Simple derivation of exponential approximation

I was watching one of Brian Douglas’ videos on control theory (Discrete Control #5) and ran into a simple derivation of an approximation I presented earlier.

Back in April I wrote several post on simple approximations for log, exp, etc. In this post I gave an approximation for the exponential function:

\exp(x) \approx \frac{2 + x}{2 - x}

The control theory video arrives at the same approximation as follows.

\begin{align*} \exp(x) &= \exp(x/2)\, \exp(x/2) \\ &= \frac{\exp(x/2)}{\exp(-x/2)} \\ &\approx \frac{1 + x/2}{1 - x/2} \\ &= \frac{2 + x}{2-x} \end{align*}

As I believe I’ve suggested before here, in a derivation like the one above, where you have mostly equalities and one or two approximations, pay special attention to the approximation steps. The approximation step above uses a first order Taylor approximation in the numerator and denominator.

The plot below shows that the approximation above (the bilinear approximation) is more accurate than doing a single Taylor approximation, approximating exp(x) by 1 + x (linear approximation).

exp, linear approx, bilinear approx

Here’s a plot focusing on the error in the bilinear and linear approximations.

error in linear and bilinear approximations to exp

The bilinear approximation is hard to tell from 0 in the plot above for x up to 0.5.

The derivation above is simple, but why is the result so good? An explanation in terms of Padé approximation is given here.

4 thoughts on “Simple derivation of exponential approximation

  1. Not math, but poetically: perhaps alternating “x” signs on the otherwise similar bilinear terms drive error in counter-balancing directions, cancelling out first order error terms.

    If so, maybe this could be generalized, and splitting a function into even/odd symmetric parts (x —> -x) before approximation has similar value.

  2. A more fair comparison might be to a 2nd order Taylor polynomial, since both have 3 degrees of freedom.

  3. Here is a “derivation” that is not black magic. Using some plain Tex notation:
    The exponential is the solution of the initial value problem
    u'(x) = u(x), u(0) = 1.
    We can approximate the solution with the trapezoidal formula
    u(x) = u(0) + \int_0^x u'(x) dx \approx u(0) + {x \over 2} (u'(0) + u'(x))
    so that
    u(x) \approx u(0) + {x \over 2} (u(0) + u(x)) = 1 + {x \over 2} (1 + u(x)) .
    Solve for u(x) and we have
    u(x) \approx {1 + x/2 \over 1 – x/2},
    which is the desired approximation.

  4. Diethard Michaelis

    @Symmetry of Approximation Error.
    By (my) intuition the above derivation seemed to enforce
    a rather symmetric approximation error.
    The absolute error, however, is quite asymmetric.
    But both the relative error
    (exp(x) – (x+2)/(x-2))/exp(x)
    and the absolute log scale error
    x – ln((x+2)/(x-2))
    are very symmetric and nearly coincide over [-1,1].

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