What is the meaning of “True” in the following result?Question about sum and list in a functionProblem evaluating a complicated integralSumConvergence[((-1)^n)/(Sqrt[n] + (-1)^n), n] returns True in Version 10.2?Infinite Sum: $sum_k (ak^2+bk+c)^-1$Numerical result using summation limitLeave Kronecker deltas unevaluatedReturn analytic expressionComputing numerically infinite sum of some double seriesEfficiently define a function as the numerical result of infinite sumsWhy is the result of a Cesaro regularized sum dependant on a mathematically redundant parameter?
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What is the meaning of “True” in the following result?
Question about sum and list in a functionProblem evaluating a complicated integralSumConvergence[((-1)^n)/(Sqrt[n] + (-1)^n), n] returns True in Version 10.2?Infinite Sum: $sum_k (ak^2+bk+c)^-1$Numerical result using summation limitLeave Kronecker deltas unevaluatedReturn analytic expressionComputing numerically infinite sum of some double seriesEfficiently define a function as the numerical result of infinite sumsWhy is the result of a Cesaro regularized sum dependant on a mathematically redundant parameter?
$begingroup$
When I do the sum
Sum[(a + (b + [Pi] n)^2)^(-1), n, -[Infinity], [Infinity]]
the result reads
$$beginarraycc
{ &
beginarraycc
fraccoth left(sqrta+i bright)+coth left(sqrta-i bright)2 sqrta & arg left(sqrta-i bright)geq -fracpi 2land arg left(sqrta-i bright)leq fracpi 2 \
fraccoth left(sqrta+i bright)2 sqrta+fraccoth left(sqrta-i bright)2 sqrta-frac1sqrta & textTrue \
endarray
\
endarray
$$
What is the meaning of "True" in the second result?
If I get rid of "pi", the result is simply:
$$ frac-pi cot left(pi sqrt-a+pi bright)-pi cot left(pi sqrt-a-pi bright)2 sqrt-a$$
summation
asked 5 hours ago
wonderingwondering
278113
$endgroup$
add a comment |
$begingroup$
When I do the sum
Sum[(a + (b + [Pi] n)^2)^(-1), n, -[Infinity], [Infinity]]
the result reads
$$beginarraycc
{ &
beginarraycc
fraccoth left(sqrta+i bright)+coth left(sqrta-i bright)2 sqrta & arg left(sqrta-i bright)geq -fracpi 2land arg left(sqrta-i bright)leq fracpi 2 \
fraccoth left(sqrta+i bright)2 sqrta+fraccoth left(sqrta-i bright)2 sqrta-frac1sqrta & textTrue \
endarray
\
endarray
$$
What is the meaning of "True" in the second result?
If I get rid of "pi", the result is simply:
$$ frac-pi cot left(pi sqrt-a+pi bright)-pi cot left(pi sqrt-a-pi bright)2 sqrt-a$$
summation
asked 5 hours ago
wonderingwondering
278113
$endgroup$
add a comment |
$begingroup$
When I do the sum
Sum[(a + (b + [Pi] n)^2)^(-1), n, -[Infinity], [Infinity]]
the result reads
$$beginarraycc
{ &
beginarraycc
fraccoth left(sqrta+i bright)+coth left(sqrta-i bright)2 sqrta & arg left(sqrta-i bright)geq -fracpi 2land arg left(sqrta-i bright)leq fracpi 2 \
fraccoth left(sqrta+i bright)2 sqrta+fraccoth left(sqrta-i bright)2 sqrta-frac1sqrta & textTrue \
endarray
\
endarray
$$
What is the meaning of "True" in the second result?
If I get rid of "pi", the result is simply:
$$ frac-pi cot left(pi sqrt-a+pi bright)-pi cot left(pi sqrt-a-pi bright)2 sqrt-a$$
summation
asked 5 hours ago
wonderingwondering
278113
$endgroup$
When I do the sum
Sum[(a + (b + [Pi] n)^2)^(-1), n, -[Infinity], [Infinity]]
the result reads
$$beginarraycc
{ &
beginarraycc
fraccoth left(sqrta+i bright)+coth left(sqrta-i bright)2 sqrta & arg left(sqrta-i bright)geq -fracpi 2land arg left(sqrta-i bright)leq fracpi 2 \
fraccoth left(sqrta+i bright)2 sqrta+fraccoth left(sqrta-i bright)2 sqrta-frac1sqrta & textTrue \
endarray
\
endarray
$$
What is the meaning of "True" in the second result?
If I get rid of "pi", the result is simply:
$$ frac-pi cot left(pi sqrt-a+pi bright)-pi cot left(pi sqrt-a-pi bright)2 sqrt-a$$
summation
summation
asked 5 hours ago
wonderingwondering
278113
asked 5 hours ago
wonderingwondering
278113
asked 5 hours ago
wonderingwondering
278113
asked 5 hours ago
wonderingwondering
278113
asked 5 hours ago
wonderingwondering
278113
278113
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
See Piecewise. It is like the cases environment in $LaTeX$. In native math, one would use "else" instead of True.
By the way, version 12 returns
$$begincases
fraccoth left(sqrta-i bright)+coth left(sqrta+i
bright)2 sqrta & arg left(b+i sqrtaright)geq 0 \
fraccoth left(sqrta-i bright)2 sqrta+fraccoth
left(sqrta+i bright)2 sqrta-frac1sqrta &
textTrue
endcases$$
to be interpreted as
$$begincases
fraccoth left(sqrta-i bright)+coth left(sqrta+i
bright)2 sqrta & arg left(b+i sqrtaright)geq 0 \
fraccoth left(sqrta-i bright)2 sqrta+fraccoth
left(sqrta+i bright)2 sqrta-frac1sqrta &
textelse
endcases$$
answered 5 hours ago
Henrik SchumacherHenrik Schumacher
63.1k587176
$endgroup$
1
$begingroup$
Ok, so it really simply means "otherwise" or "else" or "in all other cases"?
$endgroup$
– wondering
5 hours ago
$begingroup$
Yes. That is correct.
$endgroup$
– Henrik Schumacher
5 hours ago
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
See Piecewise. It is like the cases environment in $LaTeX$. In native math, one would use "else" instead of True.
By the way, version 12 returns
$$begincases
fraccoth left(sqrta-i bright)+coth left(sqrta+i
bright)2 sqrta & arg left(b+i sqrtaright)geq 0 \
fraccoth left(sqrta-i bright)2 sqrta+fraccoth
left(sqrta+i bright)2 sqrta-frac1sqrta &
textTrue
endcases$$
to be interpreted as
$$begincases
fraccoth left(sqrta-i bright)+coth left(sqrta+i
bright)2 sqrta & arg left(b+i sqrtaright)geq 0 \
fraccoth left(sqrta-i bright)2 sqrta+fraccoth
left(sqrta+i bright)2 sqrta-frac1sqrta &
textelse
endcases$$
answered 5 hours ago
Henrik SchumacherHenrik Schumacher
63.1k587176
$endgroup$
1
$begingroup$
Ok, so it really simply means "otherwise" or "else" or "in all other cases"?
$endgroup$
– wondering
5 hours ago
$begingroup$
Yes. That is correct.
$endgroup$
– Henrik Schumacher
5 hours ago
add a comment |
$begingroup$
See Piecewise. It is like the cases environment in $LaTeX$. In native math, one would use "else" instead of True.
By the way, version 12 returns
$$begincases
fraccoth left(sqrta-i bright)+coth left(sqrta+i
bright)2 sqrta & arg left(b+i sqrtaright)geq 0 \
fraccoth left(sqrta-i bright)2 sqrta+fraccoth
left(sqrta+i bright)2 sqrta-frac1sqrta &
textTrue
endcases$$
to be interpreted as
$$begincases
fraccoth left(sqrta-i bright)+coth left(sqrta+i
bright)2 sqrta & arg left(b+i sqrtaright)geq 0 \
fraccoth left(sqrta-i bright)2 sqrta+fraccoth
left(sqrta+i bright)2 sqrta-frac1sqrta &
textelse
endcases$$
answered 5 hours ago
Henrik SchumacherHenrik Schumacher
63.1k587176
$endgroup$
1
$begingroup$
Ok, so it really simply means "otherwise" or "else" or "in all other cases"?
$endgroup$
– wondering
5 hours ago
$begingroup$
Yes. That is correct.
$endgroup$
– Henrik Schumacher
5 hours ago
add a comment |
$begingroup$
See Piecewise. It is like the cases environment in $LaTeX$. In native math, one would use "else" instead of True.
By the way, version 12 returns
$$begincases
fraccoth left(sqrta-i bright)+coth left(sqrta+i
bright)2 sqrta & arg left(b+i sqrtaright)geq 0 \
fraccoth left(sqrta-i bright)2 sqrta+fraccoth
left(sqrta+i bright)2 sqrta-frac1sqrta &
textTrue
endcases$$
to be interpreted as
$$begincases
fraccoth left(sqrta-i bright)+coth left(sqrta+i
bright)2 sqrta & arg left(b+i sqrtaright)geq 0 \
fraccoth left(sqrta-i bright)2 sqrta+fraccoth
left(sqrta+i bright)2 sqrta-frac1sqrta &
textelse
endcases$$
answered 5 hours ago
Henrik SchumacherHenrik Schumacher
63.1k587176
$endgroup$
See Piecewise. It is like the cases environment in $LaTeX$. In native math, one would use "else" instead of True.
By the way, version 12 returns
$$begincases
fraccoth left(sqrta-i bright)+coth left(sqrta+i
bright)2 sqrta & arg left(b+i sqrtaright)geq 0 \
fraccoth left(sqrta-i bright)2 sqrta+fraccoth
left(sqrta+i bright)2 sqrta-frac1sqrta &
textTrue
endcases$$
to be interpreted as
$$begincases
fraccoth left(sqrta-i bright)+coth left(sqrta+i
bright)2 sqrta & arg left(b+i sqrtaright)geq 0 \
fraccoth left(sqrta-i bright)2 sqrta+fraccoth
left(sqrta+i bright)2 sqrta-frac1sqrta &
textelse
endcases$$
answered 5 hours ago
Henrik SchumacherHenrik Schumacher
63.1k587176
answered 5 hours ago
Henrik SchumacherHenrik Schumacher
63.1k587176
answered 5 hours ago
Henrik SchumacherHenrik Schumacher
63.1k587176
answered 5 hours ago
Henrik SchumacherHenrik Schumacher
63.1k587176
63.1k587176
1
$begingroup$
Ok, so it really simply means "otherwise" or "else" or "in all other cases"?
$endgroup$
– wondering
5 hours ago
$begingroup$
Yes. That is correct.
$endgroup$
– Henrik Schumacher
5 hours ago
add a comment |
1
$begingroup$
Ok, so it really simply means "otherwise" or "else" or "in all other cases"?
$endgroup$
– wondering
5 hours ago
$begingroup$
Yes. That is correct.
$endgroup$
– Henrik Schumacher
5 hours ago
1
1
$begingroup$
Ok, so it really simply means "otherwise" or "else" or "in all other cases"?
$endgroup$
– wondering
5 hours ago
$begingroup$
Ok, so it really simply means "otherwise" or "else" or "in all other cases"?
$endgroup$
– wondering
5 hours ago
$begingroup$
Yes. That is correct.
$endgroup$
– Henrik Schumacher
5 hours ago
$begingroup$
Yes. That is correct.
$endgroup$
– Henrik Schumacher
5 hours ago
add a comment |
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