/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 15 Let \(n \in \mathbb{N}\) and let... [FREE SOLUTION] | 91影视

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Let \(n \in \mathbb{N}\) and let \(a_{1}, \ldots, a_{n}\) and \(b_{1}, \ldots, b_{n}\) be any real numbers. Assume that not all \(a_{1}, \ldots, a_{n}\) are zero. Consider the quadratic polynomial $$ q(x)=\sum_{i=1}^{n}\left(x a_{i}+b_{i}\right)^{2} . $$ Show that the discriminant \(\Delta\) of \(q(x)\) is nonpositive, and \(\Delta=0\) if and only if there is \(c \in \mathbb{R}\) such that \(b_{i}=c a_{i}\) for all \(i=1, \ldots, n\). Use this to give an alternative proof of Proposition \(1.12 .\)

Short Answer

Expert verified
We find the discriminant of the quadratic polynomial \(q(x) = \sum_{i=1}^{n} (a_{i}^2 x^2 + 2a_{i}b_{i}x + b_{i}^2)\) as \(\Delta = B^2 - 4AC\), where \(A = \sum_{i=1}^{n}a_{i}^2, B = \sum_{i=1}^{n}2a_{i}b_{i}\), and \(C = \sum_{i=1}^{n}b_{i}^2\). Using the Cauchy-Schwarz inequality, we show that \(\Delta \leq 0\). We also prove that \(\Delta = 0\) if and only if there exists a c 鈭 鈩 such that \(b_{i} = c a_{i}\) for all i = 1, ..., n. This result can be used to provide an alternative proof of Proposition 1.12.

Step by step solution

01

Write out the quadratic polynomial q(x)

Given the quadratic polynomial, \[q(x) = \sum_{i=1}^{n}\left(xa_{i}+b_{i}\right)^{2}\]
02

Expand the quadratic polynomial

Expand the sum to get the coefficients of the quadratic polynomial: \[q(x) = \sum_{i=1}^{n}(a_{i}^2x^2 + 2a_{i}b_{i}x + b_{i}^2)\] Now, collect similar terms together: \[q(x) = \left(\sum_{i=1}^{n}a_{i}^2\right)x^2 + \left(\sum_{i=1}^{n}2a_{i}b_{i}\right)x + \left(\sum_{i=1}^{n}b_{i}^2\right)\] This allows us to write the polynomial as: \[q(x) = Ax^2 + Bx + C\] where \[A = \sum_{i=1}^{n}a_{i}^2\] \[B = \sum_{i=1}^{n}2a_{i}b_{i}\] \[C = \sum_{i=1}^{n}b_{i}^2\]
03

Find the discriminant 螖

Using the alternative definition of discriminant, we have: \[\Delta = B^2 - 4AC\] Substitute A, B, and C from the previous step: \[\Delta = \left(\sum_{i=1}^{n}2a_{i}b_{i}\right)^{2} - 4\left(\sum_{i=1}^{n}a_{i}^2\right)\left(\sum_{i=1}^{n}b_{i}^2\right)\]
04

Prove 螖 is nonpositive

We can use the Cauchy-Schwarz inequality to show that the discriminant is nonpositive: \[\left(\sum_{i=1}^{n}2a_{i}b_{i}\right)^{2} \leq \left(\sum_{i=1}^{n}4a_{i}^2\right)\left(\sum_{i=1}^{n}b_{i}^2\right)\] Thus, \[\Delta \leq 0\]
05

Prove 螖 = 0 if and only if bi = cai for all i

The discriminant is equal to zero if and only if the Cauchy-Schwarz inequality holds as an equality: \[\left(\sum_{i=1}^{n}2a_{i}b_{i}\right)^{2} = \left(\sum_{i=1}^{n}4a_{i}^2\right)\left(\sum_{i=1}^{n}b_{i}^2\right)\] This happens if and only if the vectors (2ai) and (bi) are proportional, i.e., there exists a constant c such that: \[b_{i} = c a_{i}\] for all i = 1, ..., n. We have now shown that the discriminant 螖 of the polynomial q(x) is nonpositive and that 螖 = 0 if and only if there exists a c 鈭 鈩 such that bi = cai for all i = 1, ..., n. This can be used to provide an alternative proof of Proposition 1.12.

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Key Concepts

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

Discriminant of a Quadratic Polynomial
The discriminant of a quadratic polynomial is a key concept that helps us understand the nature of its roots. In the particular case where we have a sum of squares like in the exercise, the discriminant, denoted by \( \Delta \), can tell us if the solutions to the polynomial equation are real or not. For a general quadratic polynomial \( Ax^2 + Bx + C \), the discriminant is calculated as \( \Delta = B^2 - 4AC \).

If \( \Delta > 0 \), the polynomial has two distinct real roots. If \( \Delta = 0 \), it has exactly one real root, and if \( \Delta < 0 \), it has no real roots, only complex ones. The exercise challenges students to show that for a specific form of a quadratic polynomial, the discriminant is always nonpositive (i.e., \( \Delta \leq 0 \) ), which implies that there are either one or no real roots at all.

The proof involves expanding the polynomial, finding an expression for the discriminant, and then showing that it is always less than or equal to zero using the Cauchy-Schwarz inequality鈥攁 fundamental inequality in vector analysis and many areas of mathematics.
Cauchy-Schwarz Inequality and Its Role
The Cauchy-Schwarz inequality is an essential tool in mathematics, particularly in linear algebra and analysis. It provides a bound on the dot product of vectors and asserts that for any vectors \( \mathbf{u} \) and \( \mathbf{v} \) in an inner product space, the absolute value of the dot product is at most the product of the vectors' magnitudes.

\[|\mathbf{u} \cdot \mathbf{v}| \leq ||\mathbf{u}|| \cdot ||\mathbf{v}||\]

Translating this into the language of summations for real numbers, as seen in this exercise, it becomes:

\[\left(\sum_{i=1}^{n} a_i b_i \right)^2 \leq \left(\sum_{i=1}^{n} a_i^2 \right)\left(\sum_{i=1}^{n} b_i^2 \right)\]

This inequality is a pivotal step in proving why the discriminant is nonpositive. By applying it to the terms involved in expressing the discriminant of the quadratic polynomial, we show that the square of the sum of products of corresponding \( a_i \) and \( b_i \) is less than or equal to the product of the sum of squares of \( a_i \) and \( b_i \) terms. This relationship demonstrates that the discriminant \( \Delta \) cannot be positive. This use of the Cauchy-Schwarz inequality is a practical demonstration of how abstract mathematical principles apply to concrete problems.
Real Numbers Vector Proportionality
Proportionality in the context of vectors derived from sets of real numbers can be interpreted through the lens of the Cauchy-Schwarz inequality. As established, when the Cauchy-Schwarz inequality holds with equality, say for sequences \( (a_1, \ldots, a_n) \) and \( (b_1, \ldots, b_n) \) as in our exercise, the vectors formed by these sequences are proportional. This means there exists a constant \( c \) such that:

\[b_i = c \cdot a_i\]

For all indices \( i \) within the sequences. This condition for proportionality is essential because it characterizes when the discriminant \( \Delta \) of our quadratic polynomial is exactly zero. The equation \( q(x) \) represents, geometrically, a parabola, and when \( \Delta = 0 \) there is exactly one point where this parabola touches the x-axis鈥攊ndicating a single, repeated real root.

This relationship between the discriminant and proportionality is not just a mathematical curiosity but an insight into the structure of solutions to polynomial equations, which is a cornerstone of algebra. By understanding this principle, students are better equipped to analyze and solve even more complex systems that involve this kind of relationship.

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Most popular questions from this chapter

Let \(p(x), q(x) \in \mathbb{R}[x] .\) Show that \(p(x)\) and \(q(x)\) are relatively prime if and only if \(u(x) p(x)+v(x) q(x)=1\) for some \(u(x), v(x) \in \mathbb{R}[x] .\) Is it true that if a nonzero polynomial \(d(x) \in \mathbb{R}[x]\) satisfies \(u(x) p(x)+v(x) q(x)=d(x)\) for some \(u(x), v(x) \in \mathbb{R}[x]\), then \(d(x)\) is a GCD of \(p(x)\) and \(q(x)\) ?

Given any \(a \in \mathbb{R}\) and \(k \in \mathbb{Z}\), the binomial coefficient associated with \(a\) and \(k\) is defined by $$ \left(\begin{array}{l} a \\ k \end{array}\right)=\left\\{\begin{array}{ll} \frac{a(a-1) \cdots(a-k+1)}{k !} & \text { if } k \geq 0 \\ 0 & \text { if } k<0 \end{array}\right. $$ where for \(k \geq 0\), we denote by \(k !\) (read as \(k\) factorial) the product of the first \(k\) positive integers. Note that \(0 !=1\) and \(\left(\begin{array}{c}a \\ 0\end{array}\right)=1\) for every \(a \in \mathbb{R}\). (i) Show that if \(a, k \in \mathbb{Z}\) with \(0 \leq k \leq a\), then $$ \left(\begin{array}{l} a \\ k \end{array}\right)=\frac{a !}{k !(a-k) !}=\left(\begin{array}{c} a \\ a-k \end{array}\right) $$ (ii) If \(a \in \mathbb{R}\) and \(k \in \mathbb{Z}\), then show that $$ \left(\begin{array}{l} a \\ k \end{array}\right)=\left(\begin{array}{c} a-1 \\ k \end{array}\right)+\left(\begin{array}{l} a-1 \\ k-1 \end{array}\right) $$ [Note: This identity is sometimes called the Pascal triangle identity. If we compute the values of the binomial coefficients \(\left(\begin{array}{c}n \\\ k\end{array}\right)\) for \(n \in \mathbb{N}\) and \(0 \leq k \leq n\), and write them in a triangular array such that the \(n\) th row consists of the numbers \(\left(\begin{array}{c}n \\ 0\end{array}\right),\left(\begin{array}{l}n \\\ 1\end{array}\right), \ldots,\left(\begin{array}{c}n \\ n\end{array}\right)\), then this array is called the Pascal triangle. It may be instructive to write the first few rows of the Pascal triangle and see what the identity means pictorially.] (iii) Use the identity in (ii) and induction to prove the Binomial Theorem (for positive integral exponents). In other words, given \(x, y \in \mathbb{R}\), prove that \((x+y)^{n}=\sum_{k=0}^{n}\left(\begin{array}{c}n \\\ k\end{array}\right) x^{k} y^{n-k}\) for all \(n \in \mathbb{N}\). [Note: Proving a statement defined for \(n \in \mathbb{N}\) such as the above identity by induction means that we should prove it for the initial value \(n=1\), and further prove it for every \(n \in \mathbb{N}\) with \(n>1\), by assuming either that it holds for \(n-1\) or that it holds for all positive integers smaller than \(n .\) The technique of induction also works when \(\mathbb{N}\) is replaced by any subset \(S\) of \(\mathbb{Z}\) such that \(S\) is bounded below; the only difference would be that the initial value 1 would have to be changed to the least element of \(S .\)

Given any \(p(x) \in \mathbb{R}[x]\) and \(\alpha \in \mathbb{R}\), show that there is a unique polynomial \(q(x) \in \mathbb{R}[x]\) such that \(p(x)=(x-\alpha) q(x)+p(\alpha) .\) Deduce that \(\alpha\) is a root of \(p(x)\) if and only if the polynomial \((x-\alpha)\) divides \(p(x)\).

Let \(S\) be a nonempty subset of \(\mathbb{R}\). If \(S\) is bounded above and \(M \in \mathbb{R}\), then show that \(M=\sup S\) if and only if \(M\) is an upper bound of \(S\) and for every \(\epsilon>0\), there exists \(x \in S\) such that \(M-\epsilon0\), there exists \(x \in S\) such that \(m \leq x

Let \(D\) be a bounded subset of \(\mathbb{R}\) and let \(f: D \rightarrow \mathbb{R}\) be a polynomial function. Prove that \(f\) is bounded on \(D\).

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