need help!

• May 2nd 2006, 11:00 AM
bobby77
need help!
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
• May 2nd 2006, 11:11 AM
TD!
The Taylor series of a polynomial is the polynomial itself!!
You can still write the summation and possibly verify this of course.
• May 2nd 2006, 01:13 PM
topsquark
Quote:

Originally Posted by TD!
The Taylor series of a polynomial is the polynomial itself!!
You can still write the summation and possible verify this of course.

Actually, no it isn't quite the same.

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
TD!
Quote:

Originally Posted by topsquark
Actually, no it isn't quite the same.
[...]
In this case your function is going to be:
$\displaystyle (x-1)^3+3(x-1)^2+2(x-1)+1$

But then we have:

$\displaystyle \begin{gathered} \left( {x - 1} \right)^3 + 3\left( {x - 1} \right)^2 + 2\left( {x - 1} \right) + 1 \hfill \\ \\ = \left( {x^3 - 3x^2 + 3x - 3} \right) + 3\left( {x^2 - 2x + 1} \right) + 2x - 2 + 1 \hfill \\ \\ = x^3 - x + 1 = g\left( x \right) \hfill \\ \end{gathered}$

So we find the same after all. This will always be the case for polynomials.
• May 2nd 2006, 01:25 PM
topsquark
Quote:

Originally Posted by TD!
But then we have:

$\displaystyle \begin{gathered} \left( {x - 1} \right)^3 + 3\left( {x - 1} \right)^2 + 2\left( {x - 1} \right) + 1 \hfill \\ \\ = \left( {x^3 - 3x^2 + 3x - 3} \right) + 3\left( {x^2 - 2x + 1} \right) + 2x - 2 + 1 \hfill \\ \\ = x^3 - x + 1 = g\left( x \right) \hfill \\ \end{gathered}$

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

True, but for the result to be a Taylor polynomial at c = 1, then it ought to be of the appropriate form. That's how I learned it, anyway. I admit the whole question is rather like splitting hairs. :)

-Dan
• May 2nd 2006, 01:27 PM
TD!
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.