Analysis Proof: Sum of odd and even functions.

[The_Iceflash]

Homework Statement

Show f(x) can be expressed as the sum of E(x) and an odd function O(x).

f(x) is defined for all x (assume domain D symmetric about 0)

f(x) = \frac{f(x)+f(-x}{2}
Then:
How does it look if f(x) = e^x?

Homework Equations

N/A

The Attempt at a Solution

So I got \frac{f(x)+f(-x)+f(x)-f(-x)}{2}

as my solution for that.

From that I need to answer: How does it look if f(x) = e^x ? I'm not sure what to do to show that.

 

[Char. Limit]

Homework Statement

Show f(x) can be expressed as the sum of E(x) and an odd function O(x).

f(x) is defined for all x (assume domain D symmetric about 0)

f(x) = \frac{f(x)+f(-x}{2}
Then:
How does it look if f(x) = e^x?

Homework Equations

N/A

The Attempt at a Solution

So I got \frac{f(x)+f(-x)+f(x)-f(-x)}{2}

as my solution for that.

From that I need to answer: How does it look if f(x) = e^x ? I'm not sure what to do to show that.

First: Is that a sum of odd and even functions?

Second: Well, try sticking e^x in for f(x) and see what functions you get.

 

[The_Iceflash]

First: Is that a sum of odd and even functions?

Second: Well, try sticking e^x in for f(x) and see what functions you get.

I stuck in e^x for f(x) and got e^x. Is this supposed to represent something?

 

[Char. Limit]

You should get...

f(x)=\frac{e^x+e^{-x}}{2}

 

[Office_Shredder]

Well, since the two sides are equal I would hope you get e^x back when you cancel everything. But notice without canceling what you get

e^x = \frac{e^x+e^{-x}}{2} + \frac{e^x-e^{-x}}{2} = cosh(x)+sinh(x) which is a way of writing e^x as a sum of an odd and an even function