Timeline for Examples of common false beliefs in mathematics
Current License: CC BY-SA 2.5
10 events
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May 9, 2011 at 2:53 | comment | added | roy smith | In a related vein, before discussing improper integrals, some books ask students to evaluate the riemann integral of functions like 1/sqrt(x), on [0,1], without noting that the function is unbounded hence not riemann integrable. the fact that the antiderivative exists and is finite on [0,1] masks this problem. | |
Jun 12, 2010 at 22:13 | comment | added | JBL | $\mathbb{R}^\times \to \mathbb{R}$ other than $\ln |x| + c$ with derivative $\frac{1}{x}$, I also agree with you; I just happen to think that the actual statement you wrote down is not incorrect but rather has an unwritten assumption built into the word "antiderivative," namely that such a thing is only defined for an interval on which the supposed antiderivative is differentiable. I hope this is clearer (and also correct!). | |
Jun 12, 2010 at 22:05 | comment | added | JBL | Now, if your point is that most students (and perhaps many instructors) in calculus classes often don't realize that this condition is necessary when they write $\int \frac{1}{x} \, dx = \ln |x| + C$, I agree with you; and if your point is that it would be less misleading to say something like, "For $x > 0$, the antiderivative of $\frac{1}{x}$ is $\ln x + C$, while for $x < 0$ the antiderivative of $\frac{1}{x}$ is $\ln (-x) + C$," I also agree with you; and if your point is that many people might not realize that there are functions (cont'd) | |
Jun 12, 2010 at 22:02 | comment | added | JBL | Sorry, I'm surely communicating less clearly than would be ideal. Starting over from scratch: suppose we have some differential equation, with unknown function $y(x)$. We say that for some function $f(x)$, $y = f(x)$ is a solution of the differential equation if there exists an interval $I$ on which $f(x)$ has all the requisite derivatives and the equation is satisfied on this interval. Antidifferentiation is the particular case $y' = g(x)$; implicit in the statement "$F(x)$ is an antiderivative of $f(x)$" is the condition "on some interval for which $F(x)$ is differentiable." | |
Jun 12, 2010 at 20:45 | comment | added | T.. | This false belief is self-consistent, hence irrefutable, as far as antidifferentiation is concerned, because one can always choose the (true) different local constants of integration to match the (false but notationally implied) single global constant. It is difficult or impossible to find an elementary function with singularities, that can be formally anti-differentiated in two different ways (incorrectly using a single "global" integration constant at each step), where subtracting one antiderivative from the other produces a locally-but-not-globally constant function. | |
Jun 12, 2010 at 19:33 | comment | added | JBL | That function is not "nice" on any interval containing 0; on any interval not containing 0, it is of the form you are complaining about. This is exactly my point -- the word "interval" is important to what I wrote! | |
Jun 12, 2010 at 4:29 | comment | added | Daniel Asimov | In case that wasn't clear: F(x) = ln(x) + C_1 for x > 0, and F(x) = ln(-x) + C_2 for x < 0, where C_1 and C_2 are arbitrary real constants. | |
Jun 12, 2010 at 4:25 | comment | added | Daniel Asimov | Really? What about the function F(x) given by ln(x) + C_1, x > 0 F(x) = ln(-x) + C_2, x < 0 for arbitrary reals C_1, C_2 ? (The appropriate technical condition is that an antiderivative be differentiable on the same domain as the function it's the antiderivative of is defined on.) | |
Jun 12, 2010 at 0:57 | comment | added | JBL | Well, the false belief is correct under the (frequently unspoken) condition that we only speak of antiderivatives over intervals on which the function we're antidifferentiating is "well-behaved" (and I'm not 100% sure what the right technical condition there is; "continuous"?). | |
Jun 12, 2010 at 0:40 | history | answered | Daniel Asimov | CC BY-SA 2.5 |