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Exercise 2.6.10

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1.

Since we can check that

f_{i} (p (x) + cq (x)) = p (c_{i}) + cq (c_{i}) = f_{i} (p (x)) + c f_{i} (q (x))

f_{i}

is linear and hence in

V^{∗}

. And we know that

\dim (V^{∗}) = \dim (V) = \dim (P_{n} (𝔽)) = n + 1

. So now it’s enough to show that

{f_{0}, f_{1}, \dots, f_{n}}

is independent. So assume that

\sum_{i = 1}^{n} a_{i} f_{i} = 0

for some

a_{i}

. We may define polynomials

p_{i} (x) = \prod_{j \neq i} (x - c_{j})

such that we know

p_{i} (c_{i}) \neq 0

but

p_{i} (c_{j}) = 0

for all

j \neq i

. So now we have that

\sum_{i = 1}^{n} a_{i} f_{i} (p_{1}) = a_{1} f_{1} (p_{1}) = 0

implies $a_{1} = 0$ . Similarly we have $a_{i} = 0$ for all proper $i$ .

2.

By the Corollary after Theorem 2.26 we have an ordered basis

β = {p_{0}, p_{1}, \dots, p_{n}}

for $V$ such that ${f_{1}, f_{2}, \dots, f_{n}}$ defined in the previous exercise is its dual basis. So we know that $p_{i} (c_{j}) = δ_{ij}$ . Since $β$ is a basis, every polynomial in $V$ is linear combination of $β$ . If a polynomial $q$ has the property that $q (c_{j}) = δ_{0 j}$ , we can assume that $q = \sum_{i = 0}^{n} a_{i} p_{i}$ . Then we have

1 = q (c_{0}) = \sum_{i = 0}^{n} a_{i} p_{i} (c_{0}) = a_{1}

and

0 = q (c_{j}) = \sum_{i = 0}^{n} a_{i} p_{i} (c_{j}) = a_{j}

for all $j$ other than $1$ . So actually we know $q = p_{0}$ . This means $p_{0}$ is unique. And similarly we know all $p_{i}$ is unique. Since the Lagrange polynomials,say ${r_{i}}_{i = 1, 2, \dots n}$ , defined in Section 1.6 satisfy the property $r_{i} (c_{j}) = δ_{ij}$ , by uniqueness we have $r_{i} = p_{i}$ for all $i$ .

3.

Let

β = {p_{0}, p_{1}, \dots, p_{n}}

be those polynomials defined above. We may check that

q (x) = \sum_{i = 0}^{n} a_{i} p_{i} (x)

has the property $q (c_{i}) = a_{i}$ for all $i$ , since we know that $p_{i} (c_{j}) = δ_{ij}$ . Next if $r (x) \in V$ also has the property, we may assume that

r (x) = \sum_{i = 0}^{n} b_{i} p_{i} (x)

since $β$ is a basis for $V$ . Similarly we have that

a_{i} = r (c_{i}) = \sum_{i = 0}^{n} b_{i} p_{i} (c_{i}) = b_{i} .

So we know $r = q$ and $q$ is unique.

4.

This is the instant result of 2.6.10(a) and 2.6.10(b) by setting

a_{i} = p (c_{i})

5.

Since there are only finite term in that summation, we have that the order of integration and summation can be changed. So we know

\int_{a}^{b} p (t) dt = \int_{a}^{b} (\sum_{i = 0}^{n} p (c_{i}) p_{i} (t)) dt

= \sum_{i = 0}^{n} \int_{a}^{b} p (c_{i}) p_{i} (t) dt .

jephianlin

2011-06-27 00:00

Exercise 2.6.10

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