Determining the Number of Distinct Terms in a Multivariate Expansion

Determining the Number of Distinct Terms in a Multivariate Expansion

When dealing with polynomial expansions, it is important to understand how to determine the number of distinct terms in an expression. This article focuses on a specific case: the expansion of the expression (1 - 2x x^{2^5} 1 - 4y 4y^{2^5}). We will apply the multinomial theorem and analyze the terms to find the total number of distinct terms in the expression.

Expression Analysis

The expression in question is:

[1 - 2x x^{2^5} 1 - 4y 4y^{2^5}]

This can be broken down into two parts for analysis:

Part 1: Analysis of (1 - 2x x^{2^5})

Let's rewrite the first part as:

[1 - 2x x^2 1]

For simplicity, let's denote (a 1), (b -2x x^2), and (c 1). We are expanding the expression ([a b c]^5]. Using the multinomial theorem, the expansion will be of the form:

[frac{5!}{k_1! k_2! k_3!} a^{k_1} b^{k_2} c^{k_3}]

where (k_1 k_2 k_3 5).

Determining the Ranges of Exponents

Let's determine the exponents of (x). The exponent of (x) in each term is given by (k_2 2k_3).

The ranges for (k_2) and (k_3) are:

(k_1) can range from 0 to 5 (k_2) can range from 0 to 5 (k_3) can range from 0 to 5

By substituting the possible values, we can determine the range of the exponents (k_2 2k_3):

If (k_3 0), (k_2) can be 0 to 5, giving exponents 0 to 5. If (k_3 1), (k_2) can be 0 to 4, giving exponents 2 to 8. If (k_3 2), (k_2) can be 0 to 3, giving exponents 4 to 10. If (k_3 3), (k_2) can be 0 to 2, giving exponents 6 to 12. If (k_3 4), (k_2) can be 0 to 1, giving exponents 8 to 14. If (k_3 5), (k_2) can only be 0, giving exponent 10.

The distinct exponents from (1 - 2x x^{2^5}) are: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14. This gives us 15 distinct terms.

Part 2: Analysis of (1 - 4y 4y^{2^5})

Now let's analyze the second part:

[1 - 4y 4y^2 1]

For simplicity, let's denote (a 1), (b -4y 4y^2), and (c 1). We are expanding the expression ([a b c]^5]). Using the multinomial theorem, the expansion will be of the form:

[frac{5!}{k_1! k_2! k_3!} a^{k_1} b^{k_2} c^{k_3}]

where (k_1 k_2 k_3 5).

Determining the Ranges of Exponents

Let's determine the exponents of (y). The exponent of (y) in each term is given by (k_2 2k_3).

The ranges for (k_2) and (k_3) are the same as before:

(k_1) can range from 0 to 5 (k_2) can range from 0 to 5 (k_3) can range from 0 to 5

By substituting the possible values, we can determine the range of the exponents (k_2 2k_3):

If (k_3 0), (k_2) can be 0 to 5, giving exponents 0 to 5. If (k_3 1), (k_2) can be 0 to 4, giving exponents 2 to 8. If (k_3 2), (k_2) can be 0 to 3, giving exponents 4 to 10. If (k_3 3), (k_2) can be 0 to 2, giving exponents 6 to 12. If (k_3 4), (k_2) can be 0 to 1, giving exponents 8 to 14. If (k_3 5), (k_2) can only be 0, giving exponent 10.

The distinct exponents from (1 - 4y 4y^{2^5}) are: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14. This gives us 15 distinct terms.

Final Count of Distinct Terms in the Entire Expression

The overall expression is:

[1 - 2x x^{2^5} 1 - 4y 4y^{2^5}]

Since the terms involving (x) and (y) are distinct, we can add the distinct terms from both expansions:

From (1 - 2x x^{2^5}): 15 distinct terms in (x) From (1 - 4y 4y^{2^5}): 15 distinct terms in (y)

Thus, the total number of distinct terms in the combined expression is:

15 15 30

Conclusion

Therefore, the total number of distinct terms in the expansion of (1 - 2x x^{2^5} 1 - 4y 4y^{2^5}) is 30.