Algebra

Recurring Series

𝑎₀+𝑎₁𝑥+𝑎₂𝑥²+𝑎₃𝑥³+⋯ is a recurring series if the coefficients are connected by the relation. 𝑎_𝑛=𝑝₁𝑎_𝑛−1+𝑝₂𝑎_𝑛−2+⋯+𝑝_𝑚𝑎_𝑛−𝑚. 251 The Scale of Relation is 1−𝑝₁𝑥−𝑝₂𝑥²−⋯−𝑝_𝑚𝑥^𝑚 252 [The first 𝑚 terms −𝑝₁𝑥 (first 𝑚−1 terms + the last term) −𝑝₂𝑥² (first 𝑚−2 terms + the last 2 terms) −𝑝₃𝑥³ (first 𝑚−3 terms + the last 3 terms) ⋯ −𝑝_𝑚−1𝑥^𝑚−1 (first term + the last 𝑚−1 terms) −𝑝_𝑚𝑥^𝑚 (the last 𝑚 terms)]÷[1−𝑝₁𝑥−𝑝₂𝑥²−⋯−𝑝_𝑚𝑥^𝑚] 253 If the series converges, and the sum to infinity is required, omit all "the last terms" from the formula.254 Example: Required the Scale of Relation, the general term, and the apparent sum to infinity, of the series 4𝑥+14𝑥²+40𝑥³+110𝑥⁴+304𝑥⁵−854𝑥⁶+⋯ Observe that six arbitrary terms given are sufficient to determine a Scale of Relation of the form 1−𝑝𝑥−𝑞𝑥²−𝑟𝑥³, involving three constants 𝑝, 𝑞, 𝑟, for, by (251), we can write three equations to determine these constants; namely, 110=40𝑝+14𝑞+4𝑟304=110𝑝+40𝑞+14𝑟854=304𝑝+110𝑞+40𝑟} The solution gives 𝑝=6, 𝑞=−11, 𝑟=6. Hence the Scale of Relation is 1−6𝑥+11𝑥²−6𝑥³. The sum of the series without limit will be found from (254), by putting 𝑝₁=6, 𝑝₂=−11, 𝑝₃=6, 𝑚=3. The first three terms=4𝑥+14𝑥²+40𝑥³ −6× the first two terms= −24𝑥²−84𝑥³ +11𝑥× the first term=   +44𝑥³  ⇒4𝑥−10𝑥² ∴ 𝑆=4𝑥−10𝑥²1−6𝑥+11𝑥²−6𝑥³ the meaning of which is that, if this fraction be expanded in ascending powers of 𝑥, the first six terms will be those given in the question.255 To obtain more terms of the series, we may use the Scale of Relation; thus the 7th term will be (6×854−11×304+6×110)𝑥⁷=2440𝑥⁷256 To find the general term, 𝑆 must be decomposed into partial fractions; thus, by the method of (235), 4𝑥−10𝑥²1−6𝑥+11𝑥²−6𝑥³=11−3𝑥+21−2𝑥−31−𝑥 By the Binomial Theorem (128), 11−3𝑥=1+3𝑥+32𝑥2+⋯+3𝑛𝑥𝑛 21−2𝑥=2+22𝑥+23𝑥2+⋯+3𝑛+1𝑥𝑛 −31−𝑥=−3−3𝑥−3𝑥2−⋯−3𝑥𝑛 Hence the general term involving 𝑥^𝑛 is (3^𝑛+2^𝑛+1−3)𝑥^𝑛. And by this formula we can write the "last terms" required in (253), and so obtain the sum of any finite number of terms of the given series. Also, by the same formula we can calculate the successive terms at the beginning of the series. In the present case this mode will be more expeditious than that of employing the Scale of Relation.257 If, in decomposing 𝑆 into partial fractions for the sake of obtaining the general term, a quadratic factor with imaginary roots should occur as a denominator, the same method must be pursued for the imaginary quantities will disappear in the final result. In this case, however, it is more convenient to employ a general formula. Suppose the fraction which gives rise to the imaginary roots to be 𝐿+𝑀𝑥𝑎+𝑏𝑥+𝑥²=𝐿+𝑀𝑥(𝑝−𝑥)(𝑞−𝑥) 𝑝 and 𝑞 being the imaginary roots of 𝑎+𝑏𝑥+𝑥²=0. Suppose {𝑝=𝛼+𝑖𝛽𝑞=𝛼−𝑖𝛽, where 𝑖=-1 If, now, the above fraction be resolved into two partial fractions in the ordinary way, and if these fractions be expanded separately by the Binomial Theorem, and that part of the general term furnished by these two expansions written out, still retaining 𝑝 and 𝑞, and if the imaginary values of 𝑝 and 𝑞 be then substituted, it will be found that the factor will disappear, and that the result may be enunciated as follows.258 The coefficient of 𝑥^𝑛-1 in the expansion of 𝐿+𝑀𝑥(𝛼²+𝛽²)^𝑛-2𝛼𝑥+𝑥² will be 𝐿𝛽(𝛼2+𝛽2){𝑛𝛼𝑛-1𝛽-𝐶(𝑛,3)𝛼𝑛-3𝛽3+𝐶(𝑛,5)𝛼𝑛-5𝛽5-⋯}+𝑀𝛽(𝛼2+𝛽2)𝑛-1{(𝑛-1)𝛼𝑛-2𝛽-𝐶(𝑛-1,3)𝛼𝑛-4𝛽3+𝐶(𝑛-1,5)𝛼𝑛-6𝛽5-⋯} 259 With the aid of the known expansion of sin 𝑛𝜃 in Trigonometry, this formula for the 𝑛^th term may be reduced to (𝐿+𝑀𝛼)²+𝑀²𝛽²𝛽²(𝛼²+𝛽²)^𝑛⋅sin(𝑛𝜃-𝜙), in which 𝜃=tan^-1𝛽𝛼, 𝜙=tan^-1𝑀𝛽𝐿+𝑀𝛼 If 𝑛 be not greater than 100, sin(𝑛𝜃-𝜙) may be obtained from the tables correct to about six places of decimals, and accordingly the 𝑛^th term of the expansion may be found with corresponding accuracy. As an example, the 100^th term in the expansion of 1+𝑥5-2𝑥+𝑥² is readily found by this method to be 4182410⁴¹𝑥⁹⁹. 260

To determine whether a given Series is recurring or not

If certain first terms only of the series be given, a scale of relation may be found which shall produce a recurring series whose first terms are those given. The method is exemplified in (255). The number of unknown coefficients 𝑝, 𝑞, 𝑟, ⋯ to be assumed for the scale of relation must be equal to half the number of the given terms of the series, if that number be even. If the number of given terms be odd, it may be made even by prefixing zero for the first term of the series.261 Since this method may, however, produce zero values for one or more of the last coefficients in the scale of relation, it may be advisable in practice to determine a scale from the first two terms of the series, and if that scale does not produce the following terms, we may try a scale determined from the first four terms, and so on until the true scale is arrived at. If an indefinite number of terms of the series be given, we may find whether it is recurring or not by a rule of Lagrange's.262 Let the series be 𝑆=𝐴+𝐵𝑥+𝐶𝑥²+𝐷𝑥³+⋯ Divide unity by 𝑆 as far as two terms of the quotient, which will be of the form 𝑝+𝑞𝑥, and write the remainder in the form 𝑆'𝑥², 𝑆' being another indefinite series of the same form as 𝑆.
Next, divide 𝑆 by 𝑆' as far as two terms of the quotient, and write the remainder in the form 𝑆″𝑥².
Again, divide 𝑆' by 𝑆″, and proceed as before, and repeat this process until there is no remainder after one of the divisions. The series will then be proved to be a recurring series, and the order of the series, that is, the degree of the scale of relation, will be the same as the number of divisions which have been effected in the process.

Example

To determine whether the series 1, 3, 6, 10, 15, 21, 28, 36, 45, ⋯ is recurring or not. Introducing 𝑥, we may write 𝑆=1+3𝑥+6𝑥²+10𝑥³+15𝑥⁴+21𝑥⁵+28𝑥⁶+36𝑥⁷+45𝑥⁸+⋯ Then we shall have 𝑆4=1-3𝑥+⋯? with a remainder 3𝑥²+8𝑥³+15𝑥⁴+24𝑥⁵+35𝑥⁶+⋯ Therefore 𝑆'=3+8𝑥+15𝑥²+24𝑥³+35𝑥⁴+⋯ 𝑆𝑆'=13+𝑥9 with a remainder 19(𝑥²+3𝑥³+6𝑥⁴+10𝑥⁵+⋯) Therefore we may take 𝑆″=1+3𝑥+6𝑥²+10𝑥³+⋯ Lastly 𝑆'𝑆″=3-𝑥 without any remainder. Consequently the series is a recurring series of the third order. It is, in fact, the expansion of 11-3𝑥+3𝑥²-𝑥³ 262

Sources and References

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