[tex]E_{1} = x + \dfrac{1}{x} = t[/tex]

[tex]t^{2} = \bigg(x + \dfrac{1}{x}\bigg)^{2} = x^{2} + 2 \cdot x \cdot \dfrac{1}{x} + \dfrac{1}{x^{2}} = x^{2} + \dfrac{1}{x^{2}} + 2 = E_{2} + 2[/tex]

[tex]\implies E_{2} = x^{2} + \dfrac{1}{x^{2}} = t^{2} - 2[/tex]

[tex]t^{3} = \bigg(x + \dfrac{1}{x}\bigg)^{3} = x^{3} + 3 \cdot x^{2} \cdot \dfrac{1}{x} + 3 \cdot x \cdot \dfrac{1}{x^{2}} + \dfrac{1}{x^{3}} = x^{3} + 3 \cdot \bigg(x + \dfrac{1}{x} \bigg) + \dfrac{1}{x^{3}} = E_{3} + 3t[/tex]

[tex]t^{3} = E_{3} + 3E_{1} = E_{3} + 3t[/tex]

[tex]\implies \boldsymbol {E_{3} = t^{3} - 3t} \\ [/tex]

Generalizarea se obține cu ajutorul binomului lui Newton

Binomul lui Newton:

[tex]\boxed{ \boldsymbol{ (x + y)^{n} = \sum^{n}_{k = 0} C^{k}_{n} \cdot {x}^{n - k} \cdot {y}^{k} }}[/tex]

[tex](x + x^{-1})^{n} = \sum^{n}_{k = 0} C^{k}_{n} \cdot {x}^{n - k} \cdot {x}^{-k} = \sum^{n}_{k = 0} C^{k}_{n} \cdot {x}^{n - 2k} \\[/tex]

Pentru n par (în acest caz avem termenul liber, pe care îl obținem în mijlocul dezvoltării):

[tex]\bigg(x + \dfrac{1}{x}\bigg)^{4} = x^{4} + 4 \cdot x^{3} \cdot \dfrac{1}{x} + 6 \cdot x^{2} \cdot \dfrac{1}{x^{2}} + 4 \cdot x \cdot \dfrac{1}{x^{3}} + \dfrac{1}{x^{4}} = \\ [/tex]

[tex]= x^{4} + 4 \cdot x^{2} + 6 + 4 \cdot \dfrac{1}{x^{2}} + \dfrac{1}{x^{4}}[/tex]

[tex]t^{4} = E_{4} + 4E_{2} + 6 = E_{4} + 4(t^{2} - 2) + 6 \\ [/tex]

[tex]\implies \boldsymbol{ E_{4} = t^{4} - 4t^{2} + 2} \\ [/tex]

[tex]\bigg(x + \dfrac{1}{x}\bigg)^{6} = x^{6} + 6 \cdot x^{5} \cdot \dfrac{1}{x} + 15 \cdot x^{4} \cdot \dfrac{1}{x^{2}} + 20 \cdot x^{3} \cdot \dfrac{1}{x^{3}} + 15 \cdot x^{2} \cdot \dfrac{1}{x^{4}} + 6 \cdot x \cdot \dfrac{1}{x^{5}} + \dfrac{1}{x^{6}} = \\ [/tex]

[tex]= x^{6} + 6 \cdot x^{4} + 15 \cdot x^{2} + 20 + 15 \cdot \dfrac{1}{x^{2}} + 6 \cdot \dfrac{1}{x^{4}} + \dfrac{1}{x^{6}} \\ [/tex]

[tex]t^{6} = E_{6} + 6E_{4} + 15E_{2} + 20 = E_{6} + 6(t^{4} - 4t^{2} + 2) + 15(t^{2} - 2) + 20 = \\ [/tex]

[tex]= E_{6} + 6t^{4} - 24t^{2} + 12 + 15t^{2} - 30 + 20 = E_{6} + 6t^{4} - 9t^{2} + 2 \\ [/tex]

[tex]\implies \boldsymbol {E_{6} = t^{6} - 6t^{4} + 9t^{2} - 2}\\[/tex]

iar pentru n impar:

[tex]\bigg(x + \dfrac{1}{x}\bigg)^{5} = x^{5} + 5 \cdot x^{4} \cdot \dfrac{1}{x} + 10 \cdot x^{3} \cdot \dfrac{1}{x^{2}} + 10 \cdot x^{2} \cdot \dfrac{1}{x^{3}} + 5 \cdot x \cdot \dfrac{1}{x^{4}} + \dfrac{1}{x^{5}} = \\ [/tex]

[tex]= x^{5} + 5 \cdot x^{3} + 10 \cdot x + 10 \cdot \dfrac{1}{x} + 5 \cdot \dfrac{1}{x^{3}} + \dfrac{1}{x^{5}} \\ [/tex]

[tex]t^{5} = E_{5} + 5E_{3} + 10E_{1} = E_{5} + 5(t^{3} - 3t) + 10t = E_{5} + 5t^{3} - 15t + 10t \\ [/tex]

[tex]\implies \boldsymbol {E_{5} = t^5 - 5t^3 + 5t}[/tex]

[tex]\bigg(x + \dfrac{1}{x}\bigg)^{7} = x^{7} + 7 \cdot x^{6} \cdot \dfrac{1}{x} + 21 \cdot x^{5} \cdot \dfrac{1}{x^{2}} + 35 \cdot x^{4} \cdot \dfrac{1}{x^{3}} + 35 \cdot x^{3} \cdot \dfrac{1}{x^{4}} + 21 \cdot x^{2} \cdot \dfrac{1}{x^{5}} + 7 \cdot x \cdot \dfrac{1}{x^{6}} + \dfrac{1}{x^{7}} = \\ [/tex]

[tex]= x^{7} + 7 \cdot x^{5} + 21 \cdot x^{3} + 35 \cdot x + 35 \cdot \dfrac{1}{x} + 21 \cdot \dfrac{1}{x^{3}} + 7 \cdot \dfrac{1}{x^{5}} + \dfrac{1}{x^{7}} \\ [/tex]

[tex]t^{7} = E_{7} + 7E_{5} + 21E_{3} + 35E_{1} = E_{7} + 7(t^5 - 5t^3 + 5t) + 21(t^{3} - 3t) + 10t =\\[/tex]

[tex]= E_{7} + 7t^5 - 35t^3 + 35t + 21t^{3} - 63t + 10t = E_{7} + 7t^5 - 14t^3 - 18t \\ [/tex]

[tex]\implies \boldsymbol {E_{7} = t^{7} - 7t^5 + 14t^3 + 18t}\\[/tex]