Bounding at Newton Divided Difference Formula












1












$begingroup$


Let $a=x_0,x_1,...,x_n=b$ are $n+1$ points which are equally spaced in $[a,b]$. The distance between consecutive terms is $h= frac{b-a}{n}$ and $x in [a,b]$



Show that $ biggr|prod_{i=}^n (x-x_i) biggr| leq frac{h^{n+1}.n!}{4} $



I have written



$ biggr|prod_{i=}^n (x-x_i) biggr| = biggr| (x-x_0)(x-x_0-h)(x-x_0-2h)...(x-x_0-nh) biggr|$



I know it is so easy but I couldn’t see in no way. I think I have written something unnecessary. Thanks for any help..






numerical-methods interpolation numerical-calculus







share|cite|improve this question











$endgroup$












  • $begingroup$
    $x$ is any real number? or $xin [a,b]$?
    $endgroup$
    – Ricardo Largaespada
    Dec 16 '18 at 0:53










  • $begingroup$
    @RicardoLargaespada sorry I’ve forgotten to write it. $in [a,b]$
    $endgroup$
    – user519955
    Dec 16 '18 at 1:01
















1












$begingroup$


Let $a=x_0,x_1,...,x_n=b$ are $n+1$ points which are equally spaced in $[a,b]$. The distance between consecutive terms is $h= frac{b-a}{n}$ and $x in [a,b]$



Show that $ biggr|prod_{i=}^n (x-x_i) biggr| leq frac{h^{n+1}.n!}{4} $



I have written



$ biggr|prod_{i=}^n (x-x_i) biggr| = biggr| (x-x_0)(x-x_0-h)(x-x_0-2h)...(x-x_0-nh) biggr|$



I know it is so easy but I couldn’t see in no way. I think I have written something unnecessary. Thanks for any help..






numerical-methods interpolation numerical-calculus







share|cite|improve this question











$endgroup$












  • $begingroup$
    $x$ is any real number? or $xin [a,b]$?
    $endgroup$
    – Ricardo Largaespada
    Dec 16 '18 at 0:53










  • $begingroup$
    @RicardoLargaespada sorry I’ve forgotten to write it. $in [a,b]$
    $endgroup$
    – user519955
    Dec 16 '18 at 1:01














1












1








1





$begingroup$


Let $a=x_0,x_1,...,x_n=b$ are $n+1$ points which are equally spaced in $[a,b]$. The distance between consecutive terms is $h= frac{b-a}{n}$ and $x in [a,b]$



Show that $ biggr|prod_{i=}^n (x-x_i) biggr| leq frac{h^{n+1}.n!}{4} $



I have written



$ biggr|prod_{i=}^n (x-x_i) biggr| = biggr| (x-x_0)(x-x_0-h)(x-x_0-2h)...(x-x_0-nh) biggr|$



I know it is so easy but I couldn’t see in no way. I think I have written something unnecessary. Thanks for any help..






numerical-methods interpolation numerical-calculus







share|cite|improve this question











$endgroup$




Let $a=x_0,x_1,...,x_n=b$ are $n+1$ points which are equally spaced in $[a,b]$. The distance between consecutive terms is $h= frac{b-a}{n}$ and $x in [a,b]$



Show that $ biggr|prod_{i=}^n (x-x_i) biggr| leq frac{h^{n+1}.n!}{4} $



I have written



$ biggr|prod_{i=}^n (x-x_i) biggr| = biggr| (x-x_0)(x-x_0-h)(x-x_0-2h)...(x-x_0-nh) biggr|$



I know it is so easy but I couldn’t see in no way. I think I have written something unnecessary. Thanks for any help..






numerical-methods interpolation numerical-calculus




numerical-methods interpolation numerical-calculus






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 16 '18 at 1:02







user519955

















asked Dec 16 '18 at 0:30









user519955user519955

312111




312111












  • $begingroup$
    $x$ is any real number? or $xin [a,b]$?
    $endgroup$
    – Ricardo Largaespada
    Dec 16 '18 at 0:53










  • $begingroup$
    @RicardoLargaespada sorry I’ve forgotten to write it. $in [a,b]$
    $endgroup$
    – user519955
    Dec 16 '18 at 1:01


















  • $begingroup$
    $x$ is any real number? or $xin [a,b]$?
    $endgroup$
    – Ricardo Largaespada
    Dec 16 '18 at 0:53










  • $begingroup$
    @RicardoLargaespada sorry I’ve forgotten to write it. $in [a,b]$
    $endgroup$
    – user519955
    Dec 16 '18 at 1:01
















$begingroup$
$x$ is any real number? or $xin [a,b]$?
$endgroup$
– Ricardo Largaespada
Dec 16 '18 at 0:53




$begingroup$
$x$ is any real number? or $xin [a,b]$?
$endgroup$
– Ricardo Largaespada
Dec 16 '18 at 0:53












$begingroup$
@RicardoLargaespada sorry I’ve forgotten to write it. $in [a,b]$
$endgroup$
– user519955
Dec 16 '18 at 1:01




$begingroup$
@RicardoLargaespada sorry I’ve forgotten to write it. $in [a,b]$
$endgroup$
– user519955
Dec 16 '18 at 1:01










1 Answer
1






active

oldest

votes


















0












$begingroup$

Argue that the maximal error occurs in the boundary intervals. Set $x=sh$. Then
$$prod_{k=0}^n|x-x_k|=h^{n+1}|s(1-s)|cdot|2-s|cdots|n-s|$$
and the bound should be obvious.



More generally, for $x=mh+s$, $m=0,...,n$, you get
$$h^{n+1}|m+1-(1-s)|⋯|2-(1-s)|⋅|s(1-s)|⋅|1-s|⋯|n-m-s|le frac14h^{n+1}(m+1)!(n-m)!$$




Take the next term into the maximum calculation, thus maximizing $s(1-s)(2-s)$ and find the maximum for $sin [0,frac32]$ at the interval ends or the root of
$$
0=3s^2-6s+2iff s=1-frac1{sqrt3}
$$
so that the bound can be decreased to
$$
h^{n+1}frac{n!}{3sqrt3}.
$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    How can I get $frac{h^{n+1}n!}{4}$ with your writtens? I’m sorry I could not understand
    $endgroup$
    – user519955
    Dec 16 '18 at 9:20












  • $begingroup$
    With the easy estimates $k-sle k$ and $s(1-s)=frac14(1-(2s-1)^2$.
    $endgroup$
    – LutzL
    Dec 16 '18 at 9:31













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1 Answer
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1 Answer
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0












$begingroup$

Argue that the maximal error occurs in the boundary intervals. Set $x=sh$. Then
$$prod_{k=0}^n|x-x_k|=h^{n+1}|s(1-s)|cdot|2-s|cdots|n-s|$$
and the bound should be obvious.



More generally, for $x=mh+s$, $m=0,...,n$, you get
$$h^{n+1}|m+1-(1-s)|⋯|2-(1-s)|⋅|s(1-s)|⋅|1-s|⋯|n-m-s|le frac14h^{n+1}(m+1)!(n-m)!$$




Take the next term into the maximum calculation, thus maximizing $s(1-s)(2-s)$ and find the maximum for $sin [0,frac32]$ at the interval ends or the root of
$$
0=3s^2-6s+2iff s=1-frac1{sqrt3}
$$
so that the bound can be decreased to
$$
h^{n+1}frac{n!}{3sqrt3}.
$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    How can I get $frac{h^{n+1}n!}{4}$ with your writtens? I’m sorry I could not understand
    $endgroup$
    – user519955
    Dec 16 '18 at 9:20












  • $begingroup$
    With the easy estimates $k-sle k$ and $s(1-s)=frac14(1-(2s-1)^2$.
    $endgroup$
    – LutzL
    Dec 16 '18 at 9:31


















0












$begingroup$

Argue that the maximal error occurs in the boundary intervals. Set $x=sh$. Then
$$prod_{k=0}^n|x-x_k|=h^{n+1}|s(1-s)|cdot|2-s|cdots|n-s|$$
and the bound should be obvious.



More generally, for $x=mh+s$, $m=0,...,n$, you get
$$h^{n+1}|m+1-(1-s)|⋯|2-(1-s)|⋅|s(1-s)|⋅|1-s|⋯|n-m-s|le frac14h^{n+1}(m+1)!(n-m)!$$




Take the next term into the maximum calculation, thus maximizing $s(1-s)(2-s)$ and find the maximum for $sin [0,frac32]$ at the interval ends or the root of
$$
0=3s^2-6s+2iff s=1-frac1{sqrt3}
$$
so that the bound can be decreased to
$$
h^{n+1}frac{n!}{3sqrt3}.
$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    How can I get $frac{h^{n+1}n!}{4}$ with your writtens? I’m sorry I could not understand
    $endgroup$
    – user519955
    Dec 16 '18 at 9:20












  • $begingroup$
    With the easy estimates $k-sle k$ and $s(1-s)=frac14(1-(2s-1)^2$.
    $endgroup$
    – LutzL
    Dec 16 '18 at 9:31
















0












0








0





$begingroup$

Argue that the maximal error occurs in the boundary intervals. Set $x=sh$. Then
$$prod_{k=0}^n|x-x_k|=h^{n+1}|s(1-s)|cdot|2-s|cdots|n-s|$$
and the bound should be obvious.



More generally, for $x=mh+s$, $m=0,...,n$, you get
$$h^{n+1}|m+1-(1-s)|⋯|2-(1-s)|⋅|s(1-s)|⋅|1-s|⋯|n-m-s|le frac14h^{n+1}(m+1)!(n-m)!$$




Take the next term into the maximum calculation, thus maximizing $s(1-s)(2-s)$ and find the maximum for $sin [0,frac32]$ at the interval ends or the root of
$$
0=3s^2-6s+2iff s=1-frac1{sqrt3}
$$
so that the bound can be decreased to
$$
h^{n+1}frac{n!}{3sqrt3}.
$$






share|cite|improve this answer











$endgroup$



Argue that the maximal error occurs in the boundary intervals. Set $x=sh$. Then
$$prod_{k=0}^n|x-x_k|=h^{n+1}|s(1-s)|cdot|2-s|cdots|n-s|$$
and the bound should be obvious.



More generally, for $x=mh+s$, $m=0,...,n$, you get
$$h^{n+1}|m+1-(1-s)|⋯|2-(1-s)|⋅|s(1-s)|⋅|1-s|⋯|n-m-s|le frac14h^{n+1}(m+1)!(n-m)!$$




Take the next term into the maximum calculation, thus maximizing $s(1-s)(2-s)$ and find the maximum for $sin [0,frac32]$ at the interval ends or the root of
$$
0=3s^2-6s+2iff s=1-frac1{sqrt3}
$$
so that the bound can be decreased to
$$
h^{n+1}frac{n!}{3sqrt3}.
$$







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 16 '18 at 9:34

























answered Dec 16 '18 at 8:54









LutzLLutzL

58.7k42055




58.7k42055












  • $begingroup$
    How can I get $frac{h^{n+1}n!}{4}$ with your writtens? I’m sorry I could not understand
    $endgroup$
    – user519955
    Dec 16 '18 at 9:20












  • $begingroup$
    With the easy estimates $k-sle k$ and $s(1-s)=frac14(1-(2s-1)^2$.
    $endgroup$
    – LutzL
    Dec 16 '18 at 9:31




















  • $begingroup$
    How can I get $frac{h^{n+1}n!}{4}$ with your writtens? I’m sorry I could not understand
    $endgroup$
    – user519955
    Dec 16 '18 at 9:20












  • $begingroup$
    With the easy estimates $k-sle k$ and $s(1-s)=frac14(1-(2s-1)^2$.
    $endgroup$
    – LutzL
    Dec 16 '18 at 9:31


















$begingroup$
How can I get $frac{h^{n+1}n!}{4}$ with your writtens? I’m sorry I could not understand
$endgroup$
– user519955
Dec 16 '18 at 9:20






$begingroup$
How can I get $frac{h^{n+1}n!}{4}$ with your writtens? I’m sorry I could not understand
$endgroup$
– user519955
Dec 16 '18 at 9:20














$begingroup$
With the easy estimates $k-sle k$ and $s(1-s)=frac14(1-(2s-1)^2$.
$endgroup$
– LutzL
Dec 16 '18 at 9:31






$begingroup$
With the easy estimates $k-sle k$ and $s(1-s)=frac14(1-(2s-1)^2$.
$endgroup$
– LutzL
Dec 16 '18 at 9:31




















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