Anonymous
Anonymous asked in Science & MathematicsMathematics · 5 months ago

Pls express in Partial Fraction: x⁴/[(x+1)⁴(x²+1)]?

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    a / (x + 1) + b / (x + 1)^2 + c / (x + 1)^3 + d / (x + 1)^4 + ex / (x^2 + 1) + f / (x^2 + 1) = x^4 / ((x + 1)^4 * (x^2 + 1))

    a * (x + 1)^3 * (x^2 + 1) + b * (x + 1)^2 * (x^2 + 1) + c * (x + 1) * (x^2 + 1) + d * (x^2 + 1) + ex * (x + 1)^4 + f * (x + 1)^4 = x^4

    a * (x^3 + 3x^2 + 3x + 1) * (x^2 + 1) + b * (x^2 + 2x + 1) * (x^2 + 1) + c * (x + 1) * (x^2 + 1) + d * (x^2 + 1) + ex * (x^4 + 4x^3 + 6x^2 + 4x + 1) + f * (x^4 + 4x^3 + 6x^2 + 4x + 1) = x^4

    a * (x^5 + x^3 + 3x^4 + 3x^2 + 3x^3 + 3x + x^2 + 1) + b * (x^4 + x^2 + 2x^3 + 2x + x^2 + 1) + c * (x^3 + x^2 + x + 1) + d * (x^2 + 1) + e * (x^5 + 4x^4 + 6x^3 + 4x^2 + x) + f * (x^4 + 4x^3 + 6x^2 + 4x + 1) = x^4

    a * (x^5 + 3x^4 + 4x^3 + 4x^2 + 3x + 1) + b * (x^4 + 2x^3 + 2x^2 + 2x + 1) + c * (x^3 + x^2 + x + 1) + d * (x^2 + 1) + e * (x^5 + 4x^4 + 6x^3 + 4x^2 + x) + f * (x^4 + 4x^3 + 6x^2 + 4x + 1) = x^4

    x^5 * (a + e) + x^4 * (3a + b + 4e + f) + x^3 * (4a + 2b + c + 6e + 4f) + x^2 * (4a + 2b + c + d + 4e + 6f) + x * (3a + 2b + c + e + 4f) + (a + b + c + d + f) = 0x^5 + x^4 + 0x^3 + 0x^2 + 0x + 0

    a + e = 0

    3a + b + 4e + f = 1

    4a + 2b + c + 6e + 4f = 0

    4a + 2b + c + d + 4e + 6f = 0

    3a + 2b + c + e + f = 0

    a + b + c + d + f = 0

    You have 6 equations and 6 unknowns. You should be able to solve

    a + e = 0

    a = -e

    -3e + b + 4e + f = 1

    -4e + 2b + c + 6e + 4f = 0

    -4e + 2b + c + d + 4e + 6f = 0

    -3e + 2b + c + e + f = 0

    -e + b + c + d + f = 0

    b + e + f = 1

    2b + c + 2e + 4f = 0

    2b + c + d + 6f = 0

    2b + c - 2e + f = 0

    b + c + d - e + f = 0

    b + c + d - e + f - (b + e + f) = 0 - 1

    b - b + c + d - e - e + f - f = -1

    c + d - 2e = -1

    2b + c + d + 6f = 0

    2b + 2e - 1 + 6f = 0

    2b + 2e + 6f = 1

    b + e + 3f = 1/2

    b + e + f = 1

    b + e + 3f = 1/2

    2f = 1/2 - 1

    2f = -1/2

    f = -1/4

    b + e + f = 1

    b + e - 1/4 = 1

    b + e = 5/4

    2b + 2e + c + 4 * (-1/4) = 0

    2b + c + d + 6 * (-1/4) = 0

    2b - 2e + c + (-1/4) = 0

    b - e + c + d + (-1/4) = 0

    2 * (b + e) + c - 1 = 0

    2b + c + d - 3/2 = 0

    2 * (b - e) + c = 1/4

    (b - e) + c + d = 1/4

    2 * (5/4) + c - 1 = 0

    5/2 + c - 1 = 0

    c = 1 - 5/2

    c = -3/2

    2 * (b - e) + c = 1/4

    2 * (b - e) - 3/2 = 1/4

    2 * (b - e) - 6/4 = 1/4

    2 * (b - e) = 7/4

    b - e = 7/8

    (b - e) + c + d = 1/4

    7/8 + (-3/2) + d = 1/4

    7/8 - 12/8 + d = 2/8

    -5/8 + d = 2/8

    d = 7/8

    2 * (b + e) + c - 1 = 0

    2 * (b + e) - 3/2 - 1 = 0

    2 * (b + e) - 5/2 = 0

    2 * (b + e) = 5/2

    b + e = 5/4

    4b + 4e = 5

    8b + 8e = 10

    b - e = 7/8

    8b - 8e = 7

    8b + 8e + 8b - 8e = 10 + 7

    16b = 17

    b = 17/16

    b - e = 7/8

    17/16 - e = 7/8

    17/16 - 14/16 = e

    3/16 = e

    a = -e

    a = -3/16

    a = -3/16 , b = 17/16 , c = -3/2 , d = 7/8 , e = 3/16 , f = -1/4

  • Indica
    Lv 7
    5 months ago

    .............................

    If you sub u=x+1 for the repeated factor, finding the constants in the PF representation by equating

    coefficients of u rather than x reduces to a simple chained calculation viz …

    (u−1)⁴ / (u⁴(u²−2u+2)) = A/u + B/u² + C/u³ + D/u⁴ + (Eu+F)/(u²−2u+1)

    Clear denominators and group like terms for

    (A+E)u⁵ + (B−2A+F−1)u⁴ + (C−2B+2A+4)u³ + (D−2C+2B−6)u² + (−2D+2C+4)u + (2D−1) = 0

    constant term ⟹ D=1/2

    u terms ⟹ −2D+2C+4=0 ⟹ C = (2D−4)/2 = −3/2

    u² terms ⟹ D−2C+2B−6=0 ⟹ B = (−D+2C+6)/2 = 5/4

    u³ terms ⟹ C−2B+2A+4=0 ⟹ A = (−C+2B−4)/2 = 0

    u⁴ terms ⟹ B−2A+F−1=0 ⟹ F = −B+2A+1 = −1/4

    u⁵ terms ⟹ A+E=0 ⟹ E =0

    PF representation is (5/4)/(x+1)² − (3/2)/(x+1)³ + (1/2)/(x+1)⁴ − (1/4)/(x²+1)

  • 5 months ago

    x⁴/[(x+1)⁴(x²+1)]

    x^4\/((x + 1)^4 (x^2 + 1)) = -i\/(8 (x + i)) + 5\/(4 (x + 1)^2) - 3\/(2 (x + 1)^3) + 1\/(2 (x + 1)^4) + i\/(8 (x - i))

  • Anonymous
    5 months ago

    You know you could always use MATLAB?

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  • Anonymous
    5 months ago

    Why the hell should i?

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