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How to visualize $ \mathcal P \ ( \ \mathcal P \ ( \ \mathbb R ^2 \ ) ) $? So,$ \ \mathcal P \ ( \ \mathbb R ^2 \ ) $ , the power set of the set of all ordered pairs of real numbers, contains every imaginable (2D) function, black and white image and text as per its elements.
But how do we go a step further in understanding what the power set of that, $ \mathcal P \ ( \ \mathcal P \ ( \ \mathbb R ^2 \ ) ) $ is about?
|
For any set of points of $\mathbb R^2$ we can draw the set in 2-D by visualizing $\mathbb R^2$ as a sheet of paper and printing a pixel black if it belongs to the set or leaving it white if it doesn't belong (keep extra toner handy, as every page is infinite in size and there are uncountably many sheets). Print every set in this fashion and you'll have quite a large stack of paper; this is $\mathcal P(\mathbb R^2)$.
Now you're going to make portfolios by compiling every conceivable combination of drawings from your stack (don't ask where the binding goes). That is, your sets are 'substacks' of paper, or sets of sets of points. The resulting collection of portfolios is $\mathcal P\bigl(\mathcal P(\mathbb R^2)\bigr)$.
Now the publisher wants an anthology of all possible collections of portfolios: $\mathcal P\bigl[\mathcal P\bigl(\mathcal P(\mathbb R^2)\bigr)\bigr]$. And so on.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Evaluating a double integral from zero to infinity How do I evaluate this integral? I don't understand at which point the limit notation should set in? And my method yields $0$ in the end. The integral is:
$$
\int_0^{\infty} \int_0^{\infty} c\,x\,y\,e^{-(x+y)} \;\mathrm{d}y\;\mathrm{d}x
$$
|
You can separate $x$ and $y$ and integrate successively with respect to each variable since $$cxye^{-(x+y)}=c(xe^{-x})(ye^{-y})$$ Now $$\int_{0}^{\infty}xe^{-x}dx$$ is equal to the expectation of an exponentially distributed random variable with parameter $1$. Therefore it is also equal to $1$. Thus $$
\int_0^{\infty} \int_0^{\infty} cxye^{-(x+y)} \;\mathrm{d}y\mathrm{d}x
=c
\int_0^{\infty}ye^{-y} \left(\int_0^{\infty}xe^{-x} \;\mathrm{d}x\right)\mathrm{d}y=c\cdot1\cdot1=c
$$
|
{
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"url": "https://math.stackexchange.com/questions/1013524",
"timestamp": "2023-03-29T00:00:00",
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|
Proving of the multiplication theorem for Bernoulli polynomial How the expression below can be proven:
$$B_n(mx) = m^{n−1} \sum\limits_{k=0}^{m-1}B_n\left(x+\frac{k}{m}\right)$$
Where $B_n(x)$ is Bernoulli polynomial
I know it is already proved by Joseph Ludwig Raabe, but I don`t know how exactly.
|
By induction over $n$.
If $n=1$ we have from definition of $B_1(x)=x-1/2$,
\begin{eqnarray*}
B_1(mx) = mx - \frac{1}{2},
\end{eqnarray*}
On the other hand, the right hand side with $n=1$ is
\begin{eqnarray*}
\sum_{k=0}^{m-1} B_1 \left ( x +\frac{k}{m} \right )
&=& \sum_{k=0}^{m-1} \left ( x+ \frac{k}{m} - \frac{1}{2} \right ) \\
&=& mx + \frac{1}{m} \frac{m(m-1)}{2} - \frac{m}{2} \\
&=& mx - \frac{1}{2}
\end{eqnarray*}
Assume that for $n-1$ the equality holds.
Define $f_n(x) = B_n(mx) - m^{n-1} \sum_{k=0}^{m-1} B_n \left ( x + \frac{k}{m} \right )$.
Since $dB_n(x)/dx= n B_{n-1}(x)$,
The derivative of $f_n$ is given by
\begin{eqnarray*}
f'_n(x) &=& m n B_{n-1}(mx) - m^{n-1} \sum_{k=0}^{m-1} \frac{n}{m} B_{n-1} \left (
x + \frac{k}{m} \right ) \\
&=& m n f_{n-1}(x) \\
&=& 0 \quad \text { by the induction hypothesis on $n-1$ }.
\end{eqnarray*}
Hence $f_n(x)=c_n$ for $c_n$ constant. To find the constant
note that since $\int_0^1 B_n(x) dx = 0$, then
\begin{eqnarray*} \int_0^{1/m} f_n(x) &=&
\int_0^{1/m} \left [ B_n(mx) - m^{n-1} \sum_{k=0}^{m-1} B_n \left ( x + \frac{k}{m} \right )
\right ] dx \\
&=& \frac{1}{m} \int_0^1 B_n(y) dy -
m^{n-1} \sum_{k=0}^{m-1} \int_0^{1/m} B_n \left ( x + \frac{k}{m} \right ) dx \\
&=& 0 + m^{n-1} \sum_{k=0}^{m-1} \int_{k/m}^{(k+1)/m} B_n \left ( y \right ) dy \\
&=& 0 + m^{n-1} \sum_{k=0}^{m-1} \int_0^1 B_n \left ( y \right ) dy \\
&=& 0.
\end{eqnarray*}
So, since the integral of a constant in the interval $[0,1/m]$, $m > 0$, is 0,
that constant has to be zero. Hence $f_n(x)=0$ as desired.
|
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|
How do I compute the kernel of this map? How do I compute $\ker{(\mathbb{Z} \otimes A \longrightarrow \mathbb{Q} \otimes A)}$ where this map comes from the short exact sequence:
$0 \rightarrow Tor(\mathbb{Q}/\mathbb{Z}, A) \rightarrow \mathbb{Z} \otimes A \rightarrow \mathbb{Q} \otimes A \rightarrow \mathbb{Q}/\mathbb{Z} \otimes A \rightarrow 0$. Where $A$ is an abelian group.
|
The map from $\mathbb{Z} \otimes A \to \mathbb{Q} \otimes A$ is induced by the canonical inclusion, so is defined on generators by $n\otimes a \mapsto \frac{n}{1}\otimes a$ (for any $n \in \mathbb{Z}, a \in A$).
First, the torsion subgroup of $A$ is contained in the kernel.
Recall that there is a canonical identification $A \simeq \mathbb{Z} \otimes A$, via the map $a \mapsto 1 \otimes a$. Let $a \in A$ be torsion, so that there is an integer $n \neq 0$ with $na =0$. Identify $a$ with $1 \otimes a$ and then note that $1 \otimes a \mapsto \frac{1}{1}\otimes a = \frac{n}{n}\otimes a = \frac{1}{n}\otimes na = \frac{1}{n}\otimes 0 = 0$, so $1 \otimes a$ is in the kernel.
For the opposite inclusion, consider why this doesn't work if $a$ isn't torsion.
|
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Do the $n\times n$ matrices over a division ring $D$ form a free $D$-module? Let $D$ be a division ring. Then the set of $n\times n$ matrices over $D$ is free as a
$D$-module.
I think this is wrong because they are linearly dependent, right?
But what does the given of $D$ as a division ring changes ?
|
It's easy to see that $M_n(R)$ is isomorphic to the direct sum of $n^2$ copies of $R$ as an $R$ module for any ring $R$.
It's clearly free and has a basis consisting of the "unit matrices" $E_{ij}$ that are $1$ on the $i,j$ entry and zero elsewhere.
|
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Proving various relations are partial orders I am given these relations, in which I have to prove or disprove each and every one.
a. The relation $\trianglelefteq$ defined on ℕ by a $\trianglelefteq$ b if a ≤ b²
b. The relation $\preceq$ defined on ℤ by m $\preceq$ n if m ≤ n + 5.
c. The relation $\ll$ on the set of continuous functions defined by f(x) $\ll$ g(x) if $\int^1_0f(x)dx \leqslant \int^1_0g(x)dx$
So, for a, I have the following: To prove reflexivity, it follows that a $\leqslant$ a², and therefore, a $\trianglelefteq$ a, so $\trianglelefteq$ is reflexive. For transitivity, for all a, b, and c in ℕ, if a ≤ b², and b ≤ c², then a ≤ c², and therefore the relation is transitive.
For b, I have that for reflexivity, m ≤ m+5, so the relation is reflexive. For transitivity,for all m,n, and p in ℤ, if m ≤ n + 5, and n ≤ p + 5, then m ≤ p + 5, so the relation is transitive.
For c, I wrote for reflexivity that since $\int^1_0f(x)dx \leqslant \int^1_0f(x)dx$, then it the relation is reflexive. For transitivity, if $\int^1_0f(x)dx \leqslant \int^1_0g(x)dx$, and $\int^1_0g(x)dx \leqslant \int^1_0h(x)dx$, then $\int^1_0f(x)dx \leqslant \int^1_0h(x)dx$, therefore, the relation is transitive.
As you can see, I left out antisymmetry, because I am not sure how to prove that. I think that my reflexive and transitive proofs are correct, but if anyone could help me out with these problems by helping me with the antisymmetry proofs and verifying/correcting my reflexive and transitive proofs, that would be great.
|
The first proof is not correct. $a\leq b^2$ and $b \leq c^2$ does not imply $a \leq c^2$. Consider $a=15, b=4, c=3$.
The second proof is also incorrect because again transitivity fails. Consider $m=10,n=5,p=0$.
Your problem is that you state, for example, that "if m ≤ n + 5, and n ≤ p + 5, then m ≤ p + 5" but do not give any justification. And the statement is simply false. If you work more carefully, you would do something like this:
If $m \leq n+5$ and $n \leq p+5$, then substituting $p+5$ for $n$ in the first inequality, we see $m \leq (p+5)+5 = p + 10$.
You would then notice that this statement is not strong enough, and so you would start to look for a counterexample.
The last proof is correct, but you still need antisymmetry. To prove this property, you would assume $f \ll g$ and $g\ll f$ and then show $f = g$. However, it isn't true! Try $f(x) = 1-x$ and $g(x) = x$. Their integrals from $0$ to $1$ are equal, but the functions are not.
|
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Is it true that an equivalent 'absolute value' is an absolute value? I've a very basic question on absolute values on fields. If $K$ is a valued field with absolute value $|- |:K\to \mathbb R_{\geq0}$ then is the map $|-|':K\to \mathbb R_{\geq0}$ defined by $|x|'=|x|^r$ for some $r> 0$ also an absolute value? How do I show that the triangle inequality still holds for $|-|'$? Maybe this is not possible?
Many thanks.
|
The triangle inequality holds if $r\leq 1$, but may not hold if $r>1$. To see that it may not hold if $r>1$, you may use $\mathbb{Q}$ with the standard absolute value as an example. If $r<1$, then note that for $x$ and $y$ positive with $y\leq x$ we have that
for some $c$ with $x\leq c\leq x+y$
$$(x+y)^r-x^r=ryc^{r-1}\leq ryx^{r-1}\leq ry^r$$
by the mean value theorem, so
$$(x+y)^r\leq x^r+ry^r\leq x^r+y^r\mathrm{.}$$
Thus
$$|a+b|^r\leq (|a|+|b|)^r\leq |a|^r+|b|^r$$
|
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Show that $n! = O(n^{n})$ As the title says, how would you show that $n! = O(n^{n})$? I'm not really understanding how one "shows" the Big O notation of a function mathematically (at least when you're dealing with things that aren't polynomials).
|
Note that $1\cdot 2\cdot \ldots \cdot n\le n\cdot n\cdot \ldots \cdot n$, so we even have $n!\le n^n$.
|
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Linear Indepence Question involving an arbitrary vector Suppose $u, v, w$ are linearly independent in $V$ and $x \in V$.
Prove $u + x, v + x, w + x$ are always linearly independent.
I'm stuck on this problem because I don't know what to do with $x$. Any hints or help would be appreciated.
|
You're stuck because the result is false as it is stated now. Let ${\bf x} = -{\bf u}$, or the opposite of any of the other two vectors. If $\{{\bf u},{\bf v}, {\bf w}, {\bf x}\}$ is linearly independent, for example, it seems true, and follows straight from definition. Maybe there is some other hypothesis we can add to fix this.
|
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Determine whether there is linear independence or not Determine if the following set of vectors is linearly independent:
$$S = \{e^x, e^{2x}\}$$ of continuous functions.
So my problem isn't in not knowing what linearly independence is, it is setting up a linear combination of this form:
$$a_1e^x + a_2e^{2x} = ????$$
What do I set it equal to to determine the coefficients? For example if it was in terms of vectors it would be something like:
$$a_1V_1 + a_2V_2 = (0,0,0,)$$ if I was in $R^3$ for instance.
|
your matrix will be in a 2x2 form where row 1 is ae^x + be^2x = 0 and row 2 = e^x + 2e^2x =0, you find the determinant by factoring out e^xe^(2x) multiplied by your simplified matrix with (1,1) and (1,2) as rows 1 and 2. Hence determinant is e^xe^(2x)* (2-1)= e^xe^(2x)(1) which is not zero, meaning you have a set of linearly independent vectors.. Sorry it is my first time using this and I dont know how to write properly yet.
|
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Sum of 6 cards being multiple of 6 I pick 6 cards from a set of 13 (ace-king). If ace = 1 and jack,queen,king = 10 what is the probability of the sum of the cards being a multiple of 6?
Tried so far:
I split the numbers into sets with values:
6n, 6n+1, 6n+2, 6n+3
like so:
{6}{1,7}{2,8}{3,9}{4,10,j,q,k}{5}
and then grouped the combinations that added to a multiple of 6:
(5c4)(1c1)(2c1) + (2c2)(2c2)(2c2) + (5c4)(2c2) + (5c2)(1c1)(1c1)(2c1)(2c1)
/ (13c6)
= 10/1716
I am almost certain I am missing combinations but am having trouble finding out which.
|
Your combinations are $4\cdot 4+1\cdot 0+1\cdot 2, 2\cdot 1+2\cdot2+2\cdot3, 4 \cdot 4+2\cdot 1,+$ something that doesn't make sense because there are three places you choose from $1$ and only $0,5$ qualify. The left number is the number of cards of that value $\pmod 6$ . You could also have $3\cdot 4+1 \cdot 5+1 \cdot 1+1 \cdot 0, 3 \cdot 4+ 2 \cdot 2 + 1 \cdot 0$ and others
|
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How many distinct binary bit strings of length fifteen are there? I know this is a simple question but i'm not sure of which combinatorial selection equation to use.
How many distinct binary bit strings of length fifteen are there?
Using a simple example, would someone be able to explain the difference between
*
*orded with repetition
*orded without repetition
*unorded with repetition
*unorderd without repetition
I have had a look around but none of the examples are making much sense.
|
With binary strings of length $15$ you can't have without repetition. There are only two characters, so by the third character you will have repetition. If the string is not ordered you only care how many $0$s and $1$s there are. The number of $0$s can range from $0$ to $15$-how many is that? Then the number of $1$s is determined. For ordered strings, each bit has two choices, so there are $2^{15}$
|
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Are epimorphisms in the category of magmas surjective? The question says it all, but let me recall the definitions.
*
*A magma $(X, \cdot)$ is a set $X$ with a binary operation $\cdot \colon X \times X \to X$ (without any further assumptions like associativity).
*A morphism between two magmas $X$ and $Y$ is a map $f \colon X \to Y$ with $f(a \cdot b) = f(a) \cdot f(b)$ for all $a, b\in X$.
*A morphism $f$ is epi, respectively mono, if one can cancel it from the right, respectively left: $g \circ f = h \circ f \implies g=h$, respectively $f \circ g = f \circ h \implies g = h$.
Using free magmas with one generator shows that monomorphisms are injective.
The question is whether epimorphisms are always surjective (in many other categories this fails, e.g., in the category of monoids).
|
Yes. Suppose $f:X\to Y$ is not surjective, and let $Z$ be the complement of $f(X)$ in $Y$. Then the disjoint union $f(X)\cup Z\cup Z'$ of $f(X)$ and two copies of $Z$ can be given a magma structure so that the two obvious embeddings of $Y$ (as $f(X)\cup Z$ and $f(X)\cup Z'$) are inclusions of submagmas, and have the same composition with $f$. Products of elements of $Z$ with elements of $Z'$ can be defined arbitrarily.
|
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A function is convex and concave, show that it has the form $f(x)=ax+b$ A function is convex and concave, it is called affine function. That is the function:
$$f(tx+(1-t)y)=tf(x)+(1-t)f(y),\, \, t\in (0,1) $$
Force $y=0$(suppose $0$ is in the domain of $f(x)$), we obtain:
$$f(tx)-f(0)=tf(x)-tf(0)=t[f(x)-f(0)]$$
So
$$F(x):=f(x)-f(0)$$ is linear, but we have a constraint here: $t\in(0,1)$.
How can we expand this to arbitrary $t$, and assert it is really linear?
Thanks.
|
First show it's homogenous. That is, $F(ax) = aF(x)$ for any $a\in\mathbb{R},x\in\mathbb{R}^n$. If $a \in (0,1)$, we have:
$$
F(ax) = F(ax + (1-a)0) = aF(x) + (1-a)F(0)=aF(x)
$$
If $a\ge 1$, he have:
$$
F(x) = F((1/a)ax + (1-1/a)0) = (1/a)F(ax) + (1-1/a)F(0)=F(ax)/a
$$
If $a<0$, we use the above cases combined with the fact that $F(-x)=-F(x)$ since
$$
0 = F(0) = F((1/2)x +(1/2)(-x))=(1/2)F(x)+(1/2)F(-x).
$$
Lastly, to show linearity, we use homogeneity:
$$
\begin{align}
F(x+y) &= F((1/2)(2x) +(1/2)(2y))\\
&= (1/2)F(2x) +(1/2)F(2y)\\
&=(1/2)(2F(x)) +(1/2)(2F(y))\\
&=F(x) +F(y)
\end{align}
$$
|
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How can we use $(im(A))^\perp = \ker(A^T)$ to prove $rank(A)=rank(A^T)$? Why is it that these two statements are essentially equivalent? $(im(A))^\perp$ represents all vectors orthogonal to the $im(A)$. Yet I'm not sure what this being equal to $\ker(A^T)$ exactly means, nor how it relates to the rank of $A$ and $A^T$. I feel like some of my confusion may be stemming from my shoddy understanding of $A^T$.
|
There are a few things we need to prove the equivalence here:
*
*the rank of a matrix is the dimension of its image
$\DeclareMathOperator{\rank}{rank} \DeclareMathOperator{\im}{im}
\DeclareMathOperator{\null}{null}$
*The rank nullity theorem: $\rank(A) + \null(A) = \rank(A^T) + \null(A^T) = n$, where $\null(A)$ means $\dim \ker(A)$
*for any subspace $U$, $\dim(U) + \dim(U^\perp) = n$
From there, we have
$$
\rank(A^T) = \dim(\im(A^T)) = \dim(\ker(A)^\perp) =\\
n - \dim(\ker(A)) = n - \null(A) =\\
n- (n - \rank(A)) = \\
\rank(A)
$$
|
{
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When does a ring map $R\to S$ produce a group epimorphism $GL_n(R)\to GL_n(S)$? Let $R$ and $S$ be rings with $1$ (not necessarily commutative) and $f:R\to S$ a ring homomorphism preserving $1$. Let $\bar{f}$ be the ring map $M_n(R)\to M_n(S)$ given by $f$ acting on the matrix elements.
My question is, what is the most general condition on $f$ for $\bar{f}$ to be a surjective group homomorphism $GL_n(R)\twoheadrightarrow GL_n(S)$? I think this is true if $f$ is an isomorphism but I can't be sure.
Many thanks for your help.
|
Here are some standard examples: Consider $a$, $b$ integers $>0$ so that $b \mid a \ $. We have the canonical surjective morphism of rings
$$\mathbb{Z}/(a) \to \mathbb{Z}/(b)$$
and the induced morphism of groups
$$GL_n(\mathbb{Z}/(a)) \to GL_n(\mathbb{Z}/(b))$$
is surjective.
Even for $n=1$ this is a good exercise.
The general case follows from the surjectivity of
$$SL_n(\mathbb{Z}) \to SL_n(\mathbb{Z}/(b))$$
and the $GL_1 (\cdot )$ case.
Also see http://en.wikipedia.org/wiki/Approximation_in_algebraic_groups
|
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Show that $(A\setminus B) \cup (A\setminus C) = B \Leftrightarrow A=B \wedge (B \cap C) = \emptyset$ I believe there are 3 parts to this.
1) $(A\setminus B) \cup (A\setminus C) = B \Rightarrow A=B $
2) $(A\setminus B) \cup (A\setminus C) = B \Rightarrow (B \cap C) = \emptyset$
3) $A=B \wedge (B \cap C) = \emptyset \Rightarrow (A\setminus B) \cup (A\setminus C) = B$
I can do the the parts labelled 1 and 3 but cannot show part 2. Anyone who can explain how you do the 2nd part ie show $(A\setminus B) \cup (A\setminus C) = B \Rightarrow (B \cap C) = \emptyset$ ?
|
Hint: Let $(A-B)\cup (A-C)=B$ hold true and assume that there is $x \in B \cap C$. This means in particular that $x\in B$ and $x \in C$. Hence $$x \notin A-B$$ and for the same reason $$x\notin A-C$$ Therefore $$x\notin (A-B)\cup (A-C)$$ but on the other hand $x\in B$ which a contradiction to the assumption that $$(A-B)\cup (A-C)=B$$
|
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|
Linear Algebra quadratic forms diagonalization I have a question that reads:
Diagonalize the quadratic form $A(x,x) = 2x^2 - 1/2 y^2 -2xy - 4xz$ by completing the squares, and find the change of basis matrix and the new basis in which A will be diagonalized.
HINT: The change of basis matrix is the inverse of the change of coordinates matrix.
Can someone give me the definition of "diagonalize" in this context and walk me through this problem? I would really prefer to understand it.
|
Note: My answer below is more than a bit disorganized; if any one cares to clean it up a bit, I welcome the effort.
It seems that diagonalize in this context would mean something like writing
$$
2x_1^2 - (1/2) x_1^2 -2x_1x_2 - 4x_1x_3 =
A(x,x) = x^TU^T DUx = (Ux)^TD(Ux)
$$
for some diagonal matrix $D$ and invertible matrix $U$, where $x = (x_1,x_2,x_3)^T$.
In particular, defining $y = Ux$, we should have
$$
A(x,x) = y^TDy = d_1 y_1^2 + d_2 y_2^2 + d_3 y_3 ^2
$$
More specifically, we have
$$
2x^2 - (1/2)y^2 -2xy - 4xz = \\
(1/2)(y^2 - 4xy + 4x^2) - 4xz =\\
(1/2)(y - 2x)^2 - 4xz =\\
(1/2)(y - 2x)^2 + [(x - z)^2 - (x+z)^2] =\\
(1/2)(y - 2x)^2 + (x - z)^2 - (x+z)^2
$$
Now, let $y = (y_1,y_2,y_3)^T$ be a coordinate vector such that
$$
\pmatrix{y_1\\y_2\\y_3} =
\pmatrix{x_2 - 2x_1\\x_1 - x_3\\x_1 + x_3}
$$
This can be written as $y = Mx$, where
$$
M = \pmatrix{-2&1&0\\1&0&-1\\1&0&1}
$$
From before, we have
$$
A(x,x) = 2x_1^2 - (1/2)x_2^2 -2x_1x_2 - 4x_1x_3 =
(1/2)(y_1)^2 + (y_2)^2 - (y_3)^2
$$
We now have $A(x,x) = D(Mx,Mx)$ where
$$
D(y,y) = (1/2)(y_1)^2 + (y_2)^2 - (y_3)^2
$$
|
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|
System of equations, limit points This is a worked out example in my book, but I am having a little trouble understanding it:
Consider the system of equations:
$$x'=y+x(1-x^2-y^2)$$
$$y'=-x+y(1-x^2-y^2)$$
The orbits and limit sets of this example can be easily determined by using polar coordinates. (My question: what is the motivation for that thinking? What should clue me in to thinking that I should use polar coordinates?)
The polar coordinate satisfies $r^2=x^2+y^2$ so by differentiating with respect to t and using the differential equations we get:
$r\cdot r'=x\cdot x'+y\cdot y'$ (I am unclear about how the book even got this first equation from $r^2=x^2+y^2$)
$=x\cdot y+x^2(1-r^2)-x\cdot y+y^2(1-r^2)$ Substitute in $x'$ and $y'$ and then multiple out and replace with $r$, I get this step
$=r^2(1-r^2)$ cancel terms, I get this step too
$r'=r(1-r)$
similarly, the angle variable $\theta$ satisfies $\tan\theta=\frac yx$, so the derivative with respect to $t$ yields $\sec^2(\theta)\theta'=x^{-2}[x^2+xy(1-r^2)-y^2-xy(1-r^2)]=-\frac{r^2}{x^2}$ so $\theta=1$
Thus the solution goes clockwise around the origin at unit angular speed.
I don't understand the $\theta$ step at all or how they reached the conclusion of clockwise around the origin with unit angular speed..
But then it just jumps to saying "the origin is a fixed point, so α(0)=ω(0)={0} but I have no idea how they reached this conclusion..
|
When you see an $x^2 + y^2$ in a problem that is probably going to be tractable, one good first thing to try is to see how the problem looks transformed to polar coordinates. (Even in a problem that comes about in real life, where you don't know the solution will be possible to obtain, this is a good first shot.)
When $r^2 = x^2 + y^2$ it is valid to take the derivative of each side with respect to $t$.
For example,
$${d \over dt} (r^2) = 2r {dr \over dt} = 2rr'
$$
The book got the equation you present by doing this on both sides of the equation, and dividing by 2.
I see another answer and that you are unclear on how $\tan \theta$ comes in. That is easy: In polar coordinates, $y/x = \tan \theta$. Draw yourself a right triangle with one point at the origin, the right angle on the X axis and the third point at $(x,y)$ and ask yourself what $\tan \theta$ would be in that picture.
By the way, the solution that approaches $r=1$ is in fact a stable orbit, since if $r = 1+ \epsilon$, $r' = -\epsilon r$ so the solution will approach $r=1$ in an exponential fachion.
Then you have seen how to get from that to
$$
r' = r(1-r) $$
The first thing to notice about that equation is that considering $r$ alone (without $\theta$) it will have fixed points wherever $r' = 0$, that is, at $r=0$ and at $r=1$. But then $r=0$ is a fixed point of the whole system since at $r=0$, $\theta$ is irrelevant.
However, that fixed point is an unstable fixed point, since near the origin with $r(t=0) = \epsilon$, the solution behaves like $r = \epsilon e^t$, exploding away from the origin.
|
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|
Probability of coins and boys example is unsolvable I have a couple of problems that are giving a headache.
P1)
A coin is tossed three times. Let $A = \{\text{Three head occurs}\}$, $B = \{\text{At least one head occurs}\}$
Find $P(A\cup B)$.
Now $P(A) = 1/8$ , $P(B) = 1-1/8 = 7/8$ so
$$P(A \cup B) = P(A)+P(B)-P(A\cap B) = 1/8+7/8 - (1/8)\cdot(7/8)=57/64$$
However, the textbook answer is 7/8. Where am i messing up?
P2)
A class has $10$ boys and $4$ girls. If $3$ students are selected at random, what is the probability that all will be boys?
I'm totally clueless with these problems.
|
You calculated $P(A)$ and $P(B)$ correctly. But you cannot in general calculate $P(A\cup B)$ as $P(A) + P(B)$, because there might be events where both $A$ and $B$ have occurred, and $P(A) + P(B) $ counts those events twice, once as part of $P(A)$ and once as part of $P(B)$.
Suppose, for example, that the cafeteria serves cake on four days of each week, and serves soup on four days of each week. You cannot conclude from this that the cafeteria serves cake or soup on eight days of each week; that is absurd, because weeks have only seven days. To get the right answer you also have to know how many days the cafeteria serves cake and soup.
In your question you have $A = \text{three heads}$ and $B = \text{at least one head}$. But in this case event $B$ contains event $A$; every instance of event $A$ is also an instance of event $B$, because you cannot have three heads without also having at least one head. So in this case $P(A\cup B) = P(B) = \frac78$, which matches the answer in your book.
In this case the problem was easy, because event $A$ was a subset of event $B$. In general it may not be so simple. A technique that always works is to calculate $$P(A \cup B) = P(A) + P(B) - P(A\cap B)\tag{$\star$}$$ where the $P(A) + P(B)$ counts the overlapping events twice, and then subtracting $P(A\cap B)$ subtracts the overlapping events once; then the overlapping events are counted only once in total. If you apply that formula in this case, you have:
$$\begin{array}{clc}
P(A) & \text{three heads} & \frac18 \\
P(B) & \text{at least one head} & \frac78\\
P(A\cap B) & \text{three heads }\textbf{and}\text{ at least one head} & \frac18
\end{array}
$$
and using $(\star)$, you get $\frac 18+\frac78 - \frac18 = \frac78$, which again matches the answer in your book.
Note that you cannot in general calculate $P(A\cap B) $ as $P(A)\cdot P(B)$; that only applies when $A$ and $B$ are independent events.
|
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|
convergence of the derivatives I am trying to solve the question:
Let $u_n$ a sequence converging uniformly to $u$ where $u_n\in C^3(\Omega)$ for each $n$ and $\Omega$ is a subset limited of $\mathbb{R^n}$. Suppose $u_n=0$ on
$\partial\Omega$. To show that
$$
\displaystyle\lim_{n\rightarrow0}\int_{\Omega}|\nabla u_n|^2=\int_{\Omega}|\nabla u|^2.
$$
|
You can suppose, without loss of generality, that $u \equiv 0$ (if not, take $v_n = u_n - u$, and prove the statement for $v_n$).
Then, you want to show that $\lim_{n\to\infty}\int_\Omega |\nabla u_n|^2 = 0$.
Use integration by parts:
$$\int_\Omega |\nabla u_n|^2 = \int_{\partial \Omega} u_n (\nabla u_n \cdot \hat n) - \int_\Omega u_n \Delta u_n$$
Now use the facts that $u_n = 0$ on $\partial\Omega$, that $u_n \to 0$ uniformly, and that $u_n \in C^3$.
|
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|
Determine the Number of Integer Solutions $x_1 + x_2 + x_3 + x_4 = 32$ with restrictions The Question
My Problem
Part a is straight forward, just $C(35,32)$. I'm having a little difficulty with the restrictions and understanding what they mean. $x_1 > 0$ means we shouldn't have any solutions of the form $(0 + 32)$? If that's the case how would I factor that into my answer? I'm thinking taking the answer to part a and subtracting all the solutions which contain $32+0$ but I don't really know how to count those.
|
For part (b), $x_i > 0$ for all $i$, let $y_i = x_i - 1$. Then the problem is equivalent to problem (a), only with a sum of 28 instead of 32 -- the answer is $C(31, 28) == C(31,3)$.
For part (c) let $y_i = x_i-5$ for $i\in \left\{ 1,2 \right\} $ and $y_i = x_i-7 $ otherwise. Then you have problem (a) again, with a sum of 8; the answer is $C(7,3)$.
For part (d) the same reasoning gives $C(3,3) = 1$.
For part (e) let $y_i = x_i +2$ to get $C(39,3)$.
For part (f), you can start with the solution to (b), and subtract the cases where $x_4 > 25$. To get the latter, do $y_4 = x_4 -25$, and find an answer of $C(10,3)$ so the answer to part (f) is $C(35,3) - C(10,3)$.
|
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|
Proving any odd number is a factor of $2^n -1$ for some $n$ I'm struggling with a proof of the following. I feel like it should be a one-liner or something simple but I'm just not grasping the idea:
Suppose that m is an odd natural number. Prove that there is a natural number $n$ such that $m$ divides $2^n -1$
Any help would be much appreciated, thanks.
|
$$(2k+1) \mid 2^{\phi(2k+1)} -1 $$
by http://en.wikipedia.org/wiki/Euler%27s_theorem
|
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|
implications, equivalence, disjunction I might not be very clear with this but i hope someone gets it
Prove that $f : X→Y$ is surjective then and only then when $g_1, g_2$ which $Y → Z$ we have $g_1 \circ f= g_2 \circ f \Rightarrow g_1 = g_2$
it would be very helpful for me if someone could break this down cause i don't really have any clue
|
*
*Assume $f$ is surjective, and that $g_1 \circ f = g_2 \circ f$. Want to show $g_1 = g_2$
Assume $g_1 \neq g_2$. Then, there is some $y \in Y$ so that $g_1(y) = z_1 \neq z_2 = g_2(y)$. Since $f$ is surjective, there is an $x \in X$ so that $f(x) = y$. What does this tell you about $g_1(f(x))$ and $g_2(f(x))$?
*
*Assume $g_1 \circ f = g_2 \circ f$ implies $g_1 = g_2$ for any $g_1,g_2$. Want to show $f$ is surjective.
Assume $f$ is not surjective. Let $A = \textrm{Im}(f) = \{ f(x) : x \in X\} \subset Y$. Because $f$ is not surjective, $Y \backslash A$ is non-empty. Choose $g_1,g_2$ so that $g_1(y) = g_2(y)$ for all $y \in A$, but $g_1 \neq g_2$ for some $y \in Y \backslash A \neq \emptyset$. Then $g_1 \neq g_2$, but $g_1\circ f = g_2 \circ f$, and you have a contradiction.
(Implicit in this is that $Z$ consists of more than one point, otherwise such $g_1,g_2$ cannot be chosen)
|
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Inclusion-exclusion principle (?) in a counting problem A group of pre-school children is drawing pictures ( one child is making one picture ) using 12-colours pencil set. Given that
(i) each pupil employed 5 or more different colours to make his drawing;
(ii) there was no identical combination of colours in the different drawings;
(iii) the same colour appeared in no more than 20 drawings,
find the maximum number of children who have taken part in this drawing activity.
( As each child can be identified with his/her unique combination of colours, the number of children can not exceed
C(5,12) + C(6,12) + C(7,12) +...+ C(12,12)
But how to NARROW it using the condition (iii) ? )
|
HINT: Each color appears in at most $20$ drawings, and there are $12$ colors, so there are at most $12\cdot20=240$ drawings. However, each drawing uses at least $5$ colors, so ... ? See if you can finish it from here; I’ve left the conclusion in the spoiler-protected block below.
So each drawing is counted at least $5$ times, and there are therefore at most $\frac{240}5=48$ drawings (and hence at most $48$ children).
|
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Show $n^{\frac{1}{n}}$ is decreasing for $n \ge 3$ How to show $\displaystyle n^{\frac{1}{n}}$ is decreasing for $n \ge 3$ ?
|
Late answer but $n>e$ so $$n^{1/n} > e^{1/n}>1 + 1/n = \frac{n+1}{n} \implies n^{1+1/n}>n+1 \implies n^{1/n} > (n+1)^{1/n+1}$$
|
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A quick question on the subring generated by a finite set Let $R$ be a commutative unital ring and let $r_1,\ldots,r_n \in R$.
Let $S$ be the unital subring of $R$ generated by $r_1,\ldots,r_n$.
Let $\varphi:\mathbb{Z}[X_1,\ldots,X_n]\to R$ be the unique homomorphism sending $1\mapsto 1_R$ and each $X_i\mapsto r_i$.
Am I right in thinking that $S=\text{im}(\varphi)$?
Many thanks!
|
Yes. It is clear that $\text{im}(\phi) \subseteq S$. On the other hand, $\text{im}(\phi)$ is a subring of $R$ that contains all the $r_i$'s, and since $S$by definition is the smallest subring of $R$ containing the $r_i$'s, we must have $S \subseteq \text{im}(\phi)$. Hence, $S = \text{im}(\phi)$.
|
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limits question with radicals, rationalizing Find the limit value
Here's what I did (Above)
I think I can rationalize the numerator to solve it,
but I'm having trouble rationalizing numerator, when I'm usually rationalizing the denominator.
How do I rationalize the numerator? (If I'm on the right track for solution)
|
$$\begin{align}\lim_{x\to0}\frac{\sqrt{x+2}-\sqrt2}{x}&=
\lim_{x\to0}\frac{\sqrt{x+2}-\sqrt2}{x}\cdot\frac{\sqrt{x+2}+\sqrt2}{\sqrt{x+2}+\sqrt2}\\
&=\lim_{x\to0}\frac{x+2-2}{x(\sqrt{x+2}+\sqrt2)}\\
&=\lim_{x\to0}\frac{x}{x(\sqrt{x+2}+\sqrt2)}\\
&=\lim_{x\to0}\frac{1}{\sqrt{x+2}+\sqrt2}\\
&=\frac{1}{2\sqrt2}\\
\end{align}$$
|
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complex eigenvalues and invariant spaces I am currently reading Guillemin and Pollack's Differential Topology, and the following claim is made without proof:
Given a linear isomorphism $E: \mathbb{R}^k \to \mathbb{R}^k$, with $k>2$ and such that $E$ can be represented by a matrix with real entries, $E$ has a one- or two-dimensional invariant space.
I understand that the Fundamental Theorem of Algebra implies that $E$ has at least one real or complex eigenvalue; if it is real, then $E$ clearly has a one-dimensional fixed space. If it is complex, however, I don't see how there needs to be a two-dimensional invariant space.
If $E$ has complex eigenvalue $a+bi$, then $a-bi$ must also be an eigenvalue (as $E$ contains real entries). These eigenvalues correspond to eigenvectors $v_1$ and $v_2$. I assume that the subspace spanned by $v_1$ and $v_2$ is the desired invariant space, but can't figure out how to prove it. Any help would be most appreciated.
|
I think here is another way. A has eigenvalue $\lambda$ and eigenvector $\textbf{v}\in \mathbb{C}^n$. We can say $\lambda = a+bi$ and $v_j=x_j+iy_j \text{ for }j=1,2,...,n\Rightarrow \textbf{v}={\textbf{x}}+i\textbf{y} \text{ where }\textbf{x,y}\in \mathbb{R}^n$ . So now we have: $$(1)\quad A\textbf{v}=\lambda \textbf{v}=\lambda (\textbf{x}+\textbf{y}i)=(a+bi)(\textbf{x}+\textbf{y}i)=(a\textbf{x}-b\textbf{y})+i(b\textbf{x}+a\textbf{y})\\$$
But now you also know that $$(2) \quad A\textbf{v}=A(\textbf{x}+\textbf{y}i)=A\textbf{x}+A\textbf{y}i.\\$$
So I can set (1) and (2) equal:
$$A\textbf{x}+A\textbf{y}i=(a\textbf{x}-b\textbf{y})+i(b\textbf{x}+a\textbf{y})$$
But now we can set the real and imaginary parts equal. We have:
$$A\textbf{x}=(a\textbf{x}-b\textbf{y}) \text{ and } A\textbf{y}=(b\textbf{x}+a\textbf{y}), \text{ where }A\textbf{x}, A\textbf{y} \in \mathbb{R}^n$$
So the invariant subspace is the plane spanned by $\textbf{x}$ and $\textbf{y}$ and is two dimensional.
Could you help me understand the connection between my x and y and your $w_1$ and $w_2$. I think my $2x=w_1$ and $2y=w_2$ so the span is the same, but then when we multiply by A the stuff is different? How?
|
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|
Study the character of the series $\sum_{n=1}^{\infty} \left|\frac{1}{n^{2z}}\right|^2$ Discuss the character of the series
$$\sum_{n=1}^{\infty} \left|\frac{1}{n^{2z}}\right|^2$$
where $z\in \mathbb C$ and $|z|=\frac{1}{4}$.
Any suggestions please?
Thank you very much
|
Note that
$$
\left|n^{-2z}\right|^2=\left|n^{-2\Re (z)-2 i \Im (z)}\right|^2=\left(n^{-2 \Re(z)}\right)^2=n^{-4\Re(z)}
$$
for $n\ge 1$, so the sum is
$$
\sum_{n=1}^{\infty}n^{-4\Re(z)}=\zeta\left(4\Re(z)\right),
$$
with convergence only for $4\Re(z) > 1$. If we are given that $\left|z\right|=1/4$, clearly $\Re(z) \le 1/4$, and so the series diverges on the specified circle in the complex plane.
|
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turning a map into a fibration In Allen Hatcher's book Spectral Sequence page 29 Example 1.18,
What means "turning the map into a fibration" and convert a map into a fibration"? Given a map $f:X\to Y$, $f$ is not necessarily a fibration. How can we turn it into a fibration?
|
Given a map $f:A\to B$, let $B^I$ denote the set of continuous functions $I \to B$ endowed with the compact open topology. Define $E_f=\{(a,\gamma)\in A \times B^I | \gamma(0)=f(a)\}$
One can show that the map $p:E_f \to B$, $(a,\gamma)\mapsto \gamma(1)$ is a fibration. We can view $A$ as a subspace of $E_f$ consisting of tuples $(a,\gamma)$ with $\gamma$ the constant path at $f(a)$, and we can also check that $E_f$ deformation retracts onto this subspace.
Therefore the map $f:A\to B$ factors as a composition $A\hookrightarrow E_f\twoheadrightarrow B$ of a homotopy equivalence and a fibration.
For details, refer to the subsection "Pathspace constructions" in Hatcher's Algebraic topology at page 407.
|
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Prove $(|x| + |y|)^p \le |x|^p + |y|^p$ for $x,y \in \mathbb R$ and $p \in (0,1]$. Prove $(|x| + |y|)^p \le |x|^p + |y|^p$ for $x,y \in \mathbb R$ and $p \in (0,1]$.
A hint is given that for $0 \le \alpha \le \beta$ there exist a number $\xi \in (\beta, \alpha + \beta)$ s.t. $$(\alpha + \beta)^p - \beta^p = p \xi^{p-1}\alpha \le \alpha^{p}$$.
I see that by proving the above hint the result holds for positive numbers.
In order to prove the above hint I've considered the Intermediate value theorem, however I've not yet succeeded.
|
We want to show
$$1\le \left(\frac{|x|}{|x|+|y|}\right)^p+\left(\frac{|y|}{|x|+|y|}\right)^p,$$
or
$$1\le a^p+b^p\tag{1}$$
for $a,b\in [0,1]$, $a+b=1$, and $0< p\le1$. But (1) is clear since
$$a\le a^p,b\le b^p.$$
|
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Hypergeometric function integral representation How to prove the following relation?
$$ \, _2{F}_1(K,K;K+1;1-m) = \frac{\Gamma (K+1)}{\Gamma (K)} \int_0^{\infty } \frac{1}{(1+x) (m+x)^K} \, dx $$
where $_2{F}_1(.,.;.;.)$ is the hypergeometric function, $m\in\mathbb{R}^+$ , and $K \in\mathbb{N}$.
|
Refer to the integral definition of $ _2F_1(a,b;c;z)$ in the general case :
$$ \, _2{F}_1(a,b;c;z) = \frac{\Gamma (c)}{\Gamma (b) \Gamma (c-b)} \int_0^{\infty} t^{-b+c-1} \, (t+1)^{a-c} \, (t-z+1)^{-a} \, dt $$
In the particular case : $a=K$ ; $b=K$ ; $c=K+1$ ; $z=1-m$ ; $t=x$
and with $\Gamma(c-b)=\Gamma(1)=1$ leads to :
$$ \, _2{F}_1(K,K;K+1;1-m) = \frac{\Gamma (K+1)}{\Gamma (K)} \int_0^{\infty } \frac{1}{(1+x) (m+x)^K} \, dx $$
Note that $\frac{\Gamma(K+1)}{\Gamma(K)}=K$
|
{
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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|
Asymptotic relation between specific binomial coefficient and exponential function I need to determine the asymptotic relationship between the functions:
$$f_1(n)={n\choose{\lfloor{n\over{2}}\rfloor}}, f_2(n)=7^{\sqrt{n}}$$
(I'm going to just assume $n$ is always even.)
I've convinced myself that $f_2(n) = o\left(f_1(n)\right)$ using the following argument:
$${n\choose{n\over 2}}={{n-2}\choose{{n\over 2}-1}}⋅{{n(n-1)}\over{\left({n\over 2}\right)^2}}=f_1(n-2)⋅4\left(1-{1\over n}\right)$$
The expression $\left(1 - {1 \over n}\right)$ tends to $1$ as $n→∞$, so $f_1(n)$ eventually grows about as fast as $2^n$ (because it grows by 4 every 2 steps; however, it still grows asymptotically slower) which is easily shown to be $ω(f_2)$.
How do I prove this relationship?
|
Express combinations number via factorials, and then use Stirling's Approximation for factorials. Then you'll be able to prove what you want.
|
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|
How find this maximum of this $(1-x)(1-y)(10-8x)(10-8y)$
let $x,y\in (0,1)$, and such
$$(1+x)(1+y)=81(1-x)(1-y)$$
Prove
$$(1-x)(1-y)(10-8x)(10-8y)\le\dfrac{9}{16}$$
I ask $\dfrac{9}{16}$ is best constant?
PS:I don't like Lagrange Multipliers,becasue this is Hight students problem.
My idea:
$$(1-x)(1-y)(10-8x)(10-8y)=\dfrac{1}{64\cdot 81}(8+8x)(8+8y)(10-8x)(10-8y)$$
since
$$(8+8x)(10-8x)\le\dfrac{18^2}{4}=81,(8+8y)(10-8y)\le 81$$
if and only if $$8+8x=10-8x,8+8y=10-8y\Longrightarrow x=y=\dfrac{1}{8}$$
but this not such
$$(1+x)(1+y)=81(1-x)(1-y)$$
so How find this maximum?
|
Let $P=xy$, $S=x+y$. Rewrite both the condition and the other formula in terms of $P$ and $S$. Use the condition to eliminate $P$, and maximize as a function of $S$. Also check the edges of the square.
|
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|
Alternative ways to evaluate $\int^1_0 \frac{\text{Li}_2(x)^3}{x}\,dx$ In the following link here I found the integral & the evaluation of
$$\displaystyle \int^1_0 \frac{\text{Li}_2(x)^3}{x}\,dx$$
I'll also include a simpler version together with the question: is it possible to find some easy
ways of computing both integrals without using complicated sums that require multiple zeta
formulae and "never-ending long" generating functions?
$$i). \displaystyle \int^1_0 \frac{\text{Li}_2(x)^2}{x}\,dx$$
$$ii). \displaystyle \int^1_0 \frac{\text{Li}_2(x)^3}{x}\,dx$$
|
By series expansion
$$\displaystyle \int^1_0 \frac{\text{Li}_2(x)^2}{x}\,dx=\sum_{k,n\geq 1}\frac{1}{(nk)^2}\int^1_0x^{n+k-1}\,dx =\sum_{k,n\geq 1}\frac{1}{(nk)^2(n+k)}$$
By some manipulations
$$\sum_{k\geq 1}\frac{1}{k^3}\sum_{n\geq 1}\frac{k}{n^2(n+k)}= \sum_{k\geq 1}\frac{1}{k^3}\sum_{n\geq 1}\frac{1}{n^2}-\sum_{k\geq 1}\frac{1}{k^3}\sum_{n\geq 1}\frac{1}{n(n+k)}$$
Now use that
$$\frac{H_k}{k} = \sum_{n\geq 1}\frac{1}{n(n+k)}$$
Hence we conclude that
$$\int^1_0 \frac{\mathrm{Li}^2_2(x)}{x}\,dx = \zeta(2)\zeta(3)-\sum_{k\geq 1}\frac{H_k}{k^4}
$$
The euler some is known
$$\sum_{k\geq 1}\frac{H_k}{k^4} = 3\zeta(5)-\zeta(2)\zeta(3)$$
Finally we get
$$\int^1_0 \frac{\mathrm{Li}^2_2(x)}{x}\,dx = 2\zeta(2)\zeta(3)-3\zeta(5)$$
The other integral is very complicated to evaluate. I obtained formula using non-linear euler some here.
$$ \int^1_0\frac{\mathrm{Li}_{2}(x)^3}{x}\, dx = \zeta(3)\zeta(2)^2- \zeta(2) S_{3,2} +\sum_{k\geq 1} \frac{H_k^{(3)} H_k}{k^3}\\-\mathscr{H}(3,3)+\zeta(3) \zeta(4)-\zeta(3)\mathscr{H}(2,1)$$
where
$$ S_{p \, , \, q} = \sum_{n\geq 1} \frac{H^{(p)}}{n^q}$$
$$\begin{align}\mathscr{H}(p,q) = \int^1_0 \frac{\mathrm{Li}_p(x)\mathrm{Li}_q(x)}{x}\,dx \end{align}$$
|
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|
Cauchy-Schwarz inequality with zero angle?
Cauchy-Schwarz Inequality:
If $\textbf{u}$ and $\textbf{v}$ are vectors in a real inner product space $V$, then $$|\left\langle\textbf{u},\textbf{v}\right\rangle|\leq||\textbf{u}||\ ||\textbf{v}||$$
What will happen with the Cauchy-Schwarz inequality if the angle between the two vectors is zero?
|
the definition of the angle $\alpha\in[0,\pi]$ between $u,v$ is:
$$
\cos\alpha = \frac{\langle u,v\rangle}{\|u\| \|v\|}
$$
so $\alpha =0 $ iff
$$
1=\frac{\langle u,v\rangle}{\|u\| \|v\|}\iff
\langle u,v\rangle = \|u\| \|v\|
$$
|
{
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|
Inequality of sides of triangle If a,b,c are the sides of a triangle ABC then prove $$ a/(b+c) + b/(a+c) + c/(a+b) < 2 $$I tried to solve this by triangle inequality but i was not able to get to the solution.
|
Hint: let $x=a+b-c$, $y=a+c-b$, $z=b+c-a$. Clearly
$$a=\frac{x+y}2,\ b=\frac{x+z}{2},\ c=\frac{y+z}{2},\ a+b=\frac{2x+y+z}{2},\cdots$$
hence the original inequality reads
$$\frac{x+y}{x+y+2z}+ \frac{x+z}{x+2y+z}+\frac{y+z}{2x+y+z}<2.$$
Note that
$$\frac{x+y}{x+y+2z}+ \frac{x+z}{x+2y+z}+\frac{y+z}{2x+y+z}<\frac{x+y}{x+y+z}+
\frac{x+z}{x+y+z}+\frac{y+z}{x+y+z}\cdots$$
|
{
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|
Prove: For any integer $n \geq 2$, there is an odd number P such that $2n \lt P \lt 3n$ I am in high school and had this for a homework problem. I got it wrong, but the teacher did not post the correct answer. Any help would be appreciated. It is about writing proofs.
Prove that for any integer n greater than or equal to 2, there is always an odd number P between 2n and 3n.
|
Write numbers between $2n$ and $3n$
$$
2n<2n+1 ,2n+1 ,2n+3 ,...,2n+(n-1)<3n
$$
if $n>2$
there is $n-1$ numbers between $2n$, $3n$
number of terms=
$$ \frac{\text{last} -\text{first}}{\text{step}}+1=\\\frac{3n -2n}{1}+1=n+1\\$$
between $2n$, $3n$ are $(n+1)-2$ terms.
|
{
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|
Probability of going from node a to node b in an undirected graph. I have a graph with n nodes. Each of which represents an activity (play, walk, sleep, etc). If I'm standing at node 1 (any), what is the probability of going from 1 to j (another node) if probability is a number between 0 and 1 (closed) between 2 nodes that refers to the chance of going in exactly one step from one to the other. And in general the probability of going from i to j is the multiplication of the probability of the nodes involved in the transition.
I was thinking about using Dijkstra's algorithm to calculate the shortest path taking into account that by shortest I mean taking the higher probability edge from the successors of the actual node and avoiding the edges with 0 values.
What would you recommend me to do ?
|
Your question seems to be a bit ill-posed (it's not really clear what question you're trying to answer), but what you've got here is a Markov chain, and there are standard techniques for answering questions about them.
|
{
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|
Are two matrices similar iff they have the same Jordan Canonical form? Are two matrices similar if and only if they have the same Jordan Canonical form?
Does the Jordan form have to have ordered eigenvalues?
For example, if $\lambda_1$ and $\lambda_2$ are eigenvalues of $A$, are $\begin{pmatrix}\lambda_1&0\\0&\lambda_2\end{pmatrix}$ and $\begin{pmatrix}\lambda_2&0\\0&\lambda_1\end{pmatrix}$ both Jordan forms of $A$?
|
*
*Up to arbitrary ordering of Jordan blocks, yes
*No
*Yes
|
{
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|
Find a positive integer with prime factors of at most 2, 3, 5, 7 and ends in the digits 11 Does there exist a positive integer whose prime factors include at most 2, 3, 5, and 7, and ends in the digits 11? If so find the smallest positive integer. If not, show why none exists.
My professor gave us this question to think over. She also gave us a hint to work in mod20.
I'm not sure why we are working in mod20. I thought it might be because anything that ends in the digits 11 will be congruent to 11 (mod20). I had a thought that the prime factorization cannot include 2 because the number is not even, nor can it include 5 since the number is not a multiple of 5. Otherwise, I am at a loss...
Any help or guidance would be greatly appreciated!
|
Clearly, $2$ and $5$ cannot divide the number. Hence, let the number be of the form $3^a7^b$. We have
$$3^a \equiv 1,3,9,7 \pmod{20} \text{ and }7^b \equiv 1,7,9,3\pmod{20}$$
Hence,
$$3^a7^b \equiv 1,3,7,9\pmod{20}$$ However, the number ending in $11$ is $\equiv 11\pmod{20}$.
|
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|
Direction of unit vector that maximize directional derivative Firstly, I am aware that there are quite a few question regarding with "maximizing direction derivative" already being asked. But after scanning through, I am still not able to figure out my question thus posting it here.
Let's say we have a function of a surface, $f(x,y)$: then the gradient is ,$\bigtriangledown f = (\dfrac{\partial{f}}{\partial{x}},\dfrac{\partial{f}}{\partial{y}})$
From my understanding, the directional derivative at Point $P$ , $P(x,y)$ in the direction of $\hat{i}$ will be:
$D_{\hat{i}}(P) = \bigtriangledown{f}(P) \bullet \left(\begin{array}{cc} 1 \\ 0 \end{array}\right)$ (dot product of gradient at point P and vector pointing in $\hat{i}$ direction)
To find the direction of $\hat{i}$ that maximize $D_{\hat{i}}$, I use their dot product,
$D_{\hat{i}}(P) = \bigtriangledown{f}(P) \bullet \left(\begin{array}{cc} 1 \\ 0 \end{array}\right)= \|\bigtriangledown{f}\| \|\hat{i}\| cos(\theta)$
from the above equation, $D_{\hat{i}}(P)$ is larget when $\theta$ is $0$ ($cos(0) = 1$).
So would I be correct to say that when vector $\hat{i}$ is parallel , (ie,$cos(\theta)=1)$ it will give the maximum directional derivative? Then, it makes no sense to find out what is the direction of vector $\hat{i}$ that yield maximum $D_{\hat{i}}$ as everytime it will simply just be $0$ (Parallel)?
EDIT: my attempt to find $u$ which maximize $D_{u}(P)$
$\bigtriangledown{f} \bullet u = \|\bigtriangledown{f}\| \|u\| cos(\theta)$
Since we are talking about unit vector, $\|u\|$ will be $1$. Also $\theta=0$ for maximum directional derivative. Thus,
$\left(\begin{array}{cc} f_x \\ f_y \end{array}\right) \bullet \left(\begin{array}{cc} u_1 \\ u_2 \end{array}\right) = \sqrt{(f_x)^2+(f_y)^2}$
$f_x u_1 + f_y u_2 = \sqrt{(f_x)^2+(f_y)^2}$
but since I only have 1 equation and two unknowns $(u_1,u_2)$ how am I gonna get the components of $u$ $(u_1 and u_2)$ that gives maximum directional derivative?
|
Yes it will be the (unit) vector that is parallel to the derivative itself but you still have the task of finding that vector don't you?
|
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|
Proving the expected value of the square root of X is less than the square root of the expected value of X How do I show that $E(\sqrt{X}) \leq \sqrt{E(X)}$ for a positive random variable $X$?
I may be intended to use the Cauchy-Schwarz Inequality, $[E(XY)]^2 \leq E(X^2)E(Y^2)$, but I'm not sure how.
|
$\sqrt{x}$, $x\geq 0$ is a concave function so Jensen's inequality gives us the result without further effort.
|
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|
Polynomials close to idempotents in quotient ring of $\Bbb R[x_1,x_2,\dots,x_n]$ Let $S=\Bbb R[x_1,x_2,\dots,x_n]/(x_1^2-x_1,x_2^2-x_2,\dots,x_n^2-x_n)$.
Given $t\in \Bbb N$, what are the polynomials $p\in S$ that satisfy the relation $$p^2=tp$$ modulo $x_i$ and $x_i-1$ for all $i$ (same as evaluation at $x_i\in\{0,1\}$)?
I think finding the result for $t=1$ suffices since:$$p^2=tp\iff\frac{1}{t}p^2=p\iff(\frac{p}{t})^2=\frac{p}{t}\iff q^2=q\mbox{ for some }q\in S.$$
What can the least degree of $p$ be?
Related question: https://mathoverflow.net/questions/138478/idempotent-polynomials However I am asking only $\mod x_i$ and $\mod (x_i-1)$?
$p=x_1$ suffices as minimal non-constant polynomial idempotent.
|
Since $(x_i)$ and $(x_i-1)$ are coprime for $1\le i\le n$ we have by CRT
$$\frac{k[x_1,\cdots,x_n]}{(x_1^2-x_1,\cdots,x_n^2-x_n)}\cong\frac{\displaystyle\frac{k[x_1,\cdots,x_{n-1}]}{(x_1^2-x_1,\cdots,x_{n-1}^2-x_{n-1})}[x_n]}{(x_n^2-x_n)}$$
$$\cong\frac{\displaystyle\frac{k[x_1,\cdots,x_{n-1}]}{(x_1^2-x_1,\cdots,x_{n-1}^2-x_{n-1})}[x_n]}{(x_n)}\times\frac{\displaystyle\frac{k[x_1,\cdots,x_{n-1}]}{(x_1^2-x_1,\cdots,x_{n-1}^2-x_{n-1})}[x_n]}{(x_n-1)}$$
$$\cong\frac{k[x_1,\cdots,x_{n-1}]}{(x_1^2-x_1,\cdots,x_{n-1}^2-x_{n-1})}\times\frac{k[x_1,\cdots,x_{n-1}]}{(x_1^2-x_1,\cdots,x_{n-1}^2-x_{n-1})}$$
$$\cong \cdots\cdots\cdots\cong k^{2^n}$$
for any field $k$. The idempotents in $k^{2^n}$ are those with coordinates $0$ or $1$. Therefore, the idempotents in your quotient ring are obtained via subsets $A\subseteq\{1,\cdots,n\}$, coordinates $\varepsilon_A\in\{0,1\}$ (make once choice for each subset $A$) and solutions to $f\equiv\varepsilon_A\bmod (x_a,x_b-1)_{a\in A,b\not\in A}$. This gives $2^{2^n}$ idempotents.
With some combinatorial finagling via mobius functions of posets I get
$$f=\sum_{R\subseteq[n]}(-1)^{|R|}\left(\sum_{R\subseteq A\subseteq[n]}\varepsilon_A\right)\prod_{s\not\in R}x_s $$
for every choice of $\varepsilon:{\cal P}(A)\to\{0,1\}$. Maybe later I'll elaborate on this.
|
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How to prove that $S^{2n+1}/S^1$ is homeomorphic to $\mathbb CP^n$ under a given identification We represent an element $(x_1,y_1,...,x_n,y_n,x_{n+1},y_{n+1})\in S^{2n+1}$ as an element $(z_1,...,z_{n+1})\in \mathbb C^{n+1}$ where $z_k = x_k+iy_k$. Now I'm considering the action of $S^1$ on $S^{2n+1}$ as $e^{iθ}·(z_1,...,z_{n+1}) = (e^{iθ}z_1,...,e^{iθ}z_{n+1})$. Under this action we identify any two elements as $(z_1,...,z_{n+1}) ∼ e^{iθ} ·(z_1,...,z_{n+1})\forall (z_1,...,z_{n+1})\in S^{2n+1}\subseteq \mathbb C^{n+1}$ and $e^{iθ}\in S^1$. For the resulting quotient space $S^{2n+1}/S^1$, I'm trying to prove that $S^{2n+1}/S^1$ is homeomorphic to $\mathbb CP^n$. How do I prove this? Any solution will be highly appreciated.
|
$$
S^{2n+1} \subseteq \mathbb{R}^{2n+2} = \mathbb{C}^{n+1}
$$
The usual map restricts to a surjection (check this)
$$
S^{2n+1} \rightarrow \mathbb{C}P^n
$$
Now check that two points $w,v$ on the sphere produce the same 1-dimensional subspace in
$\mathbb{C}^{n+1}$ iff they differ by an element of $S^1$:
$$
w=\lambda v \quad\Rightarrow\quad 1=\|w\|=|\lambda|\|v\|=|\lambda|
$$
|
{
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|
is $GL(n,\mathbb R)$ dense in $M(n,\mathbb R)$ Is $GL(n,\mathbb R)$ dense in $M(n,\mathbb R)$? I have proved it to be open,not closed,not connected but not sure about this property .How to do this?
|
The way I like writing this is as follows: Consider $A\in M_{n}(\mathbb R)$ and let $\varepsilon>0$. Let $\lambda _1, \ldots, \lambda_n$ be the eigenvalues of $A$. Take $\delta$ such that $0<\delta<\frac{\varepsilon}{n^{1/2}}$ and $\delta\neq \lambda_j$ for every $j\in \{1, \ldots, n\}$. Define
$$A_\delta:=A-\delta I.$$
Then $A_\delta$ is invertible. In fact, on the contrary,
$$\det(A_\delta)=\det(A-\delta I)=0$$
and therefore $\delta$ would be an eigenvalue of $A$, so that $\delta =\lambda_j$ for some $j\in \{1, \ldots, n\}$. This contradicts the choice of $\delta$. Finally,
$$d_2(A, A_\delta)=\|A-A_\delta\|_2=\|A-(A-\delta I)\|_2=|\delta|\|I\|_2=\delta {n^{1/2}}<\frac{\varepsilon}{n^{1/2}} n^{1/2}=\varepsilon.$$
I assumed here that your metric is induced by norm:
$$\|A\|_2=\left(\sum_{i=1}^n \sum_{j=1}^n a_{ij}^2\right)^{1/2}.$$
|
{
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|
Space time algebra isomorphic to matrix algebra i have the following problem:
I already know that there exists representation of the Clifford Algebra of the Minkowski space $\mathcal{C}l(M,\eta)$. Here $M$ denotes the Minkowski space and $\eta$ is the metric tensor with the signatur $(+1,-1,-1,-1)$.
I know that a representation is given by by the map
$\gamma:M\to M_4(\mathbb{C})$ by $\gamma(e_\mu)=\gamma_\mu$, where $\gamma_\mu$ denotes the gamma matrix.
Now I want to show, that the spaces are isomorphic. I've already shown the defining relation
$\gamma_\mu\gamma_\nu+\gamma_\nu\gamma_\mu=2\eta_{\mu\nu}$.
My problem is to show that the defined mapping $\gamma$ is bijective. An advice was given in the form that it is enough to show that the mapping is injective or surjective because of dimension of the spaces.
I'm happy about all your helping comments. If you need more information it is no problem to add these ones.
Greetings.
|
The gamma matrices are four $\Bbb C$-linearly independent matrices of $M_2(\Bbb C)$, a space with $\Bbb C$-dimension $4$, so the gamma matrices form a basis. Since the image of $\gamma$ contains this basis, it is the entire space.
Thus $\gamma$ is surjective and hence injective.
|
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How does does exponent property work on $\left(xe^{\frac{1}{x}}-x\right)$ How does
$\left(xe^{\frac{1}{x}}-x\right)$
become
$\frac{\left(e^{\frac{1}{x}}-1\right)}{\frac{1}{x}}$
|
With more intermediate steps :
$$\left(xe^{\frac{1}{x}}-x\right)=$$
$$=x\left(e^{\frac{1}{x}}-1\right)$$
Let $x=\frac{1}{t}$ Replace $x$ by $\frac{1}{t}$
$$=\frac{1}{t}\left(e^{\frac{1}{x}}-1\right)$$
$$=\frac{\left(e^{\frac{1}{x}}-1\right)}{t}$$
$t=\frac{1}{x}$ Replace $t$ by $\frac{1}{x}$
$$=\frac{\left(e^{\frac{1}{x}}-1\right)}{\frac{1}{x}}$$
Another nice method :
$$\left(xe^{\frac{1}{x}}-x\right)=$$
$$=\frac{x\left(e^{\frac{1}{x}}-1\right)}{1}$$
Mulptiply the numerator and the denominator by the same term $\frac{1}{x}$
$$=\frac{\frac{1}{x}x\left(e^{\frac{1}{x}}-1\right)}{\frac{1}{x}}$$
Simplify $\frac{1}{x}x=1$
$$=\frac{\left(e^{\frac{1}{x}}-1\right)}{\frac{1}{x}}$$
|
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|
System of nonlinear equations that leads to cubic equation The system of equations are:
$$\begin{align}2x + 3y &= 6 + 5x\\x^2 - 2y^2 - (3x/4y) + 6xy &= 60\end{align}$$
I can solve it through substitution but it is an arduous process to reach this cubic equation:
$$20x^3 + 56x^2 - 243x - 544 = 0$$
And I can only solve this using a computer.
Is there a simpler method?
edit: turns out there was a printing error that made the problem much harder. I posted the actual problem below if you want to see it.
edit 2: The actual problem is far less interesting, but I included it for completeness. There are some really great answers to the above "incorrect" problem however that are definitely worth a read. Thanks everyone for contributing.
|
The roots are all real:
$x_{1}=-\frac{14}{15}-\frac{\sqrt{4429}.cos\Big(\frac{acot(-f)}{3}\Big)} {15}$,
$x_{2}=-\frac{14}{15}+\frac{\sqrt{4429}.sin\Big(\frac{atan(f)}{3}+\frac{\pi}{3}\Big)} {15}$
$x_{3}=-\frac{14}{15}-\frac{\sqrt{4429}.sin\Big(\frac{atan(f)}{3}\Big)} {15}$
$f=\frac{192158\sqrt{222021105}}{3330316575}$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
question about a special case of an $n$ sided polygon Here is an interesting question that I have been thinking about for awhile now but do not know the answer to.
Suppose you have a convex polygon with $n$ sides. What would be an example of such a polygon s.t. if you randomly picked three sides of the polygon, they could not form a triangle? I am talking about an $n$ sides polygon not some polygon with sides like $5$ or $10$. Just $n$ sides.
|
Most polygons that people draw will have all the sides about the same length, and any set of three sides can form a triangle. I think it is more interesting to ask what polygons have some set of three sides that cannot form a triangle.
One kind that has a set that cannot form a triangle is a single long side, say $10$ units long, with lots of $1$ unit sides to connect the ends. If your group of three has the long side in it, you can't form a triangle. You can also have some number of long sides (of differing lengths) connected by lots of short sides. Then if you take two longs and a short you also can't get a triangle.
Added, you can make a quadrilateral, even a trapezoid, with sides $3,1,2,1$ where no three sides will form a triangle. Then a pentagon $5,3,1,2,1$ and generally an $n-$gon with the Fibonacci series.
If the sides are $s_i$ with $s_i\le s_{i+1}$, we need $s_{j-2}+s_{j-1} \le s_j \lt \sum_{k=1}^{j-1}s_k$ You can pick $s_1,s_2$ as you wish (and might as well scale to make $s_1=1$), then respect the left inequality for $s_3$, then get polygons of arbitrary size by successively picking the sides. Once you get started, there will always be room for the next side.
Added again: For a geometric series, you need $1+r \le r^2$, which says $r \ge \phi$ In fact you can use $\phi^n$, which is just about the Fibonacci series. Then you need $r^n\lt \frac {r^n-1}{r-1}$. This seems to be most trying at $n=3$, restricting us to $r \lt \frac 13 (1+(19-3 \sqrt{33})^{1/3}+(19+3 \sqrt{33})^{1/3}) \approx 1.839286$ as found by Alpha. I checked it in Excel and it works at $1.83928$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Computing a tricky limit $\lim_{n\to\infty} \sqrt{n}\int_0^{\infty} \cos^{2n-1}(x) e^{- \pi x} \ dx$ I'm interested in some neat approaches for
$$\lim_{n\to\infty} \sqrt{n}\int_0^{\infty} \cos^{2n-1}(x) e^{- \pi x} \ dx$$
Since I suspect my approach is wrong, and I don't wanna influence you in any way, I'll add it
in a comment after the correct approach is posted. I also plan to ofer 100 bounty for the nicer, simpler approach.
|
Since
$$\int_{0}^{+\infty}\cos(n x)\,e^{-\pi x}\,dx = \frac{\pi}{n^2+\pi^2}$$
it is sufficient to compute the Fourier cosine series of $\cos^{2n-1}x.$ We have:
$$\cos^{2n-1}x = \frac{2}{4^n}(e^{ix}+e^{-ix})^{2n-1} = \frac{1}{4^{n-1}}\sum_{k=0}^{n-1}\binom{2n-1}{k}\cos((2n-2k-1)x)$$
hence:
$$\int_{0}^{+\infty}\cos^{2n-1}(x)\,e^{-\pi x}=\frac{1}{4^{n-1}}\sum_{k=1}^{n}\binom{2n-1}{n-k}\frac{\pi}{(2k-1)^2+\pi^2}.$$
Since for any fixed $k\geq 1$ we have:
$$ \lim_{n\to +\infty}\frac{\sqrt{n}}{4^{n-1}}\binom{2n-1}{n-k}=\frac{2}{\sqrt{\pi}}$$
the value of the limit equals:
$$ L = 2\sqrt{\pi}\sum_{k=1}^{+\infty}\frac{1}{\pi^2+(2k-1)^2} = \frac{\sqrt{\pi}}{2}\tanh\frac{\pi^2}{2},$$
where the last identity follows from considering the logarithmic derivatives of the Weierstrass product for the $\cosh$ function.
Notice that this problem in disguise already appeared here.
|
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|
Evaluation of $\int_{0}^{\infty}\frac{1}{\sqrt{x^4+x^3+x^2+x+1}}dx$ Evaluation of Integral $\displaystyle \int_{0}^{\infty}\frac{1}{\sqrt{x^4+x^3+x^2+x+1}}dx$
$\bf{My\; Try::}$ First we will convert $x^4+x^3+x^2+x+1$ into closed form, which is $\displaystyle \left(\frac{x^5-1}{x-1}\right)$
So Integral is $\displaystyle \int_{0}^{\infty}\frac{\sqrt{x-1}}{\sqrt{x^5-1}}dx$
now i did not understand how can i solve it
Help me
Thanks
|
The answer in terms of elliptic integrals turns out to be very simple – but it requires a very sneaky trick.
As pointed out in Jack's answer, the integrals over $[0,1]$ and $[1,\infty]$ are the same, so
$$I=\int_0^\infty\frac1{\sqrt{x^4+x^3+x^2+x+1}}\,dx=2\int_0^1\frac1{\sqrt{x^4+x^3+x^2+x+1}}\,dx$$
The sneaky trick is to substitute $u=x+\frac1x$, which gives
$$I=2\int_2^\infty\frac1{\sqrt{(u^2-4)(u^2+u-1)}}\,du$$
This changes all the denominator quartic's roots from complex to real, which will make the sequel easier. Furthermore, the roots are very simple: $-2,-\varphi,1/\varphi,2$ where $\varphi$ is the golden ratio. We can now apply Byrd and Friedman 258.00 directly, which says that if $d<c<b<a<y$,
$$\int_a^y\frac1{\sqrt{(t-a)(t-b)(t-c)(t-d)}}\,dt=gF(\psi,m)\\
\text{where }g=\frac2{\sqrt{(a-c)(b-d)}},\psi=\sin^{-1}\sqrt{\frac{(b-d)(y-a)}{(a-d)(y-b)}},m=\frac{(b-c)(a-d)}{(a-c)(b-d)}$$
($m=k^2$ is the parameter, as used by Mathematica and mpmath.) We have $d=-2,c=-\varphi,b=1/\varphi,a=2$, but since $y=\infty$ we have to take the limit of the expression for $\psi$ as $y\to\infty$, which therefore becomes
$$\psi=\sin^{-1}\sqrt{\frac{b-d}{a-d}}=\sin^{-1}\sqrt{\frac{1/\varphi+2}4}=\sin^{-1}\frac\varphi2=\frac{3\pi}{10}$$
The other key numbers follow:
$$g=\frac2{\sqrt{(2+\varphi)(1/\varphi+2)}}=\frac2{\sqrt{4\varphi+3}}$$
$$m=\frac{4(1/\varphi+\varphi)}{(2+\varphi)(1/\varphi+2)}=\frac{4(2\varphi-1)}{4\varphi+3}=8\varphi-12$$
Finally we have the answer in terms of elliptic integrals:
$$\boxed{I=\frac4{\sqrt{4\varphi+3}}F\left(\frac{3\pi}{10},8\varphi-12\right)}$$
#!/usr/bin/env python3
from mpmath import *
mp.dps = 150
print(quad(lambda t: 1/sqrt(polyval((1,1,1,1,1),t)), [0, inf]))
g = 4/sqrt(4*phi+3)
m = 8*phi-12
print(g*ellipf(3*pi/10, m))
P.S. A certain Dr. Sonnhard Graubner gave an answer here, but it's incorrect.
|
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|
Disproving existence of real root in some interval for a quintic equation Disprove the statement: There is a real root of equation $\frac{1}{5}x^5+\frac{2}{3}x^3+2x=0$ on the interval (1,2).
I am not sure whether to prove by counter-example or by assuming the statement is true and then proving by contradiction.
|
First step: what is the value of the polynomial for $x=1$?
Second step: what is the sign of the derivative of this polynomial on the interval $[1,2]$?
Third step: conclusion.
edit
Altenative: $x\ge 1$, therefore $x^n\ge 1$ for all $n\ge 1$. Hence $$\frac{1}{5}x^5+\frac{2}{3}x^3+2x\ge \frac{1}{5} +\frac{2}{3} +2 $$
whenever $x\ge 1$.
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
Energy dissipation I've been asked to prove the following, but I don't find the way..
Let $\Omega\subset\left\{0<x_n<a\right\}$ be a subset of $\mathbb{R}^n$ such that it is bounded in the $n^{th}$ coordinate. Prove that the solution of the heat equation
\begin{equation*}
\begin{array}{lc}
u_t=\Delta u && in\ \Omega \\
u=0 && on\ \partial\Omega \\
u(x,0)=u_0(x) && in\ \Omega
\end{array}
\end{equation*}
satisfies
\begin{equation*}
\int_\Omega \dfrac{1}{2}u^2\leq M\cdot e^{-\delta t},
\end{equation*}
for some $M,\delta>0$.
By showing this it is easily seen that any solution of this problem tends to $0$ as $t$ tends to $\infty$.
The problem that follows from this one is the next one:
Can one prove, with similar methods, that all the solutions of the problem
\begin{equation*}
\begin{array}{lc}
u_t=\Delta u && in\ \Omega \\
\nabla u\cdot n=0 && on\ \partial\Omega \\
u(x,0)=u_0(x) && in\ \Omega
\end{array}
\end{equation*}
where $n$ is the exterior normal vector of $\partial\Omega$, tend to steady solutions? Given an initial condition $u_0$, can one predict which is the value of the constant solution at which $u(x,t)$ tends as $t$ tends to $\infty$?
I'd be grateful for any help!
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The operator $\Delta$ should be selfadjoint on the spatial domain with the desired boundary conditions. That allows you to write the solution as $u=e^{t\Delta}u_{0}$. The spectrum of $\Delta$ is normally going to be non-positive. So all of the modes will decay in time unless you have a non-trivial mode with eigenvalue $0$--that one can be present at $t=\infty$. In the first problem, that doesn't happen, and all of the modes die out.
In the second problem, the constant functions are eigenfunctions of the Laplacian operator because the domain consists of functions with the normal derivative equal to $0$. So the constant mode can persist without being damped in time. It fact, you know what the ultimate constant value will be: it will be the projection onto the eigenfunction which is the constant function. That projection is
$$
\frac{(u_{0},1)}{(1,1)}1,
$$
assuming the spatial region is finite. If the spatial region is finite, then the above is nothing more than the average of the initial date $u_{0}$ over the whole region, which is probably what you would expect for a heat solution inside an insulated body, which is what normal derivative $0$ means. Such a mode is unchanged throughout time.
|
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|
AM-GM inequality proof Let $a_1,...,a_n>0$. The arithmetic mean is defined by $A(a_1,...,a_n) =\frac{a_1+...+a_n}{n}$ and the geometric mean by $G(a_1,...,a_n)=\sqrt[n]{a_1\cdot ...\cdot a_n}$.
Let $S(n)$ be the statement:
$$\forall a_1,...,a_n >0: G(a_1,...,a_n) \leq A(a_1,...,a_n)$$
a) Prove $S(2)$ is true
b) Let $n\geq 2$. Prove that if $S(2)$ and $S(n)$ are true, then also $S(2n)$.
c) Prove for all $n\geq 2$ that if $S(n+1)$ is true, then $S(n)$ is true. [Hint: use $G(a_1,...,a_n,G(a_1,...,a_n))=G(a_1,...,a_n)$].
d) Explain why this shows that $S(n)$ is true for all $n\geq 2$
a) This is easy to prove from $(a_1 - a_2)^2\geq 0$
After this, I am completely stumped. Any hints or ideas in which to build on?
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To $b):$
Assume $S(2)$ and $S(n)$ hold. Then,
$$\sqrt[2n]{a_1\cdots a_{2n}}=\sqrt{\sqrt[n]{a_1\cdots a_n} \sqrt[n]{a_{n+1}\cdots a_{2n}}}\underbrace{\le}_{S(2)} \frac{\sqrt[n]{a_1\cdots a_n }+\sqrt[n]{a_{n+1}\cdots a_{2n} }}{2} \\ \underbrace{\le}_{S(n)} \frac{\displaystyle \frac{a_1+\cdots +a_n }{n}+\frac{a_{n+1}+\cdots +a_{2n}}{n} }{2} =\frac{a_1+\cdots +a_{2n}}{2n},$$ which shows that $S(2n)$ holds.
To $c):$
Assume that $S(n+1)$ holds. Then, using the hint,
$$\sqrt[n]{a_1\cdots a_n}=\sqrt[n+1]{a_1\cdots a_n \sqrt[n]{a_1\cdots a_n}}\underbrace{\le}_{S(n+1)} \frac{a_1+\cdots a_n +\sqrt[n]{a_1\cdots a_n}}{n+1} \\ \implies (n+1)\sqrt[n]{a_1\cdots a_n}\le a_1+\cdots a_n +\sqrt[n]{a_1\cdots a_n}\\ \implies \sqrt[n]{a_1\cdots a_n}\le \frac{a_1+\cdots +a_n }{n},$$ which shows that $S(n)$ holds.
To $d):$
Since $S(2)$ holds it follows from $b)$ that $S(4)$ holds. Prove by induction that $S(2^n)$ holds for any $n.$ Then conclude from $c).$
|
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|
Laplace transform of a function divided by t Using the formula
$$\mathcal{L}\left\{\frac{f(t)}{t}\right\}=\int_s^\infty F(u)~du$$
I'm trying to determine the transform with $f(t)=1-e^{-t}$.
The formula gives me
$$\mathcal{L}\left\{\frac{1-e^{-t}}{t}\right\}=\int_s^\infty \left(\frac{1}{u}-\frac{1}{u+1}\right)~du$$
which doesn't seem to converge, yet Mathematica insists the answer is
$$\ln\left(1+\frac{1}{s}\right)$$
Am I using the formula incorrectly, or did I some silly mistake in my computation?
|
This certainly converges. Write the integral as
$$\begin{align}\lim_{N \to \infty} \int_s^N du \left( \frac1{u}-\frac1{u+1} \right ) &= \lim_{N \to \infty} \left [\log{N}-\log{s} - \log{(N+1)}+\log{(s+1)}\right ]\\ &= \log{\left (1+\frac1{s}\right)} - \lim_{N\to\infty} \log{\left (1+\frac1{N}\right)}\end{align}$$
This approaches the correct limit as $N \to \infty$.
|
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|
How is $\ln(-1) = i\pi$? How do I derive:
$\ln(-1)=i\pi$ and
$\ln(-x)=\ln(x)+i\pi$ for $x>0$ and $x \in\mathbb R$
Thanks for any and all help!
|
take the exponential to get $$e^{ln(-1)}=e^{\pi i}$$ $$-1=e^{\pi i}$$ $$e^{\pi i}+1=0$$ this is Euler's identity
|
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|
Reference for principal bundles and related concepts I am looking for a good reference for fibre bundles on differential manifolds, Ehresmann connections, principal $G$-bundles and principal Ehresmann connections (the $G$-equivariant version of Ehresmann connections). Could anyone advise me on this? I have looked at the book Fibre Bundles by Hausmöller, but it isn't quite what I want.
Thanks in advance.
|
Late to the party, but: Chapter 6 in Loring Tu's Differential Geometry: Connections, Curvature, and Characteristic Classes seems friendly enough, and well explained. I think sections §27 and §28 are what you're looking for.
|
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|
Why can't a value for this definite integral be found? I was trying to find out if $\int _0^{\infty }\:\frac{1}{\left(1+x^3\right)^{\frac{1}{2}}}dx$ converges or diverges. I split it into a sum, that is
$\int _0^{1 }\:\frac{1}{\left(1+x^3\right)^{\frac{1}{2}}}dx$ + $\int _1^{y }\:\frac{1}{\left(1+x^3\right)^{\frac{1}{2}}}dx$
where $y>1$ . I found that the second integral of the sum converges as y tends to infinity, but I can not evaluate the first integral. When I plot the graph, I can see clearly that the area under the curve of $f(x)=\frac{1}{\left(1+x^3\right)^{\frac{1}{2}}}$ with x in $[0,1]$ is finite. What is wrong with that integral?
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Hint: Let $~t=\dfrac1{1+x^3}~$ and then recognize the expression of the beta function in the new integral.
|
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|
Why $f(t)= t+ 2t^2\ sin (\frac 1t) , t\neq 0$ & $f(t)=0 , t=0$. Prove that this function is not $1-1$ in any neighbourhood of zero. Why $f(t)= t+ 2t^2 \sin (\frac 1t) , t\neq 0$ & $f(t)=0 , t=0.\,$Prove that this function is not $1-1$ in any neighbourhood of zero.
It is not possible with $\frac 1 {n \pi}$.
|
Hint: $f$ is continuously differentiable for all $t>0$; you can check that
$$f'\Bigl(\frac{1}{n\pi}\Bigr)$$
is positive if $n$ is odd and negative if $n$ is even.
|
{
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"url": "https://math.stackexchange.com/questions/1019419",
"timestamp": "2023-03-29T00:00:00",
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|
Mathematical Science Writers without PhD Is there science writers who have written on mathematics without holding a ph.d in the subject? I am aware that Robert Kanigle is one such but does there exist any other?
|
Nate Silver who wrote "The Signal and the Noise" holds a BA in Economics,
The book describes methods of mathematical model-building using
probability and statistics. Silver takes a big-picture approach to
using statistical tools, combining sources of unique data (e.g.,
timing a minor league ball player's fastball using a radar gun), with
historical data and principles of sound statistical analysis;
https://en.wikipedia.org/wiki/Nate_Silver
Peter L. Bernestein who wrote "Against The Gods: The Remarkable Story of Risk" also graduated in Economics,
In 1998, it was awarded the Clarence Arthur Kulp/Elizur Wright
Memorial Book Award from The American Risk and Insurance Association
(ARIA) as an outstanding original contribution to the literature of
risk and insurance.
https://en.wikipedia.org/wiki/Peter_L._Bernstein
|
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|
Parameterizing a surface The question I was asked goes like this:
The part of the hyperboloid $5x^2 − 5y^2 − z^2 = 5$ that lies in front of the yz-plane. Let x, y, and z be in terms of u and/or v. Find a parametric representation for the surface.
So, by fixing $x$ as a constant $u$, I find that the slice is an ellipse which can be parameterized as $y = \sqrt{\frac{5u^2-5}{5}}\cos{v}$ and $z = \sqrt{5u^2-5}\sin{v}$, where $u$ ranges from $1$ to infinity, and $v$ ranges from $0$ to $2\pi$.
However, I was told that this is not the correct parameterization. The correct answer is
$$
\begin{align*}
x&=\sqrt{1+u^2+\frac{1}{5}v^2} \\
y&= u \\
z&= v
\end{align*}
$$
I believe that this is just another way of parameterizing the same surface, and I don't see why my answer is wrong. Any idea?
|
I interpret being 'in front of the $yz$-plane' as the first coordinate being non-negative, thus the surface $S$ in question $\{(x,y,z)\in \mathbb R^3\colon x\ge 0\land 5x^2 − 5y^2 − z^2 = 5\}$.
It can easily be proved that this set equals $\left\{\left(\sqrt{1+y^2+\dfrac{z^2}5}, y,z\right)\colon y,z\in \mathbb R\right\}$, therefore $S=\vec r\left[\mathbb R^2\right]$, where and $\vec r\colon \mathbb R^2\to \mathbb R^3, (u,v)\mapsto \left(\sqrt{1+u^2+\dfrac{v^2}5}, u,v\right)$. So this $\vec r$ is certainly a suitable candidate for parametrization (you need to check that it satisfies the other properties for being one). In my mind this is the most natural option.
You suggest taking $\vec s\colon D\to \mathbb R^3, (\rho,\theta)\mapsto \left(\rho, \sqrt{\rho ^2-1}\cos(\theta), \sqrt 5\sqrt{\rho^2-1}\sin(\theta)\right)$, where $D=[1,+\infty[\times [0,2\pi[$.
I don't see how to make any variation of this parametrization become injective (I think it's not possible) and that might be why yours was marked wrong.
Anyway, I'll prove that $S=\vec s[D]$. The inclusion $\supseteq$ is the easy one. I'll help out with the other one.
Let $(x,y,z)\in S$. Due to $x^2=1+y^2+\dfrac{z^2}5$ it's easy to conclude that $x\ge 1$.
One wishes to find $(\rho, \theta)\in [1,+\infty[\times[0,2\pi[$ such that $(x,y,z)= \left(\rho, \sqrt{\rho ^2-1}\cos(\theta), \sqrt 5\sqrt{\rho^2-1}\sin(\theta)\right)$.
If $x=1$, then $y=0=z$ and $(x,y,z)=\vec s(1,\theta)$, for any $\theta$.
Suppose $x>1$. If such $(\rho, \theta)$ exists, necessarily $\rho =x$ and $y=\sqrt{\rho ^2-1}\cos(\theta)$. This ensues the equality $\cos(\theta)=\dfrac{y}{\sqrt{x^2-1}}$. So it's necessary to check that $-1\leq \dfrac{y}{x^2-1} \leq 1$ or equivalently that$\dfrac{y^2}{x^2-1}\leq 1$, and this in fact holds because $x^2=1+y^2+\dfrac{z^2}5$.
So take $\theta=\arccos\left(\dfrac y{\sqrt{x^2-1}}\right)\lor \theta=2\pi-\arccos\left(\dfrac y{\sqrt{x^2-1}}\right)$, depending on whether you're in $[0,\pi]$ or $[\pi, 2\pi]$.
All that is left is to check that this choice of $(\rho, \theta)$ works. I'll check the first one.
In fact
$$\require{cancel}
\begin{align} \vec s(\rho, \theta)&=\left(\rho, \sqrt{\rho ^2-1}\cos(\theta), \sqrt 5\sqrt{\rho^2-1}\sin(\theta)\right)\\
&=\left(x, \sqrt{x ^2-1}\cos\left(\arccos\left(\dfrac y{\sqrt{x^2-1}}\right)\right), \sqrt 5\sqrt{x^2-1}\sin\left(\arccos\left(\dfrac y{\sqrt{x^2-1}}\right)\right)\right)\\
&=\left(x,y,\sqrt 5\sqrt{x^2-1}\sqrt{1-\left(\dfrac y{\sqrt{x^2-1}}\right)^2}\right)\\
&=\left(x,y,\sqrt 5\cancel{\sqrt{x^2-1}}\sqrt{\dfrac{x^2-1}{\cancel{x^2-1}}-\dfrac{y^2}{\cancel{x^2-1}}}\right)\\
&=\left(x,y,\sqrt 5\sqrt{y^2+\frac{z^2}5-y^2}\right)\\
&=(x,y,z).
\end{align}
$$
To ensure that you get an appropriate $z$ just choose one of the two possible values of $\theta$.
|
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|
Rotation matrix in spherical coordinates When given arbitrary point on a unit sphere $a = (\theta, \phi)$ and an arbitrary axis $\vec{A}=(\Theta, \Phi)$, can we have an algebraic expression for $a_1=(\theta_1, \phi_1)$ which is a rotation of $a$ around $\vec{A}$ to the angle $\beta$?
Points and axes are not on the coordinate planes, values are not trivial: $\theta \neq 0$, $\phi \neq 0$, $\Theta \neq 0$, $\Phi \neq 0$, $\beta \neq 0$.
Can this be done without transformation through Cartesian? Otherwise the analytic form becomes too complicated. If there is a particular case for $\Phi \rightarrow 0$ ($\sin{\Phi} \approx \Phi$, but not for the other values) it is also fine.
Thanks
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I think what you might be looking for is Rodrigues' Rotation Formula. Using spherical coordinates:
Your arbitrary point on the unit sphere is:
$$
\mathbf{a} = (\sin\theta\cos\phi, \sin\theta\sin\phi, \cos\theta)
$$
Your arbitrary axis is represented by the unit vector:
$$
\hat{\mathbf{k}} = (\sin\Theta\cos\Phi, \sin\Theta\sin\Phi, \cos\Theta)
$$
Then the result of rotating $\mathbf{a}$ around $\hat{\mathbf{k}}$ by the angle $\beta$, using the right-hand-rule, is given by
$$
\mathbf{b} = \cos\beta\,\mathbf{a} + \sin\beta\,(\hat{\mathbf{k}} \times \mathbf{a}) + (\hat{\mathbf{k}}\cdot\mathbf{a})(1-\cos\beta)\,\hat{\mathbf{k}}
$$
Of course, now $\mathbf{b}$'s Cartesian coordinates need to be converted to spherical:
$$
\tan\phi' = \frac{b_y}{b_x}
\qquad\mbox{and}\qquad
\tan\theta' = \frac{\sqrt{b_x^2 + b_y^2}}{b_z}
$$
so that
$$
\mathbf{b} = (\sin\theta'\cos\phi', \sin\theta'\sin\phi', \cos\theta')
$$
The same article on Rodrigues' Formula also discusses a matrix representation of the rotation operation in question.
|
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$A\in M_2(\mathbb C)$ and $A $ is nilpotent then $A^2=0$. How to prove this? $A\in M_2(\mathbb C)$ and $A $ is nilpotent then $A^2=0$. How to prove this?
I am not getting enough hints to start.
|
$A$ can be put into upper triangular form, and since $A$ is nilpotent the its eigenvalues are both zero and the upper triangular form will have zeroes on its diagonal. Squaring this matrix gives zero, and so $A^2 = 0$ also.
Note that this same argument gives that any $n$ by $n$ nilpotent matrix over ${\mathbb C}$, when raised to the $n$th power, gives the zero matrix.
|
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|
Puzzle of gold coins in the bag At the end of Probability class, our professor gave us the following puzzle:
There are 100 bags each with 100 coins, but only one of these bags has gold coins in it. The gold coin has weight of 1.01 grams and the other coins has weight of 1 gram. We are given a digital scale, but we can only use it once. How can we identify the bag of gold coins?
After about 5 minutes waiting, our professor gave us the solution (the class had ended and he didn't want to wait any longer):
Give the bags numbers from 0 through 99, then take 0 coins from the bag number 0, 1 coin from the bag number 1, 2 coins from the bag number 2, and so on until we take 99 coins from the bag number 99. Gather all the coins we have taken together and put them on the scale. Denote the weight of these coins as $W$ and the number of bag with gold coins in it as $N$, then we can identify the bag of gold coins using formula $$N=100(W-4950)$$
For instance, if the weight of all coins gathered is $4950.25$ grams, then using the formula above the bag number 25 has the gold coins in it.
My questions are:
*
*How does the formula work? Where does it come from?
*Do we have other ways to solve this puzzle? If yes, how?
*If the digital scale is replaced by a traditional scale, the scale like symbol of libra or the scale in Shakespeare's drama: The Merchant of Venice (I don't know what is the name in English), then how do we solve this puzzle?
|
According to the given values, the gold coins have essentially the same weight (down to a single percent) as the base ones. Since gold is heavy this means that each gold coin is significantly smaller.
Forget about weighing and just take the bag whose coins are much smaller than the coins in the other bags.
|
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|
Correctly predicting salt and sugar in the proper sequence. We have $10$ canisters, $5$ containing sugar and $5$ containing salt. What is the probability of naming them all in the correct order? (example: salt, sugar, sugar, salt, sugar, salt, salt, ...)
It's different from predicting $10$ coins head or tails, because there are limited quantities of each.
Thanks.
|
If we assume that each arrangement of $5$ salt and $5$ sugar canisters are equally likely, then any given arrangement is uniquely determined by the placement of, say, the $5$ sugar canisters among the $10$ canisters total. For example, we can describe one arrangement as $(2, 3, 5, 7, 10)$, meaning that the sugar canisters are found in positions $2$, $3$, $5$, $7$, $10$ (and the salt canisters are in positions $1$, $4$, $6$, $8$, $9$). Then clearly, there are exactly $$\binom{10}{5} = 252$$ ways to select $5$ distinct numbers from the set $\{1, 2, \ldots, 10\}$, representing the positions of the sugar canisters, and since each such selection is equally likely (by assumption), the probability of randomly choosing the unique correct representation is $1/252$.
|
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|
How to prove the equality of determinant? Let A be a square matrix, and B a nilpotent. (size of A and B is same)
Assume AB=BA
Show that, det(A+B)=det(A)
|
Hint: Since they commute, they are simultaneously upper triangularizable. For some unitary $U$, we can write
$$
UAU^* = \pmatrix{\lambda_1&&&&*\\&\lambda_2\\&&\ddots\\&&&&\lambda_n}: = T_A\\
UBU^* = \pmatrix{0&&&&*\\&0\\&&\ddots\\&&&&0} := T_B
$$
Note that $\det(A + B) = \det(T_A + T_B)$
|
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|
How many valuations of these literals satisfy this expression? considering all the possible valuations of literals A, B, C, D, E, F, G and H (256 valuations in total), how would you go about finding how many of these valuations satisfy this expression:
$$ (A\rightarrow B) \wedge (B\rightarrow C) \wedge (D\rightarrow E) \wedge (F\rightarrow G) \wedge (G\rightarrow H)$$
I know I could draw the truth table with all the possible valuations, but what's a more efficient method? The solution that I calculated through software simulation should be 48.
|
Note that you can divide the expression in three parts:
$$
\underbrace{(A\rightarrow B) \wedge (B\rightarrow C)}_{P} \wedge \underbrace{(D\rightarrow E)}_{Q} \wedge \underbrace{(F\rightarrow G) \wedge (G\rightarrow H)}_{R}
$$
The value of $P$ depends only on $A,B,C$, the value of $Q$ depends only on $D,E$, and the value of $R$ depends only on $F,G,H$. So if $N_P$ is the number of choices of $A,B,C$ for which $P$ is true, $N_Q$ is the number of choices of $D,E$ for which $Q$ is true, and $N_R$ is the number of choices of $F,G,H$ for which $R$ is true, the total number is
$$
N_P N_Q N_R.
$$
|
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|
CD is height of right-angled triangle ABC, M and N are midpoints of CD and BD: prove AM⊥CN I was having some troubles proving this:
CD is the height that corresponds to the hypotenuse of right-angled triangle ABC. If M and N are midpoints of CD and BD, prove that AM is perpendicular to CN.
Here's the illustration (by me): http://i.imgur.com/JyR9Gz0.jpg
Sorry for my English. Any help/hints appreciated.
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It suffices to show $\angle MAD=\angle NCD$, which means two right triangles
$ADM$ and $CDN$ are similar. This can be proved from $AD/DM=CD/DN$, which follows from $AD/CD=CD/BD$.
|
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|
Formula for the angle of a line $y = mx$ as a function of $m$. I was wondering if there was a way to calculate the angle made by a line $(\space y=mx)$ in the Cartesian plane using only $m$. I used the Pythagorean theorem in this figure:
$$AO= \sqrt{AB^2+OB^2}=\sqrt{x^2+m^2x^2}=x \sqrt{1+m^2}$$
Now I know that $\alpha = \cos^{-1} (\cos \alpha) $.
$$\cos \alpha = \frac{OB}{OA}=\frac{x}{x \sqrt{1+m^2}}=\frac{1}{\sqrt{1+m^2}}$$
$$\alpha = \cos^{-1} \left(\frac{1}{\sqrt{1+m^2}}\right)$$
Is this correct? Is there an easy way to solve this?
|
Looks good! Alternatively, notice that:
$$
m = \frac{y}{x} = \tan \alpha
$$
So we have:
$$
\alpha = \tan^{-1}(m) = \tan^{-1}(0.5) \approx 26.57^\circ
$$
|
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Conditions for using The Rational Zeros Theorem Barron's SAT Math 2 and SparkNotes state that if p/q is a rational zero of P(x) with integral coefficients, then p is a factor of the constant term of P(x) and q is a factor of the leading coefficient of P(x) .
Then they show an example in which p/q isn't even a zero of the polynomial and continue to solve it as is (solving for 9/2 does not get you to 0):
$P(x) = 2x^4 + x^3 -19x^2 - 9x + 9$
Factors of constant term: $±1 , ±3 , ±9 $.
Factors of leading coefficient: $±1 , ±2 $.
Possible values of : $±1/1 , ±1/2 , ±3/1 , ±3/2 , ±9/1 , ±9/2 $. These can be simplified to: $±1 , ±1/2 , ±3 , ±3/2 , ±9 , ±9/2 $.
My questions are:
*
*Does p/q (where p is a factor of the constant term of P(x) and q is a factor of the leading coefficient of P(x)) need to be a zero of the polynomial in order to use this theorem? (since this obviously isn't the case in the above mentioned example and several others that I came across).
*What are the conditions for one to be able to use it?
*If anyone could provide an intuitive explanation/proof of why this theorem works in general that would be awesome.
http://www.sparknotes.com/math/algebra2/polynomials/section4.rhtml
|
The theorem says if $p/q$ is a rational zero of $P(x)$, then .... The way you usually use this is to narrow down the possibilities for rational zeros. In this case $9/2$ does not happen to be a zero, but the theorem doesn't rule it out.
Once you identify the possible rational zeros using the theorem, you can test each of them by substituting in to the polynomial and see which (if any) are actually zeros.
See e.g. Wikipedia for a proof.
|
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Is $ G \cong G/N \times N$? If G is a finite group and N is a normal subgroup in G , then can we say G $\cong$ G/N $\times$ N always? Is it true for like normal nilpotent or normal solvable or any such special classes. I couldn't help but ask for it. What if G is infinite? Please tell me if you can, the concept I am skipping here. If you can avoid using exact sequences, It will be much clearer to me. I have seen an answer which still is not so clear to me. Thanks!
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I've intended to ask a similar question. Fortunately, the community has avoided a duplicate. Moreover, I've had more information to better my proposition, which is written below.
The proposition: The necessary and sufficient conditions of group $G$ and its normal subgroup $N$ (regardless of commutativity and finiteness) to satisfy $G \sim N \times G / N$ is that:
$i) \; \; N \le Z(G)$ (center of group $G$), and
$ii)$ There is a way to choose representative elements for classes of $G/N$ so that those elements would make up a subgroup of $G$.
I think I proved this, but I'm not so sure.
|
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Traffic with Poisson distribution The number of cars that cross an intersection during any interval of length t minutes between 3:00 pm and 4:00 pm has a Poisson distribution with mean t. Let W be the time that has passed after 3:00 pm and before the first car crosses the intersection. What is the probability of W being less than two minutes?
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Hint: The probability that $W\ge 2$, or equivalently the probability that $W\gt 2$, is the probability that $0$ cars go through the intersection in $2$ minutes.
|
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Why does the higher order derivative test work? I'm an AP Calculus BC student, so all I know about derivatives is the increasing/decreasing/ relative max/min function relation with first derivative (first derivative test), concavity (second derivative test). I showed that you can create a third derivative test to my teacher (by showing that if the critical numbers of the second derivative = 0, and if the third derivative is non-zero, then it must be a point of inflection as the function is either increasing or decreasing on the second derivative graph), and he said it was right and to think about derivatives at higher orders like x^5 only has a derivative at x = 0 with the fifth derivative).
When i googled it, i found the higher order derivative test. Is there a logical explanation for why this test works? I dont understand why if n-1 derivatives = 0 and the nth is non-zero, you can find max/mins/inflection points.
|
One way to understand how the test works is by looking at the Taylor Series of the function $f(x)$ centered around the critical point, $x = c$:
$$
f(x) = f(c) + f'(c)(x-c) + \frac{f''(c)}{2}(x-c)^2 + \cdots
$$
Note: In your question you said that the n-th derivative is non-zero. Here I'm assuming the n+1-st derivative is the first to be non-zero at $x=c$. It doesn't make a difference, it's just the way I learned it.
If $f'(c) = \cdots = f^{(n)}(c) = 0$ and $f^{(n+1)} \ne 0$, then the Taylor Series ends up looking like this:
$$
f(x) = f(c) + \frac{f^{(n+1)}(c)}{(n+1)!}(x-c)^{n+1} + \frac{f^{(n+2)}(c)}{(n+2)!}(x-c)^{n+2} + \cdots
$$
Consider what happens when you move $f(c)$ to the other side of the equation:
$$
f(x) - f(c) = \frac{f^{(n+1)}(c)}{(n+1)!}(x-c)^{n+1} + \frac{f^{(n+2)}(c)}{(n+2)!}(x-c)^{n+2} + \cdots
$$
What does $f(x) - f(c)$ mean?
*
*If $f(x) - f(c) = 0$, then $f(x)$ has the same value as it does at $x = c$.
*If $f(x) - f(c) < 0$, then $f(x)$ has a value less than it has at $x = c$.
*If $f(x) - f(c) > 0$, then $f(x)$ has a value greater than it has at $x = c$.
We expect $f(x) - f(c) = 0$ at $x = c$ (the equation reflects this), but we're more interested in what it does on either side of $x = c$. When $x$ is really close to $c$, i.e. $(x-c)$ is a really small number, we can say:
$$
f(x) - f(c) \approx \frac{f^{(n+1)}(c)}{(n+1)!}(x-c)^{n+1}
$$
because the higher powers of a small number "don't matter" as much.
Concerning local extrema
If $n$ is odd, then our approximation of $f(x) - f(c)$ is an even-power polynomial. That means $f(x)$ has the same behavior - is either less than or greater than $f(c)$ - on both sides of $x = c$. Therefore it's a local extreme. If $f^{(n+1)}(c) > 0$, then $f(x)$ is greater than $f(c)$ on both sides of $x = c$. Otherwise, if $f^{(n+1)}(c) < 0$, then $f(x)$ is less than $f(c)$ on both sides of $x = c$
If, on the other hand, $n$ is even, then our approximation of $f(x) - f(c)$ is an odd-power polynomial centered around $x = c$. Therefore $f(x)$ will be greater than $f(c)$ on one side of $x = c$, and less on the other. That means $x = c$ isn't a local extreme.
Concerning saddle points
Note that if you differentiate both sides of our approximation twice, you get:
$$
f''(x) \approx \frac{f^{(n+1)}(c)}{(n-1)!}(x-c)^{n-1}
$$
If $n$ is even, this is another odd-power polynomial centered around $x = c$. It therefore has opposite behavior on each side of $x = c$, giving you a saddle point.
|
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If $f(1)=1\;,f(2)=3\;,f(3)=5\;,f(4)=7\;,f(5)=9$ and $f'(2)=2,$ Then sum of all digits of $f(6)$
$(1):$ If $P(x)$ is a polynomial of Degree $4$ such that $P(-1) = P(1) = 5$ and
$P(-2)=P(0)=P(2)=2\;,$Then Max. value of $P(x).$
$(2):$ If $f(x)$ is a polynomial of degree $6$ with leading Coefficient $2009.$ Suppose
further that $f(1)=1\;,f(2)=3\;,f(3)=5\;,f(4)=7\;,f(5)=9$ and $f'(2)=2,$
Then sum of all digits of $f(6)$ is
$\bf{My\; Try\; For\; (1):}$ Given $x=-2\;,x=0\;,x=+2$ are the roots of $P(x)=0.$
So $(x+2)\;,(x-0)\;,(x-2)$ are factors of $P(x)=0$. So we can write $P(x)$ as
$P(x) = A\cdot x\cdot (x-2)\cdot (x+2)(x-r)\;,$ So we have calculate value of $A$ and $r$
Now put $x=-1\;,$ we get $P(-1)=-3A\cdot (1+r)\Rightarrow -3A\cdot (1+r)=5............................(1)$
Now put $x=1\;,$ we get $P(1)=-3A\cdot (1-r)\Rightarrow -3A\cdot (1-r)=5..................................(2)$
So from $(1)$ and $(2)\;,$ We get $r=0$ and $\displaystyle A=-\frac{5}{3}.$
So Polynomial $\boxed{\boxed{\displaystyle P(x)=-\frac{5}{3}\cdot x^2\cdot (x^2-4)}}$
$\bf{My\; Try\; For \; (2):}$Let $f(x)=2x-1\;\forall\; x=1\;,2\;,3\;,4\;,5.$
So we can say that $(x-1)\;,(x-2)\;,(x-3)\;,(x-4)\;,(x-5)$ are the roots of $f(x)-2x+1=0$
So $f(x)-2x+1=2009\cdot \underbrace{(x-1)\cdot(x-2)\cdot (x-3)\cdot (x-4)\cdot (x-5)}\cdot\underbrace{(x-r)}$
Now How can i solve after that
Help me and plz explain me, is my $(1)$ Try is right or not
Thanks
|
For part 2, here's one way to simplify the differentiation. We have
$$f(x)-2x+1=2009\cdot (x-1)\cdot(x-2)\cdot (x-3)\cdot (x-4)\cdot (x-5)\cdot(x-r)$$
Substituting $x + 3 \to x$,
$$f(x + 3) - 2(x+3)+1 = 2009(x+2)(x+1)(x)(x-1)(x-2)(x+3-r)$$
$$f(x + 3) -2x - 5 = 2009x(x^2-1)(x^2-4)(x + 3 -r)$$
$$f(x + 3) - 2x - 5 = 2009(x^5 - 5x^3 + 4x)(x + 3 -r)$$
Applying the chain rule to LHS and product rule to RHS,
$$f'(x+3) - 2 = 2009(5x^4 - 15x^2 + 4)(x + 3 - r) + 2009(x^5 - 5x^3 + 4x)$$
Substituting $x = -1$,
$$f'(2) - 2 = 2009(5 - 15 + 4)(2-r) + 2009(1-5+4)$$
$$0 = 2009\cdot(-6)\cdot(2-r)$$
to give $r=2$. Hence,
$$f(6) - 12 + 1 = 2009\cdot5!\cdot4$$
$$f(6) = 964331$$
|
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|
calculating limit for a sech function is any answer in terms of dirac delta function?
how can i calculate this expression?
$$\lim_{x\to 0} {1\over \sqrt{x}} \text{sech} \left(\dfrac{1}{x}\right)$$
|
$\lim_{x\to 0} {1\over \sqrt{x}} \text{sech} \left(\dfrac{1}{x}\right)$
$=\dfrac{\frac{1}{\sqrt{x}}}{\cosh (\frac{1}{x})}$
$t=\frac{1}{x}$
$=\lim_{t\to \infty} \dfrac{\sqrt{t}}{\cosh (t)}$
Applying L'Hopital's rule:
$=\lim_{t\to \infty} \dfrac{1}{2\sqrt{t}\sinh{t}}$
$=\frac{1}{\infty}$
$=0$
|
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|
Approach towards second order differential equation I have the following equation to be solved. Can anybody explain to me how I am supposed to approach this problem?
$$4 \frac{d^{2}y}{dx^{2}} + 4 \frac{dy}{dx} + y = (8x^{2} + 6x + 2)e^{-x/2}$$
edited
I am supposed to find the particular integral for the same.
|
First find the general solution of the homogeneous equation $4y''+4y'+y=0$ in the usual way, using the characteristic equation $4r^2+4r+1=0$. Then use the method of undetermined coefficients to find a particular solution of the nonhomogeneous equation. In "guessing" your undetermined coefficients form for the particular solution, be sure to take into account the fact that some terms of the input function (forcing function) $(8x^2+6x+2)e^{-x/2}$ are solutions of the homogeneous equation! (Basically this means that your generic guess for $y_p$ has to be multiplied by a power of $x$.) Finally, add the general solution of the homogeneous equation (which has two arbitrary constants) to the particular solution you found using undetermined coefficients.
P.S. Since the two lowest degree terms of the input function, $2e^{-x/2}$ and $6xe^{-x/2}$, are also solutions of the homogeneous equation, you need to throw a factor of $x^2$ into your form for $y_p$, so it should be
$$y_p(x)=(Ax^4+Bx^3+Cx^2)e^{-x/2}.$$
|
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Sum of cells on infinite board is even Let $a,b,c$ be pairwise relatively prime positive integers. In an infinite checker board (infinite in all directions), each cell contains an integer. The sum of the integers in any $a\times a$ square is even, in any $b\times b$ square is even, and in any $c\times c$ square also even. Does it always follow that all numbers on the board are even?
If $a=1$ then all the numbers must be even. For higher $a,b,c$, we probably have to take overlapping squares of sizes $a,b,c$ to deduce that some cells are even. But it is not clear which ones.
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Suppose that we put a light under each cell in this infinite grid, all of the lights starting off. We can, at will, switch all the state of all the lights in any $a\times a$, $b\times b$, or $c\times c$ square - that is, any light in this square that was on is turned off, and any light that was off is turned on. So, by choosing appropriate squares in succession, we may try to obtain various complex patterns of cells. However, what is of note here, is that the sum of the values in the lighted squares is always even, since, working in $\mathbb{Z}/\mathbb{2Z}$ - that is, considering only the parity of values - adding a value is the same as subtracting it, so when we flip all the lights in a square, the change in parity is the same regardless of whether lights were flipped on or off.
Now, if we can use this process to produce a lighted $1\times 1$ square, we win, since this implies that the sum of the lighted squares - of which there is but one - is even, and hence that square is even. To do this, notice that by switching $b$ side-by-side $a\times a$ squares, we could create a $ab\times a$ rectangle of lighted points. Similarly, we can create a $ab\times b$ rectangle by switching $a$ side-by-side $b\times b$ squares. Suppose we overlayed these such that two of their corners matched up and one was within the other - now we have a $ab\times|b-a|$ rectangle, since we basically lighted the bigger rectangle, then switched off lights in the smaller rectangle, but they had the same width, and hence yielded a rectangle. If we stacked one rectangle atop the other, we would have an $ab\times a+b$ rectangle. By similar reasoning, we can construct an $ab\times |k_1a + k_2b|$ rectangle for any $k_1,k_2\in \mathbb{Z}$. Since $a$ and $b$ are coprime, there exists a pair of integers $k_1$ and $k_2$ such that $k_1a+k_2b=1$, thus, by layering a lot of $a\times a$ and $b\times b$ squares, we get a $ab\times 1$ rectangle. Similarly, we may construct $bc\times 1$ and $ca\times 1$ rectangles. Then, when we layer these, we can get any rectangle expressible as, for $k_1,k_2,k_3\in\mathbb{Z}$:
$$(k_1ab + k_2bc + k_3ca)\times 1$$
and since $\gcd(ab,bc,ca)=1$, there exist $k_1,k_2,k_3$ such that the above simplifies to
$$1\times 1$$
implying that we can, by switching lights this way, yield a $1\times 1$ lighted square which must contain an even number. By translating this solution around, we can prove this of every square.
|
{
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"url": "https://math.stackexchange.com/questions/1021375",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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|
Calculating the value of $\frac{a-d}{b-c}$ If $\frac{a-b}{c-d}=2$ and $\frac{a-c}{b-d} = 3$ then determine the value of:
$$\frac{a-d}{b-c}$$
Where $a,b,c,d$ are real numbers.
Can someone please help me with this and give me a hint? I tried substitutions and solving them simultaneously but I couldn't determine this value. Please help.
|
plugging $c=\frac{a+3b}{4}$ and $d=-\frac{a}{4}+\frac{5b}{4}$ in the given term we get $-5$ as the given result.
|
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"url": "https://math.stackexchange.com/questions/1021567",
"timestamp": "2023-03-29T00:00:00",
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|
Evaluating $\int_0^1 x \tan(\pi x) \log(\sin(\pi x))dx$ What starting point would you recommend me for the one below?
$$\int_0^1 x \tan(\pi x) \log(\sin(\pi x))dx $$
EDIT
Thanks to Felix Marin, we know the integral evaluates to
$$\displaystyle{\large{\ln^{2}\left(\, 2\,\right) \over 2\pi}}$$
|
$\def\Li{{\rm{Li}_2\,}}$Denote the considered integral as $I$ and set $y=\pi x$, we have
\begin{equation}
I=\frac{1}{\pi^2}\int_0^\pi \frac{y\sin y}{\cos y}\,\ln(\sin y)\,dy
\end{equation}
Perform integration by parts by taking $u=y$, we have
\begin{align}
I&=-\left.\frac{y}{2\pi^2}\int\frac{\ln\left(1-\cos^2y\right)}{\cos y}\,d(\cos y)\right|_0^\pi+\frac{1}{2\pi^2}\int_0^\pi\int\frac{\ln\left(1-\cos^2y\right)}{\cos y}\,d(\cos y)\,dy\\
&=\left.\frac{y}{4\pi^2}\Li\left(\cos^2y\right)\,\right|_0^\pi-\frac{1}{4\pi^2}\int_0^\pi\Li\left(\cos^2y\right)\,dy\\
&=\frac{\Li\left(1\right)}{4\pi}-\frac{1}{4\pi^2}\int_0^{\pi}\sum_{k=1}^\infty\frac{\cos^{2k}y}{k^2}\,dy\quad\Rightarrow\quad\color{red}{\mbox{use series representation of dilogarithm}}\\
&=\frac{\pi}{24}-\frac{1}{4\pi^2}\sum_{k=1}^\infty\int_0^{\pi}\frac{\cos^{2k}y}{k^2}\,dy\quad\Rightarrow\quad\color{red}{\mbox{justified by Fubini-Tonelli theorem}}\\
&=\frac{\pi}{24}-\frac{1}{2\pi^2}\sum_{k=1}^\infty\frac{1}{k^2}\int_0^{\pi/2}\cos^{2k}y\,\,dy\quad\Rightarrow\quad\color{red}{\mbox{by symmetry argument}}\\
&=\frac{\pi}{24}-\frac{1}{4\pi^2}\sum_{k=1}^\infty\frac{\Gamma\left(k+\frac{1}{2}\right)\Gamma\left(\frac{1}{2}\right)}{k^2\,\Gamma\left(k+1\right)}\quad\Rightarrow\quad\color{red}{\mbox{Wallis' integrals}}\\
&=\frac{\pi}{24}-\frac{1}{4\pi}\sum_{k=1}^\infty\frac{(2k)!}{4^k\,k^2\,(k!)^2}\tag1
\end{align}
$\def\arctanh{{\rm{\,arctanh}\,}}$Here is the tedious part (and also the hardest part). I use Mathematica to help me out to find generating function of the following series. Let us start with
\begin{equation}
\sum_{k=0}^\infty\frac{\Gamma\left(k+\frac{1}{2}\right)\Gamma\left(\frac{1}{2}\right)x^k}{\Gamma\left(k+1\right)}=\pi\sum_{k=0}^\infty\frac{(2k)!\,\,x^k}{4^k\,(k!)^2}=\frac{\pi}{\sqrt{1-x}}
\end{equation}
Divide by $x$ and then integrate it, we have
\begin{equation}
\sum_{k=1}^\infty\frac{(2k)!\,\,x^k}{4^k\,k\,(k!)^2}=\int\left[\frac{1}{x\,\sqrt{1-x}}-\frac{1}{x}\right]\,dx=-2\arctanh\left(\sqrt{1-x}\,\right)-\ln x+C_1
\end{equation}
Taking the limit as $x\to0$, we obtain
\begin{equation}
C_1=\lim_{x\to0}\left(2\arctanh\left(\sqrt{1-x}\,\right)+\ln x\right)=\ln4
\end{equation}
Hence
\begin{equation}
\sum_{k=1}^\infty\frac{(2k)!}{4^k\,k\,(k!)^2}x^k=-2\arctanh\left(\sqrt{1-x}\,\right)-\ln x+\ln4
\end{equation}
Repeat the process once more, we obtain
\begin{align}
\sum_{k=1}^\infty\frac{(2k)!\,\,x^k}{4^k\,k^2\,(k!)^2}=&\,-2\int\frac{\arctanh\left(\sqrt{1-x}\,\right)}{x}\,dx-\int\frac{\ln x}{x}\,dx+\ln4\int\frac{dx}{x}\\
=&\,\,2\arctanh\left(\sqrt{1-x}\,\right)\left[\arctanh\left(\sqrt{1-x}\,\right)-2\ln\left(\frac{\sqrt{1-x}+1}{2}\right)\right]\\&\,-2\,\Li\left(\frac{\sqrt{1-x}-1}{\sqrt{1-x}+1}\right)-\frac{\ln^2x}{2}+\ln4\ln x+C_2\\
\end{align}
Taking the limit as $x\to0$, we obtain
\begin{equation}
C_2=-2\ln^22
\end{equation}
Hence
\begin{align}
\sum_{k=1}^\infty\frac{(2k)!\,\,x^k}{4^k\,k^2\,(k!)^2}
=&\,\,2\arctanh\left(\sqrt{1-x}\,\right)\left[\arctanh\left(\sqrt{1-x}\,\right)-2\ln\left(\frac{\sqrt{1-x}+1}{2}\right)\right]\\&\,-2\,\Li\left(\frac{\sqrt{1-x}-1}{\sqrt{1-x}+1}\right)-\frac{\ln^2x}{2}+\ln4\ln x-2\ln^22\tag2
\end{align}
Thus, by putting $x=1$ to $(2)$ then $(1)$ becomes
\begin{equation}
I=\frac{\pi}{24}-\frac{1}{4\pi}\left(\frac{\pi^2}{6}-2\ln^2 2\right)=\frac{\ln^22}{2\pi}
\end{equation}
and we are done.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Expected Value of product of Ito's Integral Any idea on how to compute the expected value of product of Ito's Integral with two different upper limit?
For example:
$$\mathbb{E}\left[\int_0^r f(t)\,dB(t) \int_0^s f(t)\,dB(t)\right]$$
I only know how to compute when the upper limit r and s are the same...but don't know how when r and s are different...help.
|
You can split the integrals up into parts over their domain. The part where they overlap can use the usual formula, and the variables are independent on the part where they don't overlap, so those expectations are products of the expectation of the factors.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Hopf bifurcation and limit cycles $$dV/dt=10(V-\frac{V^3}{3}-R+I_{input})$$
$$dR/dt=0.8(-R+1.25V+1.5)$$
Use $I_{input}$ as the relative parameter to prove that there these equations undergo 2 hopf bifurcations and indicate whether each is subcritical or supercritical.
Setting both equal to 0 to find the equilibrium:
$0=V-\frac{V^3}{3}-R+I_{input}$
$0=R+1.25V+1.5$
I guess I don't really know how to continue from here..
find eigenvalues?
jacobian (not sure if I computed this correctly):
\begin{array}{lcr}
\mbox1-v^2 & -1 \\
\mbox1.25 & -1 \\
\end{array}
So I know that a supercritical hopf bifurcation is a stable LC around an unstable equilibrium and a subcritical hopf bifurcation is a an unstable LC around a stable equilibrium.
|
1) You forgot the constants when you calculated the Jacobian matrix.
2) In general, when you want to prove theorically that a certain bifurcation occurs in a system, the way is verify that your system is under the hypothesis of the associate theorem (in this case the Hopf bifurcation theorem). In the case of local bifurcations (the ones associate to equilibria points), the first step is to check what are the eigenvalues of the linear part of the system (evaluated in the equilibrium point).
|
{
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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|
How to solve a coupled differential equations I tried different ways to solve this differential equation but I did not succeed. These is the first couple ODEs I try to solve. I hope somebody can give me a hint.
\begin{eqnarray}
\ddot{x} + ax - b\dot{y} = c
\end{eqnarray}
\begin{eqnarray}
\ddot{y} + ay + b\dot{x} = 0
\end{eqnarray}
|
Go to the complex plane. Do "first" + i"second" equation and you'll get something elegant. Sum of equations with second multiplied by the imaginary unit:
$$(\ddot{x}+i\ddot{y})+a(x+iy)+b(-\dot{y}+i\dot{x})=c+0i$$
Use new complex variable $z=x+iy$:
$$\ddot{z}+a z +ib\dot{z}=c$$
This is a simple second order DE with known solutions. Even better, the particular solution is just $z=c/a$ and the homogenous part works out very elegantly with $z=Ce^{i\lambda t}$, which gets you to a completely real characteristic polynomials ($i$ multiplies out):
$$-\lambda^2-\lambda b+a=0$$
Of course, $\lambda$-s may still be a complex conjugate pair, but the fact that the polynomial is real simplifies everything.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1021984",
"timestamp": "2023-03-29T00:00:00",
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|
plotting an fft function need to plot the folowing:
$$X(j\omega)=\cal F\{{2\over\pi\lambda}\}*\cal F\{sinc(2w_m\lambda)\}-\cal F\{{4\over\pi\lambda}\}*\cal F\{sinc(w_m\lambda)\}$$
as a plot of $|X(j\omega)|$ for $|\omega|<10\pi$
the code I've wrote is:
%% define
t = -4:1/100:4;
w= -10*pi:1/100:10*pi;
wm=3*pi;
z=t-(pi/wm);
a=1./(pi*z);
b=sinc(2*wm*z);
c=sinc(wm*z);
Aa=fft(a);
Bb=fft(b);
Cc=fft(c);
Dd=fft(a);
A1=fftshift(Aa);
B1=fftshift(Aa);
C1=fftshift(Aa);
D1=fftshift(Aa);
x=2*conv(A1,B1)-4*conv(D1,C1);
mag=abs(x);
%%plot
figure;
plot(w,mag,'k','LineWidth',2);
title('Q1: |X(jw)| - frequency domain');
xlabel('w [hz]','FontSize',12);
ylabel('|X(jw)|','FontSize',12);
legend([{'|X(jw)|'}]);
I know my problem is with the w-axis, but I have no idea on how to fix it.
|
You have two arrays of different dimensions in your plot.
After running your code, I am told that mag is 1 by 1601 and w is 1 by 6000 something.
Fix 1 quick and dirty:
w = linspace(-10*pi, 10*pi, 1601);
A better solution is to define w after mag since w isn't used until plotting. Then you can do
Fix 2 (a) and (b):
w = -10*pi:length(mag):10*pi;
or
w = linspace(-10*pi, 10*pi, length(mag));
This way if you change anything in the code that affects the length of mag, then w will change accordingly.
|
{
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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|
Determine whether the series $\sum _{n=1}^{\infty \:\:}\frac{\left(-1\right)^n}{(3n)!}$ is convergent or divergent. Determine whether the series
$\sum _{n=1}^{\infty \:\:}\frac{\left(-1\right)^n}{(3n)!}$
is convergent or divergent. If it is convergent, then how many terms of the series do we need in order to find the sum to within $10^{-5}$?
I'm trying to use the Alternating Series Test
$a_n=\left(-1\right)^n\:$
$b_n=\frac{1}{\left(3n\right)!}$
need to check if
$\lim _{n\to \infty }\left(\frac{1}{\left(3n\right)!}\right)$ $=\:0$
which is does because
$\frac{1->\:1}{\left(3n\right)!\:->\infty \:}\:=0$
then, need to check if $\left\{\frac{1}{\left(3n\right)!}\right\}$ is decreasing
$b_{n+1}\frac{1}{\left(3\left(n+1\right)\right)!}$ $=\:\frac{1}{(3n+3)!}$
$b_{n+1}<\:b_n$ and therefore decreasing
so the series is convergent by Alternating Series Test
Now for the other part of the question start writing out the series
$\sum _{n=1}^{\infty }\:\frac{\left(-1\right)^n}{\left(3n\right)!}\:=\:\frac{-1}{3!}+\frac{1}{6!}-\frac{1}{9!}\:$
$\frac{1}{6!}=1.388\cdot 10^{-3}$
$-\frac{1}{9!}\:=-2.755\:\cdot 10^{-6}$
since the question is within $10^{-5}$ i would think not to include $-\frac{1}{9!}$
and say i only include 2 terms. Does my work and answers look correct?
|
Edit: This is an answer for the previous version, where an error was $10^{-15}$. According to a suggestion I am leaving it, as a hint, what one should do if the error is smaller than $10^{-5}$.
Near good, but in your question is $10^{-15}$. It is known, that the error is less than the firts of remaining terms, hence $(3n)!>10^{15}$, which is true from $n=6$, hence you need two additional terms.
|
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|
Estimations in a ordered field? My Problem: I am stuck with a proof strategy on the following:
So i have got an ordered field $ (K,+,*,<) $ given. I also have
$x,y\in K$
and $0\le y < x$
I have to proof that, for every n $\in \mathbb{N}, n \ge 2$ there is:
$ny^{n-1} < \frac{x^n-y^n}{x-y} < nx^{n-1}$
What i got so far:
It can be easily proven that
$ny^{n-1} < nx^{n-1}$. It won't help us much, i guess.
We also know that the fraction in the middle is >0 because nominator and denominator will always stay positive (because $0\le y < x$). This might be of some use.
This inequality just doesn't ring a bell. I know i have to estimate the fraction somehow. Also i wonder why the task is to prove it for a specific field, not just for $\mathbb{R}$ (What difference does it make?)
Any light you can shed on this is much appreciated.
|
Since $\;x-y>0\;$ , we have that
$$ny^{n-1}<\frac{x^n-y^n}{x-y}<nx^{n-1}\iff ny^{n-1}(x-y)<x^n-y^n<nx^{n-1}(x-y)$$
and for example (left inequality):
$$ny^{n-1}(x-y)<x^n-y^n\iff nxy^{n-1}-ny^n<x^n-y^n\iff y^n(1-n)<x^n-nxy^{n-1}$$
But
$$x^n-nxy^{n-1}>x^n-nx^n=x^n(1-n)>y^n(1-n)$$
using what you mention that "can be easily proven", and thus we're done.
Try to do the rightmost inequality in the same way.
|
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|
Prove two subgroups have the same order given an equivalence relation. Let $X$ be the set of all subgroups of a group $G$. The following is an equivalence relation $H$~$K$ where $H=xKx^{-1}$.
Prove: If $H$~$K$, then $|H|=|K|$.
I'm guessing $H$,$K$ are subgroups of a group $G$. Other than that, I'm not quite sure where to start on this one. I know that if $|H|=|K|$ then they have the same amount of elements in them but I'm not sure how to prove that given the equivalence relation.
|
Check the map $f : K \rightarrow H$ such that $f(a) = xax^{-1}$. It's obviously surjective homomorphism. So if it is injective we are done. Suppose that $a \in \operatorname{Ker}(f)$, then $xax^{-1} = e \Rightarrow xa=x $ (right multiplication by $x$) $\Rightarrow a = e$ (left multiplication by $x^{-1}$). Therefore $f$ is indeed injective, because $\operatorname{Ker}(f) = \{e\}$. Here is the desired bijection.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Given the points $A,B,C,D$ in a straight line $m$ and $A,E,F,G$ in a straight line $n$, how many triangles can be formed with these points? Given the points $A,B,C,D$ in a line $m$ and $A,E,F,G$ in a straight line $n$, how many triangles can be formed with these points?
I've done the following:
I've used the following heuristic: Chose one point on $m$, then another point on $m$ which is different of the previous chosen point. This yields $4*3$, now choose a new point on $n$, this gives me $4*3*3=36$, now choose one point on $n$ and another point on $n$ different of the last point chosen and different of $A$. This gives me $3*2*3=12$ and hence $36+12=48$.
I've also tried to do it in the following way:
$${\text{#points in m} \choose 2}*\text{#points in n (w/o A)}+ {\text{#points in n (w/o A)} \choose 2}*\text{#points in m (w/o A)}$$
Then:
$${4 \choose 2}*3 + {3 \choose 2}*3=27$$
But none of them is right. I may be missing some triangles but I cant see which ones are these.
|
Here is an example,
If you choose BEGC: one line from BE, one from GC, and one from EG--> you have one more triangle. Do the same for the rest and I think you may get the result.
|
{
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"url": "https://math.stackexchange.com/questions/1022402",
"timestamp": "2023-03-29T00:00:00",
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|
Best approximation for a normed vector space $X$ I am self-studying functional analysis. As far as I know, a best approximation of $X$ by a closed subspace $C \subseteq X$ exists and is unique if $X$ is a Hilbert space, a uniformly convex Banach space, or more generally, a reflexive space.
My question is: is there a even more general condition on the space $X$ for the existence and uniqueness of the best approximation of $X$ by a closed and convex subspace $C \subseteq X$?
Any comments or suggested readings are welcome.
|
Sharper results: Best Approximation by Closed Sets in Banach Spaces and On a sufficient condition for proximity by Ka Sing Lau.
Quote from the first:
The set $K$ is called proximinal (Chebyshev) if every point $x\in X$ has a (unique) best approximation from $X$. lt is easy to see that every closed convex set $X$ in a reflexive space $X$ is proximinal. In addition, if the norm is strictly convex, then $X$ is Chebyshev. However, if $X$ is not assumed reflexive or $X$ is not assumed convex, then the above result is false in general. In [7] Steckin introduced the concept of almost Chebyshev. A set $K$ is called almost Chebyshev if the set of $x$ in $X$ such that $K$ fails to have unique best approximation to $x$ is a first category subset of $X$. He proved that if $X$ is a uniformly convex Banach space, then every closed subset is almost Chebyshev. By using this concept, Garkavi [4] showed that for any reflexive subspace $F$ in a separable Banach space, there exists a (in fact, many) subspace $G$ which is $B$-isomorphic to $F$ and is almost Chebyshev. The author [6] showed that if $X$ is a separable Banach space which is locally uniformly convex or possesses the Radon-Nikodym property, then "almost all"
closed subspaces are almost Chebyshev. In [3], Edelstein proved that if $X$
has the Radon-Nikodym property, then for any bounded cIosed convex
subset $K$, the set of $x$ in $X$ which admit best approximations from $K$ is a
weakly dense subset in $X$.
|
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|
Certain condition on an entire function implies the function is constant ? Let $f=u+iv$ be an entire function such that $v \ge 2u+1$ , then is it true that $f$ is constant ?
|
Just to throw another (almost equivalent) method in, $0$ is not in the image of the function and there is $r>0$ such that the disc $D_r=\{z\ :\ |z|<r\}$ is not either.
For example, take $r=1/4$: if $|z|<1/4$, then, $z=x+iy$ with $|x|,|y|<1/4$, so
$3/4\geq|y-2v|$, hence it is impossible that $y\geq 2x+1$ when $x+iy=z\in D_{1/4}$.
Now, take $g=1/f$. This is a holomorphic entire function. We know that $|f|>1/4$, therefore $|g|<4$, so by Liouville's theorem, $g$ is constant and so is $f$.
|
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|
How do I prove that a matrix is a rotation-matrix? I have to prove that this matrix is a rotation-matrix
$$\begin{pmatrix} \frac12 & 0 & \frac{\sqrt{3}}{2} \\
0 & 1 & 0 \\
\frac{\sqrt{3}}{2} & 0 & \frac12
\end{pmatrix}$$
How do I do this?
My idea is to multiplicate it with $\begin{pmatrix} x \\ y \\ z\end{pmatrix}$ and show that one component will remain unchanged . Is this enough? Do non-rotational transformations exist, which leave one component unchanged ?
|
I think there is a minus sign missing. As it is, the determinant is not $1$. After fixing, this specific case is easy.
$$\begin{pmatrix} \frac12 & 0 & -\frac{\sqrt{3}}{2} \\
0 & 1 & 0 \\
\frac{\sqrt{3}}{2} & 0 & \frac12
\end{pmatrix} = \begin{pmatrix} \cos \frac{\pi}{3} & 0 & -\sin \frac{\pi}{3} \\ 0 & 1 & 0 \\ \sin \frac{\pi}{3} & 0 & \cos \frac{\pi}{3} \end{pmatrix}$$
It is a rotation of $\pi/3$ around the $y$-axis.
|
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"timestamp": "2023-03-29T00:00:00",
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|
A finite semi-group with cancellation laws is a group . I don't understand how to get this:
A finite semigroup with cancellation laws is a group.
Thanks in advance for any help.
|
Let $S$ be a such semigroup. For each $a\in S$, we can find a positive integer $n(a)$ such that $a^{n(a)+1}=a$. If such $n(a)$ does not exist, then $a^1,a^2,a^3,\cdots$ are all different and is contradicting that $S$ is finite.
We will argue that, for each $a,b\in G$, $a^{n(a)}=b^{n(b)}$. Since
$$aa^{n(a)}b=a^{n(a)+1}b=ab=ab^{n(b)+1}=ab^{n(b)}b$$
so by (left and right) cancellation law, we get $a^{n(a)}=b^{n(b)}$.
Let define $e=a^{n(a)}$ then $ea=a$ and $ae=a$ for all $a\in S$. If we define $a^{-1}:=a^{n(a)-1}$ then you can check that $a^{-1}$ is an inverse of $a$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Prove $ \lim_{x\to -\infty } \frac{x+8}{x+3} = 1 $ using only the definition of a limit I need to prove this limit:
$$ \lim_{x\to -\infty } \frac{x+8}{x+3} = 1 $$
I started with:
$|\frac{x+8}{x+3}-1|< ϵ $
$|\frac{x+8-x-3}{x+3}|=|\frac{5}{x+3}|=\frac{5}{|x+3|} <ϵ$
$\frac{5}{\epsilon} < |x+3|$
How do I proceed from here? how the hell can I extract the expression inside the absolute value? Am I in the wrong direction?
Thanks a lot!
|
$$\begin{align*}
\frac5\epsilon &< \left|x+3\right|\\
\frac5\epsilon &< -\left(x+3\right)\\
x &< -\frac5\epsilon-3
\end{align*}$$
Hence, given $\epsilon>0$, we can take $M= -\frac5\epsilon-3$ such that $x<M$ implies $$\left|\frac{x+8}{x+3}-1\right|<\epsilon$$
|
{
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"url": "https://math.stackexchange.com/questions/1022853",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Domain, range and zeros of $f(x,y)=\frac{\sqrt{4x-x^2-y^2}}{x^2y^2-4xy^2+3y^2}$ Given the following function with two variables:
\begin{equation}
\frac{\sqrt{4x-x^2-y^2}}{x^2y^2-4xy^2+3y^2}
\end{equation}
I need to find a) the domain for the above function.
Can anyone give me a hint on how to find the domain in f?
I already know that:
\begin{equation}
dom f = \{ f(x,y) ∈ \mathbb{R}^2\backslash(x^2y^2-4xy^2+3y^2=0) : \sqrt{4x-x^2-y^2} \geq 0 \}
\end{equation}
But of course this needs to be written in a more simpler form.
During class we solve simpler functions like without fractions and roots, so I don't have anything that can help me.
After this I also need to find:
*
*b) zeros of the function
*c) Calculating Algebraically, the range of the function:
\begin{equation}
T(x,y) = \sqrt{4x-x^2-y^2}
\end{equation}
*d) Extreme values of the function T
I'm of course not expecting the complete solution but something like a kick start.
|
Can anyone give me a hint on how to find the domain in f?
You need to find all the values for which $x^2 y^2-4xy^2+3y^2$ is not equal to zero and $4x-x^2-y^2$ is positive or zero.
zeros of the function
Well you just need to find where $\sqrt{4x-x^2-y^2}=0$, which is just where $4x-x^2-y^2=0$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1022924",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 2
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|
Find Z Rotation based on X and Y Vector I have a vector $(x,y) = (x_2 - x_1, y_2 - y_1)$. I have an arrow pointing to 0 degrees.
With vector $(x, y)$, how can I find the number of degrees (0 - 360) that will be the direction the arrow points in order for it to point from $v_1 = (x_1, y_1)$ to $v_2 = (x_2, y_2)$?
|
I am assuming that the arrow 'pointing to 0°' is pointing in $x$-direction of the coordinate system.
Then $\frac{y}{x} = \tan(\alpha-180°)$ where $\alpha$ is the angle you are looking for, because the vector $(x,y)$ is poiting from $v_2$ to $v_1$. That means $y/x$ is the slope of the line that goes throu $v_1$ and $v_2$. But you want the other direction thats why you have to substract (or add) 180°.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1023000",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
Calculate limit without L'Hopital I need a some help with this.
Calculate:
$$\lim_{x\to\infty}\frac{\ln(x-1)}x$$
I know the answer is zero. But dont know how to handle the $\ln(x-1)$
|
We will prove instead that $\frac{\log x}{x} \to 0$, and leave you to deduce what you need. Suppose $x \geq e$ (here $e$ is the base of the logarithm). Then
$$\frac{\log (ex)}{ex} = \frac{\log x + 1}{ex} \leq \frac{2}{e} \frac{\log x}{x}, $$
since $e \leq x$ implies $1 \leq \log x$ (since $\log x$ is monotone). Monotonicity of $\log x$ and $x$ shows that for $e \leq x \leq e^2$ we have
$$ \frac{\log x}{x} \leq \frac{\log (e^2)}{e} = \frac{2}{e}. $$
Take now an arbitrary $x$ and divide it by $e$ enough times so that you obtain something in the range $[e,e^2]$. You can express the result by $x = e^n t$, where $e \leq t \leq e^2$ and $n \geq 0$. The above two calculations show that
$$ \frac{\log x}{x} \leq \left(\frac{2}{e}\right)^{n+1}. $$
As $x\to\infty$, the corresponding power $n$ also tends to infinity, and so the ratio $\frac{\log x}{x}$ tends to zero.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1023071",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 2
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|
How to find all of generators in a finite fields How can I find all generators of a finite field?
For example in GF(2^3) and X^3 + x^2 + 1 as primitive polynomial.
I don`t want all of solutions. I need some hint and help to solve this problem.
Thanks a lot.
Ya Ali.
|
the multiplicative group of $GF_{2^3}$ has prime order, so that tells you how many generators it has. also note what is the only subfield of $GF_{2^3}$ (a generator cannot lie in a subfield). and finally note that there are precisely two irreducible polynomials of degree $3$ over $GF(2)$. try computing the first seven powers of $\alpha$ where $1+\alpha^2+\alpha^3=0$, expressing each as a linear combination of $1, \alpha, \alpha^2$ with coefficients in $GF(2)$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1023166",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
}
|
How does $\sum_{n = 1}^{\infty} \frac{1}{n(n+1)}$ simplify? $\sum_{n = 1}^{\infty} \frac{1}{n(n+1)} = \frac{1}{1\cdot2} + \frac{1}{2\cdot3} + ...$
$= 1 - \frac{1}{2} + \frac{1}{2} - \frac{1}{3} + ... = 1$
My professor wrote this the other day. But I'm wondering...how does the series become $= 1 - \frac{1}{2} + \frac{1}{2} - \frac{1}{3} + ...$?
|
It is because $\frac{1}{n\left(n+1\right)}=\frac{1}{n}-\frac{1}{n+1}$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1023259",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
Presentation $\langle x,y,z\mid xyx^{-1}y^{-2},yzy^{-1}z^{-2},zxz^{-1}x^{-2}\rangle$ of group equal to trivial group
Problem: Show that the group given by the presentation $$\langle x,y,z \mid xyx^{-1}y^{-2}\, , \, yzy^{-1}z^{-2}\, , \, zxz^{-1}x^{-2} \rangle $$ is equivalent to the trivial group.
I have tried all sorts of manners to try to show that the relations given by the presentation above imply that $x=y=z=e$. However, I am stuck and would appreciate any hints as to how I should move forward.
|
This is a very well-known presentation of the trivial group, to be compared with the presentation of Higman's infinite group with no finite quotient. I do not know of any easy proof.
The proof I'm going to give is due to Bernhard Neumann in An Essay on Free Products of Groups with Amalgamations (Philosophical Transactions of the Royal Society of London, Series A, 246, 919 (1954), pp. 503-554.)
So we have to prove that in a group satisfying
$$ xyx^{-1} = y^2 \qquad (R_1)$$
$$ yzy^{-1} = z^2 \qquad (R_2)$$
$$ zxz^{-1} = x^2 \qquad (R_3)$$
the elements $x$, $y$ and $z$ are trivial.
By inverting $(R_1)$, multiplying on the left by $y$ and on the right by $x$, we get $$yxy^{-1} = y^{-1}x.$$ This easily gives $$y^i x y^{-i} = y^{-i} x \qquad(R_1^{[i]}),$$ for every integer $i$, by induction. The same argument on the second relation gives $$z^i y z^{-i} = z^{-i} y. \qquad (R_2^{[i]})$$
If we now conjugate $(R_3)$ by $y$, the left-hand side becomes
$$\begin{align}yzxz^{-1}y^{-1} &= z^2y\cdot x \cdot y^{-1}z^{-2}\\
& = z^2y^{-1}xz^{-2}\\
& = z^2 y^{-1}z^{-2}\cdot z^2xz^{-2}\\
& = y^{-1}z^2\cdot x^4
\end{align}$$
(the first equality is a double use of the relation $yz = z^2y$, a reformulation of $(R_2)$ ; the second uses $(R_1^{[1]})$ and the last uses the inverse of $(R_2^{[2]})$ and $R_3$ twice).
On the other hand, the left side becomes
$$\begin{align} yx^2y^{-1} &= (y^{-1}x)^2 \\
&= y^{-1}xy^{-1}x \\
&= y^{-3}x^2
\end{align}$$
(the first equality uses the $(R_1^{[1]})$ twice, the last uses the inverse of $(R_1)$).
Put together, we have proven $y^{-1}z^2x^4 = y^{-3}x^2$, which gives
$$z^2 = y^{-2}x^{-2}.\qquad (R^*)$$
If we conjugate $y$ by $z^{-2}$, we now get on the one hand
$$\begin{align}z^{-2}y z^2 &= x^2 y^2 \cdot y \cdot y^{-2} x^{-2} \\
&= x^2 y x^{-2} \\
&= y^4
\end{align}$$
(the first equality uses $(R^*)$ twice, the last uses $(R_1)$ twice.) But, on the other hand, $z^{-2}yz^2 = z^2 y$ because of $(R_2^{[-2]}$). So we finally get $z^2 y = y^4$, which translates to
$$z^2 = y^3.$$
This proves that $y$ and $z^2$ commute. The relation $(R_2)$ then boils down to $z = z^2$, which gives $z = 1$. Because of the symmetries in the presentation, this proves that the group is trivial.
Not very enlightening, but the fact that the corresponding group with 4 generators is highly nontrivial somehow reduces my hopes of ever finding a "good reason" for this group to be trivial.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1023341",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "23",
"answer_count": 2,
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|
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