Book edition 5th

Author(s) Otto Bretscher

Pages 442 pages

ISBN 9780321796974

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Short Answer

Find the set of all polynomial $f (t)$ in $P_{3}$ such that $f (1) = 0$ and $\int_{- 1}^{1} f (t) d t = 0$ , and determine its dimension.

The dimension of $P_{3}$ such that $f (1) = 0$ and $\int_{- 1}^{1} f (t) d t = 0$ is 3 which is spanned by $s p a n \{1, (t^{2} - \frac{2}{3} t), (t^{3} - t)\}$ .

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: Determine the span set.

Consider the set of all polynomial such that and .

The set ${1, t, t_{2}, \dots, t}$ is linear independent set of V if there exist constant $a_{i} \in R$ such that $a_{0} + a_{1} t_{1} + a_{2} t_{2} + . . . + a_{n} t_{n} = 0$ where $a_{0} = a_{1} = a_{2} = . . . = a_{n} = 0$ .

Any polynomial $f (t) \in P_{3}$ is defined as follows.

$f (t) = b_{0} + b_{1} t + b_{2} t^{2} + b_{3} t^{3}$

Substitute the value 1 for t in the equation $f (t) = b_{0} + b_{1} t + b_{2} t^{2} + b_{3} t^{3}$ as follows.

$f (t) = b_{0} + b_{1} t + b_{2} t^{2} + b_{3} t^{3} f (1) = b_{0} + b_{1} + b_{2} + b_{3}$

As $f (1) = 0$ , substitute the value 0 for $f (1)$ in the equation $f (1) = b_{0} + b_{1} + b_{2} + b_{3}$ as follows.

$\begin{array}{l} f (1) = b_{0} + b_{1} + b_{2} + b_{3} \\ b_{1} = - (b_{0} + b_{2} + b_{3}) \end{array}$

Take integration both side in the equation $f (t) = b_{0} + b_{1} t + b_{2} t^{2} + b_{3} t^{3}$ as follows.

role="math" localid="1659407877530" $f (t) = b_{0} + b_{i} t + b_{2} t^{2} + b_{3} t^{3} \int_{- 1}^{1} f (t) d t = \int_{- 1}^{1} b_{0} + b_{1} t + b_{2} t^{2} + b_{2} t^{3} d t = {\{b_{0} t + b_{1} \frac{t^{2}}{2} + b_{2} \frac{t^{3}}{3} + b_{3} \frac{t^{4}}{4}\}}_{- 1}^{1} \int_{- 1}^{1} f (t) d t = b_{0} + b_{1} \frac{1}{2} + b_{2} \frac{1}{3} + b_{3} \frac{1}{4} + b_{0} - b_{1} \frac{1}{2} + b_{2} \frac{1}{3} - b_{3} \frac{1}{4}$

Further, simplify the equation as follows.

$\int_{- 1}^{1} f (t) d t = b_{0} + b_{2} \frac{1}{3} + b_{0} + b_{2} \frac{1}{3} \int_{- 1}^{1} f (t) d t = 2 b_{0} + b_{2} \frac{2}{3}$

As $\int_{- 1}^{1} f (t) d t = 0$ , substitute the value 0 for $\int_{- 1}^{1} f (t) t$ in the equation $\int_{1}^{1} f (t) t = 2 b_{b} + b_{2} \frac{2}{3}$ as follows.

$\int_{- 1}^{1} f (t) = 2 b_{0} + b_{2} \frac{2}{3} 0 = 2 b_{0} + b_{2} \frac{2}{3} b_{0} = - b_{2} \frac{1}{3}$

Substitute the value $- b_{2} \frac{1}{3}$ for $b_{0}$ in the equation $b_{1} = - (b_{0} + b_{2} + b_{3})$ as follows.

$b_{- 1} = - (b_{0} + b_{2} + b_{3}) = - (- b_{2} \frac{2}{3} + b_{2} + b_{3}) = - (b_{2} \frac{2}{3} + b_{3}) b_{1} = - b_{2} \frac{2}{3} - b_{3}$

Finally, substitute the value $- b_{2} \frac{2}{3} - b_{3}$ for $b_{1}$ in the equation $f (t) = b_{0} + b_{1} t + b_{2} t^{2} + b_{3} t^{3}$ as follows.

$f (t) = b_{0} + b_{1} t + b_{2} t^{2} + b_{3} t^{3} f (t) = b_{0} + (- b_{2} \frac{2}{3} - b_{3}) t + b_{2} t^{2} + b_{3} t^{3} f (t) = b_{0} + b_{2} (t^{2} - \frac{2}{3}) + b_{3} (t^{3} - t)$

From the equation, the set $\{1 (t^{2} - \frac{2}{3} t), (t^{3} - t)\}$ span $P_{3}$ if $f (1) = 0$ and $\int_{1}^{1} f (t) d t = 0$ .

Step 2: Determine the property of independency.

Compare the equations $b_{0} + b_{2} (t^{2} - \frac{2}{3} t) + b_{3} (t^{3} - t) = 0$ both side as follows.

$\begin{array}{l} b + b_{2} (t^{2} - \frac{2}{3} t)) + b_{3} (t^{3} - t) = 0 \\ b_{0} + b_{2} (t^{2} - \frac{2}{3} t) + b_{3} (t^{3} - t) = (0) + (0) (t^{2} - \frac{2}{3} t) + (0) t^{3} - t) \end{array}$

By the definition of linear independence, the subset $\{1, (t^{2} - \frac{2}{3} t), (t^{3} - t)\}$ is L.I.

Therefore, the set of all polynomial $P_{3}$ such that $f (1) = 0$ and $\int_{1}^{1} f (t) d t = 0$ has dimension 3 and spanned by $s p an \{1, (t^{2} - \frac{2}{3} t), (t^{3} - t)\}$ .

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Linear Algebra With Applications

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Short Answer

Find the set of all polynomial $f (t)$ in $P_{3}$ such that $f (1) = 0$ and $\int_{- 1}^{1} f (t) d t = 0$ , and determine its dimension.

Step by Step Solution

Step 1: Determine the span set.

Step 2: Determine the property of independency.

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