First, for each letter I thought a function that would result in a curve that sort of looks like that letter (if limited to the domain 0<x<1), then I tweaked it a bit (say, by raising it to a power or adjusting coeficients). If you want to try out the next bunch of steps on a

grapher, set the x-min to 0, ymin to -1, and the maxes to 1, since that's the relavent area.

For f, I used the curves of a three-root polynomial:

-3.85*(x+0.375)*(x-0.5)*(x-1.375)

and added a sharp fall in the midde by adding to that

50*(abs(x-0.51)-abs(x-0.49))

and then cubing the whole thing to adjust the roundness if the curves, yielding:

(-3.85*(x+0.375)*(x-0.5)*(x-1.375)+50*(abs(x-0.51)-abs(x-0.49)))^3

For u, I thought of the bouncy curve that would be given given by taking the absolute value of a sin function. I flipped that over by taking 1 minus it, and like with the f before, I cubed the whole thing to pull curves to the axis more. So that was

(1-abs(sin(2*pi*(x+0.25))))^3

For n, I again used the bounce of an absolute of a sin:

abs(sin(2*pi*(x+0.5)))

but we need the beginning of that scrunched up because that first bounce is just the line up from the f and the stem of the n. How to accelerate that part? Use the function -x^2+2*x instead of using x. That's because like the function f(x)=x, the function g(x)=-x^2+2*x starts at (0,0) and goes to 1,1) but g(x) acts more quickly at the beginning. So substituting g(x) in for x, we get

abs(sin(2*pi*(-x^2+2*x+0.5)))

Oh, and the substitution also has the convenient effect of making a gentle tail curve at the end of the n.

Ok, let's zoom out a bit now. Reset the grapher to its defaults by reloading it.

We need to shift the u and n if we're going to add this all together: the u need to be in the domain from 1 to 2, and the n needs to show up from 2 to 3. Additionally, the above three functions make the letters we want in the domain 0<x<1, but they make a bunch of irrelevant squiggles outside that domain. If we're going to add them to each other to make a word, each of the three component expressions needs to evaluate flatly to zero outside the domain where it makes its letter, so the garbage squiggles don't mess up the other letters.

The line y=(0.5*(abs(x)-abs(x-1)+1)) is like the line y=x, but only from x=0 to x=1. Outside of that domain, it just makes horizontal lines. So instead of looking like "

/", it looks like "___/ŻŻŻ". This means that any function that we subtitute it in to will be active only in the domain 0<x<1. Outside of that domain, on either side the function will just be stuck at whatever values it is at the tip of the curvy area.

So for the f we take the

(-3.85*(x+0.375)*(x-0.5)*(x-1.375)+50*(abs(x-0.51)-abs(x-0.49)))^3

we got before, and subtitute in

(0.5*(abs(x)-abs(x-1)+1))

at every x, which gives:

(-3.85*((0.5*(abs(x)-abs(x-1)+1))+0.375)*((0.5*(abs(x)-abs(x-1)+1))-0.5)*((0.5*(abs(x)-abs(x-1)+1))-1.375)+50*(abs((0.5*(abs(x)-abs(x-1)+1))-0.51)-abs((0.5*(abs(x)-abs(x-1)+1))-0.49)))^3

now it's a nicely isolated f with flat y=0 lines to the left and right (zero because that's the value at both tips of the f).

Now we want to do the same kind of thing for the u and the n, but for them we also want to shift them over to the right. So for the u we substitute in the shifted expression

(0.5*(abs(x-1)-abs(x-2)+1))

and for the n we use

(0.5*(abs(x-2)-abs(x-3)+1))

Those substitutions into the expressions we got earlier for u and n yield

(1-abs(sin(2*pi*((0.5*(abs(x-1)-abs(x-2)+1))+0.25))))^3

and

abs(sin(2*pi*(-(0.5*(abs(x-2)-abs(x-3)+1))^2+2*(0.5*(abs(x-2)-abs(x-3)+1))+0.5)))

respectively.

We now have each letter in its place, so we add them together:

(-3.85*((0.5*(abs(x)-abs(x-1)+1))+0.375)*((0.5*(abs(x)-abs(x-1)+1))-0.5)*((0.5*(abs(x)-abs(x-1)+1))-1.375)+50*(abs((0.5*(abs(x)-abs(x-1)+1))-0.51)-abs((0.5*(abs(x)-abs(x-1)+1))-0.49)))^3+(1-abs(sin(2*pi*((0.5*(abs(x-1)-abs(x-2)+1))+0.25))))^3+abs(sin(2*pi*(-(0.5*(abs(x-2)-abs(x-3)+1))^2+2*(0.5*(abs(x-2)-abs(x-3)+1))+0.5)))

But wait, we're not quite done!

We don't want all that extra line before and after the word. So lets make the function undefined everywhere except where the word is.

(x+abs(x))/(x+abs(x))

is a flat line that is undefined before zero.

(x-3.5-abs(x-3.5))/(x-3.5-abs(x-3.5)

is a flat line that is undefined after 3.5.

Add them together (and throw in some extra parentheses because I was paranoid aboud graphing applets getting order of operations wrong) and multiply it by 0.1 to make it lower and you get

0.1*(((x+abs(x))/(x+abs(x)))+((x-3.5-abs(x-3.5))/(x-3.5-abs(x-3.5))))

which makes a horizontal line that is only defined in the domain 0<x<3.5. add that to the "___fun___" function and we get "fun_" (giving the n a 0.5 unit long tail)

All done.

Wow... this took a hell of a lot longer to write up than it took to just do it. =P