Prove or disprove: $frac12-frac{psi _2^{(1)}(n+2)}{log 4}$ have infinite zeros
The notation of $psi$ means QPolyGamma symbol.
I noticed an interesting thing:
zeros that computed by Mathematica
It seems that it has a lot of zeros;
And I tried to plot the function out:
Plot of the function
From the plot, we can deduce that only on the right side (or the left side) there would be zeros.
Next, I calculated the derivative of this function: $displaystyle -frac{psi _2^{(2)}(n+2)}{log 4}$
However, the derivative of the second one is $displaystyle -frac{psi _2^{(3)}(n+2)}{log 4}$. And finally this is not always $>0$, so $displaystyle -frac{psi _2^{(2)}(n+2)}{log 4}$ may have infinite sign changes.
This seems far from proving my original question, thus I didn't find any useful references to my question.
So that does $displaystyle frac{1}{2}-frac{psi _2^{(1)}(n+2)}{log 4}$ have infinite zeros?
real-analysis complex-analysis special-functions
edited Nov 26 at 19:36
Martin Sleziak
44.6k7115270
asked Nov 18 at 12:51
Sky of war
84
add a comment |
The notation of $psi$ means QPolyGamma symbol.
I noticed an interesting thing:
zeros that computed by Mathematica
It seems that it has a lot of zeros;
And I tried to plot the function out:
Plot of the function
From the plot, we can deduce that only on the right side (or the left side) there would be zeros.
Next, I calculated the derivative of this function: $displaystyle -frac{psi _2^{(2)}(n+2)}{log 4}$
However, the derivative of the second one is $displaystyle -frac{psi _2^{(3)}(n+2)}{log 4}$. And finally this is not always $>0$, so $displaystyle -frac{psi _2^{(2)}(n+2)}{log 4}$ may have infinite sign changes.
This seems far from proving my original question, thus I didn't find any useful references to my question.
So that does $displaystyle frac{1}{2}-frac{psi _2^{(1)}(n+2)}{log 4}$ have infinite zeros?
real-analysis complex-analysis special-functions
edited Nov 26 at 19:36
Martin Sleziak
44.6k7115270
asked Nov 18 at 12:51
Sky of war
84
add a comment |
The notation of $psi$ means QPolyGamma symbol.
I noticed an interesting thing:
zeros that computed by Mathematica
It seems that it has a lot of zeros;
And I tried to plot the function out:
Plot of the function
From the plot, we can deduce that only on the right side (or the left side) there would be zeros.
Next, I calculated the derivative of this function: $displaystyle -frac{psi _2^{(2)}(n+2)}{log 4}$
However, the derivative of the second one is $displaystyle -frac{psi _2^{(3)}(n+2)}{log 4}$. And finally this is not always $>0$, so $displaystyle -frac{psi _2^{(2)}(n+2)}{log 4}$ may have infinite sign changes.
This seems far from proving my original question, thus I didn't find any useful references to my question.
So that does $displaystyle frac{1}{2}-frac{psi _2^{(1)}(n+2)}{log 4}$ have infinite zeros?
real-analysis complex-analysis special-functions
edited Nov 26 at 19:36
Martin Sleziak
44.6k7115270
asked Nov 18 at 12:51
Sky of war
84
The notation of $psi$ means QPolyGamma symbol.
I noticed an interesting thing:
zeros that computed by Mathematica
It seems that it has a lot of zeros;
And I tried to plot the function out:
Plot of the function
From the plot, we can deduce that only on the right side (or the left side) there would be zeros.
Next, I calculated the derivative of this function: $displaystyle -frac{psi _2^{(2)}(n+2)}{log 4}$
However, the derivative of the second one is $displaystyle -frac{psi _2^{(3)}(n+2)}{log 4}$. And finally this is not always $>0$, so $displaystyle -frac{psi _2^{(2)}(n+2)}{log 4}$ may have infinite sign changes.
This seems far from proving my original question, thus I didn't find any useful references to my question.
So that does $displaystyle frac{1}{2}-frac{psi _2^{(1)}(n+2)}{log 4}$ have infinite zeros?
real-analysis complex-analysis special-functions
real-analysis complex-analysis special-functions
edited Nov 26 at 19:36
Martin Sleziak
44.6k7115270
asked Nov 18 at 12:51
Sky of war
84
edited Nov 26 at 19:36
Martin Sleziak
44.6k7115270
asked Nov 18 at 12:51
Sky of war
84
edited Nov 26 at 19:36
Martin Sleziak
44.6k7115270
edited Nov 26 at 19:36
Martin Sleziak
44.6k7115270
edited Nov 26 at 19:36
Martin Sleziak
44.6k7115270
44.6k7115270
asked Nov 18 at 12:51
Sky of war
84
asked Nov 18 at 12:51
Sky of war
84
asked Nov 18 at 12:51
Sky of war
84
84
add a comment |
add a comment |
1 Answer
1
active
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Write the $q$-polygamma function as a sum:
$$Gamma_2(z) = 2^{(z - 1) (z - 2)/2} hspace{1px}Gamma_{1/2}(z), \
psi_2(z) = frac 1 {Gamma_2(z)} frac d {dz} Gamma_2(z) =
psi_{1/2}(z) + z ln 2 - frac {3 ln 2} 2 =
-ln 2 left(
frac 1 2 - z + sum_{k geq 0} frac 1 {2^{k + z} - 1} right), \
frac 1 2 - frac 1 {2 ln 2} frac d {dz} psi_2(z) =
-frac {ln 2} 2 sum_{k geq 0} frac {2^{k + z}} {(2^{k + z} - 1)^2}.$$
Setting $z = n + 2$ gives your function. It's singular at $n = -2, -3, dots$ and negative everywhere else.
answered Nov 18 at 18:09
Maxim
4,4931219
add a comment |
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1 Answer
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active
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votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
Write the $q$-polygamma function as a sum:
$$Gamma_2(z) = 2^{(z - 1) (z - 2)/2} hspace{1px}Gamma_{1/2}(z), \
psi_2(z) = frac 1 {Gamma_2(z)} frac d {dz} Gamma_2(z) =
psi_{1/2}(z) + z ln 2 - frac {3 ln 2} 2 =
-ln 2 left(
frac 1 2 - z + sum_{k geq 0} frac 1 {2^{k + z} - 1} right), \
frac 1 2 - frac 1 {2 ln 2} frac d {dz} psi_2(z) =
-frac {ln 2} 2 sum_{k geq 0} frac {2^{k + z}} {(2^{k + z} - 1)^2}.$$
Setting $z = n + 2$ gives your function. It's singular at $n = -2, -3, dots$ and negative everywhere else.
answered Nov 18 at 18:09
Maxim
4,4931219
add a comment |
Write the $q$-polygamma function as a sum:
$$Gamma_2(z) = 2^{(z - 1) (z - 2)/2} hspace{1px}Gamma_{1/2}(z), \
psi_2(z) = frac 1 {Gamma_2(z)} frac d {dz} Gamma_2(z) =
psi_{1/2}(z) + z ln 2 - frac {3 ln 2} 2 =
-ln 2 left(
frac 1 2 - z + sum_{k geq 0} frac 1 {2^{k + z} - 1} right), \
frac 1 2 - frac 1 {2 ln 2} frac d {dz} psi_2(z) =
-frac {ln 2} 2 sum_{k geq 0} frac {2^{k + z}} {(2^{k + z} - 1)^2}.$$
Setting $z = n + 2$ gives your function. It's singular at $n = -2, -3, dots$ and negative everywhere else.
answered Nov 18 at 18:09
Maxim
4,4931219
add a comment |
Write the $q$-polygamma function as a sum:
$$Gamma_2(z) = 2^{(z - 1) (z - 2)/2} hspace{1px}Gamma_{1/2}(z), \
psi_2(z) = frac 1 {Gamma_2(z)} frac d {dz} Gamma_2(z) =
psi_{1/2}(z) + z ln 2 - frac {3 ln 2} 2 =
-ln 2 left(
frac 1 2 - z + sum_{k geq 0} frac 1 {2^{k + z} - 1} right), \
frac 1 2 - frac 1 {2 ln 2} frac d {dz} psi_2(z) =
-frac {ln 2} 2 sum_{k geq 0} frac {2^{k + z}} {(2^{k + z} - 1)^2}.$$
Setting $z = n + 2$ gives your function. It's singular at $n = -2, -3, dots$ and negative everywhere else.
answered Nov 18 at 18:09
Maxim
4,4931219
Write the $q$-polygamma function as a sum:
$$Gamma_2(z) = 2^{(z - 1) (z - 2)/2} hspace{1px}Gamma_{1/2}(z), \
psi_2(z) = frac 1 {Gamma_2(z)} frac d {dz} Gamma_2(z) =
psi_{1/2}(z) + z ln 2 - frac {3 ln 2} 2 =
-ln 2 left(
frac 1 2 - z + sum_{k geq 0} frac 1 {2^{k + z} - 1} right), \
frac 1 2 - frac 1 {2 ln 2} frac d {dz} psi_2(z) =
-frac {ln 2} 2 sum_{k geq 0} frac {2^{k + z}} {(2^{k + z} - 1)^2}.$$
Setting $z = n + 2$ gives your function. It's singular at $n = -2, -3, dots$ and negative everywhere else.
answered Nov 18 at 18:09
Maxim
4,4931219
answered Nov 18 at 18:09
Maxim
4,4931219
answered Nov 18 at 18:09
Maxim
4,4931219
answered Nov 18 at 18:09
Maxim
4,4931219
4,4931219
add a comment |
add a comment |
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