Mixing
How can I solve for $a$ in $frac {k-frac 32}a+frac 1{2a^2}+digamma(a)-digamma(a+n)=0$, where $digamma$ is the...
0
$begingroup$
How can I solve for $a$ for this following equation?
$$frac {k-frac 32}a+frac 1{2a^2}+digamma(a)-digamma(a+n)=0,$$
where $digamma$ represents the digamma function, i.e. it is defined as the logarithmic derivative of the gamma function.
digamma-function
edited Jan 3 at 5:35
Blue
47.8k870152
asked Jan 3 at 3:19
userFarkilluserFarkill
61
$endgroup$
|
show 1 more comment
0
$begingroup$
How can I solve for $a$ for this following equation?
$$frac {k-frac 32}a+frac 1{2a^2}+digamma(a)-digamma(a+n)=0,$$
where $digamma$ represents the digamma function, i.e. it is defined as the logarithmic derivative of the gamma function.
digamma-function
edited Jan 3 at 5:35
Blue
47.8k870152
asked Jan 3 at 3:19
userFarkilluserFarkill
61
$endgroup$
1
$begingroup$
That's not an equation...
$endgroup$
– LordVader007
Jan 3 at 3:21
$begingroup$
forget to include a =0
$endgroup$
– userFarkill
Jan 3 at 3:24
$begingroup$
What is $k$? A constant?
$endgroup$
– Test123
Jan 3 at 5:44
$begingroup$
yes. k and n are just a constant
$endgroup$
– userFarkill
Jan 3 at 5:56
$begingroup$
You could first try using the recurrence relation of the digamma function which will give you: $F(a+n)-F(a)=frac{1}{a}+dots + frac{1}{a+n-1}$.
$endgroup$
– Test123
Jan 3 at 6:32
|
show 1 more comment
0
0
0
$begingroup$
How can I solve for $a$ for this following equation?
$$frac {k-frac 32}a+frac 1{2a^2}+digamma(a)-digamma(a+n)=0,$$
where $digamma$ represents the digamma function, i.e. it is defined as the logarithmic derivative of the gamma function.
digamma-function
edited Jan 3 at 5:35
Blue
47.8k870152
asked Jan 3 at 3:19
userFarkilluserFarkill
61
$endgroup$
How can I solve for $a$ for this following equation?
$$frac {k-frac 32}a+frac 1{2a^2}+digamma(a)-digamma(a+n)=0,$$
where $digamma$ represents the digamma function, i.e. it is defined as the logarithmic derivative of the gamma function.
digamma-function
digamma-function
edited Jan 3 at 5:35
Blue
47.8k870152
asked Jan 3 at 3:19
userFarkilluserFarkill
61
edited Jan 3 at 5:35
Blue
47.8k870152
asked Jan 3 at 3:19
userFarkilluserFarkill
61
edited Jan 3 at 5:35
Blue
47.8k870152
edited Jan 3 at 5:35
Blue
47.8k870152
edited Jan 3 at 5:35
Blue
47.8k870152
47.8k870152
asked Jan 3 at 3:19
userFarkilluserFarkill
61
asked Jan 3 at 3:19
userFarkilluserFarkill
61
asked Jan 3 at 3:19
userFarkilluserFarkill
61
61
1
$begingroup$
That's not an equation...
$endgroup$
– LordVader007
Jan 3 at 3:21
$begingroup$
forget to include a =0
$endgroup$
– userFarkill
Jan 3 at 3:24
$begingroup$
What is $k$? A constant?
$endgroup$
– Test123
Jan 3 at 5:44
$begingroup$
yes. k and n are just a constant
$endgroup$
– userFarkill
Jan 3 at 5:56
$begingroup$
You could first try using the recurrence relation of the digamma function which will give you: $F(a+n)-F(a)=frac{1}{a}+dots + frac{1}{a+n-1}$.
$endgroup$
– Test123
Jan 3 at 6:32
|
show 1 more comment
1
$begingroup$
That's not an equation...
$endgroup$
– LordVader007
Jan 3 at 3:21
$begingroup$
forget to include a =0
$endgroup$
– userFarkill
Jan 3 at 3:24
$begingroup$
What is $k$? A constant?
$endgroup$
– Test123
Jan 3 at 5:44
$begingroup$
yes. k and n are just a constant
$endgroup$
– userFarkill
Jan 3 at 5:56
$begingroup$
You could first try using the recurrence relation of the digamma function which will give you: $F(a+n)-F(a)=frac{1}{a}+dots + frac{1}{a+n-1}$.
$endgroup$
– Test123
Jan 3 at 6:32
1
1
$begingroup$
That's not an equation...
$endgroup$
– LordVader007
Jan 3 at 3:21
$begingroup$
That's not an equation...
$endgroup$
– LordVader007
Jan 3 at 3:21
$begingroup$
forget to include a =0
$endgroup$
– userFarkill
Jan 3 at 3:24
$begingroup$
forget to include a =0
$endgroup$
– userFarkill
Jan 3 at 3:24
$begingroup$
What is $k$? A constant?
$endgroup$
– Test123
Jan 3 at 5:44
$begingroup$
What is $k$? A constant?
$endgroup$
– Test123
Jan 3 at 5:44
$begingroup$
yes. k and n are just a constant
$endgroup$
– userFarkill
Jan 3 at 5:56
$begingroup$
yes. k and n are just a constant
$endgroup$
– userFarkill
Jan 3 at 5:56
$begingroup$
You could first try using the recurrence relation of the digamma function which will give you: $F(a+n)-F(a)=frac{1}{a}+dots + frac{1}{a+n-1}$.
$endgroup$
– Test123
Jan 3 at 6:32
$begingroup$
You could first try using the recurrence relation of the digamma function which will give you: $F(a+n)-F(a)=frac{1}{a}+dots + frac{1}{a+n-1}$.
$endgroup$
– Test123
Jan 3 at 6:32
|
show 1 more comment
0
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1
$begingroup$
That's not an equation...
$endgroup$
– LordVader007
Jan 3 at 3:21
$begingroup$
forget to include a =0
$endgroup$
– userFarkill
Jan 3 at 3:24
$begingroup$
What is $k$? A constant?
$endgroup$
– Test123
Jan 3 at 5:44
$begingroup$
yes. k and n are just a constant
$endgroup$
– userFarkill
Jan 3 at 5:56
$begingroup$
You could first try using the recurrence relation of the digamma function which will give you: $F(a+n)-F(a)=frac{1}{a}+dots + frac{1}{a+n-1}$.
$endgroup$
– Test123
Jan 3 at 6:32