Maclaurin expansion for lnx

[apple2357]

I was always under the impression that you cannot find a Maclaurin series expansion for lnx, presumably the reason being is that you can't define the derivative of lnx at x=0 and so the formula fails.

I was searching for something else when i came across this on a website, can anyone help me see how this expansion for lnx has been attained? Is this considered a Maclaurin expansion?




(It's off a website called math24.net)

 

[HallsofIvy]

(7) is NOT a "MacLaurin series". And it isn't just that "the derivative does not exist a x= 0". ln(x) itself does not exist at x= 0.

 

[apple2357]

So how is that expansion for lnx established? Is it around a different point? like x=1

 

[Dr.Peterson]

So how is that expansion for lnx established? Is it around a different point? like x=1

See if this helps: https://en.wikipedia.org/wiki/Natural_logarithm#Series They derive your series 7 more or less the way I imagined doing it.

 

[apple2357]

Thanks!

 

[HallsofIvy]

So how is that expansion for lnx established? Is it around a different point? like x=1

There are 3 "expansions for ln(x)" given.

 

[apple2357]

There are 3 "expansions for ln(x)" given.

I can see how 5 & 6 can be established. Using the Maclaurin formula to get 5 and use log laws to get 6 from 5 ( for example). I couldn't see how 7 could be got from either the Maclaurin formula nor from 5 & 6. However, Dr P has helped with his link, which i will have a detailed look at later!

 

[apple2357]

Actually i think i can see how 7 could be established from 6 now! you are essentially replacing (1+x)/(1-x) and doing some rearranging?

 

[Dr.Peterson]

Actually i think i can see how 7 could be established from 6 now! you are essentially replacing (1+x)/(1-x) and doing some rearranging?

Exactly. You can also think of it in terms of the inverse of f(x) = (1+x)/(1-x), which turns out to be very interesting ...

 

[Cubist]

To give an idea of the accuracy I plotted "ln(x) - <series 7>" for 3, 5, 7 and 9 terms (red to blue)...

 

[Cubist]

Exactly. You can also think of it in terms of the inverse of f(x) = (1+x)/(1-x), which turns out to be very interesting ...

I didn't initially get this. I forgot about how you find an inverse function . For us forgetful people you need to write...

y = (1 + x)/(1 - x)

And rearrange to x = <something in terms of y>

Doing this really reveals the link between series 6 and 7 in the list.