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Some recurrences are not linear: they multiply terms, raise them to powers or square them. You cannot apply the characteristic equation to uₙ = uₙ₋₁³/uₙ₋₂² directly. A well-chosen substitution, which the question will give you, turns such a relation into a linear one. This lesson shows how to transform the relation and its starting values, solve the linear recurrence, and return to the original sequence without losing a sign or a domain condition. It assumes the method for linear second-order recurrences from linear recurrence relations.
A substitution reveals a linear relation
The substitution undoes whatever makes the relation non-linear:
| The relation involves | Typical substitution | What it does |
|---|---|---|
Products, quotients and powers of terms | vₙ = ln uₙ (or log ₂uₙ) | Turns products into sums and powers into multiples |
Squares of terms | wₙ = uₙ² | Removes the squares |
Reciprocals or fractions of terms | vₙ = 1/uₙ | Clears the fractions |
The method always has the same three stages:
- Rewrite the whole relation in the new variable, and transform the starting values with the same substitution.
- Solve the linear relation with its characteristic equation.
- Return to uₙ, using any sign or domain condition given in the question.
The second stage uses the results for vₙ = pvₙ₋₁ + qvₙ₋₂. Its characteristic equation is λ² = pλ + q, and
The constants A and B come from the transformed starting values v₀ and v₁.
Worked example 1
A recurrence of powers
Worked example 1
A recurrence of powers
The positive sequence uₙ satisfies uₙ = uₙ₋₁³/uₙ₋₂² for n ≥ 2, with u₀ = 2 and u₁ = 4. Using the substitution vₙ = ln uₙ, find uₙ in terms of n.
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Rewrite the relation in v
Method
Take natural logarithms of both sides.Reason
Logarithms turn the powers 3 and 2 into coefficients. Every uₙ is positive, so every logarithm is defined.Working
Answer
ln uₙ = 3 ln uₙ₋₁-2 ln uₙ₋₂,; vₙ = 3vₙ₋₁-2vₙ₋₂,with v₀ = ln 2 and v₁ = ln 4 = 2 ln 2.
Solve the linear relation
Method
Solve λ²-3λ + 2 = 0.Reason
The roots are distinct, so vₙ = A(1)ⁿ + B(2)ⁿ.Working
Answer
(λ-1)(λ-2) = 0,; vₙ = A + B 2ⁿ.From v₀ and v₁: A + B = ln 2 and A + 2B = 2 ln 2, so B = ln 2 and A = 0. Hence vₙ = 2ⁿ ln 2.
Return to u
Working
Answer
uₙ = e^vₙ = e^(2ⁿ ln 2); = 2^2ⁿ.
Quick check
The original relation gives u₂ = 4³/(2²) = 16 = 2^2² and u₃ = 16³/(4²) = 256 = 2^2³. The exponent doubles at each step, so the sequence grows far faster than any geometric sequence.
Fractions: use reciprocals
Guided practice 1
A fraction of terms
Guided practice 1
A fraction of terms
The sequence uₙ satisfies
with u₀ = 1 and u₁ = 1/2. Use the substitution vₙ = 1/uₙ to find uₙ, and describe its behaviour as n → ∞.
Try this before viewing the solution
Hint 1
Divide first
Take the reciprocal of both sides and split the resulting fraction into two terms. Each term should contain only one of uₙ₋₁ and uₙ₋₂.
Hint 2
A repeated root
The characteristic equation is a perfect square, so use the general solution (A + Bn)αⁿ.
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Take reciprocals
Method
Invert both sides and split the fraction.Reason
Each piece then contains a single reciprocal, which is a v term.Working
Answer
1/uₙ = (2uₙ₋₂-uₙ₋₁)/(uₙ₋₁uₙ₋₂) = 2/uₙ₋₁-1/uₙ₋₂,; vₙ = 2vₙ₋₁-vₙ₋₂,with v₀ = 1 and v₁ = 2.
Solve with a repeated root
Method
λ²-2λ + 1 = (λ-1)² = 0 has the repeated root 1.Reason
A repeated root needs the factor A + Bn; A · 1ⁿ + B · 1ⁿ would have only one free constant.Working
Answer
vₙ = A + Bn. Then v₀ = A = 1 and v₁ = A + B = 2, so B = 1 and vₙ = n + 1.
Return to u and describe the behaviour
Working
Answer
uₙ = 1/(n + 1),so u₉ = 0.1. As n → ∞, uₙ → 0 through positive values, decreasing at every step.
Quick check
Check against the original relation: u₂ = (1/2 · 1)/(2-1/2) = 1/3, as the formula predicts. No denominator in the recurrence is ever zero, since 2uₙ₋₂-uₙ₋₁ = 2/(n-1)-1/n ≠ 0.
Squares: return with the correct sign
Squaring loses the sign of a term. After solving for wₙ = uₙ², the question must tell you which square root to take, usually by saying that the sequence is positive. Transform the starting values too: if u₁ = 2, then w₁ = 4, not 2.
Spot the mistake 1
The starting values were not squared
Spot the mistake 1
The starting values were not squared
The positive sequence uₙ satisfies uₙ² = 5uₙ₋₁²-6uₙ₋₂² with u₀ = 1 and u₁ = 2. A student writes wₙ = uₙ², so that wₙ = 5wₙ₋₁-6wₙ₋₂ with w₀ = 1 and w₁ = 2. This gives wₙ = 2ⁿ and uₙ = 2^(n/2). Find the error and the correct uₙ.
Write your answer, then compare it with the solution
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Transform the starting values
Method
Use w₀ = u₀² = 1 and w₁ = u₁² = 4.Reason
The relation and its starting values describe the same sequence wₙ, so both need the substitution.Working
Answer
λ²-5λ + 6 = 0 has roots 2 and 3, so wₙ = A 2ⁿ + B 3ⁿ. Then A + B = 1 and 2A + 3B = 4, giving B = 2 and A = -1:
wₙ = 2 · 3ⁿ-2ⁿ,; uₙ = square root of (2 · 3ⁿ-2ⁿ),taking the positive root because the sequence is positive.
Common mistake
The student’s formula fails its own first check: it gives u₁ = square root of 2, not 2. Substituting n = 0 and n = 1 into a final answer catches this error at once. The correct formula gives u₂ = square root of 14, and the original relation gives u₂² = 5(4)-6(1) = 14.
Check your understanding 1
A recurrence in squares
Check your understanding 1
A recurrence in squares
The positive sequence uₙ satisfies uₙ² = uₙ₋₁² + 2uₙ₋₂² with u₀ = u₁ = 1. Use wₙ = uₙ² to find uₙ.
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The relation becomes wₙ = wₙ₋₁ + 2wₙ₋₂ with w₀ = w₁ = 1. The roots of λ²-λ-2 = 0 are 2 and -1, so wₙ = A 2ⁿ + B(-1)ⁿ, with
giving A = 2/3 and B = 1/3. Since uₙ > 0,
Check: w₂ = 1 + 2 = 3, and the formula gives (8 + 1)/3 = 3.
Common mistakes
- Transforming the relation but not the starting values.
- Taking logarithms when some terms could be zero or negative. The logarithm substitution needs every uₙ > 0.
- Using Aαⁿ + Bαⁿ for a repeated root, which leaves only one free constant.
- Stopping at vₙ or wₙ instead of returning to uₙ.
- Taking ± square roots, or the wrong sign, when the question has said the sequence is positive.
- Not checking the final formula against u₀, u₁ and one term generated by the original relation.
Before you move on
Try it yourself 1
Each term the geometric mean of the last two
Try it yourself 1
Each term the geometric mean of the last two
A positive sequence has u₀ = 1, u₁ = 8 and uₙ = square root of uₙ₋₁uₙ₋₂ for n ≥ 2. Using the substitution vₙ = log ₂uₙ, find uₙ in terms of n. Hence find u₄ and the limit of uₙ as n → ∞.
Try this before viewing the solution
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Rewrite in v
Method
Take log ₂ of both sides.Reason
The square root of a product becomes half the sum of the logarithms.Working
Answer
vₙ = (1/2)vₙ₋₁ + (1/2)vₙ₋₂, v₀ = 0, v₁ = 3.Solve the linear relation
Method
Solve 2λ²-λ-1 = (2λ + 1)(λ-1) = 0.Reason
The roots 1 and -1/2 are distinct.Working
Answer
vₙ = A + B(-1/2)ⁿ, with A + B = 0 and A-(1/2)B = 3. So B = -2, A = 2 and
vₙ = 2-2(-1/2)ⁿ.Return to u and find the limit
Working
Answer
uₙ = 2^(2-2(-1/2)ⁿ),; u₄ = 2^(2-1/8) = 2^(1.875) ≈ 3.67.As n → ∞, (-1/2)ⁿ → 0, so vₙ → 2 and uₙ → 2² = 4.
Quick check
From the original relation, u₂ = square root of 8 ≈ 2.83, u₃ = square root of (8 square root of 8) ≈ 4.76 and u₄ ≈ 3.67: the terms lie alternately below and above 4, because the factor (-1/2)ⁿ changes sign, and the gap shrinks by half each time.
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