Two versions of Taylor Series approxmiations, $f(x) \approx f(x_0) + f'(x_0)(x-x_0)$ and $f(x) \approx f(x_0) + f'(x'_0)(x-x_0)$

Question

My understanding of Taylor's Series for a function $f$ with a continuous $(n+1)$st derivative in a neighborhood of a point $x_0$ is that we can approximately have:

$$ f(x) \approx f(x_0) + f'(x_0)(x-x_0) $$

However, I have also seen formulations whereby we have:

$$ f(x) \approx f(x_0) + f'(x_0')(x-x_0) $$

whereby $x_0'$ is a value between $x$ and $x_0$. I am wondering why there are two different formulations and how to go between one and the other?

Mean value theorem probably, which will give $f'(x_0')$ to be a better approximation. — Simply Beautiful Art, Mar 08 '17 at 02:07
The second is not an approximation, it is an equality, given by Taylor's Theorem. Of course, $x'$ depends on $x$. — GuPe, Mar 08 '17 at 02:08
@GuachoPerez Oh right! Wow, I am off my game XD Well, I mean, I had the intuition, and indeed if you followed my comment... — Simply Beautiful Art, Mar 08 '17 at 02:11
@MichaelMcGovern "whereby $x_0'$ is a value between $x$ and $x_0$." Last few lines — Simply Beautiful Art, Mar 08 '17 at 02:22
@SimplyBeautifulArt You were not off your game, if I remember correctly, the proof of Taylor's theorem uses the mean value theorem. So you were not far off! — GuPe, Mar 08 '17 at 02:24
@GuachoPerez Yes, that's where my instinct came from, but my brain was not into it. It appears my instinct is what I run on :D — Simply Beautiful Art, Mar 08 '17 at 02:25

score 3 · Accepted Answer · edited Jun 12 '20 at 10:38

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As per Stella Biderman's comment, I will post my comment as an answer. The second expression is an equality, not an approximation, given by Taylor's theorem. The theorem states that if $f^{(n-1)}$ is continuous on an interval $[a,b]$ and $f^{(n)}(t)$ exists for all $t \in (a,b)$, with $P(t)=\sum_{k=0}^{n-1} \frac {f^{(k)}(\alpha)}{k!}(t-\alpha)^k$, then there exists a point $x\in (\alpha,\beta)$ such that $$f(\beta)=P(\beta)+\frac {f^{(n)}(x)}{n!}(\beta-\alpha)^n$$ for $\alpha$ and $\beta$ distinct points in $[a,b]$. This $x$ is the $x_0'$ in your second expression, where the theorem is applied with $n=1$.

edited Jun 12 '20 at 10:38

Community

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answered Mar 08 '17 at 02:23

GuPe

7,318

Is this related to Taylor's Series approximation or a different thing altogether? – user321627 Mar 08 '17 at 02:25
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@user321627 It is related to the remainder formula for the Taylor polynomial. – Simply Beautiful Art Mar 08 '17 at 02:26
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@user321627 An interesting proof may be found here: http://math.stackexchange.com/a/20415/272831 – Simply Beautiful Art Mar 08 '17 at 02:27
Do you know where on the wikipedia page it states your version? I didn't see a mention of the existence part, thanks! – user321627 Mar 08 '17 at 02:29
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@user321627 https://en.wikipedia.org/wiki/Taylor's_theorem#Explicit_formulas_for_the_remainder where $\xi=x_0'$. – Simply Beautiful Art Mar 08 '17 at 02:35
Does this imply that we can choose $\xi$ to be whatever we want or only the mere existence? – user321627 Mar 08 '17 at 02:39
@user321627 Merely existence, but it can later be used to show that a Taylor polynomial converges to $f(x)$ if you prove that for all $\xi$ the remainder approaches zero. (We also overapproximate with our $\xi$ to get approximations of a function through the Taylor polynomial + Remainder) – Simply Beautiful Art Mar 08 '17 at 02:43

Two versions of Taylor Series approxmiations, $f(x) \approx f(x_0) + f'(x_0)(x-x_0)$ and $f(x) \approx f(x_0) + f'(x'_0)(x-x_0)$

1 Answers1