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I would follow the Lagrange version (Remainder term)

Let $f(x)$ be $n$th differientiable function and having a $(n+1)$th derivative

$f(x)=f(a)+f'(a)(x-a)+\frac{f''(a)}{2!}(x-a)^2+\cdots+\frac{f^{(n)}(a)}{n!}(x-a)^n+R_n(x)$ (Here the $R_n(x)$ is a remainder term)

$R_n(x) = {{f^{(n+1)}(t)} \over {(n+1)!}}(x-a)^{n+1}$ for some $t \in (a,x) $ or $(x,a)$

So my question is "If the $x \to \pm\infty$, Does $t \to \pm \infty$"?

Thanks.

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  • $\begingroup$ the $t$ is not unique, so the question does not really makes sense. $\endgroup$
    – GreginGre
    Mar 2, 2020 at 7:34
  • $\begingroup$ @GreginGre, In my thought Since $t$ are not unique, Could we make the sequence of the $t$? If so the $x$ goes to infinity for example, then there are sequence of the $t$ goes to infinity. $\endgroup$ Mar 2, 2020 at 8:37

2 Answers 2

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Not necessarily. For example, the third-order Taylor polynomial with Lagrange remainder of the inverse tangent for positive values is $$ \arctan x = x + \frac{3t^2 - 1}{3(1 + t^2 )^3}x^3 , $$ with a suitable $0<t<x$. Hence, $$ 0>\frac{\arctan x - x}{x^3 } = \frac{3t^2 - 1}{3(1 + t^2 )^3}. $$ Since $$ \mathop {\lim }\limits_{x \to + \infty } \frac{\arctan x - x}{x^3 } = 0, $$ and the rational function of $t$ is positive for $\frac{1}{\sqrt 3}<t$, we must have $$ t \to \frac{1}{\sqrt 3} $$ from the left as $x\to +\infty$ (and this $t$ is unique).

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For instance, if the following conditions are met:

  1. $f$ is $C^\infty$ with compact support $K\subseteq[-M,M]$;
  2. $n>0$;
  3. $f^{(k)}(a)\ne 0$ for some $0<k\le n$;

then for large values of $x$ the function $f(x)-P_{a,n}(x)$ diverges. A fortiori, $R_{a,n}(x)$ must be non-zero eventually as $\lvert x\rvert\to\infty$ and therefore, for such $x$-s, any point $t_{n,a,x}$ such that $\frac{f^{(n+1)}(t_{n,a,x})}{(n+1)!}(x-a)^{n+1}=R_{a,n}(x)$ must be in $K$.

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