Taylor Series for ln(1-x): Get Help Now

In summary, the conversation discusses ways to represent the function ln(1-x) as -[infinity (sum) k=1] x^k / k and suggests using either Taylor series or geometric series to achieve this representation. The use of geometric series is further explained and the need for integration is mentioned, but it is acknowledged that this may not be suitable for a pre-calculus setting.
  • #1
teng125
416
0
may i know from ln(1-x) how to become - [infinity (sum) k=1] x^k / k ?
pls help
 
Physics news on Phys.org
  • #2
You posted this in the "precalculus section" so I assume you wouldn't be able to use "Taylor Series". That's the only way I know to show it.
 
  • #3
HallsofIvy said:
You posted this in the "precalculus section" so I assume you wouldn't be able to use "Taylor Series". That's the only way I know to show it.
Uhmm, I think it can also be done with geometric series:
[tex]\sum_{i = 0} ^ \infty (a r ^ i) = \frac{a}{1 - r}, \quad |r| < 1[/tex].
So if your final sum is:
[tex]\frac{1}{1 - k}[/tex], what should your geometric series look like? Then integrating that should give you the result you want (remember to choose the appropriate C, i.e the constant of integration).
 
  • #4
VietDao29 said:
Uhmm, I think it can also be done with geometric series:
[tex]\sum_{i = 0} ^ \infty (a r ^ i) = \frac{a}{1 - r}, \quad |r| < 1[/tex].
So if your final sum is:
[tex]\frac{1}{1 - k}[/tex], what should your geometric series look like? Then integrating that should give you the result you want (remember to choose the appropriate C, i.e the constant of integration).
I thought about that but then integrating is not "pre-calculus" either!
 

1. What is the formula for the Taylor Series of ln(1-x)?

The Taylor Series for ln(1-x) is given by:
ln(1-x) = -x - x^2/2 - x^3/3 - x^4/4 - ... = ∑(-1)^n * x^n/n

2. How do I find the radius of convergence for the Taylor Series of ln(1-x)?

The radius of convergence for the Taylor Series of ln(1-x) is equal to the distance from the center of the series (x = 0) to the nearest point where the function is undefined. In this case, ln(1-x) is undefined when x = 1, so the radius of convergence is 1.

3. What is the importance of the Taylor Series of ln(1-x)?

The Taylor Series of ln(1-x) is important because it allows us to approximate the value of ln(1-x) for any value of x. This can be useful in many mathematical and scientific applications, including solving differential equations and evaluating infinite series.

4. Is the Taylor Series of ln(1-x) a convergent or divergent series?

The Taylor Series of ln(1-x) is a convergent series, meaning that it approaches a finite value as the number of terms increases. However, it is only convergent within its radius of convergence, which in this case is x = 1.

5. How accurate is the Taylor Series approximation for ln(1-x)?

The accuracy of the Taylor Series approximation for ln(1-x) depends on the value of x and the number of terms used in the series. Generally, the more terms that are included in the series, the more accurate the approximation will be. However, for values of x that are close to or outside the radius of convergence, the approximation may not be very accurate.

Similar threads

Replies
2
Views
1K
  • Calculus
Replies
10
Views
1K
Replies
2
Views
1K
Replies
3
Views
575
  • Precalculus Mathematics Homework Help
Replies
5
Views
1K
  • Precalculus Mathematics Homework Help
Replies
6
Views
583
  • Math Proof Training and Practice
Replies
6
Views
1K
  • Precalculus Mathematics Homework Help
Replies
11
Views
2K
  • Precalculus Mathematics Homework Help
Replies
4
Views
603
  • Precalculus Mathematics Homework Help
Replies
11
Views
788
Back
Top