Find the Taylor series (write a summation if possible) about the indicated center and determine the interval of convergence.

g(x)= x^3-x+1 c=1

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- May 2nd 2006, 11:00 AM #1

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- May 2nd 2006, 11:11 AM #2

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- May 2nd 2006, 01:13 PM #3Originally Posted by
**TD!**

The Taylor series for a function f(x) about a point x = c is:

$\displaystyle f(x) = \sum_{n=0}^{\infty} \frac{1}{n!} \left ( \frac{d^nf}{dx^n} \right ) (c) \, (x - c)^n$

The notation $\displaystyle \left ( \frac{d^nf}{dx^n} \right ) (c)$ is meant to represent the nth derivative of f evaluated at x = c. This form is only correct, of course, assuming the series converges for a given f(x) and c.

In this case your function is going to be:

$\displaystyle (x-1)^3+3(x-1)^2+2(x-1)+1$

-Dan

- May 2nd 2006, 01:21 PM #4

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Originally Posted by**topsquark**

$\displaystyle

\begin{gathered}

\left( {x - 1} \right)^3 + 3\left( {x - 1} \right)^2 + 2\left( {x - 1} \right) + 1 \hfill \\ \\

= \left( {x^3 - 3x^2 + 3x - 3} \right) + 3\left( {x^2 - 2x + 1} \right) + 2x - 2 + 1 \hfill \\ \\

= x^3 - x + 1 = g\left( x \right) \hfill \\

\end{gathered}

$

So we find the same after all. This will always be the case for polynomials.

- May 2nd 2006, 01:25 PM #5

- May 2nd 2006, 01:27 PM #6

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I see what you mean, but the 'form' doesn't matter since the functions (the original result and the simplification to g(x) again) are completely the same.

I don't really see the use of this question, unless its intention is that the student sees that the taylor series of a polynomial is the polynomial again; by experimentally finding this through the formula like you did.