Find the error in each of these two "paradoxes".
analysis/find.the.error
Problem
Find the error in each of these two "paradoxes".
1. Step 1. Define the function f(x) = 2 arcsin(x) + arcsin( 1 - 2 x^2 ).
So f(x) is defined for x in the interval [-1, 1].
Step 2. Differentiate and simplify: find f'(x) = ??
Step 3. Check the unexpected answer for f'(x) by calculating f(0)
and f(1). No problem so far -- these values do not
contradict a fundamental theorem of calculus.
Step 4. Recheck f'(x) by calculating f(-1). *NOW* we have a problem.
Step 5. Find your error. Don't tell me that you didn't make a
mistake in Step 2 -- everybody does.
Cute, eh?
-- Barry Wolk <wolk@ccm.UManitoba.CA> (January 1995)
2. e^(2.pi.i) = 1 (1) e^c is unique over complex numbers
e = e^(1 + 2.pi.i) (2) e^1 = e and (a^b)(a^c) = a^(b+c)
= (e^(1+2.pi.i))^(1+2.pi.i) (3) repeating the above step
= e^((1+2.pi.i)^2) (4) since (a^b)^c = a^(bc)
= e^(1 + 4.pi.i - 4.pi.pi) (5) multiplying out
= 1.94551938303e-17 (6) calculating the result
On which line is the mistake? Would you have noticed the mistake if
the "proof" had led to a correct result rather than a false one?
-- Gareth Rees <gdr11@cl.cam.ac.uk> (February 1995)
[ Editor: the opening sentence is the editor's; the two puzzles were
posted in a January 1995 thread "2=3" and as "Paradox: e =
1.94551938303e-17". ]
Show solution
1. The error is the step sqrt(x^2) = x in Step 2.
f'(x) = 2/sqrt(1 - x^2) - 4x/sqrt(1 - (1 - 2x^2)^2),
and 1 - (1 - 2x^2)^2 = 4x^2 (1 - x^2), whose square root is
2|x| sqrt(1 - x^2), not 2x sqrt(1 - x^2). So
f'(x) = 2/sqrt(1 - x^2) - 2 sgn(x)/sqrt(1 - x^2)
= 0 for 0 < x < 1,
= 4/sqrt(1 - x^2) for -1 < x < 0,
and f is not differentiable at 0 (one-sided derivatives 4 and 0).
Writing 2x instead of 2|x| gives the "unexpected answer" f' = 0
everywhere. On [0, 1] that is right: f(0) = 0 + pi/2 and f(1) = pi +
arcsin(-1) = pi/2, so Step 3 finds nothing wrong. But f(-1) =
-pi - pi/2 = -3 pi/2, which a constant f would not allow. In fact
f(x) = pi/2 for 0 <= x <= 1,
f(x) = 4 arcsin(x) + pi/2 for -1 <= x <= 0,
(checked numerically at x = -1, -1/2, 0, 1/2, 1). Geometrically:
with x = sin t, 1 - 2x^2 = cos 2t, and arcsin(cos 2t) = pi/2 - 2|t|
for |t| <= pi/2, which has a corner at t = 0.
2. The mistake is on line (4): the rule (a^b)^c = a^(bc) is false
for complex exponents. With the principal value a^b = exp(b Log a),
(a^b)^c = exp(c Log(a^b)), and Log(a^b) = b Log a only when b Log a
has imaginary part in (-pi, pi]. Here a = e and b = 1 + 2 pi i, so
b Log a = 1 + 2 pi i, but Log(e^(1+2 pi i)) = Log(e) = 1. So
(e^(1+2 pi i))^(1+2 pi i) = e^(1+2 pi i) = e,
which is what line (3) says (correctly: e^(1+2 pi i) is just the
number e), while line (4) replaces it with e^((1+2 pi i)^2) =
e^(1 - 4 pi^2) = 1.9455193833e-17 (the posting's last digits,
...8303, differ slightly from the exact value ...8331, presumably
from the calculator). Lines (1), (2), (3) and (5) are right.
Would one notice? Probably not: the rule (a^b)^c = a^(bc) does hold
for all complex a, b when c is an integer, and for real a > 0 and
real b, so it is used without a second thought and usually gives
correct results. Only the absurd conclusion draws attention to the
step; a derivation that misused it to reach a true statement would
look perfectly convincing.
-- Claude (rec.puzzles Oracle, 2026-11-02)