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By Harold Jeffreys

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First factorise Q(x), x3 − 1 = (x − 1)(x2 + x + 1). We cannot factorise x2 + x + 1 into their factors with real coefficients. Write x+5 B + Cx A + = . x3 − 1 x − 1 x2 + x + 1 Multiply with x3 − 1, x + 5 = A(x2 + x + 1) + B + Cx(x − 1), substitute x = 1: 3A = 6, or A = 2. Equate coefficients of x2 : A + C = 0, C = −2. Equate coefficients of the constant part: A − B = 5, B = −3. x+5 2 3 + 2x = − 2 . 3 x −1 x−1 x +x+1 A rational function is a function of the form f (x) = P (x)/Q(x) where P and Q are both polynomials.

X2 (x − 1)(x − 2) 3 =− 1 1 1 + − . 7. 16: Simplify Solution: x+5 x3 −1 using partial fractions. First factorise Q(x), x3 − 1 = (x − 1)(x2 + x + 1). We cannot factorise x2 + x + 1 into their factors with real coefficients. Write x+5 B + Cx A + = . x3 − 1 x − 1 x2 + x + 1 Multiply with x3 − 1, x + 5 = A(x2 + x + 1) + B + Cx(x − 1), substitute x = 1: 3A = 6, or A = 2. Equate coefficients of x2 : A + C = 0, C = −2. Equate coefficients of the constant part: A − B = 5, B = −3. x+5 2 3 + 2x = − 2 . 3 x −1 x−1 x +x+1 A rational function is a function of the form f (x) = P (x)/Q(x) where P and Q are both polynomials.

2. (a · b)c is the product of the scalar a · b with the vector c. Thus the result has the same direction as c, with magnitude (a · b)c. 3. We can divide by a · b since it is a scalar! ) 4. a · (b + c) = a · b + a · c. (Distributive law). This will not be proven here, but can easily be done using component form. 1 = (a + b) · (a + b) = (a + b) · a + (a + b) · b = a·a+b·a+a·b+b·b = a2 + b2 + 2a · b Component form of dot product Let a = a1 i + a2 j + a3 k, b = b1 i + b2 j + b3 k, then a·b a·b = (a1 i + a2 j + a3 k) · (b1 i + b2 j + b3 k) = a1 b1 + a2 b2 + a3 b3 .

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Asymptotic approximations by Harold Jeffreys


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