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" Ex x2 + bx ~ x2 + bz x2 + bx~ z2 + bx Hence ax + c = Ax + Ab + Bx ; and since this condition is to be fulfilled without reference to the value of x, the principle of indeterminate coefficients will furnish the separate equations c = Ab, and a = A + B. "
A Treatise on the Differential and Integral Calculus: And on the Calculus of ... - Page 261
by Edward Henry Courtenay - 1857 - 501 pages
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Elements of the Differential and Integral Calculus

Charles Davies - Calculus - 1836 - 296 pages
...(x — o)~(x + o)(x — o)(x — a) Let us now make a? A , A' , B (x-aY(x + a) ~(xa,y+ (xa) +x + a' Reducing the terms of the second member to a common denominator, we have and developing, and comparing the coefficients of the like powers of x, we obtain the equations A'+B=l,...
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Elements of the Differential and Integral Calculus

Charles Davies - Calculus - 1838 - 298 pages
...+bx 2 . x(a — x) (a + x) ax. Let us now make , x(a — x)(a + x) xa — xa + x' 19 =^ + -*- + .° reducing the terms of the second member to a common denominator, we have a 3 + bx 2 _Aa 2 —Aa?+ Sox + Ba?+ Pax — Ca? x(a — x)(a and, comparing the like powers of x (Alg....
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Elements of the Differential and Integral Calculus

Charles Davies - Calculus - 1838 - 300 pages
...ax. x(a — x)(a + x) Let us now make a3 + Z^ _A B , C • — r • x(a — x)(a + x) xa — xa + x reducing the terms of the second member to a common denominator, we hare cf+la? _A(f— .4-rf- Bax±Bj~ + Cax— Cx2 x(a—x)(a + x)~ x(a — x) and, comparing the like...
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A Treatise on the Differential and Integral Calculus: And on the Calculus of ...

Edward Henry Courtenay - Calculus - 1855 - 526 pages
...this case, take the example ax + c ax + c ax -f- c A JB Assume = — | -- - where A and B are unknown constants whose values are to be determined by the...the second member to a common denominator, we have ax + c _ A(x + b) Ex _ Ax + Ab + Ex x2 + bx ~ x2 + bz x2 + bx~ z2 + bx Hence ax + c = Ax + Ab + Bx...
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A Treatise on the Differential and Integral Calculus: And on the Calculus of ...

Edward Henry Courtenay - Mathematics - 1856 - 524 pages
...case, take the example ax + c ax + c ax + c A . B Assume — — = — where A and B are unknown z2 + bx xx + b constants whose values are to be determined...the second member to a common denominator, we have ax + c ~A(x + b) Bx — Ax + Ab + Bx x2 + bx~ x2 + bx x2 + bx ~ x2 + bx Hence ax + c = Ax + Ab + -Sx...
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A Treatise on the Differential and Integral Calculus: And on the Calculus of ...

Edward Henry Courtenay - Calculus - 1868 - 530 pages
...ay = — — ax = — ; rr«*xz + bx x(x + b) Assume , , , = — | r where A and B are unknown x2 + bx xx + b ! constants whose values are to be determined...Reducing the terms of the second member to a common denomi nator, we have ax + c _ A(x + b) Ex _Ax + Ab + JBx x z* + bx ~ x2 + bx xz + bx ~ x* + bx Hence...
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A General Geometry and Calculus: Including Part I. of the General Geometry ...

Edward Olney - Calculus - 1871 - 394 pages
...fractions integrated separately. A _ + ? 4 _C ÍL. To v + a)" (x + a>"~1 (* + a;1—» a; -f- a bring the terms of the second member to a common denominator, we have to multiply В by x + а, С by (x + a)» - and A" by (x + a)"-1; hence the numerators, when added...
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A University Algebra

Edward Olney - 1878 - 360 pages
...— + т н т , ж, 1 — x, and 1 + x being the unequal factors of x — x3 (115, 117)- Bringing the terms of the second member to a common denominator, we have x* —2 _ A — Ax' + Bx + Bx* + Cx — Cx* Щ J.3 "" " yj 1 _ x\l\ j. z\ • Hence x* — 2 = A +...
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A University Algebra

Edward Olney - Algebra - 1880 - 354 pages
...ж, and 1 + x being ж — хл х 1-х l + х the unequal factors of x — x3 (1.15, 117). Bringing the terms of the second member to a common denominator, we have z' — a _ A — Ax* + Bx + Bx* + Cx — Cx* яс — х* ~ x(l — x)(l + x) Hence x* — 2 = A +...
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A University Algebra: Comprising a Compendious, Yet Complete and Thorough ...

Edward Olney - Algebra - 1882 - 358 pages
...x,\ — x, and 1 + x being X — X* X 1-Х l +X the unequal factors of x — x3 (115, 117)- Bringing the terms of the second member to a common denominator, we have x* - 2 _ A - Ax* + Bx + Bxf + Ox -* Cxt xx*~ x(lx)(l + x) Hence х* — 2 = А + (В+Суе + (В—...
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