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My textbook doesn't really tell the benefits it just says "it is very useful"'

- Thread starter Austin
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- #1

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My textbook doesn't really tell the benefits it just says "it is very useful"'

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Mark44

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You can't write a Maclaurin series (i.e., a Taylor series with a = 0) for f(x) = ln(x), since the function and all of its derivatives are not defined at x = 0. You can, however, write a Taylor series in powers of, say, x - 1, though.

My textbook doesn't really tell the benefits it just says "it is very useful"'

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Mark44

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Off the top of my head I don't know what the radius of convergence is for this series. You could write it in powers of x - e

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Oh I see! The point of having a Taylor Series centered around some arbitrary a is to move your radius of convergence in a sense, is that correct?Off the top of my head I don't know what the radius of convergence is for this series. You could write it in powers of x - e^{2}, with e^{2}being about 7.39, which might be close enough to 9.

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Mark44

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Something like that. You move theOh I see! The point of having a Taylor Series centered around some arbitrary a is to move your radius of convergence in a sense, is that correct?

- #7

mathman

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The power series for ln((1+x)/(1-x)) converges for -1<x<1, which can (in principal) be used for any y = (1+x)/(1-x) > 0.

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FactChecker

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1) You can get more accurate answers from fewer terms if you can expand around point 'a' close to the x value you are interested in.

2) Others have mentioned the radius of convergence, which can not go beyond any singularity. So you may need to expand in several areas.

3) You may need to cancel one function's singularity at z=a by expanding another multiplying function that has a zero at z=a.

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HallsofIvy

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