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$$f(x) = \frac{6}{x}\,\, \mathrm{at}\,\, a = -4 .$$

Assume that $f$ has a power series expansion. Do not show that $R_n(x) -> 0$

I took the derivatives of f(x): $$f(x) = 6/x$$ $$f'(x) = -6/x^2$$ $$ f''(x) = 12/x^3 $$ $$ f^{(3)}(x) = -36/x^4 $$ but I am unable to write out the series, please help! Thank you

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Does my answer help you, or do you still need assistance? – 000 Dec 5 '12 at 6:24
thanks for your comment, but i do need a little more assistance if you do not mind. The truth is that i do not quite understand how to apply the nth derivative into the series – Jaden Q Dec 5 '12 at 6:46
No problem! I'll update my answer. :) – 000 Dec 5 '12 at 6:50
I accidentally forgot the factor of $6$. I've fixed the addendum. I hope this helps. – 000 Dec 5 '12 at 7:21
up vote 2 down vote accepted

For a function $f$, the definition of the Taylor Series of $f$ is: $$\text{Taylor}(f)=\sum_{n \ge 0}\frac{f^{(n)}(a)}{n!}(x-a)^n, $$ where $f^{(n)}$ indicates the $n$th derivative of $f$ with $f^{(0)}=f$.

The problem, then, reduces to the following: Is there an $n$th derivative of $f(x)=\frac{6}{x}$? (More aptly, since $6$ is a constant: Is there an $n$th derivative of $\frac{1}{x}$? If so, what is it?)

Let's do a table of values: $$ \begin{align} \frac{d}{dx}\frac{1}{x}&=-\frac{1}{x^2}\\ \frac{d^2}{dx^2}\frac{1}{x}&=2\frac{1}{x^3}\\ \frac{d^3}{dx^3}\frac{1}{x}&=-6\frac{1}{x^4}\\ &\ldots \end{align} $$ This pattern seems to suggest $$\frac{d^n}{dx^n}\frac{1}{x}=(-1)^{n}n!\frac{1}{x^{n+1}}.$$

Let's prove this via induction on $n$. The base case trivially true. Assume the induction hypothesis for $n=k$. $$ \begin{align} \frac{d^{k+1}}{dx^{k+1}}\frac{1}{x}&=\frac{d}{dx}\frac{d^k}{dx^k}\frac{1}{x}\\ &=\frac{d}{dx}(-1)^kk!\frac{1}{x^{k+1}}\\ &=(-1)^kk!(k+1)(-1)\frac{1}{x^{k+2}}\\ &=(-1)^{k+1}(k+1)!\frac{1}{x^{k+2}}. \end{align} $$

Thus, via induction, we see this is correct!

Can you take this from here? Apply the $n$th derivative and the definition of $\text{Taylor}(f)$.

Okay, so you have the following pieces of information: $a=-4$, $f^{(n)}(x)=(-1)^nn!\frac{1}{x^{n+1}}$, and the definition of the Taylor Series. You simply substitute these into the definition as follows:

$$ \text{Taylor}(f)=6\sum_{n \ge 0}\frac{(-1)^nn!\frac{1}{a^{n+1}}}{n!}(x-a)^n=6\sum_{n \ge 0}\frac{(-1)^nn!\frac{1}{(-4)^{n+1}}}{n!}(x-(-4))^n . . . $$

Try to go from here and simplify this. :)

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thank you so much! – Jaden Q Dec 5 '12 at 7:56
Thank you for the lovely question! :-) – 000 Dec 5 '12 at 8:01

Refer to the definition of Taylor Series. You will have to evaluate your derivatives at $a=-4$.

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