Prove that the set {x,e^x,e^-x} is linearly independent in C(R)?

We must show that, if a, b and c are real numbers such that





a x + b e^x + ce^(-x) = 0 (1) for every real x, then a = b = c = 0.





Putting x = 0, we conclude that





b + c = 0 =%26gt; c = -b.





Plugging in (1), it follows that





a x + b(e^x - e^(-x)) = 0 for every real x. This means that the function g given by





g(x) = a x + b(e^x - e^(-x)) = a x + 2b sinh(x) is identically 0. Hence, g' must be identically 0 too, so that we have





g'(x) = a + 2b cosh(x) = 0 for every real x





Suppose b ≠ 0 . Then, since cosh takes on every real value on [1, ∞), it follows g' takes on every real value on [a - 2b, ∞ ), if b %26gt; 0, or on (- ∞ a - 2b], if b %26lt; 0. In either case g' is not identically zero, so that we must have b = 0. This implies that g'(x) = a is identically zero, which, of course, implies a = 0,





So, a = 0, b= 0 and c = -b = 0. This shows {x,e^x,e^-x} is a linearly independent set in C(R).

Prove that the set {x,e^x,e^-x} is linearly independent in C(R)?
Suppose a*x+b*exp(x)+c*exp(-x)=0 for all real x. Taking the derivative twice , we get


a+b*exp(x)-c*exp(-x)=0


and


b*exp(x)+c*exp(-x)=0. With the first equation, this implies a*x=0 hence a=0. With the second equation, we get 2*b*exp(x)=0 which implies b=0. Finally, the second equation and a=b=0 implies c=0, and we are done.