What is the meaning of True in my result? [closed]What is the principal difference betwen two results of a seemingly similar sums?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?
What setting controls moving the cursor on the command line?
Cascading Switches. Will it affect performance?
Getting UPS Power from One Room to Another
Is this use of the expression "long past" correct?
Can Rydberg constant be in joules?
How can I get an unreasonable manager to approve time off?
Tabular make widths equal
Is it possible to have a wealthy country without a middle class?
What can I, as a user, do about offensive reviews in App Store?
How to trick the reader into thinking they're following a redshirt instead of the protagonist?
Is it a problem if <h4>, <h5> and <h6> are smaller than regular text?
Is it expected that a reader will skip parts of what you write?
How can I end combat quickly when the outcome is inevitable?
Extreme flexible working hours: how to control people and activities?
How to tell your grandparent to not come to fetch you with their car?
Generate basis elements of the Steenrod algebra
Union with anonymous struct with flexible array member
Which languages would be most useful in Europe at the end of the 19th century?
Pre-1972 sci-fi short story or novel: alien(?) tunnel where people try new moves and get destroyed if they're not the correct ones
Why doesn't Adrian Toomes give up Spider-Man's identity?
Why are trash cans referred to as "zafacón" in Puerto Rico?
1980s live-action movie where individually-coloured nations on clouds fight
Wooden cooking layout
How is the size of a distant planet deturmined?
What is the meaning of True in my result? [closed]
What is the principal difference betwen two results of a seemingly similar sums?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 + π n)^2)^(-1), n, -∞, ∞]
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
edited May 23 at 13:13
m_goldberg
90.5k873203
asked May 22 at 18:40
wonderingwondering
295113
$endgroup$
closed as off-topic by Bob Hanlon, Carl Lange, m_goldberg, Daniel Lichtblau, MarcoB May 23 at 14:15
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question arises due to a simple mistake such as a trivial syntax error, incorrect capitalization, spelling mistake, or other typographical error and is unlikely to help any future visitors, or else it is easily found in the documentation." – Bob Hanlon, Carl Lange, m_goldberg, Daniel Lichtblau, MarcoB
add a comment |
$begingroup$
When I do the sum
Sum[(a + (b + π n)^2)^(-1), n, -∞, ∞]
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
edited May 23 at 13:13
m_goldberg
90.5k873203
asked May 22 at 18:40
wonderingwondering
295113
$endgroup$
closed as off-topic by Bob Hanlon, Carl Lange, m_goldberg, Daniel Lichtblau, MarcoB May 23 at 14:15
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question arises due to a simple mistake such as a trivial syntax error, incorrect capitalization, spelling mistake, or other typographical error and is unlikely to help any future visitors, or else it is easily found in the documentation." – Bob Hanlon, Carl Lange, m_goldberg, Daniel Lichtblau, MarcoB
add a comment |
$begingroup$
When I do the sum
Sum[(a + (b + π n)^2)^(-1), n, -∞, ∞]
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
edited May 23 at 13:13
m_goldberg
90.5k873203
asked May 22 at 18:40
wonderingwondering
295113
$endgroup$
When I do the sum
Sum[(a + (b + π n)^2)^(-1), n, -∞, ∞]
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
edited May 23 at 13:13
m_goldberg
90.5k873203
asked May 22 at 18:40
wonderingwondering
295113
edited May 23 at 13:13
m_goldberg
90.5k873203
asked May 22 at 18:40
wonderingwondering
295113
edited May 23 at 13:13
m_goldberg
90.5k873203
edited May 23 at 13:13
m_goldberg
90.5k873203
edited May 23 at 13:13
m_goldberg
90.5k873203
90.5k873203
asked May 22 at 18:40
wonderingwondering
295113
asked May 22 at 18:40
wonderingwondering
295113
asked May 22 at 18:40
wonderingwondering
295113
295113
closed as off-topic by Bob Hanlon, Carl Lange, m_goldberg, Daniel Lichtblau, MarcoB May 23 at 14:15
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question arises due to a simple mistake such as a trivial syntax error, incorrect capitalization, spelling mistake, or other typographical error and is unlikely to help any future visitors, or else it is easily found in the documentation." – Bob Hanlon, Carl Lange, m_goldberg, Daniel Lichtblau, MarcoB
closed as off-topic by Bob Hanlon, Carl Lange, m_goldberg, Daniel Lichtblau, MarcoB May 23 at 14:15
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question arises due to a simple mistake such as a trivial syntax error, incorrect capitalization, spelling mistake, or other typographical error and is unlikely to help any future visitors, or else it is easily found in the documentation." – Bob Hanlon, Carl Lange, m_goldberg, Daniel Lichtblau, MarcoB
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 May 22 at 18:43
Henrik SchumacherHenrik Schumacher
63.8k590178
$endgroup$
1
$begingroup$
Ok, so it really simply means "otherwise" or "else" or "in all other cases"?
$endgroup$
– wondering
May 22 at 18:44
$begingroup$
Yes. That is correct.
$endgroup$
– Henrik Schumacher
May 22 at 18:45
add a comment |
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 May 22 at 18:43
Henrik SchumacherHenrik Schumacher
63.8k590178
$endgroup$
1
$begingroup$
Ok, so it really simply means "otherwise" or "else" or "in all other cases"?
$endgroup$
– wondering
May 22 at 18:44
$begingroup$
Yes. That is correct.
$endgroup$
– Henrik Schumacher
May 22 at 18:45
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 May 22 at 18:43
Henrik SchumacherHenrik Schumacher
63.8k590178
$endgroup$
1
$begingroup$
Ok, so it really simply means "otherwise" or "else" or "in all other cases"?
$endgroup$
– wondering
May 22 at 18:44
$begingroup$
Yes. That is correct.
$endgroup$
– Henrik Schumacher
May 22 at 18:45
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 May 22 at 18:43
Henrik SchumacherHenrik Schumacher
63.8k590178
$endgroup$
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 May 22 at 18:43
Henrik SchumacherHenrik Schumacher
63.8k590178
answered May 22 at 18:43
Henrik SchumacherHenrik Schumacher
63.8k590178
answered May 22 at 18:43
Henrik SchumacherHenrik Schumacher
63.8k590178
answered May 22 at 18:43
Henrik SchumacherHenrik Schumacher
63.8k590178
63.8k590178
1
$begingroup$
Ok, so it really simply means "otherwise" or "else" or "in all other cases"?
$endgroup$
– wondering
May 22 at 18:44
$begingroup$
Yes. That is correct.
$endgroup$
– Henrik Schumacher
May 22 at 18:45
add a comment |
1
$begingroup$
Ok, so it really simply means "otherwise" or "else" or "in all other cases"?
$endgroup$
– wondering
May 22 at 18:44
$begingroup$
Yes. That is correct.
$endgroup$
– Henrik Schumacher
May 22 at 18:45
1
1
$begingroup$
Ok, so it really simply means "otherwise" or "else" or "in all other cases"?
$endgroup$
– wondering
May 22 at 18:44
$begingroup$
Ok, so it really simply means "otherwise" or "else" or "in all other cases"?
$endgroup$
– wondering
May 22 at 18:44
$begingroup$
Yes. That is correct.
$endgroup$
– Henrik Schumacher
May 22 at 18:45
$begingroup$
Yes. That is correct.
$endgroup$
– Henrik Schumacher
May 22 at 18:45
add a comment |
