A general function f(x) has the property f(-x)=-f(x). State a trigonometric function with this property and explain using the Maclaurin series expansion for this function why this property holds. Write down the integral in the limits -q to q of f(x) wrt x

Note that this property is the definition of an odd function, or draw a sketch of what this looks like in general about the horizontal axis. E.g. f(x)=sin(x) which has the expansion f(x)=x-((x^3)/3!)+((x^5)/5!)-...+((-1)^(n))((x^(2n+1))/(2n+1)!) where n runs from 0 to infinity. All terms in the series are odd and so when the sign of x is reversed, the sign of each term is reversed. We can factorise the negative sign out to prove f(-x)=-x+((x^3)/3!)-((x^5)/5!)+...-((-1)^(n))((x^(2n+1))/(2n+1)!)=-(x-((x^3)/3!)+((x^5)/5!)-...+((-1)^(n))((x^(2n+1))/(2n+1)!))=-f(x), as required. Simply write the integral expression and equate to zero as all positive area in the positive x domain is cancelled by all negative area in the neative x domain for any set of symmetrical limits.