Question

How do you find the Taylor series of 1/x when it is centered at 2?

Answer 1

Without verifying \[\lim_{n \rightarrow \infty} R _{n} (x) = 0\]

Answer 2

But, I thought you have to use f^(n) (0)?

Answer 3

That's for the Maclaurin series.

Taylor series are more general; they're used for constructing series representations about \(x=c\). If \(c=0\), you have a Maclaurin series.

Answer 4

Since this one is about \(x=2\), you use \(f^{(n)}(2)\).

Answer 5

Actually, I made a minor mistake with my series. Should be
\[f(x)=\frac{1}{x}\\
f'(x)=-\frac{1}{x^2}\\
f''(x)=\frac{1\cdot2}{x^3}\\
f'''(x)=-\frac{1\cdot2\cdot3}{x^4}\\
~~~~~~\vdots\\
f^{(n)}(x)=(-1)^n\frac{n!}{x^{n+1}}\]
Which gives
\[T(x)=\sum_{n=0}^\infty \dfrac{(-1)^n\dfrac{n!}{2^{n+1}}}{n!}(x-2)^n\]

Answer 6

Equivalently
\[T(x)=\sum_{n=0}^\infty \dfrac{(-1)^n}{2^{n+1}}(x-2)^n\]
Sorry about the confusion!

Answer 7

So if you were to write out the first few terms of this series, using the taylor series format, would you use f'(2) for example? @SithsAndGiggles

Answer 8

Sorry for the delay.

Yes, you would use the function's derivatives evaluated at \(x=2\). This is actually where the \(2^{n+1}\) comes from, if you examine the derivation of the nth derivative formula.

Answer 9

Great!!! Thanks!