An Invitation to the Algebra of Canonical Commutation by Denes Petz

By Denes Petz

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However, since and f(O) = 1 f(-1) = -1 there must be a value of x between 0 and -1 for which f(x) = 0. Try x = -!. = -! ) = x +! is a factor. ) or (2x+ 1) MEMORY JOGGER You can locate a solution to f(x) = 0 by finding two value of x (a and b) which give a different ign for f(a) and f(b). ;... (2x + 1) using the division process to give x 2 +x + 1 Hence the equation is (2x + 1)(x 2 + x + 1) = 0 Now x = -! or x 2 + x + 1 = 0 If you use the quadratic formula on this, -1 ± vT="4 = -----:::--- -1±H 2 Hence this part has no real solutions, as H i s not a real number.

It can be represented by a flow chart. That is, y = (x + 3) 2 + 2. If you reverse this flow chart, you get: Hence x = -3 ± J(y- 2) You would find it impossible to rearrange the original expression y 2 + 6x + 9 to make x the subject without this technique. The technique is also particularly useful when trying to find inverse functions. ;i = x 2 - = 5x+6! - 2 (x-~) -6! -3 = 2 (x-~) -9! ] = 4(x + 1) 32 2 +1 MASTERING ADVANCED PURE MATHEMATICS MEMORY JOGGER If trying to complete the square on ax 2 + bx + c, always take out a as a factor first.

4 Solve the equation 2x 3 + 3x 2 + 3x + 1 = 0. = 2x 3 + 3x 2 + 3x + 1 =1 =2+3+3+1=9 = 16 + 12 + 6 + 1 = 35 Let f(x) f(O) f(1) f(2) Clearly, as x increases, f(x) is getting bigger and not closer to zero. Try negative values of x. f(-1) = -2 + 3-3 + 1 = -1 f(-2) = -16+ 12- 6+ 1 = -9 f(-3) = -54+27- 9+ 1 = -35 Here you can see that f(x) is becoming more negative. However, since and f(O) = 1 f(-1) = -1 there must be a value of x between 0 and -1 for which f(x) = 0. Try x = -!. = -! ) = x +! is a factor.

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