Algebraic and Analytic Methods - 1.11 Zeros of Polynomials

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DLMF Formula Constraints Maple Mathematica Symbolic
Maple 		Symbolic
Mathematica 		Numeric
Maple 		Numeric
Mathematica
1.11.E3 b k = Ξ± ⁒ b k + 1 + a k subscript 𝑏 π‘˜ 𝛼 subscript 𝑏 π‘˜ 1 subscript π‘Ž π‘˜ {\displaystyle{\displaystyle b_{k}=\alpha b_{k+1}+a_{k}}}
b_{k} = \alpha b_{k+1}+a_{k}		 k = n - 1 π‘˜ 𝑛 1 {\displaystyle{\displaystyle k=n-1}} 
b[k] = alpha*b[k + 1]+ a[k]
 
Subscript[b, k] == \[Alpha]*Subscript[b, k + 1]+ Subscript[a, k]
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1.11.E5 c k = Ξ± ⁒ c k + 1 + b k subscript 𝑐 π‘˜ 𝛼 subscript 𝑐 π‘˜ 1 subscript 𝑏 π‘˜ {\displaystyle{\displaystyle c_{k}=\alpha c_{k+1}+b_{k}}}
c_{k} = \alpha c_{k+1}+b_{k}		 k = n - 1 π‘˜ 𝑛 1 {\displaystyle{\displaystyle k=n-1}} 
c[k] = alpha*c[k + 1]+ b[k]
 
Subscript[c, k] == \[Alpha]*Subscript[c, k + 1]+ Subscript[b, k]
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1.11#E10Xa βˆ‘ 1 ≀ j < k ≀ n z j ⁒ z k = a n - 2 / a n subscript 1 𝑗 π‘˜ 𝑛 subscript 𝑧 𝑗 subscript 𝑧 π‘˜ subscript π‘Ž 𝑛 2 subscript π‘Ž 𝑛 {\displaystyle{\displaystyle\displaystyle\sum_{1\leq j<k\leq n}z_{j}z_{k}=a_{n% -2}/a_{n}}}
\displaystyle\sum_{1\leq j<k\leq n}z_{j}z_{k} = a_{n-2}/a_{n}		 

sum(sum(z[j]*z[k], k = j + 1..n), j = 1..k - 1) = a[n - 2]/a[n]
 
Sum[Sum[Subscript[z, j]*Subscript[z, k], {k, j + 1, n}, GenerateConditions->None], {j, 1, k - 1}, GenerateConditions->None] == Subscript[a, n - 2]/Subscript[a, n]
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1.11#Ex4 D = b 2 - 4 ⁒ a ⁒ c 𝐷 superscript 𝑏 2 4 π‘Ž 𝑐 {\displaystyle{\displaystyle D=b^{2}-4ac}}
D = b^{2}-4ac		 

(- 4*(p)^(3)- 27*(q)^(2)) = (b)^(2)- 4*a*c
 
(- 4*(p)^(3)- 27*(q)^(2)) == (b)^(2)- 4*a*c
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1.11#Ex10 ρ = - 1 2 + 1 2 ⁒ - 3 𝜌 1 2 1 2 3 {\displaystyle{\displaystyle\rho=-\tfrac{1}{2}+\tfrac{1}{2}\sqrt{-3}}}
\rho = -\tfrac{1}{2}+\tfrac{1}{2}\sqrt{-3}		 

rho = -(1)/(2)+(1)/(2)*sqrt(- 3)
 
\[Rho] == -Divide[1,2]+Divide[1,2]*Sqrt[- 3]
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1.11#Ex11 ρ 2 = e - 2 ⁒ Ο€ ⁒ i / 3 superscript 𝜌 2 superscript 𝑒 2 πœ‹ 𝑖 3 {\displaystyle{\displaystyle\rho^{2}=e^{-2\pi i/3}}}
\rho^{2} = e^{-2\pi i/3}		 

(rho)^(2) = exp(- 2*Pi*I/3)
 
\[Rho]^(2) == Exp[- 2*Pi*I/3]
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1.11#Ex13 p = ( - 3 ⁒ a 2 + 8 ⁒ b ) / 8 𝑝 3 superscript π‘Ž 2 8 𝑏 8 {\displaystyle{\displaystyle p=(-3a^{2}+8b)/8}}
p = (-3a^{2}+8b)/8		 

p = (- 3*(a)^(2)+ 8*b)/8
 
p == (- 3*(a)^(2)+ 8*b)/8
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1.11#Ex14 q = ( a 3 - 4 ⁒ a ⁒ b + 8 ⁒ c ) / 8 π‘ž superscript π‘Ž 3 4 π‘Ž 𝑏 8 𝑐 8 {\displaystyle{\displaystyle q=(a^{3}-4ab+8c)/8}}
q = (a^{3}-4ab+8c)/8		 

q = ((a)^(3)- 4*a*b + 8*c)/8
 
q == ((a)^(3)- 4*a*b + 8*c)/8
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1.11#Ex15 r = ( - 3 ⁒ a 4 + 16 ⁒ a 2 ⁒ b - 64 ⁒ a ⁒ c + 256 ⁒ d ) / 256 π‘Ÿ 3 superscript π‘Ž 4 16 superscript π‘Ž 2 𝑏 64 π‘Ž 𝑐 256 𝑑 256 {\displaystyle{\displaystyle r=(-3a^{4}+16a^{2}b-64ac+256d)/256}}
r = (-3a^{4}+16a^{2}b-64ac+256d)/256		 

r = (- 3*(a)^(4)+ 16*(a)^(2)* b - 64*a*c + 256*d)/256
 
r == (- 3*(a)^(4)+ 16*(a)^(2)* b - 64*a*c + 256*d)/256
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1.11.E18 z 3 - 2 ⁒ p ⁒ z 2 + ( p 2 - 4 ⁒ r ) ⁒ z + q 2 = 0 superscript 𝑧 3 2 𝑝 superscript 𝑧 2 superscript 𝑝 2 4 π‘Ÿ 𝑧 superscript π‘ž 2 0 {\displaystyle{\displaystyle z^{3}-2pz^{2}+(p^{2}-4r)z+q^{2}=0}}
z^{3}-2pz^{2}+(p^{2}-4r)z+q^{2} = 0		 

(z)^(3)- 2*p*(z)^(2)+((p)^(2)- 4*r)*z + (q)^(2) = 0
 
(z)^(3)- 2*p*(z)^(2)+((p)^(2)- 4*r)*z + (q)^(2) == 0
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1.11.E20 - ΞΈ 1 ⁒ - ΞΈ 2 ⁒ - ΞΈ 3 = - q subscript πœƒ 1 subscript πœƒ 2 subscript πœƒ 3 π‘ž {\displaystyle{\displaystyle\sqrt{-\theta_{1}}\;\sqrt{-\theta_{2}}\;\sqrt{-% \theta_{3}}=-q}}
\sqrt{-\theta_{1}}\;\sqrt{-\theta_{2}}\;\sqrt{-\theta_{3}} = -q		 

sqrt(- theta[1])*sqrt(- theta[2])*sqrt(- theta[3]) = - q
 
Sqrt[- Subscript[\[Theta], 1]]*Sqrt[- Subscript[\[Theta], 2]]*Sqrt[- Subscript[\[Theta], 3]] == - q
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1.11.E22 z n = a + i ⁒ b superscript 𝑧 𝑛 π‘Ž 𝑖 𝑏 {\displaystyle{\displaystyle z^{n}=a+ib}}
z^{n} = a+ib		 

(z)^(n) = a + I*b
 
(z)^(n) == a + I*b
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1.11#Ex20 D 1 = a 1 subscript 𝐷 1 subscript π‘Ž 1 {\displaystyle{\displaystyle D_{1}=a_{1}}}
D_{1} = a_{1}		 

D[1] = a[1]
 
Subscript[D, 1] == Subscript[a, 1]
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