Answer on Question #79909 – Math – Linear Algebra
Question
Let P 3 P_3 P 3 R \mathbb{R} R ∫ 0 1 f ( x ) g ( x ) d x \int_0^1 f(x)g(x)dx ∫ 0 1 f ( x ) g ( x ) d x P 3 P_3 P 3 { 1 − 2 x , 2 x + 6 x 2 , − 3 x 2 + 4 x 3 } \{1 - 2x, 2x + 6x^2, -3x^2 + 4x^3\} { 1 − 2 x , 2 x + 6 x 2 , − 3 x 2 + 4 x 3 }
Solution
1. Obviously these vectors are linearly independent.
2.
a. u 1 = v 1 = 1 − 2 x u_{1} = v_{1} = \mathbf{1} - 2x u 1 = v 1 = 1 − 2 x
⟨ u 1 , u 1 ⟩ = ∫ 0 1 ( 1 − 4 x + 4 x 2 ) d x = 4 / 3 − 4 / 2 − 1 = 1 / 3 \langle u _ {1}, u _ {1} \rangle = \int_ {0} ^ {1} (1 - 4 x + 4 x ^ {2}) d x = 4 / 3 - 4 / 2 - 1 = 1 / 3 ⟨ u 1 , u 1 ⟩ = ∫ 0 1 ( 1 − 4 x + 4 x 2 ) d x = 4/3 − 4/2 − 1 = 1/3
b.
⟨ u 1 , v 2 ⟩ = ∫ 0 1 ( 1 − 2 x ) ( 2 x + 6 x 2 ) d x = ∫ 0 1 ( − 12 x 3 + 2 x 2 + 2 x ) d x = − 12 / 4 + 2 / 3 + 1 = − 4 / 3 \begin{array}{l}
\langle u _ {1}, v _ {2} \rangle = \int_ {0} ^ {1} (1 - 2 x) (2 x + 6 x ^ {2}) d x = \int_ {0} ^ {1} (- 1 2 x ^ {3} + 2 x ^ {2} + 2 x) d x = - 1 2 / 4 + 2 / 3 + 1 = \\
- 4 / 3 \\
\end{array} ⟨ u 1 , v 2 ⟩ = ∫ 0 1 ( 1 − 2 x ) ( 2 x + 6 x 2 ) d x = ∫ 0 1 ( − 12 x 3 + 2 x 2 + 2 x ) d x = − 12/4 + 2/3 + 1 = − 4/3 ⟨ u 2 , u 2 ⟩ = ∫ 0 1 ( 36 x 4 − 72 x 3 + 84 x 2 − 48 x + 16 ) d x = 36 / 5 − 18 + 28 − 24 + 16 = 46 / 5 \langle u _ {2}, u _ {2} \rangle = \int_ {0} ^ {1} (3 6 x ^ {4} - 7 2 x ^ {3} + 8 4 x ^ {2} - 4 8 x + 1 6) d x = 3 6 / 5 - 1 8 + 2 8 - 2 4 + 1 6 = 4 6 / 5 ⟨ u 2 , u 2 ⟩ = ∫ 0 1 ( 36 x 4 − 72 x 3 + 84 x 2 − 48 x + 16 ) d x = 36/5 − 18 + 28 − 24 + 16 = 46/5 u 2 = v 2 − ⟨ u 1 , v 2 ⟩ ⟨ u 1 , u 1 ⟩ u 1 = 2 x + 6 x 2 + 4 − 8 x = 4 − 6 x + 6 x 2 u _ {2} = v _ {2} - \frac {\left\langle u _ {1} , v _ {2} \right\rangle}{\left\langle u _ {1} , u _ {1} \right\rangle} u _ {1} = 2 x + 6 x ^ {2} + 4 - 8 x = 4 - 6 x + 6 x ^ {2} u 2 = v 2 − ⟨ u 1 , u 1 ⟩ ⟨ u 1 , v 2 ⟩ u 1 = 2 x + 6 x 2 + 4 − 8 x = 4 − 6 x + 6 x 2
c.
⟨ u 1 , v 3 ⟩ = ∫ 0 1 ( 1 − 2 x ) ( − 3 x 2 + 4 x 3 ) d x = ∫ 0 1 ( − 8 x 4 + 10 x 3 − 3 x 2 ) d x = − 8 / 5 + 10 / 4 − 1 = − 1 / 10 \begin{array}{l}
\langle u _ {1}, v _ {3} \rangle = \int_ {0} ^ {1} (1 - 2 x) (- 3 x ^ {2} + 4 x ^ {3}) d x = \int_ {0} ^ {1} (- 8 x ^ {4} + 1 0 x ^ {3} - 3 x ^ {2}) d x = - 8 / 5 + 1 0 / 4 - \\
1 = - 1 / 1 0 \\
\end{array} ⟨ u 1 , v 3 ⟩ = ∫ 0 1 ( 1 − 2 x ) ( − 3 x 2 + 4 x 3 ) d x = ∫ 0 1 ( − 8 x 4 + 10 x 3 − 3 x 2 ) d x = − 8/5 + 10/4 − 1 = − 1/10 ⟨ u 2 , v 3 ⟩ = ∫ 0 1 ( 4 − 6 x + 6 x 2 ) ( − 3 x 2 + 4 x 3 ) d x = ∫ 0 1 ( 24 x 5 − 42 x 4 + 34 x 3 − 12 x 2 ) d x = 4 − 42 / 5 + 34 / 4 − 4 = − 2 / 5 + 2 / 4 = 1 / 10 \begin{array}{l}
\langle u _ {2}, v _ {3} \rangle = \int_ {0} ^ {1} (4 - 6 x + 6 x ^ {2}) (- 3 x ^ {2} + 4 x ^ {3}) d x = \int_ {0} ^ {1} (2 4 x ^ {5} - 4 2 x ^ {4} + 3 4 x ^ {3} - 1 2 x ^ {2}) d x = 4 - \\
4 2 / 5 + 3 4 / 4 - 4 = - 2 / 5 + 2 / 4 = 1 / 1 0 \\
\end{array} ⟨ u 2 , v 3 ⟩ = ∫ 0 1 ( 4 − 6 x + 6 x 2 ) ( − 3 x 2 + 4 x 3 ) d x = ∫ 0 1 ( 24 x 5 − 42 x 4 + 34 x 3 − 12 x 2 ) d x = 4 − 42/5 + 34/4 − 4 = − 2/5 + 2/4 = 1/10 ⟨ u 3 , u 3 ⟩ = ∫ 0 1 ( 16 x 6 − 564 / 23 x 5 + 54141 / 10580 x 4 + 28199 / 5290 x 3 − 68061 / 52900 x 2 − 7257 / 26450 x + 3481 / 52900 ) d x = 16 / 7 − 94 / 23 + 54141 / 52900 + 140995 / ( 2 ∗ 52900 ) − 22687 / 52900 − 7257 / 52900 + 3481 / 52900 = − 290 / 161 + 196351 / 2 ∗ 52900 = 1759 / 32200 \begin{array}{l}
\langle u _ {3}, u _ {3} \rangle = \int_ {0} ^ {1} (1 6 x ^ {6} - 5 6 4 / 2 3 x ^ {5} + 5 4 1 4 1 / 1 0 5 8 0 x ^ {4} + 2 8 1 9 9 / 5 2 9 0 x ^ {3} - \\
6 8 0 6 1 / 5 2 9 0 0 x ^ {2} - 7 2 5 7 / 2 6 4 5 0 x + 3 4 8 1 / 5 2 9 0 0) d x = 1 6 / 7 - 9 4 / 2 3 + 5 4 1 4 1 / 5 2 9 0 0 + \\
1 4 0 9 9 5 / (2 * 5 2 9 0 0) - 2 2 6 8 7 / 5 2 9 0 0 - 7 2 5 7 / 5 2 9 0 0 + 3 4 8 1 / 5 2 9 0 0 = - 2 9 0 / 1 6 1 + \\
1 9 6 3 5 1 / 2 * 5 2 9 0 0 = \mathbf {1 7 5 9} / 3 2 2 0 0 \\
\end{array} ⟨ u 3 , u 3 ⟩ = ∫ 0 1 ( 16 x 6 − 564/23 x 5 + 54141/10580 x 4 + 28199/5290 x 3 − 68061/52900 x 2 − 7257/26450 x + 3481/52900 ) d x = 16/7 − 94/23 + 54141/52900 + 140995/ ( 2 ∗ 52900 ) − 22687/52900 − 7257/52900 + 3481/52900 = − 290/161 + 196351/2 ∗ 52900 = 1759 /32200 u 3 = v 3 − ⟨ u 1 , v 3 ⟩ ⟨ u 1 , u 1 ⟩ u 1 − ⟨ u 2 , v 3 ⟩ ⟨ u 2 , u 2 ⟩ u 2 = − 3 x 2 + 4 x 3 + 3 / 10 − 6 / 10 x − 4 / 92 + 6 / 92 x − u _ {3} = v _ {3} - \frac {\left\langle u _ {1} , v _ {3} \right\rangle}{\left\langle u _ {1} , u _ {1} \right\rangle} u _ {1} - \frac {\left\langle u _ {2} , v _ {3} \right\rangle}{\left\langle u _ {2} , u _ {2} \right\rangle} u _ {2} = - 3 x ^ {2} + 4 x ^ {3} + 3 / 1 0 - 6 / 1 0 x - 4 / 9 2 + 6 / 9 2 x - u 3 = v 3 − ⟨ u 1 , u 1 ⟩ ⟨ u 1 , v 3 ⟩ u 1 − ⟨ u 2 , u 2 ⟩ ⟨ u 2 , v 3 ⟩ u 2 = − 3 x 2 + 4 x 3 + 3/10 − 6/10 x − 4/92 + 6/92 x − 6 / 92 x 2 = 59 / 230 − 123 / 230 x − 141 / 46 x 2 + 4 x 3 6 / 9 2 x ^ {2} = \mathbf {5 9} / 2 3 0 - \mathbf {1 2 3} / 2 3 0 x - \mathbf {1 4 1} / 4 6 x ^ {2} + 4 x ^ {3} 6/92 x 2 = 59 /230 − 123 /230 x − 141 /46 x 2 + 4 x 3 Answer:
b 1 = 3 ( 1 − 2 x ) , b 2 = 5 / 46 ( 4 − 6 x + 6 x 2 ) , b _ {1} = \sqrt {3} (1 - 2 x), b _ {2} = \sqrt {5 / 4 6} (4 - 6 x + 6 x ^ {2}), b 1 = 3 ( 1 − 2 x ) , b 2 = 5/46 ( 4 − 6 x + 6 x 2 ) , b 3 = 80 5 / 1759 ( 59 / 230 − 123 / 230 x − 141 / 46 x 2 + 4 x 3 ) . b _ {3} = 8 0 \sqrt {5 / 1 7 5 9} (5 9 / 2 3 0 - 1 2 3 / 2 3 0 x - 1 4 1 / 4 6 x ^ {2} + 4 x ^ {3}). b 3 = 80 5/1759 ( 59/230 − 123/230 x − 141/46 x 2 + 4 x 3 ) .
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