How Do You Calculate the Taylor Series for ln(1-x)?

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To calculate the Taylor series for ln(1-x), the first step involves finding the derivatives of the function, with the n-th derivative expressed as f^(n)(x) = (-1)^n((n-1)!)/(1-x)^n. The Maclaurin series is derived from these derivatives, resulting in a series that alternates signs and includes terms like -x + (x^2/2) - (2x^3/3) + ... The radius of convergence can be determined using the ratio test, which requires identifying the general term from the series. The discussion also clarifies that the radius of convergence cannot be negative, as it is derived from absolute values.
  • #31
d_leet said:
That isn't the correct general term in the first place, and to find the radius of convergence the best way for this problem is either to notice the relation between the series you found and the geometric series or just use the ratio test.

you mean between the series i found in part b and the standard taylor series equation?
 
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  • #32
jkh4 said:
you mean between the series i found in part b and the standard taylor series equation?

What is the standard taylor series equation? The best way to find the radius of convergence is probably to just use the ratio test.
 
  • #33
d_leet said:
What is the standard taylor series equation? The best way to find the radius of convergence is probably to just use the ratio test.

but what i don't understand is, when you do the ratio test

lim |An+1/An|

don't you need an equation involving n? for example like n^x/(n+1)
 
  • #34
jkh4 said:
but what i don't understand is, when you do the ratio test

lim |An+1/An|

don't you need an equation involving n? for example like n^x/(n+1)

Yes you need an equation for the general term of the series an which is one of the things should be found in the first part of this question.
 
  • #35
d_leet said:
Yes you need an equation for the general term of the series an which is one of the things should be found in the first part of this question.

okay, i think i start to get it...what you saying is using the equation derive in part a) right?
 
  • #36
jkh4 said:
okay, i think i start to get it...what you saying is using the equation derive in part a) right?

More or less, yes.
 
  • #37
d_leet said:
More or less, yes.


o...

okay so in general, if we want to know the equation for An to do the ratio test, the An equation we need is derive from the f^(n)(x) pattern right?
 
  • #38
jkh4 said:
o...

okay so in general, if we want to know the equation for An to do the ratio test, the An equation we need is derive from the f^(n)(x) pattern right?

Yes.

/*Extra Chars*/
 
  • #39
d_leet said:
Yes.

/*Extra Chars*/

thank you so much

by the way, one side question, can R be negative?
 
  • #40
jkh4 said:
thank you so much

by the way, one side question, can R be negative?

Your welcome, I'm glad to have been of some help. And no R cannot be negative since it is the limit of an absolute value which is never negative.
 
  • #41
d_leet said:
Your welcome, I'm glad to have been of some help. And no R cannot be negative since it is the limit of an absolute value which is never negative.

okay so in that case -|x| < 1 is wrong right?
 
  • #42
jkh4 said:
okay so in that case -|x| < 1 is wrong right?
Wrong for what? Since |x| is non-negative, -|x| is never positive and so -|x|< 1 is true for all x. If you are talking about using the ratio test to find the radius of convergence, you should not have any negative numbers at all. Go back and check your absolute values again. You want |anxn|, not an|xn|!
|(-1)nx^n| is |x|n, not (-1)n|x|n.
 

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