0
$\begingroup$

I'm studying from Calculus: Early Transcendentals, by Briggs & Cochran, and the authors give the definition of Taylor Polynomials as:

$p_n(x) = \sum \limits_{k=0}^n c_k(x-a)^k$, where the coefficients are $c_k = \frac{f^{(k)}(a)}{k!}$, for $k = 1, 2, \dots , n$.

Yet it seems more intuitive for me to write the formula as:

$p_n(x) = \sum \limits_{k=0}^n \dfrac{f^{(k)}(a)(x-a)^k}{k!}$

Is there any mathematical difference I'm not noticing in writing the equation in this manner (maybe having to do with the summation)? If not, I'll stick with the latter. Thanks for any input.

$\endgroup$
  • 2
    $\begingroup$ There is no difference. $\endgroup$ – brick Nov 20 '14 at 16:51
1
$\begingroup$

Since you have $c_k = \frac{f^{(k)}(a)}{k!}$ Then $$\sum \limits_{k=0}^n c_k(x-a)^k = \sum \limits_{k=0}^n \left(\frac{f^{(k)}(a)}{k!}\right)(x-a)^k = \sum \limits_{k=0}^n \dfrac{f^{(k)}(a)(x-a)^k}{k!}$$ The index of the sum doesn't change, so there is no difference. The point of introducing $c_k$ is to condense the notation. It may be a bit cumbersome to always write out $\frac{f^{(k)}(a)}{k!}$ every time you are working with Taylor Series.

$\endgroup$
  • 1
    $\begingroup$ Great, thank you for the explanation about why they are separating $c_k$ as well. $\endgroup$ – user153085 Nov 20 '14 at 17:29

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy