Question!!! (1 Viewer)

[conman]

Let

Rewrite the formulae for S as quadratic equation in 2^x.

 

[airie]

Not very sure to which extent it must go, but is S = 2^(x-1) + 2^-(x+1) = 2^-(x+1) * (2^x)² + 2^-(x+1) satisfactory?

 

[Mountain.Dew]

use this identity

1/2(2^x + 2^(-x)) = a(2^x)^2 + b(2^x) + c

sub x = 0, 1, -1 into the equation ==> then solve simultaneously for a, b and c.

 

[conman]

I find the question unclear.

It could be S = 1/2(2^x/2 + i2^-x/2)(2^x/2 - i2^-x/2)

=p

Yeah I agree with u. That's why I post it here for help!!!