Taylor series for ln (1 Viewer)

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may i know from ln(1-x) how to become - [infinity (sum) k=1] x^k / k ???
pls help
 

HallsofIvy

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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.
 

VietDao29

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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).
 

HallsofIvy

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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!
 

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