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Understanding Lagrange error bound
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Estimating $ln(0.9)$ using a Taylor polynomial about $x=1$ what is the least degree of the polynomial that assures an error smaller than $0.0010$?
(The $n^th$ derivative of $(-1)^n-1cfrac(n-1!)x^n$ for $nge1$
Then Khan Academy says, the Lagrange bound for the error assures that $left|R_n(1.4)right|leleft|cfrac(-1)^ncfracn!z^n+1(n+1)!(1.4-1)^n+1right|$
Then it says $cfrac0.4^n+1(n+1)z^n+1lecfrac0.4^n+1(n+1)$
In the sentence before this one, why do they get rid of $z^n+1$?
calculus taylor-expansion
asked Jul 28 at 1:30
Jinzu
328311
add a comment |Â
up vote
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down vote
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Estimating $ln(0.9)$ using a Taylor polynomial about $x=1$ what is the least degree of the polynomial that assures an error smaller than $0.0010$?
(The $n^th$ derivative of $(-1)^n-1cfrac(n-1!)x^n$ for $nge1$
Then Khan Academy says, the Lagrange bound for the error assures that $left|R_n(1.4)right|leleft|cfrac(-1)^ncfracn!z^n+1(n+1)!(1.4-1)^n+1right|$
Then it says $cfrac0.4^n+1(n+1)z^n+1lecfrac0.4^n+1(n+1)$
In the sentence before this one, why do they get rid of $z^n+1$?
calculus taylor-expansion
asked Jul 28 at 1:30
Jinzu
328311
Because $1leq z$, therefore if you replace it by $1$ the denominator gets smaller, and the fraction larger. Take into account that I don't know what $R_n(1.4)$ and therefore, $z$, because you haven't defined it, but it looks like after you do, that is going to be the explanation.
– user578878
Jul 28 at 1:32
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Estimating $ln(0.9)$ using a Taylor polynomial about $x=1$ what is the least degree of the polynomial that assures an error smaller than $0.0010$?
(The $n^th$ derivative of $(-1)^n-1cfrac(n-1!)x^n$ for $nge1$
Then Khan Academy says, the Lagrange bound for the error assures that $left|R_n(1.4)right|leleft|cfrac(-1)^ncfracn!z^n+1(n+1)!(1.4-1)^n+1right|$
Then it says $cfrac0.4^n+1(n+1)z^n+1lecfrac0.4^n+1(n+1)$
In the sentence before this one, why do they get rid of $z^n+1$?
calculus taylor-expansion
asked Jul 28 at 1:30
Jinzu
328311
Estimating $ln(0.9)$ using a Taylor polynomial about $x=1$ what is the least degree of the polynomial that assures an error smaller than $0.0010$?
(The $n^th$ derivative of $(-1)^n-1cfrac(n-1!)x^n$ for $nge1$
Then Khan Academy says, the Lagrange bound for the error assures that $left|R_n(1.4)right|leleft|cfrac(-1)^ncfracn!z^n+1(n+1)!(1.4-1)^n+1right|$
Then it says $cfrac0.4^n+1(n+1)z^n+1lecfrac0.4^n+1(n+1)$
In the sentence before this one, why do they get rid of $z^n+1$?
calculus taylor-expansion
asked Jul 28 at 1:30
Jinzu
328311
asked Jul 28 at 1:30
Jinzu
328311
asked Jul 28 at 1:30
Jinzu
328311
asked Jul 28 at 1:30
Jinzu
328311
328311
Because $1leq z$, therefore if you replace it by $1$ the denominator gets smaller, and the fraction larger. Take into account that I don't know what $R_n(1.4)$ and therefore, $z$, because you haven't defined it, but it looks like after you do, that is going to be the explanation.
– user578878
Jul 28 at 1:32
add a comment |Â
Because $1leq z$, therefore if you replace it by $1$ the denominator gets smaller, and the fraction larger. Take into account that I don't know what $R_n(1.4)$ and therefore, $z$, because you haven't defined it, but it looks like after you do, that is going to be the explanation.
– user578878
Jul 28 at 1:32
Because $1leq z$, therefore if you replace it by $1$ the denominator gets smaller, and the fraction larger. Take into account that I don't know what $R_n(1.4)$ and therefore, $z$, because you haven't defined it, but it looks like after you do, that is going to be the explanation.
– user578878
Jul 28 at 1:32
Because $1leq z$, therefore if you replace it by $1$ the denominator gets smaller, and the fraction larger. Take into account that I don't know what $R_n(1.4)$ and therefore, $z$, because you haven't defined it, but it looks like after you do, that is going to be the explanation.
– user578878
Jul 28 at 1:32
add a comment |Â
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Because $1leq z$, therefore if you replace it by $1$ the denominator gets smaller, and the fraction larger. Take into account that I don't know what $R_n(1.4)$ and therefore, $z$, because you haven't defined it, but it looks like after you do, that is going to be the explanation.
– user578878
Jul 28 at 1:32