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I understand that the Taylor series formula is $$\frac{f^n(a)}{n!}(x-a)^n.$$ I also know that the Taylor series expansion of $$y(x_0 +h)=y(x_0) +hy'(x_0)+\frac{h^2}{2!}y''(x_0)+ \frac{h^3}{3!}y'''(x_0)+O(h^4).$$

Is h the distance which is equal to $$(x-x_0).$$ If so what if I had to find the Taylor series expansion of $$y(x_0 -h).$$ How am I supposed to view -h?

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  • $\begingroup$ $h$ is the signed distance between the center of the Taylor series and the location where the series is being evaluated. Therefore, $-h$ makes sense as a distance to the left of $x_0$. $\endgroup$ – Michael Burr Mar 11 '15 at 15:03
  • $\begingroup$ @MichaelBurr Thanks, also does the "O(h4)" mean that the right hand side of $y(x0+h)$ mean that it is a fourth order equation? $\endgroup$ – Jed Mar 11 '15 at 16:03
  • $\begingroup$ Yes, all the remaining terms have $h^4$ (or higher powers of $h$) $\endgroup$ – Michael Burr Mar 11 '15 at 16:04
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In

$$f(x)=\sum _{k=1}^{\infty } \frac{1}{k!}\frac{d^k f }{\text{dx}^k}\left(x_0\right)\left(x-x_0\right){}^k$$

substitude

$$h=x-x_0$$

then

$$x=x_0+h$$

so,

$$f(x_0+h)=\sum _{k=1}^{\infty } \frac{1}{k!}\frac{d^k f }{\text{dx}^k}\left(x_0\right){h}^k$$

Now the variable is $h$ in an allowed range.

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