請問1~3題如何解題及證明呢？ 感激~~~?
1 則回答
最佳解答
1.
(1)
f^2 - g^2 = ( f + g )( f - g ) = 2a^x*[ - 2a^(-x) ] = - 4 ... Ans
(2)
4 = f(x)*f(y) = [ a^x - a^(-x) ]*[ a^y - a^(-y) ] = a^(x+y) - a^(x-y) - a^(y-x) + a(-x-y) ... (1式)
8 = g(x)*g(y) = [ a^x + a^(-x) ]*[ a^y + a^(-y) ] = a^(x+y) + a^(x-y) + a^(y-x) + a(-x-y) ... (2式)
(1式) + (2式) 得:
12 = 2[ a^(x+y) + a(-x-y) ] = 2*g(x+y)
g(x+y) = 6 ... Ans
(3)
(2式) - (1式) 得:
4 = 2[ a^(x-y) + a(y-x) ] = 2*g(x-y)
g(x-y) = 2 ... Ans
2.
當 x+y = 1 時:
f(x) + f(y)
= 4^x/( 4^x + 2 ) + 4^y/( 4^y + 2 )
= [ 4^x( 4^y + 2 ) + 4^y( 4^x + 2 ) ] / [ ( 4^x + 2 )( 4^y + 2 ) ]
= [ 4^(x+y) + 2*4^x + 4^(x+y) + 2*4^y ] / [ 4^(x+y) + 2*4^x + 2*4^y + 4 ]
= [ 4 + 2*4^x + 4 + 2*4^y ] / [ 4 + 2*4^x + 2*4^y + 4 ]
= 1
所求
= [ f(1/2011) + f(2010/2011) ] + [ f(2/2011) + f(2009/2011) ] + ......... + [ f(1005/2011) + f(1006/2011) ]
= 1*1005
= 1005 ... Ans
3.
(1)
f(x+y) * f(x-y)
= [ a^(x+y) + a^(-x-y) ]*[ a^(x-y) + a^(y-x) ]
= a^(2x) + a^(2y) + a^(-2y) + a^(-2x)
= [ a^(2x) + a^(-2x) ] + [ a^(2y) + a^(-2y) ]
= f(2x) + f(2y)
Q.E.D.
(2)
f^2 - 2
= [ a^x + a^(-x) ]^2 - 2
= a^(2x) + 2 + a^(-2x) - 2
= a^(2x) + a^(-2x)
= f(2x)
Q.E.D.
(3)
f^3 - 3f
= [ a^x + a^(-x) ]^3 - 3*[ a^x + a^(-x) ]
= a^(3x) + 3*a^(2x)*a^(-x) + 3*a^x*a^(-2x) + a^(-3x) - 3*a^x - 3*a^(-x)
= a^(3x) + 3*a^x + 3*a^(-x) + a^(-3x) - 3*a^x - 3*a^(-x)
= a^(3x) + a^(-3x)
= f(3x)
Q.E.D.