By denition x cos 45 and y sin 45 if we drop a

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Unformatted text preview: ich is clearly true. Next we assume P (k ) is true, that is log xk = k log(x) and try to show P (k + 1) is true. Using the Product Rule for Logarithms along with the induction hypothesis, we get log xk+1 = log xk · x = log xk + log(x) = k log(x) + log(x) = (k + 1) log(x) Hence, log xk+1 = (k + 1) log(x). By induction log (xn ) = n log(x) is true for all x > 0 and all natural numbers n ≥ 1. 3. Let A be an n × n lower triangular matrix. We proceed to prove the det(A) is the product of the entries along the main diagonal by inducting on n. For n = 1, A = [a] and det(A) = a, so the result is (trivially) true. Next suppose the result is true for k × k lower triangular matrices. Let A be a (k + 1) × (k + 1) lower triangular matrix. Expanding det(A) along the first row, we have n det(A) = a1p C1p p=1 Since a1p = 0 for 2 ≤ p ≤ k + 1, this simplifies det(A) = a11 C11 . By definition, we know that C11 = (−1)1+1 det (A11 ) = det (A11 ) where A11 is k × k matrix obtained by deleting the first row and first column of A. Since A is...
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