Taylor expansion: infinity, ln(1+x)

[spaccio]

Hello,

I would appreciate some help with these Taylor expansion related issues:

1. ln(1+x) can be expanded to x-x^2/2+x^3/3... and so on.
Now, if I calculate the limit of this expansion at some point a (>0), I get that the answer depends on which n I choose.
For example: given a (point) = 2, with n = 2, the answer is 0, while with n = 3, the answer is 8/3. So, somehow this expansion should decrease as the n increases... Could anyone explain to me how to solve it?

2. How to calculate the limit of Taylor expansion at infinity? Is it enough to change x to 1/x and calculate limit at zero?

Thank you!

 

[ksdhart2]

Ehm... sorry, but I don't think I understand much of any thing you said. It's true that the Taylor series expansion of ln(1+x) at the point x = 0 is given by:

\(\displaystyle x - \dfrac{x^2}{2} + \dfrac{x^3}{3} - \dfrac{x^4}{4} + ...\)

But nowhere in there is any terms involving a or n, so I don't know to what they refer.

 

[spaccio]

Ehm... sorry, but I don't think I understand much of any thing you said. It's true that the Taylor series expansion of ln(1+x) at the point x = 0 is given by:

\(\displaystyle x - \dfrac{x^2}{2} + \dfrac{x^3}{3} - \dfrac{x^4}{4} + ...\)

But nowhere in there is any terms involving a or n, so I don't know to what they refer.

x->a: lim (ln(1+x)) = ?

1) taylor expansion of ln (1+x) = x - x^2/2 + .... + (-1)^(n-1) * x^n/n
n - how many members of this expansion we involve into this calculation.

2)a - given limit x->a

 

[ksdhart2]

n - how many members of this expansion we involve into this calculation.

2)a - given limit x->a

I think I understand a little better, but I'm still very very confused. So, n represents the number of terms used in the Taylor Series? If that's correct, then yes, the value you get out from the series will change (i.e. become more accurate) as you add terms. However, for any finite number of terms, you'll only ever get an approximation. To get the true value of ln(1+x) at some point, you'd need to consider the limit of the series, as you use infinitely many terms.

taylor expansion of ln (1+x) = x - x^2/2 + .... + (-1)^(n-1) * x^n/n

It must be noted that this is only the Taylor series expansion at the point x = 0. This infinite series will converge to the value ln(1+x) only for |x| < 1. The specific Taylor series will differ depending on which point you're looking at. Herehttp://tutorial.math.lamar.edu/Classes/CalcII/TaylorSeries.aspx is a refresher on Taylor series and how to calculate them, if you need it.

x->a: lim (ln(1+x)) = ?

2)a - given limit x->a

I'm not quite sure what you're indicating here. You have a limit of ln(1+x), and it looks like you're indicating that you want to investigate this limit as x approaches the value a. But, without knowing more about this mysterious variable a, such as any known value it may or may not have or what it represents in the context of the problem, we can say nothing more than \(\displaystyle \displaystyle \lim_{x \to a} \ln(1 + x) = \ln(1 + a)\). Further, I'm completely at a loss for how this limit in any way relates to the topic of Taylor series.

Can you please post the full and exact problem text, ideally quoting word-for-word? I highly suspect there's a communication error as it goes from the author of the problem through you and to me, because I can't seem to make heads or tails of what you're even asking.