HOW TO DERIVE THE FORMULAS OF x + 1/x?

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by Pintir
by Pintir
                                    
                                              TUTORIAL: HOW TO DERIVE THE FORMULAS OF x + 1/x?
                          Questions like:
If x + 1/x = a
                          Then:
x² + 1/x² = a² - 2
x³ + 1/x³ = a³ - 3a
x⁴ + 1/x⁴ = a⁴ - 4a² + 2
x^5 + 1/x^5 = a^5 - 5a³ + 5a
x^6 + 1/x^6 = a^6 - 6a⁴ + 9a² - 2
                          EXAMPLE:
If x + 1/x = 3 then,
                          x² + 1/x² = (3)² - 2 = 7
x³ + 1/x³ = (3)³ - 3(3) = 18
x⁴ + 1/x⁴ = (3)⁴ - 4(3)² + 2 = 47
x^5 + 1/x^5 = (3)^5 - 5(3)³ + 5(3) = 123
x^6 + 1/x^6 = (3)^6 - 6(3)⁴ + 9(3)² - 2 = 322
                          But, do you ever wondered where did these formulas came? How did these derived? Let's find out!
                          If x + 1/x = a, then, how to find x² + 1/x²?
SOLUTION:
x + 1/x = a
Square both sides
x² + 2(x)(1/x) + 1/x² = a²
x² + 2 + 1/x² = a²
x² + 1/x² = a² - 2
                          If x + 1/x = a, then, how to find x³ + 1/x³?
SOLUTION:
x + 1/x = a
Cube both sides
x³ + 3(x²)(1/x) + 3(x)(1/x²) + 1/x³ = a³
x³ + 3x + 3/x + 1/x³ = a³
x³ + 1/x³ = a³ - (3x + 3/x)
x³ + 1/x³ = a³ - 3(x + 1/x)
Recall that x + 1/x = a
Thus,
x³ + 1/x³ = a³ - 3a
                          If x + 1/x = a, then, how to find x⁴ + 1/x⁴?
SOLUTION:
x + 1/x = a
Raise both sides to 4
x⁴ + 4(x³)(1/x) + 6(x²)(1/x²) + 4(x)(1/x³) + 1/x⁴ = a⁴
x⁴ + 4x² + 6 + 4/x² + 1/x⁴ = a⁴
x⁴ + 1/x⁴ = a⁴ - (4x² + 6 + 4/x²)
x⁴ + 1/x⁴ = a⁴ - 6 - 4(x² + 1/x²)
Recall that x² + 1/x² = a² - 2
Thus,
x⁴ + 1/x⁴ = a⁴ - 6 - 4(a² - 2)
x⁴ + 1/x⁴ = a⁴ - 6 - 4a² + 8
x⁴ + 1/x⁴ = a⁴ - 4a² + 2
                          If x + 1/x = a, then, how to find x^5 + 1/x^5?
SOLUTION:
x + 1/x = a
Raise both sides to 5
x^5 + 5(x⁴)(1/x) + 10(x³)(1/x²) + 10(x²)(1/x³) + 5(x)(1/x⁴) + 1/x^5 = a^5
x^5 + 5x³ + 10x + 10/x + 5/x³ + 1/x^5 = a^5
x^5 + 1/x^5 = a^5 - (5x³ + 10x + 10/x + 5/x³)
x^5 + 1/x^5 = a^5 - [5(x³ + 1/x³) + 10(x + 1/x)]
x^5 + 1/x^5 = a^5 - 5(x³ + 1/x³) - 10(x + 1/x)
Recall that x + 1/x = a
Recall that x³ + 1/x³ = a³ - 3a
Thus,
x^5 + 1/x^5 = a^5 - 5(a³ - 3a) - 10(a)
x^5 + 1/x^5 = a^5 - 5a³ + 15a - 10a
x^5 + 1/x^5 = a^5 - 5a³ + 5a
                          If x + 1/x = a, then, how to find x^6 + 1/x^6?
SOLUTION:
x + 1/x = a
Raise both sides to 6
x^6 + 6(x^5)(1/x) + 15(x⁴)(1/x²) + 20(x³)(1/x³) + 15(x²)(1/x⁴) + 6(x)(1/x^5) + 1/x^6 = a^6
x^6 + 6x⁴ + 15x² + 20 + 15/x² + 6/x⁴ + 1/x^6 = a^6
x^6 + 1/x^6 = a^6 - (6x⁴ + 15x² + 20 + 15/x² + 6/x⁴)
x^6 + 1/x^6 = a^6 - [6(x⁴ + 1/x⁴) + 15(x² + 1/x²) + 20]
Recall that x⁴ + 1/x⁴ = a⁴ - 4a² + 2
Recall that x² + 1/x² = a² - 2
Thus,
x^6 + 1/x^6 = a^6 - 20 - 6(a⁴ - 4a² + 2) - 15(a² - 2)
x^6 + 1/x^6 = a^6 - 20 - 6a⁴ + 24a² - 12 - 15a² + 30
x^6 + 1/x^6 = a^6 - 6a⁴ + 9a² - 2
                          NOTE: There are many other ways on how to derive these formulas.
                          ALTERNATIVE SOLUTION #1: Deriving x³ + 1/x³
Recall that m³ + n³ = (m+n)(m² - mn + n²)
Thus, x³ + 1/x³ = (x + 1/x)[x² - (x)(1/x) + 1/x²]
x³ + 1/x³ = (x + 1/x)(x² - 1 + 1/x²)
Recall that x + 1/x = a
Recall that x² + 1/x² = a² - 2
Thus,
x³ + 1/x³ = (a)(a² - 2 - 1)
x³ + 1/x³ = (a)(a² - 3)
x³ + 1/x³ = a³ - 3a
                          ALTERNATIVE SOLUTION #2: Deriving x⁴ + 1/x⁴
Recall that x² + 1/x² = (x + 1/x)² - 2
Thus,
x⁴ + 1/x⁴ = (x² + 1/x²)² - 2
Recall that x² + 1/x² = a² - 2
Thus,
x⁴ + 1/x⁴ = (a² - 2)² - 2
x⁴ + 1/x⁴ = a⁴ - 4a² + 4 - 2
x⁴ + 1/x⁴ = a⁴ - 4a² + 2
                          ALTERNATIVE SOLUTION #3: Deriving x^5 + 1/x^5
Notice that x^5 + 1/x^5 = (x² + 1/x²)(x³ + 1/x³) - (x + 1/x)
Recall that x + 1/x = a
Recall that x² + 1/x² = a² - 2
Recall that x³ + 1/x³ = a³ - 3a
Thus,
x^5 + 1/x^5 = (a² - 2)(a³ - 3a) - a
x^5 + 1/x^5 = a^5 - 3a³ - 2a³ + 6a - a
x^5 + 1/x^5 = a^5 - 5a³ + 5a
                          ALTERNATIVE SOLUTION #4: Deriving x^6 + 1/x^6
Recall that x² + 1/x² = (x + 1/x)² - 2
Thus,
x^6 + 1/x^6 = (x³ + 1/x³)² - 2
Recall that x³ + 1/x³ = a³ - 3a
Thus,
x^6 + 1/x^6 = (a³ - 3a)² - 2
x^6 + 1/x^6 = a^6 - 6a⁴ + 9a² - 2
                          ALTERNATIVE SOLUTION #5: Deriving x^6 + 1/x^6
Recall that m³ + n³ = (m+n)(m² - mn + n²)
Thus, x^6 + 1/x^6 = (x² + 1/x²)[x⁴ - (x²)(1/x²) + 1/x⁴]
x^6 + 1/x^6 = (x² + 1/x²)(x⁴ - 1 + 1/x⁴)
Recall that x² + 1/x² = a² - 2
Recall that x⁴ + 1/x⁴ = a⁴ - 4a² + 2
Thus,
x^6 + 1/x^6 = (a² - 2)(a⁴ - 4a² + 2 - 1)
x^6 + 1/x^6 = (a² - 2)(a⁴ - 4a² + 1)
x^6 + 1/x^6 = a^6 - 4a⁴ + a² - 2a⁴ + 8a² - 2
x^6 + 1/x^6 = a^6 - 6a⁴ + 9a² - 2
                          PS #2: There are still other many ways to solve for these. I just showed some. Hope you learned something! Have a good day!
                          TRY THESE:
If x + 1/x = a, then solve for the following in terms of a
1.) x^7 + 1/x^7
2.) x^20 + 1/x^20
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-shinnichi