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teng125
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may i know from ln(1-x) how to become - [infinity (sum) k=1] x^k / k ?
pls help
pls help
Uhmm, I think it can also be done with geometric series: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.
I thought about that but then integrating is not "pre-calculus" either!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).
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
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.
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.
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.
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.