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I'm studying Taylor Theorem in the videos they talk about Taylor Theorem around 0 and a. (https://www.youtube.com/watch?v=HIG8iH2dEjM) and this video (https://www.youtube.com/watch?v=_PyuMFBvj8g)

I'm very curious about why the first Nth term of Taylor series can have different centre from the N+1(remainder) term?

as you can see in the equation of Taylor series around 0: $$f(x) = (\sum^N_{n=0}\frac{f^{(n)}(0)}{n!}x^n)+R_n(x)$$ Where $$R_n(x) = \frac{f^{(N+1)}(Z)}{N+1!}x^{N+1}$$ Where Z is between X and 0.

Why the remainder term can centred at Z ???? assume Z is between X=5 and 0, assume I choose Z=3 then this Taylor series expansion is like the mixture of Taylor series centred at 0 and 3 where the first Nth term centred at 0 and the remainder term centred at 3. This makes me very confuse and why it is legitimate?

I'm newbie for Sequence and Series, please give me step by step and easy understanding the logic behind and example.

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  • $\begingroup$ The thing is that if you consider only a partial Taylor series up to the $N$-th term (say centered at $0$) then clearly your true $f(x)$ is not equal to the series. It is close but not equal. So in order to correct the difference, the theorem says that it exists a value $Z$ between $0$ and $x$ such that if you add this correction term (which is $R_n(x)$) then $f(x)$ will truly be equal to the Taylor series centered at $0$ and the correction term. However you cannot freely choose this $Z$, it is defined in itself. (And might be actually hard to compute). $\endgroup$ – Zubzub Aug 5 '16 at 8:55
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The remainder term is exact, as in that's exactly what is left to make the Taylor series exactly equal $f$ at $x$. In general it's difficult to find the relevant $Z$ (it varies depending on $x$), but it makes it possible to put bounds that are usually very good on how well the Taylor series approximates $f$ ($Z$ is always between $x$ and the centre of the expansion).

If you add the $n+1$ term of the Taylor expansion, it will not be centred around $Z$ (again, since $Z$ depends on $x$ that's really difficult to do), but rather around $0$. That means that the degree $n+1$ Taylor series is not exact, so there is again a correcting remainder term, but this time the remainder term looks like a degree $n+2$ Taylor term centred on some $Y$ (which again depends on $x$, but it's known to be between $x$ and $0$).

Note that the remainder term is not really part of the Taylor expansion. The theorem really states

For any $x$ there exists a $Z$ between $0$ and $x$ such that $$f(x) - \sum_{i = 0}^n \frac{f^{(i)}(0)}{i!}x^i = \frac{f^{(n+1)}(Z)}{(n+1)!}x^{n+1}$$

The idea is that we have a function $f(x)$ that may be difficult to calculate exactly, but we can easily find its derivatives at $0$ (or $a$). Say we want to know what $f(x_0)$ is for some $x_0$, but being exact isn't really that important, it just has to be close. Then we can calculate the Taylor series $$ f(x_0) \approx\sum_{i = 0}^n \frac{f^{(i)}(0)}{i!}x_0^i $$ which, if all is good and just in this world, will be good enough for a large enough $n$. The question remains, though: how large $n$ is large enough? In other words, we need to find some way of estimating how far off the Taylor series is. That's where the above theorem statement comes in. It says that there is a $Z$ somewhere between $0$ and $x$ that exactly describes how far off the estimate is, by way of the remainder term. We don't know exactly where that $Z$ is, but if we find $f^{(n+1)}(x)$, and try to figure out the largest value it can have between $0$ and $x_0$, then we know we're closer than that.

Take an example, with $f(x) = \sin(x)$, and say we want to calculate $\sin(1)$ to within $1/1000$. We take the Taylor series for $\sin(x)$ to sixth degree, for instance, and calculate $$ \sin(1) \approx 1 + 0 - \frac 16 + 0 + \frac 1{120} + 0 = \frac{101}{120} = 0.8416666\ldots $$ Is this good enough? Well, the error term is given from the theorem by $$ \frac{f^{(7)}(Z)}{7!} = -\frac{\cos(Z)}{5040} $$ for some $Z$ between $0$ and $1$. We don't know which $Z$, but we do know that whatever that $Z$ is, we have $0\leq \cos(Z)\leq 1$. So that means that whatever the remainder term truly is, it must be somewhere between $-\frac{1}{5040}$ and $0$. This means that not only do we know that the error in our approximation is less than $0.0002$, but we even know that our estimate is definitely too big, and not too small.

The "true" answer can be found by a calculator (actually, calculators do exactly the same as we've done here, only with a few more terms, so they are not exact either, but they are usually exact to within sixteen or so decimal places). It turns out to be $0.8414709$, which means our estimate was too big by $0.0001957\approx \frac{1}{5110}$, which is indeed within the bounds we got from Taylor's theorem. Again, using calculators, we can check that in this caze, $Z \approx 0.1661$.

Note that if we'd only taken the fifth-degree Taylor expansion, even though the only difference is whether we have a "${}+0$" at the end or not, we couldn't have directly shown that is was correct to within $1/1000$, since the error term in that case is $-\frac{\sin(Z)}{720}$, which can be less than $-1/1000$. We needed to have a sixth-degree Taylor expansion in order for the error term to be guaranteed small enough to fit within the error bound I set at the beginnning.

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  • $\begingroup$ If I want to expand around 0. Then I choose the Taylor centred at 0 formula. If I want to expand around a. Then I choose the Taylor centred at a formula. but why the remainder term can use Z??? It has to be 0 or a but not Z??? Z in this case is between 0 and x or a and x but why it is legitimate??? $\endgroup$ – user3270418 Aug 5 '16 at 9:04
  • $\begingroup$ When we deriving the Taylor series formula from f(x). We get coefficient of a_n x^n from centred at 0 or a but not Z. Z seems to come when we talk about remainder. but why we don't mention it when we derive formula at the beginning ????? $\endgroup$ – user3270418 Aug 5 '16 at 9:11
  • $\begingroup$ @user3270418 I added some clarification and an example. Maybe that helps. $\endgroup$ – Arthur Aug 5 '16 at 10:40

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