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Expert: Sherry Wallin - 4/23/2011


Question
Do you know of any function that has a series (like a Taylor series) that begins:

1/r^2 - 1/2*r^3

The minus sign for the second term is important; the 2 in the divisor of the second term is not so important.

Answer
Phillip~

Just based on what you gave me and not knowing for sure if the signs will alternate but assuming they do try this:

Using 'S' for sigma = sum and (2,oo) being the range of the sum this is what I have:

S(2,oo)(-1/2)^(n-2)*r^-n

and the first few terms would be:

(-1/2)^0 *r^-2 = 1/r^2
(-1/2)^1*r^-3 = (-1/2*r^3)
(-1/2)^2*r^-4 = (1/4*r^4)
(-1/2)^3*r^-5 = (-1/8*r^5)
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Sherry Wallin

Expertise

I can answer most questions up through Calculus and some in Number Theory and Abstract Algebra.

Experience

I have had my Bachelor's Degree since 1987 and have been a teacher since 1988. I earned my Masters Degree in Mathematics May 2010. I have been teaching at the same community college since 2002.

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I have taught 12 years at the community college level, medical college, and technical college as well as a high school instructor and alternative education instructor and charter school instructor.

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Master's GPA 3.56 Bachelor's GPA 3.34 Post grad work not degree related GPA 4.0

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