about the minimax and orthogonal polynomial Exponential function
$begingroup$
using legendre Polynomial between 0,1 for the exponential function you could get te following series
$$e^{-frac{p}{2}}=sum _{n=0}^{infty } sqrt{frac{pi }{2}} (-1)^{ n} (2 n+1) I_{n+frac{1}{2}}(1) P_nleft(frac{p}{2}right)$$ it is suppose that t is a good aproximation using only m=2 you get the error
$$e^{-frac{p}{2}}simeq -frac{5}{8} sqrt{frac{pi }{2}} left(3 p^2-4right) I_{frac{5}{2}}(1)+frac{3}{2} sqrt{frac{pi }{2}} p I_{frac{3}{2}}(1)+left(-sqrt{frac{pi }{2}}right) I_{frac{1}{2}}(1)$$ for p=1 the error it is about
0.027875
it is suppose that this series could be improve ( maybe get it better improve using chevyshef series) but de following series
$$e^{-frac{p}{2}}simeq frac{3 pi (4 p+7) I_{frac{3}{4}}left(frac{p}{2}right)}{8 p^{3/4} Gamma left(frac{1}{4}right)}-frac{15 pi sqrt[4]{p} I_{-frac{1}{4}}left(frac{p}{2}right)}{8 Gamma left(frac{1}{4}right)}$$
for m=2 (usin two terms) gives te error -0.00286246 10 times better
how it is possible? it is suppose that Legendre it is orthogonal polinomial
calculus
asked Dec 25 '18 at 12:36
CLERKRAMACLERKRAMA
84
$endgroup$
add a comment |
$begingroup$
using legendre Polynomial between 0,1 for the exponential function you could get te following series
$$e^{-frac{p}{2}}=sum _{n=0}^{infty } sqrt{frac{pi }{2}} (-1)^{ n} (2 n+1) I_{n+frac{1}{2}}(1) P_nleft(frac{p}{2}right)$$ it is suppose that t is a good aproximation using only m=2 you get the error
$$e^{-frac{p}{2}}simeq -frac{5}{8} sqrt{frac{pi }{2}} left(3 p^2-4right) I_{frac{5}{2}}(1)+frac{3}{2} sqrt{frac{pi }{2}} p I_{frac{3}{2}}(1)+left(-sqrt{frac{pi }{2}}right) I_{frac{1}{2}}(1)$$ for p=1 the error it is about
0.027875
it is suppose that this series could be improve ( maybe get it better improve using chevyshef series) but de following series
$$e^{-frac{p}{2}}simeq frac{3 pi (4 p+7) I_{frac{3}{4}}left(frac{p}{2}right)}{8 p^{3/4} Gamma left(frac{1}{4}right)}-frac{15 pi sqrt[4]{p} I_{-frac{1}{4}}left(frac{p}{2}right)}{8 Gamma left(frac{1}{4}right)}$$
for m=2 (usin two terms) gives te error -0.00286246 10 times better
how it is possible? it is suppose that Legendre it is orthogonal polinomial
calculus
asked Dec 25 '18 at 12:36
CLERKRAMACLERKRAMA
84
$endgroup$
add a comment |
$begingroup$
using legendre Polynomial between 0,1 for the exponential function you could get te following series
$$e^{-frac{p}{2}}=sum _{n=0}^{infty } sqrt{frac{pi }{2}} (-1)^{ n} (2 n+1) I_{n+frac{1}{2}}(1) P_nleft(frac{p}{2}right)$$ it is suppose that t is a good aproximation using only m=2 you get the error
$$e^{-frac{p}{2}}simeq -frac{5}{8} sqrt{frac{pi }{2}} left(3 p^2-4right) I_{frac{5}{2}}(1)+frac{3}{2} sqrt{frac{pi }{2}} p I_{frac{3}{2}}(1)+left(-sqrt{frac{pi }{2}}right) I_{frac{1}{2}}(1)$$ for p=1 the error it is about
0.027875
it is suppose that this series could be improve ( maybe get it better improve using chevyshef series) but de following series
$$e^{-frac{p}{2}}simeq frac{3 pi (4 p+7) I_{frac{3}{4}}left(frac{p}{2}right)}{8 p^{3/4} Gamma left(frac{1}{4}right)}-frac{15 pi sqrt[4]{p} I_{-frac{1}{4}}left(frac{p}{2}right)}{8 Gamma left(frac{1}{4}right)}$$
for m=2 (usin two terms) gives te error -0.00286246 10 times better
how it is possible? it is suppose that Legendre it is orthogonal polinomial
calculus
asked Dec 25 '18 at 12:36
CLERKRAMACLERKRAMA
84
$endgroup$
using legendre Polynomial between 0,1 for the exponential function you could get te following series
$$e^{-frac{p}{2}}=sum _{n=0}^{infty } sqrt{frac{pi }{2}} (-1)^{ n} (2 n+1) I_{n+frac{1}{2}}(1) P_nleft(frac{p}{2}right)$$ it is suppose that t is a good aproximation using only m=2 you get the error
$$e^{-frac{p}{2}}simeq -frac{5}{8} sqrt{frac{pi }{2}} left(3 p^2-4right) I_{frac{5}{2}}(1)+frac{3}{2} sqrt{frac{pi }{2}} p I_{frac{3}{2}}(1)+left(-sqrt{frac{pi }{2}}right) I_{frac{1}{2}}(1)$$ for p=1 the error it is about
0.027875
it is suppose that this series could be improve ( maybe get it better improve using chevyshef series) but de following series
$$e^{-frac{p}{2}}simeq frac{3 pi (4 p+7) I_{frac{3}{4}}left(frac{p}{2}right)}{8 p^{3/4} Gamma left(frac{1}{4}right)}-frac{15 pi sqrt[4]{p} I_{-frac{1}{4}}left(frac{p}{2}right)}{8 Gamma left(frac{1}{4}right)}$$
for m=2 (usin two terms) gives te error -0.00286246 10 times better
how it is possible? it is suppose that Legendre it is orthogonal polinomial
calculus
calculus
asked Dec 25 '18 at 12:36
CLERKRAMACLERKRAMA
84
asked Dec 25 '18 at 12:36
CLERKRAMACLERKRAMA
84
asked Dec 25 '18 at 12:36
CLERKRAMACLERKRAMA
84
asked Dec 25 '18 at 12:36
CLERKRAMACLERKRAMA
84
asked Dec 25 '18 at 12:36
CLERKRAMACLERKRAMA
84
84
add a comment |
add a comment |
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