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Method of Detached Coefficients

Main Arithmetic Algebra Page

Page Contents

  1. Method of Detached Coefficients 
  2. Binomial Coefficients with a Calculator
  3. Multinomial Coefficients
    1. Multinomial Coefficients for (a+b+c)
    2. Multinomials of the form (1+x+x2)


Method of Detached Coefficients 

The method of detatched coefficients is simply a way of dealing with algebra by dropping the algebraic variables and simply using the numbers, keeping them in their correct places. Some examples follow.

Binomial Coefficients with a Calculator

The Method of Detached Coefficients involves detaching the coefficients of polynomial and dealing with the coefficients only. For instance:
(1+x)2=1+2x+x2
112=121

(1+x)3=1+3x+3x2+x3
113=1331

(1+x)4=1+4x+6x2+4x3 +x4
114=14641

That is, we can drop the x's, y's etc, and use the coefficients alone. In the above cases, we find the coeffients simply by using 11 instead of (1+x), and using arithmtic to expand our binomial (or whatever).

After 114, the calculator does not give us the clear result we seek:

115=161051

This reminds us that we need to keep our numbers separate:
1015=10510100501

This means we soon exceed the capacity of the ten-digit calculator, and need to use Windows Calculator. However, adding a zero, keeps the coefficients separate, so we can read of the results, and write:
1015=10510100501, as
(a+b)5=a5+5a4b+10a3b2+10a2b3+5ab4+b5
However, an ordinary calculator gives:
1015=105101005
from which we can guess the last digit is 1.

The important thing is to keep the coefficients separate. A final example:

10019=1009036084126126084036009001
Implying correctly that:
detachedCoefficients2.gif

Multinomial Coefficients

Multinomials of the form (a+b+c)n

We cannot so easily use the calculator in the case of multinomial coefficients of the form:(a+b+c)2
(a+b+c)2=a2+b2+c2+2ab+2ac+2bc

We cannot do this because if we write it as 111, for instance, with a=100, b=10, and c=1, we get confused between b2=100 and ac=100, and
1112=12321
with the 3 being a confusion between b2 and 2ac.

To draw out the difference we can write:
100010012=100020021002001
where a=10,000,000, b=1000, and c=1.

The first 1 is a2, the first 2 is 2ab, the next 2ac. The second 1 is b2 and the final 2 is 2bc, and the final 1 c2.
(a+b+c)2=a2+b2+c2+2ab+2ac+2bc

Another example:
100010013=1000300330061033003001

This implies that
(a+b+c)3=a3+b3+c3+3a2b+3a2c+3ab2+3ac2+3b2c+3bc2+6abc

Multinomials of the form (1+x+x2)n

Multinomials of the form (1+x+x2) have a basic order, so:
1112=12321

Correctly implying that:
(1+x+x2)2=1+2x+3x2+2x3+x4








Ken Ward's Mathematics Pages


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