Exercises for 1
solutions
2
If polynomials \(f(x)\) and \(g(x)\) satisfy \(f(a) = g(a)\), show that \(f(x) - g(x) = (x - a)h(x)\) for some polynomial \(h(x)\).
Exercises [ex:6_5_2], [ex:6_5_3], [ex:6_5_4], and [ex:6_5_5] require polynomial differentiation.
[ex:6_5_2] Expand each of the following as a polynomial in powers of \(x - 1\).
- \(f(x) = x^{3} - 2x^{2} + x - 1\)
- \(f(x) = x^{3} + x + 1\)
- \(f(x) = x^{4}\)
- \(f(x) = x^{3} - 3x^{2} + 3x\)
- \(3 + 4(x - 1) + 3(x - 1)^{2} + (x - 1)^{3}\)
- \(1 + (x - 1)^{3}\)
[ex:6_5_3] Prove Taylor’s theorem for polynomials.
[ex:6_5_4] Use Taylor’s theorem to derive the binomial theorem:
\[(1 + x)^n = \binom{n}{0} + \binom{n}{1} x + \binom{n}{2} x^2 + \dots + \binom{n}{n} x^n \nonumber \]
Here the binomial coefficients \(\binom{n}{r}\) are defined by
\[\binom{n}{r} = \frac{n!}{r!(n - r)!} \nonumber \]
where \(n! = n(n - 1) \cdots 2 \cdot 1\) if \(n \geq 1\) and \(0! = 1\).
[ex:6_5_5] Let \(f(x)\) be a polynomial of degree \(n\). Show that, given any polynomial \(g(x)\) in \(\mathbf{P}_{n}\), there exist numbers \(b_{0}, b_{1}, \dots, b_{n}\) such that
\[g(x) = b_0f(x) + b_1f^{(1)}(x) + \dots + b_nf^{(n)}(x) \nonumber \]
where \(f^{(k)}(x)\) denotes the \(k\)th derivative of \(f(x)\).
Use Theorem [thm:020059] to show that the following are bases of \(\mathbf{P}_{2}\).
- \(\{x^{2} - 2x, x^{2} + 2x, x^{2} - 4\}\)
- \(\{x^{2} - 3x + 2, x^{2} - 4x + 3, x^{2} - 5x + 6\}\)
- The polynomials are \((x - 1)(x - 2)\), \((x - 1)(x - 3)\), \((x - 2)(x - 3)\). Use \(a_{0} = 3\), \(a_{1} = 2\), and \(a_{2} = 1\).
Find the Lagrange interpolation expansion of \(f(x)\) relative to \(a_{0} = 1\), \(a_{1} = 2\), and \(a_{2} = 3\) if:
\(f(x) = x^{2} + 1\) \(f(x) = x^{2} + x + 1\)
- \(f(x) = \frac{3}{2}(x - 2)(x - 3) - 7(x - 1)(x - 3) + \frac{13}{2}(x - 1)(x - 2)\).
Let \(a_{0}, a_{1}, \dots, a_{n}\) be distinct numbers. If \(f(x)\) and \(g(x)\) in \(\mathbf{P}_{n}\) satisfy \(f(a_{i}) = g(a_{i})\) for all \(i\), show that \(f(x) = g(x)\). [Hint: See Theorem [thm:020203].]
Let \(a_{0}, a_{1}, \dots, a_{n}\) be distinct numbers. If \(f(x) \in\|{P}_{n+1}\) satisfies \(f(a_{i}) = 0\) for each \(i = 0, 1, \dots, n\), show that \(f(x) = r(x - a_{0})(x - a_{1}) \cdots (x - a_{n})\) for some \(r\) in \(\mathbb{R}\). [Hint: \(r\) is the coefficient of \(x^{n+1}\) in \(f(x)\). Consider \(f(x) - r(x - a_{0}) \cdots (x - a_{n})\) and use Theorem [thm:020203].]
[ex:6_5_10] Let \(a\) and \(b\) denote distinct numbers.
- Show that \(\{(x - a), (x - b)\}\) is a basis of \(\mathbf{P}_{1}\).
- Show that \(\{(x - a)^{2}, (x - a)(x - b), (x - b)^{2}\}\) is a basis of \(\mathbf{P}_{2}\).
- Show that \(\{(x - a)^{n}, (x - a)^{n-1}(x - b), \\ \dots, (x - a)(x - b)^{n-1}, (x - b)^{n}\}\) is a basis of \(\mathbf{P}_{n}\). [Hint: If a linear combination vanishes, evaluate at \(x = a\) and \(x = b\). Then reduce to the case \(n - 2\) by using the fact that if \(p(x)q(x) = 0\) in \(\mathbf{P}\), then either \(p(x) = 0\) or \(q(x) = 0\).]
- If \(r(x - a)^{2} + s(x - a)(x - b) + t(x - b)^{2} = 0\), then evaluation at \(x = a (x = b)\) gives \(t = 0 (r = 0)\). Thus \(s(x - a)(x - b) = 0\), so \(s = 0\). Use Theorem [thm:019633].
Let \(a\) and \(b\) be two distinct numbers. Assume that \(n \geq 2\) and let
\[U_n = \{f(x) \mbox{ in }\|{P}_n \mid f(a) = 0 = f(b) \}. \nonumber \]
- Show that
\[U_n = \{(x - a)(x - b)p(x) \mid p(x) \mbox{ in }\|{P}_{n - 2} \} \nonumber \]
- [Hint: If \(p(x)q(x) = 0\) in \(\mathbf{P}\), then either \(p(x) = 0\), or \(q(x) = 0\).]
- Show \(\{(x - a)^{n-1}(x - b), (x - a)^{n-2}(x - b)^{2}, \\ \dots, (x - a)^{2}(x - b)^{n-2}, (x - a)(x - b)^{n-1}\}\) is a basis of \(U_{n}\). [Hint: Exercise [ex:6_5_10].]
- Suppose \(\{p_{0}(x), p_{1}(x), \dots, p_{n-2}(x)\}\) is a basis of \(\mathbf{P}_{n-2}\). We show that \(\{(x - a)(x - b)p_{0}(x), (x - a)(x - b)p_{1}(x), \dots, (x - a)(x - b)p_{n-2}(x)\}\) is a basis of \(U_{n}\). It is a spanning set by part (a), so assume that a linear combination vanishes with coefficients \(r_{0}, r_{1}, \dots, r_{n-2}\). Then \((x - a)(x - b)[r_{0}p_{0}(x) + \dots + r_{n-2}p_{n-2}(x)] = 0\), so \(r_{0}p_{0}(x) + \dots + r_{n-2}p_{n-2}(x) = 0\) by the Hint. This implies that \(r_{0} = \dots = r_{n-2} = 0\).
