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

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In summary: So now you can see that they're all negative, and you can also see the pattern for the coefficients./*extra characters*/Yes, sorry that is a mistake on my part. I was wrong before, your original derivatives were correct and should all be negative. I apologize for the confusion and any mistakes it may have caused. But yes, as the other poster said, if you want to use
  • #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.
 
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  • #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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