Theory of Equation

The Derived Functions of 𝑓(𝑥)

Rule for forming the derived functions 424 Multiply each term by the index of 𝑥, and reduce the index by one; that is, differentiate the function with respect to 𝑥.

Example

Take 𝑓(𝑥)=𝑥5+𝑥4+𝑥3−𝑥2−𝑥−1 𝑓1(𝑥)=5𝑥4+4𝑥3+3𝑥2−2𝑥 −1  𝑓2(𝑥)=20𝑥3+12𝑥2+6𝑥 −2   𝑓3(𝑥)=60𝑥2+24𝑥 +6   𝑓4(𝑥)=120𝑥 +24   𝑓5(𝑥)=120   𝑓¹(𝑥), 𝑓²(𝑥), ⋯ are called the first, second, ⋯ derived functions of 𝑓(𝑥). 425 To form the equation whose roots differ from those of 𝑓(𝑥) by a quantity 𝑎.
Put 𝑥=𝑦+𝑎 in 𝑓(𝑥), and expand each term by the Binomial Theorem, arranging the results in vertical columns in the following manner: 𝑓(𝑎+𝑦)= (𝑎+𝑦)5+(𝑎+𝑦)4+(𝑎+𝑦)3−(𝑎+𝑦)2−(𝑎+𝑦)−1  = 𝑎5+𝑎4+𝑎3−𝑎2−𝑎−1   +(  5𝑎4+4𝑎3+3𝑎2−2𝑎−1 )𝑦    +(    10𝑎3+6𝑎2+3𝑎−1 )𝑦2   +(      10𝑎2+4𝑎+1 )𝑦3   +(        5𝑎+1 )𝑦4   +               𝑦5 426 Comparing this result with that seen in (424), it is seen that 𝑓(𝑎+𝑦)=𝑓(𝑎)+𝑓¹(𝑎)𝑦+𝑓²(𝑎)|2𝑦²+𝑓³(𝑎)|3𝑦³+𝑓⁴(𝑎)|4𝑦⁴+𝑓⁵(𝑎)|5𝑦⁵ so that the coefficient generally of 𝑦^𝑟 in the transformed equation is 𝑓^𝑟(𝑎)|𝑟. 427 To form the equation most expeditionsly when 𝑎 has a numerical value, divide 𝑓(𝑥) continuously by 𝑥−𝑎, and the successive remainders will furnish the coefficients.

Example

To expand 𝑓(𝑦+2) when, as in (425), 𝑓(𝑥)=𝑥⁵+𝑥⁴+𝑥³−𝑥²−𝑥−1 Divide repeatedly by 𝑥−2, as follows:-  1+1+1−1−1−1 2  +2+6+14+26+50  1+3+7+13+25+49=𝑓(2) 2  +2+10+34+94  1+5+17+47+119  =𝑓1(2) 2  +2+14+62  1+7+31+109    =𝑓2(2)|2 2  +2+18  1+9+49      =𝑓3(2)|3 2  +2  1+11        =𝑓4(2)|4     1          =𝑓5(2)|5 That these remainders are the required coefficients is seen by inspecting the form of the equation (426); for if that equation be divided by 𝑥−𝑎=𝑦 repeatedly, these remainders are obviously produced when 𝑎=2.
Thus the equation, whose roots are each less by 2 than the roots of the proposed equation, is 𝑦⁵+11𝑦⁴+49𝑦³+109𝑦²+119𝑦+49=0. 428 To make any assigned term vanish in the transformed equation, 𝑎 must be so determined that the coefficient of that term shall vanish.

Example

In order that there may be no term involving 𝑦⁴ in equation (426), we must have 𝑓⁴(𝑎)=0. Find 𝑓⁴(𝑎) as in (424); thus 120𝑎+24=0; ∴𝑎=−15 The equation in (424) must now be divided repeatedly by 𝑥+15 after the manner of (427), and the resulting equation will be minus its second term. 429 Note, that to remove the second term of the equation 𝑓(𝑥)=0, the requisite value of 𝑎 is =−𝑝₁𝑛𝑝₀; that is, the coefficient of the second term, with the sign changed, divided by the coefficient of the first term, and by the number expressing the degree of the equation. 430 To transform 𝑓(𝑥) into an equation in 𝑦 so that 𝑦=𝜙(𝑥), a given function of 𝑥, put 𝑥=𝜙⁻¹(𝑦), the inverse function of 𝑦.

Example

To obtain an equation whose roots are respectively three times the roots of the equation 𝑥³−6𝑥+1=0. Here 𝑦=3𝑥; therefore 𝑥=𝑦3, and the equation becomes 𝑦³27−6𝑦3+1=0, or 𝑦³−54𝑦+27=0. 431 To transform 𝑓(𝑥)=0 into an equation in which the coefficient of the first term shall be unity, and theother coefficients the least possible integers.

Example

Take the equation 288𝑥³+240𝑥²−176𝑥−21=0 Divide by the coefficient of the first term, and reduce the fractions; the eqaution becomes 𝑥³+56𝑥²−1118𝑥−796=0 Substitute 𝑦𝑥 for 𝑥, and multiply by 𝑘; we get 𝑦³+5𝑘6𝑦²−11𝑘²18𝑦−7𝑘³96=0 Next resolve the denominators into their prime factors, 𝑦³+5𝑘2⋅3𝑦²−11𝑘²2⋅3²𝑦−7𝑘³2⁵⋅3=0 The smallest value must now be assigned to 𝑘, which will suffice to make each coefficient an integer. This is easily seen by inspection to be 2²⋅3=12, and the resulting equation is 𝑦³+10𝑦²−88𝑦−126=0 the roots of which are connected with the roots of the original equation by the relation 𝑦=12𝑥

Sources and References

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