58 Chapter 4
380 Problem
Shew that tan
π
2
100
+ 2 tan
π
2
99
+2
2
tan
π
2
298
+ ··· + 2
98
tan
π
2
2
= cot
π
2
100
.
381 Problem Shew that
n
X
k=1
k k
4
+ k
2
+ 1 =
1 2
· n
2
+ n n
2
+ n + 1 .
382 Problem Evaluate
1 · 2 · 4 + 2 · 4 · 8 + 3 · 6 · 12 + ···
1 · 3 · 9 + 2 · 6 · 18 + 3 · 9 · 27 + ···
1 3
.
383 Problem Shew that
∞
X
n=1
arctan 1
1 + n + n
2
=
π
4 .
Hint: From tan x − tan y =
tan x − tan y 1 + tan x tan y
deduce that arctan a − arctan b = arctan
a − b 1 + ab
for suitable a and b.
384 Problem
Prove the following result due to Gramm
∞
Y
n=2
n
3
− 1 n
3
+ 1 =
2 3
.
4.5 First Order Recursions
We have already seen the Fibonacci numbers, defined by the recursion f = 0
, f
1
= 1 and f
n+1
= f
n
+ f
n−1
, n ≥ 1. The order of the recurrence is the difference between the highest and the lowest subscripts. For example
u
n+2
− u
n+1
= 2 is of the first order, and
u
n+4
+ 9u
2 n
= n
5
is of the fourth order. A recurrence is linear if the subscripted letters appear only to the first power. For example
u
n+2
− u
n+1
= 2 is a linear recurrence and
x
2 n
+ nx
n−1
= 1 and x
n
+ 2
x
n−1
= 3 are not linear recurrences.
A recursion is homogeneous if all its terms contain the subscripted variable to the same power. Thus x
m+3
+ 8x
m+2
− 9x
m
= 0 is homogeneous. The equation
x
m+3
+ 8x
m+2
− 9x
m
= m
2
− 3 is not homogeneous.
A closed form of a recurrence is a formula that permits us to find the n-th term of the recurrence without having to know a priori the terms preceding it.
We outline a method for solving first order linear recurrence relations of the form
x
n
= ax
n−1
+ f n , a 6= 1,
where f is a polynomial.
First Order Recursions 59
1. First solve the homogeneous recurrence x
n
= ax
n−1
by “raising the subscripts” in the form x
n
= ax
n−1
. This we call the characteristic equation. Cancelling this gives x = a
. The solution to the homogeneous equation x
n
= ax
n−1
will be of the form x
n
= Aa
n
, where A is a constant to be determined. 2. Test a solution of the form x
n
= Aa
n
+ gn , where g is a polynomial of the same degree as f .
385 Example Let x
= 7 and x
n
= 2x
n−1
, n ≥ 1. Find a closed form for x
n
. Solution: Raising subscripts we have the characteristic equation x
n
= 2x
n−1
. Cancelling, x = 2. Thus we try a solution of the form x
n
= A2
n
, were A is a constant. But 7 = x = A2
and so A = 7 . The solution is thus x
n
= 72
n
. Aliter: We have
x =
7 x
1
= 2x
x
2
= 2x
1
x
3
= 2x
2
.. .
.. .
.. .
x
n
= 2x
n−1
Multiplying both columns, x
x
1
··· x
n
= 7 · 2
n
x x
1
x
2
···x
n−1
. Cancelling the common factors on both sides of the equality,
x
n
= 7 · 2
n
.
386 Example Let x
= 7 and x
n
= 2x
n−1
+ 1 , n ≥ 1. Find a closed form for x
n
. Solution: By raising the subscripts in the homogeneous equation we obtain x
n
= 2x
n−1
or x = 2. A solution to the homogeneous equation will be of the form x
n
= A2
n
. Now f n = 1 is a polynomial of degree 0 a constant and so we test a particular constant solution C. The general solution will have the form x
n
= A2
n
+ B. Now, 7 = x = A2
+ B = A + B. Also, x
1
= 2x + 7 = 15 and so 15 = x
1
= 2A + B. Solving the simultaneous equations A + B = 7
, 2A + B = 15
, we find A = 8
, B = −1. So the solution is x
n
= 82
n
− 1 = 2
n+3
− 1 .
Aliter: We have: x
= 7
x
1
= 2x
+ 1 x
2
= 2x
1
+ 1 x
3
= 2x
2
+ 1 ..
. ..
. ..
. x
n−1
= 2x
n−2
+ 1 x
n
= 2x
n−1
+ 1
60 Chapter 4
Multiply the kth row by 2
n−k
. We obtain 2
n
x =
2
n
· 7 2
n−1
x
1
= 2
n
x + 2
n−1
2
n−2
x
2
= 2
n−1
x
1
+ 2
n−2
2
n−3
x
3
= 2
n−2
x
2
+ 2
n−3
.. .
.. .
.. .
2
2
x
n−2
= 2
3
x
n−3
+ 2
2
2x
n−1
= 2
2
x
n−2
+ 2 x
n
= 2x
n−1
+ 1 Adding both columns, cancelling, and adding the geometric sum,
x
n
= 7 · 2
n
+ 1 + 2 + 2
2
+ ··· + 2
n−1
= 7 · 2
n
+ 2
n
− 1 = 2
n+3
− 1 .
Aliter: Let u
n
= x
n
+ 1 = 2x
n−1
+ 2 = 2x
n−1
+ 1 = 2u
n−1
. We solve the recursion u
n
= 2u
n−1
as we did on our first example: u
n
= 2
n
u = 2
n
x + 1 = 2
n
· 8 = 2
n+3
. Finally, x
n
= u
n
− 1 = 2
n+3
− 1 .
387 Example Let x
= 2 , x
n
= 9x
n−1
− 56n + 63. Find a closed form for this recursion. Solution: By raising the subscripts in the homogeneous equation we obtain the characteristic equation x
n
= 9x
n−1
or x = 9. A solution to the homogeneous equation will be of the form x
n
= A9
n
. Now f n = −56n + 63 is a polynomial of degree 1 and so we test a particular solution of the form Bn + C. The general solution will have the form x
n
= A9
n
+ Bn + C. Now x
= 2 , x
1
= 92 − 56 + 63 = 25 , x
2
= 925 − 562 + 63 = 176. We thus solve the system 2 = A + C
, 25 = 9A + B + C
, 176 = 81A + 2B + C
. We find A = 2
, B = 7,C = 0. The general solution is x
n
= 29
n
+ 7n .
388 Example Let x
= 1 , x
n
= 3x
n−1
− 2n
2
+ 6n − 3. Find a closed form for this recursion. Solution: By raising the subscripts in the homogeneous equation we obtain the characteristic equation x
n
= 3x
n−1
or x = 9. A solution to the homogeneous equation will be of the form x
n
= A3
n
. Now f n = −2n
2
+ 6n − 3 is a polynomial of degree 2 and so we test a particular solution of the form Bn
2
+ Cn + D. The general solution will have the form x
n
= A3
n
+ Bn
2
+ Cn + D. Now x
= 1 , x
1
= 31 − 2 + 6 − 3 = 4 , x
2
= 34 − 22
2
+ 62 − 3 = 13 , x
3
= 313 − 23
2
+ 63 − 3 = 36. We thus solve the system
1 = A + D ,
4 = 3A + B + C + D ,
13 = 9A + 4B + 2C + D ,
36 = 27A + 9B + 3C + D .
We find A = B = 1 ,C = D = 0. The general solution is x
n
= 3
n
+ n
2
.
Practice 61
389 Example
Find a closed form for x
n
= 2x
n−1
+ 3
n−1
, x = 2
. Solution: We test a solution of the form x
n
= A2
n
+ B3
n
. Then x = 2
, x
1
= 22 + 3 = 5
. We solve the system 2 = A + B
, 7 = 2A + 3B
. We find A = 1
, B = 1. The general solution is x
n
= 2
n
+ 3
n
. We now tackle the case when a = 1
. In this case, we simply consider a polynomial g of degree 1 higher than the degree of f .
390 Example Let x
= 7 and x
n
= x
n−1
+ n , n ≥ 1. Find a closed formula for x
n
. Solution: By raising the subscripts in the homogeneous equation we obtain the characteristic equation x
n
= x
n−1
or x = 1. A solution to the homogeneous equation will be of the form x
n
= A1
n
= A, a constant. Now f n = n is a polynomial of degree 1 and so we test a particular solution of the form Bn
2
+ Cn + D, one more degree than that of f . The general solution will have the form x
n
= A + Bn
2
+ Cn + D. Since A and D are constants, we may combine them to obtain x
n
= Bn
2
+ Cn + E . Now,
x = 7
, x
1
= 7 + 1 = 8 , x
2
= 8 + 2 = 10 . So we solve the system
7 = E ,
8 = B + C + E ,
10 = 4B + 2C + E .
We find B = C = 1
2 , E = 7. The general solution is x
n
= n
2
2 +
n 2
+ 7. Aliter: We have
x =
7 x
1
= x
+ 1 x
2
= x
1
+ 2 x
3
= x
2
+ 3 ..
. ..
. ..
. x
n
= x
n−1
+ n Adding both columns,
x + x
1
+ x
2
+ ··· + x
n
= 7 + x + x
2
+ ··· + x
n−1
+ 1 + 2 + 3 + ···+ n.
Cancelling and using the fact that 1 + 2 + ···+ n =
nn + 1 2
, x
n
= 7 + nn + 1
2 .
Some non-linear first order recursions maybe reduced to a linear first order recursion by a suitable transformation.
391 Example A recursion satisfies u
= 3 , u
2 n+1
= u
n
, n ≥ 1. Find a closed form for this recursion. Solution: Let v
n
= log u
n
. Then v
n
= log u
n
= log u
1 2
n−1
= 1
2 log u
n−1
= v
n−1
2 . As v
n
= v
n−1
2, we have v
n
= v 2
n
, that is, log u
n
= log u 2
n
. Therefore, u
n
= 3
1 2
n
.
Practice
62 Chapter 4
392 Problem
Find a closed form for x = 3
, x
n
= x
n−1
+ 4 3
.
393 Problem Find a closed form for x
= 1 , x
n
= 5x
n−1
− 20n + 25 .
394 Problem Find a closed form for x
= 1 , x
n
= x
n−1
+ 12n .
395 Problem Find a closed form for x
n
= 2x
n−1
+ 95
n−1
, x = 5
.
396 Problem Find a closed form for
a = 5
, a
j+1
= a
2 j
+ 2a
j
, j ≥ 0.
397 Problem AIME, 1994 If n
≥ 1, x
n
+ x
n−1
= n
2
. Given that x
19
= 94 , find the remainder when x
94
is divided by 1000 .
398 Problem
Find a closed form for x
= −1; x
n
= x
n−1
+ n
2
, n 0.
399 Problem If u
= 1 3 and u
n+1
= 2u
2 n
− 1 , find a closed form for u
n
.
400 Problem Let x
1
= 1 , x
n+1
= x
2 n
− x
n
+ 1 , n 0. Shew that
∞
X
n=1
1 x
n
= 1 .
4.6 Second Order Recursions
