Given the system:
                                                                        f                (                x                )                                                                                            =                                  a                                      0                                                                    x                                      n                                                  +                                  a                                      1                                                                    x                                      n                    −                    1                                                  +                ⋯                +                                  a                                      n                                                                                                                                              (                1                )                                                                                                                                  =                (                x                −                                  r                                      1                                                  )                (                x                −                                  r                                      2                                                  )                ⋯                (                x                −                                  r                                      n                                                  )                                                                                                            (                2                )                                                            
Assuming no roots of                     f        (        x        )        =        0                         lie on the imaginary axis, and letting
                    N                         = The number of roots of 
                    f        (        x        )        =        0                         with negative real parts, and
                    P                         = The number of roots of 
                    f        (        x        )        =        0                         with positive real parts
then we have
                    N        +        P        =        n                (        3        )                        
Expressing                     f        (        x        )                         in polar form, we have
                    f        (        x        )        =        ρ        (        x        )                  e                      j            θ            (            x            )                                  (        4        )                        
where
                    ρ        (        x        )        =                                                                              R                  e                                                            2                                      [            f            (            x            )            ]            +                                                            I                  m                                                            2                                      [            f            (            x            )            ]                                  (        5        )                
and
                    θ        (        x        )        =                  tan                      −            1                                                        (                                                I            m                          [        f        (        x        )        ]                  /                                      R            e                          [        f        (        x        )        ]                              )                                  (        6        )                
from (2) note that
                    θ        (        x        )        =                  θ                                    r                              1                                                    (        x        )        +                  θ                                    r                              2                                                    (        x        )        +        ⋯        +                  θ                                    r                              n                                                    (        x        )                (        7        )                        
where
                              θ                                    r                              i                                                    (        x        )        =        ∠        (        x        −                  r                      i                          )                (        8        )                        
Now if the ith root of                     f        (        x        )        =        0                         has a positive real part, then (using the notation y=(RE[y],IM[y]))
                                                                                          θ                                                            r                                              i                                                                                            (                x                )                                                                            |                                                                            x                    =                    j                    ∞                                                                                              =                ∠                (                x                −                                  r                                      i                                                  )                                                                            |                                                                            x                    =                    j                    ∞                                                                                                                                    =                ∠                (                0                −                                                      R                    e                                                  [                                  r                                      i                                                  ]                ,                ∞                −                                                      I                    m                                                  [                                  r                                      i                                                  ]                )                                                                                                  =                ∠                (                −                                                      R                    e                                                  [                                  r                                      i                                                  ]                ,                ∞                )                                                                                                  =                                  lim                                      ϕ                    →                    −                    ∞                                                                    tan                                      −                    1                                                                  ϕ                =                −                                                      π                    2                                                                  (                9                )                                                            
and
                              θ                                    r                              i                                                    (        x        )                                            |                                            x            =            −            j            ∞                          =        ∠        (        −                              R            e                          [                  r                      i                          ]        ,        −        ∞        )        =                  lim                      ϕ            →            ∞                                    tan                      −            1                                  ϕ        =                              π            2                                  (        10        )                        
Similarly, if the ith root of                     f        (        x        )        =        0                         has a negative real part,
                              θ                                    r                              i                                                    (        x        )                                            |                                            x            =            j            ∞                          =        ∠        (        −                              R            e                          [                  r                      i                          ]        ,        ∞        )        =                  lim                      ϕ            →            ∞                                    tan                      −            1                                  ϕ        =                              π            2                                          (        11        )                
and
                              θ                                    r                              i                                                    (        x        )                                            |                                            x            =            −            j            ∞                          =        ∠        (        −                              R            e                          [                  r                      i                          ]        ,        −        ∞        )        =                  lim                      ϕ            →            −            ∞                                    tan                      −            1                                  ϕ        =        −                              π            2                                          (        12        )                
Therefore,                               θ                                    r                              i                                                    (        x        )                                            |                                            x            =            −            j            ∞                                x            =            j            ∞                          =        −        π                         when the ith root of                     f        (        x        )                         has a positive real part, and                               θ                                    r                              i                                                    (        x        )                                            |                                            x            =            −            j            ∞                                x            =            j            ∞                          =        π                         when the ith root of                     f        (        x        )                         has a negative real part. Alternatively,
                    θ        (        x        )                                            |                                            x            =            j            ∞                          =        ∠        (        x        −                  r                      1                          )                                            |                                            x            =            j            ∞                          +        ∠        (        x        −                  r                      2                          )                                            |                                            x            =            j            ∞                          +        ⋯        +        ∠        (        x        −                  r                      n                          )                                            |                                            x            =            j            ∞                          =                              π            2                          N        −                              π            2                          P                (        13        )                        
and
                    θ        (        x        )                                            |                                            x            =            −            j            ∞                          =        ∠        (        x        −                  r                      1                          )                                            |                                            x            =            −            j            ∞                          +        ∠        (        x        −                  r                      2                          )                                            |                                            x            =            −            j            ∞                          +        ⋯        +        ∠        (        x        −                  r                      n                          )                                            |                                            x            =            −            j            ∞                          =        −                              π            2                          N        +                              π            2                          P                (        14        )                        
So, if we define
                    Δ        =                              1            π                          θ        (        x        )                                            |                                            −            j            ∞                                j            ∞                                  (        15        )                        
then we have the relationship
                    N        −        P        =        Δ                (        16        )                        
and combining (3) and (16) gives us
                    N        =                                            n              +              Δ                        2                                           and 
                    P        =                                            n              −              Δ                        2                                  (        17        )                        
Therefore, given an equation of                     f        (        x        )                         of degree                     n                         we need only evaluate this function                     Δ                         to determine                     N                        , the number of roots with negative real parts and                     P                        , the number of roots with positive real parts.
Equations (13) and (14) show that at                     x        =        ±        ∞                        ,                     θ        =        θ        (        x        )                         is an integer multiple of                     π                  /                2                        . Note now, in accordance with (6) and Figure 1, the graph of                     tan                (        θ        )                         vs                     θ                        , that varying                     x                         over an interval (a,b) where                               θ                      a                          =        θ        (        x        )                              |                                x            =            j            a                                           and                               θ                      b                          =        θ        (        x        )                              |                                x            =            j            b                                           are integer multiples of                     π                        , this variation causing the function                     θ        (        x        )                         to have increased by                     π                        , indicates that in the course of travelling from point a to point b,                     θ                         has "jumped" from                     +        ∞                         to                     −        ∞                         one more time than it has jumped from                     −        ∞                         to                     +        ∞                        . Similarly, if we vary                     x                         over an interval (a,b) this variation causing                     θ        (        x        )                         to have decreased by                     π                        , where again                     θ                         is a multiple of                     π                         at both                     x        =        j        a                         and                     x        =        j        b                        , implies that                     tan                θ        (        x        )        =                              I            m                          [        f        (        x        )        ]                  /                                      R            e                          [        f        (        x        )        ]                         has jumped from                     −        ∞                         to                     +        ∞                         one more time than it has jumped from                     +        ∞                         to                     −        ∞                         as                     x                         was varied over the said interval.
Thus,                     θ        (        x        )                                            |                                            −            j            ∞                                j            ∞                                           is                     π                         times the difference between the number of points at which                                           I            m                          [        f        (        x        )        ]                  /                                      R            e                          [        f        (        x        )        ]                         jumps from                     −        ∞                         to                     +        ∞                         and the number of points at which                                           I            m                          [        f        (        x        )        ]                  /                                      R            e                          [        f        (        x        )        ]                         jumps from                     +        ∞                         to                     −        ∞                         as                     x                         ranges over the interval                     (        −        j        ∞        ,        +        j        ∞                )                 provided that at                     x        =        ±        j        ∞                ,                     tan                [        θ        (        x        )        ]                         is defined.
In the case where the starting point is on an incongruity (i.e.                               θ                      a                          =        π                  /                2        ±        i        π                        , i = 0, 1, 2, ...) the ending point will be on an incongruity as well, by equation (16) (since                     N                         is an integer and                     P                         is an integer,                     Δ                         will be an integer). In this case, we can achieve this same index (difference in positive and negative jumps) by shifting the axes of the tangent function by                     π                  /                2                        , through adding                     π                  /                2                         to                     θ                        . Thus, our index is now fully defined for any combination of coefficients in                     f        (        x        )                         by evaluating                     tan                [        θ        ]        =                              I            m                          [        f        (        x        )        ]                  /                                      R            e                          [        f        (        x        )        ]                         over the interval (a,b) =                     (        +        j        ∞        ,        −        j        ∞        )                         when our starting (and thus ending) point is not an incongruity, and by evaluating
                    tan                [                  θ          ′                (        x        )        ]        =        tan                [        θ        +        π                  /                2        ]        =        −        cot                [        θ        (        x        )        ]        =        −                              R            e                          [        f        (        x        )        ]                  /                                      I            m                          [        f        (        x        )        ]                (        18        )                        
over said interval when our starting point is at an incongruity.
This difference,                     Δ                        , of negative and positive jumping incongruities encountered while traversing                     x                         from                     −        j        ∞                         to                     +        j        ∞                         is called the Cauchy Index of the tangent of the phase angle, the phase angle being                     θ        (        x        )                         or                               θ          ′                (        x        )                        , depending as                               θ                      a                                           is an integer multiple of                     π                         or not.
To derive Routh's criterion, first we will use a different notation to differentiate between the even and odd terms of                     f        (        x        )                        :
                    f        (        x        )        =                  a                      0                                    x                      n                          +                  b                      0                                    x                      n            −            1                          +                  a                      1                                    x                      n            −            2                          +                  b                      1                                    x                      n            −            3                          +        ⋯                (        19        )                        Now we have:
                                                                        f                (                j                ω                )                                                            =                                  a                                      0                                                  (                j                ω                                  )                                      n                                                  +                                  b                                      0                                                  (                j                ω                                  )                                      n                    −                    1                                                  +                                  a                                      1                                                  (                j                ω                                  )                                      n                    −                    2                                                  +                                  b                                      1                                                  (                j                ω                                  )                                      n                    −                    3                                                  +                ⋯                                                                                                            (                20                )                                                                                                  =                                  a                                      0                                                  (                j                ω                                  )                                      n                                                  +                                  a                                      1                                                  (                j                ω                                  )                                      n                    −                    2                                                  +                                  a                                      2                                                  (                j                ω                                  )                                      n                    −                    4                                                  +                ⋯                                                                                                            (                21                )                                                                                                  +                                  b                                      0                                                  (                j                ω                                  )                                      n                    −                    1                                                  +                                  b                                      1                                                  (                j                ω                                  )                                      n                    −                    3                                                  +                                  b                                      2                                                  (                j                ω                                  )                                      n                    −                    5                                                  +                ⋯                                                            Therefore, if                     n                         is even,
                                                                        f                (                j                ω                )                                                            =                (                −                1                                  )                                      n                                          /                                        2                                                                                        [                                                                    a                                      0                                                                    ω                                      n                                                  +                                  a                                      1                                                                    ω                                      n                    −                    2                                                  +                                  a                                      2                                                                    ω                                      n                    −                    4                                                  +                ⋯                                                      ]                                                                                                                                              (                22                )                                                                                                  +                j                (                −                1                                  )                                      (                    n                                          /                                        2                    )                    −                    1                                                                                        [                                                                    b                                      0                                                                    ω                                      n                    −                    1                                                  +                                  b                                      1                                                                    ω                                      n                    −                    3                                                  +                                  b                                      2                                                                    ω                                      n                    −                    5                                                  +                ⋯                                                      ]                                                                                                                                                          and if                     n                         is odd:
                                                                        f                (                j                ω                )                                                            =                j                (                −                1                                  )                                      (                    n                    −                    1                    )                                          /                                        2                                                                                        [                                                                    a                                      0                                                                    ω                                      n                                                  +                                  a                                      1                                                                    ω                                      n                    −                    2                                                  +                                  a                                      2                                                                    ω                                      n                    −                    4                                                  +                ⋯                                                      ]                                                                                                                                              (                23                )                                                                                                  +                (                −                1                                  )                                      (                    n                    −                    1                    )                                          /                                        2                                                                                        [                                                                    b                                      0                                                                    ω                                      n                    −                    1                                                  +                                  b                                      1                                                                    ω                                      n                    −                    3                                                  +                                  b                                      2                                                                    ω                                      n                    −                    5                                                  +                ⋯                                                      ]                                                                                                                                                          Now observe that if                     n                         is an odd integer, then by (3)                     N        +        P                         is odd. If                     N        +        P                         is an odd integer, then                     N        −        P                         is odd as well. Similarly, this same argument shows that when                     n                         is even,                     N        −        P                         will be even. Equation (13) shows that if                     N        −        P                         is even,                     θ                         is an integer multiple of                     π                        . Therefore,                     tan                (        θ        )                         is defined for                     n                         even, and is thus the proper index to use when n is even, and similarly                     tan                (                  θ          ′                )        =        tan                (        θ        +        π        )        =        −        cot                (        θ        )                         is defined for                     n                         odd, making it the proper index in this latter case.
Thus, from (6) and (22), for                     n                         even:
                    Δ        =                  I                      −            ∞                                +            ∞                                                              −                                                I                  m                                            [              f              (              x              )              ]                                                                        R                  e                                            [              f              (              x              )              ]                                      =                  I                      −            ∞                                +            ∞                                                                              b                                  0                                                            ω                                  n                  −                  1                                            −                              b                                  1                                                            ω                                  n                  −                  3                                            +              ⋯                                                      a                                  0                                                            ω                                  n                                            −                              a                                  1                                                            ω                                  n                  −                  2                                            +              …                                              (        24        )                        and from (18) and (23), for                     n                         odd:
                    Δ        =                  I                      −            ∞                                +            ∞                                                                                                R                  e                                            [              f              (              x              )              ]                                                                        I                  m                                            [              f              (              x              )              ]                                      =                  I                      −            ∞                                +            ∞                                                                              b                                  0                                                            ω                                  n                  −                  1                                            −                              b                                  1                                                            ω                                  n                  −                  3                                            +              …                                                      a                                  0                                                            ω                                  n                                            −                              a                                  1                                                            ω                                  n                  −                  2                                            +              …                                              (        25        )                        Thus we are evaluating the same Cauchy index for both:
                    Δ        =                  I                      −            ∞                                +            ∞                                                                              b                                  0                                                            ω                                  n                  −                  1                                            −                              b                                  1                                                            ω                                  n                  −                  3                                            +              …                                                      a                                  0                                                            ω                                  n                                            −                              a                                  1                                                            ω                                  n                  −                  2                                            +              …                                              (        26        )                        Sturm gives us a method for evaluating                     Δ        =                  I                      −            ∞                                +            ∞                                                                              f                                  2                                            (              x              )                                                      f                                  1                                            (              x              )                                                      . His theorem states as follows:
Given a sequence of polynomials                               f                      1                          (        x        )        ,                  f                      2                          (        x        )        ,        …        ,                  f                      m                          (        x        )                         where:
1) If                               f                      k                          (        x        )        =        0                         then                               f                      k            −            1                          (        x        )        ≠        0                        ,                               f                      k            +            1                          (        x        )        ≠        0                        , and                     sign                [                  f                      k            −            1                          (        x        )        ]        =        −        sign                [                  f                      k            +            1                          (        x        )        ]                        
2)                               f                      m                          (        x        )        ≠        0                         for                     −        ∞        <        x        <        ∞                        
and we define                     V        (        x        )                         as the number of changes of sign in the sequence                               f                      1                          (        x        )        ,                  f                      2                          (        x        )        ,        …        ,                  f                      m                          (        x        )                         for a fixed value of                     x                        , then:
                    Δ        =                  I                      −            ∞                                +            ∞                                                                              f                                  2                                            (              x              )                                                      f                                  1                                            (              x              )                                      =        V        (        −        ∞        )        −        V        (        +        ∞        )                (        27        )                        
A sequence satisfying these requirements is obtained using the Euclidean algorithm, which is as follows:
Starting with                               f                      1                          (        x        )                         and                               f                      2                          (        x        )                        , and denoting the remainder of                               f                      1                          (        x        )                  /                          f                      2                          (        x        )                         by                               f                      3                          (        x        )                         and similarly denoting the remainder of                               f                      2                          (        x        )                  /                          f                      3                          (        x        )                         by                               f                      4                          (        x        )                        , and so on, we obtain the relationships:
                                                                                                        f                                      1                                                  (                x                )                =                                  q                                      1                                                  (                x                )                                  f                                      2                                                  (                x                )                −                                  f                                      3                                                  (                x                )                                (                28                )                                                                                                    f                                      2                                                  (                x                )                =                                  q                                      2                                                  (                x                )                                  f                                      3                                                  (                x                )                −                                  f                                      4                                                  (                x                )                                                                                                  …                                                                                                    f                                      m                    −                    1                                                  (                x                )                =                                  q                                      m                    −                    1                                                  (                x                )                                  f                                      m                                                  (                x                )                                                            
or in general
                              f                      k            −            1                          (        x        )        =                  q                      k            −            1                          (        x        )                  f                      k                          (        x        )        −                  f                      k            +            1                          (        x        )                        
where the last non-zero remainder,                               f                      m                          (        x        )                         will therefore be the highest common factor of                               f                      1                          (        x        )        ,                  f                      2                          (        x        )        ,        …        ,                  f                      m            −            1                          (        x        )                        . It can be observed that the sequence so constructed will satisfy the conditions of Sturm's theorem, and thus an algorithm for determining the stated index has been developed.
It is in applying Sturm's theorem (28) to (26), through the use of the Euclidean algorithm above that the Routh matrix is formed.
We get
                              f                      3                          (        ω        )        =                                            a                              0                                                    b                              0                                                              f                      2                          (        ω        )        −                  f                      1                          (        ω        )                (        29        )                        
and identifying the coefficients of this remainder by                               c                      0                                          ,                     −                  c                      1                                          ,                               c                      2                                          ,                     −                  c                      3                                          , and so forth, makes our formed remainder
                              f                      3                          (        ω        )        =                  c                      0                                    ω                      n            −            2                          −                  c                      1                                    ω                      n            −            4                          +                  c                      2                                    ω                      n            −            6                          −        ⋯                (        30        )                        
where
                              c                      0                          =                  a                      1                          −                                            a                              0                                                    b                              0                                                              b                      1                          =                                                            b                                  0                                                            a                                  1                                            −                              a                                  1                                                            b                                  0                                                                    b                              0                                                    ;                  c                      1                          =                  a                      2                          −                                            a                              0                                                    b                              0                                                              b                      2                          =                                                            b                                  0                                                            a                                  2                                            −                              a                                  0                                                            b                                  2                                                                    b                              0                                                    ;        …                (        31        )                        
Continuing with the Euclidean algorithm on these new coefficients gives us
                              f                      4                          (        ω        )        =                                            b                              0                                                    c                              0                                                              f                      3                          (        ω        )        −                  f                      2                          (        ω        )                (        32        )                        
where we again denote the coefficients of the remainder                               f                      4                          (        ω        )                         by                               d                      0                                          ,                     −                  d                      1                                          ,                               d                      2                                          ,                     −                  d                      3                                          ,
making our formed remainder
                              f                      4                          (        ω        )        =                  d                      0                                    ω                      n            −            3                          −                  d                      1                                    ω                      n            −            5                          +                  d                      2                                    ω                      n            −            7                          −        ⋯                (        33        )                        
and giving us
                              d                      0                          =                  b                      1                          −                                            b                              0                                                    c                              0                                                              c                      1                          =                                                            c                                  0                                                            b                                  1                                            −                              b                                  1                                                            c                                  0                                                                    c                              0                                                    ;                  d                      1                          =                  b                      2                          −                                            b                              0                                                    c                              0                                                              c                      2                          =                                                            c                                  0                                                            b                                  2                                            −                              b                                  0                                                            c                                  2                                                                    c                              0                                                    ;        …                (        34        )                        
The rows of the Routh array are determined exactly by this algorithm when applied to the coefficients of (19). An observation worthy of note is that in the regular case the polynomials                               f                      1                          (        ω        )                         and                               f                      2                          (        ω        )                         have as the highest common factor                               f                      n            +            1                          (        ω        )                         and thus there will be                     n                         polynomials in the chain                               f                      1                          (        x        )        ,                  f                      2                          (        x        )        ,        …        ,                  f                      m                          (        x        )                        .
Note now, that in determining the signs of the members of the sequence of polynomials                               f                      1                          (        x        )        ,                  f                      2                          (        x        )        ,        …        ,                  f                      m                          (        x        )                         that at                     ω        =        ±        ∞                         the dominating power of                     ω                         will be the first term of each of these polynomials, and thus only these coefficients corresponding to the highest powers of                     ω                         in                               f                      1                          (        x        )        ,                  f                      2                          (        x        )        ,        …                , and                               f                      m                          (        x        )                        , which are                               a                      0                                          ,                               b                      0                                          ,                               c                      0                                          ,                               d                      0                                          , ... determine the signs of                               f                      1                          (        x        )                        ,                               f                      2                          (        x        )                        , ...,                               f                      m                          (        x        )                         at                     ω        =        ±        ∞                        .
So we get                     V        (        +        ∞        )        =        V        (                  a                      0                          ,                  b                      0                          ,                  c                      0                          ,                  d                      0                          ,        …        )                         that is,                     V        (        +        ∞        )                         is the number of changes of sign in the sequence                               a                      0                                    ∞                      n                                          ,                               b                      0                                    ∞                      n            −            1                                          ,                               c                      0                                    ∞                      n            −            2                                          , ... which is the number of sign changes in the sequence                               a                      0                                          ,                               b                      0                                          ,                               c                      0                                          ,                               d                      0                                          , ... and                     V        (        −        ∞        )        =        V        (                  a                      0                          ,        −                  b                      0                          ,                  c                      0                          ,        −                  d                      0                          ,        .        .        .        )                        ; that is                     V        (        −        ∞        )                         is the number of changes of sign in the sequence                               a                      0                          (        −        ∞                  )                      n                                          ,                               b                      0                          (        −        ∞                  )                      n            −            1                                          ,                               c                      0                          (        −        ∞                  )                      n            −            2                                          , ... which is the number of sign changes in the sequence                               a                      0                                          ,                     −                  b                      0                                          ,                               c                      0                                          ,                     −                  d                      0                                          , ...
Since our chain                               a                      0                                          ,                               b                      0                                          ,                               c                      0                                          ,                               d                      0                                          , ... will have                     n                         members it is clear that                     V        (        +        ∞        )        +        V        (        −        ∞        )        =        n                         since within                     V        (                  a                      0                          ,                  b                      0                          ,                  c                      0                          ,                  d                      0                          ,        …        )                         if going from                               a                      0                                           to                               b                      0                                           a sign change has not occurred, within                     V        (                  a                      0                          ,        −                  b                      0                          ,                  c                      0                          ,        −                  d                      0                          ,        …        )                         going from                               a                      0                                           to                     −                  b                      0                                           one has, and likewise for all                     n                         transitions (there will be no terms equal to zero) giving us                     n                         total sign changes.
As                     Δ        =        V        (        −        ∞        )        −        V        (        +        ∞        )                         and                     n        =        V        (        +        ∞        )        +        V        (        −        ∞        )                        , and from (17)                     P        =        (        n        −        Δ                  /                2        )                        , we have that                     P        =        V        (        +        ∞        )        =        V        (                  a                      0                          ,                  b                      0                          ,                  c                      0                          ,                  d                      0                          ,        …        )                         and have derived Routh's theorem -
The number of roots of a real polynomial                     f        (        z        )                         which lie in the right half plane                                           R            e                          (                  r                      i                          )        >        0                         is equal to the number of changes of sign in the first column of the Routh scheme.
And for the stable case where                     P        =        0                         then                     V        (                  a                      0                          ,                  b                      0                          ,                  c                      0                          ,                  d                      0                          ,        …        )        =        0                         by which we have Routh's famous criterion:
In order for all the roots of the polynomial                     f        (        z        )                         to have negative real parts, it is necessary and sufficient that all of the elements in the first column of the Routh scheme be different from zero and of the same sign.