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Simple methods to help you conquer recurrence relations
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In trying to find a formula for some mathematical sequence, a common intermediate step is to find the n th term, not as a function of n, but in terms of earlier terms of the sequence. For example, while it'd be nice to have a closed form function for the n th term of the Fibonacci sequence , sometimes all you have is the recurrence relation, namely that each term of the Fibonacci sequence is the sum of the previous two terms. This article will present several methods for deducing a closed form formula from a recurrence.
Steps
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Consider an arithmetic sequence such as 5, 8, 11, 14, 17, 20, .... [1] X Research source
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Since each term is 3 larger than the previous, it can be expressed as a recurrence as shown.Advertisement
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Recognize that any recurrence of the form a n = a n-1 + d is an arithmetic sequence. [2] X Research source
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Write the closed-form formula for an arithmetic sequence , possibly with unknowns as shown. [3] X Research source
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Solve for any unknowns depending on how the sequence was initialized. In this case, since 5 was the 0 th term, the formula is a n = 5 + 3n. If instead, you wanted 5 to be the first term, you would get a n = 2 + 3n.
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Consider a geometric sequence such as 3, 6, 12, 24, 48, ....
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Since each term is twice the previous, it can be expressed as a recurrence as shown.
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Recognize that any recurrence of the form a n = r * a n-1 is a geometric sequence.
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Solve for any unknowns depending on how the sequence was initialized. In this case, since 3 was the 0 th term, the formula is a n = 3*2 n . If instead, you wanted 3 to be the first term, you would get a n = 3*2 (n-1) . [4] X Research source
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Consider the sequence 5, 0, -8, -17, -25, -30, ... given by the recursion a n = a n-1 + n 2 - 6n. [5] X Research source
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Any recursion of the form shown, where p(n) is any polynomial in n, will have a polynomial closed form formula of degree one higher than the degree of p. [6] X Research source
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Write the general form of a polynomial of the required degree. In this example, p is quadratic, so we will need a cubic to represent the sequence a n . [7] X Research source
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Since a general cubic has four unknown coefficients, four terms of the sequence are required to solve the resulting system. Any four will do, so let's use terms 0, 1, 2, and 3. Running the recurrence backwards to find the -1 th term might make some calculations easier, but isn't necessary.
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Either Solve the resulting system of deg(p)+2 equations in deg(p)=2 unknowns or Fit a Lagrange polynomial to the deg(p)+2 known points .
- If the zeroth term was one of the terms you used to solve for the coefficients, you get the constant term of the polynomial for free and can immediately reduce the system to deg(p)+1 equations in deg(p)+1 unknowns as shown.
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Present the closed formula for a n as a polynomial with known coefficients.
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This is the first method capable of solving the Fibonacci sequence in the introduction, but the method solves any recurrence where the n th term is a linear combination of the previous k terms. So let's try it on the different example shown whose first terms are 1, 4, 13, 46, 157, .... [8] X Research source
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Write the characteristic polynomial of the recurrence. This is found by replacing each a n in the recurrence by x n and dividing by x (n-k) leaving a monic polynomial of degree k and a nonzero constant term. [9] X Research source
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Solve the characteristic polynomial . In this case, the characteristic has degree 2 so we can use the quadratic formula to find its roots. [10] X Research source
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Any expression of the form shown satisfies the recursion. The c i are any constants and the base of the exponents are the roots to the characteristic found above. This can be verified by induction. [11] X Research source
- If the characteristic has a multiple root, this step is modified slightly. If r is a root of multiplicity m, use (c 1 r n + c 2 nr n + c 3 n 2 r n + ... + c m n m-1 r n ) instead of simply (c 1 r n ). For example, the sequence starting 5, 0, -4, 16, 144, 640, 2240, ... satisfies the recursive relationship a n = 6a n-1 - 12a n-2 + 8a n-3 . The characteristic polynomial has a triple root of 2 and the closed form formula a n = 5*2 n - 7*n*2 n + 2*n 2 *2 n .
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Find the c i that satisfy the specified initial conditions. As with the polynomial example, this is done by creating a linear system of equations from the initial terms. Since this example has two unknowns, we need two terms. Any two will do, so take the 0 th and 1 st to avoid having to raise an irrational number to a high power.
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Solve the resulting system of equations.
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Plug the resulting constants into the general formula as the solution.
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Consider the sequence 2, 5, 14, 41, 122 ... given by the recursion shown. This cannot be solved by any of the above methods, but a formula can be found by using generating functions. [12] X Research source
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Write the generating function of the sequence. A generating function is simply a formal power series where the coefficient of x n is the n th term of the sequence. [13] X Research source
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Manipulate the generating function as shown. The objective in this step is to find an equation that will allow us to solve for the generating function A(x). Extract the initial term. Apply the recurrence relation to the remaining terms. Split the sum. Extract constant terms. Use the definition of A(x). Use the formula for the sum of a geometric series.
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Find the generating function A(x). [14] X Research source
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Find the coefficient of the x n in A(x). The methods for doing this will vary depending on exactly what A(x) looks like, but the method of partial fractions, combined with knowing the generating function of a geometric sequence, works here as shown.
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Write the formula for a n by identifying the coefficient of x n in A(x).
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Community Q&A
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QuestionIf a sequence is defined recursively by f(0)=2 and f(n+1)=-2f(n)+3 for n0, then f(2) is equal to what?Community AnswerFor n = 0 f(0+1) = - 2 f(0) + 3 f(1) = - 2(2) + 3 So f(1) = - 4 + 3 = -1 For n = 1 f(1+1) = -2 f(1) + 3 f(2) = -2 (-1) + 3 So f(2) = 2 + 3 = 5
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QuestionIs there a sequence that has second differences which produces a geometric sequence? If there is, what is name of the sequence and how can I derive the formula for the nth term in that sequence?Community AnswerIf you start with a geometric sequence, then all its differences will be geometric sequences (constant multiple of the original). The second differences of a linear sequence vanish, so you can add a linear sequence to any other sequence without changing its second differences. I don't believe there's a special name for the sum of a geometric and a linear sequence, but the formula is (a * b^n) + (c * n) + d for some constants a, b, c, and d, and they have your desired property.
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Tips
- Induction is also a popular technique. It's often easy to prove by induction that a specified formula satisfies a specified recursion, but the problem is this requires guessing the formula in advance.Thanks
- Some of these methods are computationally intensive with many opportunities to make a stupid mistake. It's good to check the formula against a few known terms.Thanks
- "In mathematics, the Fibonacci numbers or Fibonacci sequence are the numbers in the following integer sequence: 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, ...
- The Fibonacci spiral: an approximation of the golden spiral created by drawing circular arcs connecting the opposite corners of squares in the Fibonacci tiling; this one uses squares of sizes 1, 1, 2, 3, 5, 8, 13, 21, and 34.
- By definition, the first two numbers in the Fibonacci sequence are either 1 and 1, or 0 and 1, depending on the chosen starting point of the sequence, and each subsequent number is the sum of the previous two.
- In mathematical terms, the sequence F n of Fibonacci numbers is defined by the recurrence relation
- F n = F n-1 + F n-2 with seed values F 1 = 1, F 2 = 1 or F 0 = 0, F 1 = 1.
- The limit as n increases of the ratio F n /F n-1 is known as the Golden Ratio or Golden Mean or Phi (Φ), and so is the limit as n increases of the ratio F n-1 /F n ." 1
Thanks
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References
- ↑ http://discrete.openmathbooks.org/dmoi3/sec_recurrence.html
- ↑ http://mathworld.wolfram.com/RecurrenceEquation.html
- ↑ https://discrete.openmathbooks.org/dmoi2/sec_recurrence.html
- ↑ https://math.dartmouth.edu/archive/m19w03/public_html/Section4-2.pdf
- ↑ https://www.cs.sfu.ca/~ggbaker/zju/math/recurrence.html
- ↑ http://discrete.openmathbooks.org/dmoi2/sec_recurrence.html
- ↑ https://discrete.openmathbooks.org/dmoi2/sec_recurrence.html
- ↑ https://www.math.uci.edu/~ndonalds/math180b/4recurrence.pdf
- ↑ https://people.computing.clemson.edu/~goddard/texts/discreteMath/D2.pdf
- ↑ https://www.math.uci.edu/~ndonalds/math180b/4recurrence.pdf
- ↑ https://people.computing.clemson.edu/~goddard/texts/discreteMath/D2.pdf
- ↑ https://www.math.cmu.edu/~af1p/Teaching/Combinatorics/Slides/Generating-Functions.pdf
- ↑ https://www.math.cmu.edu/~af1p/Teaching/Combinatorics/Slides/Generating-Functions.pdf
- ↑ https://www.math.cmu.edu/~af1p/Teaching/Combinatorics/Slides/Generating-Functions.pdf
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