vb.net generate qr code Relations in Two-Space in .NET

Generator Code 39 in .NET Relations in Two-Space

and ( f3 + f2)(x) = f3 (x) + f2 (x) = x1/2 + x2 = x2 + x1/2 It s customary to write polynomials (sums or differences of a variable raised to powers) with the largest power first, and then descending powers after that That s why some of the sums and differences have been rearranged in the above examples

What about f4 The independent variable in f4 doesn t match the independent variable in any of the other three functions, so we can t combine and manipulate the equations as if the variables did match We can add the equations straightaway, but that doesn t tell us much For example, we can say that

f2 (x) + f4 (q) = x2 + 2q /p but that s all we can do with it It s like trying to add minutes to millimeters We can t get a resultant function of a single variable The same problem occurs if we try to subtract, multiply, or find a ratio involving f4 and any of the other three functions

Difference between two functions When we want to find the difference between two functions, we subtract both sides of their equations This can be done in either order, usually producing different results For f1 and f2, we have

( f1 f2)(x) = f1 (x) f2 (x) = (x 1) x2 = x2 + x 1 and ( f2 f1)(x) = f2 (x) f1 (x) = x2 (x 1) = x2 x + 1 For f1 and f3, we have ( f1 f3)(x) = f1 (x) f3 (x) = (x 1) x1/2 = x x1/2 1 and ( f3 f1)(x) = f3 (x) f1 (x) = x1/2 (x 1) = x + x1/2 + 1 For f2 and f3, we have ( f2 f3)(x) = f2 (x) f3 (x) = x2 x1/2 and ( f3 f2)(x) = f3 (x) f2 (x) = x1/2 x2 = x2 + x1/2

Product of two functions When we want to find the product of two functions, we multiply both sides of their equations This can be done in either order, producing identical results For f1 and f2, we have

( f1 f2)(x) = f1 (x) f2 (x) = (x 1) x2 = x3 x2 and ( f2 f1)(x) = f2 (x) f1 (x) = x2(x 1) = x3 x2 For f1 and f3, we have ( f1 f3)(x) = f1 (x) f3 (x) = (x 1)x1/2 = x3/2 x1/2 and ( f3 f1)(x) = f3 (x) f1 (x) = x1/2(x 1) = x3/2 x1/2 For f2 and f3, we have ( f2 f3)(x) = f2 (x) f3 (x) = x2 x1/2 = x5/2 and ( f3 f2)(x) = f3 (x) f2 (x) = x1/2 x2 = x5/2

Ratio of two functions When we want to find the ratio of two functions, we divide both sides of their equations This can be done in either order, usually producing different results For f1 and f2, we have

( f1 / f2)(x) = f1 (x) / f2 (x) = (x 1)/x2 = x 1 x 2 and ( f2 / f1)(x) = f2 (x) / f1 (x) = x2/(x 1) = x2 (x 1) 1 For f1 and f3, we have ( f1 / f3)(x) = f1(x) / f3 (x) = (x 1) / x1/2 = x1/2 x 1/2 and ( f3 / f1)(x) = f3 (x) / f1 (x) = x1/2/(x 1) = x1/2(x 1) 1 For f2 and f3, we have ( f2 / f3)(x) = f2(x) / f3 (x) = x2/x1/2 = x3/2 and (f3 / f2)(x) = f3(x) / f2(x) = x1/2/x2 = x 3/2