Let AA be a real n×nn×n matrix. We say that AA is a difference of two squares if there exist real n×nn×n matrices BB and CC with BC=CB=0BC=CB=0 and A=B2−C2A=B2−C2. Now If A is a diagonal matrix, then I have to show that it is a difference of two squares.
Let aa be a real number. if a≥0a≥0 we can write a=b2−c2a=b2−c2 with c=0c=0 and if a<0a<0, we can write a=b2−c2a=b2−c2 with b=0b=0. Thus, in general, a=b2−c2a=b2−c2 with bb and cc real and bc=0bc=0. Now if A=diag(a1,a2,...,an)A=diag(a1,a2,...,an) is a real diagonal matrix, write each ai=b2i−c2iai=bi2−ci2 as above, with bici=0bici=0, and define B=diag(b1,b2,...,bn)B=diag(b1,b2,...,bn) and C=diag(c1,c2,...,cn)C=diag(c1,c2,...,cn). Then, by the way diagonal matrices multiply, we conclude that A=B2−C2A=B2−C2 and also that BC=CB=0BC=CB=0
What is the purpose of piston ring and oil rings?
Piston forms a movable boundary into an engine cylinder. If the piston is too tight, it does not move freely inside the cylinder when the pressure force applied during the power stroke and if it is too lose it would leak the pressure force. For proper functioning of an IC engine, the piston should form an intermediate fit which provides a good sealing fit and less friction resistance between piston and cylinder wall. These requirements fulfilled by piston rings. It consists mainly three rings. The first one is known as compression rings, the middle one is known as wiper ring or intermediate ring and the last one is known as oil rings. These rings are mainly made by cast iron of fine grain and sometimes by alloy steel. The function of Piston Rings: -To provide good sealing between piston and engine cylinder. -To regulate the amount of lubricant in the engine cylinder. -To provide proper lubrication. -To transfer heat piston through engine wall and fins. -To minimize friction between piston and engine wall.
Let functions f, g and h be related as follows: g(x) = f -1(x), h(x) = (g(x)) 5, f(6) = 10 and f '(6) = 12 Calculate h '(10).
We first calculate h '(x). h '(x) = 5 g '(x) g(x) 4 Which gives h '(10) = 5 g '(10) g(10) 4 g(10) is calculated as follows g(10) = f -1(10) = 6 We now need to calculate g '(10). Apply function f to both sides of the relation g(x) = f -1(x) f (g(x)) = f(f -1(x)) = x Which gives f (g(x)) = x Differentiate both sides of the equation obtained, using the chain rule to the term on the left. f '(g) . g '(x) = 1 We now set x = 10 in the above equation f '(g(10)) . g '(10) = 1 g(10) has already been calculated and is equal to 6, hence f '(6) . g '(10) = 1 g '(10) is given by g '(10) = 1 / 12 and h '(10) is given by h '(10) = 5 (1/12) 6 4 = 540