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i think you misunderstand the error term in gauss-laguerre quadrature !

that is : $E=\frac{2^{2n+1}{(n!)}^4}{(2n+1)(2n!)^3}$$\cdot{f^{(2n)}}$

see : http://mathworld.wolfram.com/Legendre-GaussQuadrature.html

so the condition is  :

$\frac{2^{2n+1}{(n!)}^4}{(2n+1)(2n!)^3}$$\ge$$\frac{1}{\epsilon}$

where, $\epsilon=f^{(2n+1)}=C_{2n}$ is the first coefficient !

so ，

$\Longrightarrow$$\frac{2^{2n+1}{(n!)}^4}{(2n+1)(2n!)^3}\ge$$\frac{1}{f(x)-f(a)-\frac{f^{(1)}{}(a)}{1!}(x-a)-\frac{f^{(2)}{(a)}}{2!}(x-a)^{2}-......}$$\cdot$$\frac{(x-a)^{2n+1}}{(2n+1)!}=$$\frac{2}{f(x)-C_{1}-\frac{C_{1}+C_{2}}{1!}x-\frac{C_{1}+C_{2}+C_{3}}{2!}x^{2}-......}$$\cdot$$\frac{x^{2n+1}}{(2n+1)!}$$=C_{2n}$ $\Longrightarrow$$\frac{2}{\sum{C_{i}}-C_{1}-\frac{C_{1}+C_{2}}{1!}-\frac{C_{1}+C_{2}+C_{3}}{2!}-......}$$\cdot$$\frac{1}{(2n+1)!}$$\le$$\frac{2^{2n+1}{(n!)}^4}{(2n+1)(2n!)^3}$

with$,C_{2n}\longrightarrow0$

I think you misunderstand the error term in gauss-laguerre quadrature !

that is $$E=\frac{2^{2n+1}{(n!)}^4}{(2n+1)(2n!)^3}{f^{(2n)}},$$

see : http://mathworld.wolfram.com/Legendre-GaussQuadrature.html.

Hence the condition is:

$$\frac{2^{2n+1}{(n!)}^4}{(2n+1)(2n!)^3}\ge\frac{1}{\epsilon}$$

where $\epsilon=f^{(2n+1)}=C_{2n}$ is the first coefficient ! So， \begin{align*} \frac{2^{2n+1}{(n!)}^4}{(2n+1)(2n!)^3} & \ge\frac{1}{f(x)-f(a)-\frac{f^{(1)}{}(a)}{1!}(x-a)-\frac{f^{(2)}{(a)}}{2!}(x-a)^{2}-......}\frac{(x-a)^{2n+1}}{(2n+1)!} \\ & =\frac{2}{f(x)-C_{1}-\frac{C_{1}+C_{2}}{1!}x-\frac{C_{1}+C_{2}+C_{3}}{2!}x^{2}-......}\frac{x^{2n+1}}{(2n+1)!}=C_{2n}, \end{align*} and thus $$\frac{2}{\sum{C_{i}}-C_{1}-\frac{C_{1}+C_{2}}{1!}-\frac{C_{1}+C_{2}+C_{3}}{2!}-......}\frac{1}{(2n+1)!}\le\frac{2^{2n+1}{(n!)}^4}{(2n+1)(2n!)^3},$$

with $C_{2n}\to0$.

i think you misunderstand the error term in gauss-laguerre quadrature !

that is : $E=\frac{2^{2n+1}{(n!)}^4}{(2n+1)(2n!)^3}$$\cdot{f^{(2n)}}$

see : http://mathworld.wolfram.com/Legendre-GaussQuadrature.html

so the condition is :

$\frac{2^{2n+1}{(n!)}^4}{(2n+1)(2n!)^3}$$\ge$$\frac{1}{\epsilon}$

where, $\epsilon=f^{(2n)}$

so ，

$\Longrightarrow$$\frac{2^{2n+1}{(n!)}^4}{(2n+1)(2n!)^3}\ge$$\frac{1}{f(x)-f(a)-\frac{f^{(1)}{}(a)}{1!}(x-a)-\frac{f^{(2)}{(a)}}{2!}(x-a)^{2}-......}$$\cdot$$\frac{(x-a)^{2n+1}}{(2n+1)!}$

i think you misunderstand the error term in gauss-laguerre quadrature !

that is : $E=\frac{2^{2n+1}{(n!)}^4}{(2n+1)(2n!)^3}$$\cdot{f^{(2n)}}$

see : http://mathworld.wolfram.com/Legendre-GaussQuadrature.html

so the condition is :

$\frac{2^{2n+1}{(n!)}^4}{(2n+1)(2n!)^3}$$\ge$$\frac{1}{\epsilon}$

where, $\epsilon=f^{(2n+1)}=C_{2n}$ is the first coefficient !

so ，

$\Longrightarrow$$\frac{2^{2n+1}{(n!)}^4}{(2n+1)(2n!)^3}\ge$$\frac{1}{f(x)-f(a)-\frac{f^{(1)}{}(a)}{1!}(x-a)-\frac{f^{(2)}{(a)}}{2!}(x-a)^{2}-......}$$\cdot$$\frac{(x-a)^{2n+1}}{(2n+1)!}=$$\frac{2}{f(x)-C_{1}-\frac{C_{1}+C_{2}}{1!}x-\frac{C_{1}+C_{2}+C_{3}}{2!}x^{2}-......}$$\cdot$$\frac{x^{2n+1}}{(2n+1)!}$$=C_{2n}$ $\Longrightarrow$$\frac{2}{\sum{C_{i}}-C_{1}-\frac{C_{1}+C_{2}}{1!}-\frac{C_{1}+C_{2}+C_{3}}{2!}-......}$$\cdot$$\frac{1}{(2n+1)!}$$\le$$\frac{2^{2n+1}{(n!)}^4}{(2n+1)(2n!)^3}$

with$,C_{2n}\longrightarrow0$

i think you misunderstand the error term in gauss-laguerre quadrature !

that is : $E=\frac{2^{2n+1}{(n!)}^4}{(2n+1)(2n!)^3}$$\cdot{f^{(2n)}}$

see : http://mathworld.wolfram.com/Legendre-GaussQuadrature.html

so the condition is :

$\frac{2^{2n+1}{(n!)}^4}{(2n+1)(2n!)^3}$$\ge$$\frac{1}{\epsilon}$

where, $\epsilon=f^{(2n)}$

i think you misunderstand the error term in gauss-laguerre quadrature !

that is : $E=\frac{2^{2n+1}{(n!)}^4}{(2n+1)(2n!)^3}$$\cdot{f^{(2n)}}$

see : http://mathworld.wolfram.com/Legendre-GaussQuadrature.html

so the condition is :

$\frac{2^{2n+1}{(n!)}^4}{(2n+1)(2n!)^3}$$\ge$$\frac{1}{\epsilon}$

where, $\epsilon=f^{(2n)}$

so ，

$\Longrightarrow$$\frac{2^{2n+1}{(n!)}^4}{(2n+1)(2n!)^3}\ge$$\frac{1}{f(x)-f(a)-\frac{f^{(1)}{}(a)}{1!}(x-a)-\frac{f^{(2)}{(a)}}{2!}(x-a)^{2}-......}$$\cdot$$\frac{(x-a)^{2n+1}}{(2n+1)!}$