01 · Negative multiplier
State the guaranteed interval for (1 − 5x)^(−½).
Hint
Take the absolute value of −5x.
Worked solution
5|x| < 1, hence −⅕ < x < ⅕.
Understand · explore · practise
Validity of binomial expansions
Find the validity interval of a binomial expansion: linear, quadratic and reciprocal substitutions, combined conditions, endpoints and the difference between convergence and accuracy.
Before you startGeneralised binomial expansion and inequalities
01 / The condition belongs to the input
Test the quantity replacing z, not just x.
For the generalised expansion of (1 + z)ᵖ, the standard guarantee is |z| < 1. If z = 2x, this is a condition on 2x. If z = 2/x, it gives a completely different region.
The model marks the guaranteed open region and tests your chosen x. It also identifies boundary points and undefined substitutions explicitly.
Test the substituted inputExplore
Input z = 2x. Condition: |2x| < 1, so −½ < x < ½.
At x = 0, z = 0. Inside the guaranteed region.
Open circles are boundary points requiring separate analysis. A point outside the blue region does not automatically mean the original function is undefined.
Watch a transformed input become an interval
Pause, replay or seek freely. The notes explain the same idea and stay in view.
02 / Linear substitutions
Absolute values handle either sign of the multiplier.
|bx| < 1 means |x| < 1/|b|, for b ≠ 0.
For (1 − 4x)⁻²Worked example
z = −4x, so |−4x| < 1
The negative exponent does not change this input condition.
4|x| < 1, hence −¼ < x < ¼
Use |−4| = 4. A negative “radius” would make no sense.
For (1 − 2x/3)^(½)Worked example
|(−2/3)x| < 1
The magnitude of the multiplier is 2/3.
|x| < 3/2
Multiplying by the positive number 3/2 preserves the inequality.
These are strict inequalities: the standard open-interval result does not decide what happens at either endpoint.
03 / After factoring a constant
Use the new input after the bracket has been normalised.
For √(9 + 6x)Worked example
√(9 + 6x) = 3(1 + 2x/3)^(½)
The new input is z = 2x/3.
|2x/3| < 1, so −3/2 < x < 3/2
Do not use |6x| < 1: that was not the normalised input.
For a > 0 and b ≠ 0:
(a + bx)ᵖ = aᵖ(1 + bx/a)ᵖ
Guaranteed condition: |x| < a/|b|.
The positive-constant assumption makes the real fractional-power factorisation safe in this interval. Integer powers can allow other factorisations, but you must still solve the actual input condition and respect the original domain.
04 / Quadratic substitutions
Solve for x after controlling the whole square.
For (1 − x²/9)⁻¹Worked example
z = −x²/9; therefore |z| = x²/9
A square is nonnegative.
x²/9 < 1 gives x² < 9
Multiply by 9.
−3 < x < 3
Both positive and negative x are allowed.
For (1 + 2x²)^(½)Worked example
2x² < 1, hence x² < ½
Keep the factor 2.
|x| < 1/√2
The endpoint is the square root of ½, not ½.
The powers in the expansion will be 1, x², x⁴, … . The condition is still about the substituted quantity, not the highest displayed power.
05 / Reciprocal substitutions
The valid region can lie outside a central interval.
Expand (1 + 2/x)⁻¹ in powers of 1/xWorked example
Require x ≠ 0 and |2/x| < 1
The substitution itself is undefined at zero.
2/|x| < 1 gives |x| > 2
Multiply by the positive quantity |x|.
So x < −2 or x > 2
This is a union of two regions, not −2 < x < 2.
1 − 2/x + 4/x² − 8/x³ + …
This expansion is useful for large |x|, because 1/x is then small.
Although the original rational function simplifies to x/(x + 2) for x ≠ 0, that does not make this inverse-power series valid at zero. A different expansion can have a different centre or small variable.
06 / Intersect conditions when combining expansions
Every separate series you use must be justified at that x.
Expand (1 + x/2)⁻¹(1 − 2x/3)⁻¹Worked example
First factor: |x/2| < 1, so |x| < 2
Find its condition separately.
Second factor: |−2x/3| < 1, so |x| < 3/2
Find the second condition.
Together: |x| < 3/2
Both must hold: take the overlap, not the union.
This overlap justifies combining those expansions. Simplify exact cancellations before deciding whether a restriction is intrinsic to the final expression. For example (1 − x)/(1 − x) = 1 for x ≠ 1: the constant identity is valid on that original domain even though expanding the denominator separately would initially require |x| < 1.
Cancellation does not restore an excluded point in the original expression. Keep the hole at x = 1.
07 / Domain, convergence and accuracy
These answer three different questions.
Consider 1/(1 − x)Worked example
Function domain: every real x except 1
At x = 2, the function exists and equals −1.
Geometric series 1 + x + x² + …: |x| < 1
At x = 2 its terms grow, so the series cannot represent −1.
At x = 0.8, the function equals 5
The input lies inside the convergence interval.
Keeping 1 + x + x² gives only 2.44
The absolute error is 2.56, or 51.2% of the exact value.
Convergence describes what happens as the number of terms tends to infinity. It does not promise that a short truncation gives a useful numerical approximation.
08 / Finite expansions and endpoints
The standard guarantee is useful without being a universal endpoint test.
A nonnegative integer exponent produces a finite polynomial. For example (1 − 5x)² = 1 − 10x + 25x² for every real x. No |5x| < 1 restriction is needed.
If the multiplier is zero, the expression is constant wherever defined. If the leading constant is zero, normalising by dividing by it is invalid.
For an infinite expansion, do not automatically include the endpoints z = ±1. Their behaviour depends on the exponent. The geometric expansion (1 − z)⁻¹ = 1 + z + z² + … fails at both z = 1 and z = −1 because its terms do not tend to zero. Other binomial series can converge at an endpoint. Use a separate justified convergence result if an endpoint is actually requested.
In these notes, “guaranteed interval” means the standard open region |z| < 1. It deliberately makes no claim that every boundary point fails.
09 / Your turn
Give the guaranteed open region, unless a finite identity applies.
02 · Factor first
State the guaranteed interval for the binomial expansion of √(16 + 8x).
Hint
Rewrite as 4(1 + x/2)^(½).
Worked solution
The input is x/2, so |x/2| < 1 gives −2 < x < 2.
03 · Quadratic input
Find the guaranteed region for (1 − 4x²)⁻³.
Hint
Use |−4x²| = 4x².
Worked solution
4x² < 1 gives x² < ¼, so −½ < x < ½.
04 · Inverse powers
Find the guaranteed region for the expansion of (1 − 3/x)^(½) in powers of 1/x.
Hint
x ≠ 0 and 3/|x| < 1.
Worked solution
|x| > 3, so x < −3 or x > 3. On this region the square-root input is positive.
05 · Two factors
Find an interval where the separate binomial expansions of (1 + 4x)⁻¹ and (1 − x/3)^(½) may be multiplied.
Hint
Intersect |x| < ¼ and |x| < 3.
Worked solution
Both conditions hold for −¼ < x < ¼. The less restrictive second condition does not enlarge the overlap.
06 · Finite exception
Does the expansion of (2 + 3x)⁴ require |x| < ⅔?
Hint
Look at the exponent.
Worked solution
No. The exponent 4 is a nonnegative integer, so the finite polynomial identity holds for every real x.
07 · Accuracy
Approximate 1/(1 − x) at x = 0.5 using 1 + x + x². Find the absolute and percentage errors.
Hint
The exact value is 2; compare the three-term value with it.
Worked solution
The approximation is 1.75. Absolute error = 2 − 1.75 = 0.25. Percentage error = (0.25/2) × 100 = 12.5%. The series converges here, but the truncation is not especially accurate.
08 · Boundary point
Is x = −2 covered by the standard condition for √(4 + 2x)? Does the function exist there?
Hint
The normalised input is x/2.
Worked solution
No: |x/2| = 1, not less than 1. The function does exist there and equals 0. Function existence alone does not settle convergence of the series at that endpoint.
09 · Cancel before expanding
Simplify (1 + 2x)/(1 + 2x), retaining the original domain. Is its constant expansion restricted to |x| < ½?
Hint
Distinguish the denominator’s separate series from the simplified expression.
Worked solution
The expression equals 1 for x ≠ −½. Its constant identity works on that entire original domain. Expanding the reciprocal denominator separately would impose |x| < ½, but that restriction is unnecessary after exact cancellation.
10 · Shifted input
Find the guaranteed region for expanding [1 + (x − 2)/3]⁻² in powers of x − 2.
Hint
Use −1 < (x − 2)/3 < 1.
Worked solution
Multiply by 3: −3 < x − 2 < 3. Add 2: −1 < x < 5. This expansion is centred at x = 2; the small input is (x − 2)/3.
10 / Recap
State what your condition guarantees.
- Write down the full substituted quantity z.
- Solve |z| < 1, keeping strict inequalities.
- Use absolute values for signed multipliers.
- Intersect conditions when combining separate series.
- Distinguish function domain, convergence and truncation accuracy.
- Handle finite polynomials and exact cancellations separately.
- Do not assume endpoint convergence or divergence.
Section 1 of 10 · The condition belongs to the input