Richard Beattie

Any function can be expressed as the sum of an even and an odd function

I’m going through MIT’s 18.01 course on OCW to prep for the 18.01 ASE which I’m hoping to take in August. The content is great but the organisation on the website isn’t. I’m currently writing up what I hope is an improved version, but I digress.

In the Graphing Functions Notes even and odd functions are mentioned:

→ An even function is one whose graph is symmetric with respect to the y-axis:

f(x)=f(x)f(x) = f(-x)

Even Function

→ An odd function is one whose graph is symmetric with respect to the origin:

f(x)=f(x)-f(x)=f(-x)

Even Function

What’s neat is that every function f(x)f(x) can be written as the sum of an even function e(x)e(x) and an odd function o(x)o(x) where:

e(x)=12(f(x)+f(x))e(x) = \frac{1}{2}(f(x)+f(-x)) and

o(x)=12(f(x)f(x))o(x)=\frac{1}{2}(f(x)-f(-x))

Now, I only found this out while trying to do the first off the Problem Set questions, which I’ll show a solution for below. But first the proof

Proof

Suppose that the statement was true.

Let f(x)=e(x)+o(x)f(x)=e(x)+o(x) where e is even and o is odd

Then:

f(x)=e(x)+o(x)f(-x)=e(-x)+o(-x)

but we know that e(x)=e(x)e(-x)=e(x) and o(x)=o(x)o(-x)=-o(x) so:

f(x)=e(x)o(x)f(-x)=e(x)-o(x)

and:

f(x)+f(x)=2e(x)f(x)+f(-x)=2e(x)

and so:

e(x)=12(f(x)+f(x))e(x)=\frac{1}{2}(f(x)+f(-x))

Now, we just have to show that o(x)=f(x)e(x)o(x)=f(x)-e(x) is odd and we have proved this

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

and this is an odd function:

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

o(x)=f(x)f(x)2o(-x)=\frac{f(-x)-f(x)}{2}

TaDa 🎉