Algebraic and Analytic Methods - 1.2 Elementary Algebra

DLMF	Formula	Constraints	Maple	Mathematica	Symbolic Maple	Symbolic Mathematica	Numeric Maple	Numeric Mathematica
1.2.E1	${\displaystyle{\displaystyle\genfrac{(}{)}{0.0pt}{}{n}{k}=\frac{n!}{(n-k)!k!}}}$ \binom{n}{k} = \frac{n!}{(n-k)!k!}		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	${\displaystyle{\displaystyle\frac{n!}{(n-k)!k!}=\genfrac{(}{)}{0.0pt}{}{n}{n-k}}}$ \frac{n!}{(n-k)!k!} = \binom{n}{n-k}		(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	${\displaystyle{\displaystyle\frac{(-1)^{k}{\left(-z\right)_{k}}}{k!}=(-1)^{k}\genfrac{(}{)}{0.0pt}{}{k-z-1}{k}}}$ \frac{(-1)^{k}\Pochhammersym{-z}{k}}{k!} = (-1)^{k}\binom{k-z-1}{k}		((- 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	${\displaystyle{\displaystyle\genfrac{(}{)}{0.0pt}{}{z+1}{k}=\genfrac{(}{)}{0.0pt}{}{z}{k}+\genfrac{(}{)}{0.0pt}{}{z}{k-1}}}$ \binom{z+1}{k} = \binom{z}{k}+\binom{z}{k-1}		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	${\displaystyle{\displaystyle\sum^{m}_{k=0}\genfrac{(}{)}{0.0pt}{}{z+k}{k}=\genfrac{(}{)}{0.0pt}{}{z+m+1}{m}}}$ \sum^{m}_{k=0}\binom{z+k}{k} = \binom{z+m+1}{m}		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	${\displaystyle{\displaystyle na+\tfrac{1}{2}n(n-1)d=\tfrac{1}{2}n(a+\ell)}}$ na+\tfrac{1}{2}n(n-1)d = \tfrac{1}{2}n(a+\ell)		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	${\displaystyle{\displaystyle M(r)=0}}$ M(r) = 0		((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	${\displaystyle{\displaystyle M(1)=A}}$ M(1) = A		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	${\displaystyle{\displaystyle M(-1)=H}}$ M(-1) = H		M(- 1) = H	M[- 1] == H	Skipped - no semantic math	Skipped - no semantic math	-	-
1.2.E26	${\displaystyle{\displaystyle\lim_{r\to 0}M(r)=G}}$ \lim_{r\to 0}M(r) = G		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	-	-