# Rewriting a function without going piecewise

You know the ridiculous kind of pseudo-context question that makes you go 'Why doesn't Lisa get a proper hobby rather than timing her friends doing jigsaw puzzles?'? You could replace pretty much all of them with "A mathematician is trying to be clever by...".

In this particular case, a mathematician has worked out a function to be:

$$f(n) = \begin{cases} 0 & n \le 0 \\ 2n + 1 & n > 0 \end{cases}$$

Only s/he wants to rewrite it as a non-piecewise function. Preferably continuous over the real numbers. Why? S/he is trying to be clever. Cleverness is important to mathematicians.

And, of course, I like both solving puzzles and looking clever, so I naturally decided to have a go.

My first thought was, the modulus function is especially good for this sort of thing. If you think about $f(x) = x + \left|x\right|$, you find that your function works out to be $f(x) =2x$ if $x≥0$, and $f(x) = 0$ otherwise. This function is a bit trickier, though, as it's really in three parts:

$$f(x) = \begin{cases} 0 & x \le 0 \\ 3x & 0 \lt x \le 1 \\ 2x + 1 & x > 1 \end{cases}$$

So, my thinking was that I'd need to come up with two functions to add together: one that was zero for negative $x$ but $3x$ elsewhere, and one that was zero for $x \lt 1$ but $-x$ elsewhere -- so that when they're added, the gradient for $x \gt 1$ would be 2.

The first one is easy: if $x + \left|x\right|$ gives gradients of 0 and 2, multiplying it by $\frac 32$ gives gradients of 0 and 3; one part of the function will be $\frac 32 x + \left| \frac 32 x \right|$.

The other is trickier. To get the split at $x=1$, I need to use something involving $\left|x-1\right|$ (or some multiple thereof. $x - 1 + \left|x - 1\right|$ would give me a gradients of 0 and 2, but I want gradients of 0 and -1 -- so I need to multiply everything by $-\frac 12$ to get $-\frac12(x - 1) - \frac12\left|x - 1\right|$.

Adding them together gives $f(x) = \frac 32 x + \left| \frac 32 x \right| - \frac12 (x-1) - \frac 12\left|x-1\right|$, which I can simplify to $f(x) = x + \frac 12 + \frac 32 \left|x\right| - \frac 12\left|x-1\right|$.

It's not a nice function, but it is continuous and not piecewise -- I'd be curious to know if anyone has a nicer version!

* Edited 2015-11-06 to fix a sign error. Many thanks to @oldandrew for spotting the mistake. ## Colin

Colin is a Weymouth maths tutor, author of several Maths For Dummies books and A-level maths guides. He started Flying Colours Maths in 2008. He lives with an espresso pot and nothing to prove.

### 2 comments on “Rewriting a function without going piecewise”

• ##### oldandrewuk

Am I missing something here? I get f(0)=-1 and f(1)=1

• ##### Colin

Quite possible I’ve made a mistake — will look into it when I have a free half-hour.

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