Did you know? You can know the first few Maclaurin terms of both a numerator and a denominator and still miss the cubic term of their quotient. The denominator creates contributions from lower powers that must be carried forward. For example, ln(1+x)/(1−x) has no constant term, but its coefficient of x³ is 5/6 — not merely the 1/3 from the logarithm. A useful shortcut is to match coefficients rather than attempting a large algebraic division at once.
This JC2 H2 Mathematics Tutor Bukit Timah guide answers a distinct A-Level 9758 problem: how do I build a Maclaurin series for a quotient, retain all terms through a requested degree, and check the range of validity? The first few paragraphs explain one recurrence for coefficient matching. A complete worked example follows, then a comparison with ordinary products and 48 original practice clinics that vary the numerator family and denominator.
For Bukit Timah JC2 learners, Maclaurin series convert complicated functions into simpler polynomials near zero. The 2027 SEAB H2 Mathematics syllabus 9758 lists standard expansions, deriving the first few terms from standard series or repeated differentiation, convergence ranges and small-angle approximation. It excludes deriving general terms of a series. This lesson therefore works only through cubic order and emphasises mathematical checks rather than unsupported claims that a short polynomial equals the exact function everywhere.
Quick answer: divide the series by matching coefficients
Suppose a numerator has N(x)=a₀+a₁x+a₂x²+a₃x³+O(x⁴) and the denominator is 1−qx. Write the quotient as Q(x)=c₀+c₁x+c₂x²+c₃x³+O(x⁴). Then (1−qx)Q(x)=N(x). Comparing like powers gives c₀=a₀, c₁=a₁+qc₀, c₂=a₂+qc₁, c₃=a₃+qc₂. Each new coefficient inherits a contribution from the preceding one.
- Write the source series: identify the numerator coefficients through the order requested.
- Write the denominator correctly: for 1−qx, use q with its sign; 1+x corresponds to q=−1.
- Match powers: the constant coefficient first, then x, x² and x³.
- Check by remultiplying: (1−qx) times your answer must reconstruct the numerator up to the omitted order.
- Check convergence: the geometric series for 1/(1−qx) requires |qx|<1, in addition to any numerator-series restriction.
- Label the approximation: a cubic truncation is not an identity valid for every x.
Worked example: ln(1+x)/(1−x) to cubic order
The familiar logarithmic series is ln(1+x)=x−x²/2+x³/3+O(x⁴) for |x|<1. The geometric expansion of the denominator’s reciprocal is 1/(1−x)=1+x+x²+x³+O(x⁴) for |x|<1. Multiplying the two gives the constant 0, linear coefficient 1, quadratic coefficient 1−1/2=1/2 and cubic coefficient 1−1/2+1/3=5/6.
Therefore ln(1+x)/(1−x)=x+(1/2)x²+(5/6)x³+O(x⁴) for |x|<1. Use a second route to audit. Set Q=c₀+c₁x+c₂x²+c₃x³. Then (1−x)Q has coefficients c₀, c₁−c₀, c₂−c₁ and c₃−c₂. Matching with 0, 1, −1/2 and 1/3 yields c₀=0, c₁=1, c₂=1/2, c₃=5/6. Both routes agree.
At x=0.1, the cubic approximation equals 0.1058333, while the original function ln(1.1)/0.9 is approximately 0.1059002. The difference of about 0.0000669 is small near zero and consistent with neglected higher-order terms. The numerical check does not mean the cubic polynomial can replace the original function accurately near x=1, where the denominator approaches zero.
The denominator’s signs matter
For 1/(1−x), the geometric series is 1+x+x²+x³+···. For 1/(1+x), substitute −x into that same series to obtain 1−x+x²−x³+···. The signs alternate. So sin x/(1+x) has expansion (x−x³/6)(1−x+x²−x³), giving x−x²+(5/6)x³+O(x⁴). If you accidentally use 1/(1−x), the quadratic term’s sign will be wrong.
The coefficient-matching method automatically handles the sign. With denominator 1+x, write 1−qx with q=−1. Then c₁=a₁−c₀, c₂=a₂−c₁ and so on. The recurrence is not a separate identity to memorise for positive and negative denominators; it is one mechanism with a signed q.
Another check: eˣ/(1−x)
Using eˣ=1+x+x²/2+x³/6+O(x⁴), compare coefficients for Q=eˣ/(1−x). We obtain c₀=1, c₁=1+1=2, c₂=1/2+2=5/2, c₃=1/6+5/2=8/3. Thus eˣ/(1−x)=1+2x+(5/2)x²+(8/3)x³+O(x⁴), for |x|<1.
A student who copies only the numerator’s cubic coefficient 1/6 has ignored the contributions arriving through the geometric denominator. The denominator is not an innocent scaling factor; it redistributes the coefficients of every lower power. Remultiplying (1−x)(1+2x+5x²/2+8x³/3) recovers 1+x+x²/2+x³/6 through cubic order.
Quotients with a quadratic denominator
The simplest coefficient recurrence above assumes the denominator 1−qx. But the same idea works for a more general denominator. For eˣ/(1+x²), write Q=c₀+c₁x+c₂x²+c₃x³+O(x⁴). Multiplying by 1+x² gives coefficients c₀, c₁, c₂+c₀ and c₃+c₁. Matching the exponential series yields c₀=1, c₁=1, c₂=1/2−1=−1/2 and c₃=1/6−1=−5/6.
Hence eˣ/(1+x²)=1+x−x²/2−5x³/6+O(x⁴) near x=0. The lesson is transferable: multiply an unknown truncated quotient by the denominator, then equate the coefficients required by the known numerator. You need not derive a general nth-term expression, which the 2027 H2 syllabus excludes.
A denominator must not vanish at the expansion point
The quotient f(x)/g(x) has a regular Maclaurin expansion through elementary division near x=0 when the relevant functions are sufficiently smooth and the denominator g(0)≠0. If g(0)=0, ordinary coefficient matching with a nonzero constant term fails. For example, sin x/x has a removable limit at zero, and one may first simplify or reason via the known series. But 1/x is not finite at zero and has no ordinary Maclaurin power series there.
This distinction is more valuable than forcing every displayed fraction into a formula. Before setting Q=c₀+c₁x+···, inspect the value of the denominator at x=0. If the quotient has a singularity or an undefined point, the correct mathematical strategy may differ. The present exercises deliberately keep denominator constant term one so coefficient matching is straightforward and within scope.
Convergence depends on both pieces
The denominator 1−qx has geometric expansion for |qx|<1. A numerator such as ln(1+ax) has its own Maclaurin convergence restriction |ax|<1. To use both expansions safely, work in their common range. For instance, ln(1+2x)/(1−x) is represented by the combined series for |x|<1/2, a sufficient range set by the logarithmic part. For eˣ/(1−2x), the geometric denominator restricts the series to |x|<1/2 even though the exponential numerator is entire.
When a question asks only for the first few terms, the algebra can be carried out formally before discussing convergence. But the final reader-facing explanation should still say where the approximation is justified near zero. A truncated series may give a numerical value outside that range, but it is not a controlled approximation merely because a calculator returns a decimal.
Check the first two coefficients independently
If Q(x)=c₀+c₁x+··· is the Maclaurin expansion of a differentiable function, then c₀=Q(0) and c₁=Q′(0). For eˣ/(1−x), Q(0)=1 and the quotient rule gives Q′(0)=2, confirming the first two coefficients 1 and 2. These easy checks do not verify every later coefficient, but they often catch an incorrect denominator sign before more complicated arithmetic accumulates.
For ln(1+x)/(1−x), Q(0)=0 and Q′(0)=1. This agrees with the first terms x+···. An answer with a nonzero constant or negative linear coefficient should be rejected immediately. Coefficient-by-coefficient audits are often faster than recomputing a long expression after a mistake has already been carried into several terms.
The one-minute diagnostic
Ask a learner to write ln(1+x) through x³, then write 1/(1−x) through x³. Without multiplying everything, ask for the coefficients of x, x² and x³ in the quotient. The correct answers are 1, 1/2 and 5/6. Next change only the denominator to 1+x. The coefficients become 1, −3/2 and 11/6? No: calculate them carefully using the signed recurrence. With q=−1, c₁=1, c₂=−1/2−1=−3/2 and c₃=1/3−(−3/2)=11/6. The changed signs are a useful test of the actual mechanism.
Then ask for the safe near-zero range for ln(1+2x)/(1−x). The learner should identify |2x|<1 from the logarithm and |x|<1 from the reciprocal, so |x|<1/2 works for the combined series. This separates a coefficient error from a convergence error. A third change to eˣ/(1−x) tests whether the numerator’s nonzero constant affects the recurrence.
Forty-eight original coefficient-matching clinics
The clinics use familiar numerator families from the 2027 H2 syllabus and denominators of the form 1−qx. The coefficients are exact rational values through cubic order, not fitted measurements. Each case states the numerator series, computes the new quotient coefficients and indicates a range in which the expansion is justified. Work them in families, then cover the outputs and remultiply to check.
1. e^(1x) divided by 1−x
Try the expansion. Find the first four coefficients of e^(1x)/(1−x). The known numerator series through cubic order is 1 + x + 1/2x² + 1/6x³+O(x⁴). Identify the signed q=1 in 1−qx, then write an unknown quotient c₀+c₁x+c₂x²+c₃x³. Use coefficient matching in ascending powers rather than expanding many unnecessary higher terms.
Complete worked coefficients. The numerator coefficient ledger [a₀,a₁,a₂,a₃] is [1, 1, 1/2, 1/6]. Applying c₀=a₀ and cⱼ=aⱼ+q cⱼ₋₁ gives [1, 2, 5/2, 8/3]. Therefore the quotient is 1 + 2x + 5/2x² + 8/3x³+O(x⁴). Multiplying this truncated answer by (1−x) reconstructs the numerator coefficients through x³, which independently verifies the sign and every rational term.
Convergence and transfer. The reciprocal denominator is expanded only when |1x|<1. The numerator has no additional finite Maclaurin convergence restriction of this kind. A safe common range is |x|<1. Remember that all lower-degree terms can contribute to a higher quotient coefficient through the denominator. Replace q with −q and predict which coefficients change before calculating a second case.
2. ln(1+1x) divided by 1−x
Try the expansion. Find the first four coefficients of ln(1+1x)/(1−x). The known numerator series through cubic order is x − 1/2x² + 1/3x³+O(x⁴). Identify the signed q=1 in 1−qx, then write an unknown quotient c₀+c₁x+c₂x²+c₃x³. Use coefficient matching in ascending powers rather than expanding many unnecessary higher terms.
Complete worked coefficients. The numerator coefficient ledger [a₀,a₁,a₂,a₃] is [0, 1, -1/2, 1/3]. Applying c₀=a₀ and cⱼ=aⱼ+q cⱼ₋₁ gives [0, 1, 1/2, 5/6]. Therefore the quotient is x + 1/2x² + 5/6x³+O(x⁴). Multiplying this truncated answer by (1−x) reconstructs the numerator coefficients through x³, which independently verifies the sign and every rational term.
Convergence and transfer. The reciprocal denominator is expanded only when |1x|<1. The numerator’s relevant binomial or logarithmic series also needs |1x|<1. A safe common range is |x|<1. Check the sign of q before any expansion; a denominator with a plus sign corresponds to negative q. Replace q with −q and predict which coefficients change before calculating a second case.
3. sin(1x) divided by 1−x
Try the expansion. Find the first four coefficients of sin(1x)/(1−x). The known numerator series through cubic order is x − 1/6x³+O(x⁴). Identify the signed q=1 in 1−qx, then write an unknown quotient c₀+c₁x+c₂x²+c₃x³. Use coefficient matching in ascending powers rather than expanding many unnecessary higher terms.
Complete worked coefficients. The numerator coefficient ledger [a₀,a₁,a₂,a₃] is [0, 1, 0, -1/6]. Applying c₀=a₀ and cⱼ=aⱼ+q cⱼ₋₁ gives [0, 1, 1, 5/6]. Therefore the quotient is x + x² + 5/6x³+O(x⁴). Multiplying this truncated answer by (1−x) reconstructs the numerator coefficients through x³, which independently verifies the sign and every rational term.
Convergence and transfer. The reciprocal denominator is expanded only when |1x|<1. The numerator has no additional finite Maclaurin convergence restriction of this kind. A safe common range is |x|<1. A correct cubic coefficient must survive multiplication back by the original denominator. Replace q with −q and predict which coefficients change before calculating a second case.
4. cos(1x) divided by 1−x
Try the expansion. Find the first four coefficients of cos(1x)/(1−x). The known numerator series through cubic order is 1 − 1/2x²+O(x⁴). Identify the signed q=1 in 1−qx, then write an unknown quotient c₀+c₁x+c₂x²+c₃x³. Use coefficient matching in ascending powers rather than expanding many unnecessary higher terms.
Complete worked coefficients. The numerator coefficient ledger [a₀,a₁,a₂,a₃] is [1, 0, -1/2, 0]. Applying c₀=a₀ and cⱼ=aⱼ+q cⱼ₋₁ gives [1, 1, 1/2, 1/2]. Therefore the quotient is 1 + x + 1/2x² + 1/2x³+O(x⁴). Multiplying this truncated answer by (1−x) reconstructs the numerator coefficients through x³, which independently verifies the sign and every rational term.
Convergence and transfer. The reciprocal denominator is expanded only when |1x|<1. The numerator has no additional finite Maclaurin convergence restriction of this kind. A safe common range is |x|<1. Do not confuse a finite cubic approximation with equality to the exact function for every real x. Replace q with −q and predict which coefficients change before calculating a second case.
5. (1+1x)^(1/2) divided by 1−x
Try the expansion. Find the first four coefficients of (1+1x)^(1/2)/(1−x). The known numerator series through cubic order is 1 + 1/2x − 1/8x² + 1/16x³+O(x⁴). Identify the signed q=1 in 1−qx, then write an unknown quotient c₀+c₁x+c₂x²+c₃x³. Use coefficient matching in ascending powers rather than expanding many unnecessary higher terms.
Complete worked coefficients. The numerator coefficient ledger [a₀,a₁,a₂,a₃] is [1, 1/2, -1/8, 1/16]. Applying c₀=a₀ and cⱼ=aⱼ+q cⱼ₋₁ gives [1, 3/2, 11/8, 23/16]. Therefore the quotient is 1 + 3/2x + 11/8x² + 23/16x³+O(x⁴). Multiplying this truncated answer by (1−x) reconstructs the numerator coefficients through x³, which independently verifies the sign and every rational term.
Convergence and transfer. The reciprocal denominator is expanded only when |1x|<1. The numerator’s relevant binomial or logarithmic series also needs |1x|<1. A safe common range is |x|<1. If the numerator has no constant term, the quotient also has zero constant term when the denominator equals one at zero. Replace q with −q and predict which coefficients change before calculating a second case.
6. (1+1x)^(−1/2) divided by 1−x
Try the expansion. Find the first four coefficients of (1+1x)^(−1/2)/(1−x). The known numerator series through cubic order is 1 − 1/2x + 3/8x² − 5/16x³+O(x⁴). Identify the signed q=1 in 1−qx, then write an unknown quotient c₀+c₁x+c₂x²+c₃x³. Use coefficient matching in ascending powers rather than expanding many unnecessary higher terms.
Complete worked coefficients. The numerator coefficient ledger [a₀,a₁,a₂,a₃] is [1, -1/2, 3/8, -5/16]. Applying c₀=a₀ and cⱼ=aⱼ+q cⱼ₋₁ gives [1, 1/2, 7/8, 9/16]. Therefore the quotient is 1 + 1/2x + 7/8x² + 9/16x³+O(x⁴). Multiplying this truncated answer by (1−x) reconstructs the numerator coefficients through x³, which independently verifies the sign and every rational term.
Convergence and transfer. The reciprocal denominator is expanded only when |1x|<1. The numerator’s relevant binomial or logarithmic series also needs |1x|<1. A safe common range is |x|<1. Whenever the numerator is trigonometric, all angles in the standard series are in radians. Replace q with −q and predict which coefficients change before calculating a second case.
7. (1+1x)² divided by 1−x
Try the expansion. Find the first four coefficients of (1+1x)²/(1−x). The known numerator series through cubic order is 1 + 2x + x²+O(x⁴). Identify the signed q=1 in 1−qx, then write an unknown quotient c₀+c₁x+c₂x²+c₃x³. Use coefficient matching in ascending powers rather than expanding many unnecessary higher terms.
Complete worked coefficients. The numerator coefficient ledger [a₀,a₁,a₂,a₃] is [1, 2, 1, 0]. Applying c₀=a₀ and cⱼ=aⱼ+q cⱼ₋₁ gives [1, 3, 4, 4]. Therefore the quotient is 1 + 3x + 4x² + 4x³+O(x⁴). Multiplying this truncated answer by (1−x) reconstructs the numerator coefficients through x³, which independently verifies the sign and every rational term.
Convergence and transfer. The reciprocal denominator is expanded only when |1x|<1. The numerator has no additional finite Maclaurin convergence restriction of this kind. A safe common range is |x|<1. The recurrence uses exact rational arithmetic; early decimal rounding can hide the final coefficient. Replace q with −q and predict which coefficients change before calculating a second case.
8. 1+1x divided by 1−x
Try the expansion. Find the first four coefficients of 1+1x/(1−x). The known numerator series through cubic order is 1 + x+O(x⁴). Identify the signed q=1 in 1−qx, then write an unknown quotient c₀+c₁x+c₂x²+c₃x³. Use coefficient matching in ascending powers rather than expanding many unnecessary higher terms.
Complete worked coefficients. The numerator coefficient ledger [a₀,a₁,a₂,a₃] is [1, 1, 0, 0]. Applying c₀=a₀ and cⱼ=aⱼ+q cⱼ₋₁ gives [1, 2, 2, 2]. Therefore the quotient is 1 + 2x + 2x² + 2x³+O(x⁴). Multiplying this truncated answer by (1−x) reconstructs the numerator coefficients through x³, which independently verifies the sign and every rational term.
Convergence and transfer. The reciprocal denominator is expanded only when |1x|<1. The numerator has no additional finite Maclaurin convergence restriction of this kind. A safe common range is |x|<1. Both the numerator and denominator convergence restrictions matter for a valid combined series. Replace q with −q and predict which coefficients change before calculating a second case.
9. e^(2x) divided by 1+x
Try the expansion. Find the first four coefficients of e^(2x)/(1+x). The known numerator series through cubic order is 1 + 2x + 2x² + 4/3x³+O(x⁴). Identify the signed q=-1 in 1−qx, then write an unknown quotient c₀+c₁x+c₂x²+c₃x³. Use coefficient matching in ascending powers rather than expanding many unnecessary higher terms.
Complete worked coefficients. The numerator coefficient ledger [a₀,a₁,a₂,a₃] is [1, 2, 2, 4/3]. Applying c₀=a₀ and cⱼ=aⱼ+q cⱼ₋₁ gives [1, 1, 1, 1/3]. Therefore the quotient is 1 + x + x² + 1/3x³+O(x⁴). Multiplying this truncated answer by (1+x) reconstructs the numerator coefficients through x³, which independently verifies the sign and every rational term.
Convergence and transfer. The reciprocal denominator is expanded only when |1x|<1. The numerator has no additional finite Maclaurin convergence restriction of this kind. A safe common range is |x|<1. Remember that all lower-degree terms can contribute to a higher quotient coefficient through the denominator. Replace q with −q and predict which coefficients change before calculating a second case.
10. ln(1+2x) divided by 1+x
Try the expansion. Find the first four coefficients of ln(1+2x)/(1+x). The known numerator series through cubic order is 2x − 2x² + 8/3x³+O(x⁴). Identify the signed q=-1 in 1−qx, then write an unknown quotient c₀+c₁x+c₂x²+c₃x³. Use coefficient matching in ascending powers rather than expanding many unnecessary higher terms.
Complete worked coefficients. The numerator coefficient ledger [a₀,a₁,a₂,a₃] is [0, 2, -2, 8/3]. Applying c₀=a₀ and cⱼ=aⱼ+q cⱼ₋₁ gives [0, 2, -4, 20/3]. Therefore the quotient is 2x − 4x² + 20/3x³+O(x⁴). Multiplying this truncated answer by (1+x) reconstructs the numerator coefficients through x³, which independently verifies the sign and every rational term.
Convergence and transfer. The reciprocal denominator is expanded only when |1x|<1. The numerator’s relevant binomial or logarithmic series also needs |2x|<1. A safe common range is |x|<1/2. Check the sign of q before any expansion; a denominator with a plus sign corresponds to negative q. Replace q with −q and predict which coefficients change before calculating a second case.
11. sin(2x) divided by 1+x
Try the expansion. Find the first four coefficients of sin(2x)/(1+x). The known numerator series through cubic order is 2x − 4/3x³+O(x⁴). Identify the signed q=-1 in 1−qx, then write an unknown quotient c₀+c₁x+c₂x²+c₃x³. Use coefficient matching in ascending powers rather than expanding many unnecessary higher terms.
Complete worked coefficients. The numerator coefficient ledger [a₀,a₁,a₂,a₃] is [0, 2, 0, -4/3]. Applying c₀=a₀ and cⱼ=aⱼ+q cⱼ₋₁ gives [0, 2, -2, 2/3]. Therefore the quotient is 2x − 2x² + 2/3x³+O(x⁴). Multiplying this truncated answer by (1+x) reconstructs the numerator coefficients through x³, which independently verifies the sign and every rational term.
Convergence and transfer. The reciprocal denominator is expanded only when |1x|<1. The numerator has no additional finite Maclaurin convergence restriction of this kind. A safe common range is |x|<1. A correct cubic coefficient must survive multiplication back by the original denominator. Replace q with −q and predict which coefficients change before calculating a second case.
12. cos(2x) divided by 1+x
Try the expansion. Find the first four coefficients of cos(2x)/(1+x). The known numerator series through cubic order is 1 − 2x²+O(x⁴). Identify the signed q=-1 in 1−qx, then write an unknown quotient c₀+c₁x+c₂x²+c₃x³. Use coefficient matching in ascending powers rather than expanding many unnecessary higher terms.
Complete worked coefficients. The numerator coefficient ledger [a₀,a₁,a₂,a₃] is [1, 0, -2, 0]. Applying c₀=a₀ and cⱼ=aⱼ+q cⱼ₋₁ gives [1, -1, -1, 1]. Therefore the quotient is 1 − x − x² + x³+O(x⁴). Multiplying this truncated answer by (1+x) reconstructs the numerator coefficients through x³, which independently verifies the sign and every rational term.
Convergence and transfer. The reciprocal denominator is expanded only when |1x|<1. The numerator has no additional finite Maclaurin convergence restriction of this kind. A safe common range is |x|<1. Do not confuse a finite cubic approximation with equality to the exact function for every real x. Replace q with −q and predict which coefficients change before calculating a second case.
13. (1+2x)^(1/2) divided by 1+x
Try the expansion. Find the first four coefficients of (1+2x)^(1/2)/(1+x). The known numerator series through cubic order is 1 + x − 1/2x² + 1/2x³+O(x⁴). Identify the signed q=-1 in 1−qx, then write an unknown quotient c₀+c₁x+c₂x²+c₃x³. Use coefficient matching in ascending powers rather than expanding many unnecessary higher terms.
Complete worked coefficients. The numerator coefficient ledger [a₀,a₁,a₂,a₃] is [1, 1, -1/2, 1/2]. Applying c₀=a₀ and cⱼ=aⱼ+q cⱼ₋₁ gives [1, 0, -1/2, 1]. Therefore the quotient is 1 − 1/2x² + x³+O(x⁴). Multiplying this truncated answer by (1+x) reconstructs the numerator coefficients through x³, which independently verifies the sign and every rational term.
Convergence and transfer. The reciprocal denominator is expanded only when |1x|<1. The numerator’s relevant binomial or logarithmic series also needs |2x|<1. A safe common range is |x|<1/2. If the numerator has no constant term, the quotient also has zero constant term when the denominator equals one at zero. Replace q with −q and predict which coefficients change before calculating a second case.
14. (1+2x)^(−1/2) divided by 1+x
Try the expansion. Find the first four coefficients of (1+2x)^(−1/2)/(1+x). The known numerator series through cubic order is 1 − x + 3/2x² − 5/2x³+O(x⁴). Identify the signed q=-1 in 1−qx, then write an unknown quotient c₀+c₁x+c₂x²+c₃x³. Use coefficient matching in ascending powers rather than expanding many unnecessary higher terms.
Complete worked coefficients. The numerator coefficient ledger [a₀,a₁,a₂,a₃] is [1, -1, 3/2, -5/2]. Applying c₀=a₀ and cⱼ=aⱼ+q cⱼ₋₁ gives [1, -2, 7/2, -6]. Therefore the quotient is 1 − 2x + 7/2x² − 6x³+O(x⁴). Multiplying this truncated answer by (1+x) reconstructs the numerator coefficients through x³, which independently verifies the sign and every rational term.
Convergence and transfer. The reciprocal denominator is expanded only when |1x|<1. The numerator’s relevant binomial or logarithmic series also needs |2x|<1. A safe common range is |x|<1/2. Whenever the numerator is trigonometric, all angles in the standard series are in radians. Replace q with −q and predict which coefficients change before calculating a second case.
15. (1+2x)² divided by 1+x
Try the expansion. Find the first four coefficients of (1+2x)²/(1+x). The known numerator series through cubic order is 1 + 4x + 4x²+O(x⁴). Identify the signed q=-1 in 1−qx, then write an unknown quotient c₀+c₁x+c₂x²+c₃x³. Use coefficient matching in ascending powers rather than expanding many unnecessary higher terms.
Complete worked coefficients. The numerator coefficient ledger [a₀,a₁,a₂,a₃] is [1, 4, 4, 0]. Applying c₀=a₀ and cⱼ=aⱼ+q cⱼ₋₁ gives [1, 3, 1, -1]. Therefore the quotient is 1 + 3x + x² − x³+O(x⁴). Multiplying this truncated answer by (1+x) reconstructs the numerator coefficients through x³, which independently verifies the sign and every rational term.
Convergence and transfer. The reciprocal denominator is expanded only when |1x|<1. The numerator has no additional finite Maclaurin convergence restriction of this kind. A safe common range is |x|<1. The recurrence uses exact rational arithmetic; early decimal rounding can hide the final coefficient. Replace q with −q and predict which coefficients change before calculating a second case.
16. 1+2x divided by 1+x
Try the expansion. Find the first four coefficients of 1+2x/(1+x). The known numerator series through cubic order is 1 + 2x+O(x⁴). Identify the signed q=-1 in 1−qx, then write an unknown quotient c₀+c₁x+c₂x²+c₃x³. Use coefficient matching in ascending powers rather than expanding many unnecessary higher terms.
Complete worked coefficients. The numerator coefficient ledger [a₀,a₁,a₂,a₃] is [1, 2, 0, 0]. Applying c₀=a₀ and cⱼ=aⱼ+q cⱼ₋₁ gives [1, 1, -1, 1]. Therefore the quotient is 1 + x − x² + x³+O(x⁴). Multiplying this truncated answer by (1+x) reconstructs the numerator coefficients through x³, which independently verifies the sign and every rational term.
Convergence and transfer. The reciprocal denominator is expanded only when |1x|<1. The numerator has no additional finite Maclaurin convergence restriction of this kind. A safe common range is |x|<1. Both the numerator and denominator convergence restrictions matter for a valid combined series. Replace q with −q and predict which coefficients change before calculating a second case.
17. e^(3x) divided by 1−2x
Try the expansion. Find the first four coefficients of e^(3x)/(1−2x). The known numerator series through cubic order is 1 + 3x + 9/2x² + 9/2x³+O(x⁴). Identify the signed q=2 in 1−qx, then write an unknown quotient c₀+c₁x+c₂x²+c₃x³. Use coefficient matching in ascending powers rather than expanding many unnecessary higher terms.
Complete worked coefficients. The numerator coefficient ledger [a₀,a₁,a₂,a₃] is [1, 3, 9/2, 9/2]. Applying c₀=a₀ and cⱼ=aⱼ+q cⱼ₋₁ gives [1, 5, 29/2, 67/2]. Therefore the quotient is 1 + 5x + 29/2x² + 67/2x³+O(x⁴). Multiplying this truncated answer by (1−2x) reconstructs the numerator coefficients through x³, which independently verifies the sign and every rational term.
Convergence and transfer. The reciprocal denominator is expanded only when |2x|<1. The numerator has no additional finite Maclaurin convergence restriction of this kind. A safe common range is |x|<1/2. Remember that all lower-degree terms can contribute to a higher quotient coefficient through the denominator. Replace q with −q and predict which coefficients change before calculating a second case.
18. ln(1+3x) divided by 1−2x
Try the expansion. Find the first four coefficients of ln(1+3x)/(1−2x). The known numerator series through cubic order is 3x − 9/2x² + 9x³+O(x⁴). Identify the signed q=2 in 1−qx, then write an unknown quotient c₀+c₁x+c₂x²+c₃x³. Use coefficient matching in ascending powers rather than expanding many unnecessary higher terms.
Complete worked coefficients. The numerator coefficient ledger [a₀,a₁,a₂,a₃] is [0, 3, -9/2, 9]. Applying c₀=a₀ and cⱼ=aⱼ+q cⱼ₋₁ gives [0, 3, 3/2, 12]. Therefore the quotient is 3x + 3/2x² + 12x³+O(x⁴). Multiplying this truncated answer by (1−2x) reconstructs the numerator coefficients through x³, which independently verifies the sign and every rational term.
Convergence and transfer. The reciprocal denominator is expanded only when |2x|<1. The numerator’s relevant binomial or logarithmic series also needs |3x|<1. A safe common range is |x|<1/3. Check the sign of q before any expansion; a denominator with a plus sign corresponds to negative q. Replace q with −q and predict which coefficients change before calculating a second case.
19. sin(3x) divided by 1−2x
Try the expansion. Find the first four coefficients of sin(3x)/(1−2x). The known numerator series through cubic order is 3x − 9/2x³+O(x⁴). Identify the signed q=2 in 1−qx, then write an unknown quotient c₀+c₁x+c₂x²+c₃x³. Use coefficient matching in ascending powers rather than expanding many unnecessary higher terms.
Complete worked coefficients. The numerator coefficient ledger [a₀,a₁,a₂,a₃] is [0, 3, 0, -9/2]. Applying c₀=a₀ and cⱼ=aⱼ+q cⱼ₋₁ gives [0, 3, 6, 15/2]. Therefore the quotient is 3x + 6x² + 15/2x³+O(x⁴). Multiplying this truncated answer by (1−2x) reconstructs the numerator coefficients through x³, which independently verifies the sign and every rational term.
Convergence and transfer. The reciprocal denominator is expanded only when |2x|<1. The numerator has no additional finite Maclaurin convergence restriction of this kind. A safe common range is |x|<1/2. A correct cubic coefficient must survive multiplication back by the original denominator. Replace q with −q and predict which coefficients change before calculating a second case.
20. cos(3x) divided by 1−2x
Try the expansion. Find the first four coefficients of cos(3x)/(1−2x). The known numerator series through cubic order is 1 − 9/2x²+O(x⁴). Identify the signed q=2 in 1−qx, then write an unknown quotient c₀+c₁x+c₂x²+c₃x³. Use coefficient matching in ascending powers rather than expanding many unnecessary higher terms.
Complete worked coefficients. The numerator coefficient ledger [a₀,a₁,a₂,a₃] is [1, 0, -9/2, 0]. Applying c₀=a₀ and cⱼ=aⱼ+q cⱼ₋₁ gives [1, 2, -1/2, -1]. Therefore the quotient is 1 + 2x − 1/2x² − x³+O(x⁴). Multiplying this truncated answer by (1−2x) reconstructs the numerator coefficients through x³, which independently verifies the sign and every rational term.
Convergence and transfer. The reciprocal denominator is expanded only when |2x|<1. The numerator has no additional finite Maclaurin convergence restriction of this kind. A safe common range is |x|<1/2. Do not confuse a finite cubic approximation with equality to the exact function for every real x. Replace q with −q and predict which coefficients change before calculating a second case.
21. (1+3x)^(1/2) divided by 1−2x
Try the expansion. Find the first four coefficients of (1+3x)^(1/2)/(1−2x). The known numerator series through cubic order is 1 + 3/2x − 9/8x² + 27/16x³+O(x⁴). Identify the signed q=2 in 1−qx, then write an unknown quotient c₀+c₁x+c₂x²+c₃x³. Use coefficient matching in ascending powers rather than expanding many unnecessary higher terms.
Complete worked coefficients. The numerator coefficient ledger [a₀,a₁,a₂,a₃] is [1, 3/2, -9/8, 27/16]. Applying c₀=a₀ and cⱼ=aⱼ+q cⱼ₋₁ gives [1, 7/2, 47/8, 215/16]. Therefore the quotient is 1 + 7/2x + 47/8x² + 215/16x³+O(x⁴). Multiplying this truncated answer by (1−2x) reconstructs the numerator coefficients through x³, which independently verifies the sign and every rational term.
Convergence and transfer. The reciprocal denominator is expanded only when |2x|<1. The numerator’s relevant binomial or logarithmic series also needs |3x|<1. A safe common range is |x|<1/3. If the numerator has no constant term, the quotient also has zero constant term when the denominator equals one at zero. Replace q with −q and predict which coefficients change before calculating a second case.
22. (1+3x)^(−1/2) divided by 1−2x
Try the expansion. Find the first four coefficients of (1+3x)^(−1/2)/(1−2x). The known numerator series through cubic order is 1 − 3/2x + 27/8x² − 135/16x³+O(x⁴). Identify the signed q=2 in 1−qx, then write an unknown quotient c₀+c₁x+c₂x²+c₃x³. Use coefficient matching in ascending powers rather than expanding many unnecessary higher terms.
Complete worked coefficients. The numerator coefficient ledger [a₀,a₁,a₂,a₃] is [1, -3/2, 27/8, -135/16]. Applying c₀=a₀ and cⱼ=aⱼ+q cⱼ₋₁ gives [1, 1/2, 35/8, 5/16]. Therefore the quotient is 1 + 1/2x + 35/8x² + 5/16x³+O(x⁴). Multiplying this truncated answer by (1−2x) reconstructs the numerator coefficients through x³, which independently verifies the sign and every rational term.
Convergence and transfer. The reciprocal denominator is expanded only when |2x|<1. The numerator’s relevant binomial or logarithmic series also needs |3x|<1. A safe common range is |x|<1/3. Whenever the numerator is trigonometric, all angles in the standard series are in radians. Replace q with −q and predict which coefficients change before calculating a second case.
23. (1+3x)² divided by 1−2x
Try the expansion. Find the first four coefficients of (1+3x)²/(1−2x). The known numerator series through cubic order is 1 + 6x + 9x²+O(x⁴). Identify the signed q=2 in 1−qx, then write an unknown quotient c₀+c₁x+c₂x²+c₃x³. Use coefficient matching in ascending powers rather than expanding many unnecessary higher terms.
Complete worked coefficients. The numerator coefficient ledger [a₀,a₁,a₂,a₃] is [1, 6, 9, 0]. Applying c₀=a₀ and cⱼ=aⱼ+q cⱼ₋₁ gives [1, 8, 25, 50]. Therefore the quotient is 1 + 8x + 25x² + 50x³+O(x⁴). Multiplying this truncated answer by (1−2x) reconstructs the numerator coefficients through x³, which independently verifies the sign and every rational term.
Convergence and transfer. The reciprocal denominator is expanded only when |2x|<1. The numerator has no additional finite Maclaurin convergence restriction of this kind. A safe common range is |x|<1/2. The recurrence uses exact rational arithmetic; early decimal rounding can hide the final coefficient. Replace q with −q and predict which coefficients change before calculating a second case.
24. 1+3x divided by 1−2x
Try the expansion. Find the first four coefficients of 1+3x/(1−2x). The known numerator series through cubic order is 1 + 3x+O(x⁴). Identify the signed q=2 in 1−qx, then write an unknown quotient c₀+c₁x+c₂x²+c₃x³. Use coefficient matching in ascending powers rather than expanding many unnecessary higher terms.
Complete worked coefficients. The numerator coefficient ledger [a₀,a₁,a₂,a₃] is [1, 3, 0, 0]. Applying c₀=a₀ and cⱼ=aⱼ+q cⱼ₋₁ gives [1, 5, 10, 20]. Therefore the quotient is 1 + 5x + 10x² + 20x³+O(x⁴). Multiplying this truncated answer by (1−2x) reconstructs the numerator coefficients through x³, which independently verifies the sign and every rational term.
Convergence and transfer. The reciprocal denominator is expanded only when |2x|<1. The numerator has no additional finite Maclaurin convergence restriction of this kind. A safe common range is |x|<1/2. Both the numerator and denominator convergence restrictions matter for a valid combined series. Replace q with −q and predict which coefficients change before calculating a second case.
25. e^(1x) divided by 1+2x
Try the expansion. Find the first four coefficients of e^(1x)/(1+2x). The known numerator series through cubic order is 1 + x + 1/2x² + 1/6x³+O(x⁴). Identify the signed q=-2 in 1−qx, then write an unknown quotient c₀+c₁x+c₂x²+c₃x³. Use coefficient matching in ascending powers rather than expanding many unnecessary higher terms.
Complete worked coefficients. The numerator coefficient ledger [a₀,a₁,a₂,a₃] is [1, 1, 1/2, 1/6]. Applying c₀=a₀ and cⱼ=aⱼ+q cⱼ₋₁ gives [1, -1, 5/2, -29/6]. Therefore the quotient is 1 − x + 5/2x² − 29/6x³+O(x⁴). Multiplying this truncated answer by (1+2x) reconstructs the numerator coefficients through x³, which independently verifies the sign and every rational term.
Convergence and transfer. The reciprocal denominator is expanded only when |2x|<1. The numerator has no additional finite Maclaurin convergence restriction of this kind. A safe common range is |x|<1/2. Remember that all lower-degree terms can contribute to a higher quotient coefficient through the denominator. Replace q with −q and predict which coefficients change before calculating a second case.
26. ln(1+1x) divided by 1+2x
Try the expansion. Find the first four coefficients of ln(1+1x)/(1+2x). The known numerator series through cubic order is x − 1/2x² + 1/3x³+O(x⁴). Identify the signed q=-2 in 1−qx, then write an unknown quotient c₀+c₁x+c₂x²+c₃x³. Use coefficient matching in ascending powers rather than expanding many unnecessary higher terms.
Complete worked coefficients. The numerator coefficient ledger [a₀,a₁,a₂,a₃] is [0, 1, -1/2, 1/3]. Applying c₀=a₀ and cⱼ=aⱼ+q cⱼ₋₁ gives [0, 1, -5/2, 16/3]. Therefore the quotient is x − 5/2x² + 16/3x³+O(x⁴). Multiplying this truncated answer by (1+2x) reconstructs the numerator coefficients through x³, which independently verifies the sign and every rational term.
Convergence and transfer. The reciprocal denominator is expanded only when |2x|<1. The numerator’s relevant binomial or logarithmic series also needs |1x|<1. A safe common range is |x|<1/2. Check the sign of q before any expansion; a denominator with a plus sign corresponds to negative q. Replace q with −q and predict which coefficients change before calculating a second case.
27. sin(1x) divided by 1+2x
Try the expansion. Find the first four coefficients of sin(1x)/(1+2x). The known numerator series through cubic order is x − 1/6x³+O(x⁴). Identify the signed q=-2 in 1−qx, then write an unknown quotient c₀+c₁x+c₂x²+c₃x³. Use coefficient matching in ascending powers rather than expanding many unnecessary higher terms.
Complete worked coefficients. The numerator coefficient ledger [a₀,a₁,a₂,a₃] is [0, 1, 0, -1/6]. Applying c₀=a₀ and cⱼ=aⱼ+q cⱼ₋₁ gives [0, 1, -2, 23/6]. Therefore the quotient is x − 2x² + 23/6x³+O(x⁴). Multiplying this truncated answer by (1+2x) reconstructs the numerator coefficients through x³, which independently verifies the sign and every rational term.
Convergence and transfer. The reciprocal denominator is expanded only when |2x|<1. The numerator has no additional finite Maclaurin convergence restriction of this kind. A safe common range is |x|<1/2. A correct cubic coefficient must survive multiplication back by the original denominator. Replace q with −q and predict which coefficients change before calculating a second case.
28. cos(1x) divided by 1+2x
Try the expansion. Find the first four coefficients of cos(1x)/(1+2x). The known numerator series through cubic order is 1 − 1/2x²+O(x⁴). Identify the signed q=-2 in 1−qx, then write an unknown quotient c₀+c₁x+c₂x²+c₃x³. Use coefficient matching in ascending powers rather than expanding many unnecessary higher terms.
Complete worked coefficients. The numerator coefficient ledger [a₀,a₁,a₂,a₃] is [1, 0, -1/2, 0]. Applying c₀=a₀ and cⱼ=aⱼ+q cⱼ₋₁ gives [1, -2, 7/2, -7]. Therefore the quotient is 1 − 2x + 7/2x² − 7x³+O(x⁴). Multiplying this truncated answer by (1+2x) reconstructs the numerator coefficients through x³, which independently verifies the sign and every rational term.
Convergence and transfer. The reciprocal denominator is expanded only when |2x|<1. The numerator has no additional finite Maclaurin convergence restriction of this kind. A safe common range is |x|<1/2. Do not confuse a finite cubic approximation with equality to the exact function for every real x. Replace q with −q and predict which coefficients change before calculating a second case.
29. (1+1x)^(1/2) divided by 1+2x
Try the expansion. Find the first four coefficients of (1+1x)^(1/2)/(1+2x). The known numerator series through cubic order is 1 + 1/2x − 1/8x² + 1/16x³+O(x⁴). Identify the signed q=-2 in 1−qx, then write an unknown quotient c₀+c₁x+c₂x²+c₃x³. Use coefficient matching in ascending powers rather than expanding many unnecessary higher terms.
Complete worked coefficients. The numerator coefficient ledger [a₀,a₁,a₂,a₃] is [1, 1/2, -1/8, 1/16]. Applying c₀=a₀ and cⱼ=aⱼ+q cⱼ₋₁ gives [1, -3/2, 23/8, -91/16]. Therefore the quotient is 1 − 3/2x + 23/8x² − 91/16x³+O(x⁴). Multiplying this truncated answer by (1+2x) reconstructs the numerator coefficients through x³, which independently verifies the sign and every rational term.
Convergence and transfer. The reciprocal denominator is expanded only when |2x|<1. The numerator’s relevant binomial or logarithmic series also needs |1x|<1. A safe common range is |x|<1/2. If the numerator has no constant term, the quotient also has zero constant term when the denominator equals one at zero. Replace q with −q and predict which coefficients change before calculating a second case.
30. (1+1x)^(−1/2) divided by 1+2x
Try the expansion. Find the first four coefficients of (1+1x)^(−1/2)/(1+2x). The known numerator series through cubic order is 1 − 1/2x + 3/8x² − 5/16x³+O(x⁴). Identify the signed q=-2 in 1−qx, then write an unknown quotient c₀+c₁x+c₂x²+c₃x³. Use coefficient matching in ascending powers rather than expanding many unnecessary higher terms.
Complete worked coefficients. The numerator coefficient ledger [a₀,a₁,a₂,a₃] is [1, -1/2, 3/8, -5/16]. Applying c₀=a₀ and cⱼ=aⱼ+q cⱼ₋₁ gives [1, -5/2, 43/8, -177/16]. Therefore the quotient is 1 − 5/2x + 43/8x² − 177/16x³+O(x⁴). Multiplying this truncated answer by (1+2x) reconstructs the numerator coefficients through x³, which independently verifies the sign and every rational term.
Convergence and transfer. The reciprocal denominator is expanded only when |2x|<1. The numerator’s relevant binomial or logarithmic series also needs |1x|<1. A safe common range is |x|<1/2. Whenever the numerator is trigonometric, all angles in the standard series are in radians. Replace q with −q and predict which coefficients change before calculating a second case.
31. (1+1x)² divided by 1+2x
Try the expansion. Find the first four coefficients of (1+1x)²/(1+2x). The known numerator series through cubic order is 1 + 2x + x²+O(x⁴). Identify the signed q=-2 in 1−qx, then write an unknown quotient c₀+c₁x+c₂x²+c₃x³. Use coefficient matching in ascending powers rather than expanding many unnecessary higher terms.
Complete worked coefficients. The numerator coefficient ledger [a₀,a₁,a₂,a₃] is [1, 2, 1, 0]. Applying c₀=a₀ and cⱼ=aⱼ+q cⱼ₋₁ gives [1, 0, 1, -2]. Therefore the quotient is 1 + x² − 2x³+O(x⁴). Multiplying this truncated answer by (1+2x) reconstructs the numerator coefficients through x³, which independently verifies the sign and every rational term.
Convergence and transfer. The reciprocal denominator is expanded only when |2x|<1. The numerator has no additional finite Maclaurin convergence restriction of this kind. A safe common range is |x|<1/2. The recurrence uses exact rational arithmetic; early decimal rounding can hide the final coefficient. Replace q with −q and predict which coefficients change before calculating a second case.
32. 1+1x divided by 1+2x
Try the expansion. Find the first four coefficients of 1+1x/(1+2x). The known numerator series through cubic order is 1 + x+O(x⁴). Identify the signed q=-2 in 1−qx, then write an unknown quotient c₀+c₁x+c₂x²+c₃x³. Use coefficient matching in ascending powers rather than expanding many unnecessary higher terms.
Complete worked coefficients. The numerator coefficient ledger [a₀,a₁,a₂,a₃] is [1, 1, 0, 0]. Applying c₀=a₀ and cⱼ=aⱼ+q cⱼ₋₁ gives [1, -1, 2, -4]. Therefore the quotient is 1 − x + 2x² − 4x³+O(x⁴). Multiplying this truncated answer by (1+2x) reconstructs the numerator coefficients through x³, which independently verifies the sign and every rational term.
Convergence and transfer. The reciprocal denominator is expanded only when |2x|<1. The numerator has no additional finite Maclaurin convergence restriction of this kind. A safe common range is |x|<1/2. Both the numerator and denominator convergence restrictions matter for a valid combined series. Replace q with −q and predict which coefficients change before calculating a second case.
33. e^(2x) divided by 1−3x
Try the expansion. Find the first four coefficients of e^(2x)/(1−3x). The known numerator series through cubic order is 1 + 2x + 2x² + 4/3x³+O(x⁴). Identify the signed q=3 in 1−qx, then write an unknown quotient c₀+c₁x+c₂x²+c₃x³. Use coefficient matching in ascending powers rather than expanding many unnecessary higher terms.
Complete worked coefficients. The numerator coefficient ledger [a₀,a₁,a₂,a₃] is [1, 2, 2, 4/3]. Applying c₀=a₀ and cⱼ=aⱼ+q cⱼ₋₁ gives [1, 5, 17, 157/3]. Therefore the quotient is 1 + 5x + 17x² + 157/3x³+O(x⁴). Multiplying this truncated answer by (1−3x) reconstructs the numerator coefficients through x³, which independently verifies the sign and every rational term.
Convergence and transfer. The reciprocal denominator is expanded only when |3x|<1. The numerator has no additional finite Maclaurin convergence restriction of this kind. A safe common range is |x|<1/3. Remember that all lower-degree terms can contribute to a higher quotient coefficient through the denominator. Replace q with −q and predict which coefficients change before calculating a second case.
34. ln(1+2x) divided by 1−3x
Try the expansion. Find the first four coefficients of ln(1+2x)/(1−3x). The known numerator series through cubic order is 2x − 2x² + 8/3x³+O(x⁴). Identify the signed q=3 in 1−qx, then write an unknown quotient c₀+c₁x+c₂x²+c₃x³. Use coefficient matching in ascending powers rather than expanding many unnecessary higher terms.
Complete worked coefficients. The numerator coefficient ledger [a₀,a₁,a₂,a₃] is [0, 2, -2, 8/3]. Applying c₀=a₀ and cⱼ=aⱼ+q cⱼ₋₁ gives [0, 2, 4, 44/3]. Therefore the quotient is 2x + 4x² + 44/3x³+O(x⁴). Multiplying this truncated answer by (1−3x) reconstructs the numerator coefficients through x³, which independently verifies the sign and every rational term.
Convergence and transfer. The reciprocal denominator is expanded only when |3x|<1. The numerator’s relevant binomial or logarithmic series also needs |2x|<1. A safe common range is |x|<1/3. Check the sign of q before any expansion; a denominator with a plus sign corresponds to negative q. Replace q with −q and predict which coefficients change before calculating a second case.
35. sin(2x) divided by 1−3x
Try the expansion. Find the first four coefficients of sin(2x)/(1−3x). The known numerator series through cubic order is 2x − 4/3x³+O(x⁴). Identify the signed q=3 in 1−qx, then write an unknown quotient c₀+c₁x+c₂x²+c₃x³. Use coefficient matching in ascending powers rather than expanding many unnecessary higher terms.
Complete worked coefficients. The numerator coefficient ledger [a₀,a₁,a₂,a₃] is [0, 2, 0, -4/3]. Applying c₀=a₀ and cⱼ=aⱼ+q cⱼ₋₁ gives [0, 2, 6, 50/3]. Therefore the quotient is 2x + 6x² + 50/3x³+O(x⁴). Multiplying this truncated answer by (1−3x) reconstructs the numerator coefficients through x³, which independently verifies the sign and every rational term.
Convergence and transfer. The reciprocal denominator is expanded only when |3x|<1. The numerator has no additional finite Maclaurin convergence restriction of this kind. A safe common range is |x|<1/3. A correct cubic coefficient must survive multiplication back by the original denominator. Replace q with −q and predict which coefficients change before calculating a second case.
36. cos(2x) divided by 1−3x
Try the expansion. Find the first four coefficients of cos(2x)/(1−3x). The known numerator series through cubic order is 1 − 2x²+O(x⁴). Identify the signed q=3 in 1−qx, then write an unknown quotient c₀+c₁x+c₂x²+c₃x³. Use coefficient matching in ascending powers rather than expanding many unnecessary higher terms.
Complete worked coefficients. The numerator coefficient ledger [a₀,a₁,a₂,a₃] is [1, 0, -2, 0]. Applying c₀=a₀ and cⱼ=aⱼ+q cⱼ₋₁ gives [1, 3, 7, 21]. Therefore the quotient is 1 + 3x + 7x² + 21x³+O(x⁴). Multiplying this truncated answer by (1−3x) reconstructs the numerator coefficients through x³, which independently verifies the sign and every rational term.
Convergence and transfer. The reciprocal denominator is expanded only when |3x|<1. The numerator has no additional finite Maclaurin convergence restriction of this kind. A safe common range is |x|<1/3. Do not confuse a finite cubic approximation with equality to the exact function for every real x. Replace q with −q and predict which coefficients change before calculating a second case.
37. (1+2x)^(1/2) divided by 1−3x
Try the expansion. Find the first four coefficients of (1+2x)^(1/2)/(1−3x). The known numerator series through cubic order is 1 + x − 1/2x² + 1/2x³+O(x⁴). Identify the signed q=3 in 1−qx, then write an unknown quotient c₀+c₁x+c₂x²+c₃x³. Use coefficient matching in ascending powers rather than expanding many unnecessary higher terms.
Complete worked coefficients. The numerator coefficient ledger [a₀,a₁,a₂,a₃] is [1, 1, -1/2, 1/2]. Applying c₀=a₀ and cⱼ=aⱼ+q cⱼ₋₁ gives [1, 4, 23/2, 35]. Therefore the quotient is 1 + 4x + 23/2x² + 35x³+O(x⁴). Multiplying this truncated answer by (1−3x) reconstructs the numerator coefficients through x³, which independently verifies the sign and every rational term.
Convergence and transfer. The reciprocal denominator is expanded only when |3x|<1. The numerator’s relevant binomial or logarithmic series also needs |2x|<1. A safe common range is |x|<1/3. If the numerator has no constant term, the quotient also has zero constant term when the denominator equals one at zero. Replace q with −q and predict which coefficients change before calculating a second case.
38. (1+2x)^(−1/2) divided by 1−3x
Try the expansion. Find the first four coefficients of (1+2x)^(−1/2)/(1−3x). The known numerator series through cubic order is 1 − x + 3/2x² − 5/2x³+O(x⁴). Identify the signed q=3 in 1−qx, then write an unknown quotient c₀+c₁x+c₂x²+c₃x³. Use coefficient matching in ascending powers rather than expanding many unnecessary higher terms.
Complete worked coefficients. The numerator coefficient ledger [a₀,a₁,a₂,a₃] is [1, -1, 3/2, -5/2]. Applying c₀=a₀ and cⱼ=aⱼ+q cⱼ₋₁ gives [1, 2, 15/2, 20]. Therefore the quotient is 1 + 2x + 15/2x² + 20x³+O(x⁴). Multiplying this truncated answer by (1−3x) reconstructs the numerator coefficients through x³, which independently verifies the sign and every rational term.
Convergence and transfer. The reciprocal denominator is expanded only when |3x|<1. The numerator’s relevant binomial or logarithmic series also needs |2x|<1. A safe common range is |x|<1/3. Whenever the numerator is trigonometric, all angles in the standard series are in radians. Replace q with −q and predict which coefficients change before calculating a second case.
39. (1+2x)² divided by 1−3x
Try the expansion. Find the first four coefficients of (1+2x)²/(1−3x). The known numerator series through cubic order is 1 + 4x + 4x²+O(x⁴). Identify the signed q=3 in 1−qx, then write an unknown quotient c₀+c₁x+c₂x²+c₃x³. Use coefficient matching in ascending powers rather than expanding many unnecessary higher terms.
Complete worked coefficients. The numerator coefficient ledger [a₀,a₁,a₂,a₃] is [1, 4, 4, 0]. Applying c₀=a₀ and cⱼ=aⱼ+q cⱼ₋₁ gives [1, 7, 25, 75]. Therefore the quotient is 1 + 7x + 25x² + 75x³+O(x⁴). Multiplying this truncated answer by (1−3x) reconstructs the numerator coefficients through x³, which independently verifies the sign and every rational term.
Convergence and transfer. The reciprocal denominator is expanded only when |3x|<1. The numerator has no additional finite Maclaurin convergence restriction of this kind. A safe common range is |x|<1/3. The recurrence uses exact rational arithmetic; early decimal rounding can hide the final coefficient. Replace q with −q and predict which coefficients change before calculating a second case.
40. 1+2x divided by 1−3x
Try the expansion. Find the first four coefficients of 1+2x/(1−3x). The known numerator series through cubic order is 1 + 2x+O(x⁴). Identify the signed q=3 in 1−qx, then write an unknown quotient c₀+c₁x+c₂x²+c₃x³. Use coefficient matching in ascending powers rather than expanding many unnecessary higher terms.
Complete worked coefficients. The numerator coefficient ledger [a₀,a₁,a₂,a₃] is [1, 2, 0, 0]. Applying c₀=a₀ and cⱼ=aⱼ+q cⱼ₋₁ gives [1, 5, 15, 45]. Therefore the quotient is 1 + 5x + 15x² + 45x³+O(x⁴). Multiplying this truncated answer by (1−3x) reconstructs the numerator coefficients through x³, which independently verifies the sign and every rational term.
Convergence and transfer. The reciprocal denominator is expanded only when |3x|<1. The numerator has no additional finite Maclaurin convergence restriction of this kind. A safe common range is |x|<1/3. Both the numerator and denominator convergence restrictions matter for a valid combined series. Replace q with −q and predict which coefficients change before calculating a second case.
41. e^(3x) divided by 1+3x
Try the expansion. Find the first four coefficients of e^(3x)/(1+3x). The known numerator series through cubic order is 1 + 3x + 9/2x² + 9/2x³+O(x⁴). Identify the signed q=-3 in 1−qx, then write an unknown quotient c₀+c₁x+c₂x²+c₃x³. Use coefficient matching in ascending powers rather than expanding many unnecessary higher terms.
Complete worked coefficients. The numerator coefficient ledger [a₀,a₁,a₂,a₃] is [1, 3, 9/2, 9/2]. Applying c₀=a₀ and cⱼ=aⱼ+q cⱼ₋₁ gives [1, 0, 9/2, -9]. Therefore the quotient is 1 + 9/2x² − 9x³+O(x⁴). Multiplying this truncated answer by (1+3x) reconstructs the numerator coefficients through x³, which independently verifies the sign and every rational term.
Convergence and transfer. The reciprocal denominator is expanded only when |3x|<1. The numerator has no additional finite Maclaurin convergence restriction of this kind. A safe common range is |x|<1/3. Remember that all lower-degree terms can contribute to a higher quotient coefficient through the denominator. Replace q with −q and predict which coefficients change before calculating a second case.
42. ln(1+3x) divided by 1+3x
Try the expansion. Find the first four coefficients of ln(1+3x)/(1+3x). The known numerator series through cubic order is 3x − 9/2x² + 9x³+O(x⁴). Identify the signed q=-3 in 1−qx, then write an unknown quotient c₀+c₁x+c₂x²+c₃x³. Use coefficient matching in ascending powers rather than expanding many unnecessary higher terms.
Complete worked coefficients. The numerator coefficient ledger [a₀,a₁,a₂,a₃] is [0, 3, -9/2, 9]. Applying c₀=a₀ and cⱼ=aⱼ+q cⱼ₋₁ gives [0, 3, -27/2, 99/2]. Therefore the quotient is 3x − 27/2x² + 99/2x³+O(x⁴). Multiplying this truncated answer by (1+3x) reconstructs the numerator coefficients through x³, which independently verifies the sign and every rational term.
Convergence and transfer. The reciprocal denominator is expanded only when |3x|<1. The numerator’s relevant binomial or logarithmic series also needs |3x|<1. A safe common range is |x|<1/3. Check the sign of q before any expansion; a denominator with a plus sign corresponds to negative q. Replace q with −q and predict which coefficients change before calculating a second case.
43. sin(3x) divided by 1+3x
Try the expansion. Find the first four coefficients of sin(3x)/(1+3x). The known numerator series through cubic order is 3x − 9/2x³+O(x⁴). Identify the signed q=-3 in 1−qx, then write an unknown quotient c₀+c₁x+c₂x²+c₃x³. Use coefficient matching in ascending powers rather than expanding many unnecessary higher terms.
Complete worked coefficients. The numerator coefficient ledger [a₀,a₁,a₂,a₃] is [0, 3, 0, -9/2]. Applying c₀=a₀ and cⱼ=aⱼ+q cⱼ₋₁ gives [0, 3, -9, 45/2]. Therefore the quotient is 3x − 9x² + 45/2x³+O(x⁴). Multiplying this truncated answer by (1+3x) reconstructs the numerator coefficients through x³, which independently verifies the sign and every rational term.
Convergence and transfer. The reciprocal denominator is expanded only when |3x|<1. The numerator has no additional finite Maclaurin convergence restriction of this kind. A safe common range is |x|<1/3. A correct cubic coefficient must survive multiplication back by the original denominator. Replace q with −q and predict which coefficients change before calculating a second case.
44. cos(3x) divided by 1+3x
Try the expansion. Find the first four coefficients of cos(3x)/(1+3x). The known numerator series through cubic order is 1 − 9/2x²+O(x⁴). Identify the signed q=-3 in 1−qx, then write an unknown quotient c₀+c₁x+c₂x²+c₃x³. Use coefficient matching in ascending powers rather than expanding many unnecessary higher terms.
Complete worked coefficients. The numerator coefficient ledger [a₀,a₁,a₂,a₃] is [1, 0, -9/2, 0]. Applying c₀=a₀ and cⱼ=aⱼ+q cⱼ₋₁ gives [1, -3, 9/2, -27/2]. Therefore the quotient is 1 − 3x + 9/2x² − 27/2x³+O(x⁴). Multiplying this truncated answer by (1+3x) reconstructs the numerator coefficients through x³, which independently verifies the sign and every rational term.
Convergence and transfer. The reciprocal denominator is expanded only when |3x|<1. The numerator has no additional finite Maclaurin convergence restriction of this kind. A safe common range is |x|<1/3. Do not confuse a finite cubic approximation with equality to the exact function for every real x. Replace q with −q and predict which coefficients change before calculating a second case.
45. (1+3x)^(1/2) divided by 1+3x
Try the expansion. Find the first four coefficients of (1+3x)^(1/2)/(1+3x). The known numerator series through cubic order is 1 + 3/2x − 9/8x² + 27/16x³+O(x⁴). Identify the signed q=-3 in 1−qx, then write an unknown quotient c₀+c₁x+c₂x²+c₃x³. Use coefficient matching in ascending powers rather than expanding many unnecessary higher terms.
Complete worked coefficients. The numerator coefficient ledger [a₀,a₁,a₂,a₃] is [1, 3/2, -9/8, 27/16]. Applying c₀=a₀ and cⱼ=aⱼ+q cⱼ₋₁ gives [1, -3/2, 27/8, -135/16]. Therefore the quotient is 1 − 3/2x + 27/8x² − 135/16x³+O(x⁴). Multiplying this truncated answer by (1+3x) reconstructs the numerator coefficients through x³, which independently verifies the sign and every rational term.
Convergence and transfer. The reciprocal denominator is expanded only when |3x|<1. The numerator’s relevant binomial or logarithmic series also needs |3x|<1. A safe common range is |x|<1/3. If the numerator has no constant term, the quotient also has zero constant term when the denominator equals one at zero. Replace q with −q and predict which coefficients change before calculating a second case.
46. (1+3x)^(−1/2) divided by 1+3x
Try the expansion. Find the first four coefficients of (1+3x)^(−1/2)/(1+3x). The known numerator series through cubic order is 1 − 3/2x + 27/8x² − 135/16x³+O(x⁴). Identify the signed q=-3 in 1−qx, then write an unknown quotient c₀+c₁x+c₂x²+c₃x³. Use coefficient matching in ascending powers rather than expanding many unnecessary higher terms.
Complete worked coefficients. The numerator coefficient ledger [a₀,a₁,a₂,a₃] is [1, -3/2, 27/8, -135/16]. Applying c₀=a₀ and cⱼ=aⱼ+q cⱼ₋₁ gives [1, -9/2, 135/8, -945/16]. Therefore the quotient is 1 − 9/2x + 135/8x² − 945/16x³+O(x⁴). Multiplying this truncated answer by (1+3x) reconstructs the numerator coefficients through x³, which independently verifies the sign and every rational term.
Convergence and transfer. The reciprocal denominator is expanded only when |3x|<1. The numerator’s relevant binomial or logarithmic series also needs |3x|<1. A safe common range is |x|<1/3. Whenever the numerator is trigonometric, all angles in the standard series are in radians. Replace q with −q and predict which coefficients change before calculating a second case.
47. (1+3x)² divided by 1+3x
Try the expansion. Find the first four coefficients of (1+3x)²/(1+3x). The known numerator series through cubic order is 1 + 6x + 9x²+O(x⁴). Identify the signed q=-3 in 1−qx, then write an unknown quotient c₀+c₁x+c₂x²+c₃x³. Use coefficient matching in ascending powers rather than expanding many unnecessary higher terms.
Complete worked coefficients. The numerator coefficient ledger [a₀,a₁,a₂,a₃] is [1, 6, 9, 0]. Applying c₀=a₀ and cⱼ=aⱼ+q cⱼ₋₁ gives [1, 3, 0, 0]. Therefore the quotient is 1 + 3x+O(x⁴). Multiplying this truncated answer by (1+3x) reconstructs the numerator coefficients through x³, which independently verifies the sign and every rational term.
Convergence and transfer. The reciprocal denominator is expanded only when |3x|<1. The numerator has no additional finite Maclaurin convergence restriction of this kind. A safe common range is |x|<1/3. The recurrence uses exact rational arithmetic; early decimal rounding can hide the final coefficient. Replace q with −q and predict which coefficients change before calculating a second case.
48. 1+3x divided by 1+3x
Try the expansion. Find the first four coefficients of 1+3x/(1+3x). The known numerator series through cubic order is 1 + 3x+O(x⁴). Identify the signed q=-3 in 1−qx, then write an unknown quotient c₀+c₁x+c₂x²+c₃x³. Use coefficient matching in ascending powers rather than expanding many unnecessary higher terms.
Complete worked coefficients. The numerator coefficient ledger [a₀,a₁,a₂,a₃] is [1, 3, 0, 0]. Applying c₀=a₀ and cⱼ=aⱼ+q cⱼ₋₁ gives [1, 0, 0, 0]. Therefore the quotient is 1+O(x⁴). Multiplying this truncated answer by (1+3x) reconstructs the numerator coefficients through x³, which independently verifies the sign and every rational term.
Convergence and transfer. The reciprocal denominator is expanded only when |3x|<1. The numerator has no additional finite Maclaurin convergence restriction of this kind. A safe common range is |x|<1/3. Both the numerator and denominator convergence restrictions matter for a valid combined series. Replace q with −q and predict which coefficients change before calculating a second case.
Ten short answer checks
- 1. 1/(1−x) through cubic order: 1+x+x²+x³+O(x⁴).
- 2. 1/(1+x): 1−x+x²−x³+O(x⁴).
- 3. ln(1+x)/(1−x): x+x²/2+5x³/6+O(x⁴).
- 4. ln(1+x)/(1+x): x−3x²/2+11x³/6+O(x⁴).
- 5. eˣ/(1−x): 1+2x+5x²/2+8x³/3+O(x⁴).
- 6. sin x/(1+x): x−x²+5x³/6+O(x⁴).
- 7. eˣ/(1+x²): 1+x−x²/2−5x³/6+O(x⁴).
- 8. A geometric denominator 1−2x requires |x|<1/2 for its standard series.
- 9. For Q=N/(1−x), coefficient c₂=a₂+c₁.
- 10. A valid cubic truncation must multiply back to the numerator through all powers 0,1,2,3.
An error-first five-day practice plan
Day one: expand 1/(1±x) and 1/(1±2x) using a geometric series, identifying the radius of convergence. Day two: attach a simple polynomial numerator and match coefficients instead of distributing blindly. Day three: use eˣ, sin x and cos x numerators, with special attention to missing even or odd terms. Day four: use logarithmic and binomial numerators and combine two convergence restrictions. Day five: complete mixed quotients without revealing whether their denominator is linear or quadratic, then verify each result by remultiplication.
For three learners sharing a tutorial, one can prepare the numerator coefficients, another can perform the quotient recurrence and the third can reverse-check by multiplication. Rotate roles and change a sign or coefficient each time. This keeps the mathematical reasoning visible: a wrong cubic term is more easily repaired by identifying the earlier coefficient that fed into it than by copying a correct final series from an answer sheet.
Frequently asked questions
Why does the cubic coefficient include lower-order numerator terms?
Because dividing by a nonconstant denominator is equivalent to multiplying by its reciprocal series. Lower-order terms combine with powers of the denominator’s reciprocal to reach cubic order.
Can I just divide the cubic coefficients?
No. Quotient coefficients come from matching the full truncated product (denominator)×(quotient) to the numerator, not from coefficient-wise division.
Does a plus sign in 1+x make every coefficient positive?
No. The reciprocal 1/(1+x) has alternating coefficients, so the sign must be carried through the recurrence.
Why check Q(0) and Q′(0)?
They are the Maclaurin constant and linear coefficients. These simple checks can quickly reveal a wrong expansion.
Is O(x⁴) a claim of equality?
It denotes the size of the omitted remainder near zero for the analytic functions used in these examples. A finite polynomial is not identical to the original quotient everywhere.
When does the geometric reciprocal series converge?
For 1/(1−qx), its familiar expansion converges when |qx|<1. Combine this with any restriction from the numerator’s standard series.
Is every quotient expandable at x=0?
No. An ordinary power-series quotient requires a valid local expansion. If the denominator vanishes at zero, separate analysis is needed and some expressions have genuine singularities.
Does the 2027 H2 syllabus ask for a general nth coefficient?
It includes deriving the first few Maclaurin terms using standard series or differentiation, but excludes problems requiring derivation of a general series term. This guide stays with the first four coefficients.
Related H2 guides and official source
- Normal Conditional Tail Probabilities
- Trigonometric General Solutions and Interval Checks
- Maclaurin Quotient Series and Missing Coefficients: current guide
- Infinite Geometric Series and Convergence
For the neighbouring mechanism, read Maclaurin Products: The Cross-Term Trap or the broader Bukit Timah H2 Maclaurin topic guide. For location context, visit the Bukit Timah JC2 Tuition hub and the original Clementi small-group Mathematics tutoring reference for the teaching format at another level.
Official curriculum: SEAB H2 Mathematics 9758 syllabus, examination in 2027, Maclaurin series. All worked examples are original fictional mathematics problems, not actual past-year questions or recorded examination results. The approximations are local near zero and must be interpreted within the stated convergence ranges.