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The first thing to say is that this is not the same as the question about interesting mathematical mistakes. I am interested about the type of false beliefs that many intelligent people have while they are learning mathematics, but quickly abandon when their mistake is pointed out -- and also in why they have these beliefs. So in a sense I am interested in commonplace mathematical mistakes.

Let me give a couple of examples to show the kind of thing I mean. When teaching complex analysis, I often come across people who do not realize that they have four incompatible beliefs in their heads simultaneously. These are

(i) a bounded entire function is constant;
(ii) $\sin z$ is a bounded function;
(iii) $\sin z$ is defined and analytic everywhere on $\mathbb{C}$;
(iv) $\sin z$ is not a constant function.

Obviously, it is (ii) that is false. I think probably many people visualize the extension of $\sin z$ to the complex plane as a doubly periodic function, until someone points out that that is complete nonsense.

A second example is the statement that an open dense subset $U$ of $\mathbb{R}$ must be the whole of $\mathbb{R}$. The "proof" of this statement is that every point $x$ is arbitrarily close to a point $u$ in $U$, so when you put a small neighbourhood about $u$ it must contain $x$.

Since I'm asking for a good list of examples, and since it's more like a psychological question than a mathematical one, I think I'd better make it community wiki. The properties I'd most like from examples are that they are from reasonably advanced mathematics (so I'm less interested in very elementary false statements like $(x+y)^2=x^2+y^2$, even if they are widely believed) and that the reasons they are found plausible are quite varied.

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    $\begingroup$ I have to say this is proving to be one of the more useful CW big-list questions on the site... $\endgroup$ May 6, 2010 at 0:55
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    $\begingroup$ The answers below are truly informative. Big thanks for your question. I have always loved your post here in MO and wordpress. $\endgroup$
    – Unknown
    May 22, 2010 at 9:04
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    $\begingroup$ wouldn't it be great to compile all the nice examples (and some of the most relevant discussion / comments) presented below into a little writeup? that would make for a highly educative and entertaining read. $\endgroup$
    – Suvrit
    Sep 20, 2010 at 12:39
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    $\begingroup$ It's a thought -- I might consider it. $\endgroup$
    – gowers
    Oct 4, 2010 at 20:13
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    $\begingroup$ Meta created tea.mathoverflow.net/discussion/1165/… $\endgroup$
    – user9072
    Oct 8, 2011 at 14:27

292 Answers 292

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A common false assumption is that that two non-orthogonal pure states of a quantum mechanical system may never be unambiguously distinguished by a measurement. (See https://arxiv.org/pdf/quant-ph/9807022.pdf)

Another false belief is that a quantum computer is similar to an analogue computer, in that large computations will necessarily fail because of accumulated error. (See, for example, https://arxiv.org/abs/quant-ph/9712048)

For that matter, another common false believe is that Bell Inequalities aren't violated, although it is mostly held by people who have never heard of Bell Inequalities.

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    $\begingroup$ I'm not sure how you can believe that something you have never heard of isn't violated. $\endgroup$ Apr 10, 2016 at 17:55
  • $\begingroup$ @GeoffRobinson: I have on remarkably many occasions introduced good mathematicians to the Bell inequalities and been told that it's clearly impossible for them to be violated. $\endgroup$ Jan 5, 2022 at 22:12
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    $\begingroup$ @StevenLandsburg: But by then they have heard of them! $\endgroup$ Jan 5, 2022 at 23:02
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A dense subspace of a Hilbert space $H$ must contain an ONB for $H$.

This is, of course, true if $H$ is separable, but false in general. See, for example, this answer in MSE: https://math.stackexchange.com/a/201149/465145.

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I don't know how common this is, but it occurs as a corollary of a theorem in the fine, and widely used, text by Shafarevich on algebraic geometry: namely, if $f \colon X \longrightarrow Y$ is a surjective algebraic map of varieties, then 1) for all $y \in Y$, the fiber over $y$ has dimension $≥ \dim(X)-\dim(Y)$; 2) on some non empty open set in $Y$ the dimension of the fibers equals $\dim(X)-\dim(Y)$; 3) for all $r$, the set of $y \in Y$ such that the fiber over $y$ has dimension $≥ r$, is closed in $Y$.

The first two are true, but the third is false. Upper semicontinuity of fiber dimension is true on the source, not the target. For the conclusion as stated to hold, one can add properness to the hypothesis on the map. I think this is not at all widely believed by experts, but for some reason it persists in the text, hence may be believed by students.

Since I have myself written notes in which blatantly false statements occur, I do not think for a moment that Shafarevich himself believed this false statement. But such things do slip by, and may mislead beginners. In fact I believed it for some time until enlightened by a friend.

In keeping with the OP's desire to know the psychological reason for the error, it seems for some reason common in my experience for people to assume unconsciously that maps are proper.

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False belief: a subgroup isomorphic to a quotient is a retract.

Formally: Let $H,N$ be subgroups of $G$ with $N$ normal and $H \simeq G/N$, then $H$ is a retract of $G$.

It is false, because otherwise $C_2$ would be a retract of $C_4$, but it is not.

In fact, $H$ is a retract of $G$ if and only if $G$ is isomorphic to $H \ltimes N$ (semidirect product).

This false belief caused this post.

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If an Abelian category $\mathcal{A}$ is a full subcategory of an Abelian category $\mathcal{B}$, then for all objects $M,N$ of $\mathcal{A}$, we have an injection $$\operatorname{Ext}^i_{\mathcal{A}}(M,N) \hookrightarrow \operatorname{Ext}^i_{\mathcal{B}}(M,N).$$

As an example, let $G$ be the free group on $2$ letters, $A$ its abelianization, $\mathcal{B} = G-mod$, $\mathcal{A}=A-mod$, and $M=N=\mathbb{Z}$ with the trivial action. Then $\operatorname{Ext}^i_{\mathcal{A}}(M,N) \cong \mathbb{Z}$, while $\operatorname{Ext}^i_{\mathcal{B}}(M,N) \cong 0$.

(This example comes from the topological fact that a torus has nontrivial $H^2$, while a punctured torus has trivial $H^2$. In algebra, it's related to the idea that group homology $H_1$ is space of generators for a group while $H_2$ is a space of relations.)

This false belief came up a context where $\mathcal{B}$ was the category of all Galois representations while $\mathcal{A}$ was a certain subcategory. See the comments to Status of the conjectured vanishing of Bloch-Kato H^2.

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Initially when I started studying sequences I believed that:

Consider $(x_n)$ and $(y_n)$ are two convergent real sequences having limits $x$ and $y$ as limits respectively. If $ x_n < y_n \quad \forall n\in \mathbb{N}$ then $x < y$

which turned out to be false.

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    $\begingroup$ but the question is "less interested in very elementary false statement like $(x+y)^2=x^2+y^2$ " $\endgroup$ Aug 16, 2021 at 9:21
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Common false belief:

Consider a non-linear system $\dot{x} = f(x)$ such that $f(0) = 0$. Suppose that $V : D \rightarrow \mathbb{R}$ is a Lyapunov function, i.e. $V$ is positive-definite in $D$ and $\dot{V}$ is negative-definite in $D$. Then $D$ is contained in the basin attraction of the equilibrium point $0$.

Here is a example of a system with a Lyapunov function that is positive definite everywhere, its time-derivative is negative definite everywhere and there are trajectories that does not converge to $0$.

This misconception is common among students in some engineering courses where the proof of the Lyapunov stability theorem is sometimes omitted. Many students rely on their intuition from simple examples and skip or forget the proof, leading to this false belief.

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“There are no trigonometric proofs, because all the fundamental formulae of trigonometry are themselves based upon the truth of the Pythagorean Theorem.”

that quote is from Elisha Scott Loomis' book, "The Pythagorean Proposition: Its Demonstrations Analyzed and Classified, and Bibliography of Sources for Data of the Four Kinds of Proofs" and is cited in the abstract of Ne'Kiya D Jackson and Calcea Rujean Johnson's latest American Mathematical Society paper, "An Impossible Proof of Pythagoras":

In the 2000 years since trigonometry was discovered it's always been assumed that any alleged proof of Pythagoras’s Theorem based on trigonometry must be circular. In fact, in the book containing the largest known collection of proofs (The Pythagorean Proposition by Elisha Loomis) the author flatly states that “There are no trigonometric proofs, because all the fundamental formulae of trigonometry are themselves based upon the truth of the Pythagorean Theorem.” But that isn’t quite true: in our lecture we present a new proof of Pythagoras’s Theorem which is based on a fundamental result in trigonometry—the Law of Sines—and we show that the proof is independent of the Pythagorean trig identity $\sin^2x + \cos^2x = 1$.

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Root(s) of a $3^{rd}$ degree polynomial over $ \Bbb Q$ are expressible using radicals with the imaginary $i$. If a root $r$ is real, by taking only the real part, $r$ is expressible using radicals over the rational numbers.

Why not? See here on Wikipedia.

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Another common mistake. If $W = _P(e_1,\ldots, e_{n})$ is a vector space and $V$ is a subspace of $W$ of dimension $k$, then $V = _P(e_{i_1},\ldots, e_{i_k})$.

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  • $\begingroup$ What does that little subscript $p$ on the equals sign mean? $\endgroup$ Feb 11, 2016 at 21:36
  • $\begingroup$ $V$ is a vector space over field $P$. $\endgroup$ Feb 12, 2016 at 6:52
  • $\begingroup$ So, what does "$V$ is a vector space over field $P$ $(e_1,\dots,e_n)$" mean? $\endgroup$ Feb 12, 2016 at 8:37
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    $\begingroup$ $W$ is a vector space over field $P$, $(e_1,\ldots, e_n)$ is a basis of $W$. $V$ is a subspace of $W$. $\endgroup$ Feb 12, 2016 at 8:40
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I cannot believe this example was not yet given (but, if my belief is false, I will happily delete this answer):

It is very common among "lay people" (who do not understand what it means for lines to be parallel) to believe that "in some kind of geometry" (frequently described as non-euclidean) parallel lines can intersect.

One finds many instances (my guess, the count is in hundreds of thousands) of this false believe just by searching the internet. Here is a random example, the article "How Looking At A Basketball Disproves Something Everybody Learns In High School Geometry" from "Business Insider", 2014. The article concludes with

And voila! We’ve successfully disproven “parallel lines never intersect” using just a basketball.

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    $\begingroup$ i.redd.it/r7etb5kayl961.jpg $\endgroup$ Feb 9, 2021 at 4:39
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    $\begingroup$ I feel like this is a confusion of terminology and not mathematics. The definition of parallel being used in the Business Insider article is clearly not 'lines that do not meet and are a constant distance apart' or whatever the official definition of parallel is. It is closer to 'distinct lines that go in the same direction'. Insofar as there is a false belief, it is that there is a coherent notion of 'same direction'! But this is not essential to the belief that lines that go in the 'same direction' can meet in some kind of geometry, which is a belief that is closer to true than false. $\endgroup$
    – Solveit
    Feb 9, 2021 at 13:37
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    $\begingroup$ @ChanBae Mathematics starts by establishing common terminology and axioms. This is what Greeks realized over 2000 years ago. Sadly, this understanding was lost with changes in math education in the last century. The thing is, math is part language and part science. You cannot separate the two and claim that inability to understand definitions is just a matter of terminological disagreements. $\endgroup$ Feb 10, 2021 at 1:37
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    $\begingroup$ @MoisheKohan : It is not unusual in mathematics for the same word to have several definitions depending on context, and I think all that is happening here is that people are using the word "parallel" in a perfectly well defined way that they fully understand and that happens to be different than the way you had in mind. Some people (including me) believe that a tangent to a circle is normal to the diameter. I do not think that counts as a false belief in mathematics just because "normal" also means "fixed under conjugations". $\endgroup$ Jan 6, 2022 at 1:48
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Characters of a Hopf algebra form a group under convolution .-

It is, of course, the set of non-commutative (or vector valued) characters as it is well-known that, a Hopf algebra $(\mathcal{H},\mu_\mathcal{H},1_\mathcal{H},\Delta,\epsilon,S)$ [1] and a commutative algebra $(\mathcal{A},\mu_\mathcal{A},1_\mathcal{A})$ being given (all over the same commutative ring $k$), the set $Hom_{k_AAU}(\mathcal{H},\mathcal{A})$ is a group under convolution (the inverse being performed through precomposition with $S$). The story is the following.

Until yesterday and based on hasty readings (or rather non-revisited readings of hasty writers :), I believed that the above was true even if the algebra $\mathcal{A}$ was non-commutative. Which is false as shows the counterexample below.

Take any commutative ring $k$ and two letters (or non-commuting variables), $\{a,b\}$, then, with the Hopf algebra $\mathcal{H}=(k<a,b>,conc,1_{\{a,b\}^*},\Delta,\epsilon,S)$ and the (non-commutative) algebra $\mathcal{A}=(k<a,b>,conc,1_{\{a,b\}^*})$, we get a $\mathcal{A}$-valued character $Id:\ \mathcal{H}\to \mathcal{H}$ such that its convolutional square $Id^{*\,2}$ is NOT a character.

If one prefers a more standard world and matrix-valued characters, one can consider the following (and essentially the same counterexample)

  1. $k$, a field
  2. a two-dimensional vector space $V=k.a\oplus k.b$
  3. The Hopf algebra $T(V)$
  4. The algebra of 2x2 matrices over $k$
  5. The character on $T(V)$, given by $a\to \Big(\begin{matrix}0 & 1\\ 0 & 0\end{matrix}\Big)$ and $b\to \Big(\begin{matrix}0 & 0\\ 1 & 0\end{matrix}\Big)$.

[1] The order is always (space, product,unit,coproduct,counit,antipode).

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Way late to the party...

"$ \mathrm{polymod}\ p$ and $\mathrm{mod}\ p$ are the same thing."

And it's cousin: "$\forall{x}, f(x) \cong g(x) \pmod{q} \implies f(x) = g(x)$"

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    $\begingroup$ What does polymod mean? $\endgroup$ Oct 20, 2010 at 11:47
  • $\begingroup$ Either the cousin needs a bit more detail if it is to be false, it is quite naive! $\endgroup$ Oct 20, 2010 at 18:25
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    $\begingroup$ Probably I understand what this means: if $f(x)=0\pmod 2$ for all $x$, then $f=0$ over $\mathbb F_2$. This is similar to my second example: mathoverflow.net/questions/23478/… $\endgroup$
    – zhoraster
    Oct 20, 2010 at 18:33
  • $\begingroup$ Consequently, there are only $4$ polynomials over $\mathbb F_2$ Isn't this convenient? :-) $\endgroup$
    – zhoraster
    Oct 20, 2010 at 18:40
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    $\begingroup$ $\mathrm{polymod}$ is "polynomial mod". Two polynomials are congruent $\mathrm{polymod} p$ iff the coefficients each power of the variable are congruent $\pmod{p}$. The equivalence classes are sets of polynomials where each coefficient ranges over an equivalence class $\pmod{p}$. For the cousin, there are many local/globals but they all seem to require additional conditions (q.v. Hensel lifting). I think the set from which $x$ was chosen was left unspecified because this "imprecise mental abbreviation" pops up at various levels of sophistication each with a different such set. $\endgroup$
    – Anonymous
    Oct 23, 2010 at 15:22
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From Keith Devlin

"Multiplication is not the same as repeated addition", as put forward in Devlin's MAA column.

I'm not really sure how I feel about this one; I might be one of the unfortunate souls who are still prey to that delusion.

Caution

In case you missed it, the column ended up spilling a lot of electronic ink (as evidenced in this follow-up column), so I don't believe it would be wise to start yet a new one on MO. Thanks in advance!

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  • $\begingroup$ I followed your link, and I cannot even tell what is wrong about attaching helium balloons to both sides of a balance to model substraction on both sides of an equation. $\endgroup$
    – user11235
    Apr 10, 2011 at 20:32
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    $\begingroup$ The more I think about this "error", the less I am convinced. It's like saying that you cannot say that $\binom n k$ is the number of $k$-element sets in an $n$-element set because then you will be unable to generalize to complex values of $n$. Or you cannot define the chromatic polynomial as the function counting the colourings and then plug in $-1$ to get the acyclic orientations of the graph. Also, I think it is perfectly understandable what it means to add something halfways. $\endgroup$
    – user11235
    Apr 10, 2011 at 20:50
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    $\begingroup$ It's not a "false belief". It's a false heuristic. And it's actually here: mathoverflow.net/questions/2358/most-harmful-heuristic $\endgroup$ Apr 10, 2011 at 21:17
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    $\begingroup$ When I taught elementary teachers the course on arithmetic, they all had been taught that multiplication is repeated addition, but I myself thought it was the cardinality of the cartesian product. We enjoyed discussing this difference in point of view. $\endgroup$
    – roy smith
    May 9, 2011 at 3:06
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    $\begingroup$ The "repeated addition" characterization has an advantage over the "cardinality of the Cartesian product" characterization (which possibly in some contexts could be considered a disadvantage). That is that it's not self-evident that it's commutative, and so one has a useful exercise for certain kinds of students: figure out why it's commutative. $\endgroup$ May 20, 2011 at 2:28
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In algebraic topology, I thought for a while:

  • "For $k \geq 2$, $H_k$ is the abelianization of $\pi_k$." False. True for $k = 1$. Also for all $k$ up to $n-1$ if the space is $(n-1)$-connected for $n \geq 2$ (vacuously, since this says the first $n-1$ homotopy groups are trivial and for these, the Hurewicz homomorphism is the isomorphism, $\pi_k \cong H_k$). See the Hurewicz theorem for more.
  • "Generically, all the $\pi_k$ are nonabelian." False. For $k \geq 2$, $\pi_k$ is abelian.

Edit: I had a third error in thinking when I first posted this, mangling the above into something further from true. Which I suppose makes the first version of this post meta-appropriate for this thread (but I've fixed it anyway). Thankfully, user Michael gently pointed out my mangling.

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  • $\begingroup$ First bullet: did you mean "True for $n=1$"? $\endgroup$
    – Michael
    Jan 15, 2019 at 23:06
  • $\begingroup$ @Michael : It's not always true for $n=1$, $\pi_1$ can be abelian, e.g. the fundamental group of the circle. For $n > 1$, $H_n \cong \pi_n$. It's easy to imagine "$\pi_n$s are (usually) nonabelian monsters and their associated homology groups are friendly abelian groups", but this difference *only* happens for $n=1$. $\endgroup$ Jan 15, 2019 at 23:31
  • $\begingroup$ I think you are confusing a few things here. Compare $H_2$ of the 2-dimensional torus with its $\pi_2$, for example. $\endgroup$
    – Michael
    Jan 15, 2019 at 23:34
  • $\begingroup$ @Michael : After actually looking up what I was talking about, I find that I have mashed together (at least) two errors to make another. Yay? $\endgroup$ Jan 16, 2019 at 4:42
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    $\begingroup$ @Michael : I think I've disentangled my mangling. I may still have a fumble-thought in the first bullet that I'm just not seeing. $\endgroup$ Jan 16, 2019 at 5:11
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For $p$ prime and the chain of embeddings $\mathbb{Z}/p\mathbb{Z} \hookrightarrow \mathbb{Z}/p^2\mathbb{Z} \hookrightarrow \cdots$ given by multiplication by $p$, then $\bigcup_n \mathbb{Z}/p^n\mathbb{Z}$ is not the group of $p$-adic integers $\mathbb{Z}_p$, but its Pontryagin dual, the Prüfer $p$-group $\mathbb{Z}(p^{\infty})$.

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    $\begingroup$ Is that actually a common false belief? After all, $\mathbb{Z}_p$ is uncountable, as everyone realizes! $\endgroup$
    – Todd Trimble
    Mar 5, 2015 at 14:25
  • $\begingroup$ "$\mathbb{Z}_p$ is countable" is also a false belief for people who didn't really read the definition of $\mathbb{Z}_p$, but I don't know how much it is common. $\endgroup$ Mar 5, 2015 at 14:34
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    $\begingroup$ It's hard for me to believe it's at all common. I wasn't the downvoter, but I think it would be better if answers were rooted either in instances that can be found in the literature, or widely encountered in one's experience as an instructor. $\endgroup$
    – Todd Trimble
    Mar 5, 2015 at 14:52
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I once very briefly thought that:

Given a vector space $V$ and a sub-space $U \subset V$ that $V-U$ is also a subspace.

I've heard this several times as a TA also.

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  • $\begingroup$ Why the downvote! I heard this from more than one student in introductory linear algebra classes and when marking. $\endgroup$
    – Benjamin
    May 12, 2015 at 22:21
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    $\begingroup$ I think this falls under $(x+y)^2=x^2+y^2$, $\endgroup$
    – Thomas Rot
    Aug 10, 2015 at 12:48
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    $\begingroup$ It always fails... But I don't think this is a common held belief. $\endgroup$
    – Thomas Rot
    Aug 10, 2015 at 21:40
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    $\begingroup$ @ThomasRot But it always fails, while $(x+y)^2=x^2+y^2$ sometimes holds, especially in characteristic 2. $\endgroup$
    – ACL
    Apr 21, 2016 at 6:37
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    $\begingroup$ I meant that $V-U$ cannot be a subspace since it doesn't contain 0. On the other hand, in any commutative ring where $1+1=0$, then the formula $(x+ y )^2=x^2+y^2$ holds. $\endgroup$
    – ACL
    Apr 21, 2016 at 10:02
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The sigma function

$$\sigma_{1}({p_i}^{\alpha_i}) = \displaystyle\sum_{j=0}^{\alpha_i}{{p_i}^j}$$

satisfies the inequalities

$$\sigma_{1}({p_i}^{\alpha_i}) \gt (\alpha_i + 1)(\sqrt{p_i})^{\alpha_i}$$

$$\sigma_{1}({p_i}^{\alpha_i}) \gt 1 + \alpha_i(\sqrt{p_i})^{1 + \alpha_i}$$

for prime $p_i$ and $\alpha_i \ge 1$.

The "proof" uses the Arithmetic Mean-Geometric Mean Inequality.

As a particular application of this result, Sorli's Conjecture implies the OPN Conjecture.

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    $\begingroup$ Is this really a common belief? $\endgroup$ Jan 28, 2011 at 18:21
  • $\begingroup$ Not so really, just pure pun intended... =) $\endgroup$ Jan 29, 2011 at 0:48
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    $\begingroup$ @Arnie Your first equation makes no sense. Presumably you want to define $\sigma_1(n)$ for every positive integer $n$, hence a product is missing on the LHS. And the sum on the RHS cannot end at $\alpha_i$. Also, I wonder what is the use of the subscript $i$ in the two inequalities. $\endgroup$
    – Did
    Feb 6, 2011 at 10:36
  • $\begingroup$ @Didier, I get your point -- thanks, I overlooked that one. Editing the first equation now. $\endgroup$ Feb 6, 2011 at 10:50
  • $\begingroup$ @Didier, nope - I didn't want to define $\sigma_{1}(n)$ for every positive integer $n$, I just needed the equation for $n = {p_i}^{\alpha_i}$. $\endgroup$ Feb 6, 2011 at 10:53
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One common misconception is that Category Theory is abstract. It is in fact, no more abstract than set theory or calculus. This common complaint is the usual complaint that people have when they first meet set theory or category theory. Since people interested in category theory are usually already interested in mathematics, it is strange - but perhaps not that strange - that they voice this complaint. It's underlying reason is that it's notions are unfamiliar, rather than abstract. And it doesn't help that there isn't a bestiary of easy examples that we would have for algebra and calculus.

This needs to be developed.

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    $\begingroup$ Set theory is considered more abstract than calculus. $\endgroup$
    – Asaf Karagila
    Apr 3, 2021 at 10:29
  • $\begingroup$ @Asaf Karaglia: From the point of view of the informed everyman, they are all as abstract as each other. The point I'm making, is that in the usual pedagogy of mathematics, geometry, algebra and logic are threaded through each other and inform each other. However, this is not as true for Category Theory, which is seen as the new kid on the block and hence its unfamiliarity to mathematicians and mathematically informed disciplines like physics. $\endgroup$ Apr 3, 2021 at 10:49
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    $\begingroup$ I think I qualify as an 'informed everyman' on these subjects (I took undergraduate calculus up through real analysis out of Baby Rudin, and taught myself set theory and category theory). I would say there was about a 3-4 month period of intense headaches and confusion moving from real analysis in Rudin to set theory in Monk, because the level of abstraction was so much higher. You work directly with primitive notions and axioms, as opposed to 'concrete mathematical entities' like the reals together with algebraic operations etc. (I didn't downvote, but I did upvote Asaf's comment) $\endgroup$
    – Alec Rhea
    Apr 4, 2021 at 0:23
  • $\begingroup$ @Alec Rhea: I wouldn't count you as an informed everyman but as a mathematician. Especially as you know enough to discriminate between 'baby Rudin' and 'adult Rudin'. $\endgroup$ Apr 4, 2021 at 6:48
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I'm not sure how common it is but I've certainly been able to trick a few people into answering the following question wrong:

Given $n$ identical and independently distributed random variables, $X_k$, what is the limiting distribution of their sum, $S_n = \sum_{k=0}^{n-1} X_k $, as $n \to \infty$?

Most (?) people's answer is the Normal distribution when in actuality the sum is drawn from a Levy-stable distribution. I've cheated a little by making some extra assumptions on the random variables but I think the question is still valid.

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  • $\begingroup$ I don't understand your third paragraph. Are you saying that under the assumptions in the 2nd paragraph, the limiting distribution (rescaling if necessary) is always Levy-stable? $\endgroup$
    – Yemon Choi
    Apr 12, 2011 at 1:28
  • $\begingroup$ @Yemon, Yes, this is what I was implying. Perhaps I was a little too cavalier? Certainly the sum of (well enough behaved) i.i.d. r.v.'s with power law tails converge to a Levy-Stable distribution... $\endgroup$ Apr 12, 2011 at 23:53
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    $\begingroup$ Generally such a limiting distribution doesn't exist. Perhaps you need to divide your sum by the square root of $n$? $\endgroup$ Dec 29, 2011 at 13:56
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When I was a kid (8th grade), I solved a bunch of math problems in an exam using the ``well-known identity'' that $(x+y)^2=x^2+y^2$, which I was sure I had been taught the year before. It was of course way before I heard about characteristic two and I didn't get a good grade that day!

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    $\begingroup$ Quoth the question, "The properties I'd most like from examples are that they are from reasonably advanced mathematics (so I'm less interested in very elementary false statements like $(x+y)^2=x^2+y^2$, even if they are widely believed)". $\endgroup$
    – JBL
    Dec 1, 2010 at 23:39
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    $\begingroup$ Also, this is of course just a special case of the more general “law of universal linearity”, which iirc was mentioned in earlier answers… $\endgroup$ Dec 2, 2010 at 0:40
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I don't know if this is what you are looking for, but I keep hearing that "a differentiable function is one that is locally linear", not one whose local variation can be approximated linearly. No one stops to think about e.g, $x^2$, and the fact that its graph does not look like a line at any value of $x$.

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    $\begingroup$ I would say this is more a heuristic than a false statement; as such, it would be more appropriate as an answer to mathoverflow.net/questions/2358/most-harmful-heuristic (although I do not think anyone interprets it the way you apparently do). $\endgroup$ May 5, 2010 at 4:53
  • $\begingroup$ Yes, I did not read the question very carefully. I realize it is not a good comment, and, yes, it is more of a abd heuristic than anything else. $\endgroup$
    – Herb
    May 25, 2010 at 23:59
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    $\begingroup$ it is also a comment on the imprecision of the words locally, infinitesimally,.... This once led Oort-Steenbrink to give some careful restatements of results previously called as "local Torelli theorems"... $\endgroup$
    – roy smith
    Apr 14, 2011 at 19:02
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