write f(x) = x ln (1 - x^2) as a power series
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Count Iblis
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Re: power series
WHere you begin depends on what you know. If you know the series expansion of Ln(1+x), then you replace x by -x^2 in here and multiply the series by x. if you don't know the series of Ln(1+x), you should derive that first. A simple way is to write down the series expansion of 1/(1-x), which is given by the well known geometric series and integrate that term by term to obtain the series expansion of Ln(1-x)
mindy88 said:the question is
write f(x) = x ln (1 - x^2) as a power series
where do i begin?
i know you use this series to find the radius and interval but i don't know where to begin with this one.
WHere you begin depends on what you know. If you know the series expansion of Ln(1+x), then you replace x by -x^2 in here and multiply the series by x. if you don't know the series of Ln(1+x), you should derive that first. A simple way is to write down the series expansion of 1/(1-x), which is given by the well known geometric series and integrate that term by term to obtain the series expansion of Ln(1-x)
Count Iblis
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mindy88 said:
That's right! There are many techniques you can use to calculate series expansions. The most straightforward way would be to use the formula for Taylor expansions, but then you need to repeatedly differentiate the function, which can be extremely cumbersome. This is why students are required to learn the series expansions of the standard functions like exp(x), sin(x), cos(x), log(1+x),
arctan(x) etc. If you then encounter some combination of these functions you can usually compute the series expansion pretty fast by using these standard series expansions.
Because you are still learning this stuff, it is instructive to derive series expansions using different techniques including less efficient techniques. So, why not also do the Taylor expansion by taking derivatives at zero? After that you can do the substitution trick I explained earlier.
Computer algebra systems such as Mathematica or Maple use sophisticated algorithms when doing series expansions. If you evaluate derivatives, then they will actually do a series expansion and extract the derivatives (using Taylor's theorem) from that. E.g. to compute the millionth derivative of some complicated function at some point y it is easier to compute the coefficient of
(x-y)^million of the series expansion. That can be done in a few times
Log(Million)/Log(2) (a few dozen) steps.