x+1)}## isn't equal to ##\frac {1}{x}##, it's ##\frac {1}{-x}##. And the integral of that is -log(|x|)+C. It's probably worth checking that the integration constant is zero. Since x is in (-2,0) to make the geometric series converge, you can write that as -log(-x). That's the function you are deriving a series for.

Ahhh okay. So I would have to add a constant of integration and my expression is for ln(-x) on (-2,0).

 

[Dick]

Ahhh okay. So I would have to add a constant of integration and my expression is for ln(-x) on (-2,0).

The constant of integration turns out to be 0. And your expression isn't for ln(-x). Can you try to be careful for a change and show what you mean?