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This is from my engineering mathematics textbook. Is this version of taylor's theorem correct ?

Successive Differentiation, Maclaurin's and Taylor's Expansion of Function $-147$

TAYLOR'S THEOREM

Let $f(x)$ be a function of $x$ and $h$ be small. If the function $f(x+h)$ is capable of being expanded in a convergent series of terms of positive integral powers of $h$, then this expansion is given by $$f(x+h)=f(x)+hf^\prime(x)+\dfrac{h^2}{2!}f^{\prime\prime}(x)+\dfrac{h^3}{3!}f^{\prime\prime\prime}(x)+\ldots+\dfrac{x^n}{n!}f^{(n)}(x)+\ldots$$

PROOF: Assume that $$f(x+h)=A_0+A_1h+A_2h^2+A_3h^3+\ldots+A_nh^n+\ldots\tag1$$ where $A$'s are functions of $x$.
Differentiating successively w.r.t $h$, we get $$\begin{align}f^\prime(x+h)&=A_1+2A_2h+3A_3h^2+4A_4h^3+\ldots+ nA_nh^{n-1}+\ldots \\ f^{\prime\prime}(x+h)&=2\cdot A_2+3\cdot 2\cdot A_3 h+4\cdot 3\cdot A_4 h^2+\ldots+n(n-1)A_n h^{n-2}+\ldots\\f^{\prime\prime\prime}(x+h)&=3\cdot2\cdot A_3+4\cdot3\cdot2\cdot A_4 h+\ldots+n(n-1)(n-2)A_nh^{n-3}+\ldots\\\end{align}$$ and, in general, $$f^{(n)}(x+h)=n(n-1)(n-2)\ldots3\cdot2 A_n+\text{ terms in ascending powers of $h$}$$ Putting $h=0$, we get $f^{(n)}(x)=n!\,A_n$ so that $A_n=\dfrac{f^{(n)}(x)}{n!}$
Substituting these value of $A_0,A_1,A_2,\ldots$ in $(1)$, we get $$f(x+h)=f(x)+\dfrac{h}{1!}f^\prime(x)+\dfrac{h^2}{2!}f^{\prime\prime}(x)+\dfrac{h^3}{3!}f^{\prime\prime\prime}(x)+\ldots+\dfrac{h^n}{n!}f^{(n)}(x)+\ldots\tag2$$ Its another form can be obtained by replacing $x$ by $a$ and $h$ by $x-a$, so as to get $$f(x)=f(a)+(x-a)f^\prime(a)+\dfrac{(x-a)^2}{2!}f^{\prime\prime}(a)+\ldots+\dfrac{(x-a)^n}{n!}f^{(n)}(a)+\ldots\tag3$$ The conditions under which the above expansion is valid, are

$ \ $ (i) the function $f(x)$ and its derivatives must be finite and continuous in the range of definition of $f(x)$.
$ \ $ (ii) the series on the right hand side of $(2)$ must be convergent for which the remainder term $R_n\to0$ as $n\to\infty$
where $R^n=\dfrac{h^n}{n!}f^{(n)}(x+\theta h)$ and $0\lt\theta\lt1$.

In the form $(3)$ of Taylor's expansion, if we take $a=0$ then we have $$f(x)=f(0)+xf^\prime(0)+\dfrac{x^2}{2!}f^{\prime\prime}(0)+\ldots$$ which is nothing but the Maclaurin's expansion of $f(x)$. Thus Macluaurin's expansion is a particular case of Taylor's expansion.
With a slightly different approach we can show here that Taylor's series can be derived from Maclaurin's series.

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Based on a comment, it seems like your trouble is with equation (3)?

I'm guessing your confusion is stemming from the authors use of variable names.

Here I'll try to use variable names in a way that makes equation (3) clear.

We know that for any $a$ and $h$ small

$$f(a + h) = f(a) + h f'(a) + \frac{h^2}{2} f''(a) + \cdots$$

Now, lets define a new variable $a + h = x$ then we get

$$f(x) = f(a) + hf'(a) + \frac{h^2}{2} f''(a) + \cdots$$

but we know that $h = x - a$ so we can eliminate $h$ from this equation to obtain

$$f(x) = f(a) + (x-a)f'(a) + \frac{(x-a)^2}{2} f''(a) + \cdots$$

which is exactly equation (3).

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This is fine. They are using alternative notation though. Usually x is your variable and the other variable shows is where its centered. See the "alternative" notation a little down for a more usual formulation. (Eqn 3 in your book)

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    $\begingroup$ I've never seen $h$ used as the centre of the Taylor expansion. Where is that the usual notation? $\endgroup$ – Daniel Fischer Oct 11 '13 at 18:35
  • $\begingroup$ The "non-x" constant is the center when presented in most books, and engineers learn the usual "centered" form. My text show it as "a" like in (eqn 3) but I think the OP is confused by the apparent similarity. Here, the h is actilng like a $\Delta x$ $\endgroup$ – user76844 Oct 11 '13 at 18:37
  • $\begingroup$ Apparently, we don't read the same books ;) In all the books I know, when you have $x$ and $h$, then $h$ is the small quantity wobbling around $0$ in $x+h$. $\endgroup$ – Daniel Fischer Oct 11 '13 at 18:41
  • $\begingroup$ Yes, but in eqn 3 , he is replacing x by a and then h by x-a and that means he can't treat '-' as subtraction, also look at the left hand side of the equation. $\endgroup$ – Isomorphic Oct 11 '13 at 18:44

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