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Chapter 1: Problem 5

Consider the polynomial $$ \left(1-x^{2}+x^{4}\right)^{2003}=a_{0}+a_{1} x+a_{2} x^{2}+\cdots+a_{8012} x^{8012} $$ Find (1) \(a_{0}\) (2) \(a_{0}+a_{1}+a_{2}+\cdots+a_{8012}\) 3 \(a_{0}-a_{1}+a_{2}-a_{3}+\cdots-a_{8011}+a_{8012}\) (4) \(a_{0}+a_{2}+a_{4}+\cdots+a_{8010}+a_{8012}\) 5 \(a_{1}+a_{3}+\cdots+a_{8009}+a_{8011}\)

Short Answer

Expert verified
1) 1 2) 1 3) 1 4) \(3^{2003}/2\) 5) \(3^{2003}/2\)

Step by step solution

01

Find the Constant Term (a0)

The constant term in the expansion \((1-x^2+x^4)^{2003}\) will be when the polynomial doesn't involve any of the variables raised to a power, i.e., \(x^0\). This occurs when all terms chosen from the expansion have their positive and negative \(x^2\) terms canceled out, giving a result as choosing \((1)^n\), i.e., 1 for every choice of \(x^2eq0\). Thus, the constant term is simply 1, i.e., \(a_0 = 1\).
02

Calculate the Sum of Coefficients

To find the sum of all coefficients \(a_0 + a_1 + a_2 + \cdots + a_{8012}\), substitute \(x = 1\) in the polynomial. The expression becomes \((1-1^2+1^4)^{2003}=(1-1+1)^{2003}=1^{2003}\), which equals 1. Thus, the sum of all coefficients is 1.
03

Find the Alternating Sum

For the alternating sum \(a_0 - a_1 + a_2 - a_3 + \cdots - a_{8011} + a_{8012}\), substitute \(x = -1\) in the polynomial. The expression becomes \((1-(-1)^2+(-1)^4)^{2003}=(1-1+1)^{2003}=1^{2003}\), which is also 1. Thus, the alternating sum is 1.
04

Sum Even Indexed Coefficients

To find \(a_0 + a_2 + a_4 + \cdots + a_{8010} + a_{8012}\), substitute \(x = i\) where \(i\) is the imaginary unit \(i = \sqrt{-1}\). The expression becomes: \((1-(i)^2+i^4)^{2003}=(1+1+1)^{2003}=3^{2003}\). Thus, the sum of even-indexed coefficients is \(3^{2003}/2\) (taking into account symmetry with odd-indexed where each pair sums to half of the full).
05

Sum Odd Indexed Coefficients

For \(a_1 + a_3 + a_5 + \cdots + a_{8011}\), note that this is the net coefficients of odd powers of \(x\) alone. By symmetry and from earlier steps, the sum of odd-indexed coefficients should equal the sum of even-indexed assuming a symmetric distribution. Thus, this sum is \(3^{2003}/2\) (considering full sum of terms is raised by 3 when using the imaginary unit).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Coefficients
In the polynomial expansion of \((1-x^2+x^4)^{2003}\), the coefficients \(a_0, a_1, a_2, \ldots, a_{8012}\) are the numerical factors of each term in the polynomial. Each coefficient corresponds to a specific power of\(x\) in the expanded form. Understanding coefficients is essential because they tell us how much each term contributes to the overall polynomial when combining like terms.
  • The constant term or \(a_0\) represents the coefficient of \(x^0\), which is simply 1 in this case.
  • To find the sum of all coefficients, we substitute \(x = 1\) into the polynomial. This trick allows us to sum all terms as powers of 1 remain the same.
  • The alternating sum and other specific combinations also rely on the calculation of these coefficients in special arrangements.
Alternating Sum
The notion of an alternating sum is a fascinating concept in polynomial analysis. For the given polynomial \((1-x^2+x^4)^{2003}\), calculating the alternating sum \(a_0 - a_1 + a_2 - a_3 + \cdots - a_{8011} + a_{8012}\) requires evaluating the polynomial at \(x = -1\).By doing this substitution, the polynomial becomes \((1 - (-1)^2 + (-1)^4)^{2003}\). This simplifies to \(1^{2003}\) as the negative terms cancel each other. Therefore, the alternating sum of the coefficients is simply 1.
  • Alternating sums help highlight certain symmetries or properties within polynomials that might otherwise be hard to analyze directly.
  • This technique is especially useful in problems dealing with series or patterns.
Imaginary Unit
The imaginary unit, denoted as \(i\), satisfies \(i^2 = -1\). It is an essential tool for understanding complex numbers and is quite useful in polynomial expansion problems. In the given problem, substituting \(x = i\) in the polynomial \((1-x^2+x^4)^{2003}\) helps us examine the sum of even-indexed coefficients.
  • On substitution, each occurrence of \(i^2\) is replaced by \(-1\), and terms such as \(i^4\) simplify back to 1 because \(i^4 = (i^2)^2 = (-1)^2 = 1\).
  • The expression simplifies to \((1+1+1)^{2003} = 3^{2003}\), demonstrating the power of using complex numbers to simplify polynomial problems.
Even and Odd Indexed Coefficients
The idea of even and odd indexed coefficients in polynomial expansions deal specifically with the terms whose powers are either even or odd numbers. For the polynomial \((1-x^2+x^4)^{2003}\), these groups of coefficients can be individually summed by substituting special values of \(x\) as observed earlier.
  • For even indexed coefficients like \(a_0, a_2, a_4, \ldots, a_{8012}\), the substitution \(x = i\) leads us to the conclusion that their sum is \(\frac{3^{2003}}{2}\).
  • Similarly, the sum of odd indexed coefficients \(a_1, a_3, a_5, \ldots, a_{8011}\) also results in \(\frac{3^{2003}}{2}\) due to symmetrical properties.
  • This balance between even and odd indexed coefficients reveals an inherent symmetry in the structure of the polynomial.

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