C3 THEORY OF COMPUTATIONAL DYNAMICS
HANDOUT 1
SOME PROBLEMS WITH TAYLOR SERIES
This handout answers Q4 on Example Sheet 1.
EXPONENTIAL GROWTH AND DECAY
The basic fact required to answer this question is that exponentials grow faster than any power of x.
In particular consider
h(x)
=
xn exp(-λx)
for λ > 0. We claim that h(x) → 0 as x → ∞. This will be used at several other points in the course. To
show this fact we evaluate the derivative of log h:
g(x)
Thus for x ≥
=
2n
λ
f(x)
d log h(x )
dx
=
n
x - λ
1
we have g(x) ≤ -2λ and hence logh(x) → -∞ as x → ∞, as required. Now let
=
( )
exp − x1
2
Set y = x-2, so that f(x) = exp(-y), then by above x-2n f(x) = y n exp(-y) → 0 as y → ∞, i.e. as x → 0. Since
for |x| ≤ 1, we have |x-n| = |x-2n||xn| ≤ |x-2n|, we conclude that x-n f(x) → 0 as x → 0. Note that this also
means that if p is some polynomial, then p(x-1)f(x) → 0 as x → 0.
TAYLOR EXPANSION OF f
First assume that the Taylor expansion of f at x = 0 exists to all orders. We shall show that all the
Taylor coefficients must be zero. Thus suppose that to rth order we have
f(x)
=
a 0 + a1x + … + arxr + O(xr+1)
Since x-(r+1) f(x) → 0 as x → 0, we see that f(x) = O(xr+1), and hence a 0 + a 1x + … + arx r = O(xr+1). The
only way this can hold is for a 0 = a1 = … = ar = 0. But this holds for any r, and hence ar = 0 for all r. The
Taylor series thus converges to the function F(x) = 0, and yet obviously f itself is non-zero for x ≠ 0.
This is thus an example of a function whose Taylor series does not converge to itself, no matter how
many terms we take.
EXISTENCE OF THE TAYLOR EXPANSION
The above argument is based on the assumption that the Taylor expansion exists to all orders. For
those of you who are interested, one proves this as follows. Denote the rth derivative of f by f (r) (with
f (0) = f). We shall show by induction that f (r)(x) = pr(x-1)f(x) for x ≠ 0, where pr is some polynomial.
This clearly holds for r = 0. Suppose it holds for some r > 0. Then f (r+1)(x) = -x-2 p'r(x-1)f(x) + 2x-3p r(x-1)
f(x), where p'r is the derivative of pr (and hence a polynomial). Setting pr+1(x) = -x2p'r(x) + 2x3pr(x), we
see that f (r+1)(x) = pr+1(x-1) f(x), and pr+1 is a polynomial, as required.
Hence by above f (r)(x) → 0 as x → 0, and so if we set f (r)(0) = 0, then f (r) is continuous at x = 0.
Furthermore since x-2f (r-1)(x) → 0 as x → 0, we have
f (r-1)(x)
=
f (r-1)(0) + f (r)(0)x + O(x2)
and hence f (r)(0) = 0 satisfies the conditions to be the derivative of f (r-1) at x = 0.