Recurrence Relations
A reduction formula is a recurrence in disguise: $J_n = f(n) \cdot J_{n-1} + g(n)$, or sometimes $J_n$ in terms of $J_{n-2}$. Today you will derive recurrences for genuinely new integral families ($x^n e^{-x}$, $x^n \ln x$, $(\ln x)^n$), evaluate specific $J_n$ values by descent, and recognise the deep connection between recurrences and the reduction formulae of lesson 11.
For $J_n = \int_0^1 x^n e^{-x}\,dx$, evaluate $J_0$ directly. Before checking — apply parts once with $u = x^n$, $dv = e^{-x}dx$. Identify which boundary term survives (and which one vanishes). Sketch your derivation below.
Recurrence problems reward two habits. Identify the right parts split: usually the polynomial-style factor ($x^n$ or $(\ln x)^n$) becomes $u$ so that differentiating drops the index. Then track the boundary term carefully — unlike the $[0,\pi/2]$ Wallis case, the boundary may contribute a non-zero constant that appears in the recurrence as $g(n)$.
The differentiate-down read: (1) put the polynomial-style factor on $u$, (2) compute the boundary term — it may not vanish, (3) rearrange the result into the standard form $J_n = f(n)J_{n-1} + g(n)$ or $J_n = f(n)J_{n-2} + g(n)$.
Example: $J_n = \int_0^1 x^n e^{-x}dx \Rightarrow J_n = -e^{-1} + nJ_{n-1}$
Key facts
- A reduction formula is a recurrence on the index $n$ of an integral family
- For $J_n = \int_0^1 x^n e^{-x}dx$: $J_n = nJ_{n-1} - e^{-1}$
- For $K_n = \int_0^1 x^n \ln x\,dx$: $K_n = -\tfrac{1}{(n+1)^2}$ (closed form, derived by parts)
- Base cases are always evaluated directly by elementary integration
Concepts
- Why "polynomial as $u$" almost always works in integral recurrences
- Why the boundary term $[uv]_a^b$ may contribute a $g(n)$ piece
- How a recurrence with linear $f(n)$ leads to factorial-style closed forms
Skills
- Derive a recurrence $J_n = f(n)J_{n-1} + g(n)$ from a definite integral by integration by parts
- Apply the recurrence repeatedly to evaluate $J_n$ for a specific $n$
- Recognise the link between Module 15's recurrences and the reduction formulae of lesson 11
Take $u = x^n$, $dv = e^{-x}dx$, so $du = nx^{n-1}dx$ and $v = -e^{-x}$. Then $$J_n = \bigl[-x^n e^{-x}\bigr]_0^1 + n\int_0^1 x^{n-1}e^{-x}\,dx.$$
Evaluate the boundary: at $x = 1$, $-x^n e^{-x} = -e^{-1}$; at $x = 0$, $-x^n e^{-x} = 0$ (provided $n \geq 1$). So $\bigl[-x^n e^{-x}\bigr]_0^1 = -e^{-1}$.
Therefore $\boxed{J_n = nJ_{n-1} - e^{-1}}$ for $n \geq 1$, with $J_0 = \int_0^1 e^{-x}dx = 1 - e^{-1}$.
Worked through the hook. $J_1 = 1 \cdot J_0 - e^{-1} = (1 - e^{-1}) - e^{-1} = 1 - 2e^{-1}$. Then $J_2 = 2J_1 - e^{-1} = 2(1 - 2e^{-1}) - e^{-1} = 2 - 5e^{-1}$. Each step uses the same one-line recurrence.
Take $u = x^n$, $dv = e^{-x}dx$; the polynomial goes on $u$ to differentiate down · Boundary $\bigl[-x^n e^{-x}\bigr]_0^1 = -e^{-1}$ (only the $x = 1$ end contributes for $n \geq 1$) · Recurrence: $J_n = nJ_{n-1} - e^{-1}$ for $n \geq 1$, with $J_0 = 1 - e^{-1}$ · Non-homogeneous recurrence — the forcing term comes from the surviving boundary value
Pause — copy the recurrence $J_n = nJ_{n-1}-e^{-1}$, its derivation ($u = x^n$, $dv = e^{-x}dx$, boundary $= -e^{-1}$), and the base $J_0 = 1-e^{-1}$ into your book.
Quick check: For $J_n = \int_0^1 x^n e^{-x}dx$ satisfying $J_n = nJ_{n-1} - e^{-1}$ with $J_0 = 1 - e^{-1}$, what is $J_2$?
We just saw the model recurrence $J_n = nJ_{n-1} - e^{-1}$ for $J_n = \int_0^1 x^n e^{-x}\,dx$, where the boundary term $-e^{-1}$ (from the $x=1$ end) creates a non-homogeneous forcing term. That raises a question: how does this non-homogeneous recurrence fit the broader framework of reduction formulas? This card answers it → homogeneous recurrences arise when the boundary term vanishes; single-step index drop is typical for $x^n \times$ exponential; two-step drop occurs when a Pythagorean identity is needed.
Every reduction formula from lesson 11 is a recurrence; every integral recurrence in this lesson is a reduction formula. The difference is rhetorical, not mathematical.
- Homogeneous form $J_n = f(n)J_{n-k} + 0$: arises when the boundary term vanishes (e.g. Wallis integrals on $[0,\pi/2]$).
- Non-homogeneous form $J_n = f(n)J_{n-k} + g(n)$: arises when the boundary term contributes a non-zero value (e.g. $J_n = nJ_{n-1} - e^{-1}$ above).
Choice between $J_{n-1}$ and $J_{n-2}$ depends on what comes out of the parts identity. Single-step recurrences (involving $J_{n-1}$) are typical when the integrand splits neatly into polynomial $\times$ elementary; two-step recurrences (involving $J_{n-2}$) are typical when a Pythagorean identity must be used to recover the same family on the right.
Reduction formula $=$ integral recurrence (same idea) · Homogeneous: boundary term vanishes; non-homogeneous: boundary term contributes $g(n)$ · Single-step ($J_{n-1}$) is typical for $x^n \times$ elementary integrands · Two-step ($J_{n-2}$) is typical when a Pythagorean identity is needed
Pause — copy the homogeneous vs non-homogeneous distinction (boundary vanishes or contributes $g(n)$), the single-step ($J_{n-1}$) vs two-step ($J_{n-2}$) rule, and the Pythagorean-identity trigger for two-step into your book.
Did you get this? True or false: the recurrence $I_n = \tfrac{n-1}{n}I_{n-2}$ for the Wallis integral is homogeneous (i.e., has no forcing term $g(n)$).
Worked examples · 3 in a row, reveal as you go
Let $J_n = \int_0^1 x^n e^{-x}\,dx$ for $n \geq 1$. Show that $J_n = nJ_{n-1} - e^{-1}$.
Use the recurrence $J_n = nJ_{n-1} - e^{-1}$ with $J_0 = 1 - e^{-1}$ to evaluate $J_3 = \int_0^1 x^3 e^{-x}\,dx$.
Let $L_n = \int_0^1 x^n (1-x)^n\,dx$. Show $L_n = \tfrac{n}{n+1}\cdot L_{n-1}$ is wrong by direct check, then derive the correct relation linking $L_n$ to $L_{n-1}$.
Fill the gap: For $J_n = \int_0^1 x^n e^{-x}dx$, applying integration by parts with $u = $ gives the recurrence $J_n = nJ_{n-1} + g$, where the forcing term is $g = $ .
Misconceptions to fix · the 3 traps that cost marks
Did you get this? True or false: in the recurrence $J_n = nJ_{n-1} - e^{-1}$ for $J_n = \int_0^1 x^n e^{-x}dx$, the term $-e^{-1}$ arises from the boundary $[-x^n e^{-x}]_0^1$ evaluated at $x = 1$.
Activities · practice with the ideas
Use the recurrence $J_n = nJ_{n-1} - e^{-1}$ with $J_0 = 1 - e^{-1}$ to evaluate $J_4$.
Let $K_n = \int_1^e (\ln x)^n\,dx$ for $n \geq 1$. Derive the recurrence $K_n = e - nK_{n-1}$ and state $K_0$.
Using the recurrence from question 2, evaluate $K_2 = \int_1^e (\ln x)^2\,dx$.
For $K_n = \int_0^1 x^n \ln x\,dx$ ($n \geq 0$), use integration by parts to show $K_n = -\tfrac{1}{(n+1)^2}$. (Hint: $u = \ln x$, $dv = x^n dx$. Justify why the boundary at $0$ vanishes.)
Connect to lesson 11: rewrite the Wallis recurrence $I_n = \tfrac{n-1}{n}I_{n-2}$ in the standard recurrence form $J_n = f(n)J_{n-k} + g(n)$, identifying $f$, $k$ and $g$.
Odd one out: Three of these recurrences have a vanishing boundary term ($g = 0$). Which one does NOT?
Earlier you computed $J_0 = 1 - e^{-1}$ and tracked which boundary term survived when applying parts to $\int_0^1 x^n e^{-x}\,dx$.
The non-vanishing boundary at $x = 1$ is precisely what makes this recurrence non-homogeneous: it contributes the $-e^{-1}$ forcing term in $J_n = nJ_{n-1} - e^{-1}$. Compare with the lesson 11 Wallis case, where the boundary term vanishes at both endpoints and the recurrence is homogeneous. Recognising which type of recurrence you are deriving — homogeneous or non-homogeneous — is the single most useful diagnostic in Module 15.
Pick your answer, then rate your confidence — that tells the system what to drill next. Each retry pulls a fresh mix from the bank.
Q1. Given $J_n = nJ_{n-1} - e^{-1}$ with $J_0 = 1 - e^{-1}$, evaluate $J_2 = \int_0^1 x^2 e^{-x}\,dx$. Show each step of the chain. (2 marks)
Q2. Let $J_n = \int_0^1 x^n e^{-x}\,dx$ for $n \geq 1$. Use integration by parts to derive $J_n = nJ_{n-1} - e^{-1}$. Justify the value of each boundary contribution. (3 marks)
Q3. For $K_n = \int_1^e (\ln x)^n\,dx$ with $n \geq 1$, show that $K_n = e - nK_{n-1}$. Hence evaluate $K_3$. (3 marks)
Comprehensive answers (click to reveal)
Activity answers:
1. $J_1 = 1 - 2e^{-1}$, $J_2 = 2 - 5e^{-1}$, $J_3 = 6 - 16e^{-1}$, $J_4 = 4J_3 - e^{-1} = 24 - 65e^{-1}$.
2. $K_0 = e - 1$. With $u = (\ln x)^n$, $dv = dx$: $K_n = [x(\ln x)^n]_1^e - n\int_1^e (\ln x)^{n-1}dx = e \cdot 1 - 0 - nK_{n-1} = e - nK_{n-1}$.
3. $K_1 = e - 1\cdot(e-1) = 1$; $K_2 = e - 2\cdot 1 = e - 2$.
4. With $u = \ln x$, $dv = x^n dx$: $K_n = \bigl[\tfrac{x^{n+1}\ln x}{n+1}\bigr]_0^1 - \tfrac{1}{n+1}\int_0^1 x^n\,dx = 0 - \tfrac{1}{(n+1)^2} = -\tfrac{1}{(n+1)^2}$. The boundary at $x = 0$ vanishes because $x^{n+1}\ln x \to 0$ as $x \to 0^+$.
5. $f(n) = \tfrac{n-1}{n}$, $k = 2$, $g(n) = 0$ — homogeneous, two-step.
Q1 (2 marks): $J_1 = 1 - 2e^{-1}$ [1]; $J_2 = 2J_1 - e^{-1} = 2 - 5e^{-1}$ [1].
Q2 (3 marks): Parts with $u = x^n$, $dv = e^{-x}dx$ [1]; identify boundary $[-x^n e^{-x}]_0^1$ — at $x = 1$ this is $-e^{-1}$, at $x = 0$ this is $0$ for $n \geq 1$ [1]; the remaining integral is $nJ_{n-1}$, giving $J_n = nJ_{n-1} - e^{-1}$ [1].
Q3 (3 marks): Derivation $K_n = e - nK_{n-1}$ as in activity 2 [1]; $K_0 = e-1$, $K_1 = 1$, $K_2 = e-2$ [1]; $K_3 = e - 3K_2 = e - 3(e-2) = 6 - 2e$ [1].
Five timed questions on deriving and applying integral recurrences. Beat the boss to bank a tier — gold (90% + speed), silver (75%), or bronze (50%). Replays welcome.
⚔ Enter the arenaClimb platforms by answering quick recurrence questions. Lighter alternative to the boss.
Mark lesson as complete
Tick when you've finished the practice and review.