Download STUDENT`S COMPANIONS IN BASIC MATH: THE SECOND Basic

STUDENT’S COMPANIONS IN BASIC MATH: THE SECOND
Basic Identities in Algebra
Let us start with a basic identity in algebra:
a2 − b2 = (a − b)(a + b).
(1)
Indeed, multiplying out the right hand side, we get a2 +ab−ba−b2 . Removing the “mixed
terms”, which cancel each other, we get the left hand side.
QUESTION 1. What is the result of multiplying out (a1 + a2 )(b1 + b2 )?
QUESTION 2. Same question for (x − a)(x − b).
QUESTION 3. Same question for (a1 + a2 + a3 )(b1 + b2 ).
QUESTION 4. Same question for (a1 + a2 )(b1 + b2 )(c1 + c2 ).
Similar to (1), we have, a3 − b3 = (a − b)(a2 + ab + b2 ).
EXERCISE 5. Check the last identity.
In general, for any positive integer n, we have
an − bn = (a − b)(an−1 + an−2 b + · · · + abn−2 + bn−1 )
(2)
QUESTION 6. What does (2) give for n = 4, 5, 6?
EXERCISE 7. Prove (2) by multiplying out its right hand side.
Switching n to n + 1, (2) becomes an+1 − bn+1 = (a − b)(an + an−1 b + · · · + abn−1 + bn ).
Pn
n−k k
b .
Using the summation notation, we write an + an−1 b + · · · + abn−1 + bn as
k=0 a
So we get another form of (2):
Xn
k=0
an−k bk =
1
an+1 − bn+1
.
a−b
(3)
Letting a = 1 and b = x, (3) becomes
Xn
k=0
xk ≡ 1 + x + x2 + · · · + xn =
1 − xn+1
,
1−x
(4)
This identity tells us how to find the sum of a geometric progression. It should be
memorized. Notice that, when |x| < 1, xn+1 converges to zero as n → ∞ and hence
the right hand side of (4) converges to 1/(1 − x). Thus we have
X∞
n=0
xn ≡ 1 + x + x2 + · · · =
1
,
1−x
for |x| < 1.
EXERCISE 8. Check the following identities
a3 + b3 = (a + b)(a2 − ab + b2 )
a5 + b5 = (a + b)(a4 − a3 b + a2 b2 − ab3 + b4 ).
Make a guess of a similar identity for an + bn where n is odd.
EXERCISE 9. Show that, when n is odd,
an + bn = (a + b)(an−1 − an−2 b + · · · + a2 bn−3 − abn−2 + bn−1 )
(5)
(Hint: Replace b by −b in (2).)
Identities (2) and (5) are often used in factorization.
EXAMPLE 10. We are asked to factorize a12 − b12 . We should make a wise choice in our
first step in order not to jeopardize the rest of our work. Here is out first step:
a12 − b12 = (a6 )2 − (b6 )2 = (a6 − b6 )(a6 + b6 ).
Now we treat a6 − b6 and a6 + b6 separately:
a6 − b6 = (a3 )2 − (b3 )2 = (a3 − b3 )(a3 + b3 )
= (a − b)(a2 + ab + b2 )(a + b)(a2 − ab + b2 ).
and
a6 + b6 = (a2 )3 + (b2 )3 = (a2 + b2 )(a4 − a2 b2 + b4 ).
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(If we accept the appearance of irrational numbers, we can go further by “completing
square”:
√
a4 − a2 b2 + b4 = a4 + 2a2 b2 + b2 − 3a2 b2 = (a2 + b2 )2 − ( 3 ab)2
√
√
= (a2 + b2 − 3 ab)(a2 + b2 + 3 ab).
But let’s not take this step.) So, finally,
a12 − b12 = (a − b)(a + b)(a2 + b2 )(a2 + ab + b2 )(a2 − ab + b2 )(a4 − a2 b2 + b4 ).
EXERCISE 11. Factorize a4 + a2 b2 + b4 .
(Hint: Rewrite it as a4 + 2a2 b2 + b4 − a2 b2 .)
QUESTION 12. From (5) we see that, when n is odd, a + b is a factor of an + bn . When
n is even, why is a + b not a factor of an + bn ?
EXERCISE 13. Factorize (a + b)5 − a5 − b5 by extracting a factor of a + b first.
Our next topic is the binomial formula. Let’s begin with
(a + b)2 = a2 + 2ab + b2
(a + b)3 = a3 + 3a2 b + 3ab2 + b3 .
(a + b)4 = a4 + 4a3 b + 6a2 b2 + 4ab3 + b4 .
EXERCISE 14. Verify the above identities.
EXERCISE 15. Verify (a + b + c)2 = a2 + b2 + c2 + 2ab + 2ac + 2bc. Establish a similar
identity for (a + b + c + d)2 .
For any positive integer n, we have
n
an−k bk
k
k=0
µ ¶
µ ¶
µ
¶
n n−1
n n−2 2
n
n
≡a +
a
b+
a
b + ··· +
abk−1 + bn ,
1
2
n−1
(a + b) =
where
n µ ¶
X
n
µ ¶
n!
n
=
k
k!(n − k)!
with
k! = 1 × 2 × 3 × · · · × (k − 1) × k
3
(6)
(7)
and, by convention 0! = 1. This is called the binomial formula. We give a very brief
reasoning instead of a proof. (For a complete proof, see the EIGHTH COMPANION.)
¡n¢
The expression
k
is read as “n choose k”. It stands for the number of ways of
choosing k objects from given n objects. Now, multiplying out (a + b)n , we get 2n
terms. Each of them is a product of the form c1 c2 · · · cn , where each factor cj (1 ≤ j ≤ n)
is either a or b. If there are exactly k factors equal to b, then all of the rest are equal to
a and hence c1 c2 · · · cn becomes an−k bk . Let us fix some k. Among these 2n terms,
how many are equal to an−k bk ? Well, it is the number of ways of assigning k factors in
¡ ¢
c1 c2 · · · cn to b and assigning the rest to a. So there are exactly nk of 2n terms equal
to an−k bk . For each k, collect all terms of the form an−k bk in the sum of 2n terms
obtained by multiplying out (a + b)n . Then you get (6).
About “n choose k”
µ
n
n−k
¡n¢
k , the following two identities are basic:
¶
µ ¶
n
=
,
k
µ
¶ µ ¶ µ
¶
n
n
n+1
+
=
.
k−1
k
k
(8)
(The second identity tells us how to construct the Pascal triangle.) To see the validity
of the first identity, let us consider the number of ways to divide n objects between Liz
and Larry, k of them for Liz and n − k of them for Larry. We can let Liz to choose her
¡ ¢
¡ ¢
k objects first. There are nk ways to do so. This tells us that there are nk ways to
¡ n ¢
¡ ¢
divide the objects. If we let Larry choose first, we have n−k
ways. Since both nk and
¡ n ¢
can represent the number of ways of dividing objects between Liz and Larry, these
n−k
two numbers must be equal.
QUESTION 16. Consider the number of ways of choosing k objects from n + 1 objects,
¡
¢
which is equal to n+1
k . Take any one from these n + 1 objects and put a tag on it.
¡
¢
Consider two groups of n+1
choices: the first group consists of choices including the
k
tagged one and the second group consists of choices excluding it. How many choices are
in each group? Now, do you see why the second identity in (8) is valid?
QUESTION 17. Why is the identity (k + 1)! = k! (k + 1) valid?
EXERCISE 18. Verify that
¡n¢
k
given in (7) indeed satisfies the identities in (8). (You
need the identity mentioned in the QUESTION 17.)
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We mention the following multinomial formula:
n
(a1 + a2 + · · · + ap ) =
µ
X
k1 +k2 +···+kp =n
where
µ
n
k1 k2 · · · kp
¶
=
n
k1 k2 · · · kp
¶
ak11 ak22 · · · akpp ,
n!
.
k1 !, k2 ! · · · kp !
(You will be asked to prove this by your SEVENTH COMPANION as a problem.)
Occasionally we encounter some simple but truly beautiful identities; for example
(a2 + b2 )(c2 + d2 ) = (ac + bd)2 + (ad − bc)2 .
(9)
This is very easy to verify. (An expression of the form a2 + b2 is called a “sum of two
squares” for obvious reason. Identity (9) says the product of sums of two squares is also a
sum of two squares. There are similar but much more intriguing identities about “sums of
four squares” and “sums of eight squares”.) The question is, how can one conjure up such
a neat identity? Or, what is behind this inspiring someone to write this down? (See the
SEVENTH COMPANION on complex numbers to find the answer.)
EXERCISE 19. Check (9).
Next item: algebra of fractions. The sum of two fractions is given by:
a
c
ad + bc
+ =
.
b
d
bd
(10)
On the right hand side, the terms of the numerator are obtained by “cross multiplication”.
To see this, notice that bd is a multiple for both denominators on the left hand side. This
fact suggests bd for the common denominator. In detail,
a
c
a d
c b
ad
bc
ad + bc
+ =
+
=
+
=
.
b
d
b d d b
bd
bd
bd
For the difference of two fractions, we have something similar to (10):
c
ad − bc
a
− =
.
b
d
bd
5
(11)
In practice, we often use some common multiple of b and d to simplify our steps. We should
try to get a common multiple as small as possible, or get the least common multiple if
possible. Assume that m is a common multiple of b and d. That m is a multiple of
b means that we can write m = bp for some p. Similarly we have m = dq for some q.
So, finding the sum of two fractions a/b and c/d is carried out like this:
a
c
a p
c q
ap
cq
ap bq
ap + bq
+ =
+
=
+
=
+
=
.
b
d
b p d q
bp
dq
m
m
m
The last expression SHOULD NOT be memorized!
EXAMPLE 20. We are asked to simplify tan x − 2 csc 2x. First, we convert this expression
to one which only involves sin x and cos x, which are easier to deal with. Write tan x as
sin x/ cos x, and csc 2x as 1/ sin 2x and Use the “trig” identity sin 2x = 2 sin x cos x:
2
sin x
1
sin x
−
=
−
cos x 2 sin x cos x
cos x sin x cos x
2
2
− cos x
cos x
sin x − 1
=
=−
= − cot x.
=
sin x cos x
sin x cos x
sin x
tan x + 2 csc 2x =
EXERCISE 21. Simplify the following expressions. (NO CALCULATOR!)
(a)
5
7
−
;
48 72
(b) 1 +
x2
1
x
+
.
− 1 (x − 1)2
Multiplication for fractions is easy:
a c
ac
= .
b d
bd
(12)
Division of fractions involves a “flip” in the denominator:
a
a d
ad
b
c = b c ≡ bc .
d
(13)
We mention the following two special cases
a
b
c
=
a
bc
and
a
c
d
=
ad
.
c
We must be very careful in manipulating complicated fractions. Pitfalls are everywhere.
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EXERCISE 22. Given a sequence {an }n≥1 ≡ {a1 , a2 , a3 , . . . }, we define a sequence of
continued fractions {P (a1 , a2 , . . . , an )}n≥1 inductively as follows:
P (a1 ) =
1
,
a1
P (a1 , a2 , . . . , an ) =
1
.
a1 + P (a2 , a3 , . . . , an )
Check that
P (a1 , a2 ) =
a2
a1 a2 + 1
and
P (a1 , a2 , a3 ) =
a2 a3 + 1
.
a1 a2 a3 + a1 + a3
A similar expression for P (a1 , a2 , a3 , a4 ) is quite complicated. Find it.
QUESTION 23. In the previous exercise, let an = 1 for all n. What do you think the
limit L ≡ limn→∞ P (a1 , a2 , . . . , an ) should be ? Compute P (a1 , . . . , a4 ) and compare
it with the numerical value of L.
By a rational number we mean a number which can be written in the form a/b,
where a and b are integers with b 6= 0. Identities (10), (11), (12) and (13) tell us that,
if r and s are rational numbers, then so are sum r + s, difference r − s, product
rs, as well as their quotient r/s, provided that s is nonzero. Because of this property,
we say that rational numbers form a field. The standard notation for the field of rational
numbers is Q. All real numbers also form a field, denoted by R.
√
PROBLEM 24. Show that all numbers which can be written as r + s 2, where r and
s are rational numbers, form a field. (Remember: in math, “show” means “prove”.)
Functions such as 2x2 + x − 1, x3 − 5x2 + 6x + 9 are called polynomials. In general,
a polynomial is a function of the form
p(x) = a0 xn + a1 xn−1 + a2 xn−2 + · · · + an−1 x + an .
where a0 , a1 , a2 , . . . , an are some constants. When n ≥ 1 and a0 6= 0, we say that the
degree of the polynomial p is n. (If p is a nonzero constant, we say that the degree of
p is zero. The zero function as a polynomial has degree ∞.) If f and g are polynomials
where g is nonconstant, we can use the usual method of long division to divide f by g
for obtaining the quotient q and the remainder r such that
f (x) = q(x)g(x) + r(x).
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Here, if r is not zero, then the degree of r is less than the degree of g.
EXERCISE 25. Divide x4 − 2x3 + 3x2 + x + 2 by x2 + x + 2 using long division.
A function f is called a rational function if there exist polynomials p and q such
that f (x) = p(x)/q(x) for all x. Certainly identities (10) to (13) above also holds when
a, b, c, d are interpreted as polynomials. Hence all rational functions form a field. It is
called the field of rational functions. (What else can you call?)
x3 − x
?
x2 + x
1
1
1
EXERCISE 27. Simplify
+
+ 2
.
x x+1 x +x
QUESTION 26. How do you simplify
Next, we deal with roots (square roots, cube roots, etc.) and fractional powers. A
√
common mistake is writing a2 = a. The correct one is
√
a2 = |a| (the absolute value of a).
√ 2
a, we tacitly assume a ≥ 0. It is correct to put
a = a,
√
√ √
keeping in mind that this is only for a ≥ 0. It is also correct to write a b = ab. But
√
√ √
writing ab = a b raises some concern about the signs of a and b.
Normally, when we write
√
QUESTION 28. Why is that so?
q
QUESTION 29. Why is
√
2 3 − 5 not a legitimate expression?
s√
EXERCISE 30. Simplify
√
3− 2
√ .
√
3+ 2
When n is a positive integer, the nth power un of u is defined recursively by
u1 = u;
un+1 = un u (n ≥ 1).
Next, we extend un for any integer n. For n = 0, we define u0 = 1, provided that
u is nonzero. The expression 00 is undefined; (it is called an indeterminate form
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in calculus, which is to be used to indicate a type of limits). For negative n, we write
n = −k where k > 0 and define
un ≡ u−k =
1
uk
(for u 6= 0).
For example, u−1 = 1/u, u−2 = 1/u2 etc. For any positive integer n, the expression
√
u1/n stands for the nth root of u. For example, u1/2 = u. When n is an even number,
we have to assume tacitly that u ≥ 0. When a is a rational number, say a = m/n,
where m, n are integers with n > 0, ua is defined for u > 0 via
ua = (u1/n )m ≡ (um )1/n .
For the general ua where u > 0 and a is any real number, we need the help of the
exponential function ex and the logarithm function ln x to assign a proper meaning:
ua (= (eln u )a ) = ea ln u .
It follows from the basic “additive formula” es et = es+t and the “multiplicative formula”
(es )t = est for the exponential function that
ua ub = ua+b ,
(ua )b = uab
(14)
for all real numbers a, b, u with u > 0. Let me emphasize again the importance of the
restriction u > 0 here. Unrestricted usage of (12) is a “weapon of math destruction”. It
is TERRIFYING.
QUESTION 31. What is the mistake made in the following “proof” of −1 = 1?
1 = 11 = 12×1/2 = (12 )1/2 = ((−1)2 )1/2 = (−1)2×1/2 = (−1)1 = −1.
EXERCISE 32. Derive (14) from es et = es+t and (es )t = est .
Sometimes the answer to an algebraic expression depends on cases. For example, if
n is an integer, then (−1)n is 1 or −1, depending on whether n is even or odd. So
we write
n
(−1) =
n
1 if n is even;
−1 if n is odd.
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This format for a mathematical expression is not rare.
EXAMPLE 33. Find (−1)(p−1)/2 , where p is an odd integer.
To answer this question, we first notice that, when an integer is divided by 4, its possible
remainders are 0, 1, 2 and 3. So each integer can be written in one of the following
forms: 4k, 4k + 1, 4k + 2, 4k + 3. Clearly 4k and 4k + 2 are even numbers. So we
are left with two cases to consider: p = 4k + 1 and p = 4k + 3. When p = 4k + 1, we
have (p − 1)/2 = (4k + 1 − 1)/2 = 2k, which is an even number and hence (−1)(p−1)/2 =
(−1)2k = 1. When p = 4k + 3, we have (p − 1)/2 = (4k + 3 − 1)/2 = 2k + 1 which is
clearly an odd number and hence (−1)(p−1)/2 = (−1)2k+1 = −1. We conclude:
½
(p−1)/2
(−1)
=
1 if p = 4k + 1 for some integer k;
−1 if p = 4k + 3 for some integer k.
EXERCISE 34. Find (−1)mn+n+m , where m and n are integers.
The so-called Kronecker’s delta δjk , commonly used in the science literature, is an
expression with double subscripts defined by
½
δjk =
1 if j = k;
0 if j 6= k.
As you can tell, the (j, k)-entry of the n × n identity matrix In is δjk (1 ≤ j, k ≤ n).
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