The discussion centers on the advantages of using Taylor series centered at a point \( a \) instead of at zero when solving differential equations, particularly for functions like \( \ln(x) \) that cannot be expressed as a Maclaurin series. Key benefits include improved accuracy with fewer terms when the expansion point is close to the desired value, the ability to manage singularities, and the convenience of matching initial conditions in linear differential equations. The power series for \( \ln\left(\frac{1+x}{1-x}\right) \) converges within the interval \(-1 < x < 1\), demonstrating the utility of shifting the center of expansion.
PREREQUISITESMathematicians, students of calculus, and anyone involved in solving differential equations or analyzing series expansions will benefit from this discussion.
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.Austin said:I have just started learning about series and I don't see the benefit of shifting the series by using some "a" other than 0?
My textbook doesn't really tell the benefits it just says "it is very useful"'
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 - e2, with e2 being about 7.39, which might be close enough to 9.Austin said:Would you be able to write a power series to estimate any value of lnx? I cannot think of a way to write a series to estimate ln9 for example... Is it possible? Every example I see is estimating ln2 and I feel like there must be some way to estimate other values but I cannot think of a way since the series diverges after 1 in each direction (from what I've seen)
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?Mark44 said: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 - e2, with e2 being about 7.39, which might be close enough to 9.
Something like that. You move the interval of convergence. The radius doesn't change.Austin said: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?
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.Austin said:Would you be able to write a power series to estimate any value of lnx? I cannot think of a way to write a series to estimate ln9 for example... Is it possible? Every example I see is estimating ln2 and I feel like there must be some way to estimate other values but I cannot think of a way since the series diverges after 1 in each direction (from what I've seen)