pls help
Use the error formulae for Taylor polynom to determine in advance the degree of taylor polynomial at a=0 that would achieve the indicated accuracy on the interval [0,2].
f(x)=e^x
x=2
error <10^-3
fasit: n=10
ma0001 taylorpolynom
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Gjest
$E_n(x) = \dfrac{f^{n+1}(c)}{(n+1)!}(x-a)^{n+1}, 0\leq c\leq2$
$E_n(2) = \dfrac{f^{n+1(2)}}{(n+1)!}(2-0)^{n+1}$
$10^{-3}=\dfrac{e^2}{(n+1)!}2^{n+1}$
Nå kan du løse med kalkulator eller bare prøve deg fram med ulike $n \in \mathbb{N}$
n=9:
$\dfrac{e^2 \cdot 2^{9+1}}{(9+1)!} = 0.0021 > 10^{-3}$
n=10:
$\dfrac{e^2 \cdot 2^{10+1}}{(10+1)!} = 0.00038 < 10^{-3}$
altså må n = 10.
$E_n(2) = \dfrac{f^{n+1(2)}}{(n+1)!}(2-0)^{n+1}$
$10^{-3}=\dfrac{e^2}{(n+1)!}2^{n+1}$
Nå kan du løse med kalkulator eller bare prøve deg fram med ulike $n \in \mathbb{N}$
n=9:
$\dfrac{e^2 \cdot 2^{9+1}}{(9+1)!} = 0.0021 > 10^{-3}$
n=10:
$\dfrac{e^2 \cdot 2^{10+1}}{(10+1)!} = 0.00038 < 10^{-3}$
altså må n = 10.