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Chapter 13: Problem 26

Carry out the indicated expansions. $$\left(1-z^{2}\right)^{7}$$

Short Answer

Expert verified
\((1-z^2)^7 = 1 - 7z^2 + 21z^4 - 35z^6 + 35z^8 - 21z^{10} + 7z^{12} - z^{14}\).

Step by step solution

01

Recognize the Binomial Expression

The expression \((1-z^2)^7\) is a binomial expression. We will expand it using the binomial theorem.
02

Binomial Theorem Formula

The binomial theorem states that \((a + b)^n = \sum_{k=0}^{n}\binom{n}{k} a^{n-k} b^{k}\). For our expression, set \(a=1\), \(b=-z^2\), and \(n=7\).
03

Apply the Binomial Theorem

Use the binomial theorem to expand \((1-z^2)^7\): \[(1-z^2)^7 = \sum_{k=0}^{7} \binom{7}{k} (1)^{7-k} (-z^2)^k\].
04

Simplify Each Term

Calculate each term in the expansion:* For \(k=0\): \(\binom{7}{0}(1)^{7-0}(-z^2)^0 = 1\).* For \(k=1\): \(\binom{7}{1}(1)^{7-1}(-z^2)^1 = -7z^2\).* For \(k=2\): \(\binom{7}{2}(1)^{7-2}(-z^2)^2 = 21z^4\).* For \(k=3\): \(\binom{7}{3}(1)^{7-3}(-z^2)^3 = -35z^6\).* For \(k=4\): \(\binom{7}{4}(1)^{7-4}(-z^2)^4 = 35z^8\).* For \(k=5\): \(\binom{7}{5}(1)^{7-5}(-z^2)^5 = -21z^{10}\).* For \(k=6\): \(\binom{7}{6}(1)^{7-6}(-z^2)^6 = 7z^{12}\).* For \(k=7\): \(\binom{7}{7}(1)^{7-7}(-z^2)^7 = -z^{14}\).
05

Final Expanded Form

Summing all the terms yields the expanded form:\[1 - 7z^2 + 21z^4 - 35z^6 + 35z^8 - 21z^{10} + 7z^{12} - z^{14}\].

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Key Concepts

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

Polynomial Expansion
Polynomial expansion involves expressing a polynomial in a scattered, expanded form. When we started with \((1-z^2)^7\), this was a compact form of a polynomial that had to be expanded. The process of going from the compact to expanded form uses the Binomial Theorem.

Expanding a polynomial makes complex expressions easier to manage, especially if you need to multiply the expression with others or plug in values to find numerical results. Most polynomial expansions arise from binomials, which are two-term expressions like \((a + b)^n\).

To proceed with polynomial expansion:
  • Identify the terms in the expression to plug into the Binomial Theorem.
  • Calculate the coefficients using combinations (from combinatorics).
  • Gather your terms after applying the power and product rules for exponents.
This transforms your polynomial into a manageable series of terms.
Combinatorics
Combinatorics is all about counting and arranging possibilities. In our polynomial expansion using the Binomial Theorem, we need to calculate combinations to find the coefficients of each term. The combination formula, \(\binom{n}{k}\), represents the number of ways to choose \(k\) elements from \(n\) available elements, regardless of the order.

In the binomial expansion of \((1 - z^2)^7\), each coefficient is calculated by a particular combination:
  • \(\binom{7}{0}\) for the first term.
  • \(\binom{7}{1}\) for the second term.
  • And so on, until \(\binom{7}{7}\) for the last term.
This methodology allows us to systematically generate the required coeffiicients for the expansion without manually doing the complex mathematical operations.
Mathematical Exponents
Understanding exponents is a key part of using the Binomial Theorem effectively. Exponents tell us how many times to multiply a number by itself. In our expansion, each term of the expression involves exponents on \(z^2\).

When using the Binomial Theorem, the exponents follow a specific pattern:
  • The power of the second term \((-z^2)^k\) increases as \(k\) increases.
  • The highest power of \(z^2\) occurs at \(k = 7\), giving us \(-z^{14}\).
Applying this correctly is essential for each step:
  • For each \(k\), raise \(-z^2\) to the power \(k\).
  • Remember that \(-z^2\) involves both negative signs and powers, affecting the resulting signs and values in each term.
This management of exponents ensures that each component of the polynomial is correctly expressed in the final expanded form.

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Most popular questions from this chapter

Verify that the given equation holds for \(n=1, n=2,\) and \(n=3 ;\) and (b) use mathematical induction to show that the equation holds for all natural numbers. $$F_{1}+F_{2}+F_{3}+\cdots+F_{n}=F_{n+2}-1$$

Use DeMoivre's theorem to find the indicated roots. Express the results in rectangular form. Cube roots of 64

Convert from rectangular to trigonometric form. (In each case, choose an argument \theta such that \(0 \leq \theta<2 \pi .)\) $$\frac{1}{4} \sqrt{3}-\frac{1}{4} i$$

The sum of the first \(n\) terms in a certain arithmetic sequence is given by \(S_{n}=3 n^{2}-n .\) Show that the \(r\) th term is given by \(a_{r}=6 r-4\).

We've seen that the Fibonacci sequence is defined recursively; each term after the second is the sum of the previous two terms. There is, in fact, an explicit formula for the \(n\) th term of the Fibonacci sequence:$$F_{n}=\frac{(1+\sqrt{5})^{n}-(1-\sqrt{5})^{n}}{2^{n} \sqrt{5}}$$ This formula was discovered by the French-English mathematician Abraham deMoivre more than 500 years after Fibonacci first introduced the sequence in 1202 in his book Liber Abaci. (For a proof of this formula, see Exercises 74 and \(75 .)\) (a) Using algebra (and not your calculator), check that this formula gives the right answers for \(F_{1}\) and \(F_{2}\) (b) Use this formula and your calculator to computer \(F_{24}\) and \(F_{25}\) (c) Use the formula and your calculator to compute \(F_{26}\) Then check your answer by using the results in part (b).
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