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

1. Jan 30, 2010

### 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

2. Jan 30, 2010

### JSuarez

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

3. Jan 30, 2010

### 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?

4. Jan 30, 2010

### uart

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.

Last edited: Jan 30, 2010
5. Jan 31, 2010

### 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

6. Feb 1, 2010

### Landau

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

7. Feb 2, 2010

### HallsofIvy

Staff Emeritus
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, ex, is neither even nor odd. Its even and odd parts are $\frac{e^x+ e^{-x}}{2}= cosh(x)$ and $\frac{e^x- e^{-x}}{2}= sinh(x)$.

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