Finding Polynomial Roots Analytically
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Let us consider the following function: $$f(s_1,s_2,...s_n,n)=frac{1}{1+s_1}+frac{1}{(1+s_2)^2}+...+frac{1}{(1+s_n)^n},$$ where $s_iin(0,1), i=1,...,n.$
Let us introduce the following function: $$g(x,n)=frac{1}{1+x}+frac{1}{(1+x)^2}+...+frac{1}{(1+x)^n}.$$
We require that $f(s_1,s_2,...s_n,n)=g(x,n)$ and from this condition it is needed to find $x$. In other words it is needed to construct $x(s_1,...,s_n)$. Since $n$ attains large numbers (e.g. 25) according Abel-Ruffini theorem this problem isn't possible to solve analytically.
Please guide me how to find some approximation. Does iteration method a good way to solve the problem? (I mean saying iteration following: assign to $s_i$'s values, find $f(s_1,...s_n)$, therefore estimate $x$).
calculus roots
asked Dec 1 '18 at 21:28
DavidDavid
408
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add a comment |
$begingroup$
Let us consider the following function: $$f(s_1,s_2,...s_n,n)=frac{1}{1+s_1}+frac{1}{(1+s_2)^2}+...+frac{1}{(1+s_n)^n},$$ where $s_iin(0,1), i=1,...,n.$
Let us introduce the following function: $$g(x,n)=frac{1}{1+x}+frac{1}{(1+x)^2}+...+frac{1}{(1+x)^n}.$$
We require that $f(s_1,s_2,...s_n,n)=g(x,n)$ and from this condition it is needed to find $x$. In other words it is needed to construct $x(s_1,...,s_n)$. Since $n$ attains large numbers (e.g. 25) according Abel-Ruffini theorem this problem isn't possible to solve analytically.
Please guide me how to find some approximation. Does iteration method a good way to solve the problem? (I mean saying iteration following: assign to $s_i$'s values, find $f(s_1,...s_n)$, therefore estimate $x$).
calculus roots
asked Dec 1 '18 at 21:28
DavidDavid
408
$endgroup$
$begingroup$
$g(x, n)$ is a geometric progression, isn't it?
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– user58697
Dec 1 '18 at 23:24
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@user58697 Yes, $g(x,n)$ is a geometric progression.
$endgroup$
– David
Dec 3 '18 at 5:58
add a comment |
$begingroup$
Let us consider the following function: $$f(s_1,s_2,...s_n,n)=frac{1}{1+s_1}+frac{1}{(1+s_2)^2}+...+frac{1}{(1+s_n)^n},$$ where $s_iin(0,1), i=1,...,n.$
Let us introduce the following function: $$g(x,n)=frac{1}{1+x}+frac{1}{(1+x)^2}+...+frac{1}{(1+x)^n}.$$
We require that $f(s_1,s_2,...s_n,n)=g(x,n)$ and from this condition it is needed to find $x$. In other words it is needed to construct $x(s_1,...,s_n)$. Since $n$ attains large numbers (e.g. 25) according Abel-Ruffini theorem this problem isn't possible to solve analytically.
Please guide me how to find some approximation. Does iteration method a good way to solve the problem? (I mean saying iteration following: assign to $s_i$'s values, find $f(s_1,...s_n)$, therefore estimate $x$).
calculus roots
asked Dec 1 '18 at 21:28
DavidDavid
408
$endgroup$
Let us consider the following function: $$f(s_1,s_2,...s_n,n)=frac{1}{1+s_1}+frac{1}{(1+s_2)^2}+...+frac{1}{(1+s_n)^n},$$ where $s_iin(0,1), i=1,...,n.$
Let us introduce the following function: $$g(x,n)=frac{1}{1+x}+frac{1}{(1+x)^2}+...+frac{1}{(1+x)^n}.$$
We require that $f(s_1,s_2,...s_n,n)=g(x,n)$ and from this condition it is needed to find $x$. In other words it is needed to construct $x(s_1,...,s_n)$. Since $n$ attains large numbers (e.g. 25) according Abel-Ruffini theorem this problem isn't possible to solve analytically.
Please guide me how to find some approximation. Does iteration method a good way to solve the problem? (I mean saying iteration following: assign to $s_i$'s values, find $f(s_1,...s_n)$, therefore estimate $x$).
calculus roots
calculus roots
asked Dec 1 '18 at 21:28
DavidDavid
408
asked Dec 1 '18 at 21:28
DavidDavid
408
edited Dec 1 '18 at 22:04
David
asked Dec 1 '18 at 21:28
DavidDavid
408
asked Dec 1 '18 at 21:28
DavidDavid
408
asked Dec 1 '18 at 21:28
DavidDavid
408
408
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$g(x, n)$ is a geometric progression, isn't it?
$endgroup$
– user58697
Dec 1 '18 at 23:24
$begingroup$
@user58697 Yes, $g(x,n)$ is a geometric progression.
$endgroup$
– David
Dec 3 '18 at 5:58
add a comment |
$begingroup$
$g(x, n)$ is a geometric progression, isn't it?
$endgroup$
– user58697
Dec 1 '18 at 23:24
$begingroup$
@user58697 Yes, $g(x,n)$ is a geometric progression.
$endgroup$
– David
Dec 3 '18 at 5:58
$begingroup$
$g(x, n)$ is a geometric progression, isn't it?
$endgroup$
– user58697
Dec 1 '18 at 23:24
$begingroup$
$g(x, n)$ is a geometric progression, isn't it?
$endgroup$
– user58697
Dec 1 '18 at 23:24
$begingroup$
@user58697 Yes, $g(x,n)$ is a geometric progression.
$endgroup$
– David
Dec 3 '18 at 5:58
$begingroup$
@user58697 Yes, $g(x,n)$ is a geometric progression.
$endgroup$
– David
Dec 3 '18 at 5:58
add a comment |
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$begingroup$
$g(x, n)$ is a geometric progression, isn't it?
$endgroup$
– user58697
Dec 1 '18 at 23:24
$begingroup$
@user58697 Yes, $g(x,n)$ is a geometric progression.
$endgroup$
– David
Dec 3 '18 at 5:58