Mixing
Could the polynomial $f(t,c)=sum_{i=1}^{c} (2i-1)t^{i-1}$ be written as a rational function in $t$ and...
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Could the polynomial $f(t,c)=sum_{i=1}^{c} (2i-1)t^{i-1}$ be written as a rational function in $t$ and $q=t^c$ of the form $frac{g(t,q)}{h(t,q)}$, where $g(t,q)$, $h(t,q)$ are polynomials in $t,q$ and the coefficients of $g, h$ do not involve $c$? I think that this is impossible. But I want to make sure. Thank you very much.
Edit:
begin{align}
(1-t)f(t,c) & = 1+sum_{i=1}^{c-1}2t^i-(2c-1)t^c \
& = frac{1+t-q-2cq}{1-t}.
end{align}
Therefore $f(t,c) = frac{1+t-q-2cq}{(1-t)^2}$. The coefficient of $q$ involves $c$. Therefore I think that it is impossible to write $f(t,c)$ as a rational function of $t, q$ whose coefficients do not involve $c$. Is this correct? Thank you very much.
calculus combinatorics
asked Feb 1 at 20:25
LJRLJR
6,66641850
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add a comment |
$begingroup$
Could the polynomial $f(t,c)=sum_{i=1}^{c} (2i-1)t^{i-1}$ be written as a rational function in $t$ and $q=t^c$ of the form $frac{g(t,q)}{h(t,q)}$, where $g(t,q)$, $h(t,q)$ are polynomials in $t,q$ and the coefficients of $g, h$ do not involve $c$? I think that this is impossible. But I want to make sure. Thank you very much.
Edit:
begin{align}
(1-t)f(t,c) & = 1+sum_{i=1}^{c-1}2t^i-(2c-1)t^c \
& = frac{1+t-q-2cq}{1-t}.
end{align}
Therefore $f(t,c) = frac{1+t-q-2cq}{(1-t)^2}$. The coefficient of $q$ involves $c$. Therefore I think that it is impossible to write $f(t,c)$ as a rational function of $t, q$ whose coefficients do not involve $c$. Is this correct? Thank you very much.
calculus combinatorics
asked Feb 1 at 20:25
LJRLJR
6,66641850
$endgroup$
$begingroup$
Why do you think it's impossible?
$endgroup$
– Carl Schildkraut
Feb 1 at 20:34
1
$begingroup$
Hint: $(1-t)f(t,c)$ is almost a finite geometric series.
$endgroup$
– Mike Earnest
Feb 1 at 21:56
add a comment |
$begingroup$
Could the polynomial $f(t,c)=sum_{i=1}^{c} (2i-1)t^{i-1}$ be written as a rational function in $t$ and $q=t^c$ of the form $frac{g(t,q)}{h(t,q)}$, where $g(t,q)$, $h(t,q)$ are polynomials in $t,q$ and the coefficients of $g, h$ do not involve $c$? I think that this is impossible. But I want to make sure. Thank you very much.
Edit:
begin{align}
(1-t)f(t,c) & = 1+sum_{i=1}^{c-1}2t^i-(2c-1)t^c \
& = frac{1+t-q-2cq}{1-t}.
end{align}
Therefore $f(t,c) = frac{1+t-q-2cq}{(1-t)^2}$. The coefficient of $q$ involves $c$. Therefore I think that it is impossible to write $f(t,c)$ as a rational function of $t, q$ whose coefficients do not involve $c$. Is this correct? Thank you very much.
calculus combinatorics
asked Feb 1 at 20:25
LJRLJR
6,66641850
$endgroup$
Could the polynomial $f(t,c)=sum_{i=1}^{c} (2i-1)t^{i-1}$ be written as a rational function in $t$ and $q=t^c$ of the form $frac{g(t,q)}{h(t,q)}$, where $g(t,q)$, $h(t,q)$ are polynomials in $t,q$ and the coefficients of $g, h$ do not involve $c$? I think that this is impossible. But I want to make sure. Thank you very much.
Edit:
begin{align}
(1-t)f(t,c) & = 1+sum_{i=1}^{c-1}2t^i-(2c-1)t^c \
& = frac{1+t-q-2cq}{1-t}.
end{align}
Therefore $f(t,c) = frac{1+t-q-2cq}{(1-t)^2}$. The coefficient of $q$ involves $c$. Therefore I think that it is impossible to write $f(t,c)$ as a rational function of $t, q$ whose coefficients do not involve $c$. Is this correct? Thank you very much.
calculus combinatorics
calculus combinatorics
asked Feb 1 at 20:25
LJRLJR
6,66641850
asked Feb 1 at 20:25
LJRLJR
6,66641850
edited Feb 2 at 9:39
LJR
asked Feb 1 at 20:25
LJRLJR
6,66641850
asked Feb 1 at 20:25
LJRLJR
6,66641850
asked Feb 1 at 20:25
LJRLJR
6,66641850
6,66641850
$begingroup$
Why do you think it's impossible?
$endgroup$
– Carl Schildkraut
Feb 1 at 20:34
1
$begingroup$
Hint: $(1-t)f(t,c)$ is almost a finite geometric series.
$endgroup$
– Mike Earnest
Feb 1 at 21:56
add a comment |
$begingroup$
Why do you think it's impossible?
$endgroup$
– Carl Schildkraut
Feb 1 at 20:34
1
$begingroup$
Hint: $(1-t)f(t,c)$ is almost a finite geometric series.
$endgroup$
– Mike Earnest
Feb 1 at 21:56
$begingroup$
Why do you think it's impossible?
$endgroup$
– Carl Schildkraut
Feb 1 at 20:34
$begingroup$
Why do you think it's impossible?
$endgroup$
– Carl Schildkraut
Feb 1 at 20:34
1
1
$begingroup$
Hint: $(1-t)f(t,c)$ is almost a finite geometric series.
$endgroup$
– Mike Earnest
Feb 1 at 21:56
$begingroup$
Hint: $(1-t)f(t,c)$ is almost a finite geometric series.
$endgroup$
– Mike Earnest
Feb 1 at 21:56
add a comment |
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$begingroup$
Why do you think it's impossible?
$endgroup$
– Carl Schildkraut
Feb 1 at 20:34
1
$begingroup$
Hint: $(1-t)f(t,c)$ is almost a finite geometric series.
$endgroup$
– Mike Earnest
Feb 1 at 21:56