# Taylor's Polynomial with Lagrange's form of remainder.

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 $$n$$

• For one $x$ or every $x?$ What are you taking the $\sup$ over? Where is $f$ defined?
– zhw.
Commented Nov 12, 2018 at 18:45
• @zhw. For any general function. We are talking for all those x at which taylor's polynomial converges to the function.
– Shak
Commented Nov 12, 2018 at 18:49
• Let $f(x)= e^x.$ Then $\sup_{x\in \mathbb R} |e^x-P_n(x)| = \infty$ for any $n.$
– zhw.
Commented Nov 12, 2018 at 19:09

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.

• Can we claim that $g(n)$ will be decreasing with $n$ only for those functions whose Taylor's polynomial will converge to the function.
– Shak
Commented Nov 12, 2018 at 18:34
• 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 $n.$
– zhw.
Commented Nov 12, 2018 at 19:46
• That is before the OP added the restriction $R_n(x)\to 0$.
– xbh
Commented Nov 13, 2018 at 1:17