0
$\begingroup$

Suppose our function be written as $$f(x)=P_n(x)+R_n(x)$$ where $P_n(x)$ is n-th Taylor's Polynomial about $x_0$ and $R_n(x)$ is associated remainder/error term both of which are given as $$P_n(x)= \sum _ { k = 0 } ^ { n } \frac { f ^ { ( k ) } \left( x _ { 0 } \right) } { k ! } \left( x - x _ { 0 } \right) ^ { k }$$ $$R _ { n } ( x ) = \frac { f ^ { ( n + 1 ) } ( \xi ( x ) ) } { ( n + 1 ) ! } \left( x - x _ { 0 } \right) ^ { n + 1 }\quad \text{for } \xi(x)\in [x,x_0]$$

Suppose $\lim\limits_{n\to\infty}R_n(x)=0$ or in other words $\lim\limits_{n\to\infty}P_n(x)=f(x)$

Define $$g(n)=\sup |R_n(x)| = \sup |f(x)-P_n(x)|$$

Can it be proven that $g(n)=\sup |R_n(x)|$ is decreasing function of $n$

$\endgroup$
  • $\begingroup$ For one $x$ or every $x?$ What are you taking the $\sup$ over? Where is $f$ defined? $\endgroup$ – zhw. Nov 12 '18 at 18:45
  • $\begingroup$ @zhw. For any general function. We are talking for all those x at which taylor's polynomial converges to the function. $\endgroup$ – Lord KK Nov 12 '18 at 18:49
  • $\begingroup$ Let $f(x)= e^x.$ Then $\sup_{x\in \mathbb R} |e^x-P_n(x)| = \infty$ for any $n.$ $\endgroup$ – zhw. Nov 12 '18 at 19:09
2
$\begingroup$

Consider $f(x) = 1/(1-x)$ on $[-2,0]$. Expand it at $x =0$, then $$ f^{(n)}(0) = n!(1-x)^{-(n+1)}|_{x=0} = n!, $$ and the Taylor formula is $$ f(x) = \sum_0^n x^j + (1- \xi)^{-(n+2)} x^{n+1}. $$ Now $$ \sup |R_n(x)| = \sup\left| \frac 1{1-x}- \sum_0^n x^j\right| = \sup\left|\frac {1-(1-x^{n+1})}{1-x}\right| = \sup \left|\frac{x^{n+1}} {1-x}\right| =\sup \frac {(-x)^{n+1}} {1-x} \geqslant \frac {2^{n+1}} {1+2} \nearrow +\infty, $$ so the sequence $g(n)$ cannot be decreasing.

UPDATE

Inspired by @zhw., if the interval is $(-1,1)$ instead, then $$ g(n) \geqslant \frac {(1-1/n)^{n+1}}{1 - 1+1/n} \xrightarrow{n\to \infty} +\infty, $$ and $g(n)$ still cannot be decreasing.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Can we claim that $g(n)$ will be decreasing with $n$ only for those functions whose Taylor's polynomial will converge to the function. $\endgroup$ – Lord KK Nov 12 '18 at 18:34
  • $\begingroup$ On $[-2,0]?$ i would think the natural domain here is $(-1,1).$ Why did you take the absolute values off at the end there? Actually the $\sup = \infty$ for every $n.$ $\endgroup$ – zhw. Nov 12 '18 at 19:46
  • $\begingroup$ That is before the OP added the restriction $R_n(x)\to 0$. $\endgroup$ – xbh Nov 13 '18 at 1:17

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.