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Is there a way to know which symbolic expression mathematica used
Given a symbolic expression how to find if starts with a minus or not?(Symbolic) LinearSolve is slower/different after upgrading to Mathematica 9Symbolic matrice or vectors in Mathematica?How to define the symbolic expression $V = frac1N sum_i=1^N ( y_i(g) - z_i)^2$ in Mathematica?Root: Simplifying symbolic expressionHow can I find the formal derivative of a polynomial w.r.t. to its coefficientsSymbolic Integration gives wrong answerSymbolic manipulation of expression with undefined functionIs it possible to get a symbolic a concise expression for my function?Trying to apply Jensen's inequality to a complicated symbolic expression
.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;
$begingroup$
Take the following symbolic sum:
Sum[(-1)^n/Factorial[2*n]*(Pi/2)^(2*n)*n, n, 1, Infinity]
Mathematica answers me: $- fracpi4$
It is great, I have an exact expression for this complicated sum. But I would like to know how mathematica knows this result.
Is there a way to ask him which function he used to compute it ?
symbolic
asked 8 hours ago
StarBucKStarBucK
8653 silver badges13 bronze badges
$endgroup$
add a comment |
$begingroup$
Take the following symbolic sum:
Sum[(-1)^n/Factorial[2*n]*(Pi/2)^(2*n)*n, n, 1, Infinity]
Mathematica answers me: $- fracpi4$
It is great, I have an exact expression for this complicated sum. But I would like to know how mathematica knows this result.
Is there a way to ask him which function he used to compute it ?
symbolic
asked 8 hours ago
StarBucKStarBucK
8653 silver badges13 bronze badges
$endgroup$
add a comment |
$begingroup$
Take the following symbolic sum:
Sum[(-1)^n/Factorial[2*n]*(Pi/2)^(2*n)*n, n, 1, Infinity]
Mathematica answers me: $- fracpi4$
It is great, I have an exact expression for this complicated sum. But I would like to know how mathematica knows this result.
Is there a way to ask him which function he used to compute it ?
symbolic
asked 8 hours ago
StarBucKStarBucK
8653 silver badges13 bronze badges
$endgroup$
Take the following symbolic sum:
Sum[(-1)^n/Factorial[2*n]*(Pi/2)^(2*n)*n, n, 1, Infinity]
Mathematica answers me: $- fracpi4$
It is great, I have an exact expression for this complicated sum. But I would like to know how mathematica knows this result.
Is there a way to ask him which function he used to compute it ?
symbolic
symbolic
asked 8 hours ago
StarBucKStarBucK
8653 silver badges13 bronze badges
asked 8 hours ago
StarBucKStarBucK
8653 silver badges13 bronze badges
asked 8 hours ago
StarBucKStarBucK
8653 silver badges13 bronze badges
asked 8 hours ago
StarBucKStarBucK
8653 silver badges13 bronze badges
asked 8 hours ago
StarBucKStarBucK
8653 silver badges13 bronze badges
8653 silver badges13 bronze badges
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
You can use a Wolfram|Alpha query and then look at the Series Representation results.
In Mathematica, the easiest way to do this is start your input with == and it will change it to a spikey input symbol.
==Sum[(-1)^n/Factorial[2*n]*(Pi/2)^(2*n)*n, n, 1, Infinity]
Result:
answered 5 hours ago
kickertkickert
1,0351 silver badge18 bronze badges
$endgroup$
add a comment |
$begingroup$
In general, symbolic sums are evaluated by using a combination of internal methods and it is usually difficult to give insight into the evaluation process.
However, this sum is evaluated by using a representation in terms of HypergeometricPFQ (strictly speaking, this is a Hypergeometric0F1).
This may be seen by using the (undocumented) Method option setting "InactivePFQ" as shown below.
In[1]:= Sum[(-1)^n/Factorial[2*n]*(Pi/2)^(2*n)*n, n, 1, Infinity,
Method -> "InactivePFQ"] // InputForm
Out[1]//InputForm=
-(Pi^2*Inactive[HypergeometricPFQ][, 3/2, -Pi^2/16])/8
In[2]:= Activate[%] // FullSimplify // InputForm
Out[2]//InputForm=
-Pi/4
In[3]:= N[%]
Out[3]= -0.785398
In[4]:= NSum[(-1)^n/Factorial[2*n]*(Pi/2)^(2*n)*n, n, 1, Infinity]
Out[4]= -0.785398
The representation in terms of HypergeometricPFQ can be understood by defining a sequence corresponding to the first argument of Sum as follows (the sequence has been shifted so that it starts at 0, rather than 1).
a[n_] = (-1)^n/Factorial[2*n]*(Pi/2)^(2*n)*n /. n -> n + 1;
The factor of-Pi^2/8 in the answer from Sum is the zeroth term of this sequence.
In[6]:= a[0] // InputForm
Out[6]//InputForm=
-Pi^2/8
The arguments of HypergeometricPFQ in this case can be understood by computing the ratio of the adjacent terms of a[n].
In[7]:= DiscreteRatio[a[n], n] // InputForm
Out[7]//InputForm=
-Pi^2/(8*(1 + n)*(3 + 2*n))
In[8]:=
Simplify[% == -(Pi^2/16)*(1/((n + 1)*(n + 3/2)))]
Out[8]= True
The factor of (n+1) in the denominator of the above comes from the definition of Hypergoemetric0F1 and can be ignored.
Hope this helps in understanding the result from Sum.
edited 3 hours ago
Community♦
1
answered 3 hours ago
Devendra KapadiaDevendra Kapadia
1,1596 silver badges9 bronze badges
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
You can use a Wolfram|Alpha query and then look at the Series Representation results.
In Mathematica, the easiest way to do this is start your input with == and it will change it to a spikey input symbol.
==Sum[(-1)^n/Factorial[2*n]*(Pi/2)^(2*n)*n, n, 1, Infinity]
Result:
answered 5 hours ago
kickertkickert
1,0351 silver badge18 bronze badges
$endgroup$
add a comment |
$begingroup$
You can use a Wolfram|Alpha query and then look at the Series Representation results.
In Mathematica, the easiest way to do this is start your input with == and it will change it to a spikey input symbol.
==Sum[(-1)^n/Factorial[2*n]*(Pi/2)^(2*n)*n, n, 1, Infinity]
Result:
answered 5 hours ago
kickertkickert
1,0351 silver badge18 bronze badges
$endgroup$
add a comment |
$begingroup$
You can use a Wolfram|Alpha query and then look at the Series Representation results.
In Mathematica, the easiest way to do this is start your input with == and it will change it to a spikey input symbol.
==Sum[(-1)^n/Factorial[2*n]*(Pi/2)^(2*n)*n, n, 1, Infinity]
Result:
answered 5 hours ago
kickertkickert
1,0351 silver badge18 bronze badges
$endgroup$
You can use a Wolfram|Alpha query and then look at the Series Representation results.
In Mathematica, the easiest way to do this is start your input with == and it will change it to a spikey input symbol.
==Sum[(-1)^n/Factorial[2*n]*(Pi/2)^(2*n)*n, n, 1, Infinity]
Result:
answered 5 hours ago
kickertkickert
1,0351 silver badge18 bronze badges
answered 5 hours ago
kickertkickert
1,0351 silver badge18 bronze badges
answered 5 hours ago
kickertkickert
1,0351 silver badge18 bronze badges
answered 5 hours ago
kickertkickert
1,0351 silver badge18 bronze badges
1,0351 silver badge18 bronze badges
add a comment |
add a comment |
$begingroup$
In general, symbolic sums are evaluated by using a combination of internal methods and it is usually difficult to give insight into the evaluation process.
However, this sum is evaluated by using a representation in terms of HypergeometricPFQ (strictly speaking, this is a Hypergeometric0F1).
This may be seen by using the (undocumented) Method option setting "InactivePFQ" as shown below.
In[1]:= Sum[(-1)^n/Factorial[2*n]*(Pi/2)^(2*n)*n, n, 1, Infinity,
Method -> "InactivePFQ"] // InputForm
Out[1]//InputForm=
-(Pi^2*Inactive[HypergeometricPFQ][, 3/2, -Pi^2/16])/8
In[2]:= Activate[%] // FullSimplify // InputForm
Out[2]//InputForm=
-Pi/4
In[3]:= N[%]
Out[3]= -0.785398
In[4]:= NSum[(-1)^n/Factorial[2*n]*(Pi/2)^(2*n)*n, n, 1, Infinity]
Out[4]= -0.785398
The representation in terms of HypergeometricPFQ can be understood by defining a sequence corresponding to the first argument of Sum as follows (the sequence has been shifted so that it starts at 0, rather than 1).
a[n_] = (-1)^n/Factorial[2*n]*(Pi/2)^(2*n)*n /. n -> n + 1;
The factor of-Pi^2/8 in the answer from Sum is the zeroth term of this sequence.
In[6]:= a[0] // InputForm
Out[6]//InputForm=
-Pi^2/8
The arguments of HypergeometricPFQ in this case can be understood by computing the ratio of the adjacent terms of a[n].
In[7]:= DiscreteRatio[a[n], n] // InputForm
Out[7]//InputForm=
-Pi^2/(8*(1 + n)*(3 + 2*n))
In[8]:=
Simplify[% == -(Pi^2/16)*(1/((n + 1)*(n + 3/2)))]
Out[8]= True
The factor of (n+1) in the denominator of the above comes from the definition of Hypergoemetric0F1 and can be ignored.
Hope this helps in understanding the result from Sum.
edited 3 hours ago
Community♦
1
answered 3 hours ago
Devendra KapadiaDevendra Kapadia
1,1596 silver badges9 bronze badges
$endgroup$
add a comment |
$begingroup$
In general, symbolic sums are evaluated by using a combination of internal methods and it is usually difficult to give insight into the evaluation process.
However, this sum is evaluated by using a representation in terms of HypergeometricPFQ (strictly speaking, this is a Hypergeometric0F1).
This may be seen by using the (undocumented) Method option setting "InactivePFQ" as shown below.
In[1]:= Sum[(-1)^n/Factorial[2*n]*(Pi/2)^(2*n)*n, n, 1, Infinity,
Method -> "InactivePFQ"] // InputForm
Out[1]//InputForm=
-(Pi^2*Inactive[HypergeometricPFQ][, 3/2, -Pi^2/16])/8
In[2]:= Activate[%] // FullSimplify // InputForm
Out[2]//InputForm=
-Pi/4
In[3]:= N[%]
Out[3]= -0.785398
In[4]:= NSum[(-1)^n/Factorial[2*n]*(Pi/2)^(2*n)*n, n, 1, Infinity]
Out[4]= -0.785398
The representation in terms of HypergeometricPFQ can be understood by defining a sequence corresponding to the first argument of Sum as follows (the sequence has been shifted so that it starts at 0, rather than 1).
a[n_] = (-1)^n/Factorial[2*n]*(Pi/2)^(2*n)*n /. n -> n + 1;
The factor of-Pi^2/8 in the answer from Sum is the zeroth term of this sequence.
In[6]:= a[0] // InputForm
Out[6]//InputForm=
-Pi^2/8
The arguments of HypergeometricPFQ in this case can be understood by computing the ratio of the adjacent terms of a[n].
In[7]:= DiscreteRatio[a[n], n] // InputForm
Out[7]//InputForm=
-Pi^2/(8*(1 + n)*(3 + 2*n))
In[8]:=
Simplify[% == -(Pi^2/16)*(1/((n + 1)*(n + 3/2)))]
Out[8]= True
The factor of (n+1) in the denominator of the above comes from the definition of Hypergoemetric0F1 and can be ignored.
Hope this helps in understanding the result from Sum.
edited 3 hours ago
Community♦
1
answered 3 hours ago
Devendra KapadiaDevendra Kapadia
1,1596 silver badges9 bronze badges
$endgroup$
add a comment |
$begingroup$
In general, symbolic sums are evaluated by using a combination of internal methods and it is usually difficult to give insight into the evaluation process.
However, this sum is evaluated by using a representation in terms of HypergeometricPFQ (strictly speaking, this is a Hypergeometric0F1).
This may be seen by using the (undocumented) Method option setting "InactivePFQ" as shown below.
In[1]:= Sum[(-1)^n/Factorial[2*n]*(Pi/2)^(2*n)*n, n, 1, Infinity,
Method -> "InactivePFQ"] // InputForm
Out[1]//InputForm=
-(Pi^2*Inactive[HypergeometricPFQ][, 3/2, -Pi^2/16])/8
In[2]:= Activate[%] // FullSimplify // InputForm
Out[2]//InputForm=
-Pi/4
In[3]:= N[%]
Out[3]= -0.785398
In[4]:= NSum[(-1)^n/Factorial[2*n]*(Pi/2)^(2*n)*n, n, 1, Infinity]
Out[4]= -0.785398
The representation in terms of HypergeometricPFQ can be understood by defining a sequence corresponding to the first argument of Sum as follows (the sequence has been shifted so that it starts at 0, rather than 1).
a[n_] = (-1)^n/Factorial[2*n]*(Pi/2)^(2*n)*n /. n -> n + 1;
The factor of-Pi^2/8 in the answer from Sum is the zeroth term of this sequence.
In[6]:= a[0] // InputForm
Out[6]//InputForm=
-Pi^2/8
The arguments of HypergeometricPFQ in this case can be understood by computing the ratio of the adjacent terms of a[n].
In[7]:= DiscreteRatio[a[n], n] // InputForm
Out[7]//InputForm=
-Pi^2/(8*(1 + n)*(3 + 2*n))
In[8]:=
Simplify[% == -(Pi^2/16)*(1/((n + 1)*(n + 3/2)))]
Out[8]= True
The factor of (n+1) in the denominator of the above comes from the definition of Hypergoemetric0F1 and can be ignored.
Hope this helps in understanding the result from Sum.
edited 3 hours ago
Community♦
1
answered 3 hours ago
Devendra KapadiaDevendra Kapadia
1,1596 silver badges9 bronze badges
$endgroup$
In general, symbolic sums are evaluated by using a combination of internal methods and it is usually difficult to give insight into the evaluation process.
However, this sum is evaluated by using a representation in terms of HypergeometricPFQ (strictly speaking, this is a Hypergeometric0F1).
This may be seen by using the (undocumented) Method option setting "InactivePFQ" as shown below.
In[1]:= Sum[(-1)^n/Factorial[2*n]*(Pi/2)^(2*n)*n, n, 1, Infinity,
Method -> "InactivePFQ"] // InputForm
Out[1]//InputForm=
-(Pi^2*Inactive[HypergeometricPFQ][, 3/2, -Pi^2/16])/8
In[2]:= Activate[%] // FullSimplify // InputForm
Out[2]//InputForm=
-Pi/4
In[3]:= N[%]
Out[3]= -0.785398
In[4]:= NSum[(-1)^n/Factorial[2*n]*(Pi/2)^(2*n)*n, n, 1, Infinity]
Out[4]= -0.785398
The representation in terms of HypergeometricPFQ can be understood by defining a sequence corresponding to the first argument of Sum as follows (the sequence has been shifted so that it starts at 0, rather than 1).
a[n_] = (-1)^n/Factorial[2*n]*(Pi/2)^(2*n)*n /. n -> n + 1;
The factor of-Pi^2/8 in the answer from Sum is the zeroth term of this sequence.
In[6]:= a[0] // InputForm
Out[6]//InputForm=
-Pi^2/8
The arguments of HypergeometricPFQ in this case can be understood by computing the ratio of the adjacent terms of a[n].
In[7]:= DiscreteRatio[a[n], n] // InputForm
Out[7]//InputForm=
-Pi^2/(8*(1 + n)*(3 + 2*n))
In[8]:=
Simplify[% == -(Pi^2/16)*(1/((n + 1)*(n + 3/2)))]
Out[8]= True
The factor of (n+1) in the denominator of the above comes from the definition of Hypergoemetric0F1 and can be ignored.
Hope this helps in understanding the result from Sum.
edited 3 hours ago
Community♦
1
answered 3 hours ago
Devendra KapadiaDevendra Kapadia
1,1596 silver badges9 bronze badges
edited 3 hours ago
Community♦
1
edited 3 hours ago
Community♦
1
edited 3 hours ago
Community♦
1
1
answered 3 hours ago
Devendra KapadiaDevendra Kapadia
1,1596 silver badges9 bronze badges
answered 3 hours ago
Devendra KapadiaDevendra Kapadia
1,1596 silver badges9 bronze badges
answered 3 hours ago
Devendra KapadiaDevendra Kapadia
1,1596 silver badges9 bronze badges
1,1596 silver badges9 bronze badges
add a comment |
add a comment |
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