1.2: Difference between revisions
Jump to navigation
Jump to search
Admin moved page Main Page to Verifying DLMF with Maple and Mathematica |
Admin moved page Main Page to Verifying DLMF with Maple and Mathematica |
||
Line 1: | Line 1: | ||
{{DISPLAYTITLE:Algebraic and Analytic Methods - 1.2 Elementary Algebra}} | |||
<div style="width: 100%; height: 75vh; overflow: auto;"> | <div style="width: 100%; height: 75vh; overflow: auto;"> | ||
{| class="wikitable sortable" style="margin: 0;" | {| class="wikitable sortable" style="margin: 0;" |
Revision as of 16:24, 25 May 2021
DLMF | Formula | Constraints | Maple | Mathematica | Symbolic Maple |
Symbolic Mathematica |
Numeric Maple |
Numeric Mathematica |
---|---|---|---|---|---|---|---|---|
1.2.E1 | Failed to parse (LaTeXML (experimental; uses MathML): Invalid response ("") from server "http://latexml:8080/convert/":): {\displaystyle \binom{n}{k} = \frac{n!}{(n-k)!k!}}
\binom{n}{k} = \frac{n!}{(n-k)!k!} |
Failed to parse (LaTeXML (experimental; uses MathML): Invalid response ("") from server "http://latexml:8080/convert/":): {\displaystyle } | binomial(n,k) = (factorial(n))/(factorial(n - k)*factorial(k))
|
Binomial[n,k] == Divide[(n)!,(n - k)!*(k)!]
|
Successful | Successful | - | Successful [Tested: 9] |
1.2.E1 | Failed to parse (LaTeXML (experimental; uses MathML): Invalid response ("") from server "http://latexml:8080/convert/":): {\displaystyle \frac{n!}{(n-k)!k!} = \binom{n}{n-k}}
\frac{n!}{(n-k)!k!} = \binom{n}{n-k} |
Failed to parse (LaTeXML (experimental; uses MathML): Invalid response ("") from server "http://latexml:8080/convert/":): {\displaystyle } | (factorial(n))/(factorial(n - k)*factorial(k)) = binomial(n,n - k)
|
Divide[(n)!,(n - k)!*(k)!] == Binomial[n,n - k]
|
Successful | Successful | - | Successful [Tested: 9] |
1.2.E6 | Failed to parse (LaTeXML (experimental; uses MathML): Invalid response ("") from server "http://latexml:8080/convert/":): {\displaystyle \frac{(-1)^{k}\Pochhammersym{-z}{k}}{k!} = (-1)^{k}\binom{k-z-1}{k}}
\frac{(-1)^{k}\Pochhammersym{-z}{k}}{k!} = (-1)^{k}\binom{k-z-1}{k} |
Failed to parse (LaTeXML (experimental; uses MathML): Invalid response ("") from server "http://latexml:8080/convert/":): {\displaystyle } | ((- 1)^(k)* pochhammer(- z, k))/(factorial(k)) = (- 1)^(k)*binomial(k - z - 1,k)
|
Divide[(- 1)^(k)* Pochhammer[- z, k],(k)!] == (- 1)^(k)*Binomial[k - z - 1,k]
|
Successful | Successful | - | Successful [Tested: 21] |
1.2.E7 | Failed to parse (LaTeXML (experimental; uses MathML): Invalid response ("") from server "http://latexml:8080/convert/":): {\displaystyle \binom{z+1}{k} = \binom{z}{k}+\binom{z}{k-1}}
\binom{z+1}{k} = \binom{z}{k}+\binom{z}{k-1} |
Failed to parse (LaTeXML (experimental; uses MathML): Invalid response ("") from server "http://latexml:8080/convert/":): {\displaystyle } | binomial(z + 1,k) = binomial(z,k)+binomial(z,k - 1)
|
Binomial[z + 1,k] == Binomial[z,k]+Binomial[z,k - 1]
|
Successful | Successful | - | Successful [Tested: 21] |
1.2.E8 | Failed to parse (LaTeXML (experimental; uses MathML): Invalid response ("") from server "http://latexml:8080/convert/":): {\displaystyle \sum^{m}_{k=0}\binom{z+k}{k} = \binom{z+m+1}{m}}
\sum^{m}_{k=0}\binom{z+k}{k} = \binom{z+m+1}{m} |
Failed to parse (LaTeXML (experimental; uses MathML): Invalid response ("") from server "http://latexml:8080/convert/":): {\displaystyle } | sum(binomial(z + k,k), k = 0..m) = binomial(z + m + 1,m)
|
Sum[Binomial[z + k,k], {k, 0, m}, GenerateConditions->None] == Binomial[z + m + 1,m]
|
Successful | Successful | - | Successful [Tested: 21] |
1.2.E10 | Failed to parse (LaTeXML (experimental; uses MathML): Invalid response ("") from server "http://latexml:8080/convert/":): {\displaystyle na+\tfrac{1}{2}n(n-1)d = \tfrac{1}{2}n(a+\ell)}
na+\tfrac{1}{2}n(n-1)d = \tfrac{1}{2}n(a+\ell) |
Failed to parse (LaTeXML (experimental; uses MathML): Invalid response ("") from server "http://latexml:8080/convert/":): {\displaystyle } | n*a +(1)/(2)*n*(n - 1)*d = (1)/(2)*n*(a + ell) |
n*a +Divide[1,2]*n*(n - 1)*d == Divide[1,2]*n*(a + \[ScriptL]) |
Skipped - no semantic math | Skipped - no semantic math | - | - |
1.2.E22 | Failed to parse (LaTeXML (experimental; uses MathML): Invalid response ("") from server "http://latexml:8080/convert/":): {\displaystyle M(r) = 0}
M(r) = 0 |
Failed to parse (LaTeXML (experimental; uses MathML): Invalid response ("") from server "http://latexml:8080/convert/":): {\displaystyle } | ((p[1]*(a[1])^(r)+ p[2]*(a[2])^(r)+ .. + p[n]*(a[n])^(r))^(1/r)) = 0 |
((Subscript[p, 1]*(Subscript[a, 1])^(r)+ Subscript[p, 2]*(Subscript[a, 2])^(r)+ \[Ellipsis]+ Subscript[p, n]*(Subscript[a, n])^(r))^(1/r)) == 0 |
Skipped - no semantic math | Skipped - no semantic math | - | - |
1.2#Ex1 | Failed to parse (LaTeXML (experimental; uses MathML): Invalid response ("") from server "http://latexml:8080/convert/":): {\displaystyle M(1) = A}
M(1) = A |
Failed to parse (LaTeXML (experimental; uses MathML): Invalid response ("") from server "http://latexml:8080/convert/":): {\displaystyle } | M(1) = ((a[1]+ a[2]+ .. + a[n])/(n)) |
M[1] == (Divide[Subscript[a, 1]+ Subscript[a, 2]+ \[Ellipsis]+ Subscript[a, n],n]) |
Skipped - no semantic math | Skipped - no semantic math | - | - |
1.2#Ex2 | Failed to parse (LaTeXML (experimental; uses MathML): Invalid response ("") from server "http://latexml:8080/convert/":): {\displaystyle M(-1) = H}
M(-1) = H |
Failed to parse (LaTeXML (experimental; uses MathML): Invalid response ("") from server "http://latexml:8080/convert/":): {\displaystyle } | M(- 1) = H |
M[- 1] == H |
Skipped - no semantic math | Skipped - no semantic math | - | - |
1.2.E26 | Failed to parse (LaTeXML (experimental; uses MathML): Invalid response ("") from server "http://latexml:8080/convert/":): {\displaystyle \lim_{r\to 0}M(r) = G}
\lim_{r\to 0}M(r) = G |
Failed to parse (LaTeXML (experimental; uses MathML): Invalid response ("") from server "http://latexml:8080/convert/":): {\displaystyle } | limit((p[1]*(a[1])^(r)+ p[2]*(a[2])^(r)+ .. + p[n]*(a[n])^(r))^(1/r), r = 0) = ((a[1]*a[2] .. a[n])^(1/n)) |
Limit[(Subscript[p, 1]*(Subscript[a, 1])^(r)+ Subscript[p, 2]*(Subscript[a, 2])^(r)+ \[Ellipsis]+ Subscript[p, n]*(Subscript[a, n])^(r))^(1/r), r -> 0, GenerateConditions->None] == ((Subscript[a, 1]*Subscript[a, 2] \[Ellipsis]Subscript[a, n])^(1/n)) |
Skipped - no semantic math | Skipped - no semantic math | - | - |