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3 Responses
Your shift is spot on: find the fixed point a* from a = 3a + 4 to get a* = −2, set b_n = a_n + 2 so b_{n+1} = 3b_n and b_n = 4·3^{n−1}, hence a_n = 4·3^{n−1} − 2. Quick check (so I stop second-guessing!): a_2 = 3·2 + 4 = 10, and the formula gives 4·3^1 − 2 = 10; a friendly walkthrough of this trick: https://brilliant.org/wiki/linear-recurrences/
Great question! You’re absolutely on the right track with the “shift to the fixed point” idea. It’s a lovely little trick for linear recurrences like a_{n+1} = 3a_n + 4. Let me walk you through why it works and pin down the closed form cleanly.
Step 1: Find the fixed point (equilibrium)
A fixed point L is a value that would stay constant under the recursion if the sequence ever hit it. So it solves:
L = 3L + 4
This gives (1 − 3)L = 4 → −2L = 4 → L = −2.
Step 2: Shift the sequence to measure “deviation from equilibrium”
Define b_n = a_n − L = a_n + 2. This measures how far a_n is from the fixed point.
Now plug into the recurrence:
b_{n+1} = a_{n+1} + 2 = (3a_n + 4) + 2 = 3a_n + 6 = 3(a_n + 2) = 3 b_n.
Boom! The shifted sequence b_n satisfies the simple geometric recurrence b_{n+1} = 3 b_n. That’s why the shift is valid and powerful: subtracting the fixed point converts the affine recurrence into a homogeneous linear one.
Step 3: Solve the geometric recurrence
From b_{n+1} = 3 b_n, we get b_n = b_1 · 3^{n−1}.
Compute b_1 from the initial condition a_1 = 2:
b_1 = a_1 + 2 = 4.
So b_n = 4 · 3^{n−1}.
Step 4: Shift back to a_n
a_n = b_n − 2 = 4 · 3^{n−1} − 2.
That matches your guess exactly!
Simple worked example (just to see it in action)
Let’s compute a few terms two ways.
– Using the recurrence:
a_1 = 2
a_2 = 3·2 + 4 = 10
a_3 = 3·10 + 4 = 34
a_4 = 3·34 + 4 = 106
– Using the formula a_n = 4·3^{n−1} − 2:
a_1 = 4·3^0 − 2 = 4 − 2 = 2
a_2 = 4·3^1 − 2 = 12 − 2 = 10
a_3 = 4·3^2 − 2 = 36 − 2 = 34
a_4 = 4·3^3 − 2 = 108 − 2 = 106
Perfect match!
Why the “shift” idea works in general (slightly nerdy meta-version)
For any recurrence of the form a_{n+1} = r a_n + c with r ≠ 1:
– The fixed point is L = c/(1 − r).
– If you define b_n = a_n − L, then b_{n+1} = r b_n, so b_n = b_1 r^{n−1}.
– Hence a_n = (a_1 − L) r^{n−1} + L.
Plugging r = 3 and c = 4 gives L = 4/(1 − 3) = −2 and yields exactly a_n = 4·3^{n−1} − 2.
Final answer
a_n = 4 · 3^{n−1} − 2.
Shift to the fixed point −2 by letting b_n = a_n + 2; then b_{n+1} = 3 b_n so b_n = 3^n and hence a_n = 4·3^{n-1} − 2 (the shift “cancels the +4” each step-I sometimes hesitate if it should be a_n−2, but it’s +2 here). Does this match your checks for n=2 and n=3?