# How "same" are two isomorphic groups?

From what I understand about isomorphisms is that two isomorphic groups are the same groups. They may have different names for the same elements and the operation. But the point is that the groups are same since their elements combine the same way.

So if $$G \cong G'$$ then every group property about $$G$$ also holds for $$G'$$, am I correct?

Now my question is— Can I interchange $$G$$ and $$G'$$ whenever and wherever I want? I thought the answer was obviously yes but... now I'm not sure.

For example: $$\mathbb{Z} \cong \mathbb{2Z}$$ so shouldn't $$\mathbb{Z}/\mathbb{Z} = \mathbb{Z}/\mathbb{2Z}$$ under the group operation $$(a+H) + (b +H) = (a+b) + H$$? where $$H$$ is either $$\mathbb{Z}$$ or $$\mathbb{2Z}$$ since both are the same groups.

But that's clearly not the case as $$\mathbb{Z}/\mathbb{Z} = \{0\}$$ but $$\mathbb{Z}/\mathbb{2Z}=\{0,1\}$$ So where does one draw the line between two isomorphic groups? How "same" are two isomorphic groups? I'm very confused.

• It is true that $\mathbb{Z} \cong \mathbb{2Z}$. However, the issue for the quotient groups here is that $2 \mathbb Z \subsetneq \mathbb Z$. Dec 27, 2021 at 16:42
• The freedom to rename is not a freedom to be inconsistent. If you want to rename the elements in $\mathbb Z$ so that $1$ is renamed to $2$, $2$ to $4$, $3$ to $6$ etc. - then of course the elements of $2\mathbb Z\subseteq \mathbb Z$ must be renamed using the same procedure. You will end up with $4\mathbb Z\subseteq 2\mathbb Z$, and indeed $\mathbb Z/2\mathbb Z\cong 2\mathbb Z/4\mathbb Z$. Dec 27, 2021 at 16:43
• Why is this closed under duplicate? I'm asking about a particular example and the difficulty I'm facing with it. How is the question same as the other? I'd already read that one. It's where I came from. The accepted answers says exactly the same thing as my first paragraph. But my question is different. Dec 27, 2021 at 16:46
• Presentations of the same group (up to isomorphism) can have different properties. Dec 27, 2021 at 16:54
• Frankly, I regard the "s" word as a 4-letter word in mathematics. I would suggest that you learn to love the 10-letter "i" word in your title. And then if you can extend that to loving the 11-letter word "isomorphism", and to think about actual isomorphisms as useful and interesting mathematical objects themselves, even better! Dec 27, 2021 at 18:51

This is a great question and your example with group quotients perfectly highlights why one has to be careful about notions of identity when doing maths. To answer your question,

Can I interchange G and G′ whenever and wherever I want?

The answer is yes, as long as you are 100% sure that you are thinking of both groups as literally just groups and nothing more (people will sometimes use the phase 'up to isomorphism' to convey this idea). This resolves the problem you had with quotients: you can't quotient an arbitrary group by another arbitrary group, the group by which you are quotienting has to be a subgroup of the other. This is extra 'data' which can be encoded in multiple ways, one nice way is to consider the special injection that embeds the one group inside the other (which doesn't exist for two arbitrary groups). So to recap, subgroups aren't just groups, they're groups with some extra data and $$2\mathbb{Z}$$ and $$\mathbb{Z}$$ are different subgroups of $$\mathbb{Z}$$ despite being 'the same' groups (i.e. isomorphic).

• "the group by which you are quotienting has to be a subgroup of the other" but $Z$ and $2Z$ are both subgroups of $Z$ no? and I don't get the part where you say they are the same groups but different subgroups? I'm sure I'm missing something here but I don't get it. What is the extra data about subgroup? Dec 27, 2021 at 17:14
• Oh you mean they're different subgroups because they are made up of different subsets of the group? Oh okay. Fair enough. But what exactly go wrong here in my example? I'm not able to "see" it still :( Dec 27, 2021 at 17:26
• The problem in your example is that, to conclude that the quotient is the same, you have to start with 'the same' subgroups, but you didn't, you started with two different subgroups (that just so happen to be 'the same' as groups). Dec 27, 2021 at 17:28
• So in total, there are four descriptions of mathematical objects: the group $\mathbb{Z}$, the group $\mathbb{2Z}$, the subgroup $\mathbb{Z}\leq\mathbb{Z}$ and the subgroup $\mathbb{2Z}\leq\mathbb{Z}$. The first two descriptions define the same mathematical object, the second two don't. Dec 27, 2021 at 17:46
• Oh! Oh! Oh!! I get it now. If $H,H' ≤G$ and $H \cong H'$ then as groups, $H$ and $H'$ are same. But "subgroup" is also a relation and not an independent concept. So you always have subgroups of some group. So as subgroups of $G$, $H$ and $H'$ are different. So $G/H ≠ G/H'$ because quotient groups aren't just about $H$ but rather $H$ wrt $G$. If it were about only $H$ then one can interchange it with $H'$ is what you're saying. Dec 27, 2021 at 18:12

Here is a question that is similar: when can replace equivalent things with one another?

So if we remember that an isomorphism of groups is a function $$\phi:G\rightarrow G'$$. With the example you gave you have $$\phi:\mathbb{Z}\rightarrow 2\mathbb{Z}$$, then your issue was that $$\mathbb{Z}/\mathbb{Z}\not\cong\mathbb{Z}/2\mathbb{Z}$$. However, the issue is where are you applying the homomorphism. You are saying that $$G/G\not\cong G/\phi(G)$$; however, we have that $$\phi(G)\subset G'$$ and $$\phi(G)\not\subset G$$ (although it may be isomorphic to some subgroup of $$G$$ like in your example).

The key use of an isomorphism is that you can take a problem in one setting and view it in another setting through the isomorphism. Thus, what would be true is that if $$\phi:G\rightarrow G'$$ is an isomorphism, then if $$H\lhd G$$, then it will be the case that $$G/H\cong \phi(G)/\phi(H)$$.

I think the long story short is that when you have an isomorphism between two objects you must think of them as living in different spaces. If you are familiar with the lattice of subgroups I think this is a nice way to view it, since if you have $$\phi: G\rightarrow G'$$ where $$G=\mathbb{Z}$$ and $$G'=2\mathbb{Z}$$ since these are isomorphic groups their lattice of subgroups are isomorphic and we should think of $$\mathbb{Z}$$ to be the maximal element in the lattice of $$G$$ and $$2\mathbb{Z}$$ to be the maximal element in the lattice of $$G'$$. Even though we have that $$2\mathbb{Z}$$ appears in the lattice of $$G$$, the corresponding element in $$G'$$ would be the subgroup $$4\mathbb{Z}$$, so under the isomorphism it would not make sense to "replace $$2\mathbb{Z}$$ in the $$G$$ world with $$2\mathbb{Z}$$ in the $$G'$$ world" as they aren't the same under the isomorphism. What we would do is replace the $$2\mathbb{Z}$$ with the corresponding subgroup in the isomorphism namely $$4\mathbb{Z}$$, and when we do so we do have that $$\mathbb{Z}/2\mathbb{Z}\cong 2\mathbb{Z}/4\mathbb{Z}$$.

There is a precise way to formalise this.

We want to speak of sets not as collections of particular mathematical objects but as a bag of featureless dots. These bags of dots are related to each other by functions, which also relate dots in one bag to dots in another.

In order to do this, we propose a language for discussing sets. Capital letters will represent sets, and lowercase variables will represent functions between sets.

We define a "nice proposition about sets" as any proposition that can be generated from the following rules:

• For any sets $$A, B$$ and any function variables $$f, g: A \to B$$, $$f = g$$ is a nice proposition
• We can combine nice propositions $$\phi, \psi$$ using the operators $$\land$$, $$\lor$$, $$\neg$$, and $$\implies$$ to get another nice proposition
• $$\top$$ and $$\bot$$ (that is, true and false) are nice propositions
• If $$\phi(A)$$ is a nice proposition, where $$A$$ is a set variable, then $$\forall A (\phi(A))$$ is a nice proposition (and similarly for $$\exists$$)
• For all set variables $$A, B$$, if $$\phi(f)$$ is a nice proposition (where $$f : A \to B$$ is a function variable), then $$\forall f : A \to B (\phi(f))$$ is a nice proposition (and similarly for $$\exists$$)

Finally, in our language, we can form new functions from old ones using function composition and also discuss identity functions. Note that we will only discuss function composition when we know that the functions are composable syntactically.*

Note that we very deliberately avoided two things. First, there is no mention of the elementhood relation at all. Second, there is no way to compare two sets for equality, nor is there a way to compare two functions for equality unless the two have the same domain and codomain. This is deliberate. Also note that all variables here have specific types, and that we rely on these types to determine whether we can discuss equality.

Despite this avoidance, it is both an empirical fact and a (rather complicated) theorem that all propositions with no free variables can be translated to a "nice proposition about sets". For most mathematically useful propositions, the translation is relatively intuitive for those comfortable with category theory.

To get around the restriction that we not discuss elements directly, we can fix some 1-element set $$1$$ and discuss elements of $$A$$ as functions $$1 \to A$$. We denote the situation $$x : 1 \to A$$ as $$x :\in A$$. When we have $$x :\in A$$ and $$f : A \to B$$, we write $$f \circ x$$ as $$f(x)$$ (note that $$f(x) :\in B$$). Note that this is enough to define, for example, a group (as a set $$G$$, together with a function $$- \cdot - : G^2 \to G$$ which satisfies the group laws**).

We thus have the following theorem:

Big Theorem. Consider some nice proposition $$\phi(G)$$ - that is, some statement $$\phi$$ where the group variable $$G$$ occurs free. Then we can prove the following statement: "For all groups $$G_1$$, $$G_2$$, if $$G_1$$ and $$G_2$$ are isomorphic and $$\phi(G_1)$$, then $$\phi(G_2)$$."

Let us discuss how this relates to your example of $$\mathbb{Z}$$ and $$2 \mathbb{Z}$$.

We may try to define $$\phi(G)$$ as the statement "$$\mathbb{Z} / G$$ is the zero group". Then taking $$G_1 = \mathbb{Z}$$ and $$G_2 = 2 \mathbb{Z}$$, we see that $$\phi(G_1)$$ holds but $$\phi(G_2)$$ is false. This would appear to violate our Big Theorem.

To understand what's going on, we need to understand exactly what is going on with the statement $$\phi(G)$$. Note that in this statement, it is necessary for $$G$$ to be a subgroup of $$\mathbb{Z}$$ - in particular, it must be a subset. That is, we must have $$\forall x \in G (x \in \mathbb{Z})$$.

Of course, we cannot express such a statement as a nice proposition about sets, since it relies on the proposition $$x \in \mathbb{Z}$$, which is not a nice proposition. So we have to approach things a bit differently.

Rather than forcing $$G$$ to be a subgroup in the traditional, literal sense, we instead make $$G$$ a subgroup of $$\mathbb{Z}$$ in a more general sense. That is, we discuss a group $$G$$, together with some injective group homomorphism $$i : G \to \mathbb{Z}$$. In other words, we discuss $$G$$ together with a specific way that $$G$$ is a subgroup of $$\mathbb{Z}$$.

This makes it clear that $$\phi(G)$$ is not really just a proposition about $$G$$ - it's also about the way that $$G$$ is a subgroup of $$\mathbb{Z}$$. So it's really a proposition $$\phi(G, i : G \to \mathbb{Z})$$. Because $$\phi$$ depends on a secondary variable $$i$$, we see that we cannot apply our Big Theorem to conclude that $$\phi(\mathbb{Z}) \iff \phi(2 \mathbb{Z})$$.

If it were the case that the isomorphism $$w : \mathbb{Z} \to 2 \mathbb{Z}$$ "played nicely" with the inclusion maps $$i_1 : \mathbb{Z} \to \mathbb{Z}$$, $$i_2 : 2 \mathbb{Z} \to \mathbb{Z}$$ (that is, if $$i_1 = i_2 \circ w$$), then we would be able to conclude that $$\phi(\mathbb{Z}, i_1) \iff \phi(2 \mathbb{Z}, i_2)$$ using a generalisation of our Big Theorem. But of course the isomorphism does not play nicely with the inclusion maps.

*An astute observer will note that this is exactly the language of category theory. A "nice proposition about sets" is just some statement about the category of sets (which avoids discussing equality of objects).

**Technically, we need a way to define $$G^2$$ first. This is done using the categorical definition of the universal property of the product.

• Why is this answer being downvoted? Dec 27, 2021 at 18:31

This is just an elaboration of Arthur's answer elsewhere in this thread. I thought you might like to see a specific example.

Consider $$G = \color{maroon}{\Bbb Z_2}\times \color{darkblue}{\Bbb Z_4}.$$

$$G$$ has a subgroup $$A$$ that is generated by $$\langle 1, 0\rangle$$:

$$\def\elt#1#2{\langle{#1},{#2}\rangle}A = \{\elt10, \elt 00\}$$

and another subgroup $$B$$ that is generated by $$\langle 0, 2\rangle$$:

$$B=\{\elt00, \elt 02\}$$

Both $$A$$ and $$B$$ are isomorphic to $$\Bbb Z_2$$, and so to each other. But the quotients $$G/A$$ and $$G/B$$ are not isomorphic. $$G/A$$ is like taking $$G$$, ignoring the first component, and keeping the second component intact. The result is $$G/A \simeq \color{maroon}{\Bbb Z_1}\times\color{darkblue}{\Bbb Z_4}.$$ $$G/B$$ is like taking $$G$$ and keeping the first component but ignoring everything but the parity of the second component. The result is $$G/B \simeq \color{maroon}{\Bbb Z_2}\times\color{darkblue}{\Bbb Z_2}.$$ Even though $$A$$ and $$B$$ are isomorphic, $$G/A$$ and $$G/B$$ are not isomorphic. (In particular, $$G/A$$ is cyclic and $$G/B$$ isn't.)

Each component of $$G = \color{maroon}{\Bbb Z_2}\times \color{darkblue}{\Bbb Z_4}$$ contains a factor of $$\Bbb Z_2$$. Each of $$A$$ and $$B$$ is isomorphic to $$\Bbb Z_2$$. When we construct the quotient of $$G$$ by $$A$$ or $$B$$, we get a different result depending on which component we divide the factor from.

Two groups that are isomorphic share the same internal structure. But a quotient $$G/N$$ doesn't depend only on the internal structures of $$G$$ and $$N$$. The quotient also depends on the relationship between $$G$$ and $$N$$. So even though $$A$$ and $$B$$ have the same internal structure, $$G/A$$ and $$G/B$$ are different, because those structures reside inside of $$G$$ in different ways.

Topological and illustrated version of Arthur's nice answer:

A trefoil knot $$K_3$$ in $$\Bbb R^3$$ cannot be unknotted but it can be in $$\Bbb R^4$$. Topologically a trefoil knot $$K_3$$ and a circle are homeomorphic i.e. one can deform one to other without cutting or gluing. So these two (one in $$\Bbb R^3$$ and one in $$\Bbb R^4$$) are the same but one uncovers a branch of mathematics known as Knot_theory while other one is fruitless from this point of view and application.

I think theses type of "same things" is a categorical notion but I am not sure.

same is true for $$y=x$$ and $$y=\exp(x)$$; both topologically are homeomorphic to $$\Bbb R$$, but one contains only positive numbers. One can say similar properties by considering them as additive and multiplicative groups.

These might be of interest to you:

Let me for convenience, first, assume all subgroups are normal, or that we are in an abelian group.

We form quotients of a group by its SUBgroups, not by another abstract GROUP. Subgroup means a SPECIFIC subset (which is closed ....) As groups Z and 2Z are "same". But as subsets of Z they are different.

Now to non-abelian case. Take $$G= \{+1,-1\}\times S_3$$ where the first factor $$H$$ is a group (of order 2) wrt multiplication of numbers.

This group $$G$$ has more subgroups of order 2 (isomorphic to H as a group) namely ones generated by a transposition in the second factor $$S_3$$. These subgroups are NOT even normal in $$G$$. SO quotient does not even exist by these subgroups.


Seperating $2n$ elements with subsets of size $n$

Suppose I have a set $X$ with $2n$ elements ($n$ nonzero natural number). I now want to find a collection of subsets $X_1, \ldots, X_k$ of $X$ such that

• Every $X_i$ contains $n$ elements.
• For every $x, y \in X$, there exists an $X_i$ such that $x \in X_i$ and $y \not\in X_i$ or $x \not\in X_i$ and $y \in X_i$ (the $X_i$ "seperate" the elements of $X$, in that the topology on $X$ generated by the $X_i$ is Kolmogorov)
• $k$ is minimal with respect to the above to properties.

An algorithm to find $X_1, \ldots, X_k$ would be great, but I am mostly looking for a way to calculate $k$. My intuition tells me that $k$ should be at least $\log_2(2n)$, more specifically I feel like $$\max_{x \in X} \# \lbrace y \in X \mid \not\exists i \in \lbrace 1, \ldots, j \rbrace : x \in X_i \enspace \& \enspace y \not\in X_i \rbrace \geq \log_2(\frac{2n}{j})$$ should hold for all $j \in \lbrace 1, \ldots, k \rbrace$, but I don't know how to prove this.

• Consider sets with $2^l$ elements first – Scipio Jun 4 '17 at 12:09
• I did; even if $n=4$ I don't see how to proceed. (i.e. I know that $k = 2$ is possible, but do not see how to show that $k = 1$ is impossible) – Bib-lost Jun 4 '17 at 12:11
• $X_1$ contains two elements $a,b$; clearly the second requirement doesn't hold for these two elements if you only have a single set. – Scipio Jun 4 '17 at 12:18
• By $n=4$ I meant the case where $X$ has $2*4 = 8$ elements. The case with $4$ elements is indeed clear. :) And surely, also the case with $8$ elements can be computed by hand in a reasonably short time; what I meant is that I do not see an argument which could somehow be generalised. – Bib-lost Jun 4 '17 at 12:44
• Could you add the cases $n=2,3,4$ into your question & give us the lists $X_1, X_2 , \cdots , X_k$ ? – Donald Splutterwit Jun 4 '17 at 12:50

Suppose $X$ contains $2^l$ elements. Let $(A_i^m)$ be the family of subsets of $X$ of non-separated elements after introducing $X_1, \cdots, X_m$, i.e. sets of elements for which the second requirement does not hold (check that these sets are well defined and uniquely partition $X$). Ultimately, we want that all the subsets $A_i$ contain just a single element (i.e. every element is separated from all other elements).

With each $X_i$ we want to separate as many elements as possible. Now, at first $A_1 = X$ containing $2^l$ elements and we take $X_1$ just an arbitrary set of $n$ elements. After this, we have sets $A_1$, $A_2$, both containing $2^{l-1}$ elements. Now, with $X_2$ we can again split both in half, resulting in four 'remaining sets' with $2^{l-2}$ elements. Continue recursively to see that $k=l$. Moreover, it is clear that after introducing $m$ sets, $\max_i \# A_i^m \geq 2^{l-m}$, so that our result is indeed optimal.

For example, start with $X = A_1^0 = \{1,2,3,4,5,6,7,8\}$. Introduce $X_1 = \{1,2,3,4\}$ to get $A_1^1 = \{1,2,3,4\}, A_2^1 = \{5,6,7,8\}$. Introduce $X_2 = \{1,2,5,6\}$ and $X_3 = \{1,3,5,7\}$ to finish the job. (Make sure to check the corresponding sets $(A_i^m)$.)

Now, for any other number of elements $2n$, your 'splits' wil not be optimal. You can check that $\max_i \# A_i^m \geq 2n\cdot2^{-m}$ still holds.

• So probably we have that, in general, $k$ is the unique integer satisfying $2^{k-1} < 2n \leq 2^{k}$? – Bib-lost Jun 4 '17 at 13:57
• yes exactly :) This hinges on the very last statement in my answer, so make sure to check that carefully. – Scipio Jun 4 '17 at 14:01

There should be $k=\lceil log_2(2n) \rceil$ partitions, $2k$ sets.

We have the conditions: $$\mathcal{P}_1:\forall X_i, \text{num}\{X_i\}=n$$ $$\mathcal{P}_2:\forall x_i,x_j \exists X_i, x_i \in X_i, x_j \not\in X_i$$

It is clear for sets of $2n=2^k$ elements, the partitions are binary, hence optimal. Under this case, the condition $\mathcal{P}_2$ implies the stronger: $$\mathcal{P}_{2b}:\forall x_i,x_j \exists X_i,X_j, x_i \in X_i, x_j \in X_j$$ because each partition requires its complement. If there is not complement, the property is not met. $$\forall X_i \exists Y_i=X -X_i$$

If the set contains $2n=2^k+2m$ elements, $2m<2^k$, each $2m$ remaining elements shall be included into different disjoint partitions. Again in this case the partition complement must exist and the condition $\mathcal{P}_{2}$ will not introduce any benefit against $\mathcal{P}_{2b}$.

• Don't you mean $k = \lceil \log_2(2n) \rceil$? Surely if $n=3$ you need at least $2 = \lceil log_2(6) \rceil$ partitions? – Bib-lost Jun 4 '17 at 14:14


# Special case of Bertrand Paradox or just a mistake?

I've been working on a question and it seems I have obtained a paradoxical answer. Odds are I've just committed a mistake somewhere, however, I will elucidate the question and my solution just in case anyone is interested.

I want to know what is the average distance between two points on a circle of radius 1 where we consider only the boundary points.

My attempt is as follows:

Consider a segment of the diameter x which is uniformly distributed between 0 and 2. Then you can calculate the distance between the points (2,0) and the point determined by x just by elementary geometry as this picture shows:

enter image description here

Here in the picture, the green segment is the geometric mean and the orange one is the distance whose distribution we want to know. Just by calculating the expected value, we obtain:

$$E\left(\sqrt{(4-2X)}\right) = \int_{0}^{2} \sqrt{(4-2x)}\cdot\frac{1}{2} dx = 1.333.... = \frac{4}{3}$$

Where $$\sqrt{(4-2x)}$$ is the transformation of the random variable and $$\frac{1}{2}$$ is the pdf of a uniform distribution $$[0,2]$$.

Also, if we derive the pdf of the transformation we obtain the same result:

$$y = \sqrt{(4-2x)} , x = 2- \frac{y^2}{2}, \mid\frac{d}{dy}x\mid = y$$

$$g(y)=f(x)\cdot\mid\frac{d}{dy}x\mid = \frac{1}{2}\cdot y$$

$$E(Y)= \int_{0}^{2}y\cdot\frac{1}{2}\cdot y dy = 1.333.... = \frac{4}{3}$$

I have seen a different approach somewhere else where the distribution of the angle is considered as a uniform distribution between 0 and $$\pi$$ and the final result was:

$$1.27... = \frac{4}{\pi}$$

That's pretty much the problem I found. Maybe I just did it wrong in some step but it all makes sense to me. I know this is not exactly what we refer as Bertrand paradox but it just suggests something like that because both problems handle with segments in circumference and maybe my result is wrong because it does not hold for rotations of the circle or something like that (I read a little bit about the Bertrand's Paradox).

That's pretty much it. Also sorry for my bad English and maybe I'm also wrong in something pretty elemental since I've just started learning about probability theory. It's also my first post so I will try to improve my exposition and LateX use in the following ones.

• welcome to MSE. I have attempted to edit your question. In case you find any discrepancy with your idea please check. Also, you can look up the edits I have made that will help you for your future questions. Thank you! – Mann Apr 19 at 21:57

This is a very nice question, well-written and researched before posting. You have clearly put a lot of careful into your question and it is very much appreciated on this site (and I thought your English reads perfectly naturally). Thanks so much for taking the time to ask a serious and well-considered question. I hope this answer comes close to meeting the high standard of quality set by your question.

I don’t see any calculation errors on your part, and I’ll wager you checked those thoroughly. The flaw is something much more subtle and is related to the initial framing of the problem.

You have correctly solved the problem of “what is the average distance between the rightmost point of a circle and another point projected upwards from a randomly chosen point on the diameter”. But this doesn’t quite capture the same probability distribution as “rightmost point and another randomly chosen point on the circle” (which is sufficient to capture the dynamics of “two random points on the circle”).

There is a hidden non-uniformity in the projection process, and that’s because the circle has varying slope so the projection hits the circle differently at different positions.

Here’s an experiment that should help convince you that uniform distribution on the diameter does not yield uniform distribution on the circle boundary. Fundamentally, a uniform distribution should treat all arcs of the same length equally: a random point has a $$1/4$$ chance of falling on a quarter-circular arc, regardless of which quarter circle that is.

Now compare two specific quarter-circles and their projections onto the diameter: 1) the quarter-circle adjacent to $$(2,0)$$, i.e. 12 o’clock to 3 o’clock, and 2) the quarter-circle centered around the top-most point $$(1,1)$$, i.e. 10:30 to 1:30.

These both project onto the diameter with no self-overlap (avoiding double counting). But the first one projects down to the interval $$[0,1]$$ which has length $$1$$, and the second one projects down to $$[-\sqrt{2}/2,\sqrt{2}/2]$$, which has length $$\sqrt2$$.

This should indicate to you that the two distributions are not equivalent (though they may intuitively seem so). Consequently, the random process of choosing uniformly from the diameter and projecting onto the circle results in disproportionately favoring the top and bottom sectors of the circle over the left and right.

• Thanks! I really appreciate your words and yes, it seems the problem really deals with something subtle. I'm going to add some considerations in an answer. – Hyz Apr 19 at 19:49

Thanks, Erick Wong for your feedback. After your answer, I calculated the distribution of the arc length subject to the uniform distribution of the point on the diameter. In fact: if we want to express the arc length $$l$$ as a function of $$x$$, $$l = f(x)$$ we obtain:

$$l = \arccos(1-x), x = 1-\cos{l}, |\frac{d}{dl}(x)| = \sin{l}$$

$$l_{pdf} = x_{pdf} \cdot |\frac{d}{dl}x| = \frac{1}{2}\cdot \sin{l}$$.

So the arc length does not distribute uniformly, we have "lost it", we might say. That's what was wrong. For instance, if the arc length obeys a uniform distribution [0, $$\pi$$], then we can calculate the segment $$s$$ as a function of the arc length:

We know $$l = f(x)$$ and want to know $$s = h(l)$$. If we calculate $$s = g(x)$$ we're done:

From the image on the question I post, $$s = g(x)=\sqrt{2x}$$ (or the opposite segment $$\sqrt{4-2x}$$ ) then $$h = g \circ f^{-1}$$, $$s = \sqrt{2(1-cosl)}$$

$$E(s) = \int_0^\pi{\sqrt{2(1-cosl)}\frac{1}{\pi}dl}=1.273... = \frac{4}{\pi}$$

Also the pdf:

$$s = h(l) = \sqrt{2(1-cosl)} , l=h^{-1}(s)= 2\cdot \arcsin(\frac{s}{2}), |\frac{d}{ds}h^{-1}|=\frac{2}{\sqrt{4-s^2}}$$

$$s_{pdf} = \frac{1}{\pi} \frac{2}{\sqrt{4-s^2}}$$

$$E(s) = \int_0^2{s\cdot \frac{1}{\pi} \frac{2}{\sqrt{4-s^2}}ds} = 1.273... = \frac{4}{\pi}$$

So we're done. Also, the pdf of the segment suggests something related to the Cauchy distribution. Not exactly but certainly it has to do something with it. If we read the description of Cauchy distribution in Wolfram MathWorld:

"The Cauchy distribution, also called the Lorentzian distribution or Lorentz distribution, is a continuous distribution describing resonance behavior. It also describes the distribution of horizontal distances at which a line segment tilted at a random angle cuts the x-axis."

And that's it. A really fascinating problem that introduces some subtle ideas of probability theory. If anyone knows something else please give me feedback. I really think there's a nice connection with Cauchy distribution.


# Does the sequence $x_{n+1}=x_n+x_n^2$ converge to $0$ whenever $-1\lt x_0\lt0$?

My question concerns using the fixed-point iteration to find the fixed point of the function $$f(x)=x+x^2=x(1+x)$$ (this function has a single fixed point at $$0$$).

# The problem

Given some fixed $$x_0$$, define the sequence $$x_n$$ by \begin{align*} x_{n+1}=x_n+x_n^2&&\text{for all n\geq0.} \end{align*} Does this sequence converge to $$0$$ for all $$x_0$$ in the range $$-1\lt x_0\lt0$$? After some computational numerical exploration, I think the answer might be yes but I'm not so sure how to prove this.

## Some notes

The sequence trivially converges to $$0$$ when $$x_0=-1$$ or $$x_0=0$$. It's also fairly easy to prove that it does not converge to $$0$$ whenever $$x_0$$ lies outside this range (i.e., $$x_0\lt-1$$ or $$x_0\gt0$$).

I've already proved that for any $$x_0$$ in this range, all successive $$x_n$$s remain in this range (i.e., $$-1\lt x_n\lt0$$ for all $$n\geq0$$) and that the sequence increases / gets closer to $$0$$ (i.e., $$x_{n+1}\gt x_n$$), but is there any way to prove it actually converges to $$0$$?

## Some examples

If, for example, $$-\frac{1}{2}\lt x_n\lt-\frac{1}{4}$$, we'll have $$-\frac{3}{8}\lt x_{n+1}\lt-\frac{1}{8}$$ and $$-\frac{21}{64}\lt x_{n+2}\lt-\frac{5}{64}$$, etc. So it seems as though the range of possible values gets closer to $$0$$. But, how would one prove this?

You can use the monotone convergence theorem. It tells us that a sequence $$\{x_n\}_{n=1}^\infty\subseteq \mathbb R$$ converges if it is bounded and decreasing. For you sequence we have $$|x_{n+1}| = |x_n| |1+x_n| \leq 1$$ since $$x_0 \in (-1,0)$$. And likewise we have $$x_{n+1} = x_n(1+x_n) \leq x_n$$ as $$1+x_n \in (0,1)$$.

(You can use induction to prove both these claims.)

Now you have shown that $$\lim_{n\to\infty} x_n=x$$ exists. From this we see that $$x = \lim_{n\to\infty} x_{n+1} = \lim_{n\to\infty} x_n(1+x_n) = x(1+x).$$ Hence $$x$$ satisfies $$x = x^2+x$$ so $$x=0$$.

• Great, exactly what I was looking for! Apr 16, 2021 at 7:15
• @ΑΘΩ You're right, I should have said increasing and replaced $\leq$ with $\geq$. Thanks. Apr 16, 2021 at 8:05
• I've already proved that $x_n$ is monotonic increasing and bounded above by $0$ in this case. So this is all I needed to complete the argument (from what I understand, I've never formally studied real analysis, so I certainly may be missing something). Apr 16, 2021 at 8:06
• @user178563 Glad to see that you agree. If we are to be very precise (which we should very well be, for this is what this forum is about) the part where it appears you want to show that $|x_n| \leqslant 1$ by induction on $n$ is also dubious (the induction is not spelled out clearly).
– ΑΘΩ
Apr 16, 2021 at 8:08

The answer to your question (regarding convergence to $$0$$) is indeed affirmative and for a reason that is not too difficult to ascertain. For simplicity let us introduce the quadratic polynomial function: \begin{align*} f \colon \mathbb{R} &\to \mathbb{R} \\ f(x)&=x^2+x \end{align*} and let us note that $$f[(-1, 0)] \subseteq (-1, 0)$$, in other words the interval $$(-1, 0)$$ is stable under $$f$$. To prove this latter claim it suffices to notice that $$x^2+x+1=\left(x+\frac{1}{2}\right)^2+\frac{3}{4} \geqslant \frac{3}{4}>0$$ for any $$x \in \mathbb{R}$$ (and thus in particular $$x^2+x>-1$$ always) and respectively that given $$x \in (-1, 0)$$ we have $$x+1>0$$ yet $$x<0$$ which means that $$x^2+x=x(x+1)<0$$.

The immediate conclusion that we draw from this is that for any fixed $$t \in (-1, 0)$$, the unique recursive sequence $$x \in \mathbb{R}^{\mathbb{N}}$$ defined by $$x_0=t$$ and $$x_{n+1}=f(x_n)$$ for any $$n \in \mathbb{N}$$ has the property that actually $$x \in (-1, 0)^{\mathbb{N}}$$ and is thus a bounded sequence. Furthermore, since $$x_{n+1}=x_n+x_n^2 \geqslant x_n$$ takes place for every $$n \in \mathbb{N}$$ we gather that $$x$$ is increasing (one can actually make the more precise observation that since $$x_n \neq 0$$ for all $$n \in \mathbb{N}$$ under the stated conditions, the inequality mentioned above is actually strict, rendering $$x$$ into a strictly increasing sequence, however that is not essential for the study of convergence).

According to one of the theorems of Weierstrass, any upper-bounded increasing sequence of real numbers is convergent, which applies in particular to $$x$$. Since $$x$$ is defined recursively with respect to the continuous function $$f$$, it follows with necessity that $$\displaystyle\lim_{n \to \infty} x_n$$ must be a fixed point for $$f$$. However it is obvious that there is just one such fixed point - the unique solution to the equation $$f(u)=u \Leftrightarrow u^2=0$$ - which is $$0$$.


# Q: Is 0.9999… repeating really equal to 1?

Mathematician: Usually, in math, there are lots of ways of writing the same thing. For instance:

$\frac{1}{4}$ = $0.25$ = $\frac{1}{\frac{1}{1/4}}$ = $\frac{73}{292}$ = $(\int_{0}^{\infty} \frac{\sin(x)}{\pi x} dx)^2$

As it so happens, 0.9999… repeating is just another way of writing one. A slick way to see this is to use:

$0.9999... = (9*0.9999...) / 9 = ((10-1) 0.9999...) / 9$

$= (10*0.9999... - 0.9999...) / 9$

$= (9.9999... - 0.9999...) / 9$

$= (9 + 0.9999... - 0.9999...) / 9 = 9 / 9 = 1$

One.

Another approach, that makes it a bit clearer what is going on, is to consider limits. Let’s define:

$p_{1} = 0.9$
$p_{2} = 0.99$
$p_{3} = 0.999$
$p_{4} = 0.9999$

and so on.

Now, our number $0.9999...$ is bigger than $p_{n}$ for every n, since our number has an infinite number of 9’s, whereas $p_{n}$ always has a finite number, so we can write:

$p_{n} < 0.9999... \le 1$ for all n.

Taking 1 and subtracting all parts of the equation from it gives:

$1-p_{n} > 1-0.9999... \ge 0$

Then, we observe that:

$1 - p_{n} = 1 - 0.99...999 = 0.00...001 = \frac{1}{10^n}$
and hence

$\frac{1}{10^n} > 1-0.9999... \ge 0$.

But we can make the left hand side into as small a positive number as we like by making n sufficiently large. That implies that 1-0.9999… must be smaller than every positive number. At the same time though, it must also be at least as big as zero, since 0.9999… is clearly not bigger than 1. Hence, the only possibility is that

$1-0.9999... = 0$

and therefore that

$0.9999... = 1$.

What we see here is that 0.9999… is closer to 1 than any real number (since we showed that 1-0.9999… must be smaller than every positive number). This is intuitive given the infinitely repeating 9’s. But since there aren’t any numbers “between” 1 and all the real numbers less than 1, that means that 0.9999… can’t help but being exactly 1.

Update: As one commenter pointed out, I am assuming in this article certain things about 0.9999…. In particular, I am assuming that you already believe that it is a real number (or, if you like, that it has a few properties that we generally assume that numbers have). If you don’t believe this about 0.9999… or would like to see a discussion of these issues, you can check this out.

This entry was posted in -- By the Mathematician, Equations, Math. Bookmark the permalink.

### 94 Responses to Q: Is 0.9999… repeating really equal to 1?

1. mathman says:

lol. You need to know what each “number” is before you can take the difference. Of course the difference is zero, because you say that they are the same before you subtract, assuming you are subtracting “numbers”

2. George says:

While the mathematical argument presented here is correct, I have to say that limit and/or mathematical infinity introduced created confusions for many. What I wanted to say is that math infinity is really a math trick. It cannot and should not be real in the real physical world. Today this infinity plagues physics and its calculations. Maybe we need something new…

3. EMINEM says:

go to this https://www.youtube.com/watch?v=wsOXvQn3JuE site and get a mathematical proof that 0.999… doesnot equal to 1….

4. . says:

EMINEM, the video was published on April fool’s day for a reason!

5. Josh says:

The simplest explanation I’ve seen for this is:

1/3 = 0.333…
(1/3)3=1 = (0.333…)3=0.999…
1 = 0.999…

I don’t see how anyone can come to any conclusion other than 1=0.999… after understanding this.

6. mathman says:

1 / 3 = ??
Use long division. There are 2 parts: the quotient and the remainder. For every iteration.
First iteration is 0 R 1/3
Second is 0.3 R 1/30
Third is 0.33 R 1/300
Etc etc.

Where does your remainder go, Josh?
You don’t account for it anywhere.

The remainder is never zero.

How many 9’s are there in your 0.999… ?

7. Josh says:

Hey mathman,

There are infinitely many 9s in my “0.999…”. If it were anything other than infinitely many, then it would be less than 1.

And as far as 1/3, your “etc. etc.” is an infinitely repeating “0.333…”. I’m failing to see how anything you’ve presented has disproved what I said.

8. Angel says:

@mathman:

The remainder does not need to be accounted for. The division 1/3 is never formally complete if there is a remainder. Therefore, one must keep iterating the operations in long division. As the number of iterations n approaches infinity, the remainder does approach zero. That is what matters here. The number of 3s in the decimal expansion of 1/3 is infinite, and so is the number of 9s in the figure 0.999…

9. Anthony.R.Brown says:

The Original author of the Proof below…used in this Thread 🙂

http://www.mathisfunforum.com/viewtopic.php?id=6069

It’s impossible for Infinite/Recurring 0.9 (0.999…) to ever equal 1

Because of the Infinite/Recurring 0.1 (0.001…) Difference! 🙂

Only by the using the + Calculation can it be possible (0.999…) + (0.001…) = 1

Anthony.R.Brown

10. Ángel Méndez Rivera says:

Anthony R. Brown,

Your argument is flawed for the simple reason that the “infinite difference” you describe equivalent to 0.000…1 is not a number. It doesn’t exist in mathematics. It can’t exist in mathematics. It is more of a problematic number than 0/0 or defining a left-identity of exponentiation Phi such that Phi^x=x. It is simply nonsense.

11. Mathman says:

Anthony,
Being someone who has researched and understands both sides of this issue, you will be told it’s nonsense by mathematicians. It’s their way of putting themselves above you to put you down. The issue with your argument is that 0.001… is defined as 0.00111111111111…. You are having a problem defining what the idea in your head is. 0.000…1 is also not a number. Mathematicians for some reason cannot visualize repeated division the same as they can visualize it in addition. There quick to use the word nonsense, and then in the next breath say that by adding infinite numbers they can have a finite result.

12. Anthony.R.Brown says:

Hi Ángel Méndez Rivera

Your try at a counter argument…saying (0.001…) does not exist in mathematics ? can’t exist in mathematics if (0.999…) exist’s which it does! 🙂
The two are the related from the calculation 1 – (0.999…) = (0.001…) one cannot exist without the other!

Anthony.R.Brown

13. Anthony.R.Brown says:

Anthony,
Being someone who has researched and understands both sides of this issue, you will be told it’s nonsense by mathematicians. It’s their way of putting themselves above you to put you down. The issue with your argument is that 0.001… is defined as 0.00111111111111…. You are having a problem defining what the idea in your head is. 0.000…1 is also not a number. Mathematicians for some reason cannot visualize repeated division the same as they can visualize it in addition. There quick to use the word nonsense, and then in the next breath say that by adding infinite numbers they can have a finite result.

Then when I wanted to reply the message above is not shown ? so I will Answer below…

First of all what you are showing as 0.00111111111111…. is wrong! there is only a single (1) following the Infinite/recurring calculation so your example above should look like 0.00000000000001….
The amount of zeros depends on how far one wants to show the calculation regarding decimal places!

Example 0.9… has the infinite/recurring difference 0.1…
Example 0.99… has the infinite/recurring difference 0.01…
Example 0.99… has the infinite/recurring difference 0.001…

So as you can see the infinite/recurring difference is always the same length as the infinite/recurring 0.9 🙂

Anthony.

14. Mathman says:

0.001….. = 0.0011111111111111111…..
Anthony do you know what you are taking about?

15. Anthony.R.Brown says:

Hi Mathman

Yes I fully know what I am talking about but you appear not to ?

As I pointed out in your example ? = 0.0011111111111111111….. it is wrong there can only be a single (1) otherwise when you do the calculation…

0.0011111111111111111….. + 0.0000000000000000009….. it will equal something like = 0.0011111201111111111… ?

Where as the True calculation (with both sides the 0.9… and the 0.1… the same amount of decimal places)

Calculates 0.9999999999…. + 0.0000000001…. = 1

Regards Anthony.

16. Joshua says:

The flaw in your logic is thinking that there is ever space for a 0.0…01. The number 0.999… repeating forever by definition has no room to add anything to it. The 9’s never stop to allow the insertion of any variation of 0.0001. And if there is no room to ever add anything to 0.999… infinitely repeating, then there is no space in between 0.999… infinitely repeating and 1, and therefore they are equal. If the 9’s stopped repeating at any point, you would be correct, but they do not stop repeating, so you are incorrect.

17. Mathman says:

The ellipsis “…” mean “repeating”. I know what you are trying to say but you are using symbols incorrectly.
1/1000… would be a more accurate to convey your idea.

18. Anthony.R.Brown says:

Hi Joshua

Your Quote:”The number 0.999… repeating forever by definition has no room to add anything to it”

A.R.B

The same as “The number 0.001… repeating forever by definition has no room to add anything to it”

They both run side by side! 🙂

0.99999999999999999999999999999999999999999999999999999999….
0.0000000000000000000000000000000000000000000000000001….

It’s only when one want’s to make the Value (1) that the + Calculation is needed it’s the only way!

( 0.9999999999….) + (0.0000000001….) = 1

19. Joshua says:

Sorry Anthony, you have not wrapped your head around what 0.999… repeating for infinity means. It’s the infinite part that’s really important. I promise you, you are wrong, and once you understand what infinitely repeating means, you’ll get it.

20. Ángel Méndez Rivera says:

Anthony, no. They don’t run side by side. The number you describe as the difference between 1 and 0.999… doesn’t exist, because by definition, it is impossible to have a 1 after an infinite amount of zeros. It just is. There is no such calculation as the one you performed to be made because the number you describe doesn’t exist.

21. Mathman says:

There is an undefined infinitesimal space between 0.999… and 1. Just because it’s undefined doesn’t mean it doesn’t exist.

22. Ángel says:

Mathman,

1. If it is undefined, then it does not exist. One implies the other.
2. No, there is no infinitesimal space between 0.999… = 1, and this can be proven.

0.999… = 9/10 + 9/100 + 9/1000 + … this is an infinite geometric series, which is evaluated and in fact defined by the formula a/(1 – r), so the sum 0.999.. = (9/10)/(1 – 1/10) = (9/10)/(9/10) = 1. As for the formula a/(1 – r), it can be easily derived and proven

23. Anthony.R.Brown says:

( 1.1 ) x 0.9 = 0.99 ” One Decimal Place = ” 0.01 < 1

( 1.11 ) x 0.9 = 0.999 " Two Decimal Place's = " 0.001 < 1

( 1.111 ) x 0.9 = 0.9999 " Three Decimal Place's = " 0.0001 < 1

( 1 / 0.9 ) x 0.9 = 0.9… " Infinite Decimal Place's = " 0.1… < 1 "

24. Joshua says:

Mathman,

You would be correct if the repeating 0.999… ever stopped, no matter how long it was. But the number that we are talking about literally never stops, and therefore does not leave any space. There is absolutley no number that would fit in between it and 1.

Most of the confusion over this matter seems to be about the nature of infinity, and how it applies to infinitely repeating decimals. The importance of wrapping our heads around what “infinitely repeating” means cannot be overstated.

25. Ángel Méndez Rivera says:

Anthony,

(1.1)(0.9) = 0.99

(1.11)(0.9) = 0.999

(1.111)(0.9) = 0.9999

(1.111…)(0.9) = 0.999…. = (1/0.9)(0.9), because 1.1111… = 1/0.9. But any number divided by itself equals 1, so (0.9)/(0.9) = 1, therefore 1 = 0.9999….

26. Anthony.R.Brown says:

Quote: “Joshua says:
You would be correct if the repeating 0.999… ever stopped, no matter how long it was. But the number that we are talking about literally never stops, and therefore does not leave any space. There is absolutley no number that would fit in between it and 1.”

A.R.B

Absolutely Correct!!! So now you and I agree it’s impossible for 0.999…to ever stop and become equal to 1 🙂

So how do we make 0.999… = 1 and that’s where we have the answer!…

( 0.999…) + (0.001…) = 1

{When great minds think alike Mountains can be moved!} 🙂

27. Joshua says:

No, Anthony, you still don’t understand. The fact that it never stops is the very thing that makes it equal to 1. Again, I come back to encouraging you to take the time to actually understand what an infinitely repeating decimal is.

28. Anthony.R.Brown says:

Hi Joshua

What you don’t seem to understand is that infinite/recurring 0.999… is a Calculation on a Number! and the number Starts as 0.9 (It’s not something that is just plucked from the sky from nothing!)

Your biggest mistake! is “The fact that it never stops is the very thing that makes it equal to 1” ? a Number that Starts < 1 can never be made to equal 1 unless the + Calculation is used!

29. Joshua says:

Anthony, the number 0.999… with infinite 9s is not a calculation. It is a static number. The 9s are not constantly being generated, they exist (all infinity of them) at once. It does not “start” at 0.9. It exists as 0.999… with an infinite number of 9s. You don’t have to work up to it, it just is.

It is clear that you still do not understand what an infinitely repeating decimal is. You do not understand the basic premise of what we are talking about. Therefore, you can not speak to it in an intelligent or useful manner.

I’m not trying to be mean or anything, either. I appreciate people pointing out when I am wrong so I can learn what is correct. You clearly don’t understand the basics of this topic, therefore you are wrong, and it is an opportunity for you to learn something new and broaden your knowledge base.

30. Angel says:

“What you don’t seem to understand is that recurring 0.999… is a calculation on a number!”

This is wrong. 0.999… is a number. It isn’t a process, it already is a number. It doesn’t “start” out as 0.9. It already is and exists as 0.999…, it doesn’t start anywhere or end anywhere. It isn’t a changing or moving thing. It just is a number.

31. Anthony.R.Brown says:

So Joshua & Angel…

Are both wrong! if they think (the number 0.999… with infinite 9s is not a calculation) ???

Because all the so called Proofs ? are based on some kind’s of Calculations!!!

Just to back that up…a Google search for 0.999… proofs shows (About 367,000 results) 21/09/2017

And they are all some kind’s of Calculations! 🙂

One fine example is shown below…with many Calculations!

Anthony.R.Brown

32. Mathman says:

Ángel,
a/(1-r) comes from a limit. The LIMIT of the infinite series is 1. The definition is flawed and contradicts the very theory of limits.

It may be impossible to have a 1 AFTER an “infinite amount”. That is because “infinite amount” doesn’t exist. Stop trying to quantify infinity. It is not impossible to describe the idea, however: 1/1000…

The very definition of limits show there is ALWAYS AN EPSILON.
n=1: 1-0.9= 0.1
n=2: 1-0.99 = 0.01
For ALL values n, there is a difference, or, contradictory to Anthony, there is always a space. If you travserse 90% of the remaining distance , ad infinitum, there will always be 10% of the last distance ahead of you. You’ll never complete the journey of say, 1 mile, going 90% each time.

I argue that “0.999…” is not a number at all.
Σ (n=1 to N) 9/10^n as N goes to infinity, implies there is always a rational number being added to the previous total. The number 1 has nothing being addd to it. Therefore they aren’t the same.
Also, ‘something’ that always has another addend can’t be a number. It’s contradictory because it always has another number being added to it.

33. Mathman says:

Thus: 1-0.999(n-times) = 1/10^n
You may let n get as arbitrarily large as you like.
You cannot have “infinite amount of 9’s” though.
You can have the LIMIT as the number of 9’s gets arbitrarily large. This is a number. “Endless 9’s” really is nonsense when it comes to a number which should really be finite.
So then, for ALL n in N, that is, for every integer, there is a difference which is smaller than the last and never 0.
Learn what arbitrarily large means, and how limits really work, and what infinite means, and how infinity is applied in calculus. Infinity and limits go hand in hand. You don’t plop infinity wherever you feel.

34. Anthony.R.Brown says:

LIMIT Cannot be used with Infinite/Recurring 0.9 (0.999…) because you will then Contradict the term (Infinite/Recurring) by forcing the .9’s to cut off at some point depending on where you want the LIMIT to be, regarding decimal places calculated,this is much the same as the cut off point used when Infinite/Recurring 0.9 (0.999…) is forced at some point to become equal to 1

So back to…

It’s impossible for Infinite/Recurring 0.9 (0.999…) to ever equal 1 (Unless you use a LIMIT) ? 🙁
And the Infinite/Recurring 0.1 (0.001…) Difference! (Also does not us a LIMIT)

Both are LIMITless 🙂

Only by the using the + Calculation can it be possible for (0.999…) + (0.001…) = 1

35. Ángel Méndez Rivera says:

No, we do not get to choose where the limit is. The limit is a number larger than every element of the sequence of partial evaluations of a function, which means that the limit is larger than the sequence itself. In other words, the term of the sequence which equals the limit is itself located at an infinite position of the sequence. Any finite evaluation yields less than the limit. However, 0.99… has an infinite number of 9s, so it exactly equals the limit. We’re not cutting it off to equal 1, you’re cutting it off to equal less than 1, but you’re Not actually using the sequence of infinite nines, so you’re argument is wrong.

There is no such a thing as limitless in mathematics, and 0.000…1 is nonsense and does not exist.

36. Ángel Méndez Rivera says:

Mathman,

Read my previous response to your argument. I already explained why your description of the definition of a limit is wrong and why the idea of there always being an epsilon is irrelevant when actually evaluating the limit of a function as the argument approaches c.

37. Anthony.R.Brown says:

Quote: “Ángel Méndez Rivera says:
September 22, 2017 at 1:50 pm
No, we do not get to choose where the limit is.

A.R.B
Yes! you do by using it in the first place!

Quote: “Ángel Méndez Rivera says:
“However, 0.99… has an infinite number of 9s, so it exactly equals the limit. ”

A.R.B
How can an infinite number of 9s, exactly equals the limit (You have set!) which is 1 ? which is an Integer number > 0.99…

Quote: “Ángel Méndez Rivera says:
There is no such a thing as limitless in mathematics, and 0.000…1 is nonsense and does not exist.

A.R.B
Infinite/Recurring 0.9 (0.999…)
Infinite/Recurring 0.1 (0.001…)
Are both (Individually!) LIMITless they only have a LIMIT when added! together to equal 1
(0.999…) + (0.001…) = 1

38. The Physicist says:

@Anthony.R.Brown
They’re not using the word “limit” the way any reasonable English speaking person uses the word limit, they mean it in the mathematical sense. You can read about what a mathematical limit is here.

39. Anthony.R.Brown says:

The Physicist says: Quote:”They’re not using the word “limit” the way any reasonable English speaking person uses the word limit, they mean it in the mathematical sense. You can read about what a mathematical limit is here.”

A.R.B

The word (LIMIT) however you define it ? or look it up on some Math sites ? is being used in a way… and the same way ? so as to make a Calculated Number,that starts life less than 1 somehow become equal to 1 ? 🙁

A better way to look at the problem is…

Infinite/Recurring 0.9 (0.999…) = Universe! 🙂

And if we place a Wall at the End of the Universe! ? there will still be Universe! behind the Wall…

0.9 (0.999…) Universe! { Wall = 1} 0.9 (0.999…) Universe!…

40. Ángel Méndez Rivera says:

Anthony, no. There is no such a thing as a limitless number. Please, study the topic you’re talking on before you use terminology. The Physicist already told you where you can study the idea of what a limit is. Take the suggestion and study it.

41. Ángel says:

Mathman,

There is a justified reason why challenging mathematics makes you a crackpot: we can prove that by challenging mathematics, you are wrong.

2. Euler did not define limits.

3. Also, a definition cannot be wrong. This is, ironically, the definition of a definition. So Euler’s definition of infinite summation is acceptable. Euler’s definition renders summation continuous in the entire complex plane, which is what it should be according to group theory.

4. It makes perfect sense to add an infinite number of things. We do it all the tim in physics, and we get observable results that we do observe consistently.

5. No, your term-by-term subtraction was done wrong. Subtraction is done digitwise right to left.

6. 0.999… is not a bad way to represent 9/10 + 9/100 + …, but rather it is the only to represent it since it is the very definition. It is how decimal representation works. It can’t also be represented as Sum[m = 1, m= n —> Infinity](9*10^(-m)), which can be shown to equal 1 since the sum is 0.9(1 – 10^[-n])/(1 – 1/10), and when n —> Infinity, this expression simplifies to 1. This is the limit.

7. Yes, I agree that f(x = c) does not equal lim x—>c [f(x)]. However, 0.999… isn’t equal to f(x=c), but rather equal to lim x—>c [f(x)]. 0.999…. IS the limit of the partial sums by the very definition of decimal representation, and the partial sum limits to 1 evaluated by Taylor-geometric series. By transitive property, 0.999… = 1.

Remember: infinite sums are a thing. Irrational numbers prove it. It has been proven. It isn’t a definition to say “we can add infinitely many things”. No, we can prove it mathematically. That being said, 0.999… = 9/10 + 9/100 + … by definition. Not debatable. Now, 9/10 + 9/100 + … = Sum [m = 1, lim n —> Infinity][9*10^(-m)]. This latter expression is a limit. Keep in mind that summation is a linear hyper-operator, which makes f(y) = Sum[k=x, y][g(k)] continuous for all continuous g. Now, also remember that by the very definition of continuity, lim (x —> c) f(x) = f(c). Hence, Sum [m = 1, n lim n —> Infinity][9*10^(-m)] = Sum[m = 1, Infinity][9*10^(-m)]. This implies that 0.999… is actually a valid sum and therefore a valid decimal representation. You can add infinitely many terms. Finally, the latter left hand side of the previous identity mentioned is equal to lim (n —> Infinity) 9/10[1 – 10^(-n)]/(1 – 1/10) = (9/10)/(9/10) = 1. By the transitive property, 0.999… = 1.

42. Jesse Sherer says:

0.999… is or it isn’t equal to 1. if you are x distant to our Universe you are not in our Universe. no 3 dimentional universe is infinite unless all universe could be infinite too. 1-x=0.9999…. if you are x distant from our universe you are not in our universe. you might say x doesn’t exist, but, nether does 1/9 in base ten math. base ten is a rational universe of numbers there can be no number that equates to almost 1 but not 1 ,,or almost 1 is 1 in a rational universe.. at some point you have to say that if an infinite string of numbers representing both a fraction of a number and a whole number, then there must exist an infinity small number also exists that both equals 0 and almost 0 too. so x is more rational of a number that can add to .999999… and get 1.. base ten can’t solve it but logic can.. 0.9999.. is a number or it isn’t x is a number or it isn’t 1-x=0.9999…. or it doesn’t 1=0.99999… or it doesn’t

43. Anthony.R.Brown says:

Jesse Sherer says:
December 1, 2017 at 6:41 am

A.R.B

= Nonsense! 🙁

0.9999.. is a number Less than 1 🙂

44. Anthony.R.Brown says:

Ángel Méndez Rivera says:
September 24, 2017 at 5:02 pm
Anthony, no. There is no such a thing as a limitless number. Please, study the topic you’re talking on before you use terminology. The Physicist already told you where you can study the idea of what a limit is. Take the suggestion and study it.

A.R.B

Totally Wrong! Give me any Number you want ? and I will add any Number I want to it! Which makes any Number you provide {a limitless number} 🙂


# Find the rotation between set of points

I have two sets (sourc and target) of points (x,y) that I would like to align. What I did so far is:

• find the centroid of each set of points
• use the difference between the centroids translations the point in x and y

What I would like is to find the best rotation (in degrees) to align the points.

Any idea?

M code is below (with plots to visualize the changes):

# Raw data
## Source data
sourc = matrix(
c(712,960,968,1200,360,644,84,360), # the data elements
nrow=2, byrow = TRUE)

## Target data
target = matrix(
c(744,996,980,1220,364,644,68,336), # the data elements
nrow=2, byrow = TRUE)

# Get the centroids
sCentroid <- c(mean(sourc[1,]), mean(sourc[2,])) # Source centroid
tCentroid <- c(mean(target[1,]), mean(target[2,])) # Target centroid

# Visualize the points
par(mfrow=c(2,2))
plot(sourc[1,], sourc[2,], col="green", pch=20, main="Raw Data",
lwd=5, xlim=range(sourceX, targetX),
ylim=range(sourceY, targetY))
points(target[1,], target[2,], col="red", pch=20, lwd=5)
points(sCentroid[1], sCentroid[2], col="green", pch=4, lwd=2)
points(tCentroid[1], tCentroid[2], col="red", pch=4, lwd=2)

# Find the translation
translation <- tCentroid - sCentroid
target[1,] <- target[1,] - translation[1]
target[2,] <- target[2,] - translation[2]

# Get the translated centroids
tCentroid <- c(mean(target[1,]), mean(target[2,])) # Target centroid

# Visualize the translation
plot(sourc[1,], sourc[2,], col="green", pch=20, main="After Translation",
lwd=5, xlim=range(sourceX, targetX),
ylim=range(sourceY, targetY))
points(target[1,], target[2,], col="red", pch=20, lwd=5)
points(sCentroid[1], sCentroid[2], col="green", pch=4, lwd=2)
points(tCentroid[1], tCentroid[2], col="red", pch=4, lwd=2)

• I cannot read your code, but the operation you need is called Procrustes rotation. Have you heard of it? It works when points are already paired ($x_i,y_i$). Pre-rotation optional operations include translation and scaling, and optional post-rotational isoscaling. – ttnphns Dec 10 '15 at 17:13
• A complex regression will do the job. – whuber Dec 10 '15 at 17:23
• I've seen, that, rotating the system about 180 degrees, then the pairs $(a,C),(b,D),(c,A),(d,B)$ become neighbours - and this is even a better fit than the best fit of the original $(a,A),(b,B),(c,C),(d,D)$ (where the small letters stand for vector source and capital letters for vector target) I've not seen this possibility mentioned and explicitely allowed or disallowed. Are you sure that you don't want that better fit? – Gottfried Helms Sep 4 '16 at 14:05

This can be done using the Kabsch Algorithm. The algorithm finds the best least-squares estimate for rotation of $RX-Y$ where $R$ is rotation matrix, $X$ and $Y$ are your target and source matrices with 2 rows and n columns.

In [1] it is shown that this problem can be solved using singular value decomposition. The algorithm is as follows:

1. Center the datasets so their centroids are on origin.
2. Compute the "covariance" matrix $C$=$XY^T$.
3. Obtain the Singular Value Decomposition of $C=UDV^T$.
4. Direction adjustment $d=sign(det(C))$.
5. Then the optimal rotation $R=V\left( \begin{array}{ccc} 1 & 0 \\ 0 & d \\ \end{array} \right)U^T$

I don't know of any implementation in R so wrote a small function below.

src <- matrix(c(712,960,968,1200,360,644,84,360), nrow=2, byrow=TRUE)
trg <- matrix(c(744,996,980,1220,364,644,68,336), nrow=2, byrow=TRUE)

Kabsch algorithm in an R funtion:

kabsch2d <- function(Y, X) {
X   <- X-rowMeans(X)
Y   <- Y-rowMeans(Y)
C   <- X %*% t(Y)
SVD <- svd(C)
D   <- diag(c(1, sign(det(C))))
t(SVD$v) %*% D %*% t(SVD$u)
}

Center the points:

src <- src-rowMeans(src)
trg <- trg-rowMeans(trg)

Obtain rotation:

rot <- kabsch2d(src, trg)

Result (black - original source, red - original target, green - rotated target)

plot(t(src), col="black", pch=19)
points(t(trg), col="red", pch=19)
points(t(rot %*% trg), col="green", pch=19)

• +1. However, the answer could be further much better if you included discourse about how the algo is related to well-known Procrustes rotation problem. – ttnphns Mar 19 '16 at 14:03

I've done this with an iterative optimum-search, and tested 2 versions.
I've taken the original arrays and centered them calling this arrays cSRC and cTAR . Then I've done a loop with angles $\varphi$ between $0$ and $2 \pi$ , and for each angle I computed the error-criterion using the difference between the rotated $\small D=rot(\text{cSRC} ,\varphi)- \text{cTAR}$.

1. In version 1) I took as criterion the sum-of-squares of all entries in $\small D$ as $$err_1 = \small \sum_{k=1}^4 \small((D_{k,1})^2+(D_{k,2})^2)$$and the angle $\small \varphi$ at which the minimal error occured is equivalent the kabsch2d-procedure in @Karolis' answer.

2. In version 2) I took as criterion the sum of the absolute distances, that means, the sum $$err_2=\small \sum_{k=1}^4 \small\sqrt{(D_{k,1})^2+(D_{k,2})^2}$$ and got a slightly different rotation angle $\small \varphi$ for the smallest error.

I don't know, which criterion fits your needs better.

Here are some results from the protocol.

$$\small \begin{array} {r|cc} & \text{version } 1 & \text{version } 2\\ \hline \varphi & -0.04895304& -0.05093647 \\ \text{rotation} & \begin{bmatrix} 0.99880204& -0.04893349\\ 0.04893349& 0.99880204\\ \end{bmatrix} & \begin{bmatrix} 0.99870302 & -0.05091444\\ 0.05091444 & 0.99870302\\ \end{bmatrix} \\ \text{distances} & \begin{bmatrix} -6.80077266 & -0.86209739\\ 2.79924551 & -9.33782500\\ -0.61309522 & 6.94156520\\ 4.61462237 & 3.25835719\\ \end{bmatrix} & \begin{bmatrix} -6.78017751& -0.37062404 \\ 3.35787307 & -9.36574874 \\ -1.16459115 & 6.95324527 \\ 4.58689559 & 2.78312752 \\ \end{bmatrix} \end{array}$$


# the Gaussian integers are isomorphic to $\mathbb{Z}[x]/(x^2+1)$

I am trying to prove that $\mathbb{Z}[i]\cong \mathbb{Z}[x]/(x^2+1)$.

My initial plan was to use the first isomorphism theorem. I showed that there is a map $\phi: \mathbb{Z}[x] \rightarrow \mathbb{Z}[i]$, given by $\phi(f)=f(i).$ This map is onto and homorphic. The part I have a question on is showing that the $ker(\phi) = (x^{2}+1)$.

One containment is trivial, $(x^2+1)\subset ker(\phi)$. To show $ker(\phi)\subset (x^2+1)$, let $f \in ker(\phi)$, then f has either $i$ or $-i$ as a root. Sot $f=g(x-i)(x+i)=g(x^2+1).$ How can I prove that $f \in \mathbb{Z}[x]\rightarrow g \in \mathbb{Z}[x]$?

• It is enough to have $f=g(x^2+1)$, because this says $f\in (x^2+1)$, and we are done. – Dietrich Burde Apr 22 '14 at 13:34
• By the division algorithm, $f = q\cdot (x^2+1) + r$ with $\deg r < 2$. Then you just need to check that $r(i) = 0$ implies $r = 0$ if $\deg r < 2$. (Note: $\deg 0 = -\infty$) You could also appeal to Gauß' lemma. – Daniel Fischer Apr 22 '14 at 13:35
• @DanielFischer, I think that is the answer I am looking for since $g \in \mathbb{R}[x]$ which is a euclidean domain and I am allowed to make that claim. Do you care to expand on that? As in, how can I be certain that the coefficients of $q$ are in $\mathbb{Z}$? – kslote1 Apr 22 '14 at 13:57
• Don't use $\mathbb{R}[x]$ here, use $\mathbb{Q}[x]$. But you need never leave $\mathbb{Z}[x]$ even potentially. The point is that $x^2+1$ is monic, i.e. has lead coefficient $1$. Thus if you do polynomial division, you always get an integer coefficient, and the existence of $q,r\in\mathbb{Z}[x]$ with $f = q\cdot (x^2+1) + r$ and $\deg r < 2$ follows. – Daniel Fischer Apr 22 '14 at 14:06
• As an aside, this is one of those problems where it's easier to show the isomorphism directly, by writing down a homomorphism and its inverse, rather than the indirect route of finding a surjective homomorphism with zero kernel. – Hurkyl Aug 9 '17 at 16:40

Let me elaborate on Daniel Fischers comment. You have a ring homomorphism $\phi: \mathbb Z[x]\to\mathbb Z[i]$ given by $x\mapsto i$. Take $f \in \ker \phi$. By the division algorithm, $$f = q\cdot(x^2+1) + r,$$ where $\deg r < 2$ and $q,r\in\mathbb Z[x]$, since $x^2+1$ is monic. Applying $\phi$ to this equation yields $$0 = \underbrace{\phi(q)\cdot(i^2+1)}_0 + \phi(r).$$ Since $r$ is of degree $<2$, we can write $r = ax+b$ for some $a,b\in\mathbb Z$. Then $\phi(r)=0$ gives $$ai+b=0.$$ This equation in $\mathbb Z[i]$ implies $a=b=0$, so we have $r=ax+b=0\in\mathbb Z[x]$ and therefore $$f = q\cdot (x^2+1) \in \langle x^2+1\rangle.$$ We conclude that $\ker \phi \subseteq \langle x^2+1\rangle$.

When quotient out an ideal, we consider what happens to the ring when all the elements in the ideal are considered as identity elements.

Now if $x^2+1=0\Rightarrow x=\pm i$ let us take the "+" root.

$\mathbb{Z}[x]/(x^2+1)=\{f\in\mathbb{Z}[x]\,|x^2+1=0\}=\{f\in\mathbb{Z}[x]\,|x=i\}=\{a+bi|a,b\in\mathbb{Z}\}=\mathbb{Z}[i]$

I.e they are isomorphic.

I have answered a similar question here Is this quotient Ring Isomorphic to the Complex Numbers

You can define your isomorphism as follows

$\phi:\mathbb{Z}[i]\rightarrow \mathbb{Z}[x]/\langle x^2+1 \rangle$

$\phi(a+bi)=(a+bx)+\langle x^2+1 \rangle$

If you compute $[a+bx+\langle x^2+1 \rangle]\times [c+dx+\langle x^2+1 \rangle]$

We get $ac+(ad+bc)x+bdx^2+\langle x^2+1 \rangle$. We reduce by long division to get,

$(ac-bd)+(ad+bc)x+\langle x^2+1 \rangle$.

Therefore,

$\phi([a+bi][c+di])=(ac-bd)+(ad+bc)x+\langle x^2+1 \rangle$.

Which is exactly what you want.

The map is obviously bijective and a ring homomorphism.


# What is the remainder of $16!$ is divided by 19?

Can anyone share me the trick to solve this problem using congruence?

Thanks,
Chan

If you've seen Wilson's theorem then you know the remainder when $18!$ is divided by $19$. To get $16!$ mod $19$, then, you can multiply by the multiplicative inverses of $17$ and $18$ mod $19$. Do you know how to find these? (Edit: Referring to both inverses might be a bit misleading, because you really only need to invert their product. See also Bill Dubuque's answer.)

• I knew Wilson's Theorem, but really don't know how to find the inverse of 17, 18 mod 19. What's the inverse? Can you give me one example? Thanks.
– Chan
Feb 26 '11 at 6:17
• @Chan: $\rm 17 \equiv -2,\ 18\equiv -1$ Feb 26 '11 at 6:21
• @Chan: A general technique to find the inverse of $k$ mod $n$ when $k$ and $n$ are relatively prime is to use integer division and the Euclidean algorithm to find $a$ and $b$ such that $ak + bn = 1$. Since $bn$ is a multiple of $n$, $ak$ is congruent to $1$ mod $n$, and thus the congruence class of $a$ is the inverse of the congruence class of $k$. In this particular case, a further hint to make things easier is that $18\equiv -1$ and $17\equiv -2$ mod $19$. (This makes computations easier for inverting their product.) Feb 26 '11 at 6:21
• Many thanks, I got it now ;)
– Chan
Feb 26 '11 at 6:24

Hint  By Wilson's theorem $$\bmod 19\!:\ \overbrace{{-}1}^{\large \color{#0a0}{18}} \equiv 18! \equiv\!\! \overbrace{18\cdot17}^{\large \color{#c00}{(-1)(-2)}}\!\!\cdot 16!\,$$ $$\Rightarrow\,16!\equiv \dfrac{\color{#0a0}{18}}{\color{#c00}2}\equiv 9$$

Generally (Wilson reflection formula) $$\rm\displaystyle\ (p\!-\!1\!-\!k)!\equiv\frac{(-1)^{k+1}}{k!}\!\!\!\pmod{\!p},\,$$ $$\rm\:p\:$$ prime

• @Bill Dubuque: Thank you!
– Chan
Feb 26 '11 at 6:24
• @Bill Dubuque: If I have $2.16! \equiv 18 \pmod{19}$. Can I cancel out the $2$ from both sides?
– Chan
Feb 26 '11 at 6:42
• @Chan: $\rm\ 2\:x\equiv 2\:y\ (mod\ 19)\ \iff\ 19\ |\ 2\ (x-y)\ \iff\ 19\ |\ x-y\ \iff\ x\equiv y\ (mod\ 19)$ Feb 26 '11 at 6:50
• @Bill Dubuque: Thank you. So if $gcd(a, p) = 1$, then $ax \equiv ay \pmod{p} \implies x \equiv y \pmod{p}$, right?
– Chan
Feb 26 '11 at 7:07
• @Chan: $\rm\ n,m\$ coprime $\rm\iff a\ n + b\ m = 1\$ for some $\rm\:a,b\:\ \iff\ a\ n\equiv 1\ (mod\ m)\:,\:$ for some $\rm\:a\:$ $\rm\iff\ n$ is invertible/cancellable $\rm\:(mod\ m)$ Feb 26 '11 at 7:12

I almost think in this case it is just faster to break it down than calculate the inverses: First the factors between 1 and 10:

$$10!= (2\cdot 10)\cdot (4\cdot5)\cdot(3\cdot6)\cdot(7\cdot8)\cdot 9\equiv 1\cdot 1\cdot (-1)\cdot(-1)\cdot 9\equiv9$$

Now we have

$$16!\equiv 9\cdot 11\cdot 12\cdot 13\cdot 14\cdot 15\cdot 16\equiv9\cdot(-8)\cdot(-7)\cdot(-6)\cdot(-5)\cdot(-4)\cdot(-3)$$

$$\equiv 9\cdot(4\cdot5)\cdot(6\cdot3)\cdot(7\cdot8)\equiv 9\cdot(1)\cdot(-1)\cdot(-1)\equiv 9$$

Of course this doesn't generalize, but the computation is faster than finding 3 inverses and multiplying them. (Again of course that is not true for larger $n$...)

A more explicit answer would go like this:

By Wilson's Theorem, $18!$ is congruent to $-1 \pmod {19}$

$$18! = (18\cdot 17)(16!)$$

Then $(18\cdot 17)(16!)$ is congruent to $-1 \pmod{19}$

But note that $18$ is congruent to $-1 \pmod{19}$ and $17$ is congruent to $-2 \pmod{19}$

Then it follows that $(18\cdot 17)(16!)$ is congruent to $((-1)\cdot (-2))(16!)$ is congruent to $-1 \pmod{19}$

Simplifying, we get, $2\cdot(16!)$ is congruent to $-1 \pmod{19}$. But we need the remainder of $16!$, not $2\cdot 16!$. and $-\frac12$ isn't a valid answer.

Then also notice that $18$ is congruent to $-1 \pmod{19}$. Then $2\cdot 16!$ is congruent to $18 \pmod{19}$ by the fact that if $a$ is congruent to $b \pmod{p}$, then $b$ is congruent to $a \pmod{p}$

So we have that $2\cdot 16!$ is congruent to $18 \pmod{19}$ and from there, its just a matter of dividing both sides by $2$ to get $16!$ is congruent to $9 \pmod{19}$


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# How to prove that $\sum_{n \, \text{odd}} \frac{n^2}{(4-n^2)^2} = \pi^2/16$?

The series:

$$\sum_{n \, \text{odd}}^{\infty} \frac{n^2}{(4-n^2)^2} = \pi^2/16$$

showed up in my quantum mechanics homework. The problem was solved using a method that avoids evaluating the series and then by equivalence the value of the series was calculated.

How do I prove this directly?

• integrate $\frac{z^2}{(4-z^2)^2}\tan(z)$ over a big circle in the complex plane – tired Feb 3 '17 at 8:28

Wolfram Alpha gives me the partial fraction expansion:

$$\frac{n^2}{(4 - n^2)^2} = \frac{1}{8}\left(\frac 1{n-2}-\frac{1}{n + 2}\right) + \frac{1}{4}\left(\frac{1}{(n-2)^2}+\frac{1}{(n+2)^2}\right)$$

So the first part telescopes, and the second part will be some modified version of $\zeta(2)$.

In more detail: \begin{align}\sum_{n\text{ odd}} \left(\frac 1{n-2}-\frac{1}{n + 2}\right)&=\frac{1}{-1}-\frac{1}{3}+\frac{1}{1}-\frac{1}{5}+\frac{1}{3}-\frac{1}{7}+\cdots\\ &=-1+1=0\end{align} since all the other terms cancel out.

And \begin{align}\sum_{n\text{ odd}}\left(\frac{1}{(n-2)^2}+\frac{1}{(n+2)^2}\right)&=\frac{1}{(-1)^2}+\frac{1}{3^2}+\frac{1}{1^2}+\frac{1}{5^2}+\frac{1}{3^2}+\frac{1}{7^2}+\cdots\\ &=2\left(\frac{1}{1^2}+\frac{1}{3^2}+\frac{1}{5^2}+\cdots\right) \end{align}

It's a famous result that $\frac{1}{1^2}+\frac{1}{3^2}+\frac{1}{5^2}+\cdots=\frac{\pi^2}{8}$. You can can prove it if you know:

$$\frac{\pi^2}{6}=\zeta(2) = \frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+\cdots$$

and thus $$\zeta(2)-\frac1{2^2}\zeta(2) = \frac{1}{1^2}+\frac{1}{3^3}+\frac{1}{5^2}+\cdots$$

So $$\sum \frac{n^2}{(4-n^2)^2}=0 + \frac{1}{4}\cdot 2\cdot \frac{\pi^2}{8}=\frac{\pi^2}{16}$$

• Hi Thomas. I had assumed that the sum over odd integers included the negative ones. Naturally, the answer in that case is twice the answer if the summation extends to the positive odds only. Anyway, (+1) for your well-written post. -Mark – Mark Viola Feb 3 '17 at 4:30
• @Dr.MV Yeah, I wondered about that. I might have read it your way if it had been $\sum_{n\text{ odd}}$ rather than $\sum_{n\text{ odd}}^{\infty}$. – Thomas Andrews Feb 3 '17 at 4:37
• Ah, yes. I see your point. Well, I did address the issue at the end of the post. ;-)) – Mark Viola Feb 3 '17 at 4:40

First, the partial fraction of the summand can be written

\begin{align} \frac{n^2}{(4-n^2)^2}&=\frac14\left(\frac{1}{n-2}+\frac{1}{n+2}\right)^2\\\\ &=\frac14 \left(\frac{1}{(n-2)^2}+\frac{1}{(n+2)^2}+\frac{1/2}{n-2}-\frac{1/2}{n+2}\right) \end{align}

Second, we note that

\begin{align} \sum_{n\,\,\text{odd}}\frac{1}{(n\pm 2)^2}&=\sum_{n=-\infty}^\infty \frac{1}{(2n-1)^2}\\\\ &=2\sum_{n=1}^\infty \frac{1}{(2n-1)^2}\\\\ &=2\left(\sum_{n=1}^\infty \frac{1}{n^2}-\sum_{n=1}^\infty \frac{1}{(2n)^2}\right)\\\\ &=\frac32 \sum_{n=1}^\infty \frac{1}{n^2}\\\\ &=\frac{\pi^2}{4} \end{align}

Third, it is easy to show that

$$\sum_{n=-\infty}^\infty \left(\frac{1}{2n-3}-\frac{1}{2n+1}\right)=0$$

Putting it all together we have

$$\sum_{n,\,\,\text{odd}}\frac{n^2}{(4-n^2)^2}=\frac{\pi^2}{8}$$

If we sum over the positive odd only, then the answer is $(1/2)\pi^2/8=\pi^2/16$

HINT

$$\sum_{n \, \text{odd}}^{\infty} \frac{n^2}{(n^2-4)^2}=\sum_{n=1}^{\infty} \frac{(2n-1)^2}{((2n-1)^2-4)^2}$$ Using partial fraction expansion, note $$\frac{(2n-1)^2}{((2n-1)^2-4)^2}=\left(\frac{1}{4(2n+1)^2}+\frac{1}{4(2n-3)^2}\right)-\left(\frac{1}{8(2n+1)}-\frac{1}{8(2n-3)}\right)$$ Note that the second part has cancelling terms.

• Why the downvote? – S.C.B. Feb 3 '17 at 4:40
• @ThomasAndrews I fixed the signs. – S.C.B. Feb 3 '17 at 4:40