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prove inequation a,b,c,d $\in \mathbb{R}$ $a,b,c,d \gt 0$ and $ c^2 +d^2=(a^2 +b^2)^3$ prove that $$ \frac{a^3}{c} + \frac{b^3}{d} \ge 1$$
If I rewrite the inequation like $ \frac{a^3}{c} + \frac{b^3}{d} \ge \frac{c^2 +d^2}{(a^2 +b^2)^3}$ and manage to simplfy it brings me nowhere. I try with
Cauchy-Schwarz Inequality but still can not solve it.
I would very much appreciate it if anyone could help me
|
Using Titu's Lemma, we have
$$ \dfrac{a^3}{c} + \dfrac{b^3}{d} \ge \dfrac{(a^2+b^2)^2}{ac+bd}$$
So, we are left to prove that
$$ (a^2+b^2)^2 \geq (ac+bd)\tag{1} $$
Using Cauchy-Schwarz inequality, we have
$$(a^2+b^2)(c^2+d^2) \geq (ac+bd)^2\tag{2}$$
Using the given proposition, $$c^2+d^2 =(a^2+b^2)^3$$ in $(2)$ and the fact that $a,b,c,d \in \mathbb{R^+}$, we get $(1)$, which was required to be proved.
|
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How to determine which of the following matrices are similar? If we have the following three matrices:
$$
A=\begin{bmatrix}
7 &1 \\
-5 &3
\end{bmatrix},\;\;
B=\begin{bmatrix}
5 &-1 \\
1 &5
\end{bmatrix},\;\;
C=\begin{bmatrix}
5 &1 \\
1 &5
\end{bmatrix}.
$$
What is the right procedure to determine if matrices are similar? I know that if the matrices are similar, then the matrices have the samen eigenvalues. Using the negative of this (determine the eigenvalues of each matrix, if they are not similar they cannot be similar), I got that only $A$ and $B$ potentially could be similar.
I also know that if the 2 matrices $A$ and $B$ are similar then $A=PBP^{−1}$ but I don't know how to work this out further.
So how do I show that matrices are similar, using $A=PBP^{-1}$ for example to show it for $A$ and $B$?
|
$A$ and $B$ cannot be similar to $C$, since $\det(A)=\det(B)=26\neq \det(C)=24$.
On the other hand $A$ and $B$ are similar with
$$
P=\begin{pmatrix} 2 & 2 \cr 6 & 2 \end{pmatrix}.
$$
Here we have $B=PAP^{-1}$. I do not need eigenvalues for this computation. I start with an invertible matrix $P=\begin{pmatrix} s_1 & s_3 \cr s_2 & s_4 \end{pmatrix}$ and then rewrite $BP=PA$ as
$$
2s_1 + s_2 - 5s_3=s_1 - 2s_3 + s_4= - s_1 + 2s_2 - 5s_4=s_2 - s_3 - 2s_4=0.
$$
The determinant condition is $s_1s_4-s_2s_3\neq 0$. Then we can solve the linear equations and take one with nonzero determinant.
|
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For matrices $C, D$, show that $(CD)^{100} \neq C^{100} D^{100}$ The question is to prove this is false:
$(CD)^{100} = C^{100}\cdot D^{100}$, where $C$ and $D$ are matrices.
I looked through my textbook and could not find a proof for this.
|
A matrix corresponds to some (linear) operation on a vector. Let's take 2-component vectors and actions $C$ "make first component zero" and $D$ "swap components". Then $(CD)^{100}$ is "swap components, then make first component zero, repeat $100$ times" (which gives you zero vector, obviously) while $C^{100}D^{100}$ is "swap components $100$ times, then make first component zero $100$ times" (which is the same as $C$).
Now, how to get a matrix by operation description? Easy: first column is the result of operation on vector $\pmatrix{1 \\ 0}$, second column is the result of operation on vector $\pmatrix{0 \\ 1}$. Since $C \pmatrix{1 \\ 0} = \pmatrix{0 \\ 0}$, $C \pmatrix{0 \\ 1} = \pmatrix{0 \\ 1}$, we have $C = \pmatrix{0 & 0 \\ 0 & 1}$; likewise, $D = \pmatrix{0 & 1 \\ 1 & 0}$.
Easy to see that $D^{100} = \pmatrix{1 & 0 \\ 0 & 1}$, $C^{100} = C$, $CD = \pmatrix{0 & 0 \\ 1 & 0}$ and $(CD)^{100} = \pmatrix{0 & 0 \\ 0 & 0} \neq C^{100}D^{100}$.
|
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In an $\Bbb{N}$-graded domain $A$, units are homogeneous
Let $A$ be a graded domain, with additive subgroups $A_n,\,\forall\,n\geq 0$, s.t. ${A_n\cdot A_m}\subseteq A_{n+m}\,\forall\,n,m\geq 0$, and $A=\bigoplus_{n=0}^\infty\, A_n$ as abelian groups. I wish to show that if $a, a^{-1}\in A$, then $a\in A_0$.
What I've gotten so far is that we must have $a=\sum_n\,a_n$ and $a^{-1}=\sum_n\,b_n$, $\,a_n, b_n\in A_n \,\forall\,n$, and therefore $1=\sum_n\,\delta_n = \sum_n \sum_{i+j=n} a_i b_j \implies \sum_{i+j=n} a_i b_j = \delta_n$, so $a_0, b_0$ are units, but I can't get any farther than this.
|
Since $1 \in A_0$, every $\delta_n$ with $n \ge 1$ must be zero. The highest degree $\delta_i$ will be the product of the highest degree $a_i$ and the highest degree $b_i$, and since $A$ is a domain this is non-zero. Therefore since the degrees add the highest $a_i$ and $b_i$ must both be in degree $0$, i.e. $a$ and $a^{-1}$ are homogeneous degree $0$.
|
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Question about simplification in summation I am reading a book where the following example is shown.
$$= \sum_{0\le n-j\le n} (a+b(n-j)) $$
$$= \sum_{0\le j\le n} (a+bn-bj) $$
Why is n-j being simplified to j? I don't understand why this is possible? To specify my question, what rule of simplification is being used for the part under the summation sign.
|
It should be clear that $a + b(n-j) = a +bn - bj$, so it is really just a question of how
$$
\sum_{0\leq n-j \leq n} = \sum_{0\leq j\leq n}
$$
First: I assume that $n$ is a fixed number and the the index variable is $j$. So you just have to convince yourself that when $j$ ranges over an interval such that $0\leq n-j \leq n$, then you get the same as $0\leq j \leq n$. And
$$\begin{align}
&0\leq n-j \leq n \quad &\Rightarrow \\
&0 \geq j-n \geq -n \quad&\Rightarrow \\
&n \geq j \geq 0
\end{align}
$$
|
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What is the meaning of $\Bbb{Q}[x]/f(x)$? I am very confused with the meaning of $\Bbb{Q}(x)/f(x)$. Does it mean the set of all polynomials modulo $f(x)$?
If it does then how can we say that $\Bbb{R}[x]/(x^2+1)$ is isomorphic to set of complex numbers?
|
Yes that's what it means. Take the isomorphism $\phi: \mathbb{R}[x]/(x^2+1)\rightarrow \mathbb{C}$ by $\phi: (a+bx) \mapsto (a+bi)$. I'll leave it to you to show that it is an isomorphism.
|
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Is it possible to convert the polar equation $\ r = k \cos (\theta n) + 2$ into cartesian form?
Is it possible to convert the polaer equation
$$\ r = k \cos (\theta n) + 2$$
into cartesian form? Here, $k$ is some constant and $n$ is any positive whole number greater than $2$.
The farthest that I managed to get was:
$$r^2 = kr \cos(\theta n) +2r$$
$$\to\qquad r^2 - 2r - kr\cos(\theta n) = 0$$
$$\to \qquad x^2 +y^2 - 2 \sqrt{x^2 +y^2} -kr \cos (\theta n) = 0$$
I noticed that $\cos(\theta n)$ is reminiscent of the Chebychev polynomials and figured that $r\cos(\theta n)$ could also be generalized into polyomials in terms of $x$, but that turns out to be impossible.
|
If you want an expression that holds for all $n \geq 2$ then it might be a tad ugly in Cartesian, because:
$$\cos(n\theta) = \sum_{\text{even }k} (-1)^{k/2}{n \choose k}\cos^{n-k} \theta \sin^k \theta \\ = (x^2+y^2)^{-n/2}\sum_{\text{even }k} (-1)^{k/2}{n \choose k} x^{n-k}y^k.$$
But this will get you there.
|
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Cardinality of subsets with finite intersections Let $\ F_0 $ be a family of disjoint subsets of $ C$. $\ |C|= \aleph_0$.
Prove that $\ (*) |F_0|\leq\aleph_0 $.
This part was relatively simple, in the presence of choice an injection can be defined from each set to one of its elements in C. By Cantor the conclusion is simple.
I'd like to show that $\ |F_1|\leq\aleph_0$ where $\ F_1 $ is a family of subsets of $C$ in which for any two distinct $\ A,B\in F_1$, $A\cap B \leq 1 $.
From here I'd like to inductivly prove $\ |F_n|\leq\aleph_0$ where $\ F_n $ is a family of subsets of $C$ in which for any two distinct $\ A,B\in F_n$, $A\cap B \leq n $.
I'd like to isolate the intersections and show each family as a disjoint family and use $ (*) $ from above. My problem is properly defining an injection.
|
Hint (for both the $n=1$ case and the inductive step): Suppose $F_n$ is uncountable, and let $F_n(x)=\{A\in F_n:x\in A\}$ for each $x\in C$. Show that $F_n(x)$ must be uncountable for some $x\in C$, and then apply the induction hypothesis to $\{A\setminus\{x\}:A\in F_n(x)\}$.
|
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Intermediate Value Property Justification I am working on showing the Intermediate Value Property holds for a certain function. I noticed the one page on here talks about this idea, but I can not follow with what they are trying to say. [The problem is posted below - Ed.] . Any guidance would be appreciated. I just am very unsure what the property says.
|
So if you believe the intermediate value theorem, then you know that if $x_1, x_2 > 0$ or $x_1, x_2 < 0$, the property follows. So suppose $x_1 \leq 0 < x_2$. It's enough to show that there exists $c \in (0, x_1)$. To do this, consider the sequence $$(t_n)_{n \in \mathbb{n}} = \left( \frac{1}{ 2 \pi n + \arcsin (k) } \right)_{n \in \mathbb{N}} ,$$ along which $f(t_n) = k$. Let $n = \lceil 2 \pi n / x_2 \rceil + 1$. Then $t_n = \frac{1}{ 2 \pi n + \arcsin (k) } \leq \frac{1}{ 2 \pi (n - 1) } \leq x_2$.
|
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Is the following integral identity true or not? Is the following statement true or not?$$\int_{-\infty}^\infty xf(x)\,dx = \left. {d\over{dt}} \int_{-\infty}^\infty e^{tx}f(x)\,dx\right|_{t = 0}$$
|
The statement is if Leibnitz's rule can be applied to $e^{tx}f(x)$ around $t=0$. Specifically in this case you need an integrable function $\theta$ such that $|xe^{tx}f(x)|\leq \theta(x)$ for $t$ near $0$.
|
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Regarding element-wise derivative of matrices
Let $X=(x_{ij})$ be a real n-by-n matrix where $x_{ij}$ are in a range such that $X$ is insvertible. Let $Y=X^{-1}=(y_{pq})$, and regard $y_{pq}=y_{pq} (x_{11},x_{12},\ldots,x_{nn})$ as a function of $x_{ij}$. Prove that $\frac{\partial y_{pq}}{\partial . x_{ij}}=-y_{pi}y_{jq}.$.
Any solution to prove this in a right way would be appreciated. And is there any name for this type of derivative? How can I handle it?
And the followings are some non-sense scribbles I've made
$XY=I$
${d(XY)\over dx}=IY+X{dY\over dx}=O$
${\partial Y\over \partial x_{ij}}=-X^{-1}Y=-Y^2$
$\frac{\partial y_{pq}}{\partial x_{ij}}=-y_{pi}y_{jq}.$.
Is there any why this can be justified?
|
I guess the argument is cleaner if you use:
$$
0=\frac{\partial (XY)}{\partial x_{ij}} =\frac{\partial X}{\partial x_{ij}}Y+ X\frac{\partial Y}{\partial x_{ij}}
$$
So
$$
\frac{\partial Y}{\partial x_{ij}}=-X^{-1}\frac{\partial X}{\partial x_{ij}}Y=-Y\frac{\partial X}{\partial x_{ij}}Y
$$
So taking the $(l,k)$ matrix entry in the previous equation we get:
$$
\frac{\partial y_{lk}}{\partial x_{ij}}=-\sum_{m,n} y_{ln}\left(\frac{\partial X}{\partial x_{ij}}\right)_{nm}y_{mk}
$$
And $\left(\frac{\partial X}{\partial x_{ij}}\right)_{nm}=\delta_{in}\delta_{jm}$ so:
$$
\frac{\partial y_{lk}}{\partial x_{ij}}=-\sum_{m,n} y_{ln}\delta_{in}\delta_{jm}y_{mk}=-y_{li}y_{jk}
$$
|
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Well-ordering principle and theorem Could somebody clearly explain the difference between the well-ordering principle and the well-ordering theorem? Is one of these related to the Principle of Mathematical Induction, and the other to the Axiom of Choice? Thanks.
|
The "well-ordering theorem" is the statement that for any set $X$, there is a relation $<$ on $X$ which is a well-ordering. This statement is equivalent to the axiom of choice.
The "well-ordering principle" has (at least) two different meanings. The first meaning is just another name for the well-ordering theorem. The second meaning is the statement that the usual relation $<$ on the set $\mathbb{N}$ is a well-ordering. This statement is equivalent to the statement that ordinary induction on the natural numbers works.
|
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Inequality : $\sum_{k=1}^n x_k\cdot \sum_{k=1}^n \frac{1}{x_k} \geq n^2$ I have to show the inequality of $$\left(\sum_{i=1}^n x_i\right)*\left(\sum_{i=1}^n \frac{1}{x_i}\right) \geq n^2.$$For $x_1, ... x_n \in \mathbb{R_{>0}}$ and $ n \geq 1$. I wanted to show this inequality by induction. The basis is clear, but I am not sure how to do the inductive step. Can someone help me with that?
|
Using Induction on $n$, $\sum_{i=1}^{n+1}x_i$.$\sum_{i=1}^{n+1}1/x_i$$\geq n^2+x_{n+1}$$(\frac{1}{x_1}+$$\frac{1}{x_2}+....+$$\frac{1}{x_n}$)$+$$\frac{1}{x_{n+1}}$$(x_1+x_2+......+x_n)$
$=n^2+1+(\frac{x_{n+1}}{x_1}+\frac{x_1}{x_{n+1}})+....+ (\frac{x_{n+1}}{x_n}+\frac{x_n}{x_{n+1}})$$\geq n^2+1+2n=(n+1)^2$
*
*Note: The minimum value of each term of type $(\frac{x_{n+1}}{x_i}+\frac{x_i}{x_{n+1}})$; $1\leq i\leq n$ is $2$ as you can see that $f(x,y)=\frac{x}{y}+\frac{y}{x}$ in one variable $z=\frac{x}{y}$ gives $f(z)=z+\frac{1}{z}$; $z>0$. With the help of second derivative test, you can easily verify that $f(z)$ has minimum value $2$ at $z=1$.
|
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Limit of $\frac{\pi^h-1}{h}$ as h approaches zero Can someone help me find this limit here. I only know how to use L'Hospital's rule but I want to be able to evaluate this limit without using differentiation.
$$\lim \limits_{h \to 0} \frac{\pi^h-1}{h}$$
The reason I want this limit is because just like $e$ can be expressed as $\sum_{n=0}^{\infty}\frac{1}{n!}$ I want to find a way to do the same with $\pi$ so i want to find the dervivative of $\pi^x$ without having $\pi$ in the result.
|
$$\lim_{x\to 0} \frac{e^{hln\pi}-1}{h} = (H) = $$
$$\lim_{x\to 0} \frac{ln{\pi} * lne^{hln\pi}}{1} = ln \pi $$
|
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Prove by induction $3+3 \cdot 5+ \cdots +3 \cdot 5^n = \frac{3(5^{n+1} -1)}{4}$ My question is:
Prove by induction that $$3+3 \cdot 5+ 3 \cdot 5^2+ \cdots +3 \cdot 5^n = \frac{3(5^{n+1} -1)}{4}$$ whenever $n$ is a nonnegative integer.
I'm stuck at the basis step.
If I started with $1$. I get the right hand side is $18$ which is clearly not even close. It says prove shouldn't it be always true?
|
Helping out with the problem.
I'm stuck at the basis step.
$p(n)$: $3+3 \cdot 5+ 3 \cdot 5^2+ \cdots +3 \cdot 5^n = \dfrac{3(5^{n+1} -1)}{4}$. Where $n \in \{0, 1, 2, \dots \}$.
We can rewrite the predicate. $p(n)$: $\sum_0^n3\cdot5^n = \dfrac{3(5^{n+1} -1)}{4}$
Base case:
Here you need to start at 0 because we only "induct" upwards.
$p(0): \sum_0^03\cdot5^n = \dfrac{3(5^{n+1} -1)}{4}$
$\implies 3 = \dfrac{3(5^1 -1)}{4} = 3 \checkmark$
Now it's up to you to show that $p(n) \implies p(n+1)$ and interpret the results.
|
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Finding the possible positions of chess knight mathematically relative to a given position From this website I found the following question:
A chess board’s 8 rows are labelled 1 to 8, and its 8 columns a to h. Each square of the board is described by the ordered pair (column letter, row number).
(a) A knight is positioned at (d, 3). Write down its possible positions after a single move of the knight.
(b) If R = {1, 2, ..., 8}, C = {a, b, ..., h}, and P = {coordinates of all squares on the chess board}, use set notation to express P in terms of R and C
(c) A rook is positioned at (g, 2). If T = {2} and G = {g}, express its possible positions after one move of the rook in terms of R, C, T and G.
For (a) is there anyway mathematically to list all the possible positions of the knight? I could for example, use a chess board to determine all the possible positions of the knight.
About (C) I completely don't get it.
The answers are provided in the website, however, I'm interested in how to solve them.
|
Given an initial position for the knight, the knight can move either north-east, north-west, south-east or south-west. In each of these directions, there are usually two possible positions. For example, from $(d,3)$, the knight can move north-east to either $(e,5)$ or $(f,4)$. Thus, there are a total of 8 possible positions for the knight if the knight starts at $(d,3)$.
Since a knight move consists of jumping 3 squares in the form of an L-shape, mathematically, the 8 possible new positions for the knight can be obtained by just adding or subtracting 1 to one of the two coordinates of $(d,3)$ and 2 to the other coordinate. For instance, adding $(1,2)$ to $(d,3)$ gives $(e, 5)$. Here, I assume that 'adding' 1 to a letter gives the next letter, and so adding 1 to $d$ gives $e$. Or, we can add $(2,1)$ to $(d,3)$, and we get $(f,4)$. Or, we can add $(-1,2)$ to $(d,3)$, and we get $(c,5)$. The number of different additions we can do is the size of the set $\{ (\pm 1, \pm 2), (\pm 2, \pm 1)\}$, which is 8.
Of course, if the knight is not near the center of the board, then the knight could fall off the board if it moves in some of these 8 directions. So, if the knight is near the edge or corner of the board, only some (not all 8) of the possible 'additions' should be carried out.
For part (b), $P= C \times R$ should be clear, where $C$ is the set of columns of the board and $R$ the set of rows.
For part (c), observe that a rook on $(g,2)$ can either move horizontally to one of the 7 squares $(a,2), (b,2), \ldots, (h,2)$ or vertically to one of the 7 squares $(g,1), (g,3), \ldots, (g,8)$. Hence, there are 14 possible new positions for the rook if the rook moves. To express this set of 14 elements in set-theoretic form, observe that there are 8 positions on row 2, namely the positions $(a,2), (b,2), \ldots, (h,2)$. In other words, row 2 is the set $\{(a,2), \ldots, (h,2)\} = \{a,b,\ldots,h\} \times \{2\} = C \times T$. Similarly, column 2 is the set $\{g\} \times \{1,\ldots,8\} = G \times R$. The set of positions in row 2 or column 2 is their union $(C \times T) \cup (G \times R)$. Observe that row 2 consists of 8 positions, column 2 consists of 8 positions, and their union $(C \times T) \cup (G \times R)$ consists of 15 positions (they have a common element $(g,2)$). If a rook must move, it cannot occupy its initial position $(g,2)$. This position is the unique element in the singleton set $\{ (g,2) \} = \{g\} \times \{2 \} = G \times T$. If you remove the element $(g,2)$ from the 15-element set, you get the expression $((C \times T) \cup (G \times R)) - (G \times T)$.
|
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Area of polygon of hyperbolic disc Let us consider hyperbolic disc. I use uniform tessellation {5,4}.Here 5 stands for pentagon, 4 for number of polygons sharing the same vertex.
{hyperbolic disc}
There exists formula which defines area of convex hyperbolic polygon:
$$A_{m} = \{ \pi(m-2) - (a(1)+...+a(m))\} \frac{1}{-K}$$
where $a(i)$ is interior angle. K is Gaussian curvature, which I define:
$$K=-\frac{1}{l^2}$$
If I apply this formula for pentagon polygon I obtain that area of each polygon is given by:
$$A_{5} = \frac{\pi}{2} l^2$$
So it follows that type of uniform tessellation together with Gaussian curvature define area associated with polygon. Is it correct?
Is it possible to have tessellations with {5,n} where n is not 4?
|
First I have to point you at a misconception you have, and I fear it will be confusing.
(I only understand just about half of it myself)
The disk model you use is the Poincare disk model (see https://en.wikipedia.org/wiki/Poincar%C3%A9_disk_model ) and has a fixed absolute distance and a fixed curvature of -1 (this curvature is just build in in the model)
off course you are free to use another distance scale with some formula $1a = i^2 c$ with $1a$ meaning the length of a segment of one absolute distance and $i^2 $ your distance measurement of which $c$ is an unit length.
And this will result in a change of curvature. (but I do hear different opinions on what the curvature then or even what the unit of curvature is. so hopefully somebody else will help you here.
In absolute measure the pentagon you describe has a fixed area of
$$A_{5} = \{ 3\pi - (5 \frac{\pi}{2}) \} = \frac{1}{2}\pi $$
And using the formulas at https://en.wikipedia.org/wiki/Hyperbolic_triangle#Trigonometry you can calculate the side lenghts
There is a hyperbolic tesselation {5,n} for every $n > 3 $
see for example https://en.wikipedia.org/wiki/Order-5_pentagonal_tiling and
https://en.wikipedia.org/wiki/Order-6_pentagonal_tiling
|
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Let $P(x)=ax^{2014}-bx^{2015}+1$ and $Q(x)=x^2-2x+1$ such that $Q(x)|P(x)$, find $a+b$
Let $P(x)=ax^{2014}-bx^{2015}+1$ and $Q(x)=x^2-2x+1$ be the polynomials where $a$ and $b$ are real numbers. If polynomial $P$ is divisible by $Q$, what is the value of $a+b$.
This is what I have tried so far: Since $Q(x)|P(x)$ we have $P(1)=0$, therefore $a-b+1=0$. Problem is because we cannot obtain system of equations, because polynomial $Q(x)$ has double root at $x=1$. From equation $a-b+1$ we cannot find $a+b$, so how to find out value of $a+b$ or $a^2-b^2$?
|
Since $Q$ divides $P$, we have that $x-1$ divides $P$ twice. Hence $1$ is a root of $P$ and of its derivative $P'(x)=2014ax^{2013}$. So $a-b^{2015}+1=0$ and $2014a=0$. Now we can solve to get $a=0$, but in that case $Q$ cannot divide $P$, since $P$ has degree $0$.
If we actually have $P(x)=ax^{2015}-bx^{2014}+1$ then its derivative is $2015ax^{2014}-2014bx^{2013}$ and we get instead the system $a-b+1=0, \, 2015a-2014b=0.$
|
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Feasible point of a system of linear inequalities Let $P$ denote $(x,y,z)\in \mathbb R^3$, which satisfies the inequalities:
$$-2x+y+z\leq 4$$ $$x \geq 1$$ $$y\geq2$$ $$ z \geq 3 $$ $$x-2y+z \leq 1$$ $$ 2x+2y-z \leq 5$$
How do I find an interior point in $P$?
Is there a specific method, or should I just try some random combinations and then logically find an interior point?
|
Clever guessing is ok, random guessing isn't advised. Since $x\geq 1$, $y\geq 2$ and $z\geq 3$. You should try some values that are close to the boundary, since then they are less likely to break the other conditions. So an $x$ value slightly greater than $1$, a $y$ value that is slightly greater than $2$, and a $z$ value slightly greater than $3$.
So an interior point would be $(1.5, 2.5, 3.5)$ since
$$-2(1.5)+(2.5)+(3.5)=3< 4$$ $$1.5 > 1$$ $$2.5>2$$ $$ 3.5 > 3 $$ $$(1.5)-2(2.5)+(3.5)=0 < 1$$ $$ 2(1.5)+2(2.5)-(3.5)=4.5 < 5$$
|
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Arrange black and white balls so that each pair of white balls is separated by at least two black balls I am trying to solve the following question:
How many linear arrangements of $m$ white balls and $(n-m)$ black balls are possible such that each pair of white balls is separated by at least two black balls? The white balls are indistinguishable among themselves and the black balls are indistinguishable among themselves.
Please provide the solution to this.
Thanks,
|
Let $x_0,x_1,\dots,x_m$ represent the number of black balls between the respective white balls. Specifically, $x_0$ is the number of black balls to the left of the first white ball. $x_1$ is the number of black balls between the first and second white ball, etc...
$\underbrace{\bullet}_{x_0}\circ\underbrace{\bullet\bullet\bullet}_{x_1}\circ\underbrace{\bullet\bullet\bullet}_{x_2}\circ\underbrace{\bullet\bullet\bullet\bullet}_{x_3}\circ\dots\circ\underbrace{\bullet\bullet\bullet}_{x_m}$
We know there are a total of $n-m$ black balls. Thus, $x_0+x_1+\dots+x_m=n-m$
Further, the condition that there must be at least two black balls between every white ball implies that for $i\in\{1,2,\dots,m-1\}$ you must have $x_i\geq 2$.
We recognize first that there is a bijection between valid arrangements of balls and solutions to the system. We try then to count how many integral solutions there are to the system:
$\begin{cases}x_0+x_1+\dots+x_m=n-m\\ 0\leq x_0\\ 2\leq x_1\\ 2\leq x_2\\ \vdots\\ 2\leq x_{m-1}\\ 0\leq x_m\end{cases}$
Make a change of variable: $y_0=x_0, y_m=x_m, y_i=x_i-2$ for all other $i\in\{1,2,\dots,m-1\}$
This is then the system:
$\begin{cases}y_0+y_1+\dots+y_m = n-3m+2\\ 0\leq y_0\\ 0\leq y_1\\ \vdots\\ 0\leq y_m\end{cases}$
This is of a known problem form. See for example this recent related post. Note that in this scenario it is as though there are $m+1$ buckets and $n-3m+2$ balls yet to place.
|
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Confused about notation ":=" versus plain old "=" Relating to sets, I find the following in a text book:
"...the set S := {1, 2, 3}".
The book has an extensive notation appendix, but the":=" notation is not included.
What exactly does ":=" mean, and how is it different from just "=", and how is it read? Many thanks for any help.
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The notation $A=B$ means $A$ is equal to $B$. The notation $A:=B$ means "Let $A=B$." It means you're saying what you will mean when you write $A$.
I suspect the $\text{“}{:=}\text{''}$ notation hasn't existed for more than about a half a century, so it's brand-new.
|
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Is $\frac {x^2 + 5x}{x} = x+5$? We are graphing functions in class and the function $f(x) = \frac {x^2 + 5x}{x}$, came up and our teacher simplified it to $x+5$ and graphed that with a hole in the function at $x=0$.
I started wondering, how in algebra can we say that $\frac {x^2 + 5x}{x} = x+5$, when the graphs of each function are not equal?
Thanks
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The key here is that the domain of the function $f$ is not all real numbers. It's $\Bbb R \setminus \{0\}$ (meaning all real numbers except $0$). So once you know that it should be clear that $\frac{x^2+5x}{x}$ is exactly equal to $x+5$ on that domain.
If we wanted to specify that, we'd write $f: \Bbb R \setminus \{0\} \to \Bbb R$ is given by $f(x)=x+5$. Then we'd immediately see that the domain of $f$ is $\Bbb R \setminus \{0\}$ and the codomain$^\dagger$ is $\Bbb R$. But often mathematicians and teachers are a little lazy and will just expect you to realize that sometimes when they write $A=B$, they mean $A=B$ on the largest domain where both $A$ and $B$ are defined.
$^\dagger$: If you've never heard of a codomain, don't worry about it exactly. It's related to the range of the function, but it's not super important to know to understand basic algebra.
|
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Find a six digit integer Find an integer with six different digits such that the six digit integer is divisible by each of its digits.
For example, find ABCDEF such that A, B, C, D, E and F all can divide the number ABCDEF.
Show your answer along with mathematical reasoning.
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$4\\
24\\
624\\
3624\\
183624$
2,4,8 are good to work with, because once you find three digits at the end, you can put whatever you want in the front.
3,6,9 are good to work with too, because if you find something that divides by 3 or 9, you can shuffle the digits and it still divides by 3 (or 9).
and 1 is just a no-brainer
|
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Operation of permutations on functions Let $P$ be the additive group of mappings from $\mathbf{Z}^n$ to $\mathbf{Z}$. For $f \in P$ and $\sigma \in \mathfrak{S}_n$ (the symmetric group of degree $n$) let $\sigma f$ be the element of $P$ defined by
$$\sigma f(z_1,\dots,z_n) = f(z_{\sigma(1)},\dots,z_{\sigma(n)}).$$
Is it not true that if $\sigma, \tau \in \mathfrak{S}_n$, we have
$$(\tau(\sigma f))(z_1,\dots,z_n) = \sigma f(z_{\tau(1)},\dots,z_{\tau(n)}) = f(z_{\sigma\tau(1)},\dots,z_{\sigma\tau(n)}) = (\sigma\tau)f(z_1,\dots,z_n)?$$
The book says that $\tau(\sigma f) = (\tau\sigma) f$.
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One way to dissolve the confusion here is to think of it this way. A point $(z_1,...,z_n)$ is a function $z:\{1,...,n\}\to Z$. Hence, the point $(z_{\sigma(1)},...,z_{\sigma(n)})$, is the composition $z\circ \sigma$. Now we can think of the domain of $f$ as the set of functions
$$Z^n\simeq \{z:\{1,...,n\}\to Z\}\;,$$
and this way you can rewrite the action on $f$, evaluated at a point $z$, via precomposition by $\sigma$.
$$\sigma f (z_1,...,z_n)=(\sigma f)(z):=f(z \circ \sigma).$$
Then
\begin{eqnarray} \tau(\sigma f)(z_1,...,z_n) &=& \tau(\sigma f)(z)
\\
&:=& (\sigma f)(z \circ \tau)
\\
&:=& f ((z\circ \tau)\circ \sigma)
\\
&=& f (z\circ (\tau\circ \sigma))
\\
&=:& (\tau \sigma)f(z)
\\
&=& (\tau \sigma) f(z_1,...,z_n)
\end{eqnarray}
Remark
Thanks to Malice Vidrine for pointing out a few mistakes on the original post.
Edit by Randy Randerson:
By definition, $\mathbf{Z}^n$ is the set of functions $z = (z_1,\dots,z_n)$ from $\{1,\dots,n\}$ to $\mathbf{Z}$. A permutation $\sigma \in \mathfrak{S}_n$ maps $\{1,\dots,n\}$ onto $\{1,\dots,n\}$. Therefore $z \circ \sigma \in \mathbf{Z}^n$. By definition $(\sigma f)(z) = f(z \circ \sigma)$, so $\tau(\sigma f) = (\tau \sigma)f$ as shown in your answer.
|
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Given $f(x,y)$ is a continuous function, Do these integrals equal? Given range $\{ 0 \le x \le 1, 0 \le y \le 1\}$
Do these integrals equal?
$\int_0^1(\int_0^y f(x,y)dx)dy = \int_0^1(\int_0^x f(x,y)dy)dx$
Well, the answer is no. It seems like the triangulars are different in LHS, RHS.
I don't understand $ (D_2)$. Here: $D_1 = \{(x,y) | 0 \le x \le 1, 0 \le y \le x\}$,
but why $D_2 = \{(x,y) | 0 \le x \le 1, x \le y \le 1\}$ ?
Especially this part of $D_2: x \le y \le 1$.
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"It seems like the rectangles are different in LHS, RHS."
In fact, they are not rectangles! $\int_0^1\int_0^x f(x, y)dydx$ takes $x$ to be from $x= 0 $to $x= 1$ and, for each $x, y$ from $y= 0$ to $y= x%. That is a **triangle** with vertices at %(0, 0), (1, 0), and (1, 1).$ $\int_0^1\int_0^y f(x, y)dxdy$ takes $y$ to be from $y= 0$ to $y= 1$ and, for each $y, x$ from $x= 0 $to $x= y$. That is a triangle with vertices at $(0, 0), (0, 1),$ and $(1, 1).$ The line $y= x$ is a diagonal of the square $0\le x\le 1$, $0\le y\le 1$. The first integral covers the triangle below that diagonal, the second the triangle above that diagonal.
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definition of derivative Definition:
A mapping $f:U\to \mathbb{R}^n$ from an open set $U\subset \mathbb{R}^m$ into $\mathbb{R}^n$ is differentiable at a point $a\in U$ if there is a linear mapping $A:\mathbb{R}^m\to \mathbb{R}^n$ described by an $n\times m$ matrix $A$ such that for all $h$ in an $\underline{\textbf{open}}$ neighborhood of the origin in $\mathbb{R}^m$, $$f(a+h)=f(a)+Ah+\epsilon(h)$$ where $\lim_{h\to 0}\frac{\|\epsilon(h)\|}{\|h\|}=0$
In this definition why do we need openness which I draw? Thanks!
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The idea is, if $f$ is differentiable at $a$, then "$f$ is approximately affine (linear plus a constant) close to $a$". The intuitive notion of "close to" generally translates as "in some open set". Note, incidentally, that the limit condition
$$
\lim_{h \to 0} \frac{\|\epsilon(h)\|}{\|h\|} = 0
$$
implicitly assumes $\epsilon(h)$ is defined in some open neighborhood of $0$.
If you assume less, i.e., that
$$
f(a + h) = f(a) + Ah + \epsilon(h)
\tag{1}
$$
for all $h$ in some set $V$ having $0$ as a limit point, and such that
$$
\lim_{n \to \infty} \dfrac{\|\epsilon(h_{n})\|}{\|h_{n}\|} = 0
$$
for every sequence $(h_{n})$ in $V$ that converges to $0$, you risk defining a condition that:
*
*Depends not only on $f$, but on the set $V$.
*Fails to capture the desired intuition of being "approximately affine" near $a$.
Think, for example, of $f(x, y) = |x|$ at $a = (0, 0)$. Condition (1) holds if $V$ is the $x$-axis, but $f$ isn't differentiable at $a$. "Worse" examples are easy to construct, e.g., functions that are discontinuous at $a$, but satsfy (1) if $V$ is an arbitrary algebraic curve through $a$.
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How to find operator with Fibonacci eigenvalues? How can I find the operator that satisfies this equation?
$$F_nx^n=Dx^n$$
Summing over $n$ we can rewrite this as $$\frac1{1-x-x^2}=D\frac1{1-x}$$
I am unsure whether this can be solved.
I am trying to solve it in a similar way to something like this:
$$nx^n=Dx^n$$ has solution $$D=x\frac{\mathrm d}{\mathrm dx}$$
I am not sure really how else to describe it. Other than the fact that $D$ can't really depend on $n$ which might make this equation unsolvable.
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Note: There are (at least) two ways of indexing the Fibonacci series. Below I work with the $F_0 = F_1 = 1$ convention rather than the $F_0 = 0, F_1 = F_2 = 1$ convention.
I assume you want $D$ to be a differential operator, given the tags. Then we have
$$D = \sum_{k = 0}^\infty f_k(x) \frac{d^k}{dx^k}$$
for some functions $f_k(x)$, and we want
$$F_n x^n = D x^n = \sum_{k = 0}^\infty f_k(x) \frac{d^k}{dx^k} x^n = \sum_{k = 0}^n (n \; P \; k) \; f_k(x) \; x^{n-k}$$
Let's suppose $f_k(x) = c_k x^k$ for some constant $c_k$, making this say
$$F_n x^n = \sum_{k = 0}^n (n \; P \; k) c_k x^n$$
or just
$$F_n = \sum_{k = 0}^n (n \; P \; k) c_k = \sum_{k=0}^n \frac{n!}{(n-k)!} c_k$$
This gives a recursive definition of the $c_k$. Now if we find the first few $c_k$ we see that they're fractions with denominators that look like k!, so let's try defining $a_k := k! c_k$ so that our expression becomes
$$F_n = \sum_{k=0}^n \frac{n!}{(n-k)! k!} k! c_k = \sum_{k=0}^n \binom{n}{k} a_k$$
Calculating the first several $a_k$ we see that they are
$$1, 0, 1, -1, 2, -3, 5, -8, 13, -21, \ldots$$
To me, that looks like $a_0 = 1$ and $a_k = (-1)^k F_{k-2}$ for $k > 0$. I imagine you can check this actually works using Pascal's identity or something, though I didn't actually check it. This would then give the formula
$$D = 1 + \sum_{k=1}^\infty \frac{(-1)^k F_{k-2}}{k!} \; x^k \; \frac{d^k}{dx^k},$$
at least assuming I got all the indices correct.
|
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$\binom{n}{k}$ is a "binomial coefficient;" $n \; P \; k$ is a "__________." If I want to search for information concerning $\binom{n}{k}$, I can't Google that symbol directly, nor can I search for something like "n C k" and get anything relevant, but because the term "binomial coefficient" exists it's possible to search for, say, "Catalan numbers in terms of binomial coefficients" or whatever. Conversely, if I want to write something involving $\binom{n}{k}$ in a fundamental way and make it discoverable to others' searches, I should make sure to include the term "binomial coefficient."
Is there a similar noun describing the function
$$n \; P \; k = \frac{n!}{(n-k)!}$$
Obviously one could use a rather verbose description, but that's not very useful for searchability -- imagine replacing the search described above with "Catalan numbers in terms of the number of combinations of k elements out of an n element subset," for instance, to which this would be analogous.
The best solution I can see at present is to use the term "binomial coefficient" and maybe rewrite the formula to include an extraneous $k!$ or something.
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This is a falling factorial:
$$ (n)_k = n^{\underline k} = \underbrace{n\cdot(n-1)\cdot(n-2)\cdots(n-k+1)}_{k\text{ factors}} = \frac{n!}{(n-k)!} $$
|
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Prove that if $f(x) = \int_{0}^x f(t)\,dt$, then $f(x) = 0$
Prove that if $f(x) = \displaystyle\int_{0}^x f(t)\, dt$ for all $x$, then $f(x) = 0$.
I first differentiated to get $f'(x) = f(x) - f(0)$. Then by the mean value theorem there exists a $c$ in $(0,x)$ such that $f'(c)=\dfrac{f(x)-f(0)}{x}$. Thus, $f'(x) = xf'(c)$. What do I do from here?
A summary of the deliberation in the comments about the necessity of assumptions on $f$:
There was worry that we needed assumptions on $f$ such as continuity in order to exploit FTC.
Thanks to Aloizio and Clark for pointing out that no assumptions need be placed on $f$ as the integral of a Riemann integrable function is continuous. This gives us that $f$ is continuous (since by assumption we have that $f$ is integrable) and thus the Fundamental Theorem of Calculus applies.
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Suppose $f$ is continuous. The fundamental theorem of calculus tells us that it is actually differentiable, and moreover by differentiating both sides we get
$$f'(x) = f(x)$$
So $f(x)$ is a function which is its own derivative, and hence is of the form $f(x) = ce^{x}$. What is the constant $c$? Simply compute
$$f(0) = \int_{0}^{0} f(x) = 0$$
to see that $c = 0$, and we are done.
|
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Evaluate the integral $\int_{-\infty}^{\infty}{\left| 2t\cdot\text{sinc}^2(2t)\right|^2}\,dt$ I have a question in solving the integral
$$\int_{-\infty}^{\infty}{\left| 2t\cdot\text{sinc}^2(2t)\right|^2}\,dt.$$
I know that you can use Parseval's Theorem to prove that $\int_{-\infty}^{\infty}\text{sinc}^4(kt) = \frac{2\pi}{3k}$ (I can provide the proof if you would like). While this might seem to help, I don't know that I can really use that fact with the $2t$ in the equation because you'll have to integrate by parts.
Is there another way that I could do this integral to make it easier? I feel like I might be overthinking it and this can be simplified down to something much more manageable.
Thank you so much for your help in advanced, I really appreciate it!!!
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Integration by parts is enough. Our integral equals
$$ \int_{-\infty}^{+\infty}\frac{\sin^4(2x)}{4x^2}\,dx=\frac{1}{2}\int_{-\infty}^{+\infty}\frac{\sin^4(u)}{u^2}\,du=\int_{-\infty}^{+\infty}\frac{2\cos(u)\sin^3(u)}{u}\,du $$
but:
$$ 2\cos(u)\sin^3(u)=\frac{1}{2}\sin(2u)-\frac{1}{4}\sin(4u)\,du $$
hence:
$$ \int_{-\infty}^{+\infty}\frac{\sin^4(2x)}{4x^2}\,dx = \left(\frac{1}{2}-\frac{1}{4}\right)\pi = \color{red}{\frac{\pi}{4}}.$$
|
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Formalization of an intuitive idea to construct a surjection Let $A$ be an arbitrary set and $B$ be any non-empty set. Furthermore, suppose that there is no injection from $A$ to $B$. I want to prove that it follows that there is a surjection from $A$ to $B$.
I have an intuitive argument in mind to prove this, but I do not know how to formalize it. Here is this intuitive argument:
Since $B$ is non-empty, there has to be a function from $A$ to $B$. Let us fix one such a function $f_0\colon A \to B$. The idea is to modify this function $f_0$ using the fact that there is no injection from $A$ to $B$ to get a surjective function. If $f_0$ is surjective, then we are done. So the interesting case is the case where there are some elements of $B$ that are not in the image of $f_0$. Our job is to modify $f_0$ in such a way that these elements of $B$ are hit too. Let $b_0$ be an element of $B$ which is not in the image of $f_0$. Since $f_0$ is not injective, there are two distinct elements $a_0, {a_0}'\in A$ such that $f_0(a_0)=f_0({a_0}')$. Now we can define a function $f_1\colon A\to B$ as follows: if $a\not = {a_0}'$ then $f_1(a) := f_0(a)$, and if $a={a_0}'$ then $f_1(a):=b_0$. If $f_1$ is surjective, then we are done. Otherwise we can do the same again: Let $b_1$ be an element of $B$ which is not in the image of $f_1$. Since $f_1$ is not injective, there are two distinct elements $a_1, {a_1}'\in A$ such that $f_1(a_1)=f_1({a_1}')$. Now we can define a function $f_2\colon A\to B$ as follows: if $a\not = {a_1}'$ then $f_2(a) := f_1(a)$, and if $a={a_0}'$ then $f_2(a):=b_1$. We can iterate this process a finite number of times. Thus, if the image of $f_0$ is finite, then $f_n\colon A\to B$ is surjective, where $n$ is the cardinality of the image of $f_0$.
Intuitively I think that this argument also works if the image of $f_0$ is countably infinite. Then we can carry out the same process to construct functions $f_1$, $f_2$, $f_3$, $f_4$ and so on. Then the limit $f_\omega$ of this countably infinite process is a surjective function $A\to B$.
Indeed, I also think that this kind of argument always works. We can iterate this process until every element of $B$ is hit, if necessary by constructing functions $f_{\omega + 1}$, $f_{\omega + 2}$, $f_{\omega +3}$, ...,
$f_{\omega\cdot 2}$, ..., $f_{\omega\cdot 3}$, ..., $f_{\omega\cdot\omega}$, ..., $f_{\alpha}$ (where $\alpha$ is an ordinal number).
How can one formalize this idea and turn it into a rigorous proof?
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Transfinite induction allows you to iterate the process you describe, and when combined with the axiom of choice, can be used to make something much like your argument work. (You will have to be a bit careful with how you handle limit ordinals, though.)
|
{
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|
Integral equation of the form $\int_{-\infty}^{\infty} e^{-a t^4} g(x,t) dt = e^{-b x^4}$ How to solve an integral equation of the following form
\begin{align}
\int_{-\infty}^{\infty} e^{-a t^4} g(x,t) dt = e^{-b x^4}
\end{align}
where $a$ and $b$ are some positive constants.
I am not very familiar with this subject any suggestions you might have would be great.
|
Hah! This is actually a specific example of something in my research! (My work attacks a more general set of integral equations, in some sense.) Let's go for something nontrivial (unlike previous answers/comments).
If you consider what I like to call a diagonal kernel, i.e. $g(x,t) = f(xt)$ for some $f$ and assume $g$ is real analytic, then this is very easily solvable. Since we want to get $\exp(-bx^4)$ after doing the integral, it stands to reason that $g$ should only consist of powers of the form $t^{4l}$.
Write $g(x,t) = \sum\limits_{l=0}^{\infty} c_l x^{4l} t^{4l}$, then we have
$$ e^{-bx^4} = \int_{-\infty}^{\infty} \sum_{l=0}^{\infty} c_l x^{4l} t^{4l} e^{-at^4}\,dt.$$
Interchanging sum and limit (which can be justified after we compute what $c_l$ has to be by appealing to Fubini-Tonelli) this becomes
$$ e^{-bx^4} = \sum_{l=0}^{\infty}2 c_l x^{4l} \int_0^{\infty} t^{4l}e^{-at^4}\,dt.$$
Letting $u = at^4$, $du = 4at^3\,dt$ so we have
$$ e^{-bx^4} = \sum_{l=0}^{\infty}2 c_l x^{4l} \int_0^{\infty} \frac{t^{4l}}{4at^3} e^{-u}\,du = \sum_{l=0}^{\infty} \frac{1}{2a} c_l x^{4l} \int_0^{\infty} \left(\frac{u}{a}\right)^{l-\frac{3}{4}} e^{-u}\,du.$$
The latter integral can be recognized as $\Gamma\left(l+\frac{1}{4}\right)$, yielding
$$ e^{-bx^4} = \frac{1}{2a^{\frac{1}{4}}}\sum_{l=0}^{\infty} a^{-l}c_l x^{4l} \Gamma\left(l+\frac{1}{4}\right). $$
Equating coefficients in the two power series gives that
$$ c_l = \frac{2(-1)^l a^{l+\frac{1}{4}}b^l}{l!\Gamma\left(l+\frac{1}{4}\right)}.$$
Plugging this back into $g$ we have
$$ g(x,t) = \sum_{l=0}^{\infty} \frac{2(-1)^la^{l+\frac{1}{4}}b^l}{l!\Gamma\left(l+\frac{1}{4}\right)} x^{4l}t^{4l}.$$
This looks pretty terrible, however it is not! This actually has a closed form solution in terms of a Bessel function:
$$ g(x,t) = 2 a^{\frac{5}{8}} b^{\frac{3}{8}} |xt|^{\frac{3}{2}} J_{-\frac{3}{4}}\left(2\sqrt{ab} x^2t^2\right). $$
What is nice is that if $a=b$, then you can use this to define a unitary integral transform (defined on a dense subspace of $L^2$ of course).
|
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|
Property of fractions Given two fractions $\frac{h}{k}$ and $\frac{h^{'}}{k^{'}}$ both in reduced form. I am unable to find a case when $\frac{h+h^{'}}{k+k^{'}}$ does not lie in the interval $\big[ \frac{h}{k},\frac{h^{'}}{k^{'}} \big]$. Is there such a case ?
PS: I was able two prove no such case exists for consecutive terms of Farey series. But can't prove in general.
|
Let's prove that, for positive $a,b,c,d$, with $\frac{a}{b}\le\frac{c}{d}$, it holds
$$
\frac{a}{b}\le\frac{a+c}{b+d}\le\frac{c}{d}
$$
The inequality on the left is equivalent to
$$
ab+ad\le ab+bc
$$
that is, $ad\le bc$, which is true.
The inequality on the right is equivalent to
$$
ad+cd\le bc+cd
$$
that is, $ad\le bc$, which is true.
Note that, if $a/b<c/d$, the two inferred inequalities are strict too. The hypothesis about the fractions being in reduced form is redundant.
|
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|
Prove $\frac{a+b+c}{abc} \leq \frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2}$. So I have to prove
$$ \frac{a+b+c}{abc} \leq \frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2}.$$
I rearranged it
$$ a^2bc + ab^2c + abc^2 \leq b^2c^2 + a^2c^2 + a^2b^2 .$$
My idea from there is somehow using the AM-GM inequality. Not sure how though. Any ideas?
Thanks
|
Because $$\sum_{cyc}\left(\frac{1}{a^2}-\frac{1}{ab}\right)=\frac{1}{2}\sum_{cyc}\left(\frac{1}{a^2}-\frac{2}{ab}+\frac{1}{b^2}\right)=\frac{1}{2}\sum_{cyc}\left(\frac{1}{a}-\frac{1}{b}\right)^2\geq0.$$
|
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|
$a\equiv b\pmod{n}\iff a/x\equiv b/x\pmod{n/\gcd(x,n)}$ for integers $a,b,x~(x\neq 0)$ and $n\in\Bbb Z^+$? I'm trying to prove/disprove the following:
If $a,b,x$ be three integers (where $x\neq 0$) such that $x\mid a,b$ and $n$ be a positive integer, then the following congruence holds:
$$a\equiv b\pmod{n}\iff a/x\equiv b/x\pmod{n/\gcd(x,n)}$$
My intuition says it's true and below here is my attempt at a proof. Can the community verify if it's correct? Thanks.
Proof.
Necessity: If $a/x\equiv b/x\pmod{n/\gcd(x,n)}$, then we can write $(a-b)/x=nl/\gcd(x,n)$ for some integer $l$, so that we have,
$$(a-b)/n = lx/\gcd(x,n) = l\ast (x/\gcd(x,n))$$
Since $\gcd(x,n)\mid x$ by definition of $\gcd$, we see that $(a-b)/n$ is an integer and hence $a\equiv b\pmod n$
Sufficiency: If $a\equiv b\pmod n$, then we can write $a-b=np$ for some integer $p$, so that we have,
$$\begin{align}a-b=np&\implies (a-b)\gcd(x,n)=np\gcd(x,n)\\&\implies \frac{(a-b)\gcd(x,n)}{nx}=\frac{p\gcd(x,n)}{x}\end{align}$$
By Bezout's Lemma, there exists integers $b_1,b_2$ such that $\gcd(x,n)=b_1x+b_2n$, so we have,
$$\frac{(a-b)\gcd(x,n)}{nx}=\frac{p(b_1x+b_2n)}{x}=pb_1+\frac{b_2np}{x}=pb_1+b_2\frac{a-b}{x}$$
Since $x\mid a,b$, we have $x\mid a-b$ and so we can write $a-b=xq$ for some integer $q$. Then,
$$\frac{(a-b)\gcd(x,n)}{nx}=pb_1+qb_2$$
So, we conclude that $\dfrac{(a-b)\gcd(x,n)}{nx}$ is an integer, i.e, $\dfrac{(a-b)/x}{n/\gcd(x,n)}$ is an integer, so $\frac ax-\frac bx$ is divisible by $n/\gcd(x,n)$ which shows that $a/x\equiv b/x\pmod{n/\gcd(x,n)}$
|
More simply, write $\ \bar a = a/x,\ \bar b = b/x\ $ and $y = \bar a - \bar b.\ $ Then we get a $1$-line proof:
$$\,n\mid a\!-\!b=xy\iff n\mid xy,ny\color{#0a0}\iff n\mid (xy,ny)\!\color{#c00}{\overset{\rm D}{=}}\!(x,n)y \iff n/(x,n)\mid y$$
We employed the universal property $\color{#0a0}\iff$ along with the $\,\color{#c00}{\overset{\rm D}{=}} $ Distributive Law of the gcd.
|
{
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|
What is the area of the triangle having $z_1$, $z_2$ and $z_3$ as vertices in Argand plane? What is the area of the triangle having $z_1$, $z_2$ and $z_3$ as vertices in Argand plane?
Is it
$$\frac{-1}{4i}[z_1(z_2^* - z_3^*)-z_1^*(z_2-z_3)+{z_2(z_3^*)-z_3(z_2^*)}]$$
where $w^*$ denotes the complex conjugate?
|
I assume you mean the oriented area of that triangle (since the number changes sign when you swap $z_2$ and $z_3$).
In order to verify it, allow me to write your formula as $$g(z_1,z_2,z_3)=\mathfrak{Im}\frac{\overline{z_1}(z_2-z_3)+z_3\overline{z_2}}{2}$$ It holds $$g(z_1+t,z_2+t,z_3+t)-g(z_1,z_2,z_3)= \mathfrak{Im}\frac{2\mathfrak{Re}(z_2\bar t)+\lvert t\rvert^2}{2}=0$$
So it is translation-invariant (hence, we need verify it only when $z_1=0$).
But that case is much easier.
|
{
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|
What is embedding? I am new to this so do I need to learn topology in order to understand this? Cause I come across this which says that unlike the 2D sphere, 2d saddle surface cannot be embedded in 3D Euclidean space(source: http://www.astro.yale.edu/vdbosch/astro610_lecture2.pdf - page 16) which I have trouble understanding. Cause from my understanding, embedding is a description of a space as a co-dimension n surface in higher-dimensional space.
|
In that context of that lecture, they are not talking just about the general topological concept of embedding, but instead a more special metric concept of embedding. A metric embedding is a topological embedding which preserves the metric tensor in an appropriate sense.
So yes, you do need to understand topological embeddings, but that's not enough, you also need to understand metric embeddings.
What the lecture is referring to (as noted in the answer of @Salvatore) is Hilbert's proof that there is no metric embedding from 2-dimensional hyperbolic space (the "saddle surface" in the lecture) to 3-dimensional euclidean space.
|
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|
Integrating $\cos (x+\sin (x))$ I tried to solve $$\int\cos(x+\sin(x))\,dx$$ but it seems to be way out of my league (tried u-substitution with $u=x+\sin(x)$ and couldn't find an answer). Also, no one on the Internet seems to have tried this before and Wolfram|Alpha and Symbolab aren't helping that much. If anyone can help me, I would apreciate it very much.
|
The Jacobi-Anger expansion gives
$$ e^{ix\sin\theta}=\sum_{n\in\mathbb{Z}}J_n(x) e^{ni\theta}\tag{1} $$
hence:
$$ e^{i\sin\theta+i\theta}=\sum_{n\in\mathbb{Z}} J_n(1)\,e^{(n+1)i\theta}\tag{2}$$
and by considering the real part of both sides:
$$ \cos(\theta+\sin\theta) = \sum_{n\in\mathbb{Z}} J_n(1)\,\cos((n+1)\theta)\tag{3} $$
so:
$$ \int\cos(\theta+\sin\theta)\,d\theta = \color{red}{C+\sum_{n\in\mathbb{Z}} \frac{J_{n}(1)}{n+1}\,\sin((n+1)\theta)}\tag{4} $$
where the RHS is a fast-converging series, since $J_n(1)$ behaves like $\frac{1}{2^n n!}$.
So, if you consider Bessel functions as elementary functions, there is a primitive in terms of elementary functions.
|
{
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|
Question on inverse limits
1.7. Remark. The inverse limit of an inverse system of non-empty sets might be empty as the following example
shows: Let $I:=\mathbb{N}$ and $X_n:=\mathbb{N}$ for every
$n\in\mathbb{N}$. Let $\theta_{n\leftarrow
> n+1}:\mathbb{N}\to\mathbb{N},k\mapsto k+1$. Now assume
$(x_n)_{n\in\mathbb{N}}$ is contained in the inverse limit. Then
$x_{n+1}=x_n-1$ for every $n\in\mathbb{N}$ yielding a contradiction.
Reference
My question is: How is the inverse limit the empty set? I have computed that the inverse limit is the set of all constant sequences of $\mathbb N$.
Could someone explain this? The author commented on it but I was unable to understand it.
|
The problem is that as the indices increase the numbers decrease by $1$. Eventually we will have to hit a negative number, but $\mathbb N$ does not contain any of those.
|
{
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|
Limit of fibonacci sequence Let $f_n$ be the $n$th Fibonacci number. Find constants $a$ and $b$ such that
$$\lim_{n\to\infty} \frac{f_n}{a\cdot b^n} = 1$$
I'm somewhat confused on how to approach this problem. I know the closed form of the Fibonnaci sequence, and I think it may have something to do with this problem, but I am unsure of how to proceed. Would love some help!
|
HINT: You know that
$$f_n=\frac{\varphi^n-\widehat\varphi^n}{\sqrt5}\;,$$
where $\varphi=\frac12\left(1+\sqrt5\right)$ and $\widehat\varphi=\frac12\left(1-\sqrt5\right)$. Note that $|\widehat\varphi|<1$, so $\widehat\varphi^n\to 0$ as $n\to\infty$. Thus, for large $n$ the Fibonacci number $f_n$ is approximately ... ?
|
{
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|
Eigenvectors are unique up to a scalar If $ A $ is a matrix with eigenvector $ v $ corresponding to the eigenvalue $ \lambda, $ can we prove that $ v $ is unique up to $ \lambda, $ that is if $ v $ and $ v' $ are eigenvectors corresponding to $ \lambda, $ then $ v = Cv' $ for some constant $ C. $
|
No. It is quite possible for an eigenspace to have more than one dimension. As commenters above pointed out, examples include the identity matrix or the zero matrix.
|
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|
Find all continuous functions $g(x)$ satisfying $\int_{0}^{f(x)}f(t)g(t)dt = g(f(x))-1$
Given a differentiable function $f(x)$, find all continuous functions $g(x)$ satisfying $$\int_{0}^{f(x)}f(t)g(t)dt = g(f(x))-1.$$
I differentiated both sides to get $f(f(x))f'(x)g(f(x)) = g'(f(x))f'(x)$. Thus, if $f'(x) \neq 0$ we have $f(f(x))g(f(x)) = g'(f(x))$. What do I do from here?
|
Let $a = \inf(\text{im } f)$ and $b = \sup(\text{im }f)$, where $\text{im }f$ is the image of $f$ (either $a$ or $b$ could be infinite). Notice the following: If $y\in (a,b)$, then the equality $$\int\limits_{0}^{z}{f(t)g(t)\text{ d}t} = g(z)-1$$ holds for all $z$ in some neighborhood of $y$. By the Fundamental Theorem of Calculus, this implies that $g$ is differentiable at $y$, with $g'(y) = f(y)g(y)$. Thus, if $y_1,y_2\in (a,b)$ are both in the interior of the image, then the interval between the two points is also in $(a,b)$ (since the image is connected), and it is easy to see that $$\ln\left(\frac{g(y_2)}{g(y_1)}\right) = \int\limits_{y_1}^{y_2}{f(t)\text{ d}t}.$$
Now we look at two cases:
Case 1: $0\in[a,b]$. In that case, for $f(x)=0$ we have
$$\int\limits_{0}^{0}{f(t)g(t)\text{ d}t} = g(0)-1\implies g(0)=1. $$
(If $0$ is not actually in the range, then take a sequence converging to $0$, and the same conclusion follows.)
Hence, for $y\in[a,b]$, we have
$$ g(y) = g(0)\exp\left(\int\limits_{0}^{y}{f(t)\text{ d}t}\right) = \exp\left(\int\limits_{0}^{y}{f(t)\text{ d}t}\right). $$
Outside of $[a,b]$, we can let $g$ take on any values, so long as $g$ remains continuous. This is because changing the value of $g$ outside $[a,b]$ does not affect the value of $g(f(x))$ (as $f(x)\in[a,b]$), nor would it affect the value of
$$\int\limits_{0}^{f(x)}{f(t)g(t)\text{ d}t}$$
as the limits of integration are contained in $[a,b]$. Thus, I claim that the set of continuous $g$ satisfying
$$ g(y) = \exp\left(\int\limits_{0}^{y}{f(t)\text{ d}t}\right) $$
for $y\in[a,b]$ solves the equation above. To check that these functions are indeed solutions, we just note that for any $x$ we have $f(t)g(t) = g'(t)$ for $t$ between $0$ and $f(x)$ (since that interval is contained in $[a,b]$), so
$$ \int\limits_{0}^{f(x)}{f(t)g(t)\text{ d}t} = \int\limits_{0}^{f(x)}{g'(t)\text{ d}t} = g(f(x))-g(0) = g(f(x)) - 1.$$
Case 2: $0\not\in[a,b]$. Then either $a,b>0$ or $a,b<0$. Assume $a>0$ (the other case is similar). For $y\in[a,b]$, we have
$$g(y) = g(a)\exp\left(\int\limits_{a}^{y}{f(t)\text{ d}t}\right) $$
and for $x$ such that $f(x)=a$ (or a sequence $f(x_n)$ converging to $a$), we have
$$\int\limits_{0}^{a}{f(t)g(t)\text{ d}t} = g(a) - 1.$$
These two necessary conditions turn out to be sufficient, since the first condition implies that $g'(y) = f(y)g(y)$ for $y\in(a,b)$, and hence
$$\int\limits_{0}^{f(x)}{f(t)g(t)\text{ d}t} = \int\limits_{0}^{a}{f(t)g(t)\text{ d}t} + \int\limits_{a}^{f(x)}{f(t)g(t)\text{ d}t} = \left(g(a)-1\right) + \int\limits_{a}^{f(x)}{g'(t)\text{ d}t} = g(f(x)) -1.$$
For Case 2, I don't know of a better way to phrase the second condition. It is a technical condition regarding the values of $g$ between $0$ and $a$, which matters only because the lower limit on the integral is $0$. The most important condition (regardless of case) is that $g$ solves $g'=fg$ in the interior of the image of $f$.
|
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|
Differential Equation Initial Value Problem Here is a pretty standard initial value problem that I'm having a little trouble with.
$$(\ln(y))^2\frac{\mathrm{d}y}{\mathrm{d}x}=x^2y$$
Given $y(1)=e^2$, find the constant $C$.
So I separated and integrated to get $\frac{(\ln(y))^3}{3}=\frac{x^3}{3}+C$. Multiplying $3$ to both sides yields $(ln(y))^3=x^3+C$. Here's where I struggled a little. So I take the cube root of both sides to get $\ln y=\sqrt[3]{x^3+C}$ then raised $e$ to both sides to get $y=e^\sqrt[3]{x^3+C}$...? Basically, the $C$ and the $e$ are giving me trouble. I appreciate any pointers.
|
It may be easier to solve for $C$ at the step
$$(\ln y)^3=x^3+C$$
$$2^3=1^3+C$$.
Everything else looks good.
|
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|
How to solve $x<\frac{1}{x+2}$ Need some help with:
$$x<\frac{1}{x+2}$$
This is what I have done:
$$Domain: x\neq-2$$
$$x(x+2)<1$$
$$x^2+2x-1<0$$
$$x_{1,2} = \frac{-2\pm\sqrt{4+4}}{2} = \frac{2 \pm \sqrt{8}}{2} = \frac{-2\pm2\sqrt{2}}{2}$$
What about now?
|
Multiply the inequality by $\;(x+2)^2>0\;$ ( obviously, $\;x\neq-2\;$) :
$$x(x+2)^2<x+2\iff x^3+4x^2+3x-2<0\iff$$
$$\iff (x+2)(x+1-\sqrt2)(x+1+\sqrt2)<0\iff \color{red}{x<-1-\sqrt 2}\;\;\text{or}\;\color{red}{-2<x<-1+\sqrt2}$$
|
{
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|
Show that $U(8)$ is Isomorphic to $U(12)$.
Question: Show that $U(8)$ is Isomorphic to $U(12)$
The groups are:
$U\left ( 8 \right )=\left \{ 1,3,5,7 \right \}$
$U\left ( 12 \right )=\left \{ 1,5,7,11 \right \}$
I think there is a bit of subtle point that I am not fully understanding about isomorphism which is hindering my progress. The solution mentions about the order of an element but I do not understand how that is pivotal to solving this.
Thanks in advance.
|
The preferred approach would be to explicitly construct that isomorphism, but since the problem suggests basing the proof around the concept of "order of an element", let us do so then.
The crucial thing here is that both groups are of order $4$, and a simple theorem says that there exist only two classes of groups of order $4$: those isomorphic to $\Bbb Z_2 \times \Bbb Z_2$ (also known as "the Klein group") and those isomorphic to $\Bbb Z_4$. Notice that in $\Bbb Z_2 \times \Bbb Z_2$ all the elements are of order $2$, while in $\Bbb Z_4$ there are elements of order $4$.
It is now a matter of simple calculations to check that in your groups all the elements have order $2$, therefore they cannot be isomorphic to $\Bbb Z_4$, hence by the above considerations they must be isomorphic to $\Bbb Z_2 \times \Bbb Z_2$, hence isomorphic between themselves. (Notice that we don't have an explicit formula for this isomorphism - but neither are we required to find it.)
|
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|
inequality proof I have come across a problem:
Let $a,b$ and $c$ be real numbers where $a > b$. Prove that if $ac \leq bc$, then $c \leq 0$.
I tried using the Indirect proof
If $a > b$, and $c > 0$, then by the 4th axiom of Inequality, Multiplicativity, $ac > bc$.
However, $ac \leq bc$, therefore $c \leq 0$.
I'm not sure if this is the correct way.
|
Your proof is correct.
Here's an alternative proof.
$bc \geq ac \Longleftrightarrow c(b-a) \geq 0 \Longleftrightarrow -c(a-b) \geq 0$
Since $a > b \Longleftrightarrow (a-b) >0$, so $c \leq 0$.
|
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Finding the locus of a point $P$ if the tangents drawn from $P$ to circle $x^2 + y^2 = a^2$ so that the tangents are perpendicular to each other? Question: Find the locus of a point $P$ if the tangents drawn from $P$ to circle $x^2 + y^2 = a^2$ so that the tangents are perpendicular to each other.
I tried solving this and then I got to this condition here, after I applied the formulua for finding the angle between the tangents
Formula is Angle btw tangents: $$\cos\theta = \frac{1 - \tan^2(\theta/2)}{ 1 + \tan^2(\theta/2)} $$
So, I got to this equation of locus after solving using that formula...
$$a^2 * \cos^2(\theta/2) = x_1^2 + y_1^2$$
But I am having trouble trying to figure out how to show that the tangents are perpendicular :C
so, I tried applying the trigonometric here, and then I got this answer
$$x_1^2 + y_1^2 - a^2 * \cos^2 (\theta/2)$$
But in my solutions book it's different, it's $x_1^2 + y_1^2 - 2a^2$
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Hint: find the hidden square in the picture below.
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Have historians responded to Raju's critique? C. K. Raju has made some outrageous criticisms of the traditional take on Euclid in particular and Western history in general. Yet he has a book published on the subject with an apparently respectable publisher in India. Have modern historians of the classical period responded to his critique?
Note. One of the responders mentioned a helpful review by Ferreiros here.
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The articles by Raju have a conspirational flavor. The history of Indian mathematics is still an uncharted territory. There are more informative unbiased articles, for instance, there are much deeper and less biased studies I've read:
A. Seidenberg, “The Origin of Mathematics,” Archive for History of Exact Sciences 18, 301-342 (1978).
S.C.Kak, “Science in Ancient India,” in Ananya:
A Portrait of India, Ed. by S.R.Sridhar and N.K.Matto
(AIA, New York, 1997), pp. 399–420.
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Every length minimizing $\mathcal C^1$ curve is a geodesic. Let $(M,g)$ a manifold and $\gamma (t)$ for $t\in [a,b]$ a curve $\mathcal C^1$ the is minimizing the length. Then, if $p=\gamma (t_0)$ and $q=\gamma (t_1)$, then $\gamma $ is also minimizing the length of $p$ and $q$ for all $a\leq t_0<t_1\leq b$. In polar coordinates in $B(p,r)$, we have $$\gamma |_{[t_0,t_1]}=(r(t),\gamma ^1(t),...,\gamma ^{n-1}(t))$$
and
\begin{align*}
d(p,q)&=\ell(\Gamma)|_{[t_0,t_1]}\\
&=\int_{t_0}^{t_1}\|\dot \gamma (t)\|\mathrm d t\\
&=\int_{t_0}^{t_1}\sqrt{\dot r(t)+g_{ij}(\gamma (t))\dot y^i\dot y^j}\mathrm d t\\
&\geq \int_{t_0}^{t_1}\sqrt{\dot r(t)^2}\\
&=r(t_1)-r(t_0)\\
&=d(p,q).
\end{align*}
We have equality if and only if $\dot y^i=0$ for all $i$, i.e. if $y^i=const$, which proves the claim.
Question
I don't understand the proof, why we have show that $\gamma $ is a geodesic ? To me, we only proved that $d(p,q)\geq d(p,q)$ (which is in fact obvious). Any way, any explanation is welcome.
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To answer the question why the proof shows that the curve is a geodesic:
Once we prove that $y^i$'s are constant, then $\gamma$ coincides with a radial curve, namely with $t \longrightarrow (t,y^1,...)$ which is by definition of polar coordinates a geodesic.
Therefore, we have shown that small segments (because we restricted ourselves to $p-q$ part) of the curve are geodesics. But this is true of any segment, hence proving that the whole curve is a geodesic.
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Sum of digits equality $Z$ return sum of digits of number: $Z(15)=6$.
If for some $W\in\mathbb{N}$, $Z(W)=100;~Z(44W)=800$ find
$$
ZZ(2015!)+ZZZ(2015!)+ZZZZ(2015!)+Z(3W)
$$
I don't have experience with this kind of problem, please give a hint what should I start? What approach is right?
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The sum of digits function returns about $4.5$ times the base $10$ log of a number, because the base $10$ log gives the number of digits and the average digit is $4.5$. It also maintains the value of the number $\bmod 9$ by the classic divisibility test. $2015!$ has about $5500$ digits (you can use Stirling to get the exact answer), so $Z(2015!) \lt 5500 \cdot 9 =49500$ Then $ZZ(2015!) \lt 4+8+9+9+9=49$ and since $2015!$ is a multiple of $9$, we have $ZZ(2015!)=9, 18, 27$ or $36$ and you should be able to get the next two terms of your sum. I don't see an easy way to distinguish the four cases for the first term. For the last term, multiplying $W$ by $44$ multiplied $Z(W)$ by $8=Z(44)$ so you can't afford any carries of one non-zero digit into another. What does that tell you about $Z(3W)$?
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Tricky detail in extreme value theorem proof I am reading Pugh's Real Mathematical Analysis, and in chapter 1, section 6, ``The Skeleton of Calculus,'' Pugh supplies a proof of the Extreme Value Theorem. I am having trouble understanding one particular point in the proof. Note that he proves the existence of maximums, and leaves minimums for the reader.
Theorem 23. A continuous function $f$ defined on an interval $[a, b]$ takes on absolute minimum and absolute maximum values: for some $x_0, x_1 \in [a,b]$ and for all $x \in [a,b]$,
$f(x_0) \leq f(x) \leq f(x_1)$.
Proof. Let $M = \sup f(t)$ as $t$ varies in $[a,b]$. This exists since the values of a continuous function defined on an interval $[a,b]$ form a bounded subset of $\mathbb R$ (his Theorem 22). Consider the set $X = \{x \in [a,b] : \sup V_x < M\}$ where $V_x$ is the set of values of $f(t)$ as $t$ varies on $[a,x]$.
Case 1: $f(a) = M$. Then $f$ takes on a maximum at $a$ and the theorem is proved.
Case 2: $f(a) < M$. Then $X$ is nonempty and we can consider the supremum of $X$, say $c$. Here's where I lose him---I don't seem to understand what's going on here. If $f(c) < M$, we choose $\epsilon > 0$ with $\epsilon < M - f(c)$. By continuity, there exists a $\delta > 0$ such that $|t - c| < \delta \implies |f(t) - f(c)| < \epsilon$. I understand the use of continuity and the application of the definition of a continuous function, but I don't understand the motivation for considering the continuity of $f$ at $c$. Thus, $\sup V_c < M$. I don't understand how we can deduce that $\sup V_c < M$ by considering the continuity of $f$ at $c$.
If you could help me understand this proof, and specifically how considering the continuity of $f$ at $c$ helps us determine that $\sup V_c < M$, I would greatly appreciate it.
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My guess: the idea is to prove that $f(c)=M$. He proves it by contradiction. Assume that $f(c)<M$ then by continuity we can go a bit further to $c+\delta$ and still have all functional values being under $M$. It contradicts the choice of $c$ as the supremum of $X$.
P.S. It is easy, but you have to mention also that $c\in[a,b]$.
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What does it mean to perform calculus upon functions of complex values? Complex numbers exist in a plane. This would lead me to believe that calculus views them as multivariate, but I am not real sure. How would one define a rate of change for a complex number valued function, or the area underneath it. Could someone explain this in simple terms?
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At a very simple level, any complex number $z$ can be expressed as $x + i y.$
and $f(z) = u(x,y) + i v (x,y)$
The definition of derivative is the same definition.
$f'(z) = \lim_\limits{z\to z_0} \frac {f(z) - f(z_0)}{z-z_0}$
$\frac {\partial f}{\partial x} = \frac {\partial u}{\partial x} + i \frac{\partial u}{\partial y}\\
\frac {\partial f}{\partial y} = - \frac {\partial v}{\partial y} + i\frac {\partial v}{\partial x} $
if $\frac {\partial f}{\partial x} = \frac {\partial f}{\partial y}$ then the function is said to be "holomorphic."
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Recurrence relation solution $a_{n}=\frac{a_{n-1}}{a_{n-1}-a_{n-2}}$ I want to find the analytic form of the recurrence relation $$a_{n}=\frac{a_{n-1}}{a_{n-1}-a_{n-2}},\,a_{1}=2,\,a_{2}=1 $$ but when looking at the results they seem chaotic. Is it possible that it simply doesn't have any analytic form?
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It seems to be chaotic I calculated and plotted the first 60 points of the solution.
If there was a solution it would have to have some periodic tendancies but it seems chaotic where these tendencies are and has unpredictable spikes well above the range of a normal Sin or Cos function.
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Calc 3: Calculate Work Done on Particle I've been working on this problem for a while and I'm pretty stuck. I tried it multiple different ways, by the last time I attempted it I realized that I hadn't converted kilometers to meters the entire time.
Anyway, the last approach I tried after doing the conversion was to integrate over the function from 7,000,000 to 8,300,000. I got a positive answer back, which I didn't think made sense given that the force would be opposing the movement, so I flipped signs. After integrating, I then multiplied the answer to the displacement (8,300,000 - 7,000,000) to get my final answer.
My logic was that integrating over the function would give me the total force, multiplying by the displacement would give me that total force over the distance which was the work. I don't understand what's wrong with this solution. This is a homework problem but at this point I've gotten it wrong too many times to get any points for it, I'm just itching to know how it's done.
Picture of the problem:
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Smart Method
The work done is equal to the change in potential energy. The potential of a gravitational field:
$$
v(\vec{r})=-\frac{GM}{\|\vec{r} \|}\ \phantom{aaaa}\dots(1)
$$
So the work done is
$$
m\Delta v=-\frac{GMm}{\|\vec{r}_1 \|}+\frac{GMm}{\|\vec{r}_2 \|}
$$
Where $\vec{r}_1$ is the initial radial vector and $\vec{r}_2$ the final.
This works as the potential is defined as the work done on a particle of unit mass in bringing it in from infinity to its final radial distance (which is why the sign in the potential is negative as the particle does work in moving from infinity). You can derive $(1)$ above by integrating force equation given in the question if you so desire.
Your answer in the attachment is wrong as you have the radial distances squared.
The gist of this can be found on the Wikipedia page on Gravitational Potential
Brute Force and Ignorance method:
$$
W=\int_{r=r_1}^{r_2}F\;dr=\int_{r=r_1}^{r_2}-\;\frac{GMm}{r^2}\;dr=\frac{GMm}{r_2}-\frac{GMm}{r_1}
$$
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Prove that $\sum_{n=1}^{\infty }\frac{B_{2n}}{(2n-1)!}=\frac{1}{2}-\frac{1}{(e-1)^2}$ Prove that $$\sum_{n=1}^{\infty }\frac{B_{2n}}{(2n-1)!}=\frac{1}{2}-\frac{1}{(e-1)^2}$$
My idea is to find the Taylor series of $\frac{1}{(e^x-1)^2}$, but it seems not useful.
Any helps, thanks
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An alternative approach is to use the integral representation
$$ B_{2n} = (-1)^{n}4n \int_{0}^{\infty} \frac{x^{2n-1}}{e^{2 \pi x}-1} \, dx.$$
Specifically,
$$ \begin{align}\sum_{n=1}^{\infty} \frac{B_{2n}}{(2n-1)!} &= \sum_{n=1}^{\infty} \frac{1}{(2n-1)!} (-1)^{n-1} 4n \int_{0}^{\infty} \frac{x^{2n-1}}{e^{2 \pi x}-1} \, dx \\ &= 4 \int_{0}^{\infty} \frac{1}{e^{2 \pi x}-1} \sum_{n=1}^{\infty} \frac{n (-1)^{n-1} x^{2n-1}}{(2n-1)!} \, dx. \end{align}$$
But notice that $$ \begin{align} \sin (x) + x \cos(x) &= \sum_{n=1}^{\infty} \frac{(-1)^{n-1} x^{2n-1}}{(2n-1)!} + \sum_{n=1}^{\infty} \frac{(-1)^{n-1}x^{2n-1}}{(2n-2)!} \\ &= \sum_{n=1}^{\infty} \frac{[1+(2n-1)](-1)^{n-1} x^{2n-1}}{(2n-1)!} \\ &= \sum_{n=1}^{\infty} \frac{2n(-1)^{n-1} x^{2n-1}}{(2n-1)!}. \end{align}$$
So using the fact that $$\int_{0}^{\infty} \frac{\sin ax}{e^{2 \pi x}-1} \, dx = \frac{1}{4} \, \coth \left(\frac{a}{2} \right) - \frac{1}{2a}, $$ we get
$$ \begin{align} \sum_{n=1}^{\infty} \frac{B_{2n}}{(2n-1)!} &= 2 \left[ \int_{0}^{\infty} \frac{\sin x}{e^{2 \pi x}-1} \,dx + \int_{0}^{\infty} \frac{x \cos x}{e^{2 \pi x}-1} \, dx \right] \\ &= 2 \left[\frac{1}{4} \, \coth \left(\frac{1}{2} \right) - \frac{1}{2} + \frac{\mathrm{d}}{\mathrm{d}a} \left(\frac{1}{4} \, \coth \left(\frac{a}{2} \right) - \frac{1}{2a}\right)\Bigg|_{a=1} \right] \\ &= 2 \left[\frac{1}{4} \, \coth \left(\frac{1}{2} \right) - \frac{1}{2} + \frac{1}{2} - \frac{1}{8} \, \text{csch}^{2} \left(\frac{1}{2} \right) \right] \\ &= \frac{1}{2} \, \coth \left(\frac{1}{2} \right) - \frac{1}{4} \, \text{csch}^{2} \left(\frac{1}{2} \right) \\&= \frac{1}{2} \frac{e+1}{e-1} - \frac{1}{4} \frac{4e}{(e-1)^{2}} \\ &= \frac{(e^{2}-1) -2e}{2(e-1)^{2}} \\ &= \frac{(e-1)^{2}-2}{2(e-1)^{2}} \\ &= \frac{1}{2} - \frac{1}{(e-1)^{2}}.\end{align}$$
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Show that : $\lim\limits_{h\to 0}\frac {h^5} {2h^4} \frac1{\sqrt{h^2 + h^4}} = \lim\limits_{h \to 0}\frac {h^5} {2h^5}$ In my textbook, I found the following step, but I don't understand how the author gets there.
$$\lim_{h\to 0} {{\frac {h^5} {2h^4} \over \sqrt{h^2 + h^4}}} = \lim_{h \to 0}\frac {h^5} {2h^5}$$
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The comment by Claude explains the step of that author but only for $\;h>0\;$, but the step is wrong for negative values of $\;h\;$ :
$$\frac{\frac{h^5}{2h^4}}{\sqrt{h^2+h^4}}=\frac h{2|h|\sqrt{1+h^2}}=\begin{cases} \cfrac1{2\sqrt{1+h^2}}, &h>0\xrightarrow[h\to0^-]{}-\cfrac12\\{}\\-\cfrac1{2\sqrt{1+h^2}},&h<0\xrightarrow[h\to0^+]{}\cfrac12\end{cases}$$
and thus the limit doesn't exist.
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product of integral of reciprocal functions let us consider the integral of the following question :
this question seems quit interesting for me and that why i have decided to think about, this, first of all i was thinking to use function like
$\frac{1}{x}$
because if $f(x)=\frac{1}{x}$
then definitely $\frac{1}{f(x)}$ is equal to $x$, integral of $\frac{1}{x}=\frac{1}{-x^2}+c$
and integral of $x$ is $\frac{x^2}{2}+c$, but this does not match required condition, should i use this property?
what about in general function $f(x)=e^{k*x}$ ?
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Differentiating both sides we get,$$f\int\dfrac{1}{f}dx + \frac{1}{f}\int{f}dx=0 \implies f^2\int\frac{1}{f}dx+\int fdx=0$$
Differentiate both sides again to get,$$f+ 2ff'\int\frac{1}{f}dx + f=0$$ $$\implies 2ff'\int\frac{1}{f}dx=2f\implies \int\frac{1}{f}dx=\frac{1}{f'}\implies\frac{1}{f}=\frac{f''}{f'^2}$$ $$\implies\frac{f'}{f}=\frac{f''}{f'}\implies lnf=ln(f')+A\implies\frac{f'}{f}=B \implies ln(f)=Bx+C\implies f=e^{Bx+C}$$
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Calculating $\int_0^{\infty } \frac{\ln (x)}{\sqrt{x} \left(a^2+x^2\right)^2} \, \mathrm{d}x$ using contour integration I can do this integral using the keyhole contour the answer is:$$\int_0^{\infty } \frac{\ln (x)}{\sqrt{x} \left(a^2+x^2\right)^2} \, \mathrm{d}x = -\frac{\pi (-6 \ln (a)+3 \pi +4)}{8 \sqrt{2}a^{7/2}}$$
but I want to calculate it with the substitution $x = ae^t$ which turns the integral into:
$$
\frac{1}{a^\frac{7}{2}}\times\int_{-\infty }^{\infty } \frac{e^{\frac{t}{2}} (\ln (a)+t)}{(e^{2t}+1)^2} \, \mathrm{d}t $$
now I think this integral should be done using the contour $-R$ to $R$, to $R+πi$, to $-R + πi$, to $-R$.
But at this point, I'm a bit lost as to what to do with the Residue in $z=i\pi/2$.
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This answer is based on Feynman's trick/differentiation under the integral sign.
Well, we may just compute:
$$I(\alpha)=\int_{0}^{+\infty}\frac{x^{\alpha-1/2}}{(x^2+1)^2}\,dx \tag{1}$$
through the substitution $\frac{1}{x^2+1}=u$, Euler's beta function and the $\Gamma$ reflection formula to get:
$$ I(\alpha) = \frac{\pi}{8}\cdot\frac{3-2\alpha}{\sin\left(\frac{(2\alpha+1)\pi}{4}\right)}\tag{2}$$
for any $\alpha$ such that $\text{Re}(\alpha)\in\left(-\frac{1}{2},\frac{7}{2}\right)$. However, assuming $a\in\mathbb{R}^+$
$$\begin{eqnarray*} \int_{0}^{+\infty}\frac{\log(x)}{\sqrt{x}(x^2+a^2)^2}&=&\frac{1}{a^{7/2}}\int_{0}^{+\infty}\frac{\log(a)+\log(x)}{\sqrt{x}(x^2+1)^2}\,dx\\[0.2cm]&=&\color{red}{\frac{1}{a^{7/2}}\left(I(0)\log(a)+I'(0)\right)}\tag{3}\end{eqnarray*}$$
where:
$$ I(0) = \frac{3\pi}{4\sqrt{2}},\qquad I'(0)=-\frac{\pi(4+3\pi)}{8\sqrt{2}}\tag{4}$$
are straightforward to compute through $(2)$.
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Concept of roots in Quadratic Equation $a$ , $b$, $c$ are real numbers where a is not equal to zero and the quadratic equation
\begin{align}
ax^2 + bx +c =0
\end{align}
has no real roots then prove that $c(a+ b+ c)>0$ and $a(a+ b + c) >0$
My Approach : As the equation has no real roots then its discriminant is less than zero. So the graph of the equation will be above $x$-axis or below $x$-axis .
I am able to conclude signs of $a$ , $b$, $c$ but still not getting appropriate answer.
Please explain the concept......
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For the second part, another justification is that since the discriminant is less than zero, then it must hold that $a,c$ have the same sign. Why? If $a,c$ had opposite signs then the discriminant $\Delta = b^2-4ac$ would have been positive.
Since $c,(a+b+c)$ have the same sign and $c,a$ have the same sign then $a,(a+b+c)$ have the same sign as well, thus:
$$a(a+b+c) >0.$$
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The first step in the proof of the Pólya-Vinogradov Inequality. The well-known Pólya-Vinogradov Inequality states:
$$\forall m, n \in \mathbb{N}: \displaystyle \left|{\sum_{k \mathop = m}^{m+n} \left({\frac k p}\right)}\right| < \sqrt p \ \ln p,$$
where $\left({\frac k p}\right)$ is the Legendre symbol.
I would like to write out a nice, detailed proof for personal reference, and I can follow most of the proofs I find well-enough, except for the first step. Many start with the same thing, or close to it:
"Start with the following manipulations: $$\displaystyle \sum_{k \mathop = m}^{m+n} \left({\frac k p}\right)=\displaystyle \frac 1 p \sum_{k \mathop = 0}^{p-1} \sum_{x \mathop = m}^{m+n} \sum_{a \mathop = 0}^{p-1} \left({\frac k p}\right) e^{2 \pi i a \left({x - k}\right) / p}=\displaystyle \frac 1 p \sum_{a \mathop = 1}^{p-1} \sum_{x \mathop = m}^{m+n} e^{2 \pi i a x / p} \sum_{k \mathop = 0}^{p-1} \left({\frac k p }\right) e^{-2 \pi i a t / p}.$$
I have absolutely no idea where this first step comes from. I am especially confused as to why the index of summation changes from how it is originally presented. I am somewhat new to quadratic Gauss sums, which I understand are the whole basis for this.
Thanks very much for any and all input.
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The key idea here is orthogonality: the complex number $\omega_r = e^{2\pi ir/p}$ has the property that $\omega_r^p = 1$. This makes it easy to evaluate $\sum_{a=0}^{p-1} \omega_r^a$, since
$$(1 - \omega_r)(1 + \omega_r + \omega_r^2 + \cdots + \omega_r^{p-1}) = 1 - \omega_r^p = 0.$$
As long as $\omega_r \ne 1$, we can cancel out the first factor to obtain $\sum_{a=0}^{p-1} \omega_r^a = 0$. On the other hand, if $\omega_r = 1$ it is trivial to evaluate the sum directly as $\sum_{a=0}^{p-1} 1 = p$, which gives the orthogonality relation:
$$ \sum_{a=0}^{p-1} e^{2\pi i ar/p} = \sum_{a=0}^{p-1} \omega_r^a = \begin{cases} p,&\text{if }r\equiv 0 \pmod p; \\ 0, &\text{otherwise}.\end{cases}.$$
Now it's easy to see how the sum unravels. First, the ranges of summation are independent so we may freely switch the order of the sums:
$$ \frac 1 p \sum_{k \mathop = 0}^{p-1} \sum_{x \mathop = m}^{m+n} \sum_{a \mathop = 0}^{p-1} \left({\frac k p}\right) e^{2 \pi i a \left({x - k}\right) / p} = \frac 1 p \sum_{x \mathop = m}^{m+n} \sum_{k \mathop = 0}^{p-1} \sum_{a \mathop = 0}^{p-1} \left({\frac k p}\right) e^{2 \pi i a \left({x - k}\right) / p}.$$
The innermost sum is exactly $0$ unless $k \equiv x \pmod p$, and since $k$ goes through a complete system of residues mod $p$ this occurs exactly once for each value of $x$, so we can restrict the sum to just the place where $k=x$.
$$\frac 1 p \sum_{x \mathop = m}^{m+n} \sum_{k \mathop = 0}^{p-1} \sum_{a \mathop = 0}^{p-1} \left({\frac k p}\right) e^{2 \pi i a \left({x - k}\right) / p} = \frac 1 p \sum_{x \mathop = m}^{m+n} \sum_{k=x}^{x} \left({\frac k p}\right) p = \sum_{x \mathop = m}^{m+n} \left({\frac x p}\right).$$
Since you're writing a detailed proof, note that even though $m,n$ might well be outside the range $[0,p)$, we can still justify restricting $k=x$ because both the Legendre symbol $(k/p)$ and the exponential term $e^{2\pi ia(x-k)/p}$ are periodic mod $p$.
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{
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How to rigorously deduce the Laurent series of $\log\frac{z-p}{z-q}$? Of course, the logarithm here is defined on the ring region $|z|>R\ge\max\{|p|,|q|\}$ as
$$\log\frac{z-p}{z-q}=\int_{z_0}^z \left(\frac1{w-p}-\frac1{w-q}\right)\mathrm d w. $$
Here the integral is along an arbitrary curve connecting $z_0$, a fixed point, to $z$ in the ring region. It's noteworthy that this logarithm is actually well defined, although not appearing so at first glance.
All I know is $\log(1+z)=z-\frac12z^2+\frac13z^3+\cdots$ when $|z|<1$ and $\log$ is chosen to be the principal branch. With this idea I can work out a naive argument: for our fractional logarithm, first rewrite the expression as $\log\frac{1-p/z}{1-q/z}$ by dividing both the numerator and denominator simultaneously (I don't know how to justify this from the integral definition, though); then it all reduces to $\log(1-p/z)-\log(1-q/z)$, with the two logarithm both chosen as suitable branches, to which the canonical power expansion applies.
I believe I'm almost on the right track, but can't get over that confusion. Could you help me? Thanks!
|
Yes, you can compute the Laurent series directly from the integral
definition.
For $\lvert w \rvert >\max\{\lvert p \rvert,\lvert q \rvert\}$,
$$
\frac{1}{w-p} - \frac{1}{w-q}
= \frac{1}{w(1-p/w)} - \frac{1}{w(1-q/w)}
= \sum_{n=1}^\infty \frac{p^n - q^n}{w^{n+1}}
$$
The terms for $n=0$ cancel, and all remaining terms have an antiderivate
in $\Bbb C$. Therefore
$$
\int_{z_0}^z \left(\frac1{w-p}-\frac1{w-q}\right)\, dw
= \sum_{n=1}^\infty (p^n - q^n) \left( -\frac{1}{nz^n} + \frac{1}{nz_0^n} \right)
= C + \sum_{n=1}^\infty \frac{q^n - p^n}{n z^n}
$$
for some constant $C$ depending on $z_0$.
|
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Limit with Lambert-$W$ function I have asked a similar question about this one particular limit:
\begin{equation}
A=\lim_{c\to 1}\exp\left[ -\left(\frac{1}{1-c}\right)\left(W_{0}\left[ B\left( 1+\frac{x}{rc}\right) \right]-W_{0}[B]\right)\right]
\end{equation}
where $r>0$, $x \in \mathbb{R}$, $W_0$ is the $k=0$ branch of the Lambert-$W$ function defined as:
\begin{equation}
x=W(x)e^{W(x)}
\end{equation}
and B is defined as:
\begin{equation}
B=\frac{(1-c)r}{1-(1-c)r}\exp\left[ \frac{(1-c)r}{1-(1-c)r} \right]
\end{equation}
Now, I tried with Mathematica and it is shown that $A \to e^{-x}$ when $c\to 1$ but I have trouble proving that analytically.
If anyone could assist me with this one I would be grateful.
Thank you!
|
This is not an answer but it is too long for a comment.
mjqxxxx's answer contains all the required steps.
Concerning the approximation made for $B$, consider the definition $$B=\frac{(1-c)r}{1-(1-c)r}\exp\left[ \frac{(1-c)r}{1-(1-c)r} \right]$$ and let us define $a=(1-c)r$ which makes $$B=\frac a{1-a}\exp\left[ \frac{a}{1-a} \right]$$ and develop as a Taylor series built at $a=0$; this gives $$B=a+2 a^2+O\left(a^3\right)$$
Similarly, for small values of $y$, using Taylor again $$W(B(1+y))-W(B)=\frac{ W(B)}{1+W(B)}y+O\left(y^2\right)$$ and, for small $B$ $$\frac{ W(B)}{1+W(B)}=B-2 B^2+O\left(B^3\right)$$ Combining all the above leads to
$$W_{0}\left[ B\left( 1+\frac{x}{rc}\right) \right]-W_{0}[B]=(1-c) x+(1-c)^2 \left(x-x^2\right)+O\left((c-1)^3\right)$$
$$
-\left(\frac{1}{1-c}\right)\left(W_{0}\left[ B\left( 1+\frac{x}{rc}\right) \right]-W_{0}[B]\right)=-x-(1-c) \left(x-x^2\right)+O\left((c-1)^2\right)
$$
|
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|
Why is the complex plane shaped like it is? It's always taken for granted that the real number line is perpendicular to multiples of $i$, but why is that? Why isn't $i$ just at some non-90 degree angle to the real number line? Could someone please explain the logic or rationale behind this? It seems self-apparent to me, but I cannot actually see why it is.
Furthermore, why is the real number line even straight? Why does it not bend or curve? I suppose arbitrarily it might be strange to bend it, but why couldn't it bend at 0? Is there a proof showing why?
Of course, these things seem natural to me and make sense, but why does the complex plane have its shape? Is there a detailed proof showing precisely why, or is it just an arbitrary choice some person made many years ago that we choose to accept because it makes sense to us?
|
Complex numbers can be constructed as couple of real numbers :
$
a+ib=(a,b)
$
with suitable definitions of the operations of sum and product ( see here).
With such definitions a complex number corresponds, in a natural way, to an element of $\mathbb{R}^2$ and we have : $1=(1,0)$ and $i=(0,1)$ and , using the usual representation of an orthogonal system for the coordinates in $\mathbb{R}^2$, these two vectors, are represented as points, at distance $=1$ from the origin, on two straight lines that forms a $90°$ angle. So this same representation is also adopted for the complex numbers.
|
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|
Maximal ideal in a local artinian ring. I know that an artinian ring $A$ is the union of its units and its zero-divisors.
So every non-zero-divisor is an unit.
I also know that in a local ring every element which is out from the maximal ideal is an unit.
Can I conclude that the set of zero-divisors is the maximal ideal of $A$?
|
I also know that in a local ring every element which is out from the maximal ideal is an unit.
Can I conclude that the set of zero-divisors [of an Artinian local ring] is the maximal ideal of $A$?
Yes: here is an elementary way to see it.
Suppose $M$ is the unique maximal ideal of a commutative Artinian ring $A$. Suppose $a$ is a nonzero element. Then the chain of ideals $aR\supseteq a^2R\supseteq a^3R\supseteq\ldots$ has to stabilize. At some point, $a^nR=a^{n+1}$, so that $a^n=a^{n+1}r$ for some $r\in A$.
Suppose $a$ is not a zero divisor. Then since $a^n(1-ar)=0$, we can cancel $a$'s until $1-ar=0$, and we find $a$ is a unit. Conclusion: $a$ is either a zero divisor or a unit.
Of course, if $M$ contained a unit, it would have to contain all of $A$, so $M$ contains no units. Everything inside must be a zero divisor.
This is actually true more generally for one-sided Artinian rings. Really the property that we used here is much weaker: the descending chain condition on chains of ideals of the form $aR\supseteq a^2R\supseteq a^3R\supseteq\ldots$ A ring that satisfies this chain condition is called strongly $\pi$-regular.
|
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Concerning The Number of Ways of Drawing a Full House vs. Two Pair The Wikipedia entry for "Poker probability" gives the following result for the number of ways of drawing a full house:
$$ \binom{13}{1} \binom{4}{3} \binom{12}{1} \binom{4}{2} = 3, 744. $$
The logic here is conventional: From the 13 kinds you choose 1, then choose 3 of the 4 available cards of that kind. Then, from the remaining 12 kinds, you choose 1, then choose 2 of the available cards of that kind. No problem.
However, for the number of ways of drawing a two pair it gives the following:
$$ \binom{13}{2} \binom{4}{2} \binom{4}{2} \binom{11}{1} \binom{4}{1} = 123, 552. $$
But, by the logic used for drawing a full house, shouldn't it be the following:
$$ \binom{13}{1} \binom{4}{2} \binom{12}{1} \binom{4}{2} \binom{11}{1} \binom{4}{1} = 247, 104? $$
OR, conversely, by using the logic used to get a two pair, shouldn't the number of ways of drawing a full house be the following:
$$ \binom{13}{2} \binom{4}{3} \binom{4}{2} = 2,808? $$
As you can see, this isn't just a matter of inconsistent approaches that give the same result; these give different values. I just don't know how I'm supposed to be able to intuit these sort of things, seeing as they seem like the same thing to me.
|
The problem with your $247,104$ is that it counts each two-pair hand two times, according to which of the pairs you mention first. But 5H-5D-7S-7H-9D is the same hand as 7S-7H-5H-5D-9D, so it gets counted both with fives first and with sevens first.
In contrast, for a full house it is unambiguous which value is the one with three cards and which is the one with two.
|
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|
Arranging numbers around a square In how many ways numbers 1 to 12 can be arranged on a sides of squares (5 places on each sides i.e 20 places total) leaving 8 places empty?
I am getting answer as 12c5(selecting 5 numbers)*7c5(selecting another 5 numbers)*2(remaining 2)*5!(arranging the 5 numbers amongst themselves)
But the answer given is 5*19P11..
Can someone please help me.
|
I think the reason your answer is wrong is because you assumed that two sides would be filled on the square and a third side would have two numbers. For example, imagine a square with 3 numbers filled on each side with two numbers empty on each side. You did not account for such a possibility in your computation.
The simple explanation behind the answer given is you fix one side of the square to account for rotations and then choose one of the five places on that fixed side for there to be the fixed number. The 19P11 is the number of ways to select and order the remaining 11 numbers in 19 remaining spots.
I would calculate it actually as 20P12*1/4 which is the number of ways to choose 12 spaces and order 12 numbers in those spaces divided by 4 to account for each rotation being overcounted exactly four times each.
EDIT: The difference between my calculation and the given one is the in which we choose to address rotations being the same. The given answer fixes rotations by automatically choosing a side and number to account for rotations. For the people who are not familiar with fixing rotations in such a manner for non-circle like objects, the safe route would be simply to overcount and then correct for overcounting by multiplying by 1/4, which you can be sure of.
|
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|
Customer problem in poisson process (two products)
Customers arrive at a shop according to a Poisson process at rate
$\lambda$ (/minute), where they choose to buy either product $A$ (with probability p) or product $B$ (with probability $1-p$), independently. Given that during the first hour $5$ customers chose product $B$, what is the probability all the customers that arrived at the shop within the first $10$ minutes, all bought product $A$?
Attempt. Let $N_A(t), N_B(t)$ the the Poisson processes, of rates $\lambda p, ~\lambda(1-p)$, respectively, that count the produts of type $A$ and $B$ bought, respectively, at time $t$. Then the probability we are asked to find is:
$$\sum_{k=0}^{\infty}P(N_A(10)=k~|~N_A(10)+N_B(10)=k)\cdot P(N_A(10)+N_B(10)=k~|~N_A(60)+N_B(60)=k+5)\cdot
P(N_A(60)+N_B(60)=k+5~|~N_B(60)=5),$$
where $$P(N_A(10)=k~|~N_A(10)+N_B(10)=k)=\binom{k}{k}\Big(\frac{\lambda p}{\lambda p+\lambda(1-p)}\Big)^k\Big(1-\frac{\lambda p}{\lambda p+\lambda(1-p)}\Big)^{k-k}=p^k,$$
$$P(N_A(10)+N_B(10)=k~|~N_A(60)+N_B(60)=k+5)=
\binom{k+5}{k}\Big(\frac{10}{60}\Big)^k\Big(1-\frac{10}{60}\Big)^{5}$$
and
$$P(N_A(60)+N_B(60)=k+5~|~N_B(60)=5)= P(N_A(60)=k~|~N_B(10)=5)=P(N_A(60)=k)=
e^{-60\lambda p}\frac{(60\lambda p)^k}{k!}.$$
Am I on the right path?
Thanks a lot in advance!
|
The direct approach here would be to note that all customers who arrived in the first $10$ minutes buying $A$ is the same as none of them buying $B$, so we don't have to worry about the customers who bought $A$ or about the whole Poisson business at all, since we just want the probability that of $5$ customers equidistributed over $60$ minutes none arrived in the first $10$ minutes, which is
$$
\left(\frac{50}{60}\right)^5=\left(\frac56\right)^5=\frac{3125}{7776}\approx40\%\;.
$$
|
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How much velocity can a canister of fuel give a spaceship? I've recently considered the issue of how much velocity a canister of fuel can provide a 'spaceship'. I assumed we could approximate a basic solution If we know the mass of the fuel $m$, the mass of the ship $M$ and the amount of energy in the fuel $E$. So the energy density of the fuel is, $\frac{E}{m} = p$, which is constant. So the amount of velocity gained by an increment of fuel is equal to $\sqrt{\frac{p\cdot{dm}}{M+m}} = dv$. Since $dE = p\cdot{dm} = \frac{1}{2}(m+M)dv^2$. In order to find $v$, I needed to integrate the left hand side from m equals the total mass of the fuel to $m=0$ (assuming the canister is massless).
Is this correct? How, (if at all) is it possible to evaluate this integral?
P.S. I think I'm way off from this derivation https://en.wikipedia.org/wiki/Tsiolkovsky_rocket_equation
Can someone please lead me in the right direction?
|
The equation in the mentioned Wikipedia page is a special case of the general problem of motion equation for a system with variable mass that comes from the conservation of the momentum of the system. You can see the derivation of the pertinent equation at this page.
The pure energetic approach in OP is not correct because the energy is conserved only in the full system, given by the rocket and the combusted fuels and also the heat produced in the combustion. The key parameter that determines the velocity of the roket is the expulsion velocity of the ejected propellant, that depends on the constructive characteristics of the roket.
|
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Is there a reduction formula for $I_n=\int_{0}^{n\pi}\frac{\sin x}{1+x}\,dx$? I haven't been able to manipulate this integral. I need to find the value of $I_n$ for $n=1,2,3,4$ and arrange them in ascending order.
|
Since the sine function oscillates with zeros at integer multiples of $\pi$, we introduce
$$
K_j=\int_{(j-1)\pi}^{j\pi}\frac{\sin x}{1+x}\,dx,\quad j\in\{1,2,3,4\}.
$$
Then, since $x\mapsto 1/(1+x)$ is decreasing and since $\sin(x+\pi)=-\sin(x)$ we find that
$$
K_1>-K_2>K_3>-K_4>0.
$$
The integrals we want to compare are given by
$$
I_1=K_1,\quad I_2=K_1+K_2,\quad I_3=K_1+K_2+K_3,\quad I_4=K_1+K_2+K_3+K_4,
$$
I stop here, and suggest you to try to conclude from the facts above that the correct order is
$$I_2<I_4<I_3<I_1$$
As suggested in the comments, it might be good to have a picture:
|
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Volume of the $N$-dimensional domain $\sum\limits_{k=1}^N (1 + |x_k|^a)^b\le\varepsilon$ I wish to calculate the following $N$-dimensional integral $$I = \int_0^\infty dx_1 \ldots \int_0^\infty dx_{N} \, H\left(\varepsilon - \sum_{k=1}^N (1 + x_k^a)^b\right),$$ where $a, b$ and $\varepsilon$ are positive reals with $\varepsilon > N$. Also, $H(x)$ is the Heaviside step function.
Is there a general way to attack this kind of integrals?
|
$$I = \int\limits_0^\infty dx_1 \ldots \int\limits_0^\infty dx_{N} \, H\left(\varepsilon - \sum_{k=1}^N (1 + x_k^a)^b\right)$$
$$ = \int\limits_0^{\large\sqrt[a]{{\sqrt[b]{\varepsilon-N+1}-1}}} dx_1 \int\limits_0^{\large\sqrt[a]{{\sqrt[b]{\varepsilon-(1+x_1^a)^b-N+2}-1}}} dx_2 \ldots \int\limits_0^{\large\sqrt[a]{{\sqrt[b]{\varepsilon-\sum\limits_{k=1}^{N-2}(1+x_k^a)^b-1}-1}}} dx_{N-1}\int\limits_0^{\large\sqrt[a]{{\sqrt[b]{\varepsilon-\sum\limits_{k=1}^{N-1}(1+x_k^a)^b}-1}}} dx_{N}$$
$$ = \int\limits_0^{\large\sqrt[a]{{\sqrt[b]{\varepsilon-N+1}-1}}} dx_1 \int\limits_0^{\large\sqrt[a]{{\sqrt[b]{\varepsilon-(1+x_1^a)^b-N+2}-1}}} dx_2 \ldots \int\limits_0^{\large\sqrt[a]{{\sqrt[b]{\varepsilon-\sum\limits_{k=1}^{N-2}(1+x_k^a)^b-1}-1}}} \large\sqrt[a]{{\sqrt[b]{\varepsilon-\sum\limits_{k=1}^{N-1}(1+x_k^a)^b}-1}}\,dx_{N-1},$$
and the further exact integration seems impossible.
Various substitutions also do not simplify calculations.
|
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Find point where a line of multiple vertices overlaps itself Since I'm not familiar with a lot of mathematical terminology, I will explain this problem with a little story.
Imagine you and your friend Anne have a piece of string each, and place it on a coordinate system. You decide to form a loop with your string, so the string will overlap itself at some point on the coordinate system.
Now both you travel along the string with a marker, and after every centimeter, you note down the coordinates of that Point.
Once you are done, you will have two lists of vertices that are on their respective strings. From looking just at these lists, how can I find out which one of them contains a loop using mathematics? Furthermore, how can I find the exact coordinates of the point of intersection (this point may not be part of the listed vertices, as this list has a 'resolution' of one centimeter)?
Here is a little MS Paint thing I made to illustrate the problem.
|
I think you're going to have to make some assumptions here. There can certainly be examples of lists which could represent both overlapping and non-overlapping strings, as there are lots of choices of strings to interpolate the points, and we may have that only some of them cause a self-intersection. There will be cases (such as in your illustration) in which this ambiguity does not arise, but there still may be many possible points of intersection.
If you make the assumption that the string follows a straight line between the sampled points (a reasonable assumption if your step size is sufficiently small), the above problem goes away, and there is an easy solution; for each of the line segments (consecutive pairs of sampled points), you simply test if it overlaps any other line segment. If so, you can find the point of intersection between the line segments; if not, the string does not self-intersect.
|
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Maximizing the sum of the squares of numbers whose sum is constant I wonder how one goes about to find the maximum of $\sum v_i^2$, the $v_i$'s being positive integers whose sum $\sum_i v_i$ is fixed.
|
Imagine that you're aiming to cover as much of the $\sum_i v_i$ square as possible:
The bigger the largest inner square, the closer it gets to covering more of the background square. The maximum sum of squares is reached when all but one of the $v_i$ is at the specified minimum - in this case, $1$.
|
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Let $f(x) = 5x+9$. Show that $\lim \limits_{x \to -3}f(x)=-6$
Let $f(x)=5x+9$. Show that $\lim \limits_{x \to -3}f(x)=-6$
A couple of questions about showing this and proving this. As I'm working through the problem I don't understand how I proved or showed anything as I don't understand the results I get.
I understand that $\delta$ is some range between $x$ and $L$, and $\epsilon$ is some range between $f(x)$ and $L$, so what do these proofs tell me?
An easier question, though, is while proving this, a term appears (my instructors notes) and I don't know where it came from:
We want to find $\delta$ such that $\left|f(x)-(-6)\right|=\left|(5x+9)-(-6)\right|\lt \epsilon$ whenever $0 \lt \left|x-(-3)\right| \lt \delta$
But $\left|(5x+9)-(-6)\right|=\left|5x+15\right|=5\left|x+3\right|=5\left|x-(-3)\right|\lt 5 \delta$ (where does the $5\delta$ come from?). So, we want $5\delta=\epsilon$, which implies that $\delta=\frac{\epsilon}{5}$
|
You want $\left|f(x)-(-6)\right|=5\left|x-(-3)\right|< \epsilon$, that is $\left|x-(-3)\right|\lt \epsilon/5$ and that is guaranteed if you take $\delta=\epsilon/5$.
|
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Turing Decryption Example I know this exact same question exists but I am still having problems in understanding it. The following is given in the text:
The message m can be any integer in the set {0,1,2,…,p−1}; in particular, the message is no longer required to be a prime(p is a prime). The sender encrypts the message m to produce m∗ by computing:
m∗=remainder(mk,p).
Multiplicative inverses are the key to decryption in Turing’s code. Specfically, we can recover the original message by multiplying the encoded message by the inverse of the key:
Let us name the equations from top to bottom from 1 to 3.
Now I am just writing what I understood and it maybe wrong, so I will appreciate if someone tells me where I am wrong.
We know m*=rem(mk, p)=mk%p, multiplying this equation by (k^-1) will give us equation 1.
Corollary 4.5.2 states that 'a is congruent to rem(a, n) modulo n'. Thus we obtained equation 2 by replacing equality with the congruency sign, as a=mk and rem(a,n)=m*, thus they are congruent to each other and multiplying by (k^-1) on both sides doesnt make a difference.
Now my confusion is (k^-1) != 1/k as k^-1 is not simply multiplicative inverse but multiplicative inverse modulo p, i.e., k.(k^-1) is congruent to 1 modulo p. Then how in equation 2 they replaced k.(k^-1) by 1 which resulted in equation 3.
And if k^-1=1/k, they wouldnt talk about finding k^-1 using the Pulverizer, which they later do.
|
As you said, $k\cdot k^{-1} \equiv 1 \pmod p$. Therefore, whenever we have $k\cdot k^{-1}$ in a $\pmod p$ equation, we can replace it with $1$, since they are congruent in such a system.
Here's an example:
$$2x \equiv 3 \pmod{5}$$
Now, after doing some guess and check, you can find that $2\cdot 3 \equiv 1 \pmod 5$, meaning $2^{-1} \equiv 3 \pmod 5$. Notice that here, $2^{-1} \pmod 5$ is not just the same as $\frac 1 2$: This is because $\frac 1 2$ is the multiplicative inverse of $2$ in the rational numbers while $3 \pmod 5$ is the multiplicative inverse in the $\pmod 5$ numbers.
Now, multiply both sides by $3$:
$$6x \equiv 9 \pmod{5}$$
We have $6x-5x=x$ and $9-5=4$, so:
$$x \equiv 4 \pmod 5$$
Notice how, by multiplying both sides by $3$, we were able to replace the coefficient of $6$ with $1$ since $6 \equiv 1 \pmod 5$. This will always happen whenever we get $k\cdot k^{-1}$ in a modular equation: $k\cdot k^{-1}$ can always be replaced by just $1$ in a modular equation because that's how modular inverses work.
|
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Real Analysis, Folland Proposition 2.29 Modes of Convergence Background Information:
$f_n\rightarrow f$ in $L^1$ $\Leftrightarrow$ $\forall\epsilon > 0,\exists N$ $\forall n\geq N$ $\int |f_n - f| < \epsilon$
A sequence $\{f_n\}$ of measurable complex-valued function on $(X,M,\mu)$ converges in measure to $f$ if for every $\epsilon > 0$, $$\mu\left\{x:|f_n(x) - f(x)|\geq \epsilon\}\right) \rightarrow 0 \ \ \text{as} \ \ n\rightarrow \infty$$
Question:
Proposition 2.29 - If $f_n\rightarrow f$ in $L^1$, then $f_n\rightarrow f$ in measure.
Attempted proof - Let $E_{n,\epsilon} = \{x:|f_n(x) - f(x)|\geq \epsilon\}$. Then, since $f_n\rightarrow f$ in $L^1$ we have for all $\epsilon > 0$ there exists an $N\in\mathbb{N}$ such that $$\int_{E_{n,\epsilon}}|f_n - f| \leq \int |f_n - f| < \epsilon \ \ \forall n\geq N$$
We observe that $$\int |f_n - f| \geq \int_{E_{n,\epsilon}}|f_n - f| = \int |f_n - f| \chi_{E_{n,\epsilon}}\geq \epsilon \mu(E_{n,\epsilon})$$
Hence we have $$\epsilon^{-1}\int |f_n - f| \geq \mu(E_{n,\epsilon})$$ Since $\epsilon$ is arbitrary we have $\epsilon^{-1}\int |f_n - f|\rightarrow 0$ which thus implies that $\mu(E_{n,\epsilon}) = 0$.
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I believe your approach is correct. I wrote something up before realizing you had already provided a proof (more-or-less what you have, just by proving the contrapositive):
Let $E_{n,\epsilon}=\{x\colon|f_{n}(x)-f(x)|\geq\epsilon\}$. Suppose
$f_{n}$ does not converge to $f$ in measure so that there exists
an $\epsilon>0$ and $\delta>0$ with $\mu(E_{n,\epsilon})\geq\delta$
for infinitely many $n$. This implies $$\int|f_{n}-f|\geq\int\mathcal{X}_{E_{n,\epsilon}}|f_{n}-f|\geq\epsilon\delta$$
for infinitely many $n$ so that $f_n$ does not converge to $f$ in $L^1$.
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Prove the inequalities $|e^{x}-1|\leq e^{|x|}-1\leq |x|e^{|x|}$ Prove that $|e^{x}-1|\leq e^{|x|}-1\leq |x|e^{|x|}$ for all $x\in \mathbb{C}$. I did this by Maclaurin series of $e^x$,
$$|e^{x}-1|\leq |x|+\frac{|x|^2}{2!}+\frac{|x|^3}{3!}+\mathcal{O}(|x|^{4})=e^{|x|}-1 \\
\leq |x|^2+\frac{|x|^3}{2!}+\frac{|x|^4}{3!}+\mathcal{O}(|x|^{5})=|x|e^{|x|}.$$
My teacher disliked the last inequality, and I don't know why. Is there a way to fix this problem?
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$$e^{|x|}-1\leq |x|e^{|x|}\iff 1-e^{-|x|}\le|x|\iff\int_0^{|x|}e^{-t}\,dt\le\int_0^{|x|}1\,dt$$
which is clearly true because $e^{-t}\le1$ for $t\ge0$
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Plotting an arc with no center point - a practical solution please! I need a mathematical solution to a very practical problem (laying a patio).
The attached will hopefully explain.
The center of the circle for the arc we wish to have is inaccessible (ie in the house behind walls).
There are 2 fixed points the arc must intersect.
There is an existing arc drawn using these 2 fixed points, but this circle is too small - radius 539cm
The desired arc is more shallow so would have a greater radius.
I think I need 10 or so measurements from the center of this existing circle, to the new arc.
These points can then be plotted on the ground and the dots joined up (obviously more than 10 will me more accurate, but 10 seems a sensible number)
I am sure this is possible, but my mathematical knowledge is not good enough, sorry. Answers or a simple formula I can apply would be very much appreciated.
Anyone out there to help?
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If you can measure tangent angles in the diagram the following could be of use in a scaled geometrical construction:
$$ R_{old}= \dfrac{r_2^2-r_1^2}{ 2(r_2 \sin \beta - r_1 \sin \alpha )}$$
Along perpendicular bisector of given points $(1,2)$ mark old center of circle $O$ with this radius and a new center $N$ with a new desired (increased/reduced) radius passing through $(1,2)$ on bisector. Sketch not to scale.
The above result for Irregular Pie is obtained by eliminating invariant semi-chord length $L$ from Circle property that product of two line segments is constant.
$$ r\cdot (2 R \sin \theta -r)= L^2 $$
EDIT1
Yet another practical way to find $R$ is the relation $ R = \dfrac{s}{\alpha_ d +\beta_d} $ for Diagon from the Gauss-Bonnet theorem. Arc length $s=1P2 $ and corner angles are to be measured.
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What motivated trying to express the signature of a manifold as a linear combination of pontrjagin numbers? I have tried reading a proof of the signature theorem but it is way beyond me, is there a way to motivate, in english, why anyone even started searching for such a formula? Why would one assume that the signature of a manifold had anything to do with pontrjagin numbers?
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First one notes that the signature is a bordism invariant, additive, and multiplicative, hence defines $\sigma:\Omega \otimes \mathbb Q\to \mathbb Q$.
Next you note that every ring morphism from the oriented bordism ring $\Omega \to \mathbb Q$ factors through $\Omega \otimes \mathbb Q$. But we know that on the latter space the Pontryagin classes are a complete invariant. That means that we should able to write down any such morphism in terms of these classes.
Lastly, one computes necessary conditions on how a ring morphism looks like, and one obtains that these are always given by a multiplicative sequence (plugging in the Pontryagin classes), which are in correspondence to power series. This implies that the signature needs to come from a power series.
Now, verifying the power series for the signature is easier (and more mysterious!) than you think. This is just done using a trick from complex analysis called the residue theorem.
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Are cyclotomic polynomials irreducible modulo a prime? I am actually interested on cyclotomic polynomials of the form $x^n+1$ (i.e. $n$ is a power of 2, for large $n$).
Are these polynomials irreducible modulo a prime $q$? I already saw that $X^2+1$ is not irreducible modulo 5 because $Z_5$ contains roots of -1. What about $x^{512}-1$? Is there a general criterion?
Thanks
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I find this relatively general answer from the Finite Field book by Rudolf, theorem 2.47, helpful:
The cyclotomic field $K^{(n)}$ is a simple algebraic extension of $K$. If $K=\mathbb{F}_q$ with $\gcd(q,n)=1$, then the cyclotomic polynomial $Q_n$ factors into $\phi(n)/d$ distinct monic irreducible polynomials in $K[x]$ of the same degree $d$. Here $d$ is the least positive integer such that $q^d \equiv 1 \mod n$.
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Is something wrong with this solution for $\sin 2x = \sin x$? I have this question. What are the solutions for $$
\sin 2x = \sin x; \\ 0 \le x < 2 \pi $$
My method:
$$ \sin 2x - \sin x = 0 $$
I apply the formula $$ \sin a - \sin b = 2\sin \left(\frac{a-b}{2} \right) \cos\left(\frac{a+b}{2} \right)$$
So:
$$ 2\sin\left(\frac{x}{2}\right) \cos\left(\frac{3x}{2}\right) = 0 $$
$$ \sin\left(\frac{x}{2}\right)\cos\left(\frac{3x}{2}\right) = 0 $$
Here one of the factors has to be $0$,
$$ \sin x = 0 \ \Rightarrow \ x = 0 \ or \ x = \pi $$
$$ \sin\left(\frac{x}{2}'\right) = 0 \ \Rightarrow \ x = 0 ;\ x \text{ can't be } \pi \text{ because of its range} $$
$$ \cos x = 0 \ \Rightarrow \ x = \frac{\pi}{2} \text{ or } \ x = \frac{3\pi}{2} $$
$$ \cos\left(\frac{3x}{2}\right) = 0 => x = \frac{\pi}{3} \text{ or } x = \pi $$
So the solutions are : $$ 0, \pi, \frac{\pi}{3} $$
I have seen other methods to solve this, so please don't post them. I'm really interested what's wrong with this one.
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$\sin A=0\implies A=n\pi$
$\cos B=0\implies B=(2m+1)\pi/2$
$m,n$ are arbitrary integers
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Convergent of sequence looking like Riemann zeta function I have a question that: Given a non-negative sequence {$\epsilon_n$} ($n \in \mathbb{Z+}$) such that $\lim_{n \rightarrow +\infty}\epsilon_n = 0$.
Can we conclude that $\sum_{n \in \mathbb{Z+}}\frac{\epsilon_n}{n} < +\infty$
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Not sure what this has to do with zeta functions. The answer is no: take $\epsilon_n=1/\ln(n+1)$. Use the integral test to show the series diverges.
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How many applicants need to apply in order to meet the hiring target? Cam needs to hire $30$ new employees. Ten percent $(10\%)$ of applicants do not meet the basic business requirements for the job, $12\%$ of the remaining applicants do not pass the pre-screening assessment, $23\%$ of those remaining applicants do not show up for the interview, and $5\%$ of those remaining applicants fail the background investigation. How many applicants need to apply in order to meet the hiring target?
$$A)\ 30\ \ \ \ \ \ B)\ 45\ \ \ \ \ \ C)\ 50\ \ \ \ \ \ D)\ 52\ \ \ \ \ \ E)\ 60$$
My answer:
I added $10+12+23+5=50$
That gave us $50\%$. I took a look at the answers and getting $50\%$ of $E)\ 60$ is $30.$
However, when I tried to solve it, I took $(0.50)(30)=15$ I then added $15$ to $30$ and it gave me $B)$ $45.$ Can someone please show me how to solve this?
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To elaborate on John's answer: You cannot simply add $10$, $12$, $23$, and $5$ percent, because the percentages are "in tandem". After you cut away the first $10$ percent, you only cut away $12$ percent of those who remain, not of those who started the selection process.
As a simpler example, consider the following selection process: You first eliminate those who are born in an odd-numbered month, and then you eliminate those who are born on an odd-numbered day. If we assume (for the sake of argument) that each step removes $50$ percent of the remaining candidates, it still is not true that at the end of the two steps, you have eliminated $50+50 = 100$ percent of the candidates.
Instead, what happens is that after the first step, $50$ percent have been eliminated, and $50$ percent remain. The second step then eliminates $50$ percent of the remaining $50$ percent, or $25$ percent. That leaves only the last $25$ percent.
In the same way, in your actual problem, the first step eliminates $10$ percent, leaving $90$ percent. The second step eliminates $12$ percent of that $90$ percent, leaving $88$ percent of $90$ percent, or $79.2$ percent. And so on.
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Spending and left over amount/percentages Question 1) Cam would like to spend some of his money on friends. Of the $350$ dollar budget he was given, he has to spend $75.00$ dollars on office supplies, $10.00$ dollars per person for an entrance fee to the concert that his top $3$ friends and himself will attend later this month, and $17.00$ dollars for pizza for his lunch tomorrow. How much money does that leave for me to spend on each of his $12$ friends?
A) $16.17$ B) $18.17$ C) $19.16$ D) $20.00$ E) $218.00$
I added $(75)+(10*4)(17)=132$
Then $350-132=218$ which would give us E) $218.00$
I thought about it said each of his friends. Does that mean $218.00/12=18.17$ which would give us B) $18.17$
I wanted to know is there an easy way to think about this and what would be the answer. Can someone please help me?
Question 2) If you increase your current score of $85%$ by $7%$, what score will you have?
A) $90.5$ B) $90.85$ C) $90.95$ D) $91%$ E) $91.25$
I added $(0.85) + 7%$ which gave me $90.95$ percent. I wanted to know if this answer is correct.
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Question 1. He spends more on the three friends then he does on the other 9. he spends 30 dollars on them. On top we have he spends 75 dollars + 17 dollars on himself. So we have the budjet for each friend to be $((350-75-17-30)/12)=19$, but remember that 3 of his 12 friends receive an extra 10 dollars so each friend receives $(3*29+9*19)/12=21.5$ dollars on average. Since these are neither of the options I would go with B and state the question is not well stated it is unkown who he spends the 75 or 17 dollars on.
Question 2. the proper way to do this is $(.85)*(1.07)=90.95$ i.e. current probability times the probability gained.
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Irreducible polynomial over $\mathbb{Q}$ implies polynomial is irreducible over $\mathbb{Z}$ Let $f(x) \in \mathbb{Z}[x]$ be a polynomial of degree $\geq 2$. Then choose correct
a) if $f(x)$ is irreducible in $ \mathbb{Z}[x] $ then it is irreducible in $ \mathbb{Q}[x] $.
b) if $f(x)$ is irreducible in $ \mathbb{Q}[x] $ then it is irreducible in $ \mathbb{Z}[x] $.
(1) is definitely true, for (2) $f(x)=2(x^2+2)$ clearly irreducible over $\mathbb{Q}[x]$
But I am confused about whether $f(x)$ is irreducible over $\mathbb{Z}[x]$ or not? According to Gallian, as 2 is non unit in $\mathbb{Z}$, $f(x)$ is reducible over $\mathbb{Z}[x]$, (2) is false.
But definition of irreducible polynomial on Wikipedia says a polynomial is reducible if it can be written as product of non constant polynomials hence $f(x)$ is irreducible over $\mathbb{Z}[x]$ accordingly (2) is true .
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You are totally correct, (1) is true and (2) is false. The statement you quote from Wikipedia is only true, if the coefficients come from a field.
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A weight problem
I am having a hard time solving the following puzzle. Could you please me to figure it out?
A chemist has a set of five weights. She knows that it includes one 1-gram weight, and also one each 2-, 3-, 4-, and 5-gram weights, but because they are unmarked, she has no way of telling them apart except by placing them on a balance. She may place any combination of weights on each of the two pans and determine if one side is heavier than the other or if they balance.
Show how in five weighings she can identify each of the weights.
Thank you.
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Okay, this isn't a complete solution as this does seem intricate but maybe this is a good starting point.
Note there are $5! = 120$ possible ways the weights can be determined. Each weighing will have $3$ possible outcomes. The weight balances, the left is heavier, and the right is heavier. If you can figure out a way to weigh things so that each outcome covers the same number of possibilities, the first weigh will reduce your point distributions to $40$ the second weighing to $14$ the third to $5$, the fourth to $2$ and the fifth to $1$.
That would be ideal.
But it can't be done.
So I evaluated how the distributions are accounted for by different combinations of weigh.
Example: If I weigh one weight (call it A) and another (call it B) then A < B accounts for 60 cases. B > A for 60 cases. But B = A will never occur. This is not a very desirable distribution.
If I weigh one weight (A) against two weights (BC) then, A > B account for A = 4; B= 1;C=2 OR A=4; B=2; C=1 OR A=5;B=1;C=2 OR A=5;B=2; C= 1; etc. basically A > BC has 6 possible solutions. A = BC has 8 possible solutions and A < BC has 106 solutions. This is a TERRIBLE distribution.
I won't bore you but if I weigh two against two I get $AB < CD$ has $48$ solutions. $AB = CD$ has $24$ solutions and $AB > CD$ has $48$ solutions. This is the best distribution.
Here's where I gave up and took a leap of faith. I figured that the second weighing between $AC$ and $BE$ would be the best second weighing for even distributions. It intuitively seem to redistribute the light pair among the heavy pair and to get the 5th unknown in early.
But I dunno.
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Galois extension definition. Let $L,K$ be fields with $L/K$ a field extension. We say $L/K$ is a Galois extension if $L/K$ is normal and separable.
I don't fully understand this definition, is it saying that
1) $L$ has to be the splitting field for some polynomial in $K[x]$ and that polynomial must not have any repeated roots, or is it saying that
2) $L$ has to be the splitting field for all polynomials in $K[x]$ and all polynomials must not have repeated roots?
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the extension $L/K$ is galoisien is equivalently to each $x\in L$,
$x$ is algebraic over $K$ and the minimal polynomial of $x$ has
simple roots and are all in $L$. so by primitive element theorem
we get: the extension $L/K$ finite and galoisienne is equivalent
to your proposal 1) but not to 2).
but in the infinity case; $L/K$ galoisienne is not equivalent
either to 1) or 2)
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Why every point of a function where differentiation exists has only one tangent? Can anyone help me out?
Why every point of a function where differentiation exists has only one tangent?
I know the slope at any point of any function is defined by differentiation at that point.But there may be another straight line which touches the function at that point but slope is different from the differentiation value at that point. But this does not happen(I am not considering singular point).
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It's not just about a line touching. (Every line through the point touches the graph there.) It's the best linear approximation, in the sense that the error $f(x)-L(x)$ goes to $0$ faster than $x$ approaches $a$. (Here $L(x)$ is the linear function whose graph we're discussing.)
EDIT: To be more precise, if $f$ is differentiable at $x=a$, then the linear function $L(x)=f(a)+f'(a)(x-a)$ is the unique linear function with the property that
$$\lim_{x\to a}\frac{f(x)-L(x)}{x-a} = 0.$$
Note that if we take $L(x)=f(a)+m(x-a)$, this limit becomes
$$\lim_{x\to a}\frac{f(x)-f(a)-m(x-a)}{x-a} = \lim_{x\to a}\frac{f(x)-f(a)}{x-a} -m = f'(a)-m,$$
and this is $0$ if and only if $m=f'(a)$.
If $f$ fails to be differentiable, this limit will never be 0. Try it, for example, with $f(x)=|x|$ at $x=0$. If you take a line of slope $m$ and $x>0$, this limit is $\lim_{x\to 0^+} \dfrac{x-mx}x = 1-m$, so $m$ would have to be $1$, but if $x<0$, this limit is $\lim_{x\to 0^-} \dfrac{-x-mx}x = -1-m$, so $m$ would have to be $-1$. There therefore is no tangent line at $x=0$.
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Bounds on $f(k ;a,b) =\frac{ \int_0^\infty \cos(a x) e^{-x^k} \, dx}{ \int_0^\infty \cos(b x) e^{-x^k}\, dx}$ Suppose we define a function
\begin{align}
f(k ;a,b) =\frac{ \int_0^\infty \cos(a x) e^{-x^k} \,dx}{ \int_0^\infty \cos(b x) e^{-x^k} \,dx}
\end{align}
can we show that
\begin{align}
|f(k ;a,b)| \le 1
\end{align}
for $ 0<k \le 2$ and $a\ge b$?
This question was motivated by the discussion here.
Note that for $k=1$ and $k=2$ this can be done, since
\begin{align}
\int_0^\infty \cos(a x) e^{-x^1} \,dx=\frac{1}{1+a^2}\\
\int_0^\infty \cos(a x) e^{-x^2} \,dx=\frac{\sqrt{\pi}}{2}e^{-a^2/4}\\
\end{align}
So, we have that
\begin{align}
f(1;a,b)&=\frac{1+b^2}{1+a^2} \\
f(2;a,b)&=e^{ \frac{b^2-a^2}{4}}
\end{align}
In which case, we have that the conjectured bound is true.
Edit: The bounty was posted specifically to address this question and a question raised by Jack D'Aurizio in the comments.
The question is:
Let
\begin{align}
g_k(z)=\int_0^\infty \cos(zx) e^{-x^k} dx
\end{align}
What is the largest value of $k$ such that $g_k(z)$ is non-negative and decreasing for $z\in \mathbb{R}^{+}$?
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Yes, the Fourier transform of $\exp(-|x|^k)$ is positive and decreasing
for all $k$ such that $0 < k \leq 2$.
This follows from the known case of $k=2$ (Gaussians) via an argument of
B.F.Logan cited in my 1991 paper with Odlyzko and Rush:
Noam D. Elkies, Andrew M. Odlyzko, and Jason A. Rush: On the packing densities of superballs and other bodies, Invent. Math. 105 (1991), 613-639.
See Lemma 5 on page 626 (with $k=\sigma$; in that paper
we needed only positivity, not that the Fourier transform is decreasing).
The key is that $\exp(-t^{k/2})$ is a "totally monotone" function of $t>0$
(its $n$-th derivative has sign $(-1)^n$ for all $t>0$),
and decays to zero as $t \to \infty$, whence it is
a nonnegative mixture of decreasing exponentials $e^{-ct}$ $(c>0)$ by
Bernstein's theorem.
Taking $t=x^2$ we deduce that $\exp(-|x|^k)$ is a nonnegative mixture of
Gaussians $\exp(-cx^2)$ $(c>0)$. Since the Fourier transform of $\exp(-cx^2)$
is positive and decreasing for all $c>0$, the same is true of the Fourier
transform of $\exp(-|x|^k)$, QED.
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If $x\in \left(0,\frac{\pi}{4}\right)$ then $\frac{\cos x}{(\sin^2 x)(\cos x-\sin x)}>8$
If $\displaystyle x\in \left(0,\frac{\pi}{4}\right)\;,$ Then prove that $\displaystyle \frac{\cos x}{\sin^2 x(\cos x-\sin x)}>8$
$\bf{My\; Try::}$ Let $$f(x) = \frac{\cos x}{\sin^2 x(\cos x-\sin x)}=\frac{\sec^2 x}{\tan^2 x(1-\tan x)} = \frac{1+\tan^2 x}{\tan^2(1-\tan x)}$$
Now Put $\tan x= t \in (0,1)\;,$ Then $$h(t) = \frac{1+t^2}{t^2(1-t)}\;\;, 0<t<1 $$ where $h(t)=f(x)$.
Now How can i solve it after that, Help Required ,Thanks
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We are to prove that $$\frac{\cos x}{\sin x (\cos x - \sin x)}> 8 \sin x $$
By Cauchy-Schwarz Inequality, since all quantities involved are positive
$$\bf{LHS = }\frac{1}{\cos x} + \frac{1}{\cos x-\sin x} \ge \frac{4}{\sin x + \cos x - \sin x} = \frac{4}{\sin x}$$
For the given range of x, we have $1> 2 \sin^2 x$
So $$\frac{4}{\sin x} > \frac{4}{\sin x} \times 2 \sin^2 x = 8 \sin x$$
and we are done
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Properly discontinuous group actions - Hausdorffness I was told to prove the following: If an action is free and satisfies that each point has a neighborhood $U$ satisfying $U \cap gU=\emptyset$ except for finitely many $g\in G$, and moreover the space is Hausdorff, then each point has a neighborhood such that $g_1\neq g_2\implies g_1U\cap g_2U=\emptyset$.
I'm pretty lost here - I don't have any intuition for why Hausdorfness is relevant either..
|
Let $x\in X$, consider a neigborhood $U$ of $x$ such that $V_x=\{g\in G:g(U)\cap U\neq \phi\}$ is finite. Write $V_x=\{y_1,...,y_n\}$ there exists $x\in U_i, y_i\in V_i$ such that $U_i\cap V_i$ is empty since $X$ is separated, set $W=U\cap_iU_i$, $W$ is an open subset containing $x$. Remark that if $g\in G$ and $g\neq I_d$ and $g(W)\cap W$ is not empty, then there exists $z\in W$ such that $g(z)\in W$, since $W\subset U$, $z\in V_x$, this is impossible since $V_x \cap W$ is empty since $W\cap V_i$ is empty.
let $g_1,g_2\in G, g_1\neq g_2$, consider $z\in g_1(W)\cap g_2(W)$. This is equivalent to saying that $z=g_1(w_1)=g_2(w_2)$ where $w_1,w_2\in W$, we deduce $w_1=g_1^{-1}g_2(w)$ this is a contradiction since $g_1^{-1}g_2(W)\cap W$ is empty.
|
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|
Finding the principal part for the Laurent series How can I find the principal part of the Laurent series for $f(z)=\dfrac{\pi^2}{(\sin \pi z)^2}$ centered at $k$ where $k \in \mathbb{Z}$.
I think there are two ways to do it either use the formula $a_k$ for Laurent coefficients or expand manipulate $\sin^2 \pi z$ and solve for the coefficients. I am not sure if any of these ways are efficient.
|
By setting $z:=k+\varepsilon$, $k \in \mathbb{Z}$, $\varepsilon \to 0$, one has by the Taylor series expansion:
$$
\sin^2 \pi z=\left(\sin (\pi k+\pi\varepsilon)\right)^2=\sin^2(\pi\varepsilon)=\pi^2 \varepsilon^2-\frac{\pi^4 \varepsilon^4}{3}+O(\varepsilon^6)
$$ giving
$$
f(z)=\frac{\pi^2}{\pi^2 \varepsilon^2-\frac{\pi^4 \varepsilon^4}{3}+O(\varepsilon^6)}=\frac1{\varepsilon^2(1-\frac{\pi^2 \varepsilon^2}{3}+O(\varepsilon^4))}=\frac1{\varepsilon^2}+\frac{\pi ^2}{3}+O(\varepsilon^2)
$$ that is, as $z \to k$,
$$
f(z)=\frac{\pi^2}{\sin^2 \pi z}=\frac1{(z-k)^2}+\frac{\pi ^2}{3}+O((z-k)^2),
$$
the principal part is
$$
\frac1{(z-k)^2}.
$$
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
Are there ways to solve equations with multiple variables? I am not at a high level in math, so I have a simple question a simple Google search cannot answer, and the other Stack Exchange questions does not either. I thought about this question after reading a creative math book. Here is the question I was doing, which I used the solutions manual in shame(Not exact wording, but same idea):
The question in the blockquotes below is not the question I am asking for answers to. Some misunderstood what I am asking. What I am asking is in the last single-sentence paragraph.
Suppose $a_2,a_3,a_4,a_5,a_6,a_7$ are integers, where $0\le a_i< i$.
$\frac 57 = \frac {a_2}{2!}+\frac {a_3}{3!}+\frac {a_4}{4!}+\frac {a_5}{5!}+\frac {a_6}{6!}+\frac {a_7}{7!}$
Find $a_2+a_3+a_4+a_5+a_6+a_7$.
The solution to this particular question requires that $a_7$ and the other variables in later steps of the algebra process to be remainders when both sides are to be divided by an integer. I am now wondering what if an equation ever comes up where the method to solve for the question above cannot work due to the variables not being able to return remainders. Thus, my question is whether it is possible to solve algebraic equations with more than two variables and most variables having constant coefficients, is not a system of equations, and the variables are assumed to be integers, and the solution is unique.
Does such a way to solve such equations in general exist? If so, please explain. What is this part of math called?
Thank you.
|
Multiplying by the common denominator $7!$, we get
$$ 2520 a_2 + 840 a_3 + 210 a_4 + 42 a_5 + 7 a_6 + a_7 = 3600$$
Take this mod $7$:
$$ a_7 \equiv 2 \mod 7$$
Since $0 \le a_7 < 7$, this means $a_7 = 2$, and then (substituting this and dividing by $7$):
$$ 360 a_2 + 120 a_3 + 30 a_4 + 6 a_5 + a_6 = 514$$
Taking this mod $6$:
$$ a_6 \equiv 4 \mod 6$$
So $a_6 = 4$.
Continuing the process, we get
$$ a_5 = 0,\ a_4 = 1,\ a_3 = 1,\ a_2 = 1 $$
|
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|
$\vDash \phi \Rightarrow\, \vDash \psi$ How can I prove the last part of the following exercise.
Show that $R\vDash \phi \Rightarrow R\vDash\psi$ for all structure $R$ implies $\vDash \phi \Rightarrow\, \vDash \psi$, but not vice versa.
I have the first part. Ans: If $\vDash \phi$ then for all structures $R$, $\phi$ is true, by hypothesis, it implies that $R\vDash \psi$ for all hypothesis, but that's by definition is $\vDash \psi$. Then, we conclude $\vDash \phi \Rightarrow \vDash \psi$.
But my concern is with the last one part of the question. "but no vice versa". That is,
$\vDash \phi \Rightarrow\, \vDash \psi$ doesn't implies
$R\vDash \phi \Rightarrow R\vDash\psi$ for all structure $R$.
Why? I can't see even why it is actually true, I am using the fact that $\vDash \phi$ says for all structure and for all interpretation, we have $\phi$, right? Then, why not, $R\vDash \phi \Rightarrow R\vDash\psi$ for all structure $R$.
The exercise was taken from the semantics section of book's VanDalen. (3.4.7).
|
The statement $\vDash \phi \Rightarrow\, \vDash \psi$ tells you that if $\phi$ is true in all structures, then $\psi$ is true for all structures. But if $\phi$ is not true in all structures, it tells you nothing at all. That is, if $\phi$ is any statement that is not true in all structures, then then the implication $\vDash \phi \Rightarrow\, \vDash \psi$ automatically holds no matter what $\psi$ is.
So for instance, if $\phi$ is a statement that is true in some structures but not all structures, and $\psi$ is a statement that is false in all structures, then $\vDash \phi \Rightarrow\, \vDash \psi$ is true but there exist structures $R$ such that $R\vDash \phi \Rightarrow\, R\vDash \psi$ is false (namely, any structure in which $\phi$ is true).
|
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|
Need a solution to this Integration problem How to evaluate:$\displaystyle\int_{0}^{r}\frac{x^4}{(x^2+y^2)^{\frac{3}{2}}}dx$
I have tried substituting $x =y\tan\ A$, but failed.
|
$\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle}
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\newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
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\newcommand{\Li}[1]{\,\mathrm{Li}_{#1}}
\newcommand{\ol}[1]{\overline{#1}}
\newcommand{\pars}[1]{\left(\,{#1}\,\right)}
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\newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
\newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
\newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
Note that
$\ds{I \equiv \int_{0}^{r}{x^{4} \over \pars{x^{2} + y^{2}}^{3/2}}\,\dd x =
y^{2}\ \overbrace{\int_{0}^{R}{x^{4} \over \pars{x^{2} + 1}^{3/2}}
\,\dd x}^{\ds{\equiv\ \,\mathcal{I}}}\ \,,\qquad
R \equiv {r \over \verts{y}}}$.
\begin{align}
\,\mathcal{I} & = \int_{0}^{R}{x^{4} \over \pars{x^{2} + 1}^{3/2}}\,\dd x =
\int_{0}^{R}{\bracks{\pars{x^{2} + 1} - 1}^{2}\over \pars{x^{2} + 1}^{3/2}}
\,\dd x
\\[4mm] & =
\int_{0}^{R}\pars{x^{2} + 1}^{1/2}\,\,\dd x -
2\int_{0}^{R}\pars{x^{2} + 1}^{-1/2}\,\dd x + \int_{0}^{R}\pars{x^{2} + 1}^{-3/2}\,\,\dd x
\end{align}
With $\ds{x \equiv \tan\pars{t}}$:
\begin{align}
&\int\bracks{\sec\pars{t} - 2\cos\pars{t} + \cos^{3}\pars{t}}\,\dd t =
\ln\pars{\sec\pars{t} + \tan\pars{t}} -
\int\bracks{1 + \sin^{2}\pars{t}}\cos\pars{t}\,\dd t
\\[4mm] = &\
\ln\pars{\sec\pars{t} + \tan\pars{t}} -
\sin\pars{t} - {1 \over 3}\,\sin^{3}\pars{t}
\\[4mm] & =
\ln\pars{\root{x^{2} + 1} + x} - {x \over \root{x^{2} + 1}} -
{1 \over 3}\,{x^{3} \over \pars{x^{2} + 1}^{3/2}}
\end{align}
\begin{align}
\color{#f00}{\,\mathcal{I}} & =
\color{#f00}{\ln\pars{\root{R^{2} + 1} + R} - {R \over \root{R^{2} + 1}} -
{1 \over 3}\,{R^{3} \over \pars{R^{2} + 1}^{3/2}}}
\end{align}
\begin{align}
\color{#f00}{I} & =
\color{#f00}{y^{2}\ln\pars{\root{r^{2} + y^{2}} + r} -
y^{2}\ln\pars{\verts{y}} -
{y^{2}r \over \root{r^{2} + y^{2}}} -
{1 \over 3}\,{y^{2}r^{3} \over \pars{r^{2} + y^{2}}^{3/2}}}
\end{align}
|
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|
intuition behind having a unique regression line I understand this mathematically. we have function of 2 variables represents the sum of square errors. We have to find the $a$ and $b$ that minimize the function. there is only one minimum point.
But when I think of it, I can't see why 2 different lines would not bring the value to the same minimum. we have 2 degrees of freedom, so why can't we find a new pair of $(a,b)$ that will have the same value.
|
You have two degrees of freedom (the slope $m$ and $y$-intercept $b$ of the regression line, say), but also two constraints: The partial derivatives of the squared total error $E$ with respect to $m$ and with respect to $b$ must vanish. That means you expect only finitely many (local) minima. (As A.E. says, $E$ is strictly convex, so there is at most one minimum.)
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1846712",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
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|
What is the expected return of this hypotetical lottery game? What is the expected return of this hypotetical lottery game?
There are 60 numbers and you must pick 6 different numbers.
Then they pick 6 numbers (that will be different from each other) at random and if your 6 numbers is equal to those 6 numbers (but dont need to be at the same order) you win $3,500,000$ dollars.
Each ticket cost $3.5$ dollars to buy.
This is a simplified version of mega sena from brazil
|
The probability of winning is 1 / total number of possible outcomes
You're looking for the number of ways to pick 6 numbers at random from 60, I am assuming without considering the order (that is, 1 3 5 7 9 51 is the same as 5 7 9 51 1 3)
This number is equal to $\binom{60}{6}$ so your probability of winning is $1/ \binom{60}{6}$
The expected value of the game therefore is
$$E = \frac 1{\binom{60}{6}} \cdot 3500000 - 3.5\approx -3.43$$
So in average every time you play you lose 3.43 dollars. Don't play.
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
Limit of a sequence defined as a definite integral Given two sequences $$a_n=\int_{0}^{1}(1-x^2)^ndx$$ and $$b_n=\int_{0}^{1}(1-x^3)^ndx$$ where $n$ is a natural number, then find the value of $$\lim_{n \to \infty}(10\sqrt[n]{a_n}+5\sqrt[n]b_n)$$
I have no starts. Looks good though. Some hints please. Thanks.
|
$\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle}
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\newcommand{\imp}{\Longrightarrow}
\newcommand{\Li}[1]{\,\mathrm{Li}_{#1}}
\newcommand{\ol}[1]{\overline{#1}}
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\newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
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\newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
\newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
&\mbox{Asymptotically}:
\\[3mm] &\ \braces{%
\left.\int_{0}^{1}\pars{1 - x^{\mu}}^{n}\,\dd x\,
\right\vert_{\ \mu\ >\ 0}}^{1/n} \sim {\alpha^{1/n}\pars{\mu} \over n^{1/\pars{\mu n}}}\
\stackrel{n\ \to \infty}{\longrightarrow}\
\color{#f00}{\large 1}\,,\qquad \mu = 2,3.
\\[3mm] &\ \mbox{where}\ \alpha\pars{\mu} >0\,, \forall\ \mu > 0
\\[3mm] & \mbox{}
\end{align}
In another words, $\ds{\lim_{n \to \infty}a_{n}^{1/n} = \lim_{n \to \infty}b_{n}^{1/n} = \color{#f00}{\large 1}}$.
$$
\mbox{Then,}\quad
\lim_{n \to \infty}\pars{\color{#f00}{10}\,a_{n}^{1/n} + \color{#f00}{5}\,b_{n}^{1/n}} = 10 + 5 = \color{#f00}{15}
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1846899",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 3
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|
Similar Triangles--Find the measurement of the unknown side This is a question I know I got wrong on a final exam in a very easy class for teaching elementary geometry/prep for Praxis II. I actually received a 99% average in the entire course because of the single point taken off of the final. I just want to know how to go about solving this. I feel like I overthought the question. It was a multiple choice question and none of the answers seemed to make any sense to me. The question was to find the measurement of line segment CD.
Don't quote me on this, but I believe the choices were:
A.) 5 cm
B.) 16 cm
C.) 19 cm
D.) 25 cm
Either way. We don't really need the choices to answer the question. I just want to know how to go about solving this. I know it has to do with similar triangles and creating a super easy proportion, but I tried every method I could think of. None of my answers were close to what the multiple choice answers were. So, I obviously did something wrong.
In case it's hard to see, AD=18cm AB= 7cm BC= 6cm. Triangle is not drawn to scale
|
Hint: Note that the sum of any two sides of a triangle is greater than the third side. That will be enough to eliminate $3$ of the given choices.
For example, $CD$ cannot be $5$, since $7+(6+5)$ is not greater than $18$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1847040",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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|
How to find $\int \frac{x^2-1}{x^3\sqrt{2x^4-2x^2+1}} dx$ How to find ?$$\int \frac{x^2-1}{x^3\sqrt{2x^4-2x^2+1}} dx$$
I tried using the substitution $x^2=z$.But that did not help much.
|
By setting $x^2=z$ we are left with:
$$ \frac{1}{2}\int\frac{z-1}{z^2\sqrt{2z^2-2z+1}}\,dz=C+\frac{\sqrt{2z^2-2z+1}}{2z}.$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1847140",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 5,
"answer_id": 0
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|
How can $i^2 = k^2 = j^2 = ijk = -1$ be true? I have just started to learn the basics of quaternions, but I immediately run into a wall.
Litteraly the first equation on Wikipedia is as follows
$i^2 = k^2 = j^2 = ijk = -1$
This implies
$i = \sqrt{-1}$
$j = \sqrt{-1}$
$k = \sqrt{-1}$
but now $ijk = -1$ also need to be true
$\sqrt{-1} * \sqrt{-1} * \sqrt{-1} = -1$
$-1 * \sqrt{-1} = -1$
$\sqrt{-1} = 1$
This can not be true. What am I missing here?
|
The way to think about this is not to think of these as normal multiplication, but rather rotation. To rotate by i means to take the point at 1 and sort of move it 90 degrees up to i. Rotation by j and k is completely similar. All other numbers on the unit circles of i, j, and k for their respective multiplications follow the same 90-degree rotations. What happens to the other quaternions? Well, the unit circle of j and k, for the rotation of i, follows a similar motion, where j goes to k, and k goes to -j. And, for multiplications by other quaternions, you can probably guess what happens by now. Obviously, 180 degrees of rotation starting at 1 gives -1. Now we can tackle ijk. As I already said, ij gives k. This means that ijk = k^2. Obviously, this is -1. Problem solved!
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1847218",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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|
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