HW1solutions_4

# HW1solutions_4 - MATH 115 SOLUTION SET 1 1.2 1 4 See back...

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MATH 115 SOLUTION SET 1 § 1.2: 1, 4. See back of book. 7. To show that f = g , it is enough to show that f ( x ) = g ( x ) for all x S . Since f (0) = g (0) = 1 and f (1) = g (1) = 3. we have f = g . Similarly for f + g = h . 8. By the ﬁrst distributive law, ( a + b )( x + y ) = ( a + b ) x + ( a + b ) y . By the second law, this expression is ax + bx + ay + by as required. 9. Corollary 1 says that the vector 0 is unique for having the property “There exists an element in V such that x + 0 = x for each x V .” This means that if 0 1 and 0 2 are two vectors which have this property, then we must have 0 1 = 0 2 . Let’s prove it this way. Since 0 1 has this property, we must have x +0 1 = x for all x V . In particular, 0 2 + 0 1 = 0 2 (note that 0 2 is playing the role of x ). Similarly, since 0 2 has the property, we have 0 1 + 0 2 = 0 1 . Putting these equations together, we have 0 1 = 0 1 + 0 2 = 0 2 + 0 1 = 0 2 as required. (Notice that we used commutativity of addition here.) Corollary 2 says that the vector y is unique for having the property that x + y = 0. Suppose y 1 and y 2 satisfy the same property: x + y 1 = 0 and x + y 2 = 0. Take the ﬁrst equation and add y 2 to both sides: y 2 + ( x + y 1 ) = y 2 + 0. Rearranging, we get ( x + y 2 ) + y 1 = y 2 (notice that we used the associative, commutative, and identity properties of addition). But x + y 2 = 0, so this becomes 0 + y 1 = y 2 . Use the identity property of addition once more to get y 1 = y 2 . 10. First we have to show that addition and scalar multiplication make sense in V (i.e., that these operations are closed). If f and g are two diﬀerentiable functions, and if c is a scalar, then standard theorems in calculus show that both f + g and cf are also diﬀerentiable. (It is okay to cite such theorems rather then prove them over again.) Now we have to check the axioms. For commutativity of addition, for instance, suppose f,g V . Is f + g = g + f ? To show this, it is enough to show that ( f + g )( x ) = ( g + f )( x ) for all x , but this is the same as f ( x ) + g ( x ) = g ( x ) + f ( x ), which is obvious because addition in R is commutative. Associativity of addition in V also follows directly from associativity of addition in R . There is an identity element, because the constant function 0 is diﬀerentiable. There is a notion of

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HW1solutions_4 - MATH 115 SOLUTION SET 1 1.2 1 4 See back...

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