What is the coefficient for x^27 in the power series expansion of 1/(1+x^9)?

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The function f(x) = 1/(1+x^9) can be expressed as a power series expansion. The series can be represented as Σ (-1)^n (-x^9)^n, leading to the identification of a_27 as the coefficient of x^27. This coefficient corresponds to the term where n=3, resulting in a coefficient of (-1)^3, which equals -1. The discussion clarifies the calculation process for finding the specific coefficient in the series expansion. Thus, the coefficient for x^27 in the power series is -1.
greenteacup
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Homework Statement



The function
f(x)=\frac{1}{1+x^{9}}
can be expanded in a power series
\sum^{\infty}_{0} a_{n}x^{n}
with center c = 0.
Find the coefficient
a_{27}
of
x^{27}
in this power series.

2. The attempt at a solution

I can get to:

\sum^{\infty}_{0} (-1)^{n}(-x^{9})^{n}

which I think is right, but I'm not sure how to find a_{27}. We didn't talk about it in class.
 
Last edited:
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You don't want (-1)^n and (-x^9)^n to both have a '-' in them do you? What are the first few terms in the series when you write them out? a_27 is the coefficient of x^27, which is the n=3 term in your series. What is it?
 
Ohhh, okay, I think I understand now. So the coefficient would just be (-1)^{3}=-1?
 
greenteacup said:
Ohhh, okay, I think I understand now. So the coefficient would just be (-1)^{3}=-1?

Right.
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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