By Hugo D. Junghenn
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Additional resources for A course in real analysis
But, by definition of 0 and commutativity of addition, 0 + 0 = 0 . Therefore 0 = 0. (b) By associativity and commutativity of addition, b = b + 0 = 0 + b = (−a + a) + b = −a + (a + b) = −a + 0 = −a. (c) If a · 1 = a for all a then, in particular, 1 · 1 = 1. But, by definition of the multiplicative identity and commutativity of multiplication, 1 · 1 = 1 . Therefore 1 = 1. (d) By the distributive property, a · 0 = a(0 + 0) = a · 0 + a · 0. Adding −(a · 0) to both sides of this equation and using associativity of addition produces the desired equation.
Xn ) and y = (y1 , . . , yn ). The Euclidean inner product x · y of x and y and the Euclidean norm x 2 of x are defined by n n 1/2 √ x·y = xj yj and x 2= x2j = x · x. j=1 j=1 The set R with its vector space structure and the Euclidean inner product is called n-dimensional Euclidean space. ♦ n The structure of Euclidean space allows one to define lines, planes, length, perpendicularity, angle between vectors, etc. These ideas will be useful in later chapters. 3 Theorem. The inner product in Rn has the following properties: (a) x · x = x 22 .
3(n + 1)nn > (n + 1)n+1 . 4 Example. We derive a closed formula for f (n) := − 1)2 and then verify the result by induction. A little experimentation suggests that we should try a polynomial in n of degree 3, say g(n) := An3 + Bn2 + Cn + D. Then g(n + 1) − g(n) = A (n + 1)3 − n3 + B (n + 1)3 − n2 + C (n + 1) − n = 3An2 + (3A + 2B)n + A + B + C and n+1 f (n+1)−f (n) = n (3k −1)2 − k=1 2 (3k −1)2 = 3(n+1)−1 = 9n2 +12n+4. k=1 Assuming that f (n) = g(n) for all n, we may equate coefficients to obtain A = 3, B = 3/2, and C = −1/2.
A course in real analysis by Hugo D. Junghenn