section 4 - HonorsPreCalTrig
... Section 4.4 Logarithms are used to solve for variable exponents. For example, finding “x” in the equation 4x = 41. Definition 4.4- If r is any positive number, then the unique exponent t such that b t = r is called the logarithm of r with base b and is denoted by logbr. log b r = t So.... log 2 8 = ...
... Section 4.4 Logarithms are used to solve for variable exponents. For example, finding “x” in the equation 4x = 41. Definition 4.4- If r is any positive number, then the unique exponent t such that b t = r is called the logarithm of r with base b and is denoted by logbr. log b r = t So.... log 2 8 = ...
Recitation 2
... smaller sizes (linear or tree recursion) and solve them recursively • Solve the very small sized problems directly • Usually some computations are required in order to split problem into sub-problems or /and to combine the results ...
... smaller sizes (linear or tree recursion) and solve them recursively • Solve the very small sized problems directly • Usually some computations are required in order to split problem into sub-problems or /and to combine the results ...
CS 170 Spring 2008 - Solutions to Midterm #1
... shortest paths has weight less than C. Thus, C 0 is a negative cycle. This shows that if there is a negative cycle C in the original graph, then there is a negative cycle consisting of positive edge shortest paths between endpoints of negative edges, i.e. a negative cycle in the new graph. Thus, our ...
... shortest paths has weight less than C. Thus, C 0 is a negative cycle. This shows that if there is a negative cycle C in the original graph, then there is a negative cycle consisting of positive edge shortest paths between endpoints of negative edges, i.e. a negative cycle in the new graph. Thus, our ...
Representation Theory.
... Since ρ is a homomorphism Sn → GLn (k), and det is a homomorphism GLn (k) → k ∗ , we see that sgn is a homomorphism Sn → k ∗ . One can quickly check that sgn is equal to (−1)a , where a is a number of transpositions needed to write σ. This of course gives us back the usual definition of the sign, bu ...
... Since ρ is a homomorphism Sn → GLn (k), and det is a homomorphism GLn (k) → k ∗ , we see that sgn is a homomorphism Sn → k ∗ . One can quickly check that sgn is equal to (−1)a , where a is a number of transpositions needed to write σ. This of course gives us back the usual definition of the sign, bu ...
