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Hence, we have a selection of $a$ $3$s, and $b$ $2$s, where $3a+2b\leq 100$. Formally, we have to look at the cases $3a+2b=100, 3a+2b=99$, since if the sum was smaller than $98$ we could simply add a $2$.
Thus, we can write our product as $3^a2^{\frac{100-3a}{2}}$. If we look at this function and its maximum/monotony properties the only possible values for a maximum are $a=32, a=33$. Comparison leads to the maximum being at $a=32, b=2$.
Another possibility with less calculus would be to look at $3\cdot 3\geq 6$, immediately telling us that there can be no more than $3$ $2$s, else we could, again, substitute them with $2$ $3$s. In hindsight, this way might actually be a lot quicker and requires no differentiation...
In any case, with both variants we are done, and you have your maximum being what quite some people have pointed out so far.
• +1 Very interesting proof. It's always a little surprising to see discrete theorems with proofs involving differentiation. But you're right that the latter method of proof would be more elegant. Oct 10, 2014 at 5:06
If we write a continuous version $f(x)=x^{\frac{100}{x}}$ it maximum is when $x=e$, the closest natural number is 3 so the maximum is $3^{32}*4$.
I will add that $f(x)=x^{\frac{b}{x}};\ \forall b\geq 1$ have a maximum on $x=e$.
We need to see now that any composition will be lesser number that just an unique exponential. I write $$f(x's)=x_0^{\frac{b-\sum_{i}c_i}{x_0}}\cdot x_1^{\frac{c_1}{x_1}}\cdot x_2^{\frac{c_2}{x_2}}\cdots x_n^{\frac{c_n}{x_n}};\ c_i\neq c_j,\ \sum c_i<b\\ x_i,b,c_i>1$$
From above we can see that the maximum for any multiplier is when it base is $x_i=e$ so you can see that for any $c_i$ we choose it maximum value will be related to a base $x_i=e$ so doesnt exist any composition with $x_i\neq e$ that make the function to have a greater maximum choosing any $c_i$ decomposition that you want. | {
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This translated to natural numbers put the maximum with $x=3$, i.e. no closest natural number to $e$ that $3$ than lead to a number f(n) closest to f(e):
$$|f(e)-f(3)|<|f(e)-f(n)|\ \forall n\in \Bbb N-\{3\}$$
And we can add that if $0\not\equiv b \mod 3$ then the next number close to $f(e)$ is $f(2)$ so we must compose the number with powers of base 3 and 2. I.e.: $$r\equiv b\mod 3\ \rightarrow f(b)=3^{\frac{b-c}{3}}2^{\frac{c}{2}}\begin{cases}r\equiv 0, f(b)=3^{\frac{b}{3}}\\r\equiv 1, f(b)=3^{\frac{b-4}{3}}*4\\r\equiv 2, f(b)=3^{\frac{b-2}{3}}*2\end{cases}$$ | {
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• Sorry for going into a bit more detail here, since your argument is definitely valid and leads to the correct solution. But I do not see how the conclusion "the closest natural number is 3, so the maximum is" would possibly hold, being examined more closely? The assumption is correct, and you can prove it fairly easily, but I fail to see how this is a formal proof. Oct 9, 2014 at 18:39
• Im not familiarized with the requirements of proofs but I think its work as a formal proof but we need to add the proof that the product of different and complementary product is a lesser number of an exponential, i.e., $(x-a)a < x^2$ and extend this notion for any division of x. Oct 9, 2014 at 18:45
• doesn't your function assume that all the numbers have to be the same - why can't they all be different or some be different Oct 9, 2014 at 18:55
• I do not quite see where to go from there, but that might just be me. Have added my variation of a formal proof from there - I think the problem with your first comment lies that, while we do get an upper bound for the problem, we do not know what the function does on any values other than the maximum. We might not have all the same numbers in the result. @zebra, we get an upper bound for our product, not an exact solution. The upper bound you get by assuming all numbers are equal but not necessarily in $\mathbb{N}$, since that's where in AM-GM equality holds. Oct 9, 2014 at 18:57
• @SomeMathStudent I still do not understand why making all the numbers equal gives the upper bound and how this relates to the AM GM inequality - isn't this to do with the nth root of the product Oct 9, 2014 at 19:00
Let's look at a factorisation containing the term $n$. We have $2(n-2)=2n-4$ so if $n\ge 4$ we have $2(n-2)= 2n-4 \ge n$, and replace $n$ with $2, n-2$, which does not reduce the product and may increase it.
If $1$ appears in the sum we can add it to another summand, which clearly increases the product without changing the sum. | {
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So we get a sum consisting of terms which are either $2$ or $3$. Then we note Barak's observation that $3\cdot 3 \gt 2\cdot 2 \cdot 2$
Let's say a set $S=\{x_i\}:i\in \{1\to n\}$ for some $n\in\mathbb{N}$ is the optimal set. Since $\forall x \geqslant 4 : x * 2 \geqslant x + 2$, the set doesn't have any element $x \geqslant 4$ because we can take it out of the set and add a two and a $(x-2)$ instead. now for elements strictly smaller than 4: adding a $1$ want help since $x*1 = x$. However, comparing 2 to 3, we can find that $3*2 = 2*3$ which means adding 3 two times is equal to add 2 three times but $3*3>2*2*2$ which means we should switch as many sets of (2*2*2) to (3*3) as possible. Here we have three different cases which are the values for sum modulo 3. The case of 100 is 1 but let's speak generally.
• for 0 the number can be divided by 3 and we are done.
• for 2 we can add a 2 at the end with affecting threes so the answer is a 2 and $x/3$ threes.
• The case of one is a bit tricky that when divided by 3 we have a rest of one so we even ignore it since $x*1 = x$ or we can take a three out if $x > 3 (x\neq1)$ and add two twos so we're comparing 2*2 to 1*3 and 2*2 is bigger so in this case we have 2 twos and a rest of threes.
• In conclusion the answer to your question should be 2 twos and $(96\div3=32)$ threes | {
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# Why does $\int_1^\sqrt2 \frac{1}{x}\ln\left(\frac{2-2x^2+x^4}{2x-2x^2+x^3}\right)dx$ equal to $0$?
In this question, the OP poses the following definite integral, which just happens to vanish: $$\int_1^\sqrt2 \frac{1}{x}\ln\bigg(\frac{2-2x^2+x^4}{2x-2x^2+x^3}\bigg)dx=0$$ As noticed by one commenter to the question, the only zero of the integrand is at $$x=\sqrt[3]{2}$$, meaning that the integral of the integrand from $$x=1$$ to $$x=\sqrt[3]{2}$$ is the additive inverse of the integral of the integrand from $$x=\sqrt[3]{2}$$ to $$x=\sqrt{2}$$.
This suggests some sort of symmetry obtainable by a substitution, but I cannot find an appropriate substitution or cancellation. It seems like the answer should be much simpler than those posted to the linked question.
Any ideas?
EDIT: I believe that this more general integral also vanishes: $$\int_1^{\sqrt{t}}\frac{1}{x}\ln\bigg(\frac{t-sx^2+x^4}{tx-sx^2+x^3}\bigg)dx=0$$
• @Rócherz Of course it isn't symmetric about $\sqrt[3]{2}$; I simply mean that perhaps an appropriate substitution might lead to a cancellation. – Frpzzd Nov 20 '18 at 0:47 | {
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We can indeed prove the result by symmetry. Put in other words we desire to show that: $$I=\int_1^\sqrt2 \frac{1}{x}\ln\bigg(\frac{x^4-2x^2+2}{x^2-2x+2}\bigg)dx=\color{green}{\int_1^\sqrt 2\frac{1}{x}\ln x dx=\frac18\ln^2 2}$$ First let us take the LHS integral and split it in two parts: $$I=\color{red}{\int_1^\sqrt2 \frac{\ln(x^4-2x^2+2)}{x}dx}-\color{blue}{\int_1^\sqrt2 \frac{\ln(x^2-2x+2)}{x}dx}$$ For the second one $$(I_2)$$ we will substitute $$\displaystyle{x=\frac{2}{t}\rightarrow dx=-\frac{2}{t^2}dt}$$: $$I_2=\int_{\sqrt 2}^2 \frac{\ln\left(\frac{2(t^2-2t+2)}{t^2}\right)}{\frac{2}{t}}\frac{2}{t^2}dt\overset{t=x}=\int_{\sqrt 2}^2 \frac{\ln(x^2-2x+2)}{x}dx+\int_{\sqrt 2}^2\frac{\ln 2 -2\ln x}{x}dx$$ Adding with the original $$I_2$$ leads to: $${2I_2=\int_1^2\frac{\ln(x^2-2x+2)}{x}dx+\int_{\sqrt 2}^2\frac{\ln 2 -2\ln x}{x}dx}$$ $${\Rightarrow I_2=\frac12\int_1^2 \frac{\ln(x^2-2x+2)}{x}dx-\frac{1}{8}\ln^22}\overset{x=t^2}=\color{blue}{\int_1^\sqrt{2}\frac{\ln(t^4-2t^2+2)}{t}dt-\frac{1}{8}\ln^2 2}$$ $$I=\color{red}{\int_1^\sqrt2 \frac{\ln(x^4-2x^2+2)}{x}dx}-\color{blue}{\int_1^\sqrt{2}\frac{\ln(x^4-2x^2+2)}{x}dx+\frac{1}{8}\ln^2 2}=\color{green}{\frac18\ln^2 2}$$
Your conjecture is indeed also correct since by the exact same method we can show that: $$\int_1^{\sqrt{t}}\frac{1}{x}\ln\bigg(\frac{t-sx^2+x^4}{t-sx+x^2}\bigg)dx=\int_1^\sqrt{t} \frac{1}{x}\ln xdx=\frac{1}{8}\ln^2t$$ And this time after we split the LHS integral into two parts, we will substitute in the second integral $$\displaystyle{x=\frac{t}{y}}$$ ($$t$$ is a constant here), followed by an addition with the original $$I_2$$ from there and the result follows. Of course it is valid in the following form too: $$\int_1^{\sqrt{t}}\frac{1}{x}\ln\bigg(\frac{t+sx^2+x^4}{t+sx+x^2}\bigg)dx=\int_1^\sqrt{t} \frac{1}{x}\ln xdx$$ | {
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What is the use of the Dot Product of two vectors?
Suppose you have two vectors a and b that you want to take the dot product of, now this is done quite simply by taking each corresponding coordinate of each vector, multiplying them and then adding the result together. At the end of performing our operation we are left with a constant number.
My question therefore is what can we do with this number,why do we calculate it so to speak? I mean it seems almost useless to me compared with the cross product of two vectors (where you end up with an actual vector).
• You can use it to find the angle between any two vectors. $\mathbf{a}\cdot\mathbf{b}=|\mathbf{a}||\mathbf{b}|\cos\theta$ where $\theta$ is the angle between the two vectors. This is a better approach than using the cross product as the cross product can only be defined in a few dimensions (normally only 3 dimensions). Obviously this is just one simple use of the dot product which is a special case of a more general phenomenon known as an inner product en.wikipedia.org/wiki/Inner_product_space . Jun 8, 2013 at 15:35
• You should take a look at how it is derived as it can be very enlightening. Take a look at Lang's Linear Algebra (2E) pg.19-20.
– user70962
Jun 8, 2013 at 20:18
• I always found that the dot product is a good way to measure how "parallel" two vectors are Jun 9, 2013 at 7:15
Re: "[the dot product] seems almost useless to me compared with the cross product of two vectors ".
Please see the Wikipedia entry for Dot Product to learn more about the significance of the dot-product, and for graphic displays which help visualize what the dot product signifies (particularly the geometric interpretation). Also, you'll learn more there about how it's used. E.g., Scroll down to "Physics" (in the linked entry) to read some of its uses:
Mechanical work is the dot product of force and displacement vectors.
Magnetic flux is the dot product of the magnetic field and the area vectors. | {
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You've shared the algebraic definition of the dot product: how to compute it as the sum of the product of corresponding entries in two vectors: essentially, computing $\;\mathbf A \cdot \mathbf B = {\mathbf A}{\mathbf B}^T.\;$
But the dot product also has an equivalent geometric definition:
In Euclidean space, a Euclidean vector is a geometrical object that possesses both a magnitude and a direction. A vector can be pictured as an arrow. Its magnitude is its length, and its direction is the direction the arrow points. The magnitude of a vector A is denoted by $\|\mathbf{A}\|.$ The dot product of two Euclidean vectors A and B is defined by
$$\mathbf A\cdot\mathbf B = \|\mathbf A\|\,\|\mathbf B\|\cos\theta,\quad\text{where \theta is the angle between A and B.} \tag{1}$$
With $(1)$, e.g., we see that we can compute (determine) the angle between two vectors, given their coordinates: $$\cos \theta = \frac{\mathbf A\cdot\mathbf B}{\|\mathbf A\|\,\|\mathbf B\|}$$
• I am not stating what the dot product signifies, in fact that is the essence of this question, I did not know that the dot product has an equivalent geometric definition, or that it could be used to calculate the angle between two vectors that is extremely useful. Jun 8, 2013 at 15:45
• I didn't intend to criticize. I'm sorry if that's how the post comes across! Indeed, I upvoted the question because it's a fine question, and you put thought into writing it. Jun 8, 2013 at 15:47
• The dot product is also the product $\bf A B^T$ of vector A with the transpose of vector B. Jun 8, 2013 at 15:49
• @amWhy: Thank you, I had an upvote mind - trigger finger! :-) Fixed! Jun 9, 2013 at 1:37
• @amWhy thank you for your suburb answer , and also thank you for adding that formula at the bottom for finding the cosine of theta this alone is extremely useful to me due to my pursuit of physics! Jun 9, 2013 at 21:24 | {
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The original motivation is a geometric one: The dot product can be used for computing the angle $\alpha$ between two vectors $a$ and $b$:
$a\cdot b=|a|\cdot|b|\cdot \cos(\alpha)$.
Note the sign of this expression depends only on the angle's cosine, therefore the dot product is
• $<0$ if the angle is obtuse,
• $>0$ if the angle is acute,
• $=0$ if the $a$ and $b$ are orthogonal.
Another important special case appears when $a=b$: The root of the scalar product of a vector with itself is the length of a vector:
$a\cdot a=|a|\cdot|a|\cdot1=|a|^2$.
There's another interesting application of the dot product, in combination with the cross product: If you have three vectors $a$, $b$ and $c$, they define a parallelepiped, and you can compute its (signed) volume $V$ as follows using the so-called scalar triple product:
$V=(a\times b)\cdot c$
(Note that this is a generalization of $|a\times b|$ being the area of the parallelogram defined by $a$ and $b$.)
• There's nothing special about 2d/3d here; this means of finding the angle between vectors applies to an arbitrary number of dimensions. Jun 8, 2013 at 15:50
• @Muphrid: Of course you can define something like an "angle" in arbitrary dimensions. I've changed "geometrical" to "visual", I think that makes more sense. Jun 9, 2013 at 6:12
• Perhaps. But two vectors define only a plane, so even in a 4d space or higher, the geometry basically isn't changing: you have two vectors in some common plane, and the dot product tells us how alike they are. I think you're making it out to be more complicated than it really is. Jun 9, 2013 at 6:17
• @Muprid: Good point. I've deleted the sentence now. Jun 9, 2013 at 6:19
Before addressing your question, i want to say that this is a very good question and you are right to expect that the dot product has use/significance. | {
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First, it is important that you think about vectors separate from their coordinates. While it is true that we often represent vectors as a series of coordinates along well-defined axes, this is merely for computational reasons. A vector as an idea "exists" in a space without any predefined coordinate system. I say this because there are two definitions of the dot product, one is coordinate free (i.e. $\mathbf a\cdot\mathbf b = \|\mathbf a\|\,\|\mathbf b\|\cos\theta$) and the other is based on coordinates (i.e. $\mathbf a\cdot\mathbf b = \sum_i{a_i b_i}$). Of these two, it is best to think of the dot product in terms of the former as it does not depend on a coordinate system. (It is relatively easy to show that the latter may be derived from the former, but in that derivation is an implicit assumption that the coordinate system being used to represent the dot product is orthogonal.) | {
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Second, given the coordinate-free definition, the fundamental idea of the dot product is that of projection. By this it gives a single number which indicates the component of a vector in the direction of another vector. Your observation of the dissimilarity between the dot and cross product is correct, however, the dot product is used to produce a vector as well, it just does it component-by-component. Let's suppose that we have a vector $\mathbf v$ represented by its components in a given coordinate system. Let's further suppose that we have an orthonormal basis defined in that same coordinate system as the set of column vectors $\{\mathbf u_1, \mathbf u_2, \ldots, \mathbf u_n\}$. Finally, suppose that we want to represent $\mathbf v$ in this basis as $\mathbf w$. The question is how do we do that? We use the dot product of course! So the first component of $\mathbf w$ would then be $w_1 = \mathbf u_1\cdot \mathbf v$, and the second component would be $w_2 = \mathbf u_2\cdot \mathbf v$ and so on. (Note that because $\|\mathbf u_i\| = 1$, we have $\mathbf u_1\cdot \mathbf v= \|\mathbf v\|\cos\theta_i$.) If we then think of the vector $\mathbf w$ defined as such we have
$$\mathbf w = \left[ \begin{array}{c} w_1 \\ w_2 \\ \vdots \\ w_n \end{array} \right] = \left[ \begin{array}{c} \mathbf u_1\cdot \mathbf v \\ \mathbf u_2\cdot \mathbf v \\ \vdots \\ \mathbf u_n\cdot \mathbf v \end{array} \right] = \left[ \begin{array}{c} \mathbf u_1^T \mathbf v \\ \mathbf u_2^T \mathbf v \\ \vdots \\ \mathbf u_n^T \mathbf v \end{array} \right] = \left[ \begin{array}{c} \mathbf u_1^T \\ \mathbf u_2^T \\ \vdots \\ \mathbf u_n^T \end{array} \right]\mathbf v = \left[ \begin{array}{cccc} \mathbf u_1 & \mathbf u_2 & \cdots &\mathbf u_n \end{array} \right]^T\mathbf v$$ | {
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Finally, we conclude that the dot product plays a key role in the transformation of a vector from one basis to another and that the dot product is hidden in the definition of matrix multiplication in that one view of a matrix-vector product is that each element in the product represents a dot product between a row of the left and a column of the right.
I hope this helps.
The dot product is an essential ingredient in matrix product. The product of the two matrices $A$ and $B$ (of compatible sizes, that is, the number of columns of $A$ equals the number of rows of $B$) is a matrix whose $(i, j)$ component is the dot product of the $i$-th row of $A$ and the $j$-th column of $B$.
Among the many applications, consider this simple one.
You have students $s_{1}, \dots, s_{n}$ taking courses $c_{1}, \dots, c_{m}$. Consider the matrix $A$ which has $0$ everywhere, except that the $(i,j)$ coefficient is $1$ if the student $s_{j}$ takes the course $c_{i}$. Now note that the dot product of the $i_{1}$-th row of $A$ with the $i_{2}$-th row gives the number of students that take both $c_{i_{1}}$ and $c_{i_{2}}$. In other words, the matrix $A A^{t}$ has in its $(i, j)$ position the number of students taking both $c_{i}$ and $c_{j}$.
This matrix is of course useful in building a course timetable.
The geometric idea of the dot product has been touched upon, but there is a vast generalization of this product in geometric algebra, the algebra of not only oriented lines (vectors) but planes, volumes, and more (called blades).
In geometric algebra, we have a generalized dot product of a vector $a$ and another blade $B$ denoted $a \cdot B$. This product has a simple geometric interpretation as the part of $B$ orthogonal to the projection of $a$, with magnitude $|a||B|| \cos \theta|$, where $\theta$ is the angle $a$ forms with its projection in $B$. | {
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(Note: as a point of fact, $a \cdot B$ is orthogonal to $a$ also, not just its projection into $B$, but thinking about this leads to some difficulties, while thinking about what's perpendicular to the projection does not.)
If $B$ is a 2-blade (also called a bivector), then you should be able to imagine this directly: if $a$ lies entirely in $B$, then $a \cdot B$ is just the vector perpendicular to $a$ in $B$. If $a$ does not lie entirely in $B$, then it can be decomposed into a tangential part and a normal part. We throw away the normal part, and the previous logic applies for the tangential part.
If $B$ is a 3-blade (a trivector), then in 3d space $a$ must lie in $B$ (for there is no 3d volume that a vector does not help span), and the product $a \cdot B$ is the "Hodge dual", or the plane perpendicular to $a$.
In this light, the dot product of vectors may actually be the most non-intuitive part of this reasoning. When you take the dot product, there's only a scalar left--there's no vector or other higher dimensional object left to be orthogonal to $a$. Again, this is why I emphasize that $a \cdot B$ is the part of $B$ orthogonal to the projection of $a$ onto $B$. When $B$ is a vector, it's clear there is no other vector or anything else that can be orthogonal to the projection of $a$, for $B$ and that projection are parallel, so the result is necessarily just a scalar.
When one calculates A.B, two measurements happen: measurement of how small the angle between them is, and how long A and B are. A.B basically means projection length of A on B, with this length then scaled by the absolute length of B.
One way to think about the interpretation of the dot product is to think how would one maximise or minimise the dot product between two vectors. Let's assume we are trying to maximise the dot product between two vectors that we can modify: | {
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The dot product will be grow larger as the angle between two vector decreases. The dot product A.B will also grow larger as the absolute lengths of A and B increase. This is because as A gets larger, its projected length will be longer, and as B's length gets larger, the scaling of A's projection will grow larger, given that B's absolute length will act as a scaler of A's projection length.
Hence in problems where is desirable to maximise or minimise the size of vectors and minimise the deviation or angle between them, quantifying it using the dot product can be useful.
I don’t see enough plain English answers here, to be honest. One simple example: you can determine in a stealth game whether an object is within a 90 degree line of sight of something, or not. | {
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# Find pairs $(a,b)$ with $\gcd(a,b),\gcd(a + 1, b),\ldots, \gcd(a + k, b)$ given
Given a set of GCD's, how to find a set of numbers that satisfy all their criteria?
Suppose we are given a $k$ integers $\gcd(a,b),\gcd(a + 1, b),\ldots, \gcd(a + k, b)$ for some k.
How to get a and b from this? Given the GCD's, I need to find (a,b).
Example: if 4 GCD's are given 3,2,1,6, then the pair (a,b) which satisfy the above condition is (3,6).
Any help will be useful!
• Note that many other $(a,b)$ will also satisfy your $(3,2,1,6)$ report, and in particular $(3+12k, 6+12k)$ are solutions for $k\in \mathbb{Z}$ – Joffan Jun 21 '16 at 15:38
• But you don't necessarily have $a+k=b$ or indeed $a<b$ is that correct? – almagest Jun 21 '16 at 15:39
• no, i do not have a+k=b. But from the sample cases that i have, a<b is valid for all those cases. also, i just need to find a pair which satisfies the GCD's given. I need to select just one of the many possible solutions – Ashwath Narayan Jun 21 '16 at 15:41 | {
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Hints: Let $g_0,g_1,\ldots,g_k$ be the specified GCDs. Clearly one must have $\text{LCM}(g_0,\ldots,g_k) \mid b$. Show that if $(a,b)$ is any solution then we may replace $b$ by $b_0 := \text{LCM}(g_0,\ldots,g_k)$, so that $(a,b_0)$ is also a solution.
Therefore we are free to assume $b = b_0$. For each prime power factor $p^r$ of $b_0$, there must be at least one $g_i$ that is divisible by $p^r$. This uniquely determines $a$ mod $p^r$. Now Chinese Remaindering determines $a$ mod $b_0$.
There is at most one working choice of $a$ mod $b_0$. Prove that if any one value of $a$ satisfies the full set of constraints, then so does every $a$ in that congruence class. This provides all solutions of the specific form $(a,b_0)$; there are other solutions with larger values of $b$, and these are more complicated to describe — but you only are looking for one.
• @AshwathNarayan In your example, $b_0=6$ factors into prime powers $2\cdot 3$. $g_1$ is divisible by $2$, which forces $a$ to be odd. $g_0$ is divisible by $3$, which forces $a$ to be divisible by $3$. Therefore $a \equiv 3 \pmod 6$, which turns out to satisfy all the other constraints so it is a bona fide solution. – Erick Wong Jun 21 '16 at 16:05 | {
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Parallel system functioning problem
I am currently solving the following problem about conditional probability:
"A parallel system functions whenever at least one of its components works. Consider a parallel system of n components, and suppose that each component works independently with probability $$\frac{1}{2}$$. Find the conditional probability that component 1 works given that the system is functioning."
I think that I do have the answer to this problem; however, since textbook does not contain answer to this one, I am sharing it with the community to poke holes in my logic, if any.
Let's call event "whole parallel system works" as $$W$$ and event "first component works" as $$W_1$$. Our task is to find $$P(W_1|W)$$. Let's use conditional probability definition to expand it: $$P(W_1|W) = \frac {P(W_1 \cap W)}{P(W)} = \frac {P(W | W_1)\cdot P(W_1)}{P(W)}$$ Now, $$P(W_1) = \frac{1}{2}$$. $$P(W) = 1 - P(\bar W) = 1 - (\frac{1}{2})^n = 1 - \frac{1}{2^n}$$. And $$P(W | W_1) = 1$$, since the whole system is active if first component is active. As the result, we have: $$P(W_1|W) = \frac{\frac{1}{2}}{1 - \frac {1}{2^n}} = \frac{\frac{1}{2}}{ \frac {2^n-1}{2^n}} = \frac{2^{n-1}}{2^n-1}$$
• I got the same answer, but the answer in textbook is given as $(1/2)/(1-1/2^{n-1})$. Don't know how they got $n-1$ instead of $n$. Aug 15 '21 at 11:17 | {
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# Making calculation of correlation dimension faster
I am trying to make my calculation faster. I was hoping, when I compiled it, it would work much faster, but in the end its even slower. I don't know if I am using Compile correctly. I am open to any suggestion on how to make this computation faster.
I use the data:
DD = RandomReal[{0, 1}, {1000, 3}];
I have written the following function for calculating correlation dimension (D2):
CORRDIM[data_] := (
m =
ParallelTable[{r,
Total[
Table[
Length[Drop[Nearest[Drop[data, i - 1], data[[i]], {All, r}], 1]],
{i, 1, Length[data]}]]}, {r, 0.001, 0.011, 0.001}];
m);
CORRDIM works pretty fast, but for a large dataset it is still too slow.
CORRDIM[DD]; // AbsoluteTiming
{0.483119, Null}
I was hoping that Compile could help me speed up this calculation, especially that the options
CompilationTarget -> "C",
RuntimeAttributes -> {Listable},
Parallelization -> True
would somehow run the code in multiple threads, but the compile version works even slower than the interpreted version.
fcc =
Compile[{{x, _Real, 2}},
Table[
{r,
Total[
Table[
Length[Drop[Nearest[Drop[x, i - 1], x[[i]], {All, r}], 1]],
{i, 1, Length[x]}]]}, {r, 0.001, 0.011, 0.001}],
CompilationTarget -> "C",
RuntimeAttributes -> {Listable},
Parallelization -> True]
Map[fcc, {DD}]; // AbsoluteTiming
{0.787973, Null}
Is it there any way to make it faster? Am I making some mistakes in the use of Compile? | {
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Is it there any way to make it faster? Am I making some mistakes in the use of Compile?
• Which definition of "correlation dimension" do you use? The one I found seems to be different than what you programmed. The correlation integral in that article, $C(N,r)$, can be computed with matrix arithmetic. (Much faster than using Nearest.) Dec 26, 2017 at 0:36
• Welcome to Mathematica.SE! I suggest the following: 1) As you receive help, try to give it too, by answering questions in your area of expertise. 2) Take the tour! 3) When you see good questions and answers, vote them up by clicking the gray triangles, because the credibility of the system is based on the reputation gained by users sharing their knowledge. Also, please remember to accept the answer, if any, that solves your problem, by clicking the checkmark sign! Dec 26, 2017 at 4:06
• I believe your issue of having slower compiled code comes from the fact that Nearest isn’t compilable. Dec 26, 2017 at 15:03
• Related: (125543) Dec 26, 2017 at 20:39
Here is my take on this... I am following the definitions and computational steps given in the article
[1] James Theiler, "Efficient algorithm for estimating the correlation dimension from a set of discrete points", Phys. Rev. A 36, 4456 – Published 1 November 1987. DOI:https://doi.org/10.1103/PhysRevA.36.4456 .
Make data:
SeedRandom[33435]
n = 1000;
DD = Sort@RandomReal[{0, 1}, {n, 3}];
Dimensions[DD]
(* {1000, 3} *)
Compute the distance matrix and take the upper triangular values:
AbsoluteTiming[
dmat = DistanceMatrix[DD];
dvals = SparseArray[
SparseArray[UpperTriangularize[dmat, 1]]["NonzeroValues"]];
]
(* {0.051459, Null} *)
Define the correlation dimension integral function:
Clear[CDIntegral]
CDIntegral[dvals_SparseArray, n_Integer, r_?NumericQ] :=
Block[{res},
res = Total@UnitStep[r - dvals];
res*2./(n*(n - 1))
];
Note that the function CDIntegral is defined in such a way so dvals can be reused. | {
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Note that the function CDIntegral is defined in such a way so dvals can be reused.
Compute the correlation integrals for a range of distance values.
rRange = Range[0.001, 0.011, 0.001];
AbsoluteTiming[
res =Table[CDIntegral[dvals, n, r], {r, rRange}];
]
(* {0.034141, Null} *)
This seems to be ~10 times faster, than using Nearest (on my recent MacBook Pro with \$Version 11.2.0.)
Using n=10000 the above computations are done for ~20 seconds.
Following [1] here is how the correlation dimension is estimated with the results obtained from the computations above:
lm = LinearModelFit[
Log@Select[Transpose[{rRange, res}], #[[2]] > 0 &], {1, x}, x]
lm["BestFitParameters"][[2]]
(* 2.68916 *)
Here is a plot of the found correlation integrals and the fit:
ListLogLogPlot[{Transpose[{rRange,res}], {#, Exp[lm["Function"][Log[#]]]} & /@ rRange},
Filling -> {1 -> Bottom}, PlotTheme -> "Detailed",
PlotLegends -> {"correlation integrals", "fit"}]
ClearAll[corrDIM]
corrDIM[data_] := Module[{nF = Nearest[data -> "Index"]},
Table[ {r, Tr[Table[Tr[UnitStep[nF[data[[i]], {All, r}]- i - 1]],
{i, 1, Length[data]}]]}, {r, 0.001, 0.011, 0.001}]]
Using Anton's example data:
SeedRandom[33435]
n = 1000;
DD = Sort@RandomReal[{0, 1}, {n, 3}];
First @ AbsoluteTiming[result = corrDIM[DD];]
0.044153
Comparing with OP's CORRDIM (on Wolfram Cloud where ParallelTable is not supported):
Quiet @ First @ AbsoluteTiming[result2 = CORRDIM[DD];]
0.192319
result == result2
True
Comparing with Anton's CDIntegral:
First @ AbsoluteTiming[dmat = DistanceMatrix[DD];
dvals = SparseArray[ SparseArray[UpperTriangularize[dmat, 1]]["NonzeroValues"]];
res = Table[CDIntegral[dvals, n, r], {r, rRange}];]
0.07153
2./(n*(n - 1)) result[[All, 2]] == res
True | {
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0.07153
2./(n*(n - 1)) result[[All, 2]] == res
True
• Nice, much faster than my implementation, especially with larger n, e.g. n=10000. Dec 27, 2017 at 15:41
• Thanks, can but I have a problem, I can not run your corrDIM function, ther is an error with "Index", what does this mean please ' Nearest[data -> "Index"] ' ? I couldnt find out. Dec 28, 2017 at 19:34
• Actually this helps me already, but originally I was trying to run computing of more of Datasets in threads, by using any of D2 estimating methods. I mean for example for all Datasets DD, DD1 DD2 ... DDn at once so it would take like the same time as for one DD. Is it possible ? I think using Compile it should work somehow, but I cant find out how. Dec 28, 2017 at 19:41
• @Mark, you can use Automatic in place of "Index" in versions before 10. Re parallelization / compile, Compile` might make Anton's method faster.
– kglr
Dec 28, 2017 at 20:07 | {
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# can all triangle numbers that are squares be expressed as sum of squares
I'm not sure if this is just a subset of Which integers can be expressed as a sum of squares of two coprime integers? which in turn points to http://en.wikipedia.org/wiki/Brahmagupta%E2%80%93Fibonacci_identity, but if so, I'm not seeing it.
Basically: looking at the numbers between 0 and 1000, if (but not iff) n is a square, then the nth triangle number (i.e. $$\frac{n(n+1)}{2}$$) can be expressed as the sum of two perfect squares . Does this hold for all squares, and can someone point me in the direction of why this is? (For what it's worth, I is an engineer but haven't really touched number theoryish stuff since college.)
(Also saw Prove that there are infinitely many natural numbers $n$, such that $n(n+1)$ can be expressed as sum of two positive squares in two distinct ways. which proves there are infinitely many for the similar $n(n+1)$ case but not that all squares will work unless I missed some aspect)
(Brute force approach to checking numbers, because brute force always works)
import math
maxsquare = 1001
squares = [i*i for i in xrange(maxsquare)]
for j in xrange(int(math.sqrt(maxsquare))):
i = j * j
n = i * (i+1) / 2
for s in squares:
f = n - s
if f in squares:
print "%d^2 + %d^2 = %d * %d+1 / 2" % (int(math.sqrt(f)),int(math.sqrt(s)),i,i)
break
else:
print "Could not find anything for i = %d" % i
-
I think the title of your question is phrased wrong. It says "all triangle numbers that are squares". But $\frac12 8(8+1)$ is a triangle number that is a square, but you are not interested in it. And obviously all triangle numbers that are squares are trivially sums of squares. Right? – MJD Jun 19 '13 at 2:01
correct... any suggestions on how to correct the title? – Foon Jun 19 '13 at 2:28
The answer is yes, it is true for all $n$. | {
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The answer is yes, it is true for all $n$.
Let me reformulate your question. You want to know if, whenever $n$ is a perfect square, it is the case that $\frac12 n(n+1)$ is a sum of two squares. If $n$ is a perfect square, then it has an integer square root, which we can call $m$, and write $n = m^2$. Then your question becomes whether $\frac12 m^2(m^2+1)$ is a sum of two squares for all integers $m$.
There is a very useful theorem about sums of two squares which says that a number $N$ is a sum of two squares if and only if every prime of the form $4k+3$ (such as $3, 7, 11, 19,$ etc.) appears in the prime factorization of $N$ an even number of times.
Now $m^2(m^2 + 1) = {\left(m^2\right)}^2 + m^2$ is obviously a sum of two squares. And dividing $m^2(m^2+1)$ by 2 can't possibly affect the number of times any prime of the form $4k+3$ appears in its factorization, so $\frac12 m^2(m^2+1)$ is also a sum of two squares.
(Addendum: Erick Wong points out below that we do not even need to invoke the Fermat $4k+3$ theorem.) | {
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-
$n^2(n^2+1)=(n^2)^2+1$ is a sum of two squares without the machinery from the preceding paragraph. – Andres Caicedo Jun 19 '13 at 1:52
@AndresCaicedo I think you mean $(n^2)^2 + n^2$, but your point still stands. – anorton Jun 19 '13 at 1:54
@andres Is there a trick for showing that $a$ is a sum of squares if and only if $2a$ is, that doesn't depend on the $4k+1$ theorem? Because if not, I don't think I can abbreviate my answer very much. – MJD Jun 19 '13 at 2:00
@anorton Yes, of course. (How embarrassing!) – Andres Caicedo Jun 19 '13 at 2:05
@MJD Yes, there is a very simple trick: if $2a = x^2+y^2$, then $x$ and $y$ have the same parity and so $a = (\frac{x+y}{2})^2 + (\frac{x-y}{2})^2$. You may recognize this as unique factorization in $\mathbb Z[i]$ in disguise. – Erick Wong Jun 19 '13 at 2:16
If you run your program you should find $$0^2+1^2 = \frac{1\times 2}{2}\\ 1^2+3^2 = \frac{4\times 5}{2}\\ 3^2+6^2 = \frac{9\times 10}{2} \\ 6^2+10^2 = \frac{16\times 17}{2} \\ 10^2+15^2 = \frac{25\times 26}{2}$$ from which you might recognize the numbers in the sums as the triangular numbers again, in which case you can guess that $$\frac{j^2(j^2+1)}{2} = \left(\frac{j(j-1)}{2}\right)^2+\left(\frac{j(j+1)}{2}\right)^2$$ which is true. | {
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# Factor table with sign-The useful tool to solve polynomial inequality
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Edit: How to solve quadratic equations - Factor quadratic equations? You may find this topic
First of all, let's try these questions.
Question 1.
Find the integer value of $$x$$ that $$(x-2)(x-4)<0$$
Question 2.
Solve for this inequation: $$x(x+1)(x-5)>0$$
Question 3.
Solve for this inequation: $$\frac{x(x-2)^3}{(x+1)(x+2)^2} \leq 0$$
To solve Question 1, we could quickly come to result $$x=3$$. However, to solve Question 2 and Question 3, how much time will we need? Almost much more time to solve them than to solve Question 1.
Basically, while solving polynomial inequalitie, we need to review each range value of $$x$$.
For example, to solve Question 1: $$(x-2)(x-4)<0$$, we need to review if $$x<2$$, if $$2 \leq x<4$$, and $$x \geq 4$$. Finally, we come to result $$2 <x<4$$. | {
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This method could be used with basic polynomial inequalitie, but it seems unuseful to solve complex polynomial inequalitie. Now, I'm going to introduce you an useful tool to solve these kinds of complex polynomial inequalitie, called: Factor table with sign.
These are the steps to solve polynomial inequalities like $$(x-a_1)(x-a_2)...(x-a_n) < 0$$. (note that sign "$$<$$" could be "$$>$$", "$$\leq$$", "$$\geq$$")
Step 1. Change the polynomial/expression into factorization form.
Step 2. Make factor table with sign to solve the inequatility
Step 2.1. Find all values of $$x$$ those make each factor equals to 0. List them in the first row of table in ascending order.
Step 2.2. In next rows of table, review the signs of each factor of the polynomial/expression based on each range value of $$x$$
Step 2.3. In the fianl row of table, review the sign of the polynomial/expression based on each sign of each factor in each range value of $$x$$.
Step 3. Finally, come to solution.
Now, let's solve the Question 1.
Step 1. We already have $$(x-2)(x-4)<0$$
Step 2. Make factor table with sign like this.
Attachment:
Capture question 1.PNG [ 2.18 KiB | Viewed 1988 times ]
Step 2.1. Find all values of $$x$$ those make each factor equals to 0.
$$x-2=0 \iff x=2$$
$$x-4=0 \iff x=4$$
In the first row, list all these values in ascending order: $$-\infty , 2, 4, +\infty$$.
Step 2.2. In next 2 rows, review the signs of each factor.
This step based simply on this rule: If $$x<a \implies x-a<0$$. If $$x>a \implies x-a>0$$.
For example, $$x-2=0 \implies x=2$$. Hence, any value of $$x<2$$, then $$x-2$$ will be negative (-); any value of $$x>2$$, then $$x-2$$ will be positive (+).
We dont need to care about what is the value of $$x-2$$ if $$x=4$$ or what is the value of $$x-4$$ if $$x=2$$. These values are expressed as sign "|". | {
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Step 2.3. In final row, review the sign of the expression.
The sign of the expression simply based on this rule:
$$(+) \times (+) = (+)$$
$$(-) \times (-) = (+)$$
$$(+) \times (-) = (-)$$
$$(-) \times (+) = (-)$$
Step 3. Come to solution.
We need to find sign (-) of the expression. Based on the table, we simply have $$2<x<4$$.
That's it.
Now let's solve Question 2.
$$x=0$$
$$x+1=0 \implies x=-1$$
$$x-5=0 \implies x=5$$
The factor table with sign:
Attachment:
Capture question 2.PNG [ 2.97 KiB | Viewed 1982 times ]
Hence, we simply have $$x \in (-1,0) \cup (5, +\infty)$$
Solution for Question 3.
$$x=0$$
$$x-2=0 \implies x=2$$
$$x+1=0 \implies x=-1$$
$$x+2=0 \implies x=-2$$
The factor table with sign
Attachment:
Capture question 3.PNG [ 9.54 KiB | Viewed 1977 times ]
In the table above, I've divided into 4 parts. Part 2 contains every factor without power for simplified purpose. Then I come to Part 3 that contains every factor with its power. In part 4 or the final row, I come to the sign of the expression.
Also note that the expression is undefined when $$x=-2$$ or $$x=-1$$. The solution is $$x \in (-\infty, -2) \cup (-2,-1) \cup [0,2]$$
Now come to another example.
Question 4. Find the product of the integer values of x that satisfy the inequality $$\frac{(x^2-4)^3}{(x-5)^5(x^2-9)^4}$$
This question belongs to e-GMAT http://gmatclub.com/forum/wavy-line-met ... l#p1727256
You could find the solution in http://gmatclub.com/forum/wavy-line-met ... l#p1771328
Question 5. Find the value of $$x$$ that $$x^4+5x^3-7x^2-41x-30 \leq 0$$.
Solution.
$$x^4+5x^3-7x^2-41x-30 =(x-3)(x+1)(x+2)(x+5)$$
The factor table with sign
Attachment:
Capture question 4.PNG [ 4.26 KiB | Viewed 1986 times ]
The solution is $$x \in [-5,-2] \cup [-1,3]$$ | {
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The solution is $$x \in [-5,-2] \cup [-1,3]$$
This tool is really useful when we need to find the value of x satisfied the polynomial inequalities. I hope this tool could assist you in solve DS/PS GMAT questions in the future.
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Re: Factor table with sign-The useful tool to solve polynomial inequality [#permalink] 08 Dec 2017, 06:16
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# Sets
You’ve seen sets before. {1, 2} is a set. 1..N is the set of all numbers between 1 and N. x \in S checks whether S contains x, unless you use it in a PlusCal variables statement or a with (where it creates new behaviors). Like everything else, you can nest them: {{{}}} is a set containing a set containing an empty set.
Often, we’re interested in some transformation of the set, such as “the set of male participants under 60” or “The owners of the pets.” In this section we’ll cover three ways to transform a set and how they are combined.
### Filtering
We can filter a set with {x \in S: P(x)}, which is the set of all x \in S where P(x) is true. For example, {x \in 1..8 : x % 2 = 1} is the set {1, 3, 5, 7}. P must return a boolean, so { x \in 1..4 : x * x } raises an error.
If S is a set of tuples, you can filter on some relationship between the elements of the tuple by instead using <<...>> \in S. If you want the set of ordered pairs, you could do {<<x, y>> \in S \X S : x >= y}.
As always, you can nest filters, and {x \in {y \in S : P(y)} : Q(x)} will filter on the filtered list. Generally, though, {x \in S: P(x) /\ Q(x)} is easier to understand.
### Mapping
{P(x): x \in S} applies P to every element in the set. { x * x : x \in 1..4 } is the set {1, 4, 9, 16}. { x % 2 = 1:x \in 1..8 } is the set {TRUE, FALSE}.
You can also write {P(x, y, ...) : x \in S, y \in T, ...}. { x + y : x \in 0..9, y \in { y * 10 : y \in 0..9} } is the first hundred numbers, in case you wanted to obfuscate 0..99.
Given DOMAIN Tuple is the set of numbers Tuple is defined over, write an operator that gives you the values of the Tuple, ie the range.
Range(T) == { T[x] : x \in DOMAIN T }
This is a useful enough operator that I’ll assume it’s available for all other examples and exercises.
### CHOOSE | {
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### CHOOSE
CHOOSE x \in S : P(x) is some x where P(x) is true. CHOOSE x \in 1..8 : x % 2 = 1 will be one of 1, 3, 5, 7. TLC does not branch here; while the number it chooses is arbitrary, it will always return that number. This is similar to how CASE statements work: CHOOSE x \in S : TRUE is some element of S, but TLC won’t check all of them.
TLC assumes that you always intend for there to be at least one element to choose. If there aren’t any (trivial example: CHOOSE x \in S : FALSE), it will consider this a problem in your spec and raise an error. TLC will also raise if S is infinite because TLC can’t evaluate P on an infinite number of elements. There still may be a problem with your spec, though, and it’s a good idea to try it on a finite subset.
## Set Operators
Finally, there are extra operations for working with sets:
logic operator TRUE FALSE
in set \in 1 \in {1, 2} 1 \in {{1}, 2}
not in set \notin 1 \notin {} {1} \notin {{1}}
is subset \subseteq {1, 2} \subseteq {1, 2, 3} {1, 2} \subseteq {1, 3}
Write an operator that takes two sets S1 and S2 and determines if the double of every element in S1 is an element of S2.
IsDoubleSubset(S1, S2) == {x * 2 : x \in S1} \subseteq S2. If you wanted to check both ways (doubles of S2 are in S1), you could write two expressions with \/.
operator operation example
\union Set Union {1, 2} \union {2, 3} = {1, 2, 3}
\intersect Set Intersection {1, 2} \intersect {2, 3} = {2}
S1 \ S2 The elements in S1 not in S2 {1, 2} \ {2, 3} = {1}, {2, 3} \ {1, 2} = {3}
SUBSET S The set of all subsets of S SUBSET {1, 2} = {{}, {1}, {2}, {1, 2}}
UNION S Flatten set of sets UNION {{1}, {1, 2}, {5}} = {1, 2, 5}
Given a sequence of sets, write an operator that determines if a given element is found in any of the sequence’s sets. IE Op("a", <<{"b", "c"}, {"a", "c"}>>) = TRUE.
InSeqSets(elem, Seq) == elem \in UNION Range(Seq)
If you add EXTENDS FiniteSets, you also get the following operators: | {
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If you add EXTENDS FiniteSets, you also get the following operators:
operator operation
IsFiniteSet(S) TRUE iff S is finite
Cardinality(S) Number of elements of S, if S is finite
Given a set, write an operator that returns all subsets of length two. IE Op(1..3) = {{1, 2}, {1, 3}, {2, 3}}.
Op(S) == { subset \in SUBSET S : Cardinality(subset) = 2 } | {
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# What is the symbolic form of "there does not exist a largest natural number "
Other students in office hour said this is the correct form $(\forall x)(\exists y)(y>x)$ { for all x natural number, there exists y such that y is greater than x }
But "there does not exist a largest natural number " $\neg(\exists x)(x\text{ largest natural number})$
Am I even close ?
• You have said "it is not true that for all $x$ there exists $y$ such that $y>x$". Think about whether that is an appropriate translation.
– Ian
Apr 19, 2017 at 0:17
• $\forall x \exists y (y > x)$ for $\forall x \exists y (y > x)$. Formatting tips here.
– Em.
Apr 19, 2017 at 3:30
• This was edited to say something completely different from what it originally said. Don't do that, because it makes the answers based on the original question seem nonsensical. Apr 20, 2017 at 6:31
The formula $\forall x \exists y (y > x)$ reads "for any $x$ there is a $y$ which is larger than $x$" which says that any $x$ is not largest and therefore no $x$ is largest. You can also use the formula $\forall \equiv \neg\exists\neg$ to get
\begin{align*} \forall x \exists y (y > x) &\equiv \neg\exists x \neg \exists y (y > x) \\ &\equiv \neg \exists x \forall y \neg(y > x) \\ &\equiv \neg \exists x \forall y (y \le x) \end{align*}
which says that "there is no $x$ such that every $y$ is less than or equal to $x$". This better fits with the "there does not exist a largest natural number" phrasing.
The statement as you have written it is false. If you read it out loud, $x$ has to be chosen before $y$ and for all $x$ you can find a greater $y$. You want to say that if you choose $y$. You want to choose $y$ first, then say that it is not greater than all $x$. | {
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• so, (Ay)(Ex)(y>x)??? Apr 19, 2017 at 0:26
• No, given $y$ there has to be a larger $x$, so it should be $x \gt y$ Apr 19, 2017 at 1:07
• @QuencyCaroline That statement says every natural number has one smaler than it, which is not true. Apr 19, 2017 at 1:37
• $(\forall x)(\exists y)(y>x)$ is false? Assuming the universe of discourse is the natural numbers, can you give a counterexample? Apr 19, 2017 at 17:59
• @jwodder: when the answer was written there was a not sign before the sentence. As you have written it and the question now has it, it is true. Apr 19, 2017 at 18:41
# Yes, you are close.
So far, you have:
$\neg \exists x \,\,(x\text{ largest natural number})$
Since you're looking for the "symbolic form", your next step is to convert "largest natural number" to symbols.
• Note that "largest natural number" is the same thing as "number that is greater than all other natural numbers";
• To make things easier, note that the above is the same thing as "number that is greater than or equal to all natural numbers" (therefore, including itself without a problem!);
• Further rephrase it as "number such that all numbers are less than or equal to it", so we have:
$x$ is a number such that all numbers are less than or equal to $x$
• Observe that this last sentence can be easily translated to symbols as
$$\forall n \,\,\, n \le x$$
Now, just plug that into your original sentence, obtaining:
$\neg \exists x \,\,(\forall n \,\,\, n \le x)$
Note 1: This is not the only correct way to do this. It is possible to express the same fact differently.
Note 2: in the context of this question, it is clear that we are talking about natural numbers. But generally, it would be better to specify this, by writing:
$\neg \exists x \in \mathbb{N} \,\,(\forall n \in \mathbb{N} \,\,\, n \le x)$ | {
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$\neg \exists x \in \mathbb{N} \,\,(\forall n \in \mathbb{N} \,\,\, n \le x)$
Bonus: The students you mentioned are correct too. They chose to rephrase the sentence in a different sentence (but still equivalent). Instead of saying "there is no largest natural number", they are saying "all natural numbers have the property of being smaller than some number", which is the same thing, in the end.
It depends on whether with largest natural number you mean a greatest or a maximal element, though the difference would be important only for a non-total order.
If $(X,<)$ is a partially ordered set, we say that $a\in X$ is
• a greatest element if $\forall x\in X\colon (x< a\lor x=a)$
• a maximal element if $\forall x\in X\neg(a<x)$
(In a totally ordered set, exactly one of $x<a$, $x=a$, $x>a$ must be true, hence therre the notions of greatest and maximal element coincide).
Hence "There is no $a$ such that $a$ is a greatest/maximal element" translates "literally" to either $$\neg\exists a\in \Bbb N\colon \forall x\in\Bbb N\colon (x<a\lor x=a)$$ or $$\neg\exists a\in \Bbb N\colon \forall x\in\Bbb N\colon \neg(a<x)$$
If you make use of "$\neg\exists=\forall\neg$" and "$\neg\forall=\exists\neg$", you might equivalently write $$\forall a\in \Bbb N\colon \exists x\in\Bbb N\colon (x\not <a\land x\ne a)$$ and $$\forall a\in \Bbb N\colon \exists x\in\Bbb N\colon (a<x),$$ respectively.
In English: "there is no natural n, for which all natural k would be smaller than n".
$\lnot \exists n \forall k (k, n) \in \mathbb{N}^2 \land k < n$
I was going to write this in a comment, but I think it's better if I gave an answer.
I'm going to begin by pointing out things about the two statements in the question.
The first statement does have the intended meaning. I think it can still be made more precise.
The second statement is circular, or ambiguous at best. There is no way of saying what the largest natural number is until it has been decided definitely whether it exists or not. | {
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Finally, although the domain has been described in the worded statement, for the logical statement to make sense, it must include the correct domain.
Since T. Gunn's answer very nicely covers the question, I will instead give an example of a correct statement which is useful in the study of modern algebra, how number (and other) sets are constructed, the relations between them, and the rigorous framework used to describe these ideas:
$$\forall n \in \mathbb N,n + 1 > n \land n + 1 \in \mathbb N$$
Though, to be precise this is a logically equivalent statement, not an identical one. And I have assumed, the addition operator and ordering relation ">" have already been defined.
This statement makes the intended meaning, and its demonstration, more explicit. It describes a function which guarantees the required property will be satisfied.
• If the addition operator has already been defined (as a function $\mathbb N\times\mathbb N\to\mathbb N$), then the proposition $n+1\in\mathbb N$ is redundant.
– user856
Apr 19, 2017 at 22:27
• @Rahul Fair point. I just thought writing it out would make the underlying framework more apparent. Apr 19, 2017 at 23:28
Since you are interested in formalisms, I would suggest to rephrase the statement to be more formal. In Mathematical logic, one usually uses quantors (similarly, the negation operator) and parentheses in the following way: $$\forall x (\; \text{logical statement} \;)$$ So, nesting this, your statement becomes $$\forall x \in \mathbb{N}\left(\,\exists y \in \mathbb{N}\left(\,y>x\right) \right)$$ and the other statement, the one you were asking for, becomes $$\neg (\,\exists M \in \mathbb{N} \,(\,\forall l \in \mathbb{N} \,(\,M \geq l))$$ where I replaced your variable-symbol $x$ by $M$ and used $l$ as the variable-symbol for your natural number. | {
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# pre-Algebra
Write the following as a fraction and show your work. 0.83 (the three is repeating) I don't know how to do it with only one of the numbers repeating and it is not in my book.
1. 👍 0
2. 👎 0
3. 👁 105
1. Here is a neat algorithm to change any repeating decimal into an exact fraction
Assume you have only the decimal portion of your number, ie, for 4.5676767... consider only the 5676767..
For the numerator, write down all the digits to the end of your repeat, subtract from that the part that does not repeat.
For the denominator, write down a 9 for each digit in the repeating loop, followed by the number of digits that don't repeat.
e.g. 0.45676767..
=(4567-45)/9900
=4522/9900, now reduce to
2261/9900
.833333. = (83-8)/90
= 75/90
= 5/6
1. 👍 0
2. 👎 0
posted by Reiny
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5. ### Math 115 | {
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asked by blondie on December 5, 2007
5. ### Math 115
Every fraction has a decimal equivalent that either terminates (for example,1/4=0.25 ) or repeats (for example,2/9=0.2 ). Work with a group to discover which fractions have terminating decimals and which have repeating decimals.
asked by blondie on December 5, 2007
this is from homework work book . write each fraction or mixed numberas a decimal. use bar notation if the decimal is a repeating decimal
asked by naida on September 13, 2010
7. ### math
You are given a fraction in simplest form. The numerator is not zero. When you write the fraction as a decimal, it is a repeating decimal. Which numbers from 1 to 10 could be the denominator? THANK YOU!
asked by Scotty on September 14, 2015
8. ### math
Tell me whether each fraction. 1/n, is given as a terminating decimal or a repeating decimal for the given rules of n. Write t for terminating or r for repeating. It starts at 2- 10 which are t and r.
asked by Wossum on September 3, 2016
9. ### Math
Express the repeating decimal 0.513 (the 13 is repeating) as a fraction in lowest terms using the infinite geometric series method.
asked by Haile on September 29, 2011
10. ### math
ACTIVITY 12: TERMINATE OR REPEAT? Every fraction has a decimal equivalent that either terminates (for example, ) or repeats (for example, ). Work with a group to discover which fractions have terminating decimals and which have
asked by gertrude on October 8, 2008
More Similar Questions | {
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# Integral of Gaussian function, for squared x
1. Jul 27, 2014
### Avinto
1. The problem statement, all variables and given/known data
I am trying to compute an integral, as part of the expected value formula (using a Gaussian PDF)
$$\int_{-∞}^{∞} (x)^2 p(x) dx$$
Where p(x) is the Gaussian probability density function:
$$\frac{1}{\sigma \sqrt(2 \pi)} \exp(\frac{-x^2}{2 \sigma^2})$$
My aim after this is to be able to compute for all even x^n in the above formula. For all odd x^n, the positive and negative componets cancel out, with an computation of zero for all odd functions.
2. Relevant equations
Wikipedia lists two equations that relate to this:
[1]https://en.wikipedia.org/wiki/List_of_integrals_of_Gaussian_functions
[1]
$$\int_{-∞}^{∞} (x)^2 \phi(x)^n dx = \frac{1}{\sqrt(n^3 (2 \pi)^{n-1})}$$
and
[2]https://en.wikipedia.org/wiki/List_of_integrals_of_exponential_functions
[2]
The range here is zero to infinity, however as this function is even, the result for minus infinity to infinity should be twice what the below computes.
$$\int_{0}^{∞} (x)^n \exp(-\alpha x^2) dx= \frac{(2 k -1)!!}{2^{k+1} \alpha^k} \sqrt(\frac{\pi}{\alpha})$$
Where
$$n=2 k, k integer, \alpha > 0$$
3. The attempt at a solution
Using [1], I can compute a answer equalling 1, which is apparently the right answer. However when using [2] (where k=1), I compute a different anwser:
Compute:
Let $$\alpha = \frac{1}{2 \sigma^2}$$
and
$$k=1$$
then:
$$\int_{0}^{∞} (x)^2 \exp(-\alpha x^2) dx = \frac{(2 k -1)!!}{2^{k+1} \alpha^k} \sqrt(\frac{\pi}{\alpha})$$
$$=\frac{1!!}{4 \alpha}\sqrt(\frac{\pi}{\alpha})$$
$$=\frac{\sigma^2}{2}\sqrt(2 \sigma^2 \pi)$$
$$=\frac{\sigma^2}{2} \sigma \sqrt(2 \pi)$$
$$=\frac{\sigma^3 \sqrt(2 \pi)}{2}$$
Now, this is where I think I messed up. I previously removed
$$\frac{1}{\sigma \sqrt(2 \pi)}$$
from inside the integral, and attempted to multiply it by the obtained result (and then doubled everything, as it is an even function) | {
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$$2 \frac{1}{\sigma \sqrt(2 \pi)} \frac{\sigma^3 \sqrt(2 \pi)}{2}$$
Then by cancelling:
$$=\sigma$$
I'm thinking this is because I took part of the PDF out of the integral, and then changed the limits. However, I'm not sure how to go about working this out while leaving it in there.
I will keep browsing around for a solution to this problem (and the more general x^n), and would really appreciate any hints on this.
Thanks.
2. Jul 27, 2014
### Ray Vickson
First get
$$I_n = \int_0^{\infty} x^n \phi(x) \, dx, \:\: \phi(x) = \frac{1}{\sqrt{2 \pi}} e^{-x^2/2},$$
then obtain the general result by re-scaling the variable. Note that $x \phi(x) \, dx = d(-\phi(x))$ so we can use integration by parts:
$$\int x^n \phi(x) \, dx = \int u dv,\\ u = x^{n-1}, \: dv = x \phi(x) \, dx = d(-\phi(x))$$
3. Jul 27, 2014
### Orodruin
Staff Emeritus
Alternatively, change variables to $t = x^2$ and use the definition of the Gamma function.
4. Jul 28, 2014
### Avinto
Ray, I will attempt to work it out using your way later today, and see how it goes.
$$t = x^2$$
$$\Gamma(z) = \int_0^{∞} t^{z-1} exp(-t) dt$$
$$\Gamma(3) = \int_0^{∞} t^{3-1} \exp(-t) dt$$
$$\Gamma(3) = (3-1)! = 2$$
Then for the range ${-∞,0}$, this answer should be the same, so I get an answer of 4 overall.I'm working under the assumption that the answer should be 1
Last edited: Jul 28, 2014
5. Jul 28, 2014
### Orodruin
Staff Emeritus
Well, the answer cannot be 1. It should be the variance of the distribution (since the expectation value is zero). In fact, you had it (almost) right in your first post... Also note that in the substitution, z is not 3. If you do it correctly, z should have a half-integer value (which is why you end up with a semi-factorial and 2^(k+1)).
Edit: Well, of course the answer would be one if the variance is ... :)
6. Jul 28, 2014
### Avinto | {
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Edit: Well, of course the answer would be one if the variance is ... :)
6. Jul 28, 2014
### Avinto
Ok, I see what you mean about the variance, the definition of $\phi$ on the wiki page assumes $\sigma=1$. Using $\sigma=1$, my calculation did evaluate to one, which is where I must have gotten that assumption from.
I redid my calculations with including $\sigma$, and realised I made a silly error in my cancellation.
$$2 \frac{1}{\sigma \sqrt(2 \pi)} \frac{\sigma^3 \sqrt(2 \pi)}{2}$$
Should evaluate to $\sigma^2$, not $\sigma$.
I will look more into the Gamma function as well, and try to understand it from that direction. Thanks | {
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# How to find the minimum of the function?
How to find the minimum of the following function
$${\rm f}\left(w\right) = {1 \over 2}\sum_{i = 1}^{n}\left({1 \over 1 + {\rm e}^{-x_{i}\,w}} -y_{i}\right)^{2}$$
where $x_{i}, y_{i} \in \left(0, 1\right)$ are constants, $w\in \mathbb{R}$?
Could you find a analytic or computational way to get the minimum of the function ?.
• Please try to make use of MathJax here :) Dec 5 '13 at 12:13
• Ok, I will try it next time.@Shaun Dec 5 '13 at 12:15
• Check the following link ---> dlmf.nist.gov/software Dec 6 '13 at 4:20
• Thanks for your software index.@Felix Marin Dec 6 '13 at 5:03
I suppose that this is a nonlinear least square fit problem in which you have data points [x(i) , y(i)] and you want to adjust the parameter w. If you establish the derivative of f(w) with respect to w, you have one (not too complex) equation to solve but it can easily be done using Newton method. The problem is to start with a reasonable value; you can have one rewriting x(i) as a function of y(i). Going to logarithms, you will see that x(i) is along a straight line of Log[y(i) / (1-y(i)] and the slope of this line is w. So, you have everything to start. | {
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• Thank you for your answer. With your method, I can find several points that the derivative of f(w) are 0. But can I find every point that the derivative of f(w) are 0? So that I can find the minimum of the whole function. Dec 5 '13 at 13:10
• No, this is not the way. Since your data y(i) are in error, you search for the value of "w" which minimizes your function f(w) that is to say which is the solution of f'(w)=0. The solution is unique. As I told you (with a typo), a look of the plot of Log[y(i) / (1 - y(i)] versus x(i) will give you an estimate of "w" which is the root you look for. If you want, post a few points for me and I shall enter into details in such a way you can continue by yourself. If you like my answer, you can accept it. Dec 5 '13 at 15:12
• To correct some typos (I am almost blind), plot Log[1 - 1 / y(i)] against x(i). The graph will look like a straight line the slope of which being "- w". Make a visual estimation of that (or use Excel plot with regression in the transformed space). Now, solve f'(w)=0 using Newton. You can also visualize ploting f(w) versus "w". Dec 5 '13 at 15:24
• I don't understand why you think there is only one solution of f'(w)=0. Could you proof it prove it? And I will appreciate your answer. Change y(i) into Log[1 - 1 / y(i)] can make it into a linear regression problem. But it's meaningless for me because I need to estimate the error with the form y(i).@Claude Leibovici Dec 6 '13 at 3:24
To try to clarify things, I generated 10 values [x(i) = i / 10] and the corresponding values y[i] are [0.58, 0.65, 0.72, 0.78, 0.83, 0.87, 0.90, 0.92, 0.94, 0.96]. | {
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Based on these, I wrote function f(w) as given in your post (sum of 10 terms). When plotted as a function of w, f(w) exhibits (as totally normal) a paraboloid shape with a marked minimum around 3.15 (to give you an idea, for f(2.0)=0.0337713, f(2.5)=0.00852217, f(3.0)=0.000357495, f(3.5)=0.00181296, f(4.0)=0.00843496. The absolute minimum corresponds to w=3.13938 for which f(w)=0.000031942 and this is the solution.
Using the second approach, I wrote function f'(w). Ploted against w, this function has a very nice shape and becomes exactly zero at w=3.13938 and this is the solution.
You must remember than solving an equation is much simpler than minimizing a function (almost if not constrained). For illustration purposes, to solve f'(w)=0, I used Newton method starting at w=2.0 (value which is far away from the one I suggested you to generate using some changes). The successive iterates for w are 2.45293, 2.75188, 2.93280, 3.03609, 3.09582, 3.13997. | {
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• Thanks for your example. My colleagues also tried many examples and haven't find a case that has more than one relative extremum. I wonder could it be proved analytically that the function f(w) have exactly one relative extremum, or have a sufficiently large probability that has only one relative extremum, which you proved experimentally. Dec 6 '13 at 6:50
• @YangzheLau.In the case of your model, which is extremely simple, only one extremum can exist. Forgetting mathematics, you search for a value of "w" which gives you the best compromise between the y[i] and the model. This minimum is then unique. If your model was more complex such y = a / (b + Exp[-w x]), the story would be very different and in this case, except if you can generate good estimates, several minima could exist and global optimization could be required. But, this is NOT the case for your problem. Dec 6 '13 at 7:06
• @ClaudeLeibovici As I showed in my comment below on my answer, the functional need not have a global minimum (I.e. It is at $\pm\infty$). There are also times where it is I feasible to calculate the derivative due to its complexity, let alone root find with it! There is a reason that a great deal of research has been performed on minimisation techniques. Dec 6 '13 at 7:22
• @Daryl x[i] and y[i] belong to (0,1). We didn't find a counterexample yet. Dec 6 '13 at 7:34
• @ClaudeLeibovici You are right at least in any application case. Ok, I just consider it as a theorem and continue my work. Dec 6 '13 at 7:45
From a numerical point of view, your problem looks quite iteresting and I played with it, my concern being to get a quick estimate of the parameter "w" which has to be adjusted to your data.
What I did was to start a Newton procedure at w=0. As a result, what I obtained is that a rough estimate of "w" can be obtained using the following formula for the estimate
w = 2 (2 S3 - S1) / S2
in which S1 is the sum of the x(i), S2 is the sum of x(i)^2 and S3 is the sum of x(i) * y(i). | {
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For an exact value of w = 1, 2 or 3, the corresponding estimates are is 0.95, 1.66 and 2.11. The corresponding "experimental" data were x(i) = i / 10 and the y(i) were calculated using the exact formula.
In the case I previously illustrated (with noise in the y data), this procedure would lead to an estimate equal to 2.16 for an excat value of 3.14.
• Your estimate can be used as an good start value of w for me, and I 'm surprised at the succinct form of the derivative and second derivative of the function when w=0. By the way, I'm considering a two dimensional case. $${\rm f}\left(w_{1},w_{2}\right) = {1 \over 2}\sum_{i = 1}^{n}\left({1 \over 1 + {\rm e}^{-x_{i}w_{1}-z_{i}w_{2}}} -y_{i}\right)^{2}$$ Now there are more than one local extremum. Dec 8 '13 at 12:51
• @YangzheLau. Now, we are in the serious case with two dimensions. However, you can first perform a linear regression (no intercept) for the model log(1/y - 1) = - w1 x - w2 z. This will give you very reasonable estimates and, starting from these, I am almost ready to bet that you will reach THE optimum. Dec 8 '13 at 13:30
• Thanks for your suggestion, I will do experiments on it. Dec 8 '13 at 14:43
If you are finding the minimum with respect to $w$, the domain of $w$ will effect what your minimum will be. So is $w\in \mathbb{R}$ or is it that $w$ can only take a particular set of values?
• Sorry, I forgot to notice that $w\in \mathbb{R}$. Dec 5 '13 at 13:02
• In that case you can it is clear that the minimum value should be zero. Since $w=\frac{1}{x_i}\ln\left(\frac{y_i}{1-y_i}\right)$ will make all the terms inside the summation zero.
– user112535
Dec 5 '13 at 13:06
• But there is only one w with several $x_i$ and $y_i$.@LinearAlgebra Dec 5 '13 at 13:17
• Oh I am sorry. I mixed up your data points as fixed constant values. :)
– user112535
Dec 5 '13 at 13:47 | {
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Given your function $f(w)$, its derivative wrt $w$ is given by \begin{align} \frac{df}{dw}&=\sum\limits_{i=1}^n\left(\frac{1}{1+\exp(-x_iw)}-y_i\right)\cdot\frac{x_i\exp(-x_iw)}{\left(1+\exp(-x_iw)\right)^2}\\ &=\sum\limits_{i=1}^n\frac{\left(1-y_i\left(1+\exp(-x_iw)\right)\right)\cdot x_i\exp(-x_iw)}{\left(1+\exp(-x_iw)\right)^3}, \end{align} which is a quite complicated expression when attempting to find the root analytically.
A number of computational methods exist for finding the minimum of $f(w)$. The class of solvers which is applicable to this problem solve unconstrained non-linear optimisation problems, of which a simple google search reveals many possibilities.
• Thank you for your suggestion, I'm searching on Google. It' will be appreciated if you or other mathematicians show me more clues on solving this problem as unconstrained non-linear optimization. Dec 6 '13 at 4:17
• @YangzheLau. You may adress the problem is two different manners : the first one would be an unconstrained minimization problem (minimize f(w) for w); the second is much simpler : back to definition, solve f'(w) = 0 for w. The second approach is much simpler. Dec 6 '13 at 4:39
• @ClaudeLeibovici There are multiple w where f'(w) = 0. If you can find all these w, the second approach is much simpler. Dec 6 '13 at 5:13
• @YangzheLau. I think we misunderstand : there is only one value of w which makes f'(w)=0. Please see my next answer with details on a built example. Dec 6 '13 at 5:17
• @Daryl Thank you for your link. Dec 6 '13 at 6:54 | {
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# Packing obtuse vectors in $\mathbb{R}^d$
I came across this attractive theorem:
• Theorem. In $\mathbb{R}^d$, there can be at most $d+1$ vectors that form an obtuse angle with one another.
This was proved1 as a corollary of a lemma about irreducible matrices. I am wondering if anyone knows of an alternative, more geometric proof that somehow more directly captures the sense that one cannot "pack" more than $d+1$ obtuse vectors in $\mathbb{R}^d$.
1Lipeng Ning, Tryphon T. Georgiou, Allen Tannenbaum, Stephen P. Boyd. "Linear models based on noisy data and the Frisch scheme." SIAM Review. 57(2) 2015. arXiv preprint.
• Thanks to all for the clever and educational alternative proofs. I'd like to select them all, but I must choose one, and have done so. – Joseph O'Rourke Jun 5 '15 at 19:05
You can prove it by induction. | {
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You can prove it by induction.
The case $d=1$ is obvious. Assume it's true for dimension $d \geq 1$, and consider any set of $d+3$ vectors in $\mathbb{R}^{d+1}$. Fix one of these, say $\vec{v}_0$, and consider the hyperplane $H$ perpendicular to it (which we can identify with $\mathbb{R}^d$). Any one of the remaining $d+2$ vectors must lie on the far side of H (i.e., not the same side as $\vec{v}_0$). No vector other than $\vec{v}_0$ can be perpendicular to $H$, since then no third vector could form an obtuse angle with both. Thus all of the remaining $d+2$ vectors lie on the same side of $H$ and each has a nonzero projection onto $H$. By hypothesis, two of them, say $\vec{v}_1$ and $\vec{v}_2$, have projections forming a non-obtuse angle. Since they lie on the same side of $H$ and their projections onto $H$ form a non-obtuse angle, the angle between them is non-obtuse. (To see that this last statement is true, consider the dot product $\vec{v}_1 \cdot \vec{v}_2$. If you compute this using a basis consisting of $\vec{v}_0$ and vectors in $H$, then the $\vec{v}_0$ term is positive (they're on the same side of $H$), and the rest of the dot product is non-negative (their projections form a non-obtuse angle).)
I don't know if this is exactly what you're looking for, but to me this proof is pretty intuitive: you can picture your vectors all pointing more or less away from $\vec{v}_0$, and then if there are too many two of them get crammed close together. The exact value of "too many" is then easily computed by induction.
I always give this as an exercise in my linear algebra class in the form:
If $k+1$ vectors in $\mathbb{R}^n$ form obtuse angles with one another, then after removing any one vector from that collection, the remaining vectors are linearly independent. | {
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Proof. Assume that $v_1,\ldots,v_k$ are the remaining vectors, and that they are linearly dependent. Then $c_1v_1+\cdots+c_kv_k=0$ for some $c_1,\ldots,c_k$ not all of which are equal to zero. Without the loss of generality, $c_1,\ldots,c_m$ are positive, and $c_{m+1},\ldots,c_k$ are negative (we can remove the vectors with zero coefficients in front of them). Computing the scalar product with the vector $v$ we removed, we note that we cannot have $m=k$, so $m<k$. Now, $$c_1v_1+\cdots+c_mv_m=-c_{m+1}v_{m+1}-\cdots-c_kv_k.$$ Therefore, $$(c_1v_1+\cdots+c_mv_m,c_1v_1+\cdots+c_mv_m)= (c_1v_1+\cdots+c_mv_m,-c_{m+1}v_{m+1}-\cdots-c_kv_k)<0,$$ a contradiction.
In the context of spherical codes, this is an instance of Rankin's bound. The largest minimum distance between $m$ points on the $(n-1)$-sphere is achieved by:
(1) The vertices of a $(m-1)$-simplex inscribed in a great $(m-2)$-sphere when $m\le n+1$.
(2) Any $m$ of the $2n$ vertices of a cross-polytope inscribed in the sphere, if $n+1<m\le 2n$.
Therefore, any $n+2$ points will have to have at least one pair at a right angle or smaller.
Here is the reference to the original paper of Rankin: R. A. Rankin (1955). The Closest Packing of Spherical Caps in n Dimensions. Proceedings of the Glasgow Mathematical Association, 2, pp 139-144.
The proof I know of this result is essentially the same as Vladimir Dotsenko's, but in terms of Radon's lemma.:
Assume such a set exists with at least $d+2$ points. By Radon's lemma, given any $d+2$ points in $\mathbb{R}^d$ there is a partition of them into two sets $A$, $B$ such that $conv(A) \cap conv(B) \neq \emptyset$. Let $p$ be a point of this intersection. If we write $p$ as a convex combination of $A$ and one of $B$, and using the fact that the dot product of any two points in the original set is negative, we immediately obtain $p \cdot p < 0$, a contradiction. | {
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Without loss of generality, the vectors are in general position. Take $d$ vectors with pairwise obtuse angles and rotate them so that they form the columns of a $d\times d$ upper-triangular matrix with nonnegative diagonal (à la $QR$ factorization). Then every entry on the diagonal is strictly positive, since otherwise the vectors would be linearly dependent, thereby violating general position. Next, in order for the columns to be pairwise obtuse, every entry above the diagonal must be strictly negative. At this point, observe that every vector which is obtuse with these $d$ columns necessarily has all strictly negative entries, but two vectors of this sort necessarily have a positive inner product with each other. | {
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# How to calculate variance of W? Find the probability distribution of W?
$W=Y-X$
I have figured out that $E(W)=0.3$ by using this formula $E(X+Y)=E(X)+E(Y)$.
I tried using the same formula with $E(X^2)$ and $E(Y^2)$ to find $E(W^2)$.
I also tried using $V(X+Y)=V(X)+V(Y)+2Cov(X,Y)$, but changing all the positive signs to negative, to find the Variance of W.
Here is the joint distribution of $X$ and $Y$:
$$\begin{array}{c||c|c|c} & Y=0 & Y=1 & Y=2 \\\hline X=0 & 0.1 & 0.1 & 0.2 \\ X=1& 0.3 & 0.2 & 0.1 \end{array}$$
What I think I got right.
• $E(X) = 0.6$
• $E(Y) = 0.9$
• $E(X^2) = 0.6$
• $E(Y^2) = 1.5$
• $X$ and $Y$ are not independent
I am also definitely mixing when I can use what formula and when I can't.
• $E(W^2) = 1.3$
• $Var(W)=1.21$
• Need more information about the relationship between $X$ and $Y$. Also what is the first itemized list? $\bullet XY - 0.0 - 1.0$ etc.? – jdods Jul 14 '16 at 14:09
• It is the probability distribution table with 2 variables, I think it's called. – David Lund Jul 14 '16 at 14:11
• ah. the edit messed up your text. I'll go in and fix it with proper tabular format. – jdods Jul 14 '16 at 14:12
• Does the edit now seem to reflect your wishes? Feel free to edit it more to your liking. Hope this helps. – jdods Jul 14 '16 at 14:24
• Yes, thank you :) Perfect! I am going to have to learn do edit like that myself soon. – David Lund Jul 14 '16 at 14:25
The easiest thing to do is to first compute the probability distribution of $W$ from the joint distribution of $X$ and $Y$; then use this to compute $\operatorname{Var}[W]$ directly.
to this end, simply create a table for $W$ using the table for $X$ and $Y$:
$$\begin{array}{|c|c|c|c|} \hline x & y & w & \Pr[(X,Y) = (x,y)] \\ \hline 1 & 0 & -1 & 0.3 \\ \hline 1 & 1 & 0 & 0.2 \\ \hline 1 & 2 & 1 & 0.1 \\ \hline 0 & 0 & 0 & 0.1 \\ \hline 0 & 1 & 1 & 0.1 \\ \hline 0 & 2 & 2 & 0.2 \\ \hline \end{array}$$ | {
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"url": "https://math.stackexchange.com/questions/1859269/how-to-calculate-variance-of-w-find-the-probability-distribution-of-w"
} |
Then collapse this table for distinct values of $w$, that is to say, add the rightmost column values for each row with the same value of $w$:
$$\begin{array}{|c|c|} \hline w & \Pr[W = w] \\ \hline -1 & 0.3 \\ \hline 0 & 0.2 + 0.1 = 0.3 \\ \hline 1 & 0.1 + 0.1 = 0.2 \\ \hline 2 & 0.2 \\ \hline \end{array}$$
This gives the desired probability distribution of $W$
Now the expectation and variance are trivially computed from this table: $$\operatorname{E}[W] = -1(0.3) + 0(0.3) + 1(0.2) + 2(0.2) = 0.3 \\ \operatorname{E}[W^2] = (-1)^2 (0.3) + 0^2 (0.3) + 1^2 (0.2) + 2^2 (0.2) = 1.3 \\ \operatorname{Var}[W] = \operatorname{E}[W^2] - \operatorname{E}[W]^2 = 1.21.$$
If you calculated the variance from the joint distribution of $X$ and $Y$ directly, then you'd need to go back to get the probability distribution of $W$ for the second part of your question, anyway.
• Hi, you don't think you know about any videos that can maybe explain how to make that table. I don't quite get it. My book does not explain it at all, I think. I see that you take Y-X, and you get W, literally 0-1=-1. But I don't get much more than that. – David Lund Jul 14 '16 at 17:19
• @DavidLund All I did in the first table was rearrange the table in your question. The first, second, and fourth columns in my first table are taken directly from your table. The third column is just the calculation of $w = y - x$ from the first two columns. Then, in the second table I created, I just dropped the $x$ and $y$ columns, and for the $w$ column, I added together the two rows from the first table for which $w = 0$; and I also added together the two rows from the first table for which $w = 1$. – heropup Jul 14 '16 at 17:28
• Okey, I did not realize that, but now I am so happy. Thanks. – David Lund Jul 14 '16 at 17:56
Now you calculate $\mathbb E(W^2)$
$\mathbb E(W^2)=\sum_{i=1}^2 \sum_{j=1}^3 (w_{ij})^2 \cdot p(w_{ij})$
$\mathbb E(W^2)=\sum_{i=1}^2 \sum_{j=1}^3 (x_i-y_j)^2 \cdot p(x_i,y_j)$ | {
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"url": "https://math.stackexchange.com/questions/1859269/how-to-calculate-variance-of-w-find-the-probability-distribution-of-w"
} |
$\mathbb E(W^2)=\sum_{i=1}^2 \sum_{j=1}^3 (x_i-y_j)^2 \cdot p(x_i,y_j)$
$=0^2\cdot 0.1+(0-1)^2\cdot 0.1+(0-2)^2\cdot 0.2+(1-0)^2\cdot 0.3+(1-1)^2\cdot 0.2+(1-2)^2\cdot 0.1=1.3$
And finally
$Var(X-Y)=Var(W)=\mathbb E(W^2)-[\mathbb E(W)]^2=1.3-(0.6-0.9)^2=1.21$
with $\mathbb E(W)=\mathbb E(X)-\mathbb E(Y)$
Remark
You can use the formula $V(X+Y)=V(X)+V(Y)+2Cov(X,Y)$ as well.
$Cov(aX,bY)=abCov(X,Y)$
In your case $a=1$ and $b=-1$. Thus $Cov(X,-Y)=-Cov(X,Y)$
And $Cov(X,Y)=\sum_{y=0}^2 \sum_{x=0}^1 p(x,y)\cdot (x-E(x))\cdot (y-E(y))$
$=0.1\cdot (-0.6)\cdot (-0.9)+(-0.1\cdot 0.6\cdot 0.1)+0.2\cdot (-0.6)\cdot 1.1+0.3\cdot 0.4\cdot (-0.9)+0.2\cdot 0.4\cdot 0.1+0.1\cdot 0.4\cdot 1.1=-0.14$
Thus $Cov(X,-Y)=0.14$
And $Var(X)=\mathbb E(X^2)-[\mathbb E(X)]^2=0.6-0.6^2=0.24$
$Var(Y)=\mathbb E(Y^2)-[\mathbb E(Y)]^2=1.5-0.9^2=0.69$
Therefore $Var(X-Y)=0.24+0.69+2\cdot 0.14=1.21$
• So I also got -0.14. So if one of them, either X or Y is negative, the Cov will become negative as well? Cov(X,−Y)=−Cov(X,Y) – David Lund Jul 14 '16 at 17:11
• Not in general. Here you have an example. X is postive and -Y is negative, but the covariance is positive. Maybe you have meant the following:".So if one of them, either $X$ or $Y$ is negative, the Cov will have the opposite sign of $Cov(X,Y)$". You statement $Cov(X,-Y)=-Cov(X,Y)$ is right. – callculus Jul 14 '16 at 17:17
• Yea, that is what I meant. I realized that I needed 0.14, but I got -0.14. So is it like that always? What if both are negative? I haven't been studying this for a long time. – David Lund Jul 14 '16 at 17:22
• @DavidLund Just use the formula $Cov(aX,bY)=abCov(X,Y)$. What do you get for $Cov(-X,-Y)$ ? – callculus Jul 14 '16 at 17:23
• That's a handy formula. + :D I am going to write that down, definitely. Thanks – David Lund Jul 14 '16 at 17:26 | {
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} |
# Problem of dice
The initial purpose of this post was give a proper proof of a problem posted on Twitter by @jamestanton (it is hard within the 140 char limit), but the post was later extended to cover some other problems.
Show that a four-sided and a nine-sided dice cannot be used to simulate the probability distribution of the product of outcomes when using two six-sided dice. In the original wording:
Is there a 4-sided die & a 9-sided die that together roll each of the products 1,2,3,…,30,36 w the same prob as two ordinary 6-sided dice?
We make an argument by contradiction, considering only the possible outcomes without taking the actual probabilities into account.
Obviously, to reach the same outcomes as for two normal dice $\{1,2,3,4,5,6\} \times \{1,2,3,4,5,6\}$, we need both dice to have the identity $\{1\}$ (otherwise, we will not be able to reach $1 \cdot 1 = 1$). So, $\{1,*,*,*\} \times \{1,*,*,*,*,*,*,*,*\}$.
Now, consider the prime $5$. It must be on both dice, or we would have $\mathbb{P}(5^2\cdot b)>0, b>1$. So, $\{1,5,*,*\} \times \{1,5,*,*,*,*,*,*,*\}$. Also, since $5$ appears on both dice, no dice can contain some product of the primes $\{2,3,5\}$ and their powers (e.g $2^2 \cdot 3$) that does not exist on the original dice, because then impossible products could be reached.
Hence, $6$ must be on both dice, giving $\{1,5,6,*\} \times \{1,5,6,*,*,*,*,*,*\}$. There are $6$ sides left on the larger die but we have more even products, so $2$ must also be on each die. $\{1,5,6,2\} \times \{1,5,6,2,*,*,*,*,*\}$. Now, there is no space left for $3$ on the smaller die. This means that $3^2$ must be one the larger die, but then $\mathbb{P}(3^2\cdot 5)>0$, which is a contradiction.
(@FlashDiaz gave a shorter proof)
Project Euler 205
Peter has nine four-sided (pyramidal) dice, each with faces numbered $1, 2, 3, 4$.
Colin has six six-sided (cubic) dice, each with faces numbered $1, 2, 3, 4, 5, 6$. | {
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Peter and Colin roll their dice and compare totals: the highest total wins. The result is a draw if the totals are equal.
What is the probability that Pyramidal Pete beats Cubic Colin? Give your answer rounded to seven decimal places in the form $0.abcdefg$.
The probability functions of the nine four-sided dice and the six six-sided dice are given by the generating functions $\frac{1}{4^9} \cdot (x^1+x^2+x^3+x^4)^9$ and $\frac{1}{6^6} \cdot (y^1+y^2+y^3+y^4+y^5+y^6)^6$, respectively.
Let $X_1,...,X_9$ be i.i.d random variables taking values in the range $[1,4]$ and let $Y_1,...,Y_6$ taking values in the range $[1,6]$. We want to determine the probability that $\rho = \mathbb{P}(X_1+...+X_9 > Y_1+...+Y_6)$.
The distributions can be computed as
def rec_compute_dist(sides, nbr, side_sum):
global dist
if nbr == 1:
for i in range(1, sides+1):
dist[side_sum+i] += 1
else:
for i in range(1, sides+1):
rec_compute_dist(sides, nbr-1, side_sum+i)
dist = [0]*37
rec_compute_dist(4,9,0)
dist_49 = dist
dist = [0]*37
rec_compute_dist(6,6,0)
dist_66 = dist
To determine $\rho$, we may express it as
$\begin{array}{rl} \rho = & \sum_{t=6}^{36} \mathbb{P}(X_1+...+X_9 > t| Y_1+...+Y_6 = t)\cdot \mathbb{P}(Y_1+...+Y_6 = t) \\\\ = & \sum_{t=6}^{36} \mathbb{P}(X_1+...+X_9 > t)\cdot \mathbb{P}(Y_1+...+Y_6 = t) \end{array}$.
Computing the sum using the following code,
probability = 0
for i in range(6,36+1):
for j in range(i+1,36+1):
probability += dist_66[i]*dist_49[j]
print 1.0 * probability/(6**6 * 4**9)
we obtain the answer. Great 🙂
Project Euler 240
There are $1111$ ways in which five six-sided dice (sides numbered $1$ to $6$) can be rolled so that the top three sum to $15$. Some examples are:
$\begin{array}{rcl} D_1,D_2,D_3,D_4,D_5 &=& 4,3,6,3,5\\ D_1,D_2,D_3,D_4,D_5 &=& 4,3,3,5,6\\ D_1,D_2,D_3,D_4,D_5 &=& 3,3,3,6,6\\ D_1,D_2,D_3,D_4,D_5 &=& 6,6,3,3,3 \end{array}$ | {
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In how many ways can twenty twelve-sided dice (sides numbered $1$ to $12$) be rolled so that the top ten sum to $70$?
Let us first consider the simpler problem $\left\{ d_1+..+d_{10}=70 \right\}$. If we restrict the remaining ten dice to be less than or equal to the minimum value of the ten dice, we then can compute the cardinality. Let $n_i$ denote the number of $i$‘s we got. Then, $n_1 \cdot 1 + n_2 \cdot 2 + ... + n_{12} \cdot 12 = 70$ where $n_1 + n_2 + ... + n_{12} = 10, n_i \geq 0$.
All histograms of top-ten dice can be computed with
from copy import copy
d = [0] * 12
possible = []
def rec_compute(i, j, sum):
global d
if j == 0:
if sum == 70:
possible.append(copy(d))
return
while i > 0:
if sum + i <= 70:
d[i - 1] += 1
rec_compute(i, j - 1, sum + i)
d[i - 1] -= 1
i -= 1
rec_compute(12, 10, 0)
The code exhausts all solutions in 200ms. Call any solution $H$. For instance H = [0, 0, 0, 0, 0, 0, 10, 0, 0, 0, 0, 0]. The remaining dice can take any values in the range $[1, j]$, where $j$ is the left-most non-zero index (starting from $1$). The number of configurations for this particular solution is then given by
$20! \cdot \left((10+H_7)!H_6!H_5!H_4!H_3!H_2!H_1!\right)^{-1}$, where $\sum^7_1 H_i = 10$.
Unfortunately, there is no good analytical way of computing this. So, the easiest way is to enumerate all possible $H_i$. Disregarding $H_7$, we compute all permutations of a given histogram in the same way (hence, we can make the search space a lot smaller) and the using the multiplicity to determine the exact number. All and all, the following code gives our answer:
def configurations(i, j, x, s, l):
if sum(x) == s:
# we can stop here as the sum cannot get smaller
multiplicity = fact(l) / fact(l-len(x)) / \
reduce(lambda m, n: m * n, \
[fact(y) for y in \
Counter(x).values()])
return fact(DICE) * multiplicity / \
reduce(lambda m, n: m * n, \
[fact(y) for y in x]) | {
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return fact(DICE) * multiplicity / \
reduce(lambda m, n: m * n, \
[fact(y) for y in x])
if j == 0 or i == 0: return 0
return configurations(i-1, j, x, s, l) + \
configurations(i, j-1, x + [i], s, l)
S = 0
for H in possible_top_dice:
min_index = next((i for i, \
x in enumerate(H) if x), None)
for j in range(0, REMAINING+1):
u = reduce(lambda m, n: m * n, \
[fact(y) for y in H])
if j < REMAINING:
q = configurations(REMAINING-j, min_index, \
[], REMAINING-j, min_index) / u
else:
q = fact(DICE) / u
H[min_index] += 1
S += q
print S
# Breaking affine ciphers – a matrix approach
An affine cipher is a one the remnants of classical cryptography, easily broken with todays computational power. The cipher defines a symbol mapping from $f :\{A,B,\ldots,\} \mapsto \mathbb{Z}_n$. Each cipher symbol is then computed as $a \cdot x + b \rightarrow y$, where $a \in \mathbb{Z}^*_n$ and $b \in \mathbb{Z}_n$. Decryption is then done by computing $x= (y - b) \cdot a^{-1}$.
In this blog post, I will show how to break this cipher in time faster than trying all keys. Let us first sketch the general idea. Consider an expected distribution $\hat{P}$ of the symbols and a given distribution $P$, the integral $\int (\hat{P}(x) - P(x))^2 dx$ defines a statistical distance between the distributions (this would correspond to the Euclidian distance), which we would like to minimize. Now, clearly
$(\hat{P}(x) - P(x))^2 = \hat{P}(x)^2 - \hat{P}(x)P(x) + P(x)^2$.
Trivially, $\hat{P}(x)^2$ and $P(x)^2$ remains constant over any keypair $(a,b)$, so instead of minimizing the above, we can maximize $\hat{P}(x)P(x)$. Therefore, the minimization problem can be turned into a maximization problem $\max_{a,b} \int \hat{P}(x)P_{a,b}(x) dx$. Cool. | {
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In terms of our cipher, which is discrete, the minimization problem is a sum $\max_{a,b} \sum \hat{P}(x)P_{a,b}(x)$. The observant reader may notice that this looks like a term in a matrix multiplication. There is just one caveat; the indices corresponding appear only in one term. There is an easy way to get around this. Instead of applying transformations on only $P$, we may split them among the two. So by instead computing
$\max_{a,b} \sum \hat{P}_a(x) P_{b}(x)$,
we have achieved what we desired. This means that we shuffle $\hat{P}$ with $a$ and ${P}$ with $b$. Let us interpret this as Python. The expected distribution of an alphabet ABCDEFGHIJKLMNOPQRSTUVWXYZ ,. may be as follows (depending on the observation):
P_hat = {' ': 0.05985783763561542, ',': 0.0037411148522259637, '.': 0.0028058361391694723, 'A': 0.0764122708567153, 'C': 0.02600074822297044, 'B': 0.012065095398428732, 'E': 0.11878039655817432, 'D': 0.03974934530490086, 'G': 0.018892630003741116, 'F': 0.020856715301159744, 'I': 0.0651889263000374, 'H': 0.05695847362514029, 'K': 0.00720164609053498, 'J': 0.0014029180695847362, 'M': 0.02254021698466143, 'L': 0.03769173213617658, 'O': 0.07023943135054246, 'N': 0.06313131313131314, 'Q': 0.0009352787130564909, 'P': 0.01805087916199027, 'S': 0.05920314253647587, 'R': 0.0560231949120838, 'U': 0.025813692480359144, 'T': 0.08473625140291807, 'W': 0.022072577628133184, 'V': 0.00916573138795361, 'Y': 0.01842499064721287, 'X': 0.0014029180695847362, 'Z': 0.0006546950991395436}
The transformations are done by computing the matrices
# compute first matrix for transformed P_hat
for i in range(1, N):
for element in priori_dist:
X[i, (look_up.index(element) * i) % N] = priori_dist[element]
# compute second matrix for transformed P
for j in range(N):
for element in dist:
Y[(look_up.index(element) - j) % N, j] = dist[element] | {
"domain": "grocid.net",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9848109520836027,
"lm_q1q2_score": 0.8484522660367423,
"lm_q2_score": 0.8615382112085969,
"openwebmath_perplexity": 2297.7335466269597,
"openwebmath_score": 0.8094127774238586,
"tags": null,
"url": "https://grocid.net/page/4/"
} |
Here, the $i$th row in $X$ corresponds to $\hat{P}$ transformed by $a = i$. Moreover, the $j$th row in $Y$ corresponds ${P}$ transformed by $b = j$.
For some distribution, they may look like
As we can see, $X$ is only shifted (by the addition), while in $Y$ the indices are reordered by multiplication with row index $i$. Taking advantage of the matrix multiplication property, we may now compute $Z=XY$.
Any entry in $Z$ is $Z_{a,b} = \sum_x X_{a,x} Y_{x,b}$ so finding a maximum element in $Z$ is equivalent to saying
$\max_{a,b} \sum_x X_{a,x} Y_{x,b}$.
Looks familiar? It should. This is our maximization problem, which we stated earlier.
Therefore, we may solve the problem using
Z = numpy.dot(X, Y)
a, b = numpy.unravel_index(Z.argmax(), Z.shape)
This breaks the affine cipher.
Some notes on complexity
So, what is the complexity of the matrix approach? Computing the matrices takes $O(N^2)$ modular operations. The matrix multiplication takes naively $O(N^3)$ operations, but for large $N$ this can be achieved faster. For instance Strassen takes $O(N^{2.807})$ but faster algorithms exist. Also, taking advantage of symmetry and structure could probably decrease the complexity further. This is the total complexity of this approach.
Compare this with brute-force guessing of the key (taking $O(N^2)$ guesses) and for each guess, compute the distance taking $O(N)$ operations, which in total yields $O(N^3)$. It should be noted that complexity of this approach may be reduced by picking $a,b$ in an order which minimizes the number of tries.
Example implementation for the id0-rsa.pub: github
# Custom terminal for Vagrant/SSH
Short story: I wanted to distinguish my terminal windows between local sessions, ssh sessions and vagrant sessions.
SSH_THEME="SSH"
VAGRANT_THEME="Vagrant" | {
"domain": "grocid.net",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9848109520836027,
"lm_q1q2_score": 0.8484522660367423,
"lm_q2_score": 0.8615382112085969,
"openwebmath_perplexity": 2297.7335466269597,
"openwebmath_score": 0.8094127774238586,
"tags": null,
"url": "https://grocid.net/page/4/"
} |
SSH_THEME="SSH"
VAGRANT_THEME="Vagrant"
set_th () {
osascript -e "tell app \"Terminal\" to set current settings of first window to settings set \"$1\"" } set_id () { osascript -e "tell app \"Terminal\" to set current settings of first window to$1 $2$3 $4" #$@ does not work!
}
get_id () {
cur_id=$(osascript -e "tell app \"Terminal\" to get current settings of first window") } ssh(){ #!/bin/sh get_id set_th$SSH_THEME
/usr/bin/ssh "$@" set_id$cur_id
} | {
"domain": "grocid.net",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9848109520836027,
"lm_q1q2_score": 0.8484522660367423,
"lm_q2_score": 0.8615382112085969,
"openwebmath_perplexity": 2297.7335466269597,
"openwebmath_score": 0.8094127774238586,
"tags": null,
"url": "https://grocid.net/page/4/"
} |
vagrant(){
#!/bin/sh
if [ $1 = "ssh" ]; then get_id set_th$VAGRANT_THEME
/opt/vagrant/bin/vagrant "$@" set_id$cur_id
else | {
"domain": "grocid.net",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9848109520836027,
"lm_q1q2_score": 0.8484522660367423,
"lm_q2_score": 0.8615382112085969,
"openwebmath_perplexity": 2297.7335466269597,
"openwebmath_score": 0.8094127774238586,
"tags": null,
"url": "https://grocid.net/page/4/"
} |
/opt/vagrant/bin/vagrant "$@" fi } The code creates a temporary variable of the current theme before switching. So, when ending the session, the original theme changes back instead of a fixed one. Putting the above code in your .bash_profile: gives the following nice behavior: Color coding your sessions is a great way to visualize things and make sure you do not take any unwanted action by mistake 🙂 Of course, the code can be used to wrap any application. For instance, one could use it to make the interactive interpreter of Python/Sage or terminal sessions using torsocks appear in different colors or even fonts. # Re-mapping KBT Pure Pro in OS X For my everyday-use computer, I use a modded KBT Pure Pro; this is a small mechanical keyboard with aluminium base and background lightning, perfect for programming and typing. The size of the keyboard is 60 % of a normal one, making it suitable for spatially constrained workspaces. To my experience, it is also more ergonomic. Below is a comparison of the Pure Pro and a wireless Apple keyboard. For those being the in the process of buying a keyboard, I recommend this one 🙂 For quite a while, I have used Linux on this computer. But after installing OS X, the keyboard map went wack, so to speak. Many keys were mapped incorrectly. Using Ukulele, I created a customized layout with correct mapping (don’t mind the duplicate keys): The layout covers all keys and can be found here. NOTE: this is a layout for KBT Pure Pro with British ISO layout and not ANSI. # BackdoorCTF16 – Collision Course With 350 points and a description as follows: In today’s world, hash collisions are becoming more and more popular. That is why, one must rely on standardized hashing techniques, such as bcrypt. However, n00bster shall never learn, and he has implemented his own hash function which he proudly calls foobar. Attached is an implementation of the hash function and a file with which you are supposed to find a collision. He believes that you will not | {
"domain": "grocid.net",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9848109520836027,
"lm_q1q2_score": 0.8484522660367423,
"lm_q2_score": 0.8615382112085969,
"openwebmath_perplexity": 2297.7335466269597,
"openwebmath_score": 0.8094127774238586,
"tags": null,
"url": "https://grocid.net/page/4/"
} |
function and a file with which you are supposed to find a collision. He believes that you will not be able to find a collision for the file, especially since he hasn’t even given you the hashing algorithm, but has packaged it as a black box application. Prove to him that he is wrong. Note: Multiple collisions are possible, but only one of them is a valid flag. You will realize you’ve gotten it once you do. The hash is given as follows: So, we start off by looking at the binary. Using Hopper, we obtain the following pseudo code by decompilation: int hash(int input) { eax = _rotr(input ^ 0x24f50094, (input ^ 0x24f50094) & 0xf); eax = _rotl(eax + 0x2219ab34, eax + 0x2219ab34 & 0xf); eax = eax * 0x69a2c4fe; return eax; } int main() { esp = (esp & 0xfffffff0) - 0x20; puts(0x80486d0); gets(0x804a060); stack[2039] = "\nBar:"; puts(stack[2039]); while (stack[2039] < *(esp + 0x18)) { stack[2039] = *(stack[2039] + stack[2039] * 0x4); *(esp + 0x14) = *(esp + 0x14) ^ hash(stack[2039]); eax = _rotr(stack[2039], 0x7); printf("%08lx", stack[2039]); *(esp + 0x10) = *(esp + 0x10) + 0x1; } eax = putchar(0xa); return eax; } We sketch the above code as block scheme below: The first thing to note is that we can find an infinite number of collisions just by appending arbitrary data after 10 blocks. However, this is not interesting to us, but completely defeats the conditions for a safe cryptographic hash function. This Merkle-Damgård-like structure allows us to solve blocks iteratively, starting from the first. Here is how. Starting from the first block, we can find an input to the function $H$ such that when rotated 7 steps is equal to block 0 (here, denoted $B_0$). Hence, the problem we solve is to find an $x$ such that $H(x) \ll 7 = B_0$. This is a simple thing for Z3. Then, we take the next block and solve for $(H(x) \oplus B_0) \ll 7 = B_1$ and so forth. Implemented in Python/Z3, it may look like the following: from z3 import * import binascii, string, itertools bits = 32 mask = | {
"domain": "grocid.net",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9848109520836027,
"lm_q1q2_score": 0.8484522660367423,
"lm_q2_score": 0.8615382112085969,
"openwebmath_perplexity": 2297.7335466269597,
"openwebmath_score": 0.8094127774238586,
"tags": null,
"url": "https://grocid.net/page/4/"
} |
it may look like the following: from z3 import * import binascii, string, itertools bits = 32 mask = 2**bits - 1 allowed_chars = string.printable def convert_to_hex(s): return ''.join([hex(ord(x))[2:].zfill(2) for x in s[::-1]]) def convert_to_string(h): return ''.join([chr(int(x, 16)) for x in list(map(''.join, zip(*[iter(hex(h)[2:])]*2)))[::-1]]) def rot(val, steps): return (val << (bits-steps)) | LShR(val, steps) def hash_foobar(input): eax = rot(input ^ 0x24f50094, (input ^ 0x24f50094) & 0xf) eax = rot(eax + 0x2219ab34, bits - (eax + 0x2219ab34 & 0xf)) eax = eax * 0x69a2c4fe return eax & mask def break_iteratively(hashdata, i): if i == 0: prev_block = 0 else: prev_block = hashdata[i-1] s = Solver() j = BitVec('current_block', bits) eax = rot(prev_block ^ hash_foobar(j), 7) s.add(eax == hashdata[i]) block_preimages = [] while s.check() == sat: sol = s.model() s.add(j != sol[j].as_long()) block_string = convert_to_string(sol[j].as_long()) if all(c in allowed_chars for c in block_string): block_preimages.append(block_string) return block_preimages known = '9513aaa552e32e2cad6233c4f13a728a5c5b8fc879febfa9cb39d71cf48815e10ef77664050388a3' # this the hash of the file data = list(map(''.join, zip(*[iter(known)]*8))) hashdata = [int(x, 16) for x in data] print '[+] Hash:', ''.join(data) print '[+] Found potential hashes:\n' for x in itertools.product(*[break_iteratively(hashdata, i) for i in range(10)]): print ' * ' + ''.join(x) This code is surprisingly fast, thanks to Z3, and runs in 0.3 seconds. Taking all possible collisions into consideration… [+] Hash: 9513aaa552e32e2cad6233c4f13a728a5c5b8fc879febfa9cb39d71cf48815e10ef77664050388a3 [+] Found potential hashes: * CTFEC0nstra1nts_m4keth_fl4g} * CTFEC0nstra1nts_m4keth_nl4g} * CTFEC0nstra1nws_m4keth_fl4g} * CTFEC0nstra1nws_m4keth_nl4g} * CTFEC0nstra9nts_m4keth_fl4g} * CTFEC0nstra9nts_m4keth_nl4g} * CTFEC0nstra9nws_m4keth_fl4g} * CTFEC0nstra9nws_m4keth_nl4g} * CTF{C0nstra1nts_m4keth_fl4g} * CTF{C0nstra1nts_m4keth_nl4g} | {
"domain": "grocid.net",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9848109520836027,
"lm_q1q2_score": 0.8484522660367423,
"lm_q2_score": 0.8615382112085969,
"openwebmath_perplexity": 2297.7335466269597,
"openwebmath_score": 0.8094127774238586,
"tags": null,
"url": "https://grocid.net/page/4/"
} |
* CTFEC0nstra9nws_m4keth_nl4g} * CTF{C0nstra1nts_m4keth_fl4g} * CTF{C0nstra1nts_m4keth_nl4g} * CTF{C0nstra1nws_m4keth_fl4g} * CTF{C0nstra1nws_m4keth_nl4g} * CTF{C0nstra9nts_m4keth_fl4g} * CTF{C0nstra9nts_m4keth_nl4g} * CTF{C0nstra9nws_m4keth_fl4g} * CTF{C0nstra9nws_m4keth_nl4g} …we finally conclude that the flag is the SHA-256 of C0nstra1nts_m4keth_fl4g. # BackdoorCTF16 – Baby Worth 200 points, this challenge was presented with the following: z3r0c00l has a safe repository of files. The filename is signed using z3r0c00l’s private key (using the PKCS-1 standard). Anyone willing to read a file, has to ask for a signature from z3r0c00l. But z3r0c00l is currently unavailable. Can you still access a file named “flag” on z3rc00l’s repository? nc hack.bckdr.in 9001 Let us take a look at the public key… 3072 bits and public exponent $e = 3$. Hmm… having a small exponent is usually not a good practice. First, I tried computing the roots to $x^3 - s \bmod n$, where $s$ is the signature and $n$ is the modulus, but then I realized that this was not the way to go. What if we use non-modular squareroot, plain old Babylonian style? After looking around, I also realized that this is Bleicherbacher’s $e = 3$ attack, which I probably should have known about. There is a lot of information about this attack (therefore, I will not describe it here) and, of course, lots of people have already written code for this. Being lazy/efficient, I rewrote a functional code into the the following: from libnum import * from gmpy2 import mpz, iroot, powmod, mul, t_mod import hashlib, binascii, rsa, os def get_bit(n, b): """ Returns the b-th rightmost bit of n """ return ((1 << b) & n) >> b def set_bit(n, b, x): """ Returns n with the b-th rightmost bit set to x """ if x == 0: return ~(1 << b) & n if x == 1: return (1 << b) | n def cube_root(n): return int(iroot(mpz(n), 3)[0]) snelhest = hashlib.sha256('flag') ASN1_blob = rsa.pkcs1.HASH_ASN1['SHA-256'] suffix = b'\x00' + ASN1_blob + snelhest.digest() | {
"domain": "grocid.net",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9848109520836027,
"lm_q1q2_score": 0.8484522660367423,
"lm_q2_score": 0.8615382112085969,
"openwebmath_perplexity": 2297.7335466269597,
"openwebmath_score": 0.8094127774238586,
"tags": null,
"url": "https://grocid.net/page/4/"
} |
ASN1_blob = rsa.pkcs1.HASH_ASN1['SHA-256'] suffix = b'\x00' + ASN1_blob + snelhest.digest() sig_suffix = 1 for b in range(len(suffix)*8): if get_bit(sig_suffix ** 3, b) != get_bit(s2n(suffix), b): sig_suffix = set_bit(sig_suffix, b, 1) while True: prefix = b'\x00\x01' + os.urandom(3072//8 - 2) sig_prefix = n2s(cube_root(s2n(prefix)))[:-len(suffix)] + b'\x00' * len(suffix) sig = sig_prefix[:-len(suffix)] + n2s(sig_suffix) if b'\x00' not in n2s(s2n(sig) ** 3)[:-len(suffix)]: break print hex(s2n(sig))[2:-1] Ok, so lets try it: Great! # Defcon CTF – b3s23 (partial?) The server runs a program (game of life) which has a $110 \times 110$ board with cells (bits). After a fixed number $n$ of iterations, the simulation stops and the program jumps to the first bit of the memory containing the board. We want to create an input which contains shellcode in this area after $n$ iterations. Obviously, we could choose any shellcode, and run game of life backwards. Cool, let us do that then! Uh-oh, inverting game of life is in fact a very hard problem… so it is not really feasible 😦 What to do, then? Game of life Game of life a cellular automata, found by Jon Conway, and is based on the following rules: 1. A cell is born if it has exactly 3 neighbours. Neighbors are defined as adjacent cells in vertical, horistontal and diagonal. 2. A cell dies if it has less than two or more than three neighbors. Stable code (still life) Still life consists of cell structures with repeating cycles having period 1. Here are the building blocks I used to craft the shellcode. Of course, the still life is invariant of rotation and mirroring. Shellcode So, I tried to find the shortest shellcode that would fit one line (110 bits). This one is 8 bytes. Great. 08048334 <main>: 8048334: 99 cltd 8048335: 6a 0b push$0xb | {
"domain": "grocid.net",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9848109520836027,
"lm_q1q2_score": 0.8484522660367423,
"lm_q2_score": 0.8615382112085969,
"openwebmath_perplexity": 2297.7335466269597,
"openwebmath_score": 0.8094127774238586,
"tags": null,
"url": "https://grocid.net/page/4/"
} |
8048337: 58 pop %eax
8048338: 60 pusha
8048339: 59 pop %ecx
804833a: cd 80 int \$0x80 | {
"domain": "grocid.net",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9848109520836027,
"lm_q1q2_score": 0.8484522660367423,
"lm_q2_score": 0.8615382112085969,
"openwebmath_perplexity": 2297.7335466269597,
"openwebmath_score": 0.8094127774238586,
"tags": null,
"url": "https://grocid.net/page/4/"
} |
In binary, this translates to:
000001101001100101101010000010110101100001100000010110011100110110000000
Ok, so we note that
110101 ... 01110
cannot be constructed by our building blocks (there most certainly exist such blocks, but I didn’t consider them). So, I use a padding trick. By inserting an operation which does nothing specific
10110011 00000000 mov bl,0x0
we are able to use the blocks given in previous section. This Python code gives the binary (still-life-solvable) sequence:
from pwn import *
binary_data = ''.join([bin(ord(opcode))[2:].zfill(8) for opcode in shellcode])
context(arch = 'i386', os = 'linux')
print disasm(shellcode)
print binary_data[0:110]
which is
0000011010011001101100110000000001101010000010110101100001100000010110011100110110000000
The following cellular automata is stable, and the first line contains our exploit:
As can be seen in the animation below, we have found a still-life shellcode.
When feeding it to the program, we find that it remains in memory after any number of iterations:
Nice! Unfortunately, the code did not give me a shell, but at least the intended code was executed. I had a lot of fun progressing this far 🙂
# TU CTF – Secure Auth
This was a 150 point challenge with the description:
We have set up this fancy automatic signing server!
We also uses RSA authentication, so it’s super secure!
nc 104.196.116.248 54321
Connecting to the service, we get the following
Obviously, we cannot feed the message get_your_hands_off_my_RSA! to the oracle. So, we will only receive signatures, but no way to verify them; this means we don’t know either the public modulus, nor the public exponent. But, of course, we could guess the public exponent… there are a few standard ones: $3, 17, 65537...$ | {
"domain": "grocid.net",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9848109520836027,
"lm_q1q2_score": 0.8484522660367423,
"lm_q2_score": 0.8615382112085969,
"openwebmath_perplexity": 2297.7335466269597,
"openwebmath_score": 0.8094127774238586,
"tags": null,
"url": "https://grocid.net/page/4/"
} |
First, I obtained the signatures for $3$ and $4$ from the provided service. Denote these $s_3, s_4$, respectively. We note that given a correct public exponent $e$, we may compute $s_3^e = 3 + k \cdot N$ and $s_4^e = 4 + l \cdot N$. Inevitably, $\textnormal{gcd}(s_3^e-3,s_4^e-4) = \textnormal{gcd}(k,l)\cdot N$. Hoping for $\textnormal{gcd}(k,l)$ to be small, we can use serveral pairs until we find one that works.
Trying all the listed (guessed) public exponents, we find that $e = 65537$ (this was performed surprisingly fast in Sage with my Intel hexacore). Hence, we have now determined the modulus
$\begin{array}{rl} N = & 24690625680063774371747714092931245796723840632401231916590850908498671935961736 \\ &33219586206053668802164006738610883420227518982289859959446363584099676102569045 \\ &62633701460161141560106197695689059405723178428951147009495321340395974754631827 \\ &95837468991755433866386124620786221838783092089725622611582198259472856998222335 \\ &23640841676931602657793593386155635808207524548748082853989358074360679350816769 \\ &05321318936256004057148201070503597448648411260389296384266138763684110173009876\\ &82339192115588614533886473808385041303878518137898225847735216970008990188644891 \\ &634667174415391598670430735870182014445537116749235017327.\end{array}$
Now, note that
libnum.strings.s2n('get_your_hands_off_my_RSA!') % 3 == 0
OK, so we may split this message $m$ into a product of two message factors: $m_1 = 3$ and $m_2 = 166151459290300546021127823915547539196280244544484032717734177$ and sign them. Then, we compute the final signature $s = m^d = (m_1 \cdot m_2)^d = m_1^d \cdot m_2^d = s_1 \cdot s_2 \bmod N$. Mhm, so what now?
Phew 🙂
# TU CTF – Hash’n’bake
This challenge, worth 200 points, exhibits a trivial (and, obviously, non-secure) hash function with the objective to find a keyed hash. The description:
A random goat from Boston hashed our password!
Can you find the full output?
The hash function is defined as: | {
"domain": "grocid.net",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9848109520836027,
"lm_q1q2_score": 0.8484522660367423,
"lm_q2_score": 0.8615382112085969,
"openwebmath_perplexity": 2297.7335466269597,
"openwebmath_score": 0.8094127774238586,
"tags": null,
"url": "https://grocid.net/page/4/"
} |
The hash function is defined as:
def to_bits(length, N):
return [int(i) for i in bin(N)[2:].zfill(length)]
def from_bits(N):
return int("".join(str(i) for i in N), 2)
CONST2 = to_bits(65, (2**64) + 0x1fe67c76d13735f9)
def hash_n_bake(mesg):
mesg += CONST
shift = 0
while shift < len(mesg) - 64:
if mesg[shift]:
for i in range(65):
mesg[shift + i] ^= CONST2[i]
shift += 1
return mesg[-64:]
def xor(x, y):
return [g ^ h for (g, h) in zip(x, y)]
The following computations will give the hash
PLAIN_1 = "goatscrt"
PLAIN_2 = "tu_ctf??"
def str_to_bits(s):
return [b for i in s for b in to_bits(8, ord(i))]
def bits_to_hex(b):
return hex(from_bits(b)).rstrip("L")
if __name__ == "__main__":
with open("key.txt") as f:
print PLAIN_1, "=>", bits_to_hex(hash_n_bake(xor(KEY, str_to_bits(PLAIN_1))))
print "TUCTF{" + bits_to_hex(hash_n_bake(xor(KEY, str_to_bits(PLAIN_2)))) + "}"
# Output
# goatscrt => 0xfaae6f053234c939
# TUCTF{****REDACTED****}
So, the problem is: we need to compute the hash without knowing the key (or brute forcing it). The first observation we make is that the hash function is a truncated affine function, i.e., $h(m) = f((m \cdot 2^{64} \oplus \texttt{CONST})\cdot \texttt{CONST}_2)$, with $f(a \oplus b) = f(a) \oplus f(b)$ . There is a quite simple relation emerging: $h(k \oplus m) = h(k) \oplus h(m) \oplus h(0)$ (note: $h(0)$ denotes empty input here). Using this relation, we can do the following. We know $h(k \oplus m_1)$ and $h(m_2)$ and want to determine $h(k \oplus m_2)$. Consider the following relation:
$\begin{array}{rl} h(k \oplus m_2) = & h(k) \oplus h(m_2) \oplus h(0) \\ = & h(k) \oplus h(m_1) \oplus h(0) \oplus h(m_1) \oplus h(m_2) \phantom{\bigg(} \\ = & h(k \oplus m_1) \oplus h(m_1) \oplus h(m_2). \end{array}$
All terms on the last line of the above equation are known. So, we can easily compute the hash, even without knowing the key. Ha-ha! Computing the above relation using Python can be done in the following manner: | {
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xor(xor(to_bits(64, 0xfaae6f053234c939), hash_n_bake(str_to_bits(PLAIN_1))),hash_n_bake(str_to_bits(PLAIN_2)))
This gives the flag TUCTF{0xf38d506b748fc67}. Sweet 🙂
# TU CTF – Pet Padding Inc.
A web challenge worth 150 points, with description
We believe a rouge whale stole some data from us and hid it on this website. Can you tell us what it stole?
http://104.196.60.112/
Visiting the site, we see that there is a cookie youCantDecryptThis. Alright… lets try to fiddle with it.
We run the following command
curl -v --cookie "youCantDecryptThis=aaaa" http://104.196.60.112/
and we observe that there is an error which is not present compared to
when running it with the correct cookie is set, i.e.,
curl -v --cookie "youCantDecryptThis=0KL1bnXgmJR0tGZ/E++cSDMV1ChIlhHyVGm36/k8UV/3rmgcXq/rLA==" http://104.196.60.112/
Clearly, this is a padding error (actually, there is an explicit padding error warning but it is not shown by curl). OK, so decryption can be done by a simple padding oracle attack. This attack is rather simple to implement (basically, use the relation $P_i = D_K(C_i) \oplus C_{i-1}$ and the definition of PCKS padding, see the wikipedia page for a better explanation), but I decided to use PadBuster. The following (modified example) code finds the decryption:
class PadBuster(PaddingOracle):
def __init__(self, **kwargs):
self.session = requests.Session()
self.wait = kwargs.get('wait', 2.0)
def oracle(self, data, **kwargs):
while 1:
try:
response = self.session.get('http://104.196.60.112',
stream=False, timeout=5, verify=False)
break
except (socket.error, requests.exceptions.RequestException):
logging.exception('Retrying request in %.2f seconds...', self.wait)
time.sleep(self.wait)
continue
self.history.append(response)
return
The decrypted flag we get is TUCTF{p4dding_bec4use_5ize_m4tt3rs}! | {
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# Eigenvalues of periodic lattice Laplacian?
Consider the graph given by taking a rectangular lattice with $m$ rows and $n$ columns and joining each vertex to its four nearest neighbors, where vertices on the boundary are connected periodically (for example, $(1,2)$ is connected to $(1,1),(1,3),(2,2)$ and $(m,2)$). Are the exact eigenvalues of its Laplacian matrix $L$ known?
I know how to handle the problem in the 1D case. Here except for the first and last rows, the matrix is tridiagonal, with its diagonal entries being $2$ and its superdiagonal and subdiagonal entries being $-1$. (Here I am using the positive semidefinite convention for the Laplacian, as usual in graph theory but reversed from the usual for PDE). Additionally there is a $-1$ in the top right and bottom left corners. This means that we have the circulant matrix corresponding to the vector $(2,-1,0,0,\dots,-1)$, where there are $n-3$ zeros. In this case as for any circulant matrix, the eigenvectors are the columns of the discrete Fourier transform and the eigenvalues can be read off by substitution.
This suggests that the columns of an appropriate 2D Fourier transform would be the eigenvectors of our matrix here...and in fact that is correct. To be specific, under the column major ordering of the vertices, the eigenvectors of the Laplacian matrix are the columns of $F_n \otimes F_m$ where $F_n$ and $F_m$ are the discrete Fourier matrices and $\otimes$ is the Kronecker product.
Is there an easy way to read off the actual eigenvalues from this representation of the eigenvectors? In other words, I can see that we have, for example
$$\lambda_i=\frac{\sum_{j=1}^{mn} L_{1j} (F_n \otimes F_m)_{ji}}{(F_n \otimes F_m)_{1i}}$$
but does this have some simple explicit formula? | {
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but does this have some simple explicit formula?
• Put the 1D analog in your question to see how much you can repeat the same process. – AHusain Jun 16 '16 at 21:37
• @AHusain Done. Does it help? – Ian Jun 16 '16 at 22:18
• So you have essentially answered your own question now. You have a guess for the eigenvectors and now you just check their eigenvalues. – AHusain Jun 16 '16 at 23:01
• @AHusain You are right, I have done most of the problem. But is there any chance of a simple explicit formula for the eigenvalues? The above is a formula in terms of the entries, but I am not especially comfortable with the Kronecker product so I am not so sure if there is a simple formula for the five relevant entries of $(F_n \otimes F_m)$ (for a given $i$). – Ian Jun 16 '16 at 23:07 | {
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Yes, I believe the eigenvalues are of the form $$4 - 2 \cos \left( \frac{2 \pi j}{n} \right) - 2 \cos \left( \frac{2 \pi k}{m} \right)$$ for $1 \leq j \leq n$, and $1 \leq k \leq m$.
Let's denote your graph as $G$ and let $C_n$ be the cylic graph on $n$ vertices. Note that we can write your graph as $G = C_n \ \displaystyle \square \ C_m$ where $\displaystyle \square$ denotes the Cartesian product of graphs. In general, for graphs $G$ and $H$ with Laplacian matrices $L_G$ and $L_H$, we have that $L_{G \square H} = L_G \otimes I + I \otimes L_H$ where $\otimes$ is the Kronecker product. All eigenvectors of $L_{G \square H}$ are of the form $v_G \otimes v_H$ where $v_G$ and $v_H$ are eigenvectors of $L_G$ and $L_H$, respecitvely. Moreover, the eigenvalue associated with $v_G \otimes v_H$ is $\lambda_G + \lambda_H$ where $\lambda_G$ and $\lambda_H$ are eigenvalues of $v_G$ and $v_H$ , respectively. Thus, knowing all the eigenvalues of $L_G$ and $L_H$ will exactly give us all the eigenvalues of $L_{G \square H}$. | {
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So to find the eigenvalues of $L_G$, we need only to find the eigenvalues of the Laplacian matrix of $C_n$. You can check that the Laplacian matrix of $C_n$ is a circulant matrix and that their eigenvalues are of a special form. In this case, using $\omega_j = \exp (\frac{2 \pi i j}{n})$, we have that the eigenvalues of $L_{C_n}$ are of the form, \begin{align} \lambda_j &= 2 - \omega_j - \omega_j^{n-1} \\ &= 2 - \exp \left(\frac{2 \pi i j}{n} \right) - \exp \left( \frac{2 \pi i j (n-1)}{n} \right) \\ &= 2 - \exp \left(\frac{2 \pi i j}{n} \right) - \exp \left( -\frac{2 \pi i j}{n} \right) \\ &= 2 - 2 \cos \left( \frac{2 \pi j}{n} \right) \end{align} Using this and the fact above, we have that the eigenvalues of $G$ are of the form $$4 - 2 \cos \left( \frac{2 \pi j}{n} \right) - 2 \cos \left( \frac{2 \pi k}{m} \right)$$ for $1 \leq j \leq n$, and $1 \leq k \leq m$, as desired. Please check my work as I could have easily made a mistake and please let me know if you'd like more explanation on any of the parts.
• Looks good. I think in your final simplification you made a simple mistake: you should have $\lambda_{j,k}(L_{G \square H})=\lambda_j(L_G)+\lambda_k(L_H)=4-\omega_n^j-\omega_n^{j(n-1)}-\omega_m^k-\omega_m^{k(m-1)}$, where $\omega_n=\exp(2 \pi i/n)$. Also these can be simplified to be expressed in terms of cosine: since $\omega_n^n=\omega_m^m=1$ you have $4-2\cos(2 \pi j/n)-2\cos(2 \pi k/m)$. – Ian Jun 17 '16 at 0:54
• Anyway, with this correction the result matches up with numerical tests including with a case of $m \neq n$. So once you make this modification I will accept. Thanks for your help. Also thanks for giving your help in a "generalizable" form (i.e. it is clear how to extend this to more than two dimensions). – Ian Jun 17 '16 at 0:59 | {
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• Sure thing - glad I could help! It looks like we both had the same answer for the eigenvalues but different definitions for $\omega_k$, which led to the confusion. I added a bit of working out the eigenvalue at the end, along with your nice simplification! – Chris Harshaw Jun 17 '16 at 1:16
• You still haven't adjusted the denominators to be $m$, which you do need to do in some places. (That is why I changed the notation, I needed the denominator and the exponent to both be variables.) – Ian Jun 17 '16 at 1:17 | {
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# Homework Help: How do you find the equation of a Cubic function given 5 points? (no zeros)?
1. Jan 8, 2012
### srizen
1. The problem statement, all variables and given/known data
what the questions asks is that i need to find the equation of a polynomial with these given points:
1,1
2,-3
3,5
4,37
5,105
i know that one way to solve is by creating 5 equations then solve for ax^3+bx^2+cx+d using the elimination/substitution method.
however is there another, much easier way of doing this question?
2. Relevant equations
ax^3+bx^2+cx+d
3. The attempt at a solution
1= a+b+c+d
-3=8a+4b+2c+d
5= 27a+9b+3c+d
37=64a+16b+4c+d
108=125a+25b+5c+d
fixed, yes, my mistake
Last edited: Jan 8, 2012
2. Jan 8, 2012
### eumyang
Three of the equations are wrong. They should be
1= a+b+c+d
-3=8a+4b+2c+d
5= 27a+9b+3c+d
37=64a+16b+4c+d
108=125a+25b+5c+d
Also, you don't need the last equation, because there are 4 unknowns.
As for other methods, there's the finite difference method, but I don't think it will help for this particular problem (because you already told us that this is a cubic).
Last edited: Jan 8, 2012
3. Jan 8, 2012
### srizen
fixed, i typed the equations too fast
4. Jan 8, 2012
### SammyS
Staff Emeritus
True, but I got the coefficients fairly quickly using a difference method.
Actually, after playing around a bit with this, I got the result with two different difference methods.
5. Jan 8, 2012
### srizen
what exactly is the difference method?
i ask because i've tried this question, and i kept getting it wrong.
6. Jan 8, 2012
7. Jan 8, 2012
### SammyS
Staff Emeritus
Make a table of differences | {
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6. Jan 8, 2012
7. Jan 8, 2012
### SammyS
Staff Emeritus
Make a table of differences
$$\begin{array}{|c|c|c|c|c|c|c|c|c|} \hline \quad i \quad & \quad x_i \quad & \quad f(x_i) \quad &\quad (\Delta^1)_i \quad &\quad (\Delta^2)_i \quad & \quad (\Delta^3)_i \quad & \quad (\Delta^4)_i \quad \\ \hline & & & & & & \\ 1 & 1 & 1 & -4 & 12 & 12 & 0 \\ & & & & & & \\ 2 & 2 & -3 & 8 & 24 & & -- \\ & & & & & & \\ 3 & 3 & 5 & & & -- & -- \\ & & & & & & \\ 4 & 4 & 37 & & -- & -- & -- \\ & & & & & & \\ 5 &5 & 105 & -- & -- & -- & -- \\ & & & & & & \\ \hline \end{array}$$
Where: $(\Delta^1)_i=f(x_{i+1})-f(x_{i})\,,$
$(\Delta^2)_i=(\Delta^1)_{i+1}-(\Delta^1)_i$
etc.
See if you can fill in the rest.
If f(x) is truly a cubic function then the Δ3 column will all be the same.
Fill out a similar Table for g(x) = x3 . The Δ3 column will all be 6's.
What do you suppose that means about the x3 coefficient of f(x) ?
8. Jan 9, 2012
### srizen
OMG! i love you forever, i had no idea this method existed, i already solved through almost an hour of writing matrices, with this i solved it in 3 minutes. thank you!
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook | {
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# Is it possible to find a non-zero solution of an ODE?
Is it possible to solve the 4th-order ode analytically?
op[r_] = (D[#, {r, 2}] - 1/r*D[#, r]) &;
DSolve[{op[r][op[r][f[r]]] == 0, f[b] == 0, f'[b] == 0, (f''[r] - 1/r*f'[r] - f[r]/(c - 1) /. r -> (1 + b)) == 0, (f'''[r] - 1/r*f''[r] + 1/r^2*f'[r] /. r -> (1 + b)) == 0}, f[r], r]
I am using v11 in which DSolve returns a trivial solution f[r] -> 0 only.
Any suggestion is welcome. Thank you!
For general values of $$b$$ and $$c$$, the trivial solution is indeed the only solution. However, there are special values of $$b$$ and $$c$$ which can yield non-trivial solutions.
To find these, we can start by telling Mathematica to find the general solution to the ODE:
soln = DSolve[{op[r][op[r][f[r]]] == 0}, f, r]
(* {f -> Function[{r}, (r^2 C[1])/2 - (r^2 C[2])/4 + (r^4 C[3])/4 + C[4] + 1/2 r^2 C[2] Log[r]]} *)
We can then look at the boundary conditions and see what they imply about the coefficients C[i]. We do this by storing the vanishing quantities in a list BCs; the actual equations are then BCs == 0.
BCs = { f[b],
f'[b],
(f''[r] - 1/r*f'[r] - f[r]/(c - 1) /. r -> (1 + b)),
(f'''[r] - 1/r*f''[r] + 1/r^2*f'[r] /. r -> (1 + b))};
quants = BCs /. First[soln]
The quantities in quants are four linear combinations of the coefficients C[1], C[2], C[3], and C[4], all of which must vanish. We can think of this set of simultaneous equations as the result of a matrix $$M$$ multiplying the vector $$\vec{v} = \{C_1, C_2, C_3, C_4\}$$. From basic results in linear algebra, we know that the only way for there to be a non-trivial solution for $$\vec{v}$$ is for the matrix $$M$$ to have a non-zero determinant. So we construct this matrix and take its determinant:
mat = Outer[Coefficient, quants, {C[1], C[2], C[3], C[4]}];
Simplify[Det[mat]]
(* -((b (1 + b) (-3 + 2 b + 4 c + 2 (1 + b)^2 Log[b] - 2 (1 + b)^2 Log[1 + b]))/(-1 + c)) *) | {
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(* -((b (1 + b) (-3 + 2 b + 4 c + 2 (1 + b)^2 Log[b] - 2 (1 + b)^2 Log[1 + b]))/(-1 + c)) *)
This implies that the trivial solution is the only solution to the ODE unless this last quantity (in terms of the constants $$b$$ and $$c$$) is zero, which occurs when $$b = 0$$, $$b = -1$$, or $$c = \frac{3 - 2b + 2 (1+b)^2 \ln( (1+b)/b )}{4}.$$
• thank you very much, but the question is actually not solved completely... I tried csoln = Solve[Det[mat] == 0, c] // Simplify and substituted the special c in the original system but the solution still includes one unknown constant. Please see DSolve[{op[r][op[r][f[r]]] == 0, f[b] == 0, f'[b] == 0, (f''[r] - 1/r*f'[r] - f[r]/(c - 1) /. {r -> (1 + b), csoln[[1, 1]]}) == 0, (f'''[r] - 1/r*f''[r] + 1/r^2*f'[r] /. r -> (1 + b)) == 0}, f[r], r], it also gives a warning _Unable to resolve some of the arbitrary constants _ Jan 20 at 9:20
• Yes, there will be unknown constants in the solution. They should correspond to the overall normalization of the solution; if $f(x)$ is a solution then so is $\alpha f(x)$ for any $\alpha$, because the equation is linear. Jan 20 at 12:13
• Prof. Seifert thanks a lot! Your comment is plausible. But with Det[mat]==0 we should have non-zero solution for C[i], however, when substituting the eigenvalue c back to mat, which represents vanishing quantities for the b.c.s, I got zero solution... Please try csoln = Solve[Det[mat] == 0, c] and LinearSolve[mat/.First[csoln], {0, 0, 0, 0}] Jan 21 at 2:41
A somewhat different way is as follows.
s = DSolve[{op[r][op[r][f[r]]] == 0, f[b] == 0, f'[b] == 0}, f, r]
{{f -> Function[{r}, (1/( 4 b^2))(-b^4 C[1] + 2 b^2 r^2 C[1] - r^4 C[1] + b^4 C[2] - b^2 r^2 C[2] - b^4 C[2] Log[b] - r^4 C[2] Log[b] + 2 b^2 r^2 C[2] Log[r])]}}
Resolve[Exists[{C[1], C[2]},Simplify[f'[1 + b]/(1 + b)^2 - f''[1 + b]/(1 + b) + f'''[1 + b] /.
s[[1]]]==0 && Simplify[-(f[1 + b]/(-1 + c)) - f'[1 + b]/(1 + b) + f''[1 + b] /.
s[[1]]]==0&& C[1]^2 + C[2]^2 != 0], Reals] | {
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b > 0 && c == 1/4 (3 - 2 b - 2 Log[b] - 4 b Log[b] - 2 b^2 Log[b] + 2 Log[1 + b] + 4 b Log[1 + b] + 2 b^2 Log[1 + b])
The main difference from Michael Seifert's answer consists in the use of quantifiers instead of linear algebra. This way is more automatical.
• One could also start from DSolve[{op[r][op[r][f[r]]] == 0, f, r]. Jan 16 at 17:12 | {
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Each positive number has two square roots. Adding and Subtracting Square Roots. Example 1: Add or subtract to simplify radical expression: $2 \sqrt{12} + \sqrt{27}$ Solution: Step 1: Simplify radicals Combining functions. Adding and Subtracting Like Radicals Simplify each expression. Simplify radicals. Improve your math knowledge with free questions in "Add and subtract radical expressions" and thousands of other math skills. Learn vocabulary, terms, and more with flashcards, games, and other study tools. Adding, Subtracting & Multiplying Radicals DRAFT. Break down the given radicals and simplify each term. Add or subtract the like radicals by adding or subtracting their coefficients. Simplify each radical completely before combining like terms. This calculator performs addition and subtraction of algebraic fractions. The steps in adding and subtracting Radical are: Step 1. Adding and Subtracting Radicals – Practice Problems Move your mouse over the "Answer" to reveal the answer or click on the "Complete Solution" link to reveal all of the steps required for adding and subtracting radicals. This gives mea total of five copies: That middle step, with the parentheses, shows the reasoning that justifies the final answer. Email. Learn vocabulary, terms, and more with flashcards, games, and other study tools. The number i is defined as the square root of -1, so that i squared equals -1. It includes four examples. Start studying Simplifying/Adding and Subtracting Radicals V2. Next What Are Radicals. Radical expressions may be combined by using addition or subtraction only if they are SIMILAR, that is, if they have the same radicand with the same index. 57% average accuracy. 900 seconds . If the indices or radicands are not the same, then you can not add or subtract the radicals. This video looks at adding and subtracting radical expressions (square roots). Subtraction of radicals follows the same set of rules and approaches as addition—the radicands and the indices must | {
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radicals follows the same set of rules and approaches as addition—the radicands and the indices must be the same for two (or more) radicals to be subtracted. Examples: 1. This game goes along with the game in the last section. You probably won't ever need to "show" this step, but it's what should be going through your mind. The amount of bacteria in an infection n hours after taking medication is (1/x 3) n. Write a simplified expression that represents the amount of bacteria in an infection 4 hours after taking medication. Which of the following is a like radical to 3x 5? Division of Radical Expressions with binomial divisor. SURVEY . Break down the given radicals and simplify each term. Identify the like radicals. 66 times. Q. Intro to combining functions. Any number that can be written in the form, Addition and subtraction of complex numbers. 3. In the three examples that follow, subtraction has been rewritten as addition of the opposite. Edit. If the index and radicand are exactly the same, then the radicals are similar and can be combined. How much longer is the side of a cube with a surface area of 180 square meters than a cube with the surface area of 120 square meters? Tags: Question 13 . This practice includes adding and subtracting radicals that have the same and different numbers under the radical. 2. This involves adding or subtracting only the coefficients; the radical part remains the same. Adding and subtracting rational expressions (not factored) Subtracting rational expressions. salexander10. • Sometimes it is necessary to simplify radicals first to find out if they can be added or subtracted. Multiplying functions. Start studying Adding and Subtracting Radicals. -3√75 - √27. Cube roots, fourth roots, and so on are denoted by using an index and a radical. If the divisor ( the denominator and write the result in standard form not have like radicals adding! Of the following is a like radical to 3 6x2 solve radical equations this! Not have the same, | {
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# Find all solutions, other than $2$ for $12x^3-23x^2-3x+2=0$
Find all solutions, other than $2$ for $12x^3-23x^2-3x+2=0$
I started off by taking out an $x$ and got $$x(12x^2-23x-3)+2=0$$
I do not know if this is the correct first step, if it is, then am I able to use the quadratic formula or complete the square to get the answers. Can anyone give me general hints. Please do not solve this for me in anyway. Just give me hints.
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That first step doesn't work very well. It's good if you can find something to factor out of the entire expression so that you end up with something like $AB=0$, because then you can break up the problem into "$A=0$ or $B=0$". But changing it to $AB+C=0$ does not usually simplify the problem. – Arturo Magidin Jul 20 '12 at 3:39
The step is technically not incorrect. But it is not at all useful. – André Nicolas Jul 20 '12 at 5:59
Since $12x^2-23x-3$ at $x=2$ is $-1$, then $x=2$ is a solution of the polynomial. Then divide the original polynomial by $x-2$ to get a quadratic which can be easily solved for the other roots 1/4 and -1/3. – i. m. soloveichik Jul 22 '12 at 14:33
HINT
The great bit here is to note that $2$ is a solution. By the factor theorem, we know that the linear factor $(x - 2)$ divides our polynomial.
So perhaps you should divide out $(x-2)$. You'll be left with a quadratic, which we know how to solve very quickly. | {
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So, I synthetically divide and then use the quadratic formula for the rest? – Austin Broussard Jul 20 '12 at 3:39
That sounds like a great strategy to me. Good luck! – mixedmath Jul 20 '12 at 3:40
I'll recomment with my answer and hopefully it'll be right! Thanks for the help! – Austin Broussard Jul 20 '12 at 3:40
@Austin: You don't need to ask. You could just plug in those values and check whether they satisfy your equation. – Javier Jul 20 '12 at 3:49
@Austin: Hey that's great! Although, unless my math is off, I happen to get $1/4$ and $-1/3$, as if one of us dropped a negative somewhere. – mixedmath Jul 20 '12 at 3:55
The preceding answers have provided you with sufficient information. However the following might help you with future endeavors:
How would you factor the polynomial if 2 wasn't given as a root?
There exists a useful theorem: the Rational Root Theorem, that helps factor polynomials (indirectly). It gurantees that any root of the polynomial will have a numerator that has $c$, the constant, as a factor and a denominator that has $a_n$, the leading coefficient, as a factor.
Therefore by investigating all possible conbinations of the factors of the constant and the factors of the leading coefficient you can eventually arrive at a valid root. The polynomial can then be further factored into a quadratic through synthetic division which can be factored (as you know) through the quadratic formula.
The interesting thing about this method is its large scope of applicability: it not only works on cubic polynomials but on a polynomial with any degree.
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The reciprocals of the roots $\rm\,r,s,1/2\:$ are roots of the reversed polynomial $\rm\:2\, x^3\! -\! 3\, x^2\! -\! 23\, x\! +\! 12.\:$ Thus by Vieta's Formulas $\rm\:r\!+\!s\!+\!1/2 = 3/2,\ rs/2 = -6,\:$ so $\rm\:r\!+\!s = 1,\ rs = -12,\:$ so $\rm\:r,s = \ldots$
Alternatively, by the Factor Theorem, $\rm\:f(2) = 0\:$ $\Rightarrow$ $\rm\:f(x)\:$ has $\rm\:x\!-\!2\:$ as a factor. Comparing coef's | {
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$$\rm (x-2)(a\, x^2 + b\, x + c)\ =\ 12\, x^3 - 23\,x^2 -3\,x + 2$$
$\rm\qquad x^3\:$ coef $\rm\:\Rightarrow\: a = 12$
$\rm\qquad x^0\:$ coef $\rm\:\Rightarrow\: -2\,c = 2\:\Rightarrow\: c = -1$
$\rm\qquad x^1\:$ coef $\rm\:\Rightarrow\: -3 = c-2b = -1-2b\:\Rightarrow\: b = 1$
So the quadratic factor is $\rm\: 12\,x^2 + x - 1,\:$ which can be solved by either the Quadratic Formula or Rational Root Test, or $\rm\:(c\,x\!-\!1)\,(d\,x\!+\!1) = 12\,x^2\!+\!x\!-\!1\:$ $\Rightarrow$ $\rm\:cd=12,\ c\!-\!d = 1,\:$ so $\rm\:c,d = \ldots$
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$c=4$ and $d=3$? – Austin Broussard Jul 20 '12 at 4:07
@Austin Indeed, they satisfy the equations. – Bill Dubuque Jul 20 '12 at 4:12 | {
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# Will this series with radical converge?
I'm trying to test the following series for convergence: $$\sum_{n=2}^{\infty}\frac{n}{\sqrt{n^5-n^3}}$$
I've progressed through several tests but am having trouble developing an intuition of how to approach a problem like this. I've checked the following cases so far:
• Divergence Test ($\lim a_n = 0$, so not helpful)
• Geometric Series Test (I can't find a straightforward way to find a common ratio)
• p-Series Test (it does not appear to be a p-Series)
• Limit Comparison and Comparison Tests (I can't find another series with which to prove convergence or divergence)
• Integral Test (I'm unable to integrate the expression)
Obviously I'm missing something here, but I'm just not sure which it is.
• Terms are roughly on the order of $\frac{n}{n^{5/2}} = \frac{1}{n^{3/2}}$ which should converge. How you get there is combination of comparison and $p$-series tests. Jan 23, 2018 at 14:28
• OK. It appears that I can just use the comparison test with something like $\frac{1}{x^\left(1.1\right)}$. Thank you! Jan 23, 2018 at 14:31
• Oftentimes, limit comparison is easier to use than comparison because you don't have to care if the "right" series dominates. Jan 23, 2018 at 14:33
• You should try to work it all out and answer your own question here. Would be good for you. Jan 23, 2018 at 14:33
• @Randall The trouble I have with limit comparison is that I have a hard time ending up with a constant. Seems I'm always getting infinity or infinity over infinity. Adding in the radical in the denominator makes it even easier to get lost down dead-ends. Jan 23, 2018 at 14:46
First of all, you seem to have misused some of the tests. For example, the divergence test only says that the series $\sum a_n$ diverges if $\lim a_n\neq 0$, which is not true here since $$\lim_{n\to\infty}\frac{n}{\sqrt{n^5-n^3}} = 0.$$
Second of all, you can make a comparison test: | {
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Second of all, you can make a comparison test:
$$\frac{n}{\sqrt{n^5-n^3}} = \frac{n}{\sqrt{n^5}\sqrt{1 - \frac{1}{n^2}}}=\frac{1}{\sqrt{n^3}}\cdot \frac{1}{\sqrt{1-\frac{1}{n^2}}}\leq 2\cdot \frac{1}{\sqrt{n^3}}$$
and of course, $$\sum_{n=1}^\infty \frac{2}{\sqrt{n^3}}$$ converges as it is two times a $p$-series.
• Thanks for the explanation and for pointing out my error with the divergence test. I'd just been checking if the sequence increases, but now I I see what I was doing wrong there. As for implementing the comparison test, I think I just need more practice so that I can "see" what you're seeing without having to try a dozen other strategies first. I appreciate your help. Jan 23, 2018 at 14:52
• @AlexJohnson In general, you just need to find the biggest factor of the numerator and denominator. In this case, the numerator is $n$, and the denominator, for large $n$, is dominated by $\sqrt{n^5}$. Since $\frac{5}{2} - 1> 1$, you have convergence.
– 5xum
Jan 23, 2018 at 14:54
You forgot the equivalence test: $$\sqrt{n^5-n^3\strut}\sim_\infty n^{5/2},\enspace\text{hence }\;\frac n{\sqrt{n^5-n^3\strut}}\sim_\infty \frac n{n^{5/2}}=\frac 1{n^{3/2}},$$ which is a convergent Riemann series.
Let $a_n$ be the $n$th term of your series and let $b_n= \frac{1}{n^{3/2}}$. Why is this my choice of $b_n$? See my first comment under your question.
Now, limit compare by $\lim_{n \to \infty} \frac{b_n}{a_n}$. Do the algebra and eventually get $$\frac{b_n}{a_n}= \frac{\sqrt{n^5-n^3}}{n^{5/2}} = \frac{\sqrt{n^5-n^3}}{\sqrt{n^5}} = \sqrt{\frac{n^5-n^3}{n^5}} = \sqrt{1-\frac{1}{n^2}}.$$ Now take the limit, and notice that it's easily $1$. Limit comparison now tells you that your series converges because $\sum_n \frac{1}{n^{3/2}}$ is a convergent $p$-series.
• I fudged a step or two but was able to get to the same place as you. This was all a tremendous help. Thank you again. Jan 23, 2018 at 15:00
• Excellent. You're well on your way. Jan 23, 2018 at 15:00
Simply note that | {
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Simply note that
$$\frac{n}{\sqrt{n^5-n^3}}\sim \frac{1}{n^\frac32}$$
thus $\sum_{n=2}^{\infty}\frac{n}{\sqrt{n^5-n^3}}$ converges by comparison with
$$\sum_{n=2}^{\infty}\frac{1}{n^\frac32}$$ | {
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Lemma 10.115.7. Let $R \to S$ be an injective finite type ring map. Assume $R$ is a domain. Then there exists an integer $d$ and a factorization
$R \to R[y_1, \ldots , y_ d] \to S' \to S$
by injective maps such that $S'$ is finite over $R[y_1, \ldots , y_ d]$ and such that $S'_ f \cong S_ f$ for some nonzero $f \in R$.
Proof. Pick $x_1, \ldots , x_ n \in S$ which generate $S$ over $R$. Let $K$ be the fraction field of $R$ and $S_ K = S \otimes _ R K$. By Lemma 10.115.4 we can find $y_1, \ldots , y_ d \in S$ such that $K[y_1, \ldots , y_ d] \to S_ K$ is a finite injective map. Note that $y_ i \in S$ because we may pick the $y_ j$ in the $\mathbf{Z}$-algebra generated by $x_1, \ldots , x_ n$. As a finite ring map is integral (see Lemma 10.36.3) we can find monic $P_ i \in K[y_1, \ldots , y_ d][T]$ such that $P_ i(x_ i) = 0$ in $S_ K$. Let $f \in R$ be a nonzero element such that $fP_ i \in R[y_1, \ldots , y_ d][T]$ for all $i$. Then $fP_ i(x_ i)$ maps to zero in $S_ K$. Hence after replacing $f$ by another nonzero element of $R$ we may also assume $fP_ i(x_ i)$ is zero in $S$. Set $x_ i' = fx_ i$ and let $S' \subset S$ be the $R$-subalgebra generated by $y_1, \ldots , y_ d$ and $x'_1, \ldots , x'_ n$. Note that $x'_ i$ is integral over $R[y_1, \ldots , y_ d]$ as we have $Q_ i(x_ i') = 0$ where $Q_ i = f^{\deg _ T(P_ i)}P_ i(T/f)$ which is a monic polynomial in $T$ with coefficients in $R[y_1, \ldots , y_ d]$ by our choice of $f$. Hence $R[y_1, \ldots , y_ d] \subset S'$ is finite by Lemma 10.36.5. Since $S' \subset S$ we have $S'_ f \subset S_ f$ (localization is exact). On the other hand, the elements $x_ i = x'_ i/f$ in $S'_ f$ generate $S_ f$ over $R_ f$ and hence $S'_ f \to S_ f$ is surjective. Whence $S'_ f \cong S_ f$ and we win. $\square$
Comment #4307 by Rankeya on
Does the proof use that S is a domain anywhere?
Comment #4468 by on
It looks like it doesn't. If you need this for something, then I'll change it.
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Comment #4481 by Rankeya on
I had occassion to use this lemma when $S$ is not a domain, so I would appreciate if you can change this. It allows me to refer to this lemma without having to say "the proof does not need $S$ to be a domain." Thanks!
Comment #4641 by Andy on
$R[y_1,\ldots,y_n]$ on the $4$-th line from the bottom should be a $R[y_1,\ldots,y_d]$
There are also:
• 2 comment(s) on Section 10.115: Noether normalization
In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar). | {
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# Sum of $n$ vectors in $(\mathbb Z/n)^k$
Let $n,k$ be positive integers. What is the smallest value of $N$ such that for any $N$ vectors (may be repeated) in $(\mathbb Z/(n))^k$, one can pick $n$ vectors whose sum is $0$?
My guess is $N=2^k(n-1)+1$. It is certainly sharp: one can pick our set to be $n-1$ copies of the set $(a_1,...,a_k)$, with each $a_i=0$ or $1$. The case $k=1$ is some math competition question (I think, but can't remember the exact reference). Does anyone know of some references? Thanks.
Thank you all! I wish I could accept all the answers, they are very helpful!
• In the title of the question, don't you mean the sum of $N$ (not $n$) vectors? Dec 5, 2009 at 18:18
• @José: I think the lowercase n is correct. It's slightly long, but a good title would be, "Minimum N st any size N set in (Z/n)^k has a size n subset summing to 0." Dec 5, 2009 at 18:29
• Indeed, the same example has no size n-1 subsets summing to 0 vector, which explains why the problem is about size n subsets. Dec 5, 2009 at 23:54
• I got a bit confused by the notations $\mathbb Z/n$ and $\mathbb Z/(n)$. Perhaps better: Given two strictly positive integers $n,k$, find the smallest integer $N\geq n$ such that every set of $N$ elements in $\mathbb Z^k$ contains a subset of $n$ elements with sum in $n\mathbb Z^k$. Apr 25 at 16:27
Your guess is correct for k=1 and 2, but when k is bigger, things get more complicated. For instance, when k=n=3, N=19. For a summary of some known results, see:
Elsholtz, C. Lower Bounds For Multidimensional Zero Sums. Combinatorica 24, 351–358 (2004).
The case k = 1 is the Erdős-Ginzburg-Ziv Theorem. Take a look at this Wikipedia article which has links to some surveys of the large literature of similar results. (The particular generalizations I'm aware of ask for a set of vectors summing to 0 whose size is the cardinality of the group, not its exponent.) | {
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The case k = 2 is the Kemnitz conjecture, as pointed out by Kristal, and as Ricky mentioned (and also as mentioned on the Wikipedia page I linked to) it was proved by Reiher in 2003.
There was some discussion of the case n=3 in sci.math around 1994. There is a card game called Set with an 81 card deck so that each card is naturally a point in $$(\mathbb Z/3)^4$$. Several cards are dealt out, and your task is to identify triples of cards called Sets which form a line, or equivalently, which add up to the 0 vector. A natural question is how many points you can deal out without the existence of a line. It's not too hard to construct 9 distinct points in affine 3-space, or 20 distinct points in affine 4-space over $$\mathbb Z/3$$ so that there is no line contained in the points, and these are the maximums. These correspond to $$N=19$$ for $$(n,k) = (3,3)$$ and $$N=41$$ for $$(n,k) = (3,4)$$, as in the reference Ricky Liu linked, by repeating each point twice.
The maximal configurations are highly symmetric. The 9 points in dimension 3 correspond to a nondegenerate conic, which is unique up to symmetry. The 20 points in dimension 4 actually correspond to a nondegenerate conic containing 10 points in projective 3-space viewed as lines passing through the origin in affine 4-space.
For example, there are 9 points in dimension 3 satisfying $$z=x^2 + y^2:$$ $$\{(0,0,0),(\pm1,0,1),(0,\pm1,1),(\pm1,\pm1,-1)\}$$ and this set contains no lines.
While I dont know the answer to your specific question it seems related to a well known problem where you only insist that the number of simmands is non zero. Much is known about it and there are interesting open problems. When n is a prime you need (n-1)k+1 vectors (and this is sharp). This is "Olson's theorem" and it can be proved by Chevaley's theorems on non zero solutions for polynomial equations when the number of variables exceeds the degree. Maybe similar techniques (at least for the n=prime case) will work for your problem.) | {
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The reference for Olson is: J. E. Olson, A combinatorial problem on nite abelian groups I, J. Number Theory 1 (1969), 8-10.
For the non prime power look at the paper by R. Meshulam: An uncertainty inequality and zero subsums. Discrete Math. 84 (1990), no. 2, 197--200.
For a general relevant method: N. Alon, Combinatorial Nullstellensatz. Combin. Probab. Comput. 8 (1999), no. 1-2, 7--29.
The case $$k=2$$ is handled (extremal sequences identified) in this paper of Gao and Goldinger, which was published in the journal Integers.
For $$k$$ greater than one the smallest value will be equal or less than $$n^{k-1}(2n-2)+1$$. To see this use the pigeonhole principle for the first $$k-1$$ coordinates there are only $$n^{k-1}$$ sets of possible values so one set of these coordinates must have $$2n-1$$ elements we can choose $$n$$ of these that have coordinate $$k$$ sum to zero by the Erdős–Ginzburg–Ziv theorem then since the first $$k-1$$ coordinates are fixed they will sum to zero as well and we have the desired set of vectors summing to zero.
For $$k=2$$ there is the Kemnitz conjecture that it is $$4n-3$$.
I now see this conjecture is proved. See:
Reiher, C. On Kemnitz’ conjecture concerning lattice-points in the plane. Ramanujan J 13, 333–337 (2007).
We can then apply the argument of the first paragraph and get for $$k=2$$ or more the smallest value must be less than or equal to $$n^{k-2}(4n-4)+1$$.
For the pattern to continue the next case would be for $$k=3$$ it would be $$8m-7$$ there is a counterexample in fact for all odd $$k$$ and $$n$$ greater than $$3$$ it is not true. The following paper was mentioned in another answer; the last sentence of the abstract has the general result.
Elsholtz, C. Lower Bounds For Multidimensional Zero Sums. Combinatorica 24, 351–358 (2004).
There is a factor of $$1.125$$ to the $$d/3$$ power so there is an exponent greater than two as a lower bound. | {
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You are currently browsing the tag archive for the ‘Wiki’ tag.
Wikipedia calls this a classic paradox of elementary probability theory.
Taken from Wikipedia.org:
## Card version
Suppose you have three cards:
• black card that is black on both sides,
• white card that is white on both sides, and
• mixed card that is black on one side and white on the other.
You put all of the cards in a hat, pull one out at random, and place it on a table. The side facing up is black. What are the odds that the other side is also black?
The answer is that the other side is black with probability 2/3. However, common intuition suggests a probability of 1/2 because across all the cards, there are 3 white, 3 black. However, many people forget to eliminate the possibility of the “white card” in this situation (i.e. the card they flipped CANNOT be the “white card” because a black side was turned over).
In a survey of 53 Psychology freshmen taking an introductory probability course, 35 incorrectly responded 1/2; only 3 students correctly responded 2/3.[1]
### Preliminaries
To solve the problem, either formally or informally, we must assign probabilities to the events of drawing each of the six faces of the three cards. These probabilities could conceivably be very different; perhaps the white card is larger than the black card, or the black side of the mixed card is heavier than the white side. The statement of the question does not explicitly address these concerns. The only constraints implied by the Kolmogorov axioms are that the probabilities are all non-negative, and they sum to 1.
The custom in problems when one literally pulls objects from a hat is to assume that all the drawing probabilities are equal. This forces the probability of drawing each side to be 1/6, and so the probability of drawing a given card is 1/3. In particular, the probability of drawing the double-white card is 1/3, and the probability of drawing a different card is 2/3. | {
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