Chapter 7. Induction and Recursion Part 1. Mathematical Induction ...

Chapter 7. Induction and Recursion
Part 1. Mathematical Induction
The principle of mathematical induction is this: to establish an infinite sequence of
propositions
P1, P2, P3, . . . , Pn, . . .
(or, simply put, Pn (n ≥ 1)), it is enough to verify the following two things
(1) P1, and
(2) Pk ⇒ Pk+ 1; (that is, assuming Pk vaild, we can rpove the validity of Pk+ 1).
These two things will give a “domino effect” for the validity of all Pn (n ≥ 1). Indeed,
step (1) tells us that P1 is true. Applying (2) to the case n = 1, we know that P2 is
true. Knowing that P2 is true and applying (2) to the case n = 2 we know that P3 is
true. Continue in this manner, we see that P4 is true, P5 is true, etc. This argument
can go on and on. Now you should be convinced that Pn is true, no matter how big is n.
We give some examples to show how this induction principle works.
Example 1. Use mathematical induction to show 1 + 3 + 5 + · · · + (2n − 1) = n 2 .
(Remember: in mathematics, “show” means “prove”.)
Answer: For n = 1, the identity becomes 1 = 1 2 , which is obviously true. Now
assume the validity of the identity for n = k:
1 + 3 + 5 + · · · + (2k − 1) = k 2 .
For n = k + 1, the left hand side of the identity is
1 + 3 + 5 + · · · + (2k − 1) + (2(k + 1) − 1) = k 2 + (2k + 1) = (k + 1) 2
which is the right hand side for n = k + 1. The proof is complete.
1

Example 2. Use mathematical induction to show the following formula for a geometric
series:
1 + x + x 2 + · · · + x n =
1 1 − xn+
. (x = 0) (7.1)
1 − x
Answer: For n = 1, the formula becomes 1 = 1−x
1−x , which holds trivially. Now we
assume the validity of the identity for n = k: 1 + x + x2 + · · · + xk = (1 − xk+ 1 )/(1 − x).
For n = k + 1, the left hand side of the formula is
1 + x + x 2 + · · · + x k + x k+ 1 1 1 − xk+
=
1 − x
+ xk+
1
= 1 − xk+ 1
1 − x + xk+ 1 (1 − x)
=
1 − x
1 − xk+ 1
1 − x + xk+ 1 − x
1 − x
k+ 2
= 1 − xk+ 2
1 − x
which is the right hand side for n = k + 1. So the formula is valid for general n.
Exercise 1. Prove the following identity by induction:
1 + 2 + 3 + · · · + n =
Exercise 2. Prove the following identity by induction:
1 2 + 2 2 + 3 2 + · · · + n 2 =
Exercise 3. Prove the following identity by induction:
n(n + 1)
2
1 3 + 2 3 + 3 3 + · · · + n 3 = (1 + 2 + 3 + · · · + n) 2
(7.2)
n(n + 1)(2n + 1)
. (7.3)
6
(According to Exercise 1, you only need to prove 1 3 + 2 3 + 3 3 + · · · + n 3 =
2 n(n+ 1)
2 .)
Not every identity depending on n can be handled by induction, as indicated in the
following exercise.
Exercise 4. Try to prove the following identity by induction:
a n+ 1 − b n+ 1 = (a − b)(a n + a n−1 b + a n−2 b 2 + · · · + ab n−1 + b n ). (7.4)
After failing to do so, try to prove this by multiplying out the right hand side.
2

Sometimes we can use induction to statements depending on a parameter n of natural
numbers, other than identities, as shown in the following example:
Example 3. The celebrated Cauchy-Schwarz inequality says
(a1b1 + a2b2 + · · · + anbn) 2 ≤ (a 2 1 + a 2 2 + · · · + a 2 n)(b 2 1 + b 2 2 + · · · + b 2 n) (7.5)
Check this for n = 1, 2. Then use induction to prove this inequality for general n.
Answer. When n = 1, the inequality reads (a1b1) 2 ≤ a2 1b22 . This is clear: in fact,
(a1b1) 2 = a2 1b2 2. Next we verify this inequality for n = 2. Here we rewrite this inequality
in different symbols because we need it in the inductive step:
(A1B1 + A2B2) 2 ≤ (A 2 1 + A22 )(B2 1 + B2 2 ), (∗)
that is, A 2 1B 2 1 + 2A1B1A2B2 + A 2 2B 2 2 ≤ A 2 1B 2 1 + A 2 1B 2 2 + A 2 2B 2 1 + A 2 2B 2 2. So it is enough to
check
2A1B1A2B2 ≤ A 2 1 B2 2 + A2 2 B2 1 ,
which can be recast as A2 1B2 2 − 2A1B1A2B2 + A2 2B2 1 ≥ 0. But
A 2 1 B2 2 − 2A1B1A2B2 + A 2 2 B2 1 = (A1B2 − A2B1) 2
which is clearly ≥ 0. Thus (∗) is valid. Now we assume the validity of
(a1b1 + a2b2 + · · · + anbn) 2 ≤ (a 2 1 + a22 + · · · + a2n )(b2 1 + b22 + · · · + b2n ) (∗∗)
for n = k. Notice that the left hand side of (∗∗) increases or remains the same if all
numbers in it are replaced by their absolute values, while the right hand side remains the
same. Hence we may (and we do) assume that aj ≥ 0 and bj ≥ 0 for all j = 1, . . . , n.
By the induction hypothesis, (a1b1 + · · · + akbk) 2 ≤ (a2 1 + · · · + a2 k )(b21 + · · · + b2 k ), or
a1b1 + a2b2 + · · · + akbk ≤ A1B1,
where A1 =
a 2 1 + a2 2 + · · · + a2 k , and B1 =
Now let A2 = ak+ 1 and B2 = bk+ 1 and use (∗) to obtain
(a1b1 + · · · + akbk + ak+ 1bk+ 1) 2 ≤ (A1B1 + ak+ 1bk+ 1) 2
≡ (A1B1 + A2B2) 2 ≤ (A 2 1 + A 2 2)(B 2 1 + B 2 2)
b 2 1 + b2 2 + · · · + b2 k .
= (a 2 1 + a2 2 + · · · + a2 k + a2 k+ 1 )(b2 1 + b2 2 + · · · + b2 k + b2 k+ 1 ),
3

and hence (∗∗) is also valid for n = k + 1.
Question 5. What is wrong with the following “proof” of “n = n + 1”?
“Assume that n = n + 1 holds for n = k, that is, k = k + 1. Add both sides by 1,
we get k + 1 = k + 1 + 1, which shows that the identity n = n + 1 also holds for
n = k + 1. By the principle of induction, n = n + 1 is valid for all n.”
Question 6. What is wrong with the following “proof” of the statement “every man is
bald” by induction?
“All we need to do is to establish the statement ‘a man with n hairs or less is bald’
for all n. When n = 0, 1 or 2, the statement is clearly valid. Now assume that the
statement is true for n = k, that is, assume that “a man with k hairs or less is
bald” is valid. Then the statement n = k + 1 is also valid because an additional hair
cannot change the fact of being bald. So by the principle of induction, the statement
‘a man with n hairs is bald’ is valid for all n. This proves all men are bald.”
An important role played by induction is to give proper definitions for many mathe-
matical expressions involving natural numbers. The usual pattern (or its variant) of this
definition by induction or by recursion is such: to define a sequence {f(n)}n≥1, we
take the following two steps:
(I) Specify what f(1) is,
(II) Specify how f(n + 1) can be decided from a subset of f(1), f(2), . . . , f(n).
Example 4. You know the meanings of the expressions a n , and n!. Strictly speaking,
they should be defined by induction. We write down the two steps (I), (II) described above
for defining each of them:
a 1 = a; a n+ 1 = a n a. (With f(n) = a n , f(1) = a and f(n + 1) = f(n)f(1).)
1! = 1; (n+1)! = n! (n+1). (With f(n) = n!, f(1) = 1 and f(n+1) = f(n)(n+1).)
Certainly, na is also defined by induction: 1a = a and (n + 1)a = na + a.
4

Example 5. This is an example of great importance about an expression involving the
summation symbol that can be defined recursively, given as follows
n
ak.
k= 1
We read the expression as “the sum of all terms ak, where k runs from 1 to n. (Earlier
we write such an expression simply as a1 + a2 + · · · + an.) Many identities and exercises
above can be recast nicely in the format using the summation symbol, for example,
n
k= 1 (2k − 1) = n2 , n
k= 1 k3 =
n
k= 1 k
2 etc.; (see Example 1 and Exercise 3 above). The recursive definition of n
k= 1 ak is
1
k= 1 ak = a1; n+ 1
k= 1 ak =
n
k= 1 ak
(with f(n) = n
k= 1 ak, f(1) = a1 and f(n + 1) = f(n) + an+ 1.)
,
+ an+ 1
Example 6. The famous Fibonacci numbers Fn (n = 0, 1, 2, 3, . . . ) are defined by
induction (or recursion):
Fn = Fn−1 + Fn−2, n ≥ 2, (7.6)
which says, starting from the third term, every term in the sequence of Fibonacci numbers
is the sum of the previous two terms. In order to determine this sequence, we have to
specify the first two terms. Here they are: F0 = 1 and F1 = 1. It is easy to produce a
few terms at the beginning of the sequence
1, 1, 2, 3, 5, 8, 13, 21, 34, . . .
(This sequence was originally described by the inventor as how rabbits multiply: Fn is
the number of pairs of rabbits on the nth day. There is an interesting story of this but it
does not concern us here.) The mathematical question here is, how can we find a closed
expression for Fn for general n? There are many ways to answer this question, such as:
treat it as a difference equation, or rewrite the recursion relation as a matrix identity
Fn 1 1 Fn−1
=
Fn−1 1 0 Fn−2
and use the method of diagonalization in linear algebra, or use generating functions.
5

** Appendix (may be omitted) . We briefly describe the last method here. Form the
power series f(x) = ∞
n= 0 Fnx n . It is called the generating function for the Fibonacci
sequence. It converges when |x| is small enough. Now
f(x) = F0 + F1x +
n≥2 Fnx n = F0 + F1x +
n≥2 (Fn−1 + Fn−2) x n
= 1 + x +
n≥2 Fn−1 x n +
n≥2 Fn−2 x n
= 1 + x +
k≥1 Fk x k+ 1 +
ℓ≥0 Fℓ
ℓ+ 2
x (k = n − 1, ℓ = n − 2)
= 1 + x + x
k≥1 Fk x k
2
+ x
ℓ≥0 Fℓ x ℓ
= 1 + x + x(f(x) − 1) + x 2 f(x) = 1 + xf(x) + x 2 f(x).
So we have (1 − x − x 2 )f(x) = 1. We can factorize the polynomial 1 − x − x 2 as
1 − x − x 2 = (1 − r+ x)(1 − r−x), where r± = (1 ± √ 5)/2 (with r+ + r− = 1 and
r+ r− = −1).
Now
1
1
f(x) =
=
1 − x − x2 (1 − r+ x)(1 − r−x) =
r+ r− 1
−
1 − r+ x 1 − r−x
= r+
n≥0 rn + x n
− r−
n≥0 rn − x n 1
√5 = 1
1
√ (rn+ + − r
5 n≥0
Comparing this with f(x) =
n≥0 Fn xn , we obtain
Fn = 1 n+ 1 n+ 1
√ (r+ − r− ) ≡
5 1
√
5
r+ − r−
1 + √ n+ 1
5
−
2
1
1 −
√
5
√ n+ 1
5
.
2
n+ 1
− ) x n .
The amazing thing here is the appearance of the irrational number √ 5 in this answer for
Fn, which is always an integer. (End of the appendix **)
Exercise 7. Use the recursion relation for Fibonacci numbers to prove the following
identity
F0 + F1 + F2 + · · · + Fn−1 = Fn+ 1 − 1, (n ≥ 1).
Exercise 8. Prove the amazing identity Fm+ n = FmFn + Fm−1Fn−1 by induction.
Recursion is an important notion for constructing so called recursive functions,
which is a key notion in logic and theoretical computer science.
6

Part 2. Polynomials: Taylor’s formula, binomial formula, etc.
By a polynomial we mean a function of the form
p(x) = anx n + an−1x n−1 + · · · + a2x 2 + a1x + a0 ≡ n
k= 0
akx k
where a0, a1, a2, . . . , an are some constants, called coefficients of p(x). When an = 0,
we say that the degree of p(x) is n. Now let us take an arbitrary constant a. Then
p(x + a) as a function of x is a polynomial of the same degree as that of p(x). However,
the degree of p(x + a) − p(x) is lower by 1, since the highest power terms in p(x + a)
and p(x) are canceled out.
Example 7. Consider the polynomial p(x) = 2x 2 + 3x + 1 of degree 2. Then, for
any constant a,
p(x + a) = 2(x + a) 2 + 3(x + a) + 1 = 2(x 2 + 2ax + a 2 ) + 3(x + a) + 1
= 2x 2 + (4a + 3)x + (2a 2 + 3a + 1)
which is also a polynomial of degree 2. So p(x + a) − p(x) = 4ax + (2a 2 + 3a), which is a
polynomial of degree 1.
The derivative of a polynomial p(x), denoted by p ′ d (x) or p(x), is defined to be
dx
the limit of the expression (p(x + h) − p(x))/h, called a difference quotient” as h tends
to zero. Thus we write
d
dx p(x) ≡ p′ (x) = lim
h→ 0
p(x + h) − p(x)
.
h
Suppose that the degree of p(x) is n. Then the degree of the difference p(x + h) − p(x)
drops by 1 and hence we expect that the degree of p ′ (x) is n − 1. When p(x) is of
degree zero, in other words, when p(x) is a constant function, say p(x) ≡ C, we have
d
C = 0.
dx
Simply put, the derivative of a constant function is zero.
Example 8. Consider the polynomial p(x) = 2x 2 + 3x + 1. From the previous
example we know that p(x + a) − p(x) = 4ax + (2a 2 + 3a). Replace a by h to get
p(x + h) − p(x) = 4hx + (2h 2 + 3h) = h(4x + 2h + 3). So
p ′ p(x + h) − p(x)
(x) = lim
h→ 0 h
= lim
h→ 0 (4x + 2h + 3) = 4x + 3.
7

We will develop some general rules to obtain this answer within seconds.
The following are two basic rules for computing derivatives. For simplicity, we write
u and v for two polynomials (instead of u(x) and v(x)) and a, b as two constants:
Linearity: (au + bv) ′ = au ′ + bv ′ .
Product rule: (uv) ′ = u ′ v + uv ′ .
They can be checked as follows. For linearity:
(au + bv) ′ (x) = lim
h→ 0
= lim
h→ 0
[au(x + h) + bv(x + h)] − [au(x) + bv(x)]
h
u(x + h) − u(x) v(x + h) − v(x)
a + b = au
h
h
′ (x) + bv ′ (x).
For the product rule, we need a clever trick of inserting appropriate terms used first time
by Leibnitz:
(uv) ′ (x) = lim
h→ 0
= lim
h→ 0
= lim
h→ 0
u(x + h)v(x + h) − u(x)v(x)
h
u(x + h)v(x + h) − u(x)v(x + h)
+
h
u(x)v(x + h) − u(x)v(x)
h
u(x + h) − u(x)
v(x + h) − v(x)
v(x + h) + u(x)
h
h
= u ′ (x)v(x) + u(x)v ′ (x).
(Note: In the mathematical literature this rule is commonly known as Leibnitz rule. “Prod-
uct rule” is used in textbooks for the convenience of beginners.)
The following identity is a basic formula for differentiating polynomials
d
dx xn = nx n−1
(n ≥ 0) (7.7)
Together with linearity, we can find the derivative of a given polynomial instantly. For
example, the derivative of p(x) = 2x 2 + 3x + 1 (Example 8 above) is p ′ (x) = 2(2x) +
3. 1 + 0 = 4x + 3.
Example 9. Prove (7.7) by induction.
Solution. For n = 0, x n is the constant function 1. In this case the identity becomes
d 1 = 0, which is valid – as we know that the derivative of a constant function vanishes.
dx
For n = 1, we have
dx
dx
(x + h) − x
= lim
h→ 0 h
h
= lim
h→ 0 h
8
= lim
h→ 0 1 = 1 ≡ 1. x0 .

Hence the identity is valid for n = 1 as well. Now assume the validity of the identity for
n = k, that is, d
dx xk = kx k−1 . By means of the product rule, we have
d
dx xk+ 1 = d
dx xk x =
d
dx xk
x + x k
d
dx x
= kx k−1 x + x k = (k + 1)x k ,
showing the validity of the identity for n = k + 1. The induction principle tells us that
the identity holds for all n ≥ 0.
The derivative of the derivative p ′ (x) of a polynomial p(x) is called the second
derivative of p(x), and is denoted by p ′′ (x) or p (2) (x). The derivative of p (2) (x) is
called the third derivative of p(x) and is denoted by p (3) (x). In general, we can define
the nth derivative p (n) (x) of a polynomial p(x) recursively as follows:
p (0) (x) = p(x), p (k+ 1) (x) = d
dx p(k) (x).
For example, with p(x) = 2x 2 + 3x + 1 (Example 8 above) we have p (1) (x) = 4x + 3,
p (2) (x) = 4, p (3) (x) = 0.
Exercise 9. In each of the following cases, find all derivatives of a given polynomial p(x):
(a) p(x) = 3 + 4x + 2x 2 , (b) p(x) = 7 + 3x + 2x 2 + x 3 , (c) p(x) = 1
15 x5 + 1
12 x4 .
Let f(x) be an arbitrary polynomial and let a be any number. Then f(x + a) is
a polynomial in x of the same degree and hence we can put it in the form f(x + a) =
n k= 0 ckxk . Replace x by x − a through out the last identity, we get
n
f(x) = ck(x − a) k ≡ c0 + c1(x − a) + c2(x − a) 2 + · · · + an(x − a) n . (7.8)
k= 0
To determine the constant term c0, we let x = a in (7.8) to get f(a) = c0. To determine
c1, we take the derivatives of both sides of (8) first:
f ′ (x) = c1 + c2. 2(x − a) + · · · + cn. n(x − a) n−1
and then we substitute x = a to get p ′ (a) = c1. To get an expression for cj for general j
(j ≤ n), we need to differentiate both sides of (7.8) j times. This leads us to the question
of finding the jth derivative of (x − a) k . The answer to this question is
dj dxj (x − a)k ⎧
k!
⎨ (k−j)!
=
⎩
(x − a)k−j if j < k ;
k! if j = k ;
0 if j > k .
9
(7.9)

You have to convince yourself that (7.9) is correct.
Exercise 10. Verify (7.9) for k = 4 and j = 1, 2, 3, 4.
When j = k, the right hand side of (7.9) either has a factor of x − a or is equal to zero,
and therefore its value is zero at x = a. When j = k, it becomes k! which is the same
as j!. Thus we have
dj (x − a)k
dxj
x= a
=
j! if j = k ;
0 if j = k .
(7.10)
Now take the jth derivatives of both sides of (7.8) and then evaluate at x = a. The left
hand side becomes f (j) (a). It follows from (7.10) that the right hand side becomes j!cj.
Thus we have f (j) (a) = j!cj, or cj = f (j) (a)/j!. Replace j by k and substitute the
resulting identity ck = f (k) (a)/k! back to (7.8). We finally get
f(x) =
n
k= 0
f (k) (a)
k!
(x − a) k . (7.11)
The last identity is called the Taylor expansion at x = a for the polynomial f.
Example 10. Use the Taylor expansion to find the partial fractions of x2 + x + 2
.
(x − 1) 3
Solution. Let p(x) = x 2 + x + 2. Then p (1) (x) = 2x + 1, p (2) (x) = 2, p (3) (x) = 0.
So p(1) = 4, p (1) (1) = 3, p (2) (1) = 2. The Taylor expansion of p(x) at x = 1 is
p(x) = 4 + 3(x − 1) + (x − 1) 2 . Hence
x2 + x + 2 4 + 3(x − 1) + (x − 1)2
=
(x − 1) 3 (x − 1) 3
which is the requires partial fraction decomposition.
= 1
x − 1 +
3
+
(x − 1) 2
4
.
(x − 1) 3
Now we apply (7.11) to the monomial f(x) = x n . We have to compute the kth
derivative of f(x) = x n for k ≤ n. The answer to this is supplied by (7.9). In using
(7.9), we switch k to n and j to k, and set a = 0. We obtain
f (k) (x) ≡ dk
dx k xn =
10
n!
(n − k)! xn−k

from which we get f (k) (a) = n!
(n−k)! an−k . So (7.11) gives
x n = n
k= 0
n!
k!(n − k)! an−k (x − a) k .
Substituting x = a + b to the last identity and noticing that
n!
k!(n − k)! =
n
, (7.12)
k
we obtain
(a + b) n =
n
k= 0
which is the celebrated binomial theorem.
n
a
k
n−k b k
Exercise 11. Use the binomial theorem to verify the identities
n n n n
n
≡ + + · · · + = 2
k= 0 k 0 1
n
n
and
n
k= 0 (−1)k
n
≡
k
n
−
0
n
+ · · · + (−1)
1
n
n
= 0.
n
The number n
k , read as “n choose k”, is interpreted as the number of ways to choose
k objects from n objects. For example, 4 4! = 2 2!2! = 6 tells us that there are 6 ways to
take two objects from a set of 4. Indeed, suppose that the four objects are {♥ ♣ ♦ ♠},
then the six ways of picking two are {♥ ♣}, {♥ ♦}, {♥ ♠}, {♣ ♦}, {♣ ♠}, {♦ ♠}.
The following identities concerning “n choose k” is basic:
n n n + 1 n
= ,
= +
k n − k k k − 1
n
. (7.13)
k
Both can be understood as follows. The first identity says, in dividing n objects between
you and me, k for me and n − k for you, the two methods below give the same number of
ways. One method is that I pick k objects, leaving the rest to you, and there are n
k ways
to do this. The other method is that you pick n − k objects, leaving the rest to me and
there are n
n−k ways to do this. The second identity concerns the number of ways to pick
k objects from n + 1 objects. Let us label on of these n + 1 objects and call it ♠. There
are n
n
k ways of taking k objects not including ♠ and there are k−1 ways of taking k
objects including ♠. Together they give n n
+ ways
k k−1
11

Exercise 12. Use (7.12) to verify (7.13).
*The rest material of this chapter is optional.
We use the binomial theorem to prove the following assertion inductively:
Proposition If p(x) is a polynomial of degree n, then there is a polynomial f(x)
of degree n + 1 such that p(x) = f(x + 1) − f(x).
This proposition suggests how to find a recipe for the sum
Sn = p(0) + p(1) + p(2) + p(3) + · · · + p(n),
where p(x) is a given polynomial. Indeed, if we know f(x) with p(x) = f(x+1)−f(x) ≡
−f(x) + f(x + 1), Sn becomes a “telescoping sum”:
Sn = (−f(0) + f(1)) + (−f(1) + f(2)) + (−f(2) + f(3)) + · · · + (−f(n) + f(n + 1))
= −f(0) + f(n + 1) = f(n + 1) − f(0).
Since the relation f(x + 1) − f(x) = p(x) is unchanged when we add an constant to f(x),
we may arrange f(x) in such a way that f(0) = 0 so that the above identity becomes
Sn = f(n + 1), which is the required recipe for the sum. This proposition tells us that
such f(x) exists. All we need is to look for it – that may involve some hard work!
Now we use this method to find the recipe for the sum 1 2 +2 2 +3 2 +· · ·+n 2 . Certainly
the answer is given by (7.3) in Exercise 2, which asks to verify this recipe by induction.
But how one can conjure up such a recipe is a complete myth! Our task here is to use a
systematic method to demystify this recipe. Here p(x) = x 2 . The above proposition tells
us that there is a polynomial f(x) of degree 3 such that f(x + 1) − f(x) = p(x). Let
us write f(x) = ax 3 + bx 2 + cx + d. Let us set f(0) = 0 so that d = 0. Now
So we have
f(1) = f(1) − f(0) = p(0) = 0,
f(2) = f(2) − f(1) = p(1) = 1 2 = 1,
f(3) = f(2) + p(2) = 1 + 2 2 = 5.
a + b + c = 0, 8a + 4b + 2c = 1, 27a + 9b + 3c = 4.
The solution to this system of linear equations is a = 1/3, b = −1/2, c = 1/6. So
f(x) = 1
3 x3 − 1
2 x2 + 1
6 x = (2x2 − 3x + 1)x
6
12
= (x − 1)(2x − 1)x
6

and hence Sn = f(n + 1) = n(2n + 1)(n + 1)/6.
Now we return to the proof of the proposition by induction on the degree n of p(x).
When n = 0, p(x) is a constant, say p(x) = c. In this case we just let f(x) = cx, which
is a polynomial of degree 1 ≡ 0 + 1. Indeed,
f(x + 1) − f(x) = c(x + 1) − cx = c = p(x).
Now we assume that the proposition is valid for all n ≤ k − 1 and let p(x) be a
polynomial of degree k. We can write p(x) = cx k + s(x), where s(x) is of degree
≤ k − 1. By the induction hypothesis, there is a polynomial g(x) of degree ≤ k such
that s(x) = g(x + 1) − g(x). On the other hand, the binomial theorem tells us that
(x + 1) k+ 1 = x k+ 1 + (k + 1)x k + r(x), where r(x) is a polynomial of degree k − 1 and
hence r(x) = h(x + 1) − h(x) for some polynomial h(x) of degree k. Thus we have
x k = 1 k+ 1 k+ 1
(x + 1) − x − (h(x + 1) − h(x)) = q(x + 1) − q(x)
k + 1
where q(x) = (x k+ 1 − h(x))/(k + 1) is a polynomial of degree k + 1. Hence
p(x) = c(q(x + 1) − q(x)) + (g(x + 1) − g(x)) = f(x + 1) − f(x),
where f(x) = cq(x) + g(x) is a polynomial of degree k + 1. The proof is complete.
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