Using Chinese Dumbass Notation to FindAlgebraic Identities Daniel S.. Liu February 21, 2015 Abstract In this paper we will utilize Chinese Dumbass Notation in order to find well-known an
Trang 1Using Chinese Dumbass Notation to Find
Algebraic Identities Daniel S Liu February 21, 2015
Abstract
In this paper we will utilize Chinese Dumbass Notation in order to find well-known and not-so-well-known 3-variable homogeneous algebraic identities The purpose of using Chinese Dumbass Notation is because it allows for a new perspective of manipulating messy algebra that is much neater and tidier
Trang 23.1 Identity 1 7
3.2 Identity 2 8
3.3 Identity 3 9
3.4 Identity 4 10
3.5 Identity 5 11
Trang 31 Chinese Dumbass Notation
Chinese Dumbass Notation (or CDN as I will use for the rest of this paper) is
a concise method of organizing terms of a 3-variable algebraic expression [1]
For example, a general 3rd degree homogeneous expression in x, y, z would be written:
x3
x2y x2z
As an explicit example, the expression
x3+ 2y3+ 3z3+ 4(xy2+ x2z) − 3xyz would be written in CDN as:
1
We see the real power of CDN when we seek to express a complicated and messy expression like
X
sym
3x2y +X
sym
2x3−X
sym
xyz which in CDN is simply
4
Adding and subtracting in CDN is also easy We do it just like a matrix:
a1
a2 a3
+
b1
b2 b3
b4 b5 b6
b7 b8 b9 b10
=
a1+ b1
a2+ b2 a3+ b3
a4+ b4 a5+ b5 a6+ b6
a7+ b7 a8+ b8 a9+ b9 a10+ b10
Multiplying by a scalar is also exactly the same as a matrix:
k
a1
a2 a3
=
ka1
ka2 ka3
Trang 42 CDN for Common Expressions
In order to derive algebraic identities, it is useful to list out the CDN of some expressions Here are some of the most common third-degree expressions:
(x + y + z)3 =
1
x3+ y3+ z3 =
1
0
(x + y + z)(x2+ y2+ z2) =
1
(x + y + z)(xy + yz + zx) =
0
(x + y)(y + z)(z + x) =
0
Trang 53 Deriving Algebraic Identities
The usefulness of CDN in deriving algebraic identities comes from the ability to visualize what combination of expressions equals what As an example, I will derive a well-known identity Using this example, I will show the key steps in using CDN to derive new algebraic identities
Suppose I want to find another way to express (x + y + z)(x2+ y2+ z2) The CDN of this expression is
1
I look at the list of CDN for common expressions and see that the CDN for (x + y + z)(xy + yz + zx),
0
might be useful because the middle ring of ones can cancel some terms out:
1
−
0
=
1
which corresponds to
(x + y + z)(x2+ y2+ z2) − (x + y + z)(xy + yz + zx)
= (x + y + z)(x2+ y2+ z2− xy − yz − zx)
But clearly,
1
Trang 6
1
−
0
= x3+ y3+ z3− 3xyz Thus,
(x + y + z)(x2+ y2+ z2− xy − yz − zx) = x3+ y3+ z3− 3xyz and we have derived a famous identity
In the previous example, we see some key steps in finding an identity:
1 Start out with an expression and its CDN that we want to represent in another way In the previous example, we started out with
(x + y + z)(x2+ y2+ z2) =
1
2 Using the list of CDN for common expressions, we add and subtract ex-pressions from this starting expression in smart ways In the previous example, we subtracted
(x + y + z)(xy + yz + zx) =
0
to get rid of six one’s in the original expression
3 The goal is to try to reduce the original expression using additions and subtractions into an expression with a CDN that we recognize In the previous example, we reduced it to this, which we recognized as x3+ y3+
z3− 3xyz:
1
4 Remember that the less reductions you make, the simpler the final identity will be A short and simple identity is more beautiful and more useful than
a long, contrived identity
In the next few examples, we will put these key steps to work, deriving or perhaps rederiving identities
Trang 73.1 Identity 1
Let’s change the expression (x + y)3+ (y + z)3+ (z + x)3 to CDN:
2
In order to make every term equal to three, let’s add x3+ y3+ z3to get
x3+ y3+ z3+ (x + y)3+ (y + z)3+ (z + x)3 which corresponds to
3
But we know that (x + y + z)(x2+ y2+ z2) can be expressed as
1
so 3(x + y + z)(x2+ y2+ z2) written in CDN is
3
Finally, this means that
x3+ y3+ z3+ (x + y)3+ (y + z)3+ (z + x)3= 3(x + y + z)(x2+ y2+ z2) and we have our identity
Trang 83.2 Identity 2
Let’s start with the expression (x + y + z)3:
1
We notice that the inside seven numbers resemble a multiple of the CDN for (x + y)(y + z)(z + x) Specifically, 3(x + y)(y + z)(z + x) in CDN is
3
0
=
0
Subtracting this from our original expression gives (x + y + z)3− 3(x + y)(y + z)(z + x) in CDN is
1
−
0
=
1
But this is just x3+ y3+ z3 Thus,
(x + y + z)3− 3(x + y)(y + z)(z + x) = x3+ y3+ z3
and after rearrangement we have an identity
(x + y + z)3− (x3+ y3+ z3) = 3(x + y)(y + z)(z + x)
Trang 9
3.3 Identity 3
The CDN for the expression (x + y + z)(xy + yz + zx) is
0
Subtracting xyz gives that (x + y + z)(xy + yz + zx) − xyz in CDN is
0
−
0
=
0
but this is just (x + y)(y + z)(z + x) Thus,
(x + y + z)(xy + yz + zx) − xyz = (x + y)(y + z)(z + x) Rearranging a little gives that our final identity is
(xy + yz + zx)(x + y + z) = (x + y)(y + z)(z + x) + xyz
Trang 10
3.4 Identity 4
We will try to find another way to represent (x + y)3+ (y + z)3+ (z + x)3 In CDN it is:
2
Note that the CDN for (x + y + z)3as a ring of three’s that we can use to cancel terms out:
1
Subtracting gives that (x + y)3+ (y + z)3+ (z + x)3− (x + y + z)3in CDN is
2
−
1
=
1
But that is just x3+ y3+ z3− 6xyz so
(x + y)3+ (y + z)3+ (z + x)3− (x + y + z)3= x3+ y3+ z3− 6xyz After some rearrangement we get the identity
x3+ y3+ z3− (x + y)3− (y + z)3− (z + x)3+ (x + y + z)3= 6xyz
Trang 11
3.5 Identity 5
We will represent (x + y − z)3 in CDN:
1
This is not symmetric (at least no 3-fold symmetry), which does not please us because our list of CDN for common expressions only has symmetric expressions This prompts us to add this expression cyclically and represent it in CDN We find that the CDN ofX
cyc
(x + y − z)3 is
1
+
1
+
−1
=
1
The entire outside ring looks suspiciously like the CDN for (x + y + z)3, so let’s go ahead and subtract this from (x + y + z)3 to get that the CDN of the expression (x + y + z)3−X
cyc
(x + y − z)3is
1
−
1
=
0
But this is simply just 24xyz This tells us that we have found another identity:
(x + y + z)3−X
cyc
(x + y − z)3= 24xyz
which in expanded form is
(x + y + z)3− (x + y − z)3− (y + z − x)3− (z + x − y)3= 24xyz
Trang 12
4 Conclusion
These identities are only a tiny fraction of all the hundreds of possible identities you can derive just from playing around with patterns I have only limited myself to three-variable, degree 3 homogeneous algebraic identities You may want to play around with degree 2 or degree 4 expressions, or perhaps using a tetrahedron extension of CDN to investigate 4-variable identities
Homogeneous identities aren’t the only ones you can create You can also model non-homogeneous expression by replacing a variable with 1 as shown:
x3
x2y x2
Using this new model, a homogeneous identity like (xy + yz + zx)(x + y + z) = (x + y)(y + z)(z + x) + xyz that we derived in Identity 3 becomes something completely unrecognizable:
(xy + x + y)(x + y + 1) = (x + y)(x + 1)(y + 1) + xy
The key point in this paper is the idea of how seeing these algebraic expression geometrically helps us visualize what combined with what to get what It is
a lot harder to recognize that (xy + yz + zx)(x + y + z) is in fact equal to (x + y)(y + z)(z + x) + xyz than it is to recognize that
0
=
0
+
0
In addition, just looking for relations between the CDN of different expressions can give ample inspiration for creating a new identity
The final lesson to be taken from all of this is to just let yourself loose Go crazy, get creative, and you’ll be whipping up identities you have never seen before in
no time
References
[1] ”The Art of Dumbassing” by Brian Hamrick