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H: Number of integer solutions of $2n^2=a^2+b^2$
For a given integer $n$, how many positive integer $(a,b)$ pairs exist which satisfy $2n^2=a^2+b^2$?
In particular, I'm looking for all $n$s where there are exactly 105 solutions. (One solution is $(n,n)$, and there are $2\cdot 52$ other solutions: $(a,b)$ and $(b,a)$ are two different solutions if $a\ne b$.)
I'm sure that there are theorems about the solutions of this kinds of equations. Where can I find them and read more?
AI: Here is a very closely related question. The formula for the number of solutions is the product of $2m+1$ where $m$ runs through the multiplicities of division of the prime factors of $n$ that are${}\equiv1\pmod4$. So you could take $n=5^{52}$ as one solution to you problem, but you'll get smaller solutions using the factorization $105=3\times5\times7$, the smallest one being $n=5^3\times13^2\times17=359125$. A proof of the result (which uses Gaussian integers) can be found in this answer. |
H: Primes, congruent, Fermat squares
I need some proof or procedure to solve the question:
For given n = 1 or > 1, We know by fermat, $p = x^2 + ny^2$ for an odd prime $p$. Now, can we proof the
(a) $(-n/p) = 1$ and $(p/q) = 1$ for every odd prime $q\mid n$?
(b) $p \equiv 1 \pmod{4}$ for $n \equiv 0, 1 \pmod{4}$?
(c) $p \equiv 1 \pmod{8}$ for $8\mid n$?
The proofs of the above problem will answer many questions easily, especially primes of the form $p = x^2 + ny^2$. So, kindly discuss the proofs.
AI: can you explain the notation $(-n/p)$?
By the way, is this homework?
Concerning $(b)$ and $(c)$, take a look at the groups $\mathbb{Z}/4\mathbb{Z}$ and $\mathbb{Z}/8\mathbb{Z}$, square each element, and see which values are possible. You'll see that if you also use the fact that $p$ is odd, the results $(b)$ and $(c)$ follow easily.
I leave it to you as an exercise on purpose, its really useful to find this out for yourself, but let me know if you need more hints!
Edit:
$(b)$: First of all, $\mathbb{Z}/4\mathbb{Z}$ has
\begin{align*}
0^2=2^2 = 0\\
1^2=3^2=1
\end{align*}
So an even square is zero, an odd square is one.
Two cases. If $n \equiv 0 $ (mod 4), then $ny^2 \equiv 0 $ (mod 4) and in particular $ny^2$ is even. Since $p$ is odd, $x^2$ (and hence $x$ itself) must be odd. Then $x^2 \equiv 1$. Hence
\begin{align*}
p &= x^2 + ny^2\\
&\equiv 1 + 0\\
&\equiv 1
\end{align*}
If $n \equiv 1$ (mod 4), then $p \equiv x^2 + y^2$. The sum of two odd numbers or two even numbers is even, since $p$ is odd we know either $x$ is odd and $y$ is even or the other way around. In any case we find by the inspection of squares in $\mathbb{Z}/4\mathbb{Z}$ above that $p \equiv 1 + 0 \equiv 1$ as desired.
Notice that strictly speaking i was working modulo 2 most of the time in the last argument. Formally i could have argued that $n \equiv 1$ (mod 4) implies $n \equiv 1$ (mod 2) and hence $p \equiv x^2 + y^2$ (mod 2), from which the statement of the parity of $x$ and $y$ follows. Then, we return to working modulo 4 again: the value of a number modulo 2 is of course not enough to determine its value modulo 4, but as we saw above, we are able to find the value of the square of the number modulo 4, just by knowing its value modulo 2. Since we are only interested in the values of the squares, this suffices.
Now $(c)$. We determine the squares in $\mathbb{Z}/8\mathbb{Z}$. Noticing that $5 \equiv -3$ hence $5^2 \equiv 3^2$ and similarly $7^2 \equiv (-1)^2 \equiv 1^2$, we can quickly see that all odd numbers square to one (In fact this shows that $\mathbb{Z}/8\mathbb{Z}^*$ is the Klein four group). Now $8 | n$ is a reformulation of $n \equiv 0$ (mod 8), which certainly implies $ny^2$ is even, and again since $p$ is odd we know $x^2$ and hence $x$ must be odd. Putting this together we see
\begin{align*}
p &= x^2 + ny^2\\
&\equiv 1 + 0\\
&\equiv 1
\end{align*}
Modulo 8 of course, as desired. Whereever i left out the (mod ...) notation i assume it to be clear modulo which number we work.
Hope this clears it up. |
H: Showing $L=\{uw \mid \exists v:uv\in L_{1},vw\in L_{2}\}$ is regular
Let $L_{1,}L_{2}$ be regular languages and define $L:=\{uw \mid \exists v\in\Sigma^{*}:uv\in L_{1},vw\in L_{2}\}$.
I wish to prove that $L$ is regular using only closure properties
(such as $L_{1},L_{2}$ is regular then so is $L_{1}\cap L_{2},L1\cup L_{2},L_{1}^{*}$ etc.).
My thoughts: I tried to define a homomorphism $h:\Sigma\cup\Sigma'\to\Sigma^{*}$
$$\forall\sigma\in\Sigma:h(\sigma)=\sigma$$
$$\forall\sigma'\in\Sigma':h(\sigma')=\epsilon$$
Then I tried to look at $h^{-1}(L_{1}),h^{-1}(L_{2})$, I thought
about looking at their intersection with the languages $\Sigma^{*}(\Sigma')^{*},(\Sigma')^{*}\Sigma^{*}$
etc to give the words in $h^{-1}(L_{1}),h^{-1}(L_{2})$ the form of
$ww'$s.t $w\in L_{1}$ or $w'w$ s.t $w\in L_{2}$.
I am having difficulty to continue, I thought about trying to get
to something like $L_{3}(\Sigma')^{*}L_{4}$ where $L_{3}$ is all
the prefix of words in $L_{1}$ and $L_{4}$ is all the endings of
words in $L_{2}$ and then intersect this with $\Sigma^{*}L_{5}'\Sigma^{*}$
where $L_{5}$ is some language to enforce that the words in $(\Sigma')^{*}$in
the expression$L_{3}(\Sigma')^{*}L_{4}$ will be such that if I remove
all the tags (using a homomorphism) I would get $L$.
How can I show $L$ is regular using only closure properties ?
AI: Let $\underline \Sigma := \{ \underline a : a \in \Sigma \}$ be a disjoint copy of $\Sigma$, assume $<$ and $>$ are symbols not already present in $\Sigma$ and define a homomorphism $h: \Sigma \cup \underline \Sigma \cup \{<,>\} \to \Sigma$ by
\begin{align*}
a &\mapsto a,\\
\underline a &\mapsto a,\\
< &\mapsto \varepsilon,\\
> &\mapsto \varepsilon.\\
\end{align*}
Then
\begin{align*}
K &:= \{ u \text{<}\underline v \text{>} w : uv \in L_1, vw \in L_2 \}\\
&= \Sigma^* \text{<}\underline{\Sigma}^*\text{>} \Sigma^* \cap h^{-1}(L_1)\text{>}\Sigma^* \cap \Sigma^* \text{<}h^{-1}(L_2)
\end{align*}
is regular by the closure properties of regular languages, and so is $L$ since it is the homomorphic image of $K$ by deleting the symbols from $\underline \Sigma$ and $\{<,>\}$. |
H: Finding \alpha and \beta of Beta-binomial model via method of moments
I am looking for a laymen step by step of how the process of finding the 1st and 2nd sample moments located:
http://en.wikipedia.org/wiki/Beta-binomial_distribution#Maximum_likelihood_estimation
Also it's my limited understanding that k-th sample moments are defined as $${\frac {\sum _{i=1}^{n}{x_{{i}}}^{k}}{n}}$$
For samples $x_1, x_2...x_n$ where $n$ = total number of samples. (source: http://en.wikipedia.org/wiki/Moment_(mathematics)#Sample_moments)
Given their example data:
Males 0 1 2 3 4 5 6 7 8 9 10 11 12
Families 3 24 104 286 670 1033 1343 1112 829 478 181 45 7
The first thing I don't understand is why they say $n=12$ when there are 13 data points. Wouldn't that imply $n=13$
I believe the sample moments are:
$m_1 = \frac{0+1+2+3+4+5+6+7+8+9+10+11+12}{13} = 6$
$m_2 = \frac{0^2+1^2+2^2+3^2+4^2+5^2+6^2+7^2+8^2+9^2+10^2+11^2+12^2}{13} = 50$
Yet they have
$$m_1 = 6.23$$
$$m_2=42.31$$
Even If I use $n=12$ and cut off either the first or last record I am left with different values.
Despite that, even using their values of $$m_1 = 6.23$$$$m_2=42.31$$$$n=12$$ going by the equation for the method of moments estimates:
$$\alpha= \frac{( nm_{{1}}-m_{{2}} ) }{n ( {\frac {m_{{2
}}}{m_{{1}}}}-m_{{1}}-1 ) +m_{{1}}} = 33.59257915$$
$$\beta= \frac{( n-m_{{1}} ) ( n-{\frac {m_{{2}}}{m_{{1}}}}
)}{n ( {\frac {m_{{2}}}{m_{{1}}}}-m_{{1}}-1
) +m_{{1}}} = 31.11222820 $$
which do not match his values of:
$$\alpha= 34.1350$$
$$\beta = 31.6085$$
Edit: Given this question was spawned from Rating system incorporating experience; For purposes of record keeping for later googlers, I decided to reword this question to better suit the answers. A detail explanatin of Beta-binomial model and the MLE method of finding $\alpha$ and $\beta$ are located there.
AI: The moments should be $$m_k = \frac{ \sum_{i=0}^{12} f_i \times i^k}{\sum_{i=0}^{12} f_i}$$ where $f_i$ is the number of families with $i$ males.
The calculation of $\hat{\alpha}$ and $\hat{\beta}$ require the use of $n=12$, as @did says.
Do both and you will get the stated values. |
H: Do there exist vector spaces over a finite field that have a dot product?
I've stumbled in the german wikipedia over the question if a vector space over a finite field exists, that has a dot product.
Definition of dot product
A dot product over a $\mathbb{K}$-vector space V is a mapping $F : V \times V \rightarrow \mathbb{R}$ which is...
... bilinear: $\forall a, a_1, a_2, b, b_1, b_2 \in V \forall \lambda_1, \lambda_2, \mu_1, \mu_2: $
$\begin{align*}
F(\lambda_1 \cdot a_1 + \lambda_2 \cdot a_2, b) &= \lambda_1 \cdot F(a_1, b) + \lambda_2 \cdot F(a_2, b)\\
F(a, \mu_1 \cdot b_1 + \mu_2 \cdot b_2) &= \mu_1 \cdot F(a, b_1) + \mu_2 \cdot F(a, b_2)
\end{align*}$
... symmetric: $F(a, b) = F(b,a)$
... positive definite: $\forall a \in V: F(a,a) \geq 0 \land ( F(a,a) = 0 \Leftrightarrow a = 0)$
My try
Please correct my choice of words if you know what I mean and if you know how to write it in English. I am not used to write math texts in English.
$\mathbb{K} := \mathbb{Z} / 2 \mathbb{Z}$
Let $V$ be a $\mathbb{K}$-vector space over the set $\{0,1\}$.
$\mathbb{K}$ is a finite filed.
Define $\langle 0, 0 \rangle := 0$, $\langle 1, 0 \rangle := \langle 0,1 \rangle := 1$ and $\langle 1, 1 \rangle = 1$.
symmetry
$\langle, \rangle$ is symmetric per definition
bilinear form
$f(1 + 1, 1) = f(0,1) = 1 \neq 0 = 1+1 = f(1,1) + f(1,1)$
Is it possible to define a dot product for a bigger vector space / finite field?
(If the answer is yes, please give me an example)
edit: argh - $f(a,a) = 0 \Leftrightarrow a = 0$, but $f(a,b)$ can be 0. I always forget that :-/
AI: Certainly the "obvious" inner product of $(v_1,\dots v_n)$ with $(w_1,\dots w_n)$ given by $\sum_n v_n w_n$ is defined for any (finite dimensional) vector space. It is symmetric and bilinear and fairly useful. It is used extensively, for example, when discussing linear codes over finite fields. (The codes are vector spaces!)
There isn't any reason to discouraged that the "positive definite" condition is missing. There are many interesting inner product spaces where the inner product is not positive definite... even over $\mathbb{R}$! The inner product of Minkowski space is not positive definite, for example, and this is the setting for special relativity.
Positive definiteness is kind of a "Euclidean ideal" that we would hope for, but the Minkowski space example shows that nature isn't always Euclidean, and that non-positive-definite inner products arise naturally. |
H: divisor of rational function on a surface
Help me please with this question:
Find divisor of rational function $f=(w_0,w_1)$ on a surface $X=\left \{ w_0w_1-w_2w_3 \right \}\subset \mathbb{P}^{3}$
How should I to take into account this surface?
Thanks a lot!
AI: I interpret your question as asking about the rational function $f=w_1 /w_0$ on the surface $X$.
The divisor of zeros is $div(f)_0=D_2+D_3$ where $D_2$ is the line $w_1=w_2=0$ and $D_3$ is the line $w_1=w_3=0$.
This can be checked on the affine chart $w_0=1$.
The divisor of poles is $div(f)_\infty=E_2+E_3$ where $E_2$ is the line $w_0=w_2=0$ and $E_3$ is the line $w_0=w_3=0$.
This can be checked on the affine chart $w_1=1$.
The locus of indeterminacy of $f$ consists of the two points $D_3\cap E_3=[0:0:1:0]$ and $D_2\cap E_2=[0:0:0:1]$
And finally the answer to your question is
$$ div(f)=div(f)_0-div(f)_\infty=D_2+D_3-E_2-E_3 $$
A picture
The lines $D_2,E_3$ are in one of the two rulings of the quadric $X$ and the lines $D_3,E_2$ on the other one. You might try to draw these four lines as the sides of a square and see where the indeterminacy points lie. |
H: Analogy of Binary subtraction to decimal subtraction
Possible Duplicate:
How to do +, -, *, / with number in a base b?
Subtraction of numbers with arbitrary bases
This is a very basic question. But I come from non computer science background and is having trouble understanding this . Lets say I want to subtract two binary numbers (say 100 - 001 = 011). How do I go about doing this subtraction ? Please provide an analogy for a decimal subtraction
P.S: Actually I found out that the HashMap in Java requires its bucket size to be a power of 2 and finds the bucketIndex by the following code and this works well if n is a power of 2 . h is the value of the key to be hashed and length is number of buckets in the HashMap
static int indexFor(int h, int length) {
return h & (length-1);
}
So I am assuming that that n here is 4 . So lets say we do 4-1 =3 which is 011 in binary .I am getting confused with borrowing operation here for the 2's place and 4's place during the subtraction process.
AI: You go for that subtraction exactly as you would do for decimal, except that you need to borrow already for $2=$10. So for your example, you have
100
- 001
<- place for borrows
___
So you start in the rightmost column, where you have $0-1$. Since $0<1$, you have to borrow from the left, so you have 10-1 that is $2-1$ which is $1$. So after this step, you end up with
100
- 001
1
___
1
Now to the second solumn. Again, including the borrow, you have to subtract $1$ from $0$, so you need to borrow again. Again, you'll get 10-1$=2-1=1$:
100
- 001
11
___
11
Finally, with the left column, you have $1-1$, which doesn't need a borrow and gives $0$. So you end up with
100
- 001
11
___
011
Which evidently is the correct result.
However it may be easier to first only collect the borrows and then subtract them in a separate step (continuing until no borrows are left). Note that this could be done in decimal subtraction as well. With that algorithm you collect the borrows below the preliminary result, and then restart the subtraction algorithm. This is especially useful for problems like 100-011 where the borrow gets larger than $1$. Here's how that method works:
100
- 011
___
<- Preliminaty result
0 <- Borrows (initialized with a rightmost 0)
First, subtract the right-most column. This needs a borrow, which this time is noted below.
100
- 011
___
1
10
Now we get to the second column. With the previous algorithm, you'd have to subtract $10$=2$, which is of course possibile with borrow, but you need to think a little bit more (which is especially bad if you are a computer and thus quite stupid). With this second algorithm, you just ignre the borrow for the moment, and subtract the second column normally (which again gives a borrow).
100
- 011
___
11
110
Finally you subtract the leftmost column, which works without a borrow:
100
- 011
___
111
110
Now the borrows are not empty, therefore you have to subtract them. Fortunately you know how to do that:
100
- 011
___
111
- 110
___
001
Voila, the correct result!
OK, so why is this algorithm better? Well, because you never get a borrow larger than $1$, and since computers are especially good with $0$ and $1$, this is better suited for them. Moreover, if you look at the preliminary results, you'll find that each bit is 1 exactly if the original bits are different; that operation is known as XOR and is a basic logic operation which can be easily implemented on computers. Also, it is easy to see that the borrow now occurs exactly when the bit of the first number is 0 and the bit of the second number is 1. Which is also a combination of two basic logical operations, NOT and AND.
Also note that the XOR is also valid for addition, only the borrow has to be changed to a carry. A carry happens exactly if both bits are 1. So basically the only difference between binary addition and binary subtraction is a NOT on the first number when calculating the borrow instead of the carry. |
H: A homotopy question
I already asked about the interpretation of this problem here. Now I would like to ask about the solution. The problem is
Let $A\subseteq X$ be a contractible space. Let $a_0\in A$. Is the embedding $f: X\setminus A\to X\setminus\{a_0\}$ a homotopy equivalence?
I'm not very bright, and this is very difficult for me. I think I've solved this problem though:
Let $X=[0,1]\cup\{2\},\, A=[0,1],\,a_0=0.$ Then $X\setminus A=\{2\}$ and $X\setminus\{a_0\}=(0,1]\cup\{2\}.$ There is exactly one function $g:(0,1]\cup\{2\}\to \{2\}$ so I only have to check if $f\circ g$ and $g\circ f$ are homotopic to the appropriate identity functions. $g\circ f=\mathrm{id}_{\{2\}}$, so this one is OK. I have to check if $f\circ g$ is homotopic to $\mathrm{id}_{(0,1]\cup\{2\}}.$ Suppose there is a homotopy $H(x,t).$ Then $H(1,t)$ is a path from $1$ to $2$ in $(0,1]\cup \{2\},$ which is a contradiction.
So it's not true, if that is correct. But I took a non-connected $X$. I remember that during the course some verbal agreements were made that we generally consider all spaces connected and Hausdorff, and maybe even more. Unfortunately, I don't remember it very well. And I'm not sure the agreement applied to this problem. But regardless of that I would like to know if this is still false when $X$ is connected. Or even more: what conditions on $X$ do I need for this to be true?
I'm asking this because I feel that I'm cheating in my solution. I feel I haven't really understood the problem, despite a lot of time and effort I've put into it.
AI: I think your argument is correct, indeed since path-connectedness is invariant under homotopy equivalence the two spaces $X \setminus A$ and $A \setminus a_0$ cannot be homotopy equivalent - the first is path-connected while the second is not.
A similar example when $X$ is connected is $X = [0,2]$, $a_0 = 1$ and $A = [0,1]$. Then $X \setminus A = (1,2]$, and $X \setminus a_0 = [0,1) \cup (1, 2]$ which are not homotopy equivalent, so in particular the embedding cannot be a homotopy equivalence. |
H: Whether there is a non commutative k-algebra of dimension no larger than 3
Suppose $k$ is a field and $A$ is a $k$-algebra of dimension no larger than 3. If $A$ is semi-simple, then $A$ can be written as a direct sum of simple $k$-algebras. Further one can find $A$ is commutative by exhausting all the cases.
Without the semi-simple condition, what can we say about $A$? Is it still commutative?
AI: The upper triangular $2\times 2$ matrix ring over a field is three dimensional and noncommutative.
Here's what I mean, if you're not familiar with it:
$\{\begin{bmatrix}a&b\\0&c\end{bmatrix}\mid a,b,c\in\mathbb{F}\}$ |
H: Lebesgue generalizations of Hilbert spaces?
Is an L[p] space a generalization of Hilbert spaces using Lebesgue integration?
And if this is the case, is it true that Holder's and Minkowski's Inequalities are generalizations of the Cauchy-Schwarz and triangle inequalities using Lebesgue measures?
AI: No. To start, $L^p$ is a true Banach space, not a "generalized" Banach space.
I assume you use the term Lebesgue integration as contrasted with Riemann integration. You could, if you wanted, define the normed space of all functions $f:[0,1]\to\mathbf{R}$ with $|f|^p$ Riemann-integrable, with norm $\|f\|_p = \int |f|^p$. You might then hope to prove that this is a Banach space, but you would fail (the completion of this space is precisely $L^p([0,1])$).
If you did carry out this programme, you would still want to prove Minkowski's inequality and Holder's inequality, which are generalisations of the triangle and Cauchy-Schwarz inequalities to $p\neq 2$, even though you've decided that you hate Lebesgue.
Edit: In response to your new phrasing, I think it's best to say that Banach spaces are generalized Hilbert spaces. Note that $L^2([0,1])$ is a Hilbert space, but definitely requires Lebesgue integration to ensure that the space is complete. (In general, the correct way to view Lebesgue integration is as the "completion" of Riemann integration.) |
H: $\mathbb{Z}/n\mathbb{Z}[T]$ and zero divisors
I read the following in wiki, but I can't understand what is meant by "divisor" there.
Notice that $(\mathbb{Z}/2\mathbb{Z})[T]/(T^{2}+1)$ is not a field since it admits a zero divisor $(T+1)^2=T^2+1=0$ (since we work in $\mathbb{Z}/2\mathbb{Z}$ where $2=0$)
I understand that if $a^2 \bmod p = 0$, then $p$ then isn't prime, but what did they mean about "divisor"?
AI: An nonzero element $a$ of a ring is a zero divisor if there is another nonzero element $b$ such that $ab=0$. In particular, a zero divisor can't have an inverse since otherwise $a^{-1}ab=a^{-1}*0$ hence $b=0$, a contradiction.
That being the case, in your example $a=b=T+1$ is the zero divisor hence is noninvertible and so your ring is not a field. |
H: For what values of $n$ is $n^2+n+2$ a power of $2$?
Working on isometric paths in hypercubes, I came up with the following simple, yet (imo) interesting problem. For what natural numbers $n$ exists a natural number $t$ such that $n^2+n+2=2^t$? The first few terms are $n=0,1,2,5,90$, and these are all below one million. Does someone have any idea how to approach this problem? I basically only want to know whether there exist infinitely many $n$ or not (maybe even that 0, 1, 2, 5, and 90 are the only possible ones).
Thanks,
Sacha
AI: That there are no more positive solutions was proved by Nagell, settling a conjecture of Ramanujan. The problem is discussed in this paper, and in this Wikipedia article.
To see that these solve the problem, a small preliminary transformation of your equation is useful. Rewrite it as $4n^2+4n+8=2^k$, and then as $(2n+1)^2+7=2^k$. We have arrived at the Ramanujan-Nagell equation. |
H: 5 Points uniformly placed on a sphere
I have a sphere and I have to place some points on it, the most uniformly possible.
If I have 4 points, placing them as vertices of a tetrahedron seems good.
If I have 6 points, placing them as vertices of a octahedron seems very good too.
How can I find a way (the best if it exists) to place only 5 points ?
EDIT : "Uniformly" would mean that if I draw a Voronoï diagram on the sphere, each point has a same-area cell and the diameter of a cell is minimized (they are "round" and not some thin slices of the sphere).
AI: The answer depends on what you mean by "uniform". One way of doing this is to minimize the "energy" the system would have if each of the points was a charged particle. This "Thomson's Problem" is quite a famous problem in global minimum finding algorithms.
The answer in this case for $n=5$ would be:
Two points on the poles, and three as an equilateral triangle on the equator.
More answers for other values of $n$ can be found here. |
H: "Commutative" functions
Say a function is commutative if it remains unchanged under any permutation of its arguments. E.g. $f(0,1)=f(1,0)$. (Alternatively we could describe these as functions over multi-sets, or say that they are reflective about any hyperplane $x_i=x_j$). Some examples are sum, product and average.
Is there a name for these functions? Google searches for "commutative" and "reflective" functions don't turn up anything.
Can we say anything interesting about these functions? For example, I note that any commutative function which is linear must be the sum function, multiplied by some constant. (i.e. $f(x)=c\sum x_i$). Also I see that the functions make up a field under the obvious operations.
AI: These are called symmetric functions (of two variables.) There is a large literature, that mostly concentrates on symmetric polynomials. Any symmetric polynomial in two variables $x$, $y$ is a polynomial in the variables $x+y$ and $xy$. There is an important analogue for symmetric polynomials in more variables. |
H: Stuck With The Differentiation Of A Inverse Hyperbolic Function
I'am suppose to show that $$\frac{\mathrm{d} }{\mathrm{d} x}[x \operatorname{cosh}^{-1}(\cosh x)] = 2x$$
And this is what i've tried.Upon differentiating the above function wrt $x$ using the product rule and applying the formula $\cosh^{-1}f(x) = \frac{f\prime(x)}{\sqrt{[f(x)]^2 - 1}}$, I end up getting $$x\frac{\sinh x}{\sqrt{(\cosh )^2 - 1}} + \cosh^{-1}(\cosh x)$$
How to proceed from here?
Thank You
AI: You’re working way too hard. Simplify! What is $\cosh^{-1}\big(\cosh x\big)$? |
H: Expressing $\mathbb{R}$ as the quotient of a disjoint union of unit intervals
I am trying to complete exercise 3.18(a) in Lee's Introduction to Topological Manifolds. The exercise is as follows:
Let $A \subseteq \mathbb{R}$ be the set of integers and let $X$ be the quotient space $\mathbb{R}/A$, where $A$ is collapsed down to a point.
(a) Show that $X$ is homeomorphic to a wedge sum of countably infinitely many circles.
Lee gives the hint that one should express both spaces as quotients of a disjoint union of intervals, implying that one will make use of the uniqueness of quotient spaces.
The wedge sum of countably infinitely many circles is the space $B=\bigvee_{i=0} ^\infty S_i ^1=\amalg_{i=0} ^\infty S_i ^1/\{p_i\}$, where $p_i \in S_i ^1$ and $\{p_i\}$ denotes the relation $p_i\sim p_j$ for all $i,j$. Since $S^1$ is the quotient space $I/(0\sim1)$, we can express $B$ as $B=\amalg_{i=0} ^\infty I_i/\sim$, where $\sim$ is the relationship $p_i \sim p_j$ for all $i,j$ and $0_i\sim 1_i$ for all $i$. With this one defines the quotient map $q:\amalg_{i=0} ^\infty I_i \rightarrow B$ by sending elements to their equivalence classes.
I am stuck trying come up with a similar formulation for $\mathbb{R}$. Graphically I could imagine $\mathbb{R}$ being constructed by attaching countably infinitely many of the unit intervals $I_i$ together with the identification $1_i\sim 0_{i+1}$. Formally this would be $\mathbb{R}=\amalg_{i=0} ^\infty I_i/(1_i\sim 0_{i+1})$. Using this idea though I can only see myself being able to construct the positive real numbers $\mathbb{R}^+$ which is not homeomorphic to $\mathbb{R}$. Is their some way to save this idea?
If I could figure this out, then I would have to find a quotient map $r:\amalg_{i=0} ^\infty I_i\rightarrow \mathbb{R}/A$ that makes the same identifications as $q$ to prove that the two spaces are homeomorphic by the uniqueness of the quotient space.
AI: Your idea is fine with just a small modification: $\Bbb R=\amalg_{n\in A}I_n/(1_n\sim 0_{n+1})$. To make life easier, you should also use $A$ as the index set for $\amalg_n S_n^1/P$, where $P$ is the set of distinguished points. |
H: Given known reference points, how do I calculate the approximate height of a point from a photograph
I have a photo with three towers. I know that the base of all three towers are at the same elevation, and at the top they each comes to a point. Two of the towers are the same height, though they appear to be different heights in the image, due to perspective.
I know the x/y/z coordinates of the peaks of the two known towers. I know the x/z coordinates of the third tower. I am trying to find the y (height) coordinate of the peak of the third tower. All three peaks are clearly visible within the photo, but despite knowing that the bases are all at the same eleveation (lets call it y:0) They are not visible in the photo. I don't think this should have any affect, as the are known to be equall I should only have to deal with the x/y/z of the three peaks, and I think I can safely ignore anything else.
For clarity: X = East/West, Y = Elevation and Z = North/South
Tower 1 Peak (X, Y, Z) : 0, 129, 0
Tower 2 Peak (X, Y, Z) : 16, 129, 97
Tower 3 Peak (X, Y, Z) : -40, ???, 78
The following are the X/Y pixel coordinates of the three peaks in the photo:
Tower 1 Peak (X, Y): 1235, 227
Tower 2 Peak (X, Y): 1445, 528
Tower 3 Peak (X, Y): 2042, 397
Please note that the pixel coordinates are taken with 0,0 being the top left corner of the image.
Is it even possible to calculate the approximate height of the third tower? If so, how might I go about doing do?
AI: Counting degrees of freedom, I'd say it's unlikely you can fix the camera's location in space well enough from the information you have given here. You have measured 8 quantities (namely the image coordinates of the three tower tops and the vanishing point), but the degrees of freedom you have to fix are:
Where in space was the camera located when the picture was taken? (3 degrees of freedom)
Were did the camera's optical axis point? (2 degrees of freedom)
How was the camera turned about the optical axis? (1 degree of freedom)
What is the focal length of the camera, measured in the image coordinate uints? (1 degree of freedom)
Where in the image coordinate system is the optical axis? (2 degrees of freedom)
That's nine degrees of freedom all in all, which is one more than the number of numbers you have measured. And what you actually need is a tenth degree of freedom about the situation:
Where on the vertical line representing tower 3 is your third image point? (1 degree of freedom)
If you know the towers themselves to be exactly vertical, it is possible that you can just about get a fix by measuring the coordinates of the vertical vanishing point in the image. But it's going to be a messy and possibly not very robust robust calculation. |
H: orthogonal subspaces in a Hilbert space
Is it true that if $A,B$ are closed subsets of a Hilbert space $H$, such that $A\perp B$, we have $A+B+(A\cup B)^{\perp} =H$ ? What if $A,B$ are closed subspaces ?$ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ $
AI: If $A$ and $B$ are arbitrary sets, it is not hard to construct a counterexample:
Let $\{e_i\}$ be an orthogonal basis of the Hilbert space. Let $A=\{e_1\}$ (that is, the set containing nothing but $e_1$, and $B=\{e_2\}$. Then $A\cup B=\{e_1,e_2\}$, and thus $(A\cup B)^\perp$ consists of all vectors whose first two components are $0$ (because if they aren't, they are either not orthogonal on $e_1$ or not orthogonal on $e_2$). Furthermore, $A+B+(A\cup B)^\perp$ consists of all vectors where the first two components are $1$ (because they are the sum of the only element of $A$, $e_1$, the only element of $B$, $e_2$, and any vector orthogonal to then). Note that not even $e_1$ itself fulfils that condition (because its $e_2$ component is $0$),thus clearly $A+B+(A\cup B)^\perp$ is not the full Hilbert space.
Now things are different if $A$ and $B$ are subspaces. In that case, be $P_A$ the orthogonal projector onto $A$ and $P_B$ the orthogonal projector onto $B$. Since $A$ and $B$ are orthogonal, $P_AP_B=P_BP_A=1$ and thus $P_A+P_B$ is also an orthogonal projector: $(P_A+P_B)^2 = P_A^2+P_AP_B+P_BP_A+P_B^2=P_A+P_B$. And so is $P_C:=1-P_A-P_B$. By construction, $P_A+P_B+P_C=1$.
Now consider an arbitrary vector $v\in H$. Then we can define $v_A=P_Av$, $v_B=P_Bv$ and $v_C=P_Cv$. Clearly $v_A+v_B+v_C = (P_A+P_B+P_C)v = v$. Also $v_A\in A$ and $v_B\in B$ because by definition, $P_A$ and $P_B$ are the orthogonal projectors to $A$ and $B$. So all left to prove is that $v_C\in(A\cup B)^\perp$.
Consider an arbitrary vector $w\in(A\cup B)$. Then either $w\in A$ or $w\in B$. Assume without loss of generality that $w\in A$. Then $P_Aw=w$. Now we calculate the scalar product with $v_C$:
$$\langle w,v_C\rangle = \langle P_Aw,P_Cv\rangle = \langle P_CP_Aw,v\rangle = \langle(1-P_A-P_B)P_Aw,w\rangle = \langle(P_A-P_A)w,v\rangle=0.$$
Therefore $v_C$ is orthogonal to $w$, and thus, since $w$ was chosen arbitrary from $(A\cup B)$, $v_C\in(A\cup B)^\perp$.
Thus we can decompose an arbitrary vector $v\in H$ into a sum of a vector $v_A\in A$, a vector $v_B\in B$ and a vector $v_C\in (A\cup B)^\perp$, and therefore $$A+B+(A\cup B)^\perp=H.$$ |
H: $x,y,z$ independent and uniform random on $[0,1]$, $P(x \geq yz)$?
$x,y,z$ are independent and uniform random on $[0,1]$.
I know that $P(x \geq yz) = 3/4$ by triple integrating the triple density function.
However, is there a easier way to get this solution? Without triple integration.
AI: The conditional probability given $z \in [0,1]$, $P(x \ge yz | z)$, is the area of the region of the square $[0,1]^2]$ below the line $y = x/z$. Since the region above the line
is a right triangle with sides $1$ and $z$, we have $P(x \ge y z | z) = 1 - z/2$.
Since $E[z] = 1/2$, the unconditional probability is $E[P(x \ge y z | z)] = E[1 - z/2] = 1 - E[z]/2 = 3/4$. |
H: Give an example of a noncyclic Abelian group all of whose proper subgroups are cyclic.
I've tried but I could not find a noncyclic Abelian group all of whose proper subgroups are cyclic. please help me.
AI: The simplest possible example of this would be $\mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z}$, as this is abelian and is the smallest group which is not cyclic. It is also known as the Klein four-group. |
H: Prove that any family of disjoint 8-signs on the plane is countable
I try to answer the following question from Basic Set Theory by Shen and Vereshchagin :
1.
(a).Prove that any family of disjoint 8-signs on the plane is countable.(By an 8-sign we mean a union of two tangent circles of any size; the interior part of the circles is not included).
(b)Prove a similar statement for letters T or E on the plane(but not M or O!).
AI: A good place to start is to ask yourself what the difference between the shapes of 8, T, and E on the one hand and M and O on the other. One thing that stands out is that each of 8, T, and E has a point that does not have a neighborhood that looks like a line segment: 8 has the point of tangency of the two circles, T has the intersection of the two straight lines, and E has the intersection of the vertical line and the middle horizontal line. (The other two horizontal lines of the E) can be straightened out so that it looks like $\vdash$. The M, on the other hand, can be straightened out into a line segment, and each point of the O looks like a point on C if you look at it up close, so it also looks like it’s on a line segment.
I would start by trying to show that a collection of pairwise disjoint T’s in the plane must be countable: they’re simpler than the 8, but every 8 contains a (slightly deformed) T, so if you can do it for T’s, you’ve basically done it for 8’s.
Suppose that $\mathscr{T}$ is an uncountable collection of pairwise disjoint T’s in the plane. For each $T\in\mathscr{T}$ there are rational numbers $p_T,q_T$, and $r_T$ with $r_T>0$ such that if $C_T$ is the circle of radius $r_T$ centred at $\langle p_T,q_T\rangle$, then the intersection point of $T$ is inside $C_T$, and the three endpoints of $T$ are outside $C_T$. There are only countably many triples of rational numbers, so there must be some $p,q,r\in\Bbb Q$ such that
$$\mathscr{T}_0=\{T\in\mathscr{T}:p_T=p,q_T=q,\text{ and }r_T=r\}$$ is uncountable.
Now trim the ends of each $T\in\mathscr{T}_0$ so that the three segments terminate exactly on the circle $C_T$; you now have a circle $C_T$ divided into three regions by the $T$ inside it. (If you bend the top bar of $T$ at the intersection point, this circle-with-T will look something like the Mercedes-Benz symbol.) Each of the regions must contain a point whose coordinates are both rational; call these points $x_T,y_T$, and $z_T$. There must be some $x,y,z\in\Bbb Q^2$ such that
$$\mathscr{T}_1=\{T\in\mathscr{T}_0:x_T=x,y_T=y,\text{ and }z_T=z\}$$
is uncountable. Can you show that if $T,T'\in\mathscr{T}_1$, then $T\cap T'\ne\varnothing$ and so get a contradiction? A sketch may help; note that $C_T=C_{T'}$.
I’ve added a suitable sketch below; $C=C_T=C_{T'}$, $P$ is the intersection point of $T'$, and $x,y$, and $z$ are as above. |
H: A finite sum of reciprocals of complex numbers cannot be confined to a half-plane
Let $z_1,\dots,z_n$ be non-zero complex numbers such that $\sum_{k=1}^n \frac{1}{z_k} = 0$. Prove that for any line $ax+by=0$ passing through the origin, $z_1,\dots,z_n$ cannot all lie in either of the half-spaces $\{ ax+by<0\}$, $\{ ax+by >0\}$. Any help would be greatly appreciated, thanks in advance!
AI: Note that $z_1,\ldots,z_n$ all lie in the same half-plane iff $\arg(z_1),\ldots,\arg(z_n)$ differ by less than $\pi$, and that $\arg(z)=-\arg(1/z)$. Thus we can substitute $1/z_i$ for each $z_i$, and need only show that if $\sum_i z_i=0$ then not all $z_i$ live in the same half-plane. Suppose all $z_i$ live in the half-plane defined by $ax+by>0$, and let $z_i=x_i+iy_i$. Then $0<\sum_i ax_i+by_i=a\sum_i x_i+b\sum_i y_i=0$, a contradiction. |
H: Two complex series of functions
The serie of function
$$\sum_{n\in\mathbb Z}\frac{1}{{(z-n)}^2}$$
converges normally in $\mathbb C\setminus\mathbb Z$ and it defines a meromorphic simply periodic function. Now let be $\Lambda$ a lattice, then the serie
$$\sum_{\omega\in\Lambda}\frac{1}{{(z-\omega)}^2}$$
doesn't converge normally in $\mathbb C\setminus \Lambda$ so it doesn't define an elliptic function. Geometrically, Why does this occur? If there is a discrete line of poles is ensured the convergence but if poles form a lattice the serie doesn't converge. I have just proved the above statements with analysis tools, but i don't understand why such two series have a so different behavior.
AI: Take, for example, $\Lambda=\{(n,m):n,m \in \mathbb{Z}\}$ and look at the sum for $z=0$ -- ignore $\omega=(0,0)$ for the sake of this discussion. Then you can rearrange terms so that you sum over the terms which are from the squares centered around $0$. The inner square has $8$ points contributing to the sum the next one already $16$, in general the $n$- th will contribute $2*(2n+1)+ 2*(2n-1)$ (draw a picture). So each of these squares contributes $Cn$ points, but the order of the terms is roughly $1/n^2$. In other words, the sum behaves like a harmonic series. This is basically because the number of terms of a certain magnitude is increasing, due to 2 dimensions, linearly. In one dimension the number of terms of a certain magnitude is constant.
This behaviour is similar to the behaviour you observe if you try to integrate $$\int_{|x|>1}\left(\frac{1}{|x|}\right)^k$$
in spaces of increasing dimensions. The bound for $k$, for which this converges, depends in a similar manner (for the same reason) on the dimension $n$. Technically this corresponds to a factor of $r^{n-1}$ in the volume element if written down in polar coordinates, geometrically this of course just the way the surface of the $(n-1)$-dimensional sphere scales.
In other words, this is not a question about complex analytic function. It becomes a question about these if you add terms forcing normal convergence and arrive at the Weierstrass function. But then other effects become predominant (topological ones, elliptic functions are functions on a torus). |
H: Is this 3D curve a circle?
The following is a curve in $3$ dimensions:
$$\begin{eqnarray} x & = & \cos(\theta) \\ y & = & \cos(\theta - \pi/3) \\ z & = & \cos(\theta - 2\pi/3) \end{eqnarray}$$
Is the curve a circle?
If it is, what about this curve in $4$ dimensions?
$$\begin{eqnarray} x & = & \cos(\theta) \\ y & = & \cos(\theta - \pi/4) \\
z & = & \cos(\theta - 2\pi/4) \\ w & = & \cos(\theta - 3\pi/4) \end{eqnarray}$$
I don't know if there is something like a circle in $4$-D. If there is, is this curve the $4$-D version of a circle?
P.S.: Is two-dimensional subspace the generalized plane? I want to learn more about this. What should I read?
AI: Here is a simple treatment of the $n$-dimensional case, where the point on the curve is $\vec x = (x_0, \ldots, x_{n-1})$ with $x_i = \cos(\theta-\pi i/n)$.
The curve lies on a sphere.
We have $x_i^2 = \cos^2(\theta-\pi i/n) = \frac12+\frac12\cos(2\theta-2\pi i/n)$. So $$\|\vec x\|^2 = \sum x_i^2 = \frac n2 + \frac12 \sum \cos(2\theta-2\pi i/n).$$ The latter term is zero because it is the sum of $n$ equally spaced sinusoids (it is equivalently the $x$-component of the sum of $n$ unit vectors equally spaced along the unit circle, or the real part of the sum of the $n$th roots of $e^{2n\theta\sqrt{-1}}$; in either case, the entire thing is zero by symmetry). So $\|\vec x\|^2$ is a constant, $\frac n2$, independent of $\theta$.
The curve lies on a two-dimensional subspace.
We have $x_i = \cos(\theta-\pi i/n) = a_i\cos\theta + b_i\sin\theta$ for some fixed $a_i$ and $b_i$ independent of $\theta$. Then $\vec x = \vec a\cos\theta + \vec b\sin\theta$. So $\vec x$ lies on the two-dimensional subspace spanned by $\vec a$ and $\vec b$.
Thus, $\vec x$ lies on the intersection of a sphere in $n$ dimensions and a two-dimensional subspace, i.e. a sphere in two dimensions, also known as a circle. |
H: Which parentheses are implied by $\prod$?
Which is correct?
$$ \prod_a ab = \left[ \prod_a a\right]b $$
or
$$ \prod_a ab = \prod_a \left[ ab \right] $$
I'd say the latter, but with $\sum$ we have
$$\sum_a a + b = \left [ \sum_a a \right ] + b $$
Do they work differently, or did I guess wrong? Also, are these things defined somewhere (or at least documented)? Wikipedia and Mathworld let me down on this one.
AI: If $b$ is does not depend on the index of summation, $$\sum_a(ab)=\left(\sum_aa\right)b=b\sum_aa\;,$$ so there’s no ambiguity. If $b$ does depend on the index of summation, either $\sum\limits_aab$ is to be understood as $\sum\limits_a(ab)$, or the writer made a bad mistake.
The expression $\sum\limits_aa+b$, however, is potentially ambiguous and should not be used; write $\sum\limits_a(a+b)$ if that’s the desired interpretation, and $\left(\sum\limits_aa\right)+b$ or, better, $b+\sum\limits_aa$ if that’s the desired interpretation. When reading, you have to use your head and pay attention to the context. E.g., something like $\sum\limits_{k=1}^na_k+b_k$ is surely intended to be read as $\sum\limits_{k=1}^n(a_k+b_k)$, though it’s a horribly sloppy way of writing it. $\sum\limits_{k=1}^na_k+b$, on the other hand, probably means $b+\sum\limits_{k=1}^na_k$, but if you see it being evaluated as $$\sum\limits_{k=1}^na_k+b=A(n)+nb\;,$$ it probably meant $\sum\limits_{k=1}^n(a_k+b)$, and the $nb$ is the result of summing those $n$ $b$ terms. |
H: Prove $\liminf\limits_{n\to\infty} F(x_n)\le F(x)$
Suppose $F$ is a nondecreasing and right continuous function, and the sequence $\{x_n\}_{n\geq1}$ converges to $x$. Then $\liminf\limits_{n\to\infty}F(x_n)\leq F(x)$.
How can I prove this?
AI: Hints:
For every $z\gt F(x)$, there exists $y\gt x$ such that $F(u)\leqslant z$ for every $u\leqslant y$.
Since $x_n\to x$, $x_n\leqslant y$ for every $n$ large enough.
Hence...
Edit:
... $F(x_n)\leqslant F(y)\leqslant z$ for every $n$ large enough. In particular, $\limsup\limits_{n\to\infty}F(x_n)\leqslant z$. This is valid for every $z\gt F(x)$, hence... |
H: Show a function is holomorphic
Let $\phi: [0,1] \rightarrow \mathbb{C}$ be a continuous function. For all $z \in \mathbb{C} \setminus [0,1] $ define $f(z) = \int_0^1\frac{\phi(t)}{t-z} \ dt$. Prove that f is holomorphic on $\mathbb{C} \setminus [0,1]$.
I can't express $f$ as a composition of holomorphic functions, is there another way to prove $f$ is holomorphic? Thanks in advance!
AI: By standard theorems on differentiation of parameter dependent integrals, for $z\notin [0,1]$,
(Edit: to make it a bit clearer: $f$ is differentiable in $z$ and)$$\frac{\partial}{\partial \bar{z}}f(z)=\int_0^1\left(\frac{\partial}{\partial \bar{z} }\frac{\phi(t)}{t-z}\right)dt = 0$$ |
H: Proof by induction for Division
I was kind of lost with the following example of induction:
$$ (11^{n+1} + 12^{2n-1}) \mathbin{\%} 133 = 0 $$
It shows the following steps to solve it: (I excluded base proof for n = 1)
$$ 11^{(n+1)+1} + 12^{2\cdot(n+1)-1} = $$
$$ = 11 \cdot 11^{n+1} + 12^{2} \cdot 12^{2n-1} = $$
$$ = 11 \cdot 11^{n+1} + 11 \cdot 12^{2n-1} + 133 \cdot 12^{2n-1} = $$
$$ = 11 \cdot ( 11^{n+1} + 11 \cdot 12^{2n-1}) + 133 \cdot 12^{2n-1} $$
$$ = 11 \cdot 133k + 133 \cdot 12^{2n-1} = $$
$$ = 133 (11k + 12^{2n-1}) $$
I prooven for n=1, but i am totally confused by what follows next in this example there is no description in the book why are these steps taken whats going on etc.
I get the first line expanding to n+1 but i dont know how he gets $$ 12^{2} $$ from factoring that. And i dont get the rest of it.
Thanks in advance for any help.
AI: Most of the missing manipulations are just applications of the distributive law, $a(b+c)=ab+ac$.
First,
$$\begin{align*}
12^{2(n+1)-1}&=12^{2n+2-1}=12^{(2n-1)+2}\\
&=12^{2n-1}\cdot12^2=12^2\cdot12^{2n-1}\\
&=144\cdot12^{2n-1}=(11+133)\cdot12^{2n-1}\\
&=11\cdot12^{2n-1}+133\cdot12^{2n-1}\;.
\end{align*}$$
this gets you down to the line $11 \cdot 11^{n+1} + 11 \cdot 12^{2n-1} + 133 \cdot 12^{2n-1}$. Then
$$11\cdot 11^{n+1}+11\cdot12^{2n-1}=11\left(11^{n+1}+12^{2n-1}\right)\;,$$
not $11\left(11^{n+1}+11\cdot12^{2n-1}\right)$. Your induction hypothesis was that $11^{n+1}+12^{2n-1}$ is divisible by $133$, so there is some integer $k$ such that $11^{n+1}+12^{2n-1}=133k$. Thus,
$$\begin{align*}
11^{(n+1)+1}+12^{2(n+1)-1}&=11\left(11^{n+1}+12^{2n-1}\right)+133\cdot12^{2n-1}\\
&=11\cdot133k+133\cdot12^{2n-1}\\
&=133\left(11k+12^{2n-1}\right)
\end{align*}$$
is also a multiple of $133$.
Added to answer question in comments:
$$\begin{align*}16\cdot16^{n+1}+4\cdot4^n-2&=(4+12)16^{n+1}+4\cdot4^n-2\\&=4\cdot16^{n+1}+12\cdot16^{n+1}+4\cdot4^n-2\\&=4\cdot16^{n+1}+4\cdot4^n+12\cdot16^{n+1}-8+6\\&=4\cdot16^{n+1}+4\cdot4^n-4\cdot2+12\cdot16^{n+1}+6\\&=4\left(16^{n+1}+4^n-2\right)+12\cdot16^{n+1}+6\end{align*}$$ |
H: How to generate equiprobable numbers with a random generator?
Is it possible to emulate a 2 sided coin flip (50/50) with a random number generator which outputs equiprobably the numbers 1, 2 and 3 ?
If yes, how ?
If not why ?
Is there a theorem?
Can it be expanded to any numbers x ("sides" of the coin) and y (random generator output) ?
AI: Yes, of course. The simplest way to do this is to treat $1$ as heads, $2$ as tails, and try again if you get a $3$.
The same approach works whenever $y>x$. If $y<x$, choose $n$ large enough that $y^n>x$, then generate $n$ numbers at a time and use the above approach with the resulting $n$-tuples. |
H: Combining primes
A friend felt the need to explore wikipedia and stumbled across "prime notation". I've been working with primes for awhile (cryptography programmer) and have never seen a notation for "combining prime" until today.
I'm not sure if this is an accepted notation or of it is some wikipedia art. I've searched around the web and haven't come across it. Maybe it is some obscure area of mathematics?
Q: Is "combining prime" an actual mathematical operation and does it have the following notation?
AI: Unfortunately, the notation prime (') mentioned on this Wikipedia page has nothing to do with prime numbers or cryptography. English has many confusing homographs (words spelled the same but with different meanings,) and this is one of them!
The "prime" that the Wikipedia article is talking about is a notation more often used in the past for things like feet and inches. For example 3' 6'' would be used to indicate "3 feet and 6 inches". |
H: Create Graph In Sage
I want define new graph in sage. Let $V$ be vector space over finite field $GF(q)$. The graph's vertices are 1-dimensional subspace from $V$ and $n-1$ -dimensional subspace from $V$ and two vertices are adjacent if and only if direct sum of two subspace is $V$.
I have trouble with define this graph in sage. Any suggestion?
AI: Here is a simple algorithm to create the graph you want.
p=5
n=3
V=VectorSpace(GF(p),n)
edges=[]
for i in V.subspaces(1):
for j in V.subspaces(n-1):
gens=list(j.gens())
gens.append(i.gens()[0])
H=V.subspace(gens)
if(H.dimension()==n):
edges.append((i,j))
G=Graph(edges)
I'm not sure what you want to label them, so the edges are actually the current vectorspaces, which is rather inefficient. You should probably come up with a simple naming scheme using a python dictionary, for instance. There are also likely faster ways to compute this that cut out a good deal of the tests. |
H: Prove inequality $\frac{a}{b}+\frac{b}{c}+\frac{c}{a}+\frac{9\sqrt[3]{abc}}{a+b+c}\geq 6$
Let $a,b,c>0$.
What is the proof that:
$$\frac{a}{b}+\frac{b}{c}+\frac{c}{a}+\frac{9\sqrt[3]{abc}}{a+b+c}\geq 6$$
AI: Take
$$ \frac{a}{b}+\frac{a}{b}+\frac{b}{c}\geq 3\frac{a}{ (abc)^\frac{1}{3}}$$
$$ \frac{b}{c}+\frac{b}{c}+\frac{c}{a}\geq 3\frac{b}{ (abc)^\frac{1}{3}}$$
$$ \frac{c}{a}+\frac{c}{a}+\frac{a}{b}\geq 3\frac{c}{ (abc)^\frac{1}{3}}$$
from AM-GM
and then add them and you get
$$\frac{a}{b}+\frac{b}{c}+\frac{c}{a} \geq \frac{a+b+c}{(abc)^\frac{1}{3}}$$
So it suffices to prove that
$$\left ( \frac{(a+b+c)}{(abc)^\frac{1}{3}}\, \, \, \, \, \, +\frac{9 (abc)^\frac{1}{3}}{a+b+c}\, \, \, \, \,\right )\geq 6$$
which holds
from the basic inequality $x^2+y^2 \geq 2xy$ |
H: Calculate the first derivative without the function?
I am a complete newbie when it comes to advanced mathematics, and am trying to learn calculus on my own. I wanted to know - is it possible to calculate the first derivative if you don't know the function that created a curve, but you DO have all of the points along the curve?
Edit: I created the curve using a cubic Spline interpolation
If so, can you point me to a place where I can learn how this would be accomplished?
Thanks!!
AI: If you have the curve, then geometrically, that is all you need to find a derivative value at a given point. You could estimate the direction of the tangent line at a given $x=a$. The slope of that tangent line is the value of $f'(a)$.
If you have a table of values, let's say you know $f(2.9), f(3), f(3.1)$, etc., but perhaps you have no info about $f(3.05)$. Then you can still estimate $f'(3)$ (in this what-if), by calculating the average rate of change over the smallest interval available in the data. For example, $f'(3) \approx \frac{f(3.1) - f(3)}{0.1} \approx \frac{f(3) - f(2.9)}{0.1}$. Perhaps a better estimate can be had by averaging those two to get: $f'(3) \approx \frac{f(3.1) - f(2.9)}{0.2}$.
Hope this helps! |
H: In eigenvector calculations, why does $A-\lambda I$ need to be singular?
I have this confusion.
If $A$ is a matrix, then its eigenvectors are given by $(A-\lambda I)x = 0$. Since $x$ is not equal to zero, $A-\lambda I$ needs to be singular. I don't understand why $A-\lambda I$ needs to be singular. I thought $A-\lambda I$ needed to be zero, isn't it? Any pointers?
AI: Let $(A - \lambda I)$ be an $n \times n$ matrix and $x$ be a non-zero $n \times 1$ vector. Let the lower case $c_i$ denote the columns of $A - \lambda I$. Now consider the following interpretation of $(A - \lambda I)x = \mathbf{0}$ as
$$ \pmatrix{\mid & \mid & & \mid \\
c_1 & c_2 & \dots & c_n \\
\mid & \mid & & \mid }
\pmatrix{x_1 \\ x_2 \\ \vdots \\ x_n} = \mathbf{0}$$
which is equivalent to $$x_1 c_1 + x_2 c_2 + \dots + x_n c_n = \mathbf{0}.$$ If the $x_i$'s are non-zero, then either
$c_1 = c_2 = \dots = c_n = \mathbf{0}$, and hence $A - \lambda I = \mathbf{0}$, or
The set of vectors $\{ c_1, c_2, \dots, c_n \}$ is linearly dependent. In such case, we say that the kernel of $A - \lambda I$ is non-empty (non-trivial). For more information, read about nullspaces here and here. |
H: order of a meromorphic function and relation
could any one help me to prove the following?
Let $f$ be meromorphic at $p$, whose Laurent series in local coordinate is $\sum_{n} c_n(z-z_0)^n$. The order of $f$ at $p$ denoted by $ord_p(f)$, is the minimum exponent actually appearing(with non zero coefficient) in the Laurent series expansion.$$ord_p(f)=min\{n:c_n\neq 0\}$$
Let $f$ be meromorphic at $p$. then $f$ is holomorphic at $p$ iff $ord_p(f)\ge 0$. In this case iff $f(p)=0$ iff $ord_p(f)>0$. $f$ has a pole at $p$ iff $ord_p(f)<0$. $f$ has neither $0$ nor a pole at $p$ iff $ord_p(f)=0$.
AI: These are rather straightforward results which can found in most books on Complex Analysis.
I'm going to write $m(f)$ to be $\mathrm{ord}_p(f)$.
1) $f$ is holomorphic at $p$ if and only if $m(f) \ge 0$.
This is more or less the definition of being holomorphic. If $f$ is holomorphic at $p$ then $f$ can be expanded into it's Taylor Series centered at $p$ which shows that the negative coefficients are vanishing. Conversely, if $f$ has a Laurent series expansion with vanishing negative terms, then the series is a simple power series from which we deduce that $f$ is holomorphic.
2) $f(p)=0$ if and only if $m(f) > 0$
If $m(f) < 0$ then the series isn't even convergent at $p$ (Either $f\rightarrow\infty$ in the case of a pole or $f$ experiences wild fluctuations in the case of an essential singularity (Picard/Casorati-Weirstrass Theorems)). If $m(f)=0$ then the series has a non-vanishing constant term $c_0$ from which it follows that $f(p) = c_0 \neq 0$. Finally if $m(f) > 0$ then we may write $$f(z) = (z-p)^{m(f)}g(z)$$ where $g$ is some holomorphic function (in fact we can say that $g(p)\neq 0$). It follows then that $f(p)=0$.
3) This is not quite correct as the order needs to be finite for the singularity to be classified as a pole. If $m(f) = -\infty$ then we conventionally say that $p$ is an essential singularity. So let $-\infty<m(f)<0$.
By definition we say that $f$ has an order $k$ pole at $p$ if and only if we can write $f$ as
$$f(z) = (z-p)^{-k}g(z)$$
where $g(z)$ is holomorphic. We require $k$ to be the smallest integer satisfying this condition. Equivalently we can say that $f$ has a pole of order $k$ when $|m(f)| = k$. To see this note that since $g$ is holomorphic, we can expand it into a series centered at $p$. Then the extra $(z-p)^{-k}$ term produces a Laurent series with order $m(f) = -k$.
4) $f$ has neither a zero nor a pole if and only if $m(f) = 0$.
This is essentially a special case of the above parts reworded. |
H: Formula for number of lines you can draw through $n$ points
So I've got a homework question I'm stuck on. It's asking me to develop a formula that when given $n$ points, it gives the number of straight lines that can be drawn through those points.
For example, the first two questions were "How many lines can be drawn through 3 points?" Which is 3, and "How many lines can be drawn through 4 points?" Which is 6. Now, it says "Develop a formula that gives the number of lines that can be drawn through $n$ points." I understand the relationship, and i've developed a formula, but it relies on having a predetermined list of answers.
I've found that where $a_n$ is the number of points you have, you can find the number of lines with $a_n = a_{n-1} + (n-1)$ where $a_{n-1}$ is the previous item in the list. However, I don't think that's the formula that my teacher is looking for. Is there a less complicated way to solve this problem?
AI: It says "drawn through $n$ points", but it really means "drawn through any two of $n$ points", where we assume no three points are collinear. So the number of lines is the same as the number of ways to choose two points out of $n$, where order doesn't count. Do you know about permutations, combinations, binomial coefficients? |
H: What are the subobjects of a manifold?
Categorically a subobject of an object $a$ of some category $A$ is an object $a'$ with a monic morphism to $a$, ie $a'\to a$, upto isomorphism.
When $A$ is either a Topological or Differential manifold, what are the subobjects of a manifold, and are they the same as submanifolds?
edit
My own thinking on this is that in the smooth category one ought to work with manifolds together with the tangent bundle structure and then epics turn out to be surjective submersions and monics, injective immersions. These being very useful notions in the wider context of manifold theory.
AI: Let $\mathcal{C}$ be the category of topological (resp. differential) manifolds.
The objects of $\mathcal{C}$ are topological(resp. differential) manifolds and the morphisms of $\mathcal{C}$ are continuous(resp. smooth) maps.
Let $f\colon X \rightarrow Y$ be a morphism in $\mathcal{C}$.
We claim $f$ is a monomorphism if and only if $f$ is injective.
Suppose $f$ is a monomorphism.
Let $x, y$ be distinct points of $X$.
Let $p$ be a $0$-dimensional object in $\mathcal{C}$.
There exists the unique morphism $g\colon p \rightarrow X$ such that $g(p) = x$.
Similarly, there exists the unique morphism $h\colon p \rightarrow X$ such that $h(p) = y$.
Since $g \neq h$, $fg \neq fh$.
Hence $f(x) \neq f(y)$.
Hence $f$ is an injective map.
Conversely, suppose $f$ is injective.
Clearly, $f$ is a monomorphism.
"Are they the same as submanifolds?"
Generally no.
Counter-example:
Let $f\colon \mathbb{R} \rightarrow \mathbb{R}^2$ be the map defined by $f(x) = (x^3, 0)$.
$f$ is smooth and injective but is not an immersion($f'(0) = 0$).
Hence $\mathbb{R}$ cannot be identified with a submanifold of $\mathbb{R}^2$ by $f$. |
H: Integration over a sphere
While I am aware that when integrating over a ball in $\mathbb{R}^n$, we have
$\int_{B(0,R)}f(x)dx=\int_{S^{n-1}}\int_0^Rf(\gamma r)r^{n-1}drd\sigma(\gamma)$
I cannot figure why it is true that
$\int_{S(0,r)}f(x)d\sigma(x)=\int_{S(0,1)}f(y)r^{n-1}d\sigma(y)$
I know this is very trivial but I keep thinking that on changing the variables, the Jacobian should be $r^n$ and not $r^{n-1}$, and that is clearly wrong.
Would be grateful for the help. I only require a vector calculus explanation.
AI: It is the surface integral, not the volume integral. Surface area is proportional to $r^{n - 1}$. If you want to use Jacobian determinant, I guess you might need to be able to formulate what $d\sigma$ is first. Iterated spherical coordinates (I don't know what it's actually called) should work. |
H: Differential equation involving function defined in terms of an integral
Let $$F(y)=\int_0^\infty e^{-2x}\cos(2xy)\, dx$$ for $y\in \mathbb{R}$. Show that $F$ satisfies $F'(y)+2yF(y)=0$.
I've shown that I can differentiate under the integral sign, and this gives $F'(y)=-2\int_0^\infty xe^{-2x}\sin(2xy)\,dx$, but how in the world is this equal to $-2yF(y)$? Is there some trick I'm missing here?
AI: Integration by parts:
$$u=\sin 2xy\,\,,\,\,u'=2y\cos 2xy$$
$$v'=-xe^{-2x}\,\,,\,\,v=\frac{1}{2}e^{-2x}\,\,\,,\,\,\text{so:}$$
$$\left.-2\int_0^\infty xe^{-2x}\sin 2xy\,\,dx=\sin 2xy\,\,e^{-2x}\right|_0^\infty-2y\int_0^\infty e^{-2x}\cos 2xy\,\,dx$$
Now just check the first term in the RHS above is zero and you're done.
Added: The above is wrong and the reason why is in the comment by Adrian below, yet the solution is way more involved than it'd appear and I'll try to post it later.
Further added: I already tried but I can't get what the OP asks. Putting:
$$\mathcal I:=\int_0^\infty xe^{-2x}\sin 2xy\,\,dx\,\,\,,\,\,\,\mathcal J:=\int_0^\infty e^{-2x}\cos 2xy\,\,dx$$
I get the following integrating by parts (first time $\,u=x\sin 2xy\,$ , second time $\,\sin 2xy\,$ , third time $\,u=x\cos xy\,$ , and all the times $\,v'=e^{-2x}\,$ ):
$$\mathcal I=\stackrel{\text{this equals zero}}{\overbrace{\left.-\frac{1}{2}xe^{-2x}\sin 2xy\right|_0^\infty}}+\frac{1}{2}\int_0^\infty e^{-2x}\sin 2xy\,\,dx+y\int_0^\infty xe^{-2x}\cos 2xy\,\,dx\Longrightarrow $$
$$\Longrightarrow \left.\mathcal I=-\frac{1}{4}e^{-2x}\sin 2xy\right|_0^\infty+\frac{y}{2}\mathcal J-\left.\frac{y}{2}xe^{-2x}\cos 2xy\right|_0^\infty+\frac{y}{2}\mathcal J-y^2\mathcal I$$
as the limits above equal zero, we get:
$$\mathcal I=yJ-y^2\mathcal I\Longrightarrow \mathcal I=\frac{y}{y^2+1}J$$
so I get that denominator $\,y^2+1\,$ there that shouldn't, according to the OP, be there. Either I'm mistaken or the OP is, but I already checked this several times.
Everybody is wholeheartedly invited to either confirm my calculations of refute them. Thanks. |
H: Equivalent definitions of linear function
We say a transform is linear if $cf(x)=f(cx)$ and $f(x+y)=f(x)+f(y)$. I wonder if there is another definition.
If it's relevant, I'm looking for sufficient but possibly not necessary conditions.
As motivation, there are various ways of evaluating income inequality. Say the vector $w_1,\dots,w_n$ is the income of persons $1,\dots,n$. We might have some $f(w)$ telling us how "good" the income distribution is. It's reasonable to claim that $cf(w)=f(cw)$ but it's not obvious that $f(x+y)=f(x)+f(y)$. Nonetheless, there are some interesting results if $f$ is linear. So I wonder if we could find an alternative motivation for wanting $f$ to be linear.
AI: Assume that we are working over the reals. Then the condition $f(x+y)=f(x)+f(y)$, together with continuity of $f$ (or even just measurability of $f$) is enough. This can be useful, since on occasion $f(x+y)=f(x)+f(y)$ is easy to verify, and $f(cx)=cf(x)$ is not. |
H: Show that exists $\delta >0$ such that:$d(x,y)\geq \delta$
This is the problem:
Let $X$ be a metric space with metric $d$ and $K\subset X$ compact and $F\subset X$ closed and $K\cap F=\varnothing$.
Let $x\in F,y\in K$.
Prove that there exists $\delta>0$ such that for every $x\in F$ and $y\in K$, $d(x,y)>
\delta.$
My idea was to use sequences,once I know that in a compact set we have a subsequence converging to an element in $K$ and how $F$ is closed there's a sequence converging to an element in $F$.
But ,then I got stuck...I don't know what to do now.Any hint?Much appreciated!
AI: Assume by contradiction, then for every $n$ you can find $x_n\in K, y_n\in F$ such that $d(x_n,y_n)<\frac1n$.
Since $K$ is compact there is a convergent subsequence $x_{n_k}$, whose limit is $x$. By triangle inequality $$d(x,y_{n_k})\leq d(x,x_{n_k})+d(x_{n_k},y_{n_k})$$
Juggle $\varepsilon$'s a bit and derive that $x\in F$, and the contradiction. |
H: Using adjacency matrix to calculate the number of hamiltonian paths
I heard that adjacency matrix can be used to calculate the number of k-length paths in a graph. Can't this be used as a way to calculate the number of hamiltonian paths?
AI: The adjacency matrix does not calculate the number of $k$-length paths in a graph. It calculates the number of $k$-length walks from one vertex to another. (More specifically, the entries of the $n$th power of the adjacency matrix encodes the number of walks). To see the difference between paths and walks, consider any two adjacent vertices not part of a cycle. Then there is exact one path between the two vertices i.e. the edge between them, but there exists an infinite number of walks between them. For example $v_1 \rightarrow v_2 \rightarrow v_1 \rightarrow v_2$ is a walk from $v_1$ to $v_2$ of length $3$. Specifically, paths do not allow repetition of vertices while walks do, therefore this does not in general aid in the computation of Hamiltonian paths or indeed, paths of arbitrary lengths. |
H: Manipulating Exponents
I'm doing my homework and there are a couple of things that I am having trouble grasping. All my homework asks is that I simplify the exponents. For example: 6^5 * 6^3 = 6^8
There are 2 problems I am unsure on what to do. They have to do with multiplying fractions:
(7^3)^5/6 = 7^5/2 is the answer but I don't understand how they got to that. I know there is a certain example I was shown where you had to get common demonators and such but if I apply that it doesn't turn out well. Same for this problem:
(7^3/5)^5/6 = 7^2
3^5/2 * 3^-2 = 3^1/2
AI: For any $x$, $a$, and $b$.
$(x^a)^b = x^{ab}$.
Hence for the first one, letting $x = 7$, $a = 3$, $b = \frac{5}{6}$, you have
$(7^3)^{\frac{5}{6}} = 7^{3 \cdot \frac{5}{6}} = 7^{\frac{5}{2}}$
For the second letting $x = 7$, $a = \frac{3}{5}$ and $b = \frac{5}{6}$, you have
$(7^{\frac{3}{5}})^{\frac{5}{6}} = 7^{\frac{3}{5} \cdot \frac{5}{6}} = 7^{\frac{3}{6}} = 7^{\frac{1}{2}}$.
For the last one, for any $x$, $a$ and $b$
$(x^a)(x^b) = x^{a + b}$.
Letting $x = 3$, $a = \frac{5}{2}$ and $b = -2$, you have
$(3^{\frac{5}{2}})(3^{-2}) = 3^{\frac{5}{2} + -2} = 3^{\frac{5}{2} + \frac{-4}{2}} = 3^{\frac{1}{2}}$. |
H: Silly problem with joint variation.
So, I was recently revisiting some old competition math problems and came to the jarring (and embarrassing) realization that I apparently don't understand joint variation as well as I thought. By way of example, if we have $4 apples = 3 oranges$ and $3 apples = 2 pears$, we could say that apples and oranges have a direct relationship and that apples and pears have a direct relationship, so: $apples/oranges = 3/4$ and $apples/pears = 2/3$.
My problem is that, we could also say that apples are jointly proportional to pears and oranges, which would yield: $4 apples = 6 oranges \times pears$ according to the joint variation formula. My question is how do we go from two direct variation equations to 1 joint variation equation?
This all came about when I was working through the following problem:
Given that x is directly proportional to y and to z and is inversely proportional to w and that x = 4 when (w,y,z) = (6,8,5), what is x when (w,y,z) = (4,10,9).
I know how to solve this problem, when I was doing this sort of thing, I was just taught to multiply when I saw inverse and divide for direct variation, so: $xw/yz=...$ and solve. I'm a little confused as to why this works.
AI: To say that $x$ is directly proportional to $y$ and to $z$, and inversely proportional to $w$, is to say that there is a constant $k$ such that
$$x=k\frac{yz}{w}.$$
(Here there is an implicit assumption that $y$, $z$, and $w$ are independent factors).
Since you know the value of $x$ when $(w,y,z)=(6,8,5)$, you can calculate $k$. And now that you know $k$, you can find $x$ when $(w,y,z)=(4,10,9)$.
We can short-circuit this process, at the cost of perhaps losing control over what's happening. To switch $y$ from $8$ to $10$, we multiply by $\frac{10}{8}$, that is, scale by that factor. To switch $z$ from $5$ to $9$, multiply by $\frac{9}{5}$. Finally, to switch $w$ from $6$ to $4$, we multiply by $\frac{6}{4}$ (note the changed order). So we multiply the original value of $x$ by $\frac{10}{8}\cdot\frac{9}{5}\cdot\frac{6}{4}$.
More unpleasantly still, let $(x_1,w_1,y_1,z_1)$ be the first values, and $(x_2,w_2,y_2,z_2)$ be the second values. Then
$$x_1=k\frac{y_1z_1}{w_1}\qquad\text{and}\qquad x_2=k\frac{y_2z_2}{w_2}.$$
Divide. We get after a little algebra
$$\frac{x_2}{x_1}=\frac{y_2z_2w_1}{y_1z_1w_2}.$$
To finish, substitute. |
H: How to evaluate $(-2\sqrt2)^{2/3}$?
What is the exact value of this expression?
$$ \left ( -2 \sqrt 2 \right )^{2 \over 3} $$
Isn't $2$ one of the answer? Wolfram gets $-1+i \sqrt 3$. Is root multivariable function for complex number?
AI: Let $x = (-2 \sqrt 2)^{2 \over 3}$.
$x^3 = (-2\sqrt{2})^2 = 8$.
Let $\omega \neq 1$ be a root of $x^3 = 1$.
Then, the roots of $x^3 = 8$ are $2, 2\omega, 2\omega^2$.
Let's compute $\omega$.
$x^3 - 1 = (x - 1)(x^2 + x + 1)$.
Hence $\omega = \frac{-1 + i\sqrt{3}}{2}$ or $\frac{-1 - i\sqrt{3}}{2}$.
Hence the roots of $x^3 = 8$ are $2, -1 + i\sqrt{3}, -1 - i\sqrt{3}$. |
H: How to normalize a range of values taking the difference into consideration
Given a set of numbers
e.g.
{1, 2, 3, 4, 5} or { 50, 100} or {50000, 50001}
I want to normalize these into a range with a min and max e.g. 2 >= x <= 50
My current algorithm is
$$
((range_{max} - range_{min}) / (x_{max} - x_{min})) * (x - x_{min}) + range_{min}
$$
This does result is numbers within the range however a set of numbers like {50000,500001} will result in 50000 = 2 and 50000 = 50 which is too skewed. In this case I would like a result still in the range but with 2 numbers closer together e.g. 2 & 3 or 30 & 31 .
What formula could I use to do this? I'm guessing I need to use $\log(x_{max} / x_{min})$ somewhere but I'm not sure how to work it into the equation.
AI: If what you're trying to do is pick radii of points for visualization, I would just use $$\max\left(2, 50\sqrt{x/x_\max}\right).$$ This accomplishes two things: the area of the point becomes proportional to the value, and extremely small values get clamped so their points are not too small to be visible. |
H: Expanding a basis of a subspace to a basis for the vector space
I'm not really sure how to extend a basis. I'm trying to do the following question.
Consider the subspace $ W = \{(x_1, x_2, x_3, x_4) \in \mathbb{R}^4 : x_1 = -x_4, x_2 = x_3\}$ of $ \mathbb{R}^4$. Extend the basis $\{(0,2,2,0),(1,0,0,-1)\}$ of $W$ to a basis of $ \mathbb{R}^4$.
I know I need to add another two vectors for it to be a basis of $ \mathbb{R}^4$ but I'm not sure how to pick the vectors. In general, how do you expand a basis?
AI: Hint: Any $2$ additional vectors will do, as long as the resulting $4$ vectors form a linearly independent set. Many choices! I would go for a couple of very simple vectors, check for linear independence. Or check that you can express the standard basis vectors as linear combinations of your $4$ vectors. |
H: Find $\lim_{x\to 0}x^{\frac{1}{x}}$
I tried to solve $$\lim_{x\to 0} x^{\frac{1}{x}}$$
I tried doing it like this:
$$\lim_{x\to 0} x^{\frac{1}{x}}= \lim_{x\to 0} e^{\ln(x^{\frac{1}{x}})} = \lim_{x\to 0} e^{\frac{\ln(x)}{x}} = \exp\left(\lim_{x\to 0} \frac{\ln(x)}{x}\right)$$
Then solving $\lim_{x\to 0} \frac{\ln(x)}{x}$ with l'Hospital's Rule I get
$$\lim_{x\to 0} \frac{\ln(x)}{x}=\lim_{x\to 0} \frac{1}{x}$$
Which remains indeterminate as it has no two-sided limit.
So is that the answer, that it remains indeterminate? Or would I say that it has two limits depending on the side you approach? How do I phrase this result?
AI: The expression $\lim_{x\to 0^-}x^{\frac{1}{x}}$ makes no sense, for in general $x^y$ is undefined if $x$ is negative. As a consequence, $\lim_{x\to 0}x^{\frac{1}{x}}$ does not exist.
As was pointed out by Lubin, L'Hospital's Rule is not appropriate here, since after taking the logarithm we are looking at $\frac{\ln x}{x}$. For $x$ near $0$, the numerator $\ln x$ is very large negative, but the denominator is close to $0$, so we do not have a suitable form. But luckily, the rule is not needed, since raising a number close to $0$ to a large power gives a number very close to $0$. |
H: Two subgroups being conjugate
As with the element-wise conjugation action, for $H$ and $K$ subgroups of $G$ in the same $G\text{-orbit}$ under this action, say that $H$ and $K$ are conjugate.
I know that $Gs_o = \lbrace gs_o\ :\ g ∈ G\rbrace$ can be one orbit for G-orbits, but I am not just getting what "same $g\text{-orbit}$" means.
If I am somehow mistaken, can anyone show me where I got wrong in the meaning of G-orbits and relate it to the comprehension of the quote?
AI: Let $\mathcal{S}$ be the set of subgroups of $G$.
For $H \in \mathcal{S}$ and $g \in G$, we denote $gHg^{-1}$ by $g.H$.
Then $G$ acts on $\mathcal{S}$ by $H \rightarrow g.H$.
If $H, K \in \mathcal{S}$ are in the same $G$-orbit in this action, we say $H$ and $K$ are conjugate. |
H: Ellipse with non-orthogonal minor and major axes?
If there's an ellipse with non-orthogonal minor and major axes, what do we call it?
For example, is the following curve a ellipse?
$x = \cos(\theta)$
$y = \sin(\theta) + \cos(\theta) $
curve $C=\vec(1,0)*\cos(\theta) + \vec(1,1)*\cos(\theta) $
The major and minor axes are $\vec(1,0)$ and $\vec(1,1)$. They are not orthogonal.
Is it still an ellipse?
Suppose I have a point $P(p_1,p_2)$ can I find a point Q on this curve that has shortest euclidean distance from P?
AI: More explicitly, we have the decomposition
$$\begin{pmatrix}\cos\,t\\\cos\,t+\sin\,t\end{pmatrix}=\begin{pmatrix}\cos\,\lambda&-\sin\,\lambda\\\sin\,\lambda&\cos\,\lambda\end{pmatrix}\cdot\begin{pmatrix}\sqrt{1+\phi}\cos(t+\eta)\\\sqrt{2-\phi}\sin(t+\eta)\end{pmatrix}$$
where $\tan\,\lambda=\phi$, $\tan\,\eta=1-\phi$, and $\phi=\dfrac{1+\sqrt{5}}{2}$ is the golden ratio. You can check that both your original parametric equations and the new decomposition both satisfy the Cartesian equation $2x^2-2xy+y^2=1$. What the decomposition says is that your curve is an ellipse with axes $\sqrt{1+\phi}$ and $\sqrt{2-\phi}$, with the major axis inclined at an angle $\lambda$.
If we take the linear algebraic viewpoint, as suggested by Robert in the comments, what the decomposition given above amounts to is the singular value decomposition (SVD) of the shearing matrix; i.e.,
$$\begin{pmatrix}1&0\\1&1\end{pmatrix}=\begin{pmatrix}\cos\,\lambda&-\sin\,\lambda\\\sin\,\lambda&\cos\,\lambda\end{pmatrix}\cdot\begin{pmatrix}\sqrt{1+\phi}&\\&\sqrt{2-\phi}\end{pmatrix}\cdot\begin{pmatrix}\cos\,\eta&\sin\,\eta\\-\sin\,\eta&\cos\,\eta\end{pmatrix}^\top$$
The SVD is in fact an excellent way to look at how a matrix transformation geometrically affects points: the two orthogonal matrices on the left and right can be thought of as rotation matrices, reflection matrices, or products thereof, and the diagonal matrix containing the singular values amounts to nothing more than a scaling about the axes of your coordinate system. |
H: Intuition on the sum of first (n-1) numbers is equal to the number of ways of picking 2 items out of n.
While going through an equation today i realized that sum of first (n-1) numbers is [n*(n-1)/2] which is equal to combinations of two items out of n i.e [n!/((n-2)! * 2!)]. I need some intuition on how these two things are related?
AI: Wolfram proof without words. Note that this uses Pascal's Triangle. |
H: Iterated exponent of $i$
WolframAlpha seems to tell me that $e^{e^{e^{e^{e^{e^{e^{e^{e^{e^{e^i}}}}}}}}}} = 1$, see link. Is this just an error or is it for real? Adding one more $e$ to the bottom of the tower gives me the number $e$, so it's specific to the 11 $e$'s I used in the tower.
AI: You can readily check this using an independent method. Let $x_n + i y_n\in\mathbb{C}$ be the value of a tower of $n$ copies of $e$ with a single $i$ at the top, so that $x_{n+1}+iy_{n+1}=\exp(x_n+iy_n)$. This can be rewritten as $e^{x_n}\left(\cos y_n+i \sin y_n\right)$, giving the recursion
$$
x_{n+1}=e^{x_n}\cos y_n,\qquad y_{n+1}=e^{x_n}\sin y_n.
$$
The starting values are $x_0=0$ and $y_0=1$. Evaluating this recursion numerically gives the following table:
$$
\begin{eqnarray}
(x_1,y_1) &=& (0.5403023058681398, &&0.8414709848078965) \\
(x_2,y_2) &=& (1.1438356437916404, &&1.2798830013730222) \\
(x_3,y_3) &=& (0.9002890839010574, &&3.006900083345737) \\
(x_4,y_4) &=& (-2.438030346526128, &&0.3303849520417783) \\
(x_5,y_5) &=& (0.08260952954639851, &&0.028331354522507797) \\
(x_6,y_6) &=& (1.0856817633955023, &&0.030767067267249513) \\
(x_7,y_7) &=& (2.960056578435498, &&0.09110100745978908) \\
(x_8,y_8) &=& (19.21903374615272, &&1.7557331998479278) \\
(x_9,y_9) &=& (-40856897.72613553, &&218399070.28039825) \\
(x_{10},y_{10}) &=& (-0.0, &&-0.0) \\
(x_{11},y_{11}) &=& (1.0, &&-0.0),
\end{eqnarray}
$$
which seems to confirm what Alpha says. However, it should be clear that what's actually happening is that a large negative real part is reached at $n=9$. This produces a numerical zero at $n=10$, followed by $1.0000\ldots$ at $n=11$. While the correct value at $n=11$ is close to $1$, it's not exact. The exact value will differ from $1$ somewhere around the $18$-millionth digit. |
H: Rational solutions for arbitrary p
Let p be a prime, and by experimenting with various p, how can we guess a necessary and sufficient condition for x^2 + y^2 = p to have rational solutions.
And how to prove the guess?
One thing I see is maybe using Sylvester's determinant, by how and why?
AI: Let $p$ be a prime number.
We claim that $x^2 + y^2 = p$ has a rational solution if and only if $p = 2$ or $p \equiv 1$ (mod 4).We assume we have the basic knowledge of the ideal theory of $\mathbb{Z}[\sqrt{-1}]$.
Let $K = \mathbb{Q}(\sqrt{-1})$.
Suppose $x^2 + y^2 = p$ has a rational solution.
Let $\gamma = x + \sqrt{-1}y \in K$.
Then $N(\gamma) = p$.
Let $A = \mathbb{Z}[\sqrt{-1}]$.
It is well known that $A$ is a principal ideal domain.
Suppose $\gamma = \alpha/\beta$, where $\alpha, \beta \in A$ and $\alpha$ and $\beta$ are relatively prime.
Then $N(\gamma) = N(\alpha)/N(\beta)$.
Hence $N(\alpha) = pN(\beta)$.
Since $\alpha$ and $\beta$ are relatively prime, $N(\alpha)$ and $N(\beta)$ are relatively prime.
Hence $N(\beta)$ is not divisible by $p$.
Since $N(\alpha)$ is divisible by $p$, there exists a prime ideal $P$ dividing $\alpha$ and $p$.
Then $N(P) = p$.
Hence $P \neq pA$.
Hence, by the theorem of the decomposition of a prime number in $K$, $p = 2$ or $p \equiv 1$ (mod 4).
Conversely suppose $p = 2$ or $p \equiv 1$ (mod 4).
There exists a prime ideal $P$ such that $N(P) = p$.
Hence $x^2 + y^2 = p$ has an integer solution.
QED |
H: A nontrivial subgroup of $G$ contained in every other nontrivial subgroup.
This question is from a collection of past master's exams.
Let $G$ be a group with a subgroup $H$ as described in the title. I'd like to show that $H$ is in the center of $G$.
My intuition is to choose some element $x\in H$ (and thus in every nontrivial subgroup $K$ of $G$) and then apply the counting formula; i.e. the order of $G$ is the product of the order of the centralizer of $x$ with that of its conjugacy class. I feel like this, together with the class equation, should tell me everything I need to know. It's been awhile since I've done a problem like this, so any help would be appreciated.
AI: Consider an arbitrary element $g\in G$. $g$ induces the cyclic subgroup $C:=\{g^n:n\in\mathbb Z\}$. Since $C$ is nontrivial (unless $g$ is the neutral element), $H$ must, by definition, be a subgroup of $C$. But all elements of $C$ commute with $g$, by construction. Therefore since $H$ is a subgroup of $C$, all elements of $H$ also commute with $g$. But $g$ was chosen arbitrary, thus all elements of $H$ commute with every element of $G$, thus $H$ is a subgroup of the center. |
H: Definition of Geodesic - Distance between two points on the same latitude
I do have a problem understanding the concept of geodesics. As I understand it a geodesic is the shortest distance between two points on any manifolds. Let's consider a spherical earth as depicted in this figure from the Wikipedia. As I read in the Wikipedia article about geodesics they are always part of a great circle. Though, if I am not mistaken, the shortest distance between two points on the same latitude (dashed lines in the figure) is to follow this latitude, is it not? But latitudes are not great circles (which are the solid lines in the figure).
So if following the latitude is the shortest path between two points on the same latitude, i.e. the geodesic, this would not be part of a great circle, since latitudes are not great circles which contradicts that geodesics are always part of a great circle. Can someone please point out my error or help me to understand the terminology better?
Thank you very much.
AI: Lines of latitude are not shortest paths, except for the Equator. Here is a counterexample that's easy to visualize.
Suppose you're very close to the North Pole, say at $89^\circ$ North, and you want to go to the opposite end of the same latitude. At this scale, the line of latitude looks like a small circle. It's shorter to cut straight across the diameter, passing over the North Pole, than to walk halfway along the circle's circumference.
This is one reason why long-distance flight paths look funny when you see them on a map. Most maps make lines of latitude be straight lines, so the actual shortest paths look bent. |
H: Partial recurrence relation for the number of permutations in $S_n$ which have a square root.
I ran into this problem the other day. The proof is supposed to be done by exhibiting an explicit bijection between two sets, without using induction, recurrence, or generating functions.
Denote by $\omega(n)$ the number of permutations $\sigma\in S_n$ so that $\sigma$ has a square root (that is, there exists $\tau\in S_n$ so that $\tau^2 = \sigma$). Prove that $\omega(2n+1) = (2n+1)\omega(2n)$.
AI: The problem features as exercise 59 in chapter 3 of Miklos Bona's Combinatorics of permutations.
The solution to the exercise is on page 337 of the book, which is not included in the Google preview but can be found here. The proof is due to Dennis White. |
H: What does it mean for an expression to hold "identically"?
As an example, the following expression $$\sin2x=2\sin x\cos x$$ is a trigonometric identity. Because it is an identity we can replace $x$ by $ax$ and differentiate with respect to $a$ to get $$2x\cos2ax=2x\cos^2ax-2x\sin^2ax.$$ Then, let $a=1$ to obtain another well known identity $$\cos2x = \cos^2x-\sin^2x.$$ But what does it mean to hold "identically" ? Is there a definition ?
AI: It depends on the context. Usually it means that the LHS and RHS describe functions $f, g$ from some set $X$ to some other set $Y$ and the claim is that they are precisely the same function, which is equivalent to saying that $f(x) = g(x)$ for all $x \in X$.
Sometimes it doesn't mean this. For example, if we say that $f(x) = g(x)$ identically where $f, g$ are polynomials, it means that all of their coefficients are equal. This is not equivalent to saying that $f(x) = g(x)$ for all $x$ if $f, g$ are polynomials over a finite field. In other words, the notion of equality implicit here is equality in a ring of polynomials. |
H: Finding range of a linear transformation
Define $T: \mathbb{R}^3 \rightarrow \mathbb{R}^2$ by $T(x,y,z) = (2y + z, x-z)$. Find $\mbox{ker}(T)$ and $\mbox{range}(T)$
I could find the kernel easy enough, and ended up getting $\{(-2x, x, -2x) : x \in \mathbb{R}\}$ but I don't really know how the get the range. In this case isn't the range effectively just the set with elements satisfying the equation $T$? I'm not really sure what the question is wanting to be honest. Some help would be great.
Thanks
AI: $(2y+z,x-z)=x(0,1)+y(2,0)+z(1,-1)$. Since $(0,1)$ and $(2,0)$ span $\mathbb R^2$, the range is $\mathbb R^2$. To find the kernel, set $(2y+z,x-z)=(0,0)$ so that we have $z=x=-2y$. This gives the kernel to be $\{(-2y,y,-2y):y\in\mathbb R\}$ which is what you have obtained correctly. Note the kernel is simply the line passing through the origin with direction $(-2,1,-2)$. |
H: Evaluate $\lim_{x\to \pi/2}\frac{\cos x}{x-\pi/2}$ in terms of $(\cos x)'$
Calculate $$\lim_{x\to\frac{\pi}{2}}\frac{\cos x}{x-\frac{\pi}{2}}$$ by relating it to a value of $(\cos x)'$.
The answer is available here (pdf) at 1J-2.
However, I can't seem to make sense of what is actually being done here.
AI: Applying l'Hopital rule, you have
$$
\lim_{x\to\pi/2}\frac{\cos x}{x-\pi/2}=\lim_{x\to\pi/2}(\cos x)'=\left.(\cos x)'\right|_{x=\pi/2}
$$
that in turn becomes
$$\left.(\cos x)'\right|_{x=\pi/2}=\left.(-\sin x)\right|_{x=\pi/2}=-1$$ |
H: Is this problem correct that $HG'=G$?
Here, I have the following homework:
Let $G$ is a finite $p-$group and let $H$ be a subgroup of it such that $HG'=G$. Prove that $H=G$ ($G'$ is the commutator subgroup).
I have tried to show that $G\subseteq H$ by taking an element in $G$ but this way seems to be weak here. Is it possible that this exercise is printed mistakenly? Thank you friends.
AI: It suffices to prove that $G'$ is a subgroup of $\Phi(G)$, the Frattini subgroup of $G$. In the case of $p$-groups, you can do this by showing that $G/\Phi(G)$ is elementary abelian. (This is actually true for nilpotent groups in general, in which case $G/\Phi(G)$ just has to be abelian, though not necessarily elementary.) |
H: Rearrangement of sequences with limit $0$
Is it true that every real sequence that converges to zero has the property that every rearrangement of it also converges to zero?
I have a proof in mind, but I'm not 100% sure it's correct (although I'm pretty sure), so I just want a yes/no answer.
AI: Yes. A sequence converges to $0$ if and only if all but finitely many terms in it are less than $\epsilon$ in absolute value for any $\epsilon > 0$, and this condition is manifestly invariant under rearrangement. |
H: Elementary rectangles in product measurable spaces
I am working with showing certain equivalence between two probability spaces, one of them being a countable product space of finite spaces, i.e.
$$
(\Omega,\mathscr F) = \prod_{k=0}^\infty \left(X,2^X\right)
$$
so clearly $\mathscr F$ is generated with measurable rectangles, i.e. the class $\mathscr R$ given by
$$
\mathscr R:=\left\{\left.\prod_{k=0}^n A_k\times \prod_{k=n+1}^\infty X\right|A_k\in 2^X,n\in \mathbb N_0\right\}.
$$
But I want to use that there is a simpler class that generates $\mathscr F$, namely
$$
\mathscr R_e:=\left\{\left.\prod_{k=0}^n \{x_k\}\times \prod_{k=n+1}^\infty X\right|x_k\in X,n\in \mathbb N_0\right\}.
$$
which I was going to call "elementary rectangles" - but I am not sure if it is a good name, or if it has been used already somewhere in this or another meaning. What would you advise?
AI: In topology these are often called the basic open sets of $\prod_{k=0}^\infty X$,
because they form a commonly used basis for the product topology on that space.
I personally would write $X^{\mathbb N}$ or $X^\omega$ instead $\prod_{k=0}^\infty X$, though. But of course, your notation is more flexible as you can easily use it for products that don't have the same space $X$ in every coordinate.
One common notation for your elementary rectangles is $[s]$ where $s=(x_0,\dots,x_k)$. Here $[s]$ denotes all extensions of the finite sequence (or function, if you wish) in the space $X^{\mathbb N}$.
Not directly related to your question: you are probably aware of the fact that for finite spaces $X_k$, $k\in\mathbb N$, the product $\prod_{k=0}^\infty X_k$
is homeomorphic the Cantor cube $\{0,1\}^{\mathbb N}$ (or $2^\omega$ in more set theoretic notation). The Cantor cube is homeomorphic to the usual middle thirds
Cantor set on the real line.
Finally, if $X$ is an uncountable, separable complete metric space (a Polish space), then it is Borel-isomorphic to $\mathbb R$. I.e., all your spaces together with their Borel $\sigma$-algebras are actually isomorphic (neglecting the case that $X$ could be empty or have only one point) to $\mathbb R$ with its Borel $\sigma$-algebra.
Of course, you can have various different Borel probability measures on $\mathbb R$, some of which are easier to write down if you use another incarnation of the reals, such as $X^\omega$ for some finite space $X$. |
H: For which $m \in \mathbb N$ is the ideal $(m,x^2+y^2)$ prime in $\mathbb Z[x,y]$?
Let $m \in \mathbb N$. Find a necessary and sufficient condition for
$m$ such that the ideal $(m,x^2+y^2)$ is prime in $\mathbb Z[x,y]$.
I have to find for which $m$ the quotient ring is an integral domain. I do not know how to use the isomorphism theorem: is it ture that
$$
\mathbb Z[x,y] /(m,x^2+y^2) \cong (\mathbb Z_m[x])[y]/(x^2+y^2)?
$$
What should we do? Thanks in advance.
AI: Clearly $m = p$ has to be a prime, otherwise its factors are zero divisors.
Also, $\mathbb{Z}[x,y]/(p,x^2+y^2) \cong (\mathbb{Z}_p[x])[y] / (x^2 + y^2)$ is isomorphic by the third isomorphism theorem.
With $\mathbb{Z}_p$ being a field, $\mathbb{Z}_p[x]$ is a PID and so $(x^2 + y^2)$ is a prime ideal in $\mathbb{Z}_p[x,y]$ if and only if $f(y) := x^2 + y^2$ is irreducible over $\mathbb{Z}_p[x]$ (by Gauss's lemma).
$f(y)$ is irreducible if and only if it has no zeros (being a quadratic polynomial), and any zero would have to be of the form $\sqrt{-1}x$; so $f$ is irred. if and only if $-1$ is not a square ($\bmod p$).
By quadratic reciprocity (or rather its supplement) this is if and only if $p \equiv 3 (\bmod 4)$. |
H: Having Trouble With A Simple Equation Involving Differentials
Given that $xyz = c$, show that $dz = -z\left(\frac{dx}{x}+\frac{dy}{y}\right)$. I'am not sure how to get started. How do i differentiate this equation and wrt what? I really would like to try this before posting it but unfortunately i'am unable to even get started.
Thank You
AI: Supposing $x,y,z>0$ and applying the $\log$ we have
$$\log(xyz)=\log c$$
$$\log x+\log y+\log z=\log c$$
$$\frac{1}{x}dx+\frac{1}{y}dy+\frac{1}{z}dz=0$$
$$dz=-z\left(\frac{dx}{x}+\frac{dx}{y}\right)$$
The obtained result is valid also if the above conditions on the signs are not valid. |
H: Rearranging the spectral theorem
The spectral theorem for selfadjoint compact operators $L$ with infinite range says that $$Lx=\sum_{k=1}^{\infty} \alpha_k \langle x,f_k \rangle f_k, $$ where the $f_k$'s form an orthonormal system and the $\alpha_k$'s are real nonzero eigenvalues, such that they tend to zero. Now my question is, if we rearrange this sum arbitrary, will it still converge ?
My hunch would be "yes", since the proof of this theorem seems to work for any arrangement, but I'm feeling uneasy accepting that...
AI: It's a standard fact from hilbertian analysis that if $(x_i)_{i\in I }$ is an orthogonal (not necessarily orthonormal) system, then $(x_i)_{i\in I}$ is a summable family iff $(||x_i||^2)_{i \in I}$ is summable.
An application of Bessel's inequality and the previous fact shows that $(a_k \langle x, f_k \rangle f_k)_k$ is summable, and hence you can rearrange arbitrary this series and still get a convergent series with the same sum.
Indeed, since $(\alpha_k \langle x, f_k \rangle f_k)_{k \ge 0}$ is an orthogonal system, it's summable if and only if $\sum_{k\ge 0} \| \alpha_k \langle x, f_k \rangle f_k \|^2$ is finite, that is $\sum_{k\ge 0} | \alpha_k|^2 | \langle x, f_k \rangle |^2 \| f_k \|^2$ is finite.
But $(\alpha_k)_k$ is a convergent, and hence bounded, sequence, and $\|f_k \| = 1$, so we are reduced to prove that $\sum_{k \ge 0} | \langle x, f_k \rangle |^2$ is finite; but it's exactly the content of Bessel's inequality.
EDIT : A few facts about summable families
The precise definition of a summable family is as follows : a family $(x_i)_{i \in I}$ (where the set $I$ can be uncountable) of vectors in a Banach space $E$ is summable, with sum $x$, if for every $\epsilon > 0$, there exists a finite set $J \subset I$ such that for every finite set $K \subset I$ with $J \subset K$ one has $\| \sum_{i \in K} x_i - x \| \le \epsilon$. This can be equivalently characterised by the convergence of an appropriated net. If a family is summable, its sum is unique. We donote it by $x = \sum_{i \in I} x_i$.
This notion is invariant under rearrangement of indices : if $\sigma : J \to I$ is a bijection, then $(x_i)_{i \in I}$ is summable iff $(x_{\sigma(j)})_{j \in J}$ is summable, and sums are equal.
If the family of real numbers $(\| x_i \|)_{i \in I}$ is summable, then $(x_i)_{i \in I}$ is itself summable (provided $E$ is a Banach space, completeness is the key point here). The converse is not true in general, except if $E$ is finite dimensional.
In the special case where $I = \mathbb{N}$, summability implies the convergence of the series associated, but the converse is not true, even if $E$ is finite dimensionnal.
Indeed, if $E$ is finite dimensional (which covers the case of a family of real numbers), by what was said previously, a family $(x_n)_{n \in \mathbb{N}}$ is summable iff the series $\sum_{n = 0}^{+ \infty} x_n$ is absolutely convergent, that is $\sum_{n = 0}^{+ \infty} ||x_n|| < \infty$.
You can find further details and proofs in "General Topology" by N. Bourbaki. They cover the wider case of summable families in topological groups, but you can easily specialize to Banach space if you are not interested in such a generalization.
Now, I'll turn to summable families in Hilbert space.
Bessel's inequality asserts that if $(e_i)_{i \in I}$ is an orthonormal system of an Hilbert space $H$, then for every $x \in H$, the family $( |\langle x, e_i \rangle|^2)_{i \in I}$ of real numbers is summable, and its sum is less or equal to $\|x \|^2$.
Parseval identity says that if $(e_i)_{i \in I}$ is a complete orthonormal system in $H$, then for every $x \in H$, the family $( \langle x, e_i \rangle e_i)_{i \in I}$ is summable, and its sum is exactly $x$. Moreover, Bessel's inequality turns out to be actually an equality.
For all of this, and for the fact recalled at the beginning of my answer, you can take a look at "The Elements of Operator Theory" by C.S. Kubrusly. |
H: Dimensions of symmetric and skew-symmetric matrices
Let $\textbf A$ denote the space of symmetric $(n\times n)$ matrices over the field $\mathbb K$, and $\textbf B$ the space of skew-symmetric $(n\times n)$ matrices over the field $\mathbb K$. Then $\dim (\textbf A)=n(n+1)/2$ and $\dim (\textbf B)=n(n-1)/2$.
Short question: is there any short explanation (maybe with combinatorics) why this statement is true?
EDIT: $\dim$ refers to linear spaces.
AI: All square matrices of a given size $n$ constitute a linear space of dimension $n^2$, because to every matrix element corresponds a member of the canonical base, i.e. the set of matrices having a single $1$ and all other elements $0$.
The skew-symmetric matrices have arbitrary elements on one side with respect to the diagonal, and those elements determine the other triangle of the matrix. So they are in number of $(n^2-n)/2=n(n-1)/2$, ($-n$ to remove the diagonal).
For the symmetric matrices the reasoning is the same, but we have to add back the elements on the diagonal: $(n^2-n)/2+n=(n^2+n)/2=n(n+1)/2$. |
H: In a polynomial of $n$ degree, what numbers can fill the $n$?
Until now, I've seen that the $n$ could be filled with the set $\mathbb{N}_0$ and $-\infty$ but I still didn't see mentions on other sets of numbers. As I thought that having 0 and $-\infty$ as degrees of a polynomial were unusual, I started to think if it would be possible for other numbers to also fill the gap.
AI: So far you are talking about polynomials. I assume the coefficients are in some field $k$ so you would denote them as $k[x]$. Here the degree is in $\mathbb N_0\cup\{-\infty\}$.
Common generalisations are:
The ring $k[x,x^{-1}]$. So these are polynomials in the two variables $x$ and $x^{-1}$ with the relation $x\cdot x^{-1}=1$. Then we can define the degree similar to before as the highest non-vanishing power of $x$. Thus the degree will be in $\mathbb Z\cup\{-\infty\}$. Another name for these objects is Laurent polynomials.
The ring of formal power series $k[[x]]$. Here you allow also the degree $\infty$ for an formal infinite sum $\sum_{i=0}^\infty a_ix^i$. Degrees are in $\mathbb N_0\cup\{\pm\infty\}$
The ring of Laurent series. This is a combination of 1. and 2. Polynomials may include negative powers and (possibly infinitely many) positive powers of $x$. Note that we only allow finitely many negative powers since we can't porperly define multiplication otherwise. In the non-algebraic world people sometimes allow infinitely many negative powers as well. The degree however will be in $\mathbb Z\cup\{\pm\infty\}$
The field of fractions of the polynomial ring $k(x)$. Here an element is a fraction $\frac{P}{Q}$ of two polynomials, where $Q\neq 0$. Addition and multiplication are defined similar as the same operations for ordinary fractions. The degree can be defined as $deg(P)-deg(Q)$. Check that the properties of the degree wrt to the operations are still in place. Also note that $k(x)\neq k[[x]]$, although we also have degrees in $\mathbb Z\cup\{-\infty\}$. |
H: The variance in the average of a set of normally distributed random variables
I have a set of $M$ normally distributed random variables, $r_i$, each with an associated mean $u_i$, but the same variance $\sigma^2$. What is the variance of the average of these $M$ random variables: $\frac{\sum_{i=1}^{M} u_i}{M}$? How does the variance change as $M$ increases? What if the $M$ variables have a uniform, rather than a normal distribution, over some interval $[A, B]$?
AI: Assuming the M variables are independent the average has a normal distrbution with mean equal to the average of the u$_i$s as you guessed and variance σ$^2$/M. The mean and the variance will be the same for a uniform but the average will have its distirbution on[A, B]. but if you define all the uniforms to be over the same interval [A, B] they will be IID and the distribution when the mean is appropriately normalized will converge to a normal by the central limit theorem. |
H: Probability for 3 balls of different colours
I got this question in an exam a while ago and I wasn't sure how exactly I could solve it:
An urn contains 3 red balls, 3 blue balls and 3 green balls. Three are drawn at
random from the urn. What's the probability they're all of a different colour?
It is not specified if with or without replacement. Could someone please clarify me on both variants? Thanks :)
AI: If without replacement, 1st ball could be any, 2nd could be any of 6 of the 8, 3rd any of 3 of the 7, so $(6/8)(3/7)=9/28$.
If with replacement, 1st could be any, 2nd any of 6 of the 9, 3rd any of 3 of the 9, so $(6/9)(3/9)=2/9$. |
H: Find the probability of one expression
Could you please help me to calculate the probability as follows
\begin{align}
\rho=\mathsf{Pr}\left\{\frac{X_1}{X_2} >\frac{Y_1+a}{Y_2 + a}\right\}
\end{align}
where $X_1,X_2, Y_2$, and $Y_2$ are independent exponential distributed random variables with mean $\Omega_{X_1}, \Omega_{X_2}, \Omega_{Y_1}$, and $\Omega_{Y_2}$, respectively and $a$ is a positive constant.
AI: I will change the notations. Let $\mu_k = \mathsf{E}(X_k)$ and $\lambda_k = \mathsf{E}(Y_k)$ for $k=0, 1$.
Lemma 1. For $\alpha > 0$, $\mathsf{Pr}\left( X_1 > X_2 \alpha \right) = \frac{\mu_1}{\mu_1 + \alpha \mu_2}$.
Proof:
$$
\mathsf{Pr}(X_1 > \alpha X_2) = \mathsf{E}\left( \mathsf{Pr}(X_1 > \alpha X_2 |X_2) \right) = \mathsf{E}\left(\mathrm{e}^{\frac{\alpha}{\mu_1} X_2}\right)
$$
The latter is the moment generating function of $X_2$ at $t=\frac{\alpha}{\mu_1}$, yielding the result. $\square$
Lemma 2. $\mathsf{E}\left(\mathrm{e}^{-t Y_k}\right) = \frac{1}{1 + t \lambda_k}$.
Proof:
$$
\mathsf{E}\left(\mathrm{e}^{-t Y_k}\right) = \int_0^\infty \frac{1}{\lambda_k} \mathrm{e}^{-\frac{x}{\lambda_k}} \mathrm{e}^{-t x} \mathrm{d} x = \frac{1}{\lambda_k}\frac{1}{t+ \frac{1}{\lambda_k}} = \frac{1}{1+t \lambda_k} .\quad \square
$$
Now:
$$
\mathsf{Pr}\left( \frac{X_1}{X_2} > \frac{Y_1+a}{Y_2+a} \right) = \mathsf{E}\left( \mathsf{Pr}\left( \left. X_1 > X_2 \frac{Y_1+a}{Y_2+a} \right| Y_1, Y_2 \right) \right) \stackrel{\text{Lemma 1}}{=} \mathsf{E}\left( \frac{\mu_1}{\mu_1 + \mu_2 \frac{Y_1+a}{Y_2+a} } \right)
$$
The latter expectation can be evaluated using $\frac{1}{x} = \int_0^\infty \mathrm{e}^{-x t} \mathrm{d} t$:
$$
\mathsf{E}\left( \frac{\mu_1}{\mu_1 + \mu_2 \frac{Y_1+a}{Y_2+a} } \right) =
\mathsf{E}\left( \frac{\mu_1 (Y_2+a)}{\mu_1 (Y_2+a) + \mu_2 (Y_1+a)} \right) = \int_0^\infty \mathsf{E} \left( \mu_1 (Y_2+a) \mathrm{e}^{- t (\mu_1 (Y_2+a) + \mu_2 (Y_1+a))} \right)\mathrm{d} t
$$
The expectation under the integral sign is easy to evaluate using:
$$
\mathsf{E} \left( \mu_1 (Y_2+a) \mathrm{e}^{- t (\mu_1 (Y_2+a) + \mu_2 (Y_1+a))} \right) = -\left. \frac{\mathrm{d}}{\mathrm{d} s} \mathsf{E} \left( \mathrm{e}^{- s \mu_1 (Y_2+a) - t \mu_2 (Y_1+a))} \right) \right|_{s=t} \stackrel{\text{indep.}}{=} -\left. \frac{\mathrm{d}}{\mathrm{d} s} \mathsf{E} \left( \mathrm{e}^{- s \mu_1 (Y_2+a)} \right) \mathsf{E} \left( \mathrm{e}^{- t \mu_2 (Y_1+a))} \right) \right|_{s=t}
$$
Now using Lemma 2,
$$
\mathsf{E} \left( \mu_1 (Y_2+a) \mathrm{e}^{- t (\mu_1 (Y_2+a) + \mu_2 (Y_1+a))} \right) = -\left. \frac{\mathrm{d}}{\mathrm{d} s} \left(\mathrm{e}^{- s \mu_1 a - t \mu_2 a} \frac{1}{1 + s \mu_1 \lambda_2} \frac{1}{1 + t \mu_2 \lambda_1} \right) \right|_{s=t} = \mathrm{e}^{-a t (\mu_1 + \mu_2)} \frac{\mu_1 ( a +\lambda_2 + a t \lambda_2 \mu_1)}{ (1 + t \lambda_1 \mu_2 )( 1 + t \lambda_2 \mu_1)^2}
$$
It now remains to evaluate the integral with respect to $t$:
$$
\mathsf{Pr}\left( \frac{X_1}{X_2} > \frac{Y_1+a}{Y_2+a} \right) = \int_0^\infty \mathrm{e}^{-a t (\mu_1 + \mu_2)} \frac{\mu_1 ( a +\lambda_2 + a t \lambda_2 \mu_1)}{ (1 + t \lambda_1 \mu_2 )( 1 + t \lambda_2 \mu_1)^2} \mathrm{d} t
$$
The result involves exponential integrals:
$$
\mathsf{Pr}\left( \frac{X_1}{X_2} > \frac{Y_1+a}{Y_2+a} \right) = \frac{\mu_2 \left(\lambda_1 \lambda_2 \mu_1 + a \left(\lambda_2 \mu_1 - \lambda_1 \mu_2\right)\right)}{(\lambda_1 \mu_2 - \lambda_2 \mu_1)^2} f \left( \frac{a\left(\mu_1+\mu_2\right)}{\mu_1 \lambda_2} \right) -
\frac{\mu_1 \left(\lambda_1 \lambda_2 \mu_2 + a \left(\lambda_1 \mu_2- \lambda_2 \mu_1\right)\right)}{(\lambda_1 \mu_2 - \lambda_2 \mu_1)^2} f \left( \frac{a\left(\mu_1+\mu_2\right)}{\mu_2 \lambda_1} \right) + \frac{\lambda_2 \mu_1}{\lambda_2 \mu_1 - \mu_2 \lambda_1}
$$
where $f(x) = \exp(x) \operatorname{Ei}(-x)$.
The result can be obtained in a symbolic computer algebra systems, like Mathematica. The result can be obtained by using the partial fraction decomposition for the rational factor of the integrand. Here is the result, with verification by Monte-Carlo and by quadratures: |
H: Resources for IMO
I am seeking an online resource or any book where I can find the questions of International Mathematical Olympiad questions chapter wise eg Number Theory problems grouped together.
Like this site where we can get all the programming problems of SPOJ classified under different sections.
I have googled but didn't come across any such site. So I will greatly appreciate if anybody provides me some reference.
AI: Art of Problem Solving
Also check their forum, they have lots of extra materials:
Forum |
H: Checking Uniform Convergence
Question:
Prove that the sequence of functions $f_n(x) = x^n$ converges uniformly to zero on any interval of the form $[0,\mu]$ if $\mu < 1$.
My Work:
Recalling the definition of uniform convergence, we say that $f_n(x) \to 0$ uniformly if $$\forall\epsilon>0 \ \exists N \ \forall n \ge N \ \forall x \in \mathrm{Dom}\,(f) \ \Big[\left|f_n(x) - 0 \right|\le \epsilon \Big] $$
Since $x \in [0,\mu]$, we have $x \le \mu \Rightarrow \left| f_n(x) -f(x) \right| \le \mu^N$. We want $\mu^N \le \epsilon$ for some large $N$, so taking the natural logarithm of the inequality, which we can do because the natural log is monotone increasing for $x >0$, we have $$ \mu^N \le \epsilon \Rightarrow N \ln \mu \le \ln \epsilon $$ Since $\mu, \epsilon < 0$ for $\epsilon$ sufficiently small, $\ln \mu, \ln \epsilon < 0$. So the inequality becomes $$ N \ge {\ln \epsilon \over \ln \mu}$$
This makes sense to write because ${\ln \epsilon \over \ln \mu}$ is a positive number, so we can always take $N \ge {\ln \epsilon \over \ln \mu}$.
My Question: How do I incorporate this $N$ into an appropriately worded proof? I know it would go along the lines of "Taking $N \ge {\ln \epsilon \over \ln \mu}$, we have $$ \mu^N - 0 \le \mu^{{\ln \epsilon \over \ln \mu}} $$ By virtue of my previous calculation, I know this should be less than $\epsilon$, but I am not sure how to perform the necessary algebra. Its a small, but crucial detail.
Edit:
Thanks to Nate Eldrege, who pointed out that $a^b = \exp(b \ln a)$. So we have $$ \mu^{{\ln \epsilon \over \ln \mu}} = \mu ^{\ln(\epsilon - \mu)} = e^{\ln(\epsilon - \mu)}e^{\ln \mu} = \epsilon $$ So as required for any $n \ge N$, where the value of $N$ is given above, we have $\mu^N - 0 \le \epsilon$.
AI: Seeing as I have found my answer, and this question needs no more work, I am posting Nate Eldredge's hint as an answer and accepting it so that this question moves off the "Unanswered" list:
Nate Eldredge: "Hint: $a^b = \exp(b \ln a)$" |
H: Why is this sequence of functions not uniformly convergent?
For the following sequence of functions and its limit function, we can see that $f_n(x)$ is clearly pointwise convergent
$$f_n(x) = x^n\text{ }\forall x\in[0,1]\text{ and }\forall n\in\mathbb N^*\\
f(x) = \begin{cases}0&\text{if } x\in[0,1)\\1&\text{if } x=1\end{cases}$$
However, I was wondering why this is not uniformly convergent. The condition for uniform convergence is:
$$|f_n(x) - f(x)| < \epsilon,\ \ \ \forall x \text{ when } n > N$$
Now most sources present an argument along the lines of: assume that $f_n(x)$ is uniformly convergent and that $0 < x < 1$, this means that $x^n<\epsilon$ whenever $n>N$. Specifically, this would mean $x^{N+1}<\epsilon$ for some fixed $N$. But if we now pick $x$ such that $1 > x > ε^{\frac{1}{N+1}}$, then this would lead to a contradiction, therefore $f_n(x)$ is not uniformly convergent.
However, I was wondering why couldn't we take $n$ to infinity. If $0 < x < 1$, then $\lim_{n\rightarrow \infty} |f_n(x) - f(x)| = 0$ (which is less than $\epsilon$). Now since $|f_n(x) - f(x)|$ will always be $0$ if $n > \infty$, then wouldn't this be uniformly convergent?
AI: You can't just take $n=\infty$. In effect what you have done is to verify pointwise convergence, not uniform convergence.
Uniform convergence fails because, for every $n$, $\sup_{x\in[0,1]} |f_n(x)-f(x)| = 1$, which you can see by continuity.
Alternatively, uniform convergence must fail because each $f_n$ is continuous, $f$ is not continuous, and a uniform limit of continuous functions is continuous. |
H: why don't we define vector multiplication component-wise?
I was just wondering why we don't ever define multiplication of vectors as individual component multiplication. That is, why doesn't anybody ever define $\langle a_1,b_1 \rangle \cdot \langle a_2,b_2 \rangle$ to be $\langle a_1a_2, b_1b_2 \rangle$? Is the resulting vector just not geometrically interesting?
AI: This is Hadamard product, which is defined for matrices, and hence for vector columns. See Wikipedia page : Hadamard Product |
H: What's the limit of this expression: $\lim\limits_{M \to \infty}1/(\sum_{i=0}^{\infty}\frac{M!}{\left(M+i\right)!}x^{i})$
I have a question about a limit: Assume $x$ is a positive real constant $(x>0)$, then what's the limit of the following expression?
$$
\lim_{M\rightarrow\infty}\frac{1}{\sum_{i=0}^{\infty}\frac{M!}{\left(M+i\right)!}x^{i}}
$$
Is this dependent on the value of $x$? Thank you very much....
AI: As $1/x$ is continuous, you need to calculate
$$\lim_{M\to \infty} \sum_{i=0}^{\infty}\frac{M!}{\left(M+i\right)!}x^{i}.$$
We have
$$ \frac{1}{M^i} \geq \frac{M!}{(M+i)!} $$
independent of $x$. Thus,
$$\frac{M}{M-x}=\sum_{i=0}^\infty \frac{1}{M^i} x^i \geq\sum_{i=0}^{\infty}\frac{M!}{\left(M+i\right)!}x^{i} .$$
With $M\to\infty$, we find that the limit of the sum is smaller or equal to 1 for all $x$.
To find a lower bound, we just take the term corresponding to $i=0$ (all terms are positive), and we have
$$\sum_{i=0}^{\infty}\frac{M!}{\left(M+i\right)!}x^{i} \geq 1.$$
Concluding, we have that $$\lim_{M\to \infty} \sum_{i=0}^{\infty}\frac{M!}{\left(M+i\right)!}x^{i}=1$$
so your limit is also 1 independent of $x$. |
H: Homomorphism in algebraic systems
In the problems involving two algebraic systems, for eg.,$\langle S,*\rangle$ and $\langle P,\bigoplus\rangle$ where the sets $S=\{a,b,c\}$ and $P=\{1,2,3\}$. Here we have to check whether they both are isomorphic or not. While solving, they take values as $g(a)=3$, $g(b)=1$ and $g(c)=2$ and prove the systems as isomorphic. If I try other combination of values, it doesn't satisfy isomorphism. Then, on what basis these values are chosen(a=3,b=1,c=2)?
Kindly check out this Pg. 234 for the definitions of the operations.
AI: Look at the multiplication tables at the bottom of the page in your link, and try rewriting the second one with the columns and rows in the order $3,1,2$ instead of $1,2,3$. You should see something that looks almost identical to the table on the left, but with different symbols. Specifically, $a$ is replaced by $3$, $b$ by $1$ and $c$ by $2$. This is why the two are isomorphic - the two algebraic structures are the same, just the symbols for the elements are different. If you reshuffle the columns on the right-hand side in any other way, the tables won't match up properly, which is why other definitions of $g$ won't work. |
H: Help solving differential equation
I want to solve the following differential equation:
$y[t]$ : vertical position (height) of the object at time t
$y_c$ : height of the ceiling
$y_e$ : equilibrium point, the height at which the mass will stop at the end of its movement.
$a[t]$ : acceleration at time t
$t$ : time
$k$ : spring coefficient
$m$ : mass of the object
$G$ : gravity
$$\begin{align}
&F = -ky[t] - mG \\
\Leftrightarrow &ma[t] = -ky[t] -mG \\
\Leftrightarrow&my''[t] = -ky[t] -mG\\
\Leftrightarrow&y''[t] = -\dfrac{k}{m}y[t] -G
\end{align}$$
subject to:
$$y[0] = y_c\\y'[0] = 0$$
I'm not really sure how to do so. The equation is for modeling the movement of a falling object that is attached to a spring that is attached to the ceiling, thus gravity ($G$) is involved.
appreciate your help:
Appreciate your help :)
AI: Let $\omega = \sqrt{k/m}$ so that the differential equation we wish to solve is
$$ y'' + \omega^2 y = -G $$
Given any two solutions $y_1, y_2$, their difference satisfies
$y'' + \omega^2 y = 0$,
and the general solution of this is given by $y(t) = A cos(\omega t) + B \sin(\omega t)$. Hence the general solution of the original equation is
$$ y(t) = A \cos(\omega t) + B \sin (\omega t) + y_p(t)$$
where $y_p(t)$ is any particular solution to the original equation. However, the equation is easy enough that we can guess a solution using the method of undetermined coefficients. In this case we can guess $y_p(t) = c$ where $c$ is a constant. Plugging it into the differential equation yields:
$$ \omega^2 c = - G$$
hence $c = -G / \omega^2$. So the general solution to the original equation is
$$ y(t) = A \cos(\omega t) + B \sin(\omega t) - \frac{G}{\omega^2}$$
Setting $t=0$, we find
$$ y_0 = A - \frac{G}{\omega^2} $$
and
$$ y'_0 = B \omega $$
so that
$$ y(t) = \left(y_0 + \frac{G}{\omega^2} \right) \cos(\omega t) + \frac{y'_0}{\omega} \sin(\omega t) - \frac{G}{\omega^2} $$ |
H: Weak convergence in Sobolev spaces involving time
Let $$Lu:=-\sum_{i,j=1}^n(a^{i,j}(t,x)u_{x_i}(t,x))_{x_j}+\sum_{i=1}^nb^i(t,x)u_{x_i}(t,x)+c(t,x)u(t,x)$$ and the associated bilinear form $$B[u,v;t]:=\int_U\sum_{i,j=1}^na^{i,j}(t,x)u_{x_i}(t,x)v_{x_j}(t,x)+\sum_{i=1}^nb^i(t,x)u_{x_i}(t,x)v(t,x)+c(t,x)u(t,x)v(t,x)dx$$.
The notation is from Evans "Partial Differential Equation". Now suppose that I know that $u_m$ converges weakly to $u$ in the space $L^2(0,T;H^1_0(U))$, where $U$ is nice (open, bounded, etc.). Weak convergence in this space means
$$\lim_{m\to \infty}\int_0^T\langle \phi,u_m\rangle dt= \int_0^T\langle \phi,u\rangle dt$$ for all $\phi\in L^2(0,T;H^{-1}(U))$. I know that for every $\phi\in H^{-1}(U)$ there exists $\phi^0,\dots,\phi^n\in L^2(U)$ such that $\langle \phi,u\rangle = \int_U \phi^0u+\sum_{i=1}^n \phi^iu_{x_i}dx$ for all $u\in H^1_0(U)$.
Now my question is, why does $\int_0^TB[u_m,v;t]dt$ converges to $\int_0^TB[u,v;t]$? Of course, we have to use the weak convergence, but I do not see how exactly.
Just for completeness: for a Banach space $X$, we define $L^2(0,T;X)$ as the space of all measurable function $f:[0,T]\to X$ such that $\sqrt{\int_0^T\|f\|^2_Xdt}<\infty$.
AI: Here's a sketch:
Show that the bilinear forms $B[\cdot,\cdot\;; t]$ are continuous on $H^1_0(U)$, i.e. there is a constant $C$ such that $|B[f,g;t]| \le C \|f\|_{H^1_0(U)} \|g\|_{H^1_0(U)}$. Moreover, the constant can be taken independent of $t$. (Use Cauchy-Schwarz repeatedly. Here I'm assuming that the coefficient functions $a^{i,j}, b^i, c$ are bounded.)
Conclude that the bilinear form $(u,v) \mapsto \int_0^T B[u,v;t]\,dt$ is continuous on $L^2(0,T; H^1_0(U))$. (Cauchy-Schwarz will be used again.)
In particular, $u \mapsto \int_0^T B[u,v; t]\,dt$ is a continuous linear functional on $L^2(0,T; H^1_0(U))$.
Recall the definition of weak convergence in a Banach space: $x_n \to x$ weakly iff $\ell(x_n) \to \ell(x)$ for every continuous linear functional $f$. (The "definition" you give is based on the fact that the dual space of $L^2(0,T;H^1_0(U))$ can be naturally identified with $L^2(0,T; H^{-1}(U))$.) |
H: Finding the sum of this alternating series with factorial denominator.
What is the sum of this series?
$$ 1 - \frac{2}{1!} + \frac{3}{2!} - \frac{4}{3!} + \frac{5}{4!} - \frac{6}{5!} + \dots $$
AI: Hint: We have
$$e^{-x}=1-x+\frac{x^2}{2!}-\frac{x^3}{3!}+\cdots.$$
Multiply both sides by $x$ and differentiate. |
H: Prove that if $n$ is a positive integer then $2^{3n}-1$ is divisible by $7$.
I encountered a problem in a book that was designed for IMO trainees. The problem had something to do with divisibility.
Prove that if $n$ is a positive integer then $2^{3n}-1$ is divisible by $7$.
Can somebody give me a hint on this problem. I know that it can be done via the principle of mathematical induction, but I am looking for some other way (that is if there is some other way)
AI: Hint: Note that $8 \equiv 1~~~(\text{mod } 7)$.
So,
$$2^{3n}=(2^3)^n=8^n\equiv \ldots~~~(\text{mod } 7)=\ldots~~~(\text{mod } 7)$$
Try to fill in the gaps!
Solution:
Note that $8 \equiv 1~~(\text{mod } 7)$. This means that $8$ leaves a remainder of $1$ when divided by $7$. Now assuming that you are aware of some basic modular arithmetic,
$$2^{3n}=(2^3)^n=8^n\equiv 1^n ~~(\text{mod } 7)=1~~(\text{mod } 7)$$
Now if $2^{3n}\equiv 1~~(\text{mod } 7)$ then it follows that,
$$2^{3n}-1=8^n-1\equiv (1-1)~~(\text{mod } 7)~\equiv 0~~(\text{mod } 7)\\ \implies 2^{3n}-1\equiv 0~~(\text{mod } 7)$$
Or in other words, $2^{3n}-1$ leaves no remainder when divided by $7$ (i.e. $2^{3n}-1$ is divisible by $7$). As desired |
H: Removing term from RHS of equation
Suppose that we have the following equation:
$g = c\left( {\frac{{a + {\sigma ^2}}}{{b + {\sigma ^2}}}} \right)$
Is there anything that can be done to remove the $\sigma ^2$ term from the RHS of the equation so that the RHS $\rightarrow ca/b$, and the LHS of the equation has $\sigma ^2$ instead? Why or why not?
By dividing both sides by $\sigma ^2$, we get:
$\frac{g}{{{\sigma ^2}}} = c\left( {\frac{{\frac{a}{{{\sigma ^2}}} + 1}}{{b + {\sigma ^2}}}} \right)$
But this doesn't seem to get me any closer to my goal of removing the $\sigma ^2$ from the RHS. Maybe someone could point me in the right direction?
AI: Since it's not homework, here's what you can do. You want to put the sigmas on the left, so
$$
\begin{align*}
g &= \frac{ca+c\sigma^2}{b+\sigma^2}&\text{so}\\
g(b+\sigma^2) &= ca+c\sigma^2&\text{an hence}\\
gb+g\sigma^2 &= ca+c\sigma^2&\text{collect the terms to get}\\
gb+g\sigma^2-c\sigma^2&=ca&\text{and divide both sides by }b\text{ to obtain}\\
g+\frac{g-c}{b}\sigma^2 &= \frac{ca}{b}
\end{align*}
$$
Presto! We've isolated the sigmas on one side (with a bit of extra stuff) and we've managed to get $ca/b$ on the right. Of course this works in this particular example; in general things might not be arranged so that you can always get the form you want. |
H: Noncommutative rings, finding $a$ and $b$ such that every term in the sum $(a+b)^n = a^n + a^{n-1}b + \ldots$ is distinct
This question was inspired by the binomial theorem for rings. For commutative rings, we have the identity
$$(a+b)^n = \sum_{k=0}^n {n \choose k}a^kb^{n-k}$$
which does not hold for non-commutative rings. However, we can still expand $(a+b)^n$ to get
\begin{align*}
(a+b)^2 &= a^2 + ab + ba + b^2 \\
(a+b)^3 &= a^3 + a^2b + aba + ab^2 + ba^2 + bab + b^2a + b^3 \\
\cdots \\
(a+b)^n &= aa\ldots a + aa \ldots ab + aa \ldots ba + aa \ldots bb + \ldots
\end{align*}
This motivates the following question. For every $n$, is it possible to find a ring $R$ with elements $a$ and $b$ such that the $2^n$ terms in the expansion of $(a+b)^n$ are distinct?
There is also a stronger question: is it possible to find $a$ and $b$ such that for every $n$, this is true?
For example, taking $A = \begin{bmatrix}2 &1 \\0 &1\end{bmatrix}$ and $B = \begin{bmatrix}1 &2 \\0 & 1\end{bmatrix}$ from $M_2(\mathbb{Z})$ works for $n = 1, 2, 3$ and $4$ but fails for $n = 5$ because $ABBBA = BBAAB$.
AI: One way to do this is to take $R$ to be the group ring $\mathbb{Z}[F_2]$, where $F_2$ is the free group on generators $a$ and $b$. Then the fact that the summands are all different (for any $n$) is a direct consequence of the freeness of the group.
Of course, one doesn't have to explicitly embed a free group in a ring for a free group to be present. For example, $\begin{pmatrix} 1 & 2 \\ 0 & 1\end{pmatrix}$ and $\begin{pmatrix} 1 & 0 \\ 2 & 1\end{pmatrix}$ generate a free group of rank $2$, so you can take $R$ to be the ring of two-by-two integer matrices and these matrices as $a$ and $b$. |
H: Prove that $\left (\frac{a^2 + b^2 +c^2}{a+b+c} \right) ^ {(a+b+c)} > a^a b^b c^c$
Prove that $\left (\dfrac{a^2 + b^2 +c^2}{a+b+c} \right) ^ {(a+b+c)} > a^a b^b c^c$ if $a$, $b$ and $c$ are distinct natural numbers. Is it possible using induction?
AI: I recommend to read the answer from here. section Weighted AM-GM Inequality :) it is a good one . |
H: Question about topology definition
I am reading a topology definition:
Let $X$ be a set and let $\tau$ be a family of subsets of $X$. Then $\tau$ is called a topology on $X$ if:
Both the empty set and $X$ are elements of $\tau$
Any union of elements of $\tau$ is an element of $\tau$
Any intersection of finitely many elements of $\tau$ is an element of $\tau$
If $\tau$ is a topology on $X$, then the pair $(X, \tau)$ is called a topological space. The notation $X_\tau$ may be used to denote a set $X$ endowed with the particular topology $\tau$.
The members of $\tau$ are called open sets in $X$.
My question is:
Are the members of $\tau$ open sets or only called open sets?
AI: This definition abstracts the idea of an open set in a metric space, so in a sense the members of $\tau$ are only called open sets - they aren't open in the sense that they contain a ball around every point, because there isn't any distance to say what that means. (Of course, as other people have pointed out, once you decide to call them open sets, the distinction between being called open and actually being open is somewhat unclear).
The reason this definition is chosen is that the collection of open sets in a metric space (defined in the usual metric spaces way) forms a topology in this sense. So in that case the sets in $\tau$ are open both because they are in $\tau$ and because they are open in the metric sense. |
H: Number of basis representations of Boolean vectors
Suppose I have a basis $v_1, ..., v_n$ of boolean vectors and a boolean vector $v$ that was constructed by iterated $v = v_i$ $OR$ $v_j$ statements. I want a way to figure our how many unique combinations $v_I$, $I = \alpha_1,...,\alpha_k$, $\alpha_i \neq \alpha_j \forall i \neq j$ there are that, when union/or'd, yield $v$. The only thing I know about the basis vectors is that each basis vector has an entry that is true only for that basis vector, and false for all other basis vectors (note then that the vector $(TRUE,...,TRUE)$ has exactly one representation). The order of the basis vectors does not matter - I treat $v_i$ $OR...OR$ $v_j$ and $v_j$ $OR...OR$ $v_i$ equally. An analytic method is preferred but a polynomial-time algorithm would be great too. A polynomial-time algorithm for an upper bound for the number of basis representations (over the whole space) would be incredible, but if someone answered the main question I could probably do that on my own or with an optimization suite.
AI: For each i, let S(i) be the index of the bit that is set only in the
i'th basis vector v_i
If V is the result of orring together some subset T of basis vectors,
then for each i, the S(i)'th bit is set in V iff v_i is in T.
So for any vector V there is at most one such representation. |
H: Explain the arithmetics of this trigonometrical expression
$$(\frac{d_1}{2} + \frac{d_1}{2} \cdot \cos(2x))+(\frac{d_2}{2} \cdot \sin(2x)) = \frac{d_1}{2}.$$
Can someone explain how the arithmetics works here?
(My teacher has only noted that "filtration removes $\cos(2x)$ and $\sin(2x)$")
AI: As has been pointed out in comments, for this equation to hold for all $x$, we need to have $d_1=d_2=0$. The reference to being removed by filtration doesn't refer to an arithmetic operation inherent in this equation, but to the application of a low-pass filter indicated in the second image linked to in a comment. |
H: Question about measurable sets $E_n$ such that $\lim_{n\rightarrow \infty}L^N(E_n) = 0$.
Let $E,E_n \subset \mathbb{R}^N$ be measurable such that $E_n \subset E, E$ is bounded domain, $E_{n+1} \subset E$ and $\lim_{n\rightarrow \infty}L^N(E_n) = 0$. Are there $K_n$ compact such that $E_n \subset K_n \subset E$ and $\lim_{n\rightarrow \infty}L^N(K_n)=0$?
$L^N(A)$ denotes the $N$-dimensional Lebesgue measure of $A$.
AI: No. For a counterexample, take $N=1$, enumerate the rationals in $(0,1)$ as $(q_k)_{k\geqslant1}$ and consider $E=(-3,+8)$, $A_n=\bigcup\limits_{k\geqslant1}\,(q_k-1/(2^kn),q_k+1/(2^kn))$ and $E_n=A_n\cap (0,1)$. Then each $E_n$ is dense in $[0,1]$ hence the only $K_n$ compact such that $E_n\subseteq K_n\subseteq E$ are such that $K_n\supseteq[0,1]$. And $\mathrm{Leb}(E_n)\leqslant1/n$ but $\mathrm{Leb}(K_n)\geqslant1$. |
H: How to calculate the Mahalanobis distance
I am really stuck on calculating the Mahalanobis distance. I have two vectors, and I want to find the Mahalanobis distance between them. Wikipedia gives me the formula of
$$
d\left(\vec{x}, \vec{y}\right) = \sqrt{\left(\vec{x}-\vec{y}\right)^\top S^{-1} \left(\vec{x}-\vec{y}\right) }
$$
Suppose my $\vec{y}$ is $(1,9,10)$ and my $\vec{x}$ is $(17, 8, 26)$ (These are just random), well $\vec{x}-\vec{y}$ is really easy to calculate (it would be $(16, -1, 16)$), but how in the world do I calculate $S^{-1}$. I understand that it is a covariance matrix, but that is about it. I see an "X" everywhere, but that only seems like one vector to me. What do I do with the Y vector? Can someone briefly walk me through this one example? Or any other example involving two vectors.
AI: I think, there is a misconception in that you are thinking, that simply between two points there can be a mahalanobis-distance in the same way as there is an euclidean distance. For instance, in the above case, the euclidean-distance can simply be compute if $S$ is assumed the identity matrix and thus $S^{-1}$ the same. The difference, which Mahalanobis introduced is, that he would compute a distance, where the measurements (plural!) are taken with a correlated metric, so to say. So $S$ is not assumed to be the identity matrix (which can be understood as special correlation-matrix where all correlations are zero), but where the metric itself is given in correlated coordinates, aka correlation-values in the $S $ matrix, which are also cosines betwen oblique coordinate axes (in the euclidean metric they are orthogonal and their cosine/their correlation is zero by definition of the euclidean).
But now - what correlation does Mahalanobis assume? This are the empirical correlations between the x and the y - thus we need that correlations from external knowledge or from the data itself. So I'd say in answering to your problem, that the attempt to use Mahalanobis distance requires empirical correlations, thus a multitude of x- and y measurements, such that we can compute such correlations/ such a metric: it does not make sense to talk of Mahalanobis-distance without a base for actual correlations/angles between the axes of the coordinatesystem/the measure of the obliqueness in the metric.
Edit:When all correlations are zero, $S$ is diagonal, but not necessarily identity matrix. For $S$ to be equal to identity matrix all sampled variables must have equal value of standard deviation.
The $i$-th diagonal element of the matrix $S$ represents the metric for $i$-th variable in units of $\sigma_i$ (or proportional to $\sigma_i$) |
H: Confusion about pointwise vs. uniform convergence
I'm confused about expressions like $$L=\sum_{k=1}^{\infty} R_k, \quad \quad (1)$$where $L:H\rightarrow H$ and $R_k:H \rightarrow H$ are continuous linear operators in a Hilbert space $H$. We assume $\sum_{k=1}^{\infty} R_k x$ is convergent every $x\in H$.
The thing I'm confused about, is: If we know that for every $x\in H$ we have $$Lx=\sum_{k=1}^{\infty} R_k x,$$can we that automatically conclude that $(1)$ holds ?
I would conjecture "yes" since two maps are identical if they have the same domain and range (trivial here) and are identical for every argument - which they are, as the above shows.
But if the answer is indeed "yes", why is it in sum texts shown, that the $R_k$ additionally converge in the operator norm to $L$, i.e. $$\left\|L- \sum_{k=1}^{n} R_k\right\| \rightarrow 0\quad (n\rightarrow \infty),$$ before they conclude that $(1)$ is true ? Was that really necessary ?
AI: One of the issues with convergence is that it need not preserve continuity.
The simplest example (although not too relevant here) is $x^n$ on $[0,1]$ converging to a discontinuous function.
Another example is convergence for a series of rationals in $\mathbb Q$. If the limit is irrational, then it does not exist (although it does exist in a larger space). This case is similar, if the limit operator is not continuous then it does not exist in our space -- and the sequence is not convergent.
Pointwise convergence is indeed enough to show that $L$ is well-defined, but it is not enough to show that it is continuous. If we wish to talk about convergence in a space of continuous linear operators then we must also verify that the convergence is strong enough for that, namely that the limit operator is itself continuous. |
H: Foam-like graphs
What's the "official" name of a connected planar graph consisting entirely of polygons (cycles), glued together at edges, e.g. - among other things - without "end vertices" (of degree 1) and without edges not belonging to an inner face. Maybe something like "planar foam graph"?
Has this concept been generalized to higher dimensions: "3D foam graphs" (entirely consisting of polyhedrons, glued together at faces) and so on?
AI: It took me a while to figure out what you mean. To clarify a few things first. A planar graph does not come with faces. It depends on the combinatorial embedding of the graph, which determines the faces. However, if the graph is 3-vertex-connected, the combinatorial embedding is unique (up to a global reflection). Moreover, even if you have fixed the combinatorial embedding, you can draw every face as the outer face.
The graphs you are asking for have the following property. When deleting one edge, the graph will not split into two unconnected subgraphs. This property is known as 2-edge-connected. So you are interested in the planar 2-edge-connected graphs. The "3d generalization" could be the planar 3-edge-connected graphs, but I am not sure what you mean by this. |
H: How do I solve a Continued Fraction of solution to quadratic equation?
I know that it is possible to make a CF (continued fraction) for every number that is a solution of a quadratic equation but I don't know how.
The number I'd like to write as a CF is:
$$\frac{1 - \sqrt 5}{2}$$
How do I tackle this kind of problem?
AI: Let's rewrite $\ \dfrac {1-\sqrt{5}}2=-1+\dfrac {3-\sqrt{5}}2$ to have something positive to evaluate then :
$$\frac 1{\dfrac {3-\sqrt{5}}2}=\frac 2{3-\sqrt{5}}=\frac {2(3+\sqrt{5})}{(3-\sqrt{5})(3+\sqrt{5})}=\frac {2(3+\sqrt{5})}{9-5}=\frac {3+\sqrt{5}}{2}=2+\frac {\sqrt{5}-1}{2}$$
(we want the term at the right to be between $0$ and $1$ at each stage)
You may continue this process until repetition !
You should get :
$$\dfrac {1-\sqrt{5}}2=-1+\cfrac 1{2+\cfrac 1{1+\cfrac 1{1+\ddots}}}$$ |
H: How do I find clockwise angle of a point from the negative x-axis?
Title pretty much says it. How can I rotate a point around the origin and find the clockwise angle of the point from the negative x-axis? I tried using the atan(height/width), but that gives me the angle in the specific quadrant, not from the negative x-axis.
Edit I got some good advice in the comments. What this question is really asking is, "How can I calculate the clockwise angle between a vector and the negative x-axis?". I was looking for a programming answer, but both perspectives (programming and not) are answered below.
AI: Most programming languages provide an $\operatorname{atan2}$ function that deals with quadrants correctly. If you have a point $(x,y) \ne (0,0)$, then $\operatorname{atan2}(y,x)$ gives the counter-clockwise angle from the positive $x$-axis to $(x,y)$, in the range $(-\pi,\pi]$. Since you want the clockwise angle from the negative $x$-axis, it is enough to observe that when $\operatorname{atan2}(y,x) = 0$ the angle you want is $\pi$, and when it is $\pi/2$ you want $\pi/2$, so in general what you want is $\pi-\operatorname{atan2}(y,x)$, which lies in the range $[0,2\pi)$. |
H: How to determine if this is a field?
A finite subring $R$ of a field $V$ contains $1$ (so $1$ is an element of $R$).
The question is:
True or False: The ring $R$ must be a field.
I thought that if $R$ was a field it had to be a finite field in this case (because $R$ is a finite ring). And to be able to be a finite field, it should have
$$ |R| = p^n $$
With $p$ a prime and $n$ a natural number, but this doesn't have to be the case.
I could be wrong or what I'm saying could be insufficient to prove this right or wrong.
Please help ^^
AI: Hint: It is a finite integral domain. Or else if you are not acquainted with the relevant theorem, the only issue is whether every non-zero element $a$ of $R$ has a (multiplicative) inverse.
To show that it does, consider the powers of $a$. By finiteness, there must be integers $m$ and $n$, with $0\le m \lt n$ such that $a^m=a^n$. But then $a^{n-m}=1$. From this you should be able to show that some power of $a$ is the inverse of $a$.
Remark: You are right that if $R$ is a field (which it is) then $R$ will have $p^n$ elements for some prime $p$ and some positive integer $n$. However, that does not lead to a contradiction. |
H: How do I solve this limit scenario?
If $f$ is continuous at $5$ and $f(5)=2$ and $f(4)=3$, then $\lim_{x \to 2} f(4x^2-11)=2$
Is the above statement true or false with an explanation for the answer?
I'm not sure how to approach this.
AI: Theorem: if $g$ is continuous at $x_0$ and $f$ is continuous at $g(x_0)$, then the composite function $f(g(x))=fog(x)$ is continuous at $x_0$.
As you noted $f$ is continuous at $5$ and you know that polynomials are continuous at every real numbers especially at $x_0=2$. so $$\lim_{x\to2}f(4x^2-11)=f(\lim_{x\to2}(4x^2-11))=f(4*2^2-11)=f(5)=2$$ |
H: Derivation of formula for pressure gradient (fluid mechanics)
There is a derivation of a formula in my textbook which I don't fully understand. Most of the formulas I encounter I understand without difficulty, but if someone can help me understand one step of the following derivation, I would be very grateful:
Consider a flow of fluid through an inclined core sample of length $\Delta l$, with a constant flow rate, $q$, maintained at a pressure differential $\Delta p$. The flow at an angle $\theta$ above the horizontal can be described by the following version of the Darcy equation:
$$q = -A\frac{k}{\mu}\frac{d(p + \rho g z)}{dl}$$
where $z$ is the elevation in the gravitational field. Since $z = l \sin \theta$, with $l$ as the direction of flow, the equation written for the pressure gradient, becomes:
$$\frac{dp}{dl} = - \left(\frac{q \mu}{Ak} + \rho g \sin \theta\right)$$
OK, so it is this last step here I don't quite understand. I see that from the given equation we have:
$$q = -A\frac{k}{\mu}\frac{d(p + \rho g z)}{dl}$$
$$\frac{d(p + \rho g z)}{dl} = - \frac{q \mu}{Ak}$$
$$\frac{d(p + \rho g l \sin \theta)}{dl} = -\frac{q \mu}{Ak}$$
But I don't see this last step from here on. How do you "extract" the $\sin \theta$ part from the differential expression on the left side of this equation?
If anyone can explain this to me, I would appreciate it a lot!
AI: Are you sure about your definition of $x$? If $\theta$ is given as the angle from the horizontal, and $x$ is the horizontal component, then $x = l\cos \theta$, and $z = l \sin \theta$, which looks like what you need.
In your statement $\frac{d(p+\rho g l \sin \theta)}{dl}$, we simply apply differentiation rules. Since $\theta$ is independent of $l$, then we have
$$\frac{d(p+\rho g l \sin \theta)}{dl} = \frac{dp}{dl}+\frac{d\rho g l \sin\theta}{dl} = \frac{dp}{dl} + \rho g \sin \theta \frac{dl}{dl}.$$
This brings you to the form above -- if that's what you're asking. |
H: How many ways can you arrange the letters PLAY without repetition if the first letter can't be A and the second can't be Y?
How would you do this sort of problem in general, for example, if the the letter sequence was longer and there were more conditions?
One way I used to solve this problem was to enumerate the possibilities (I counted 14):
L => PLAY, PLYA
/
P
\
A = > PALY, PLYA
P => LPAY, LPYA
/
L
\
A => LAYP, LAPY
P => YPLA, YPAL
/
Y -- L => YLPA, YLAP
\
A => YAPL, YALP
The other way I tried to think about this was to start with all possible combinations of the letters and subtract out letter combinations that didn't meet the given conditions.
4! - 3! if you start with A - 2 x 1 * 1 * 2 if you start with P or L but have Y as the second letter.
It would seem if you make the problem larger and add more conditions it becomes error prone to try to subtract out all the unacceptable combinations of letters.
AI: There are $4!=24$ possible permutations without any restrictions. How many are bad? There are $3!=6$ that have A as the first letter, and there are $3!=6$ that have Y as the second letter, so to a first approximation there are $24-6-6=12$ allowable permutations. But some of the bad permutations are bad because they have A as first letter and Y as second letter, and we subtracted these twice: once on account of the A, and once on account of the Y. To correct that first approximation, we need to add these ‘doubly bad’ permutations back into the total. There are $2!=2$ of them, so the final count is $24-6-6+2=14$.
This technique of counting is known as the method of inclusion-exclusion and can be extended to arbitrarily large problems, though the computations may become rather extensive. |
H: Sequence of $C^1[0,1]$ functions $(f_n) \to f$ but $f \notin C^1[0,1]$
Question:
Give an example of a sequence of continuously differentiable function $(f_n)$ on $[0,1]$ so that $f_n \to f$ uniformly, but $f$ is not differentiable at all points of $[0,1]$.
My Thoughts:
Would a Fourier Series be a correct answer to this question? Take the Triangle Wave for instance. Wikipedia gives me the following equation:
Here $\omega$ is the angular frequency. Instead of $\infty$ in the sum, could each of my $f_n$ be $\sum_{k=0}^n$. In the limit, this sum of continuously differentiable functions converges to a function that is not differentiable at its cusps.
AI: A simpler alternative to the triangular wave is $f_n(x)=\sqrt{(x-\frac12)^2+\frac1n}$ and $f(x)=|x-\frac12|$. |
H: Open Sets - example
I have the following exercise:
Prove that $$A=\{(x,y)\in \mathbb{R}^{2} \mid x >0\}$$ is a open set.
I try to solve that exercise with the help of definition, so :
To prove that $A$ is open, we show for every point $(x,y) \in A$ there exists an $r>0$ such that $D_{r}(x,y)\subset A$. Now I must know the definition for $D_{r}(x,y)$ and from the definition we find out that: $\displaystyle D_{r}(x,y)=\{(\alpha,\beta)\mid{(\alpha,\beta)-(x,y) <r}\}.$
My question is: How do I prove that there is an $r>0$ such that $D_{r}(x,y) \subset A$ ?
Thanks :)
AI: Hint:
Consider the point $P=(x,y)=(7,0)$, can you think of an $r$ such that $D_r(P)\subset A$?
Consider the point $P=(x,y)=(7,4)$, can you think of an $r$ such that $D_r(P)\subset A$?
Consider the point $P=(x,y)=(7,y_0)$ (for some $y_0$), can you think of an $r$ such that $D_r(P)\subset A$?
Consider a point $P=(x,y)=(x_0,y_0)$ where $x_0>0$, can you think of an $r$ such that $D_r(P)\subset A$? |
H: Differential system on the torus 2
I've got the following question related to my previous post and I was suggested to write a new question, so there you are.
The question is the following: prove that if $f$ is smooth, periodic and with zero average, then the solution to the system on $\mathbb T^2$
$$\begin{cases}\frac{\partial}{\partial y}u'-\frac{\partial}{\partial x}v'=0,\\\\ \frac{\partial}{\partial x} u'+\frac{\partial}{\partial y}v'=f(x,y),\end{cases}$$
satisfies
$$\int_{0}^{2\pi}\int_0^{2\pi}\left(-\cos (y)u'(x,y)+\sin(x)v'(x,y)\right)\mathrm dx\mathrm dy=0.$$
I tried to apply similar techniques to those explained to me in the other question, however it seems that in this case I am not supposed to solve the system explicitly. In particular I cannot see where the condition on the zero mean of $f$ comes into play. Can anybody help me?
Many thanks
Guido
AI: It seems to follow if you multiply the first equation by $\cos x + \sin y$ and integrate by parts. |
H: Showing that a function is bijective
Show that the function $f$ defined by $$f(x):=\frac{x}{\sqrt{x^2+1}}\;,$$ $x$ is an element of the reals, is a bijection of the reals onto $\{y:-1<y<1\}$.
So we need to show that it is 1-1 and onto.
I begin by trying to prove that it is 1-1:
$$\frac{x}{\sqrt{x^2+1}} = \frac{y}{\sqrt{y^2+1}}$$
solving, we get $x^2 = y^2$
How can we now prove that $x=y$? I think there must be two solutions but am not sure what to do.
Also, I am not entirely sure where I should begin to prove that it is onto.
AI: Once you know $x^2 = y^2$, you know that either $x = y$ or $x = -y$. Plug both into the original equation and you'll know that $x = -y$ is impossible. |
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