# Thread: [SOLVED] Power/Taylor series centered at x = a

1. ## [SOLVED] Power/Taylor series centered at x = a

Hi everyone. I am a senior in high school taking AP Calc BC and I have a question on Power Series. I just finished taking notes on Section 9.1 (Power Series) and 9.2 (Taylor Series) in this textbook: Amazon.com: Calculus: Graphical Numerical Algebraic (9780130631312): Ross L. Finney, Franklin Demana, Bert K. Waits, Daniel Kennedy: Books - Anyhow, my question is, why move a function's center? For example:

Find the third order Taylor polynomial for $\displaystyle f(x) = 2x^3 - 3x^2 + 4x - 5$
(a) at $\displaystyle x = 0$ (b) at $\displaystyle x = 1$

(a) is already centered at $\displaystyle x = 0$.

(b) I did Taylor series and got this: $\displaystyle P_{3}(x) = 2(x-1)^3 + 3(x-1)^2 + 4(x-1) - 2$

I graphed out the two equations and they are exactly identical. So what is the point of moving the center? Not Answered

Another question: is there a thread on using the [tex] tag? Answered

Any help is appreciated. Thanks in advance.

2. Originally Posted by lilaziz1
Another question: is there a thread on using the [tex] tag?
I dont know anything about that center thing but you can find LaTex Tutorial from here at two sticky topics. --> http://www.mathhelpforum.com/math-help/latex-help/

3. Thanks! now just 1 more question to go.

4. do u guys think my answer might be answered if i keep reading?

5. Understand that Taylor polynomials are used as approximations for non-polynomial functions.

A Taylor polynomial for a polynomial function is just that particular polynomial, no matter where it is centered, i.e nothing is gained by its representation.

Try determining a Taylor polynomial for a non-polynomial function ... you'll se that the centering point will be the only point where the Taylor polynomial = the function of interest.

6. oh! like linearization! ty