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Nontrivial solutions to $sum f(x) - int f(x) = gamma $?
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Consider
$$ T(f) = lim_t to infty sum_i =1^t f(i) - int_1^t f(x) dx $$
Notice $ T(a + b) = T(a) + T(b) $ and $T(C a) = C T(a)$ when $C $ is a constant.
Also $T(0)= 0$.
Let $T(d) = 0$.
So $T(1/x + d) = gamma $
Now i Know sophomore's dream , so that is 1 solution to $d$ that is nontrivial.
Also $T(x^-a) = zeta(a) - frac1a-1$.
I wonder , is there a solution or many such that
$T(v) = gamma $
Such that $v$ is no asymptotic of $ln $ ?
In particular a solution $v_2$ such that both the sum and integral converge ?? And another one $v_3 $ where both the sum and integral diverge.
Both existence and closed forms are considered .
Also of particular interest are those $T(g) $ solutions above where for $x ≥ 1$ we have ,$ g '(x) $ is always the same sign.
calculus real-analysis asymptotics monotone-functions eulers-constant
asked Jul 14 at 23:22
mick
4,79311961
add a comment |Â
up vote
0
down vote
favorite
Consider
$$ T(f) = lim_t to infty sum_i =1^t f(i) - int_1^t f(x) dx $$
Notice $ T(a + b) = T(a) + T(b) $ and $T(C a) = C T(a)$ when $C $ is a constant.
Also $T(0)= 0$.
Let $T(d) = 0$.
So $T(1/x + d) = gamma $
Now i Know sophomore's dream , so that is 1 solution to $d$ that is nontrivial.
Also $T(x^-a) = zeta(a) - frac1a-1$.
I wonder , is there a solution or many such that
$T(v) = gamma $
Such that $v$ is no asymptotic of $ln $ ?
In particular a solution $v_2$ such that both the sum and integral converge ?? And another one $v_3 $ where both the sum and integral diverge.
Both existence and closed forms are considered .
Also of particular interest are those $T(g) $ solutions above where for $x ≥ 1$ we have ,$ g '(x) $ is always the same sign.
calculus real-analysis asymptotics monotone-functions eulers-constant
asked Jul 14 at 23:22
mick
4,79311961
$$ |sum-int| leq max(f(1), f(t)) $$
– Vladislav Kharlamov
Jul 14 at 23:27
Yes that is true. So that shows ... Existence ??
– mick
Jul 14 at 23:34
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Consider
$$ T(f) = lim_t to infty sum_i =1^t f(i) - int_1^t f(x) dx $$
Notice $ T(a + b) = T(a) + T(b) $ and $T(C a) = C T(a)$ when $C $ is a constant.
Also $T(0)= 0$.
Let $T(d) = 0$.
So $T(1/x + d) = gamma $
Now i Know sophomore's dream , so that is 1 solution to $d$ that is nontrivial.
Also $T(x^-a) = zeta(a) - frac1a-1$.
I wonder , is there a solution or many such that
$T(v) = gamma $
Such that $v$ is no asymptotic of $ln $ ?
In particular a solution $v_2$ such that both the sum and integral converge ?? And another one $v_3 $ where both the sum and integral diverge.
Both existence and closed forms are considered .
Also of particular interest are those $T(g) $ solutions above where for $x ≥ 1$ we have ,$ g '(x) $ is always the same sign.
calculus real-analysis asymptotics monotone-functions eulers-constant
asked Jul 14 at 23:22
mick
4,79311961
Consider
$$ T(f) = lim_t to infty sum_i =1^t f(i) - int_1^t f(x) dx $$
Notice $ T(a + b) = T(a) + T(b) $ and $T(C a) = C T(a)$ when $C $ is a constant.
Also $T(0)= 0$.
Let $T(d) = 0$.
So $T(1/x + d) = gamma $
Now i Know sophomore's dream , so that is 1 solution to $d$ that is nontrivial.
Also $T(x^-a) = zeta(a) - frac1a-1$.
I wonder , is there a solution or many such that
$T(v) = gamma $
Such that $v$ is no asymptotic of $ln $ ?
In particular a solution $v_2$ such that both the sum and integral converge ?? And another one $v_3 $ where both the sum and integral diverge.
Both existence and closed forms are considered .
Also of particular interest are those $T(g) $ solutions above where for $x ≥ 1$ we have ,$ g '(x) $ is always the same sign.
calculus real-analysis asymptotics monotone-functions eulers-constant
asked Jul 14 at 23:22
mick
4,79311961
edited Jul 14 at 23:27
asked Jul 14 at 23:22
mick
4,79311961
asked Jul 14 at 23:22
mick
4,79311961
asked Jul 14 at 23:22
mick
4,79311961
4,79311961
$$ |sum-int| leq max(f(1), f(t)) $$
– Vladislav Kharlamov
Jul 14 at 23:27
Yes that is true. So that shows ... Existence ??
– mick
Jul 14 at 23:34
add a comment |Â
$$ |sum-int| leq max(f(1), f(t)) $$
– Vladislav Kharlamov
Jul 14 at 23:27
Yes that is true. So that shows ... Existence ??
– mick
Jul 14 at 23:34
$$ |sum-int| leq max(f(1), f(t)) $$
– Vladislav Kharlamov
Jul 14 at 23:27
$$ |sum-int| leq max(f(1), f(t)) $$
– Vladislav Kharlamov
Jul 14 at 23:27
Yes that is true. So that shows ... Existence ??
– mick
Jul 14 at 23:34
Yes that is true. So that shows ... Existence ??
– mick
Jul 14 at 23:34
add a comment |Â
1 Answer
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0
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This is perhaps not quite in the spirit of the question, but if $f(x)$ is a solution, so is
$$ g(x) = f(x)-f(lfloor x rfloor); $$
the sum is identically zero, the integral converges if and only if the limit exists, and therefore $g$ will not be asymptotic to $log$.
As for a divergent solution, $T(c)$ is defined and equal to zero for constant $c$, so $T(g+c)=T(g)$ for any $c$, while the sum is equal to $tc$ and the integral is $-tc + int_0^t g$, both of which diverge if $c neq 0$.
As for solutions with a derivative of constant sign, consider $f(x)=r^x$. The sum is
$$ fracr1-r(1-r^t), $$
while the integral is
$$ -fracrlogr(1-r^t-1). $$
Supposing that $0<r<1$, taking the limit gives
$$ fracr1-r + fracrlogr. $$
One can show that this is a positive increasing function that maps $(0,1) to (0,1/2)$. Hence one can find many nontrivial convergent solutions this way: for $0<r<1$,
$$ T(ar^x) = aleft(fracr1-r + fracrlogr right), $$
and one can find $a,r$ so that the right-hand side is $gamma$ provided that $a>2gamma$, and any $a$ gives exactly one such $r$.
answered Jul 15 at 0:24
Chappers
55k74191
But a and r are not closed form. They are functions of eulers gamma itself that do not reduce to standard forms. The Floor idea is ok but not analytic and trivial. +1 but No accept.
– mick
Jul 17 at 5:36
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
This is perhaps not quite in the spirit of the question, but if $f(x)$ is a solution, so is
$$ g(x) = f(x)-f(lfloor x rfloor); $$
the sum is identically zero, the integral converges if and only if the limit exists, and therefore $g$ will not be asymptotic to $log$.
As for a divergent solution, $T(c)$ is defined and equal to zero for constant $c$, so $T(g+c)=T(g)$ for any $c$, while the sum is equal to $tc$ and the integral is $-tc + int_0^t g$, both of which diverge if $c neq 0$.
As for solutions with a derivative of constant sign, consider $f(x)=r^x$. The sum is
$$ fracr1-r(1-r^t), $$
while the integral is
$$ -fracrlogr(1-r^t-1). $$
Supposing that $0<r<1$, taking the limit gives
$$ fracr1-r + fracrlogr. $$
One can show that this is a positive increasing function that maps $(0,1) to (0,1/2)$. Hence one can find many nontrivial convergent solutions this way: for $0<r<1$,
$$ T(ar^x) = aleft(fracr1-r + fracrlogr right), $$
and one can find $a,r$ so that the right-hand side is $gamma$ provided that $a>2gamma$, and any $a$ gives exactly one such $r$.
answered Jul 15 at 0:24
Chappers
55k74191
But a and r are not closed form. They are functions of eulers gamma itself that do not reduce to standard forms. The Floor idea is ok but not analytic and trivial. +1 but No accept.
– mick
Jul 17 at 5:36
add a comment |Â
up vote
0
down vote
This is perhaps not quite in the spirit of the question, but if $f(x)$ is a solution, so is
$$ g(x) = f(x)-f(lfloor x rfloor); $$
the sum is identically zero, the integral converges if and only if the limit exists, and therefore $g$ will not be asymptotic to $log$.
As for a divergent solution, $T(c)$ is defined and equal to zero for constant $c$, so $T(g+c)=T(g)$ for any $c$, while the sum is equal to $tc$ and the integral is $-tc + int_0^t g$, both of which diverge if $c neq 0$.
As for solutions with a derivative of constant sign, consider $f(x)=r^x$. The sum is
$$ fracr1-r(1-r^t), $$
while the integral is
$$ -fracrlogr(1-r^t-1). $$
Supposing that $0<r<1$, taking the limit gives
$$ fracr1-r + fracrlogr. $$
One can show that this is a positive increasing function that maps $(0,1) to (0,1/2)$. Hence one can find many nontrivial convergent solutions this way: for $0<r<1$,
$$ T(ar^x) = aleft(fracr1-r + fracrlogr right), $$
and one can find $a,r$ so that the right-hand side is $gamma$ provided that $a>2gamma$, and any $a$ gives exactly one such $r$.
answered Jul 15 at 0:24
Chappers
55k74191
But a and r are not closed form. They are functions of eulers gamma itself that do not reduce to standard forms. The Floor idea is ok but not analytic and trivial. +1 but No accept.
– mick
Jul 17 at 5:36
add a comment |Â
up vote
0
down vote
up vote
0
down vote
This is perhaps not quite in the spirit of the question, but if $f(x)$ is a solution, so is
$$ g(x) = f(x)-f(lfloor x rfloor); $$
the sum is identically zero, the integral converges if and only if the limit exists, and therefore $g$ will not be asymptotic to $log$.
As for a divergent solution, $T(c)$ is defined and equal to zero for constant $c$, so $T(g+c)=T(g)$ for any $c$, while the sum is equal to $tc$ and the integral is $-tc + int_0^t g$, both of which diverge if $c neq 0$.
As for solutions with a derivative of constant sign, consider $f(x)=r^x$. The sum is
$$ fracr1-r(1-r^t), $$
while the integral is
$$ -fracrlogr(1-r^t-1). $$
Supposing that $0<r<1$, taking the limit gives
$$ fracr1-r + fracrlogr. $$
One can show that this is a positive increasing function that maps $(0,1) to (0,1/2)$. Hence one can find many nontrivial convergent solutions this way: for $0<r<1$,
$$ T(ar^x) = aleft(fracr1-r + fracrlogr right), $$
and one can find $a,r$ so that the right-hand side is $gamma$ provided that $a>2gamma$, and any $a$ gives exactly one such $r$.
answered Jul 15 at 0:24
Chappers
55k74191
This is perhaps not quite in the spirit of the question, but if $f(x)$ is a solution, so is
$$ g(x) = f(x)-f(lfloor x rfloor); $$
the sum is identically zero, the integral converges if and only if the limit exists, and therefore $g$ will not be asymptotic to $log$.
As for a divergent solution, $T(c)$ is defined and equal to zero for constant $c$, so $T(g+c)=T(g)$ for any $c$, while the sum is equal to $tc$ and the integral is $-tc + int_0^t g$, both of which diverge if $c neq 0$.
As for solutions with a derivative of constant sign, consider $f(x)=r^x$. The sum is
$$ fracr1-r(1-r^t), $$
while the integral is
$$ -fracrlogr(1-r^t-1). $$
Supposing that $0<r<1$, taking the limit gives
$$ fracr1-r + fracrlogr. $$
One can show that this is a positive increasing function that maps $(0,1) to (0,1/2)$. Hence one can find many nontrivial convergent solutions this way: for $0<r<1$,
$$ T(ar^x) = aleft(fracr1-r + fracrlogr right), $$
and one can find $a,r$ so that the right-hand side is $gamma$ provided that $a>2gamma$, and any $a$ gives exactly one such $r$.
answered Jul 15 at 0:24
Chappers
55k74191
answered Jul 15 at 0:24
Chappers
55k74191
answered Jul 15 at 0:24
Chappers
55k74191
answered Jul 15 at 0:24
Chappers
55k74191
55k74191
But a and r are not closed form. They are functions of eulers gamma itself that do not reduce to standard forms. The Floor idea is ok but not analytic and trivial. +1 but No accept.
– mick
Jul 17 at 5:36
add a comment |Â
But a and r are not closed form. They are functions of eulers gamma itself that do not reduce to standard forms. The Floor idea is ok but not analytic and trivial. +1 but No accept.
– mick
Jul 17 at 5:36
But a and r are not closed form. They are functions of eulers gamma itself that do not reduce to standard forms. The Floor idea is ok but not analytic and trivial. +1 but No accept.
– mick
Jul 17 at 5:36
But a and r are not closed form. They are functions of eulers gamma itself that do not reduce to standard forms. The Floor idea is ok but not analytic and trivial. +1 but No accept.
– mick
Jul 17 at 5:36
add a comment |Â
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$$ |sum-int| leq max(f(1), f(t)) $$
– Vladislav Kharlamov
Jul 14 at 23:27
Yes that is true. So that shows ... Existence ??
– mick
Jul 14 at 23:34