3.10

DLMF	Formula	Constraints	Maple	Mathematica	Symbolic Maple	Symbolic Mathematica	Numeric Maple	Numeric Mathematica
3.10#Ex1	${\displaystyle{\displaystyle b_{0}=u_{0}}}$ b_{0} = u_{0}		b[0] = u[0]	Subscript[b, 0] == Subscript[u, 0]	Skipped - no semantic math	Skipped - no semantic math	-	-
3.10#Ex2	${\displaystyle{\displaystyle b_{1}=1}}$ b_{1} = 1		b[1] = 1	Subscript[b, 1] == 1	Skipped - no semantic math	Skipped - no semantic math	-	-
3.10#Ex3	${\displaystyle{\displaystyle a_{1}=u_{1}}}$ a_{1} = u_{1}		a[1] = u[1]	Subscript[a, 1] == Subscript[u, 1]	Skipped - no semantic math	Skipped - no semantic math	-	-
3.10#Ex4	${\displaystyle{\displaystyle b_{n}=1+\frac{u_{n}}{u_{n-1}}}}$ b_{n} = 1+\frac{u_{n}}{u_{n-1}}		b[n] = 1 +(u[n])/(u[n - 1])	Subscript[b, n] == 1 +Divide[Subscript[u, n],Subscript[u, n - 1]]	Skipped - no semantic math	Skipped - no semantic math	-	-
3.10#Ex5	${\displaystyle{\displaystyle a_{n}=-\frac{u_{n}}{u_{n-1}}}}$ a_{n} = -\frac{u_{n}}{u_{n-1}}	${\displaystyle{\displaystyle n\geq 2}}$	a[n] = -(u[n])/(u[n - 1])	Subscript[a, n] == -Divide[Subscript[u, n],Subscript[u, n - 1]]	Skipped - no semantic math	Skipped - no semantic math	-	-
3.10#Ex6	${\displaystyle{\displaystyle e_{0}^{n}=0}}$ e_{0}^{n} = 0		(exp(1)[0])^(n) = 0	(Subscript[E, 0])^(n) == 0	Skipped - no semantic math	Skipped - no semantic math	-	-
3.10#Ex7	${\displaystyle{\displaystyle q_{1}^{n}=c_{n+1}/c_{n}}}$ q_{1}^{n} = c_{n+1}/c_{n}		(q[1])^(n) = c[n + 1]/c[n]	(Subscript[q, 1])^(n) == Subscript[c, n + 1]/Subscript[c, n]	Skipped - no semantic math	Skipped - no semantic math	-	-
3.10#Ex8	${\displaystyle{\displaystyle e_{j}^{k}=e_{j-1}^{k+1}+q_{j}^{k+1}-q_{j}^{k}}}$ e_{j}^{k} = e_{j-1}^{k+1}+q_{j}^{k+1}-q_{j}^{k}	${\displaystyle{\displaystyle j\geq 1,k\geq 0}}$	(exp(1)[j])^(k) = (exp(1)[j - 1])^(k + 1)+ (q[j])^(k + 1)- (q[j])^(k)	(Subscript[E, j])^(k) == (Subscript[E, j - 1])^(k + 1)+ (Subscript[q, j])^(k + 1)- (Subscript[q, j])^(k)	Skipped - no semantic math	Skipped - no semantic math	-	-
3.10#Ex9	${\displaystyle{\displaystyle q_{j+1}^{k}=q_{j}^{k+1}e_{j}^{k+1}/e_{j}^{k}}}$ q_{j+1}^{k} = q_{j}^{k+1}e_{j}^{k+1}/e_{j}^{k}	${\displaystyle{\displaystyle j\geq 1,k\geq 0}}$	(q[j + 1])^(k) = (q[j])^(k + 1)*(exp(1)[j])^(k + 1)/(exp(1)[j])^(k)	(Subscript[q, j + 1])^(k) == (Subscript[q, j])^(k + 1)*(Subscript[E, j])^(k + 1)/(Subscript[E, j])^(k)	Skipped - no semantic math	Skipped - no semantic math	-	-
3.10#Ex10	${\displaystyle{\displaystyle a_{0}=c_{0}}}$ a_{0} = c_{0}		a[0] = c[0]	Subscript[a, 0] == Subscript[c, 0]	Skipped - no semantic math	Skipped - no semantic math	-	-
3.10#Ex11	${\displaystyle{\displaystyle a_{1}=q_{1}^{0}}}$ a_{1} = q_{1}^{0}		a[1] = (q[1])^(0)	Subscript[a, 1] == (Subscript[q, 1])^(0)	Skipped - no semantic math	Skipped - no semantic math	-	-
3.10#Ex12	${\displaystyle{\displaystyle a_{2}=e_{1}^{0}}}$ a_{2} = e_{1}^{0}		a[2] = (exp(1)[1])^(0)	Subscript[a, 2] == (Subscript[E, 1])^(0)	Skipped - no semantic math	Skipped - no semantic math	-	-
3.10#Ex13	${\displaystyle{\displaystyle a_{3}=q_{2}^{0}}}$ a_{3} = q_{2}^{0}		a[3] = (q[2])^(0)	Subscript[a, 3] == (Subscript[q, 2])^(0)	Skipped - no semantic math	Skipped - no semantic math	-	-
3.10#Ex14	${\displaystyle{\displaystyle a_{4}=e_{2}^{0}}}$ a_{4} = e_{2}^{0}		a[4] = (exp(1)[2])^(0)	Subscript[a, 4] == (Subscript[E, 2])^(0)	Skipped - no semantic math	Skipped - no semantic math	-	-
3.10#Ex16	${\displaystyle{\displaystyle u_{n}=b_{n}}}$ u_{n} = b_{n}		u[n] = b[n]	Subscript[u, n] == Subscript[b, n]	Skipped - no semantic math	Skipped - no semantic math	-	-
3.10#Ex17	${\displaystyle{\displaystyle u_{k}=b_{k}+\frac{a_{k+1}}{u_{k+1}}}}$ u_{k} = b_{k}+\frac{a_{k+1}}{u_{k+1}}	${\displaystyle{\displaystyle k=n-1}}$	u[k] = b[k]+(a[k + 1])/(u[k + 1])	Subscript[u, k] == Subscript[b, k]+Divide[Subscript[a, k + 1],Subscript[u, k + 1]]	Skipped - no semantic math	Skipped - no semantic math	-	-
3.10.E14	${\displaystyle{\displaystyle C=\sum_{k=0}^{\infty}t_{k}}}$ C = \sum_{k=0}^{\infty}t_{k}	${\displaystyle{\displaystyle a_{n}\neq 0}}$	(b[0]+(a[1])/(b[1]+(a[2])/(b[2]+ ..))) = sum(t[k], k = 0..infinity)	(Subscript[b, 0]+Divide[Subscript[a, 1],Subscript[b, 1]+Divide[Subscript[a, 2],Subscript[b, 2]+ \[Ellipsis]]]) == Sum[Subscript[t, k], {k, 0, Infinity}, GenerateConditions->None]	Skipped - no semantic math	Skipped - no semantic math	-	-
3.10#Ex18	${\displaystyle{\displaystyle t_{0}=a_{0}}}$ t_{0} = a_{0}		t[0] = a[0]	Subscript[t, 0] == Subscript[a, 0]	Skipped - no semantic math	Skipped - no semantic math	-	-
3.10#Ex19	${\displaystyle{\displaystyle t_{k}=\rho_{k}t_{k-1}}}$ t_{k} = \rho_{k}t_{k-1}		t[k] = rho[k]*t[k - 1]	Subscript[t, k] == Subscript[\[Rho], k]*Subscript[t, k - 1]	Skipped - no semantic math	Skipped - no semantic math	-	-
3.10#Ex20	${\displaystyle{\displaystyle\rho_{0}=0}}$ \rho_{0} = 0		rho[0] = 0	Subscript[\[Rho], 0] == 0	Skipped - no semantic math	Skipped - no semantic math	-	-
3.10#Ex21	${\displaystyle{\displaystyle\rho_{k}=\frac{a_{k}(1+\rho_{k-1})}{1-a_{k}(1+\rho_{k-1})}}}$ \rho_{k} = \frac{a_{k}(1+\rho_{k-1})}{1-a_{k}(1+\rho_{k-1})}		rho[k] = (a[k]*(1 + rho[k - 1]))/(1 - a[k]*(1 + rho[k - 1]))	Subscript[\[Rho], k] == Divide[Subscript[a, k]*(1 + Subscript[\[Rho], k - 1]),1 - Subscript[a, k]*(1 + Subscript[\[Rho], k - 1])]	Skipped - no semantic math	Skipped - no semantic math	-	-
3.10#Ex22	${\displaystyle{\displaystyle C_{0}=b_{0}}}$ C_{0} = b_{0}		C[0] = b[0]	Subscript[C, 0] == Subscript[b, 0]	Skipped - no semantic math	Skipped - no semantic math	-	-
3.10#Ex23	${\displaystyle{\displaystyle D_{1}=1/b_{1}}}$ D_{1} = 1/b_{1}		D[1] = 1/b[1]	Subscript[D, 1] == 1/Subscript[b, 1]	Skipped - no semantic math	Skipped - no semantic math	-	-
3.10#Ex26	${\displaystyle{\displaystyle D_{n}=\frac{1}{D_{n-1}a_{n}+b_{n}}}}$ D_{n} = \frac{1}{D_{n-1}a_{n}+b_{n}}		D[n] = (1)/(D[n - 1]*a[n]+ b[n])	Subscript[D, n] == Divide[1,Subscript[D, n - 1]*Subscript[a, n]+ Subscript[b, n]]	Skipped - no semantic math	Skipped - no semantic math	-	-