Theory of Equation

Incommensurable Roots

506 Sturm's Theorem: If 𝑓(𝑥), freed from equal roots, be divided by 𝑓'(𝑥), and the last divisor by the last remainder, changing the sign of each remainder before dividing by it, until a remainder independent of 𝑥 is obtained, or else a remainder which cannot change its sign; then 𝑓(𝑥), 𝑓'(𝑥), and the successive remainders constitute Sturm's functions, and are denoted by 𝑓(𝑥), 𝑓₁(𝑥), 𝑓₂(𝑥), ⋯, 𝑓_𝑚(𝑥) The operation may be exhibited as follows: 𝑓(𝑥)=𝑞1𝑓1(𝑥)−𝑓2(𝑥) 𝑓1(𝑥)=𝑞2𝑓2(𝑥)−𝑓3(𝑥) 𝑓2(𝑥)=𝑞3𝑓3(𝑥)−𝑓4(𝑥) ⋯ 𝑓𝑚−2(𝑥)=𝑞𝑚−1𝑓𝑚−1(𝑥)−𝑓𝑚(𝑥) 507 Note: Any constant factor of a remainder may be rejected, and the quotient may be set down for the corresponding function. 508 An inspection of the foregoing equations shews:

1. That 𝑓_𝑚(𝑥) cannot be zero; for, if it were, 𝑓(𝑥) and 𝑓₁(𝑥) would have a common factor, and therefore 𝑓(𝑥) would have equal roots, by (432)
2. Two consecutive functions, after the first, cannot vanish together; for this would make 𝑓_𝑚(𝑥) zero.
3. When any function, after the first, vanishes, the two adjacent ones have contrary signs

509 If, as 𝑥 increases, 𝑓(𝑥) passes through the value zero, Sturm's functions lose one change of sign.
For, before 𝑓(𝑥) takes the value zero, 𝑓(𝑥) and 𝑓₁(𝑥) have contrary signs, and afterwards they have the same sign; as may be shewn by making ℎ small, and changing its sign in the expansion of 𝑓(𝑥+ℎ), by (426). 510 If any other of Sturm's functions vanishes, there is neither loss nor gain in the number of changes of sign.
This will appear on inspecting the equations. 511 Result. The number of roots of 𝑓(𝑥) between 𝑎 and 𝑏 is equal to the difference in the number of changes of sign in Sturm;s functions, when 𝑥=𝑎 and when 𝑥=𝑏. 512 Cor. The total number of roots of 𝑓(𝑥) will be found by taking 𝑎=+∞ and 𝑏=−∞; the sign of each function will then be the same as that of its first term. 513 When the number of functions exceed the degree of 𝑓(𝑥) by unity, the two following theorems hold: If the first terms in all the functions, after the first, are positive; all the roots of 𝑓(𝑥) are real. 514 If the first terms are not all positive; then, for every change of sign, there will be a pair of imaginary roots.
For the proof put 𝑎=+∞ and 𝑏=−∞, and examine the number of changes of sign in each case, applying Descartes' rule 416 515 If 𝜙(𝑥) has no factor in common with 𝑓(𝑥), and if 𝜙(𝑥) and 𝑓'(𝑥) take the same sigtn when 𝑓(𝑥)=0; then the rest of Sturm's functions may be found from 𝑓(𝑥) and 𝜙(𝑥), instead of 𝑓'(𝑥). For the reasoning in (509) and (510) will apply to the new functions. 516 If Sturm's functions be formed without first removing equal roots from 𝑓(𝑥), the theorem will still give the number of distinct roots, without repetitions, between assigned limits.
For if 𝑓(𝑥) and 𝑓₁(𝑥) be divided by their highest common factor (see 444), and if the quotients be used instead of 𝑓(𝑥) and 𝑓₁(𝑥) to form Sturm's functions; then, by (515), the theorem will apply to the new set of functions, which will differ only from those formed from 𝑓(𝑥) and 𝑓₁(𝑥) by the absence of the same factor in every term of the series. 517 Example:- To find the position of the roots of the equation 𝑥⁴−4𝑥³+𝑥²+6𝑥+2=0 Sturm's functions, formed according to the rule given above, are here calculated. 𝑓(𝑥)=𝑥4−4𝑥3+𝑥2+6𝑥+2 𝑓1(𝑥)=  2𝑥3−6𝑥2+𝑥+3 𝑓2(𝑥)=    5𝑥2−10𝑥−7 𝑓3(𝑥)=      𝑥−1 𝑓3(𝑥)=        12 The first terms of the functions are all positive; therefore there is no imaginary root. 𝑥=−2−101234 𝑓(𝑥)=+++++++ 𝑓1(𝑥)=−−++−++ 𝑓2(𝑥)=++−−−++ 𝑓3(𝑥)=−−−++++ 𝑓4(𝑥)=+++++++ No. of changes of signs4422200 The changes of sign in the functions, as 𝑥 passes through integral values, are exhibited in the adjoining table. There are two changes of sign lost while 𝑥 passes from −1 to 0, and two more lost while 𝑥 passes from 2 to 3. There are therefore two roots lying between 0 and −1; and two roots also between 2 and 3.
These roots are all incommensurable, by (503). 518 Fourier's Theorem: Fourier's functions are the following quantities 𝑓(𝑥), 𝑓′(𝑥), 𝑓″(𝑥), ⋯, 𝑓ⁿ(𝑥) 519 Properties of Fourier's functions: As 𝑥 increases, Fourier's functions lose one change of sign for each roo of the equation 𝑓(𝑥)=0, through which 𝑥 passes, and 𝑟 changes of sign for 𝑟 repeated roots. 520 If any of the other functions vanish, an even number of changes of sign is lost. 521 Results: The number of real roots of 𝑓(𝑥) between 𝛼 and 𝛽 cannot be more than the difference between the number of changes of sign in Fourier's functions when 𝑥=𝛼, and the number of changes when 𝑥=𝛽. 522 When that difference is odd, the number of intermediate roots is odd, and therfore one at least. 523 When the same difference is even, the number of intermediate roots is either even or zero. 524 Descartes's rule of signs follows from the above for the sign of Fourier's functions, when 𝑥=0 are the signs of the terms in 𝑓(𝑥); and when 𝑥=∞, Fourier's functions are all positive. 525

Lagrange's method of approximating to the incommensurable roots of an equation.

Let 𝛼 be the greatest integer less than an incommensurable root of 𝑓(𝑥). Diminish the roots of 𝑓(𝑥) by 𝑎. Take the reciprocal of the resulting equation. Let 𝑏 be the greates integer less than a positive root of this equation. Diminish the roots of this equation by 𝑏, and proceed as before. 526 Let 𝑎, 𝑏, 𝑐, ⋯, be the quantities thus determined; then, an approximation to the incommensurable root of 𝑓(𝑥) will be the continued fraction 𝑥=𝑎+1𝑏+1𝑐+ 527 Newton's method of approximation: If 𝑐₁ be a quantity a little less than that one of the roots of the equation 𝑓(𝑥)=0, so that 𝑓(𝑐₁+ℎ)=0; then 𝑐₁ is a first approximation to thevalue of the root. Also because 𝑓(𝑐₁+ℎ)=𝑓(𝑐₁)+ℎ𝑓′(𝑐₁)+ℎ²|2𝑓″(𝑐₁)+⋯426 and ℎ is but small, a second approximation to the root will be 𝑐₁−𝑓(𝑐₁)𝑓′(𝑐₁)=𝑐₂ In the same way a third approximation may be obtained from 𝑐₂ and so on. 528 Fourier's limitation of Newton's method: To ensure that 𝑐₁, 𝑐₂, 𝑐₃, ⋯ shall successively increase up to the value 𝑐₁+ℎ without passing beyond it, it is necessary for all values of 𝑥 between 𝑐₁ and 𝑐₁+ℎ. (i.) That 𝑓(𝑥) and 𝑓′(𝑥) should have contrary signs (ii.) That 𝑓(𝑥) and 𝑓″(𝑥) should have same signs fig. A proof may be obtained from the figure. Draw the curve 𝑦=𝑓(𝑥). Let 𝑂𝑋 be a root of the equation, 𝑂𝑁=𝑐₁; draw the successive ordinates and tangents 𝑁𝑃, 𝑃𝑄, 𝑄𝑅, ⋯. Then 𝑂𝑄=𝑐₂, 𝑂𝑆=𝑐₃ and so on.
Fig. (2) represents 𝑐₂>𝑂𝑋, and the subsequent approximations decreasing towards the root. 530 Newton's Rule for Limits of the Roots: Let the coefficients of 𝑓(𝑥) be respectively divided by the Binomial coefficients, and let 𝑎₀, 𝑎₁, 𝑎₂, ⋯, 𝑎_𝑛, be the quotients, so that 𝑓(𝑥)=𝑎₀𝑥^𝑛+𝑛𝑎₁𝑥^𝑛−1+𝑛(𝑛−1)1⋅2𝑎₂𝑥^𝑛−2+⋯+𝑛𝑎_𝑛−1𝑥+𝑎_𝑛 Let 𝐴₁, 𝐴₂, 𝐴₃, ⋯, 𝐴_𝑛, be formed by the law 𝐴_𝑟=𝑎²_𝑟−𝑎_𝑟−1𝑎_𝑟+1. Write the first series of quantities over the second, in the following manner: 𝑎₀𝑎₁𝑎₂𝑎₃⋯𝑎_𝑛−1𝑎_𝑛 𝐴₀𝐴₁𝐴₂𝐴₃⋯𝐴_𝑛−1𝐴_𝑛 Whenever two adjacent terms in the first series have the same sign, and the two corresponding terms below them in the second series also the same sign; let this be called a double permancence. When the two adjacent terms above have different signs, and the two below the same sign, let this be known as a variation-permancence. 531 Rule: The number of double permancences in the associated series is a superior limit to the number of negative roots of 𝑓(𝑥).
The number of variation-permancences is a superior limit to the number of positive roots.
The number of imaginary roots cannot be less than the number of variations of sign in the second series. 532 Sylvester's Theorem: Let 𝑓(𝑥+𝜆) be expanded by (426) in powers of 𝑥, and let the two series be formed as in Newton's Rule (530).
Let 𝑃(𝜆) denote the number of double permanences.
Then 𝑃(𝜆)~𝑃(𝜇) is either equal to the number of roots of 𝑓(𝑥), or surpasses that number by an even integer.
Note: The first series may be multiplied by |𝑛, and will then stand thus, 𝑓^𝑛(𝜆), 𝑓^𝑛−1(𝜆), |2𝑓^𝑛−2(𝜆), |3𝑓^𝑛−3(𝜆), ⋯, |𝑛𝑓(𝜆) The second series may be reduced to 𝐺_𝑛(𝜆), 𝐺_𝑛−1(𝜆), 𝐺_𝑛−2(𝜆),⋯, 𝐺(𝜆), where 𝐺_𝑟(𝜆)≡{𝑓^𝑟(𝜆)}²−𝑛−𝑟+1𝑛−𝑟𝑓^𝑟−1(𝜆)𝑓^𝑟+1(𝜆) 533 Horner's Method: To find the numerical values of the roots of an equation. Take, for example, the equation 𝑥⁴−4𝑥³+𝑥²+6𝑥+2=0 and find limits of the roots by Sturm's Method or otherwise.
It has been shewn in (517) that this equation has two incommensurable roots between 2 and 3. The process of calculating the least of these roots is here exhibited.  −4     +1        +6           +2(2.414213      2     −4        −6           0               −2     −3        0          𝐴120000            2     0        −6           −19584            0     −3       𝐵1−6000       𝐴24160000       2     4        1104        −2955839       2    𝐶1100      −4806       𝐴312041610000  2     176      1872        −11437245184 𝐷140    276     𝐵2−3024000    𝐴4604364816  4    192      68161     −566003348  44    468      −2955839    𝐴538361468  4    208      68723     −28285470  48   𝐶267600   𝐵3−2887116000 𝐴610075998  4    561    27804704  − 8485368  52    68161    −2859311296 𝐴71590630  4    562    27895072 𝐷2560   68723   𝐵4−2831446224  1   563    139918   282843)1590630(562372  561  𝐶36928600  −283001674   1414215  1   22576  139970   28284)176415  562   6951176 𝐵5−282861704   169706  1   22592  700    2828)  6709  563   6973768  −28285470    5657  1   22608  700    282)  1052 𝐷35640 𝐶46996376 𝐵6−28284770    848  4  11    21     28)  204  5644  69974    −2828456     197  4  11    21     2)  7  5648  69985   𝐵7−2828435     5  14  11       2  5652 𝐶569996    4 𝐶67       𝐷45656 Root=2.414213562372 Method: Diminish the roots by 2 in the manner of (427). The resulting coefficients are indicated by 𝐴₁, 𝐵₁, 𝐶₁, 𝐷₁. By Newton's rule (527), −𝑓(𝑐)𝑓′(𝑐), that is, −𝐴₁𝐵₁ is an approximation to the remaining part of the root. This gives ⋅3 for the next figure; ⋅4 will be found to be the correct one. The highest figure must be taken which will not change the sign of 𝐴.
Diminish the roots by ⋅4. This is accomplished most easily by affixingt ciphers to 𝐴₁, 𝐵₁, 𝐶₁, 𝐷₁, in the manner shewn, and then employing 4 instead of ⋅4.
Having obtained 𝐴₂ and observing that its sign is +, retrace the steps, trying 5 instead of 4. This gives 𝐴₂ with a minus sign, thereby proving the existence of a root between 2⋅4 and 2⋅5. The new coefficients are 𝐴₂, 𝐵₂, 𝐶₂, 𝐷₂.
−𝐴₂𝐵₂ gives 1 for the next figure of the root.
Affix ciphers as before, and diminish the roots by 1, distinguishin the new coefficients as 𝐴₃, 𝐵₃, 𝐶₃, 𝐷₁₌₃.
Note that at every stage of the work 𝐴 and 𝐵 must preserve their signs unchanged. If a change of sign takes place it shews that tow large a figure has been tried.
To abridge the calculation proceed thus: After a certain number of figures of the root have been obtained (in this example four), instead of adding ciphers cut off one digit from 𝐵₄, two from 𝐶₄ and three from 𝐷₄. This amounts to the same thing as addign the ciphers, and then dividing each number by 10000.
Continue the work with the numbers so reduced, and cut off digits in like manner at each stage until the 𝐷 and 𝐶 columns have disappeared.
𝐴₇ and 𝐵₇ now alone remain, and six additional figures of the root are determined correctly by the division of 𝐴₇ and 𝐵₇.
To find the other root which lies between 2 and 3, we proceed as follows: After diminishing the roots by 2, try 6 for the next figure. This gives 𝐴₂ negative; 7 does the same, but 8 makes 𝐴₂ positive. That is to say, 𝑓(2⋅7) is negative, and 𝑓(2⋅8) positive. Therefore a root exists between 2⋅7 and 2⋅8, and its value may be approximated to, in the manner shewn.
Throughout this last calculation 𝐴 will preserve the negative sign. Observe also that the trial number for the next figure of the root given at each stage of the process by the formula −𝑓(𝑐)𝑓′(𝑐), will in this case be always too great, as in the former case it was always too small.

Sources and References

https://archive.org/details/synopsis-of-elementary-results-in-pure-and-applied-mathematics-pdfdrive