Power Series constant values

Question

I know that we could represent the function $\frac{8x}{7+x}$ as a power series $8\sum\limits_{n=0}^{\infty}(-1)^n(\frac{x}{7})^{n+1}$

Therefore the first few terms would be: $\frac{8x}{7}-\frac{8x^2}{49}+\frac{8x^3}{343}-\dots$

according to this summation, wouldn't the value of $C_0 = \frac{8}{7}$?

apparently it is not and I need help understanding why $C_1 = \frac{8}{7}$ and not $C_0$.

Thank you!

Because of the $8x$, the coefficient of $x^0$, that is, the constant term, is $0$. The coefficient of $x^1$ is $8/7$. — André Nicolas, Aug 14 '14 at 20:01

score 1 · Answer 1 · answered Aug 14 '14 at 20:05

$C_0$ is 0 because $f(0) = \frac{8(0)}{7+(0)}$ = 0. $C_k$ is, in general, the value of the kth derivative of f, at 0, divided by k!: $C_k = \frac{f^{(k)}(0)}{k!}$. The 0th derivative is simply the function itself, and this function has 0 for its value at 0.

score 1 · Answer 2 · answered Aug 14 '14 at 20:09

1

See http://en.wikipedia.org/wiki/Taylor_series.

Let $f(x)=\frac{8x}{7+x}. $ Since $f(0)=0$ the first term of Maclaurin series of the function $f(x)$ equal to zero.

answered Aug 14 '14 at 20:09

nadia-liza

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Power Series constant values

2 Answers2