Function, f, with domain (-infinity, + infinity)

[ajassat]

I was working on the following problem from a textbook. The textbook has no answer. I have included my solution - I am not sure whether it is correct Any ideas and or solutions? (guidance)

Question:
Suppose that f is any function with domain (-infinity, +infinity)

a) Does the function g defined by g(x) = f(x) + f(-x) have any special symmetry?

My solution

A function f which satisfies:
f(-x) = -f(x) is symmetrical about the horizontal axis. This is also:
f(x) + f(-x) = 0
or we could substitute values of x giving non zero solutions for:
f(x) + f(-x) = g(x) (some other function)

Therefore g(x) = f(x) + f(-x) has symmetry about horizontal axis?

Any guidance appreciated.

Regards,
Adam

 

[JSuarez]

No. f(x)+f(-x) is an even function: it's symmetric about the vertical axis.

 

[some_dude]

Try and work backwards: take an arbitrary function g defined on \mathbb{R} and try to create a function f s.t. f(x) + f(-x) = g(x) for all x. What assumptions must be made about g in order to do this?

 

[uart]

I was working on the following problem from a textbook. The textbook has no answer. I have included my solution - I am not sure whether it is correct Any ideas and or solutions? (guidance)

Question:
Suppose that f is any function with domain (-infinity, +infinity)

a) Does the function g defined by g(x) = f(x) + f(-x) have any special symmetry?

My solution

A function f which satisfies:
f(-x) = -f(x) is symmetrical about the horizontal axis. This is also:
f(x) + f(-x) = 0
or we could substitute values of x giving non zero solutions for:
f(x) + f(-x) = g(x) (some other function)

Therefore g(x) = f(x) + f(-x) has symmetry about horizontal axis?

Any guidance appreciated.

Regards,
Adam

Just use the definition of odd and even functions for this type of problem. Without making any assumptions about f(x) (apart from the given domain) simply write g(-x) and compare it with g(x). That is,

g(x) = f(x) + f(-x) : given

So g(-x) = f(-x) + f(-(-x))
= f(-x) + f(x)
= g(x).

And g(-x) = g(x) is the definition of an even function.

 

[ajassat]

I figured it was an even function after I had posted this.

Thanks to both some_dude and uart for posting two different approaches to the problem. I understand this problem a lot better and have solved some similar now.

Thanks again,
Adam

 

[Landau]

To test yourself, a new question: do the same for g(x)=f(x)-f(-x).

 

[HallsofIvy]

By the way, f(x)+ f(-x) and f(x)- f(-x) are almost the "even and odd parts" of the function f(x). e(x)= (f(x)+ f(-x))/2 and o(x)= (f(x)- f(-x))/2 are even and odd functions that add to give f(x). Of course, if f(x) is an even function to begin with, o(x) will be 0 and e(x)= f(x). If f(x) is an odd function, e(x)= 0 and o(x)= f(x).

The exponential function, e^x, is neither even nor odd. Its even and odd parts are \frac{e^x+ e^{-x}}{2}= cosh(x) and \frac{e^x- e^{-x}}{2}= sinh(x).