Using squares to prove e > 2.7

Edited to replace $$\exp(-x)$$ with $$\exp(x)$$. My apologies.

I loved this puzzle, so thought I'd submit a similar one:

The definite integral $$\int_{−\infty}^{1} \exp(x)dx$$ is equal to $$e$$ . Using two squares of side 1 and one rectangle size 1*0.7, show how to cut them and fit inside the $$\exp(x)$$ curve.
Unlike a circle, this curve has a left-hand tail with height asymptotic to zero, which may be of interest.

• @JaapScherphuis Dang! I put in an unwanted negative sign. see quora.com/… or Aug 30 '21 at 19:19
• It can definitely be done with a finite number of pieces. Cut the rectangles into teeny-tiny squares, and just pack them below the curve. If we make the squares small enough we can get as close as we like to the actual area under the curve. Aug 30 '21 at 21:17
• My suggestion would be to change the question from "show you can fit the squares and rectangle under the curve" (which you can do just by showing e > 2.7 by any method) to "show how to fit the squares and rectangle under the curve," i.e. actually exhibiting a dissection. Sep 1 '21 at 0:41
• @2012rcampion thanks; did that Sep 1 '21 at 11:48

This is mostly straight-forward, a matter of allocating the slack of

e-2.7 = 0.01828... or in relative terms 0.0067255...

For example, we can allocate roughly a third into discarding the inifinite tail. Conveniently, the integral from -inf to the cutoff equals the function value at the cutoff and we find that

c = -5 ; e^c = 0.0067378...

is in the right ballpark.

We will chop the rest into equal intervals and and approximate the integral by tangents to the midpoint (these are relatively easy to cut out of rectangular blocks). if s is half the interval length, then the relative error is

1-2s/(e^s-e^-s)

which, conveniently, doesn't depend on location, only on interval width. Further, there is a closed form expression for the integral of the approximation function

(*) 2s(e^1-e^c)/(e^s-e^-s)

because the areas under the tangents form a geometric series. Using this formula and limiting ourselves to s of the form 1/(2n) (for easy cutting) we find

s = 1/8

is just small enough.

In summary:

we cut the interval [-5:1] into bins of width 1/4 and in each bin we approximate the integral by the area under the tangent to its midpoint. Using formula (*) we find that this cutting friendly lower approximation has integral 2.704495...

exact function in red, piecewise linear lower approximation in black

and will therefore accomodate our two squares and one rectangle with some area to spare.

• Oh, but if you use the midpoint wouldn't the bars overlap the curve at the lower part of the curve? Sep 1 '21 at 12:03
• A quick estimation of the number of cuts this method needs: First cut the three pieces into strips of width 1/4, which takes 9 cuts, and lay these end to end as one long strip. Then cut this strip into 6*4=24 rectangles of the correct heights, which takes 23 more cuts. Lastly change one end of each rectangle to the correct slope, which takes a further 24 cuts. So 9+23+24=56 cuts all together. Assuming things can be arranged so that no cuts intersect previous ones, this uses 3+56=59 pieces. Sep 1 '21 at 12:09
• @justhalf not using bars, using trapezoids (bars with sloped top). so the top of the bar is not horizontal but the tangent to the midpoint. This just fits and works for bars up to width 2. Sep 1 '21 at 14:07
• @JaapScherphuis thanks! There are obvious ways of hand-optimising this. Like using adaptive interval widths or puttting one square whole under the curve (bottom left corner at the origin) But I do not see a systematic approach. Sep 1 '21 at 14:42
• By the way, you can use just 19 bins of width 1/3, with total area 2.70093... (and at most 43 cuts) Sep 3 '21 at 17:40