Hey So I am trying to prove this.

I tried using linear combinations and not sure how that would help. I am just not familiar with combinatorics and wondering if anyone would enlighten me. ]]>

I would really appreciate your folks help, thanks in advance :)

I wasn't sure wether to post this in the computer science or statistics category so I posted it in both, sorry about this.

I want to find subsets $S$ of $\mathbb{R}^2$ such that $S$ satisfies all but one axioms of subspaces.

A subset that doesn't satisfy the first axiom: We have to find a subset that doesn't contain the zero vector. Is this for example $\left \{\begin{pmatrix}x \\ y\end{pmatrix} : x,y>0\right \}$ ?

A subset that doesn't satisfy the second axiom: We have to find a subset that doesn't contain the sum of two vectors of $S$. Could you give me an example for that?

A subset that doesn't satisfy the third axiom: We have to find a subset that doesn't contain the scalar product of a vector of $S$. Is this for example $\left \{\begin{pmatrix}x \\ y\end{pmatrix} : x\geq 0\right \}$ since $\begin{pmatrix}1 \\ 0\end{pmatrix}\in S$ but $(-1)\cdot \begin{pmatrix}1 \\ 0\end{pmatrix}\notin S$.

(Wondering) ]]>

A square with the side length $2^n$ length units (LU) is divided in sub-squares with the side length $1$. One of the sub-squares in the corners has been removed. All other sub-squares should now be covered completely and without overlapping with L-stones. An L-stone consists of three sub-squares that together form an L.

I want to draw the problem for the first three cases described above ($1 \leq n \leq 3$).

Then I want to show the following using induction:

For all $n \in N$ the square with side length $2^n$ LU can be covered completely and without overlapping with L-stones, after one of the sub-squares in the corners has been removed.

For the first part:

L_stones.png

Is the drawing correct?

(Wondering)

Let $X=(0,1)$. For $x,y>0$ we consider the metrices:

- $d_1(x,y)=|x-y|$
- $d'(x,y)=\left |\frac{1}{x}-\frac{1}{y}\right |=\frac{|x-y|}{|xy|}$

I want to show that these are topologically equivalent but not strongly equivalent.

$d_1$ and $d'$ are strongly equivalent iff there are constants $k>0$ and $K>0$ such that \begin{equation*}kd_1(x,y)\le d'(x,y)\le \text{ for all $x,y$.}\end{equation*}

From there we get \begin{equation*}k\le \frac{d'(x,y)}{d_1(x,y)}\le K \Rightarrow k\le \frac{1}{|xy|}\le K\end{equation*} for all $x,y$.

It holds that $\frac{1}{|xy|}$ is not bounded for all $x,y$. If $x,y\rightarrow 0$ then $\frac{1}{|xy|}\rightarrow \infty$.

Therefore the metrices $d_1$ and $d'$ are not strongly equivalent.

Is everything correct so far? (Wondering)

Could you give me a hint how to show that these are topologically equivalent? (Wondering) ]]>

Why don't mathematicians define a multiplication operation between a pair of elements and investigate the resulting field ...

For example ... why not define multiplication as X where

$ \displaystyle (x_1, x_2, \ ... \ ... \ , x_n) \ X \ (y_1, y_2, \ ... \ ... \ , y_n) = (x_1 y_1, x_2 y_2 , \ ... \ ... \ , x_n y_n)$ ...

Peter ]]>

At a relocation we have $N$ objects with the same weight that we want to pack in boxes. We have at our disposal boxes of capacity 1,2,5,10 and 20 objects (so many that we want from each size). In order the objects not to move at the transfer, each box should be completely full. I want to write and analyze a greedy algorithm that computes the minimum number of boxes needed to pack the objects.

I have thought of the following algorithm:

Code:

`boxes=0`

remaining=N

while (remaining>=20) {

boxes=boxes+1

remaining=remaining-20

}

while (remaining>=10) {

boxes=boxes+1

remaining=remaining-10

}

while (remaining>=5) {

boxes=boxes+1

remaining=remaining-5

}

while (remaining>=2) {

boxes=boxes+1

remaining=remaining-2

}

while (remaining>=1) {

boxes=boxes+1

remaining=remaining-1

}

return remaining

Is my algorithm right? Could something be improved?

The algorithm is greedy, isn't it? (Thinking) ]]>

Ok this is considered a "hard" GRE geometry question... notice there are no dimensions

How would you solve this in the fewest steps?

I am focused on Section 3.2 The Cauchy Riemann Equations ...

I need help in fully understanding the Proof of Theorem 3.4 ...

The start of Theorem 3.4 and its proof reads as follows:

M&H - 1 - Theorem 3.1 ... .PART 1 ... .png

In the above proof by Mathews and Howell we read the following:

" ... ... The partial derivatives $ \displaystyle u_x$ and $ \displaystyle u_y$ exist, so the mean value theorem for real functions of two variables implies that a value $ \displaystyle x*$ exists between $ \displaystyle x_0$ and $ \displaystyle x_0 + \Delta x$ such that we can write the first term in brackets on the right side of equation (3-17) as

$ \displaystyle u(x_0 + \Delta x, y_0 + \Delta y) - u(x_0, y_0 + \Delta y) = u_x(x*, y_0 + \Delta y) \Delta x $ ... ... "

Can someone please explain how exactly the mean value theorem for real functions of two variables implies that a value $ \displaystyle x*$ exists between $ \displaystyle x_0$ and $ \displaystyle x_0 + \Delta x$ such that we can write the first term in brackets on the right side of equation (3-17) as

$ \displaystyle u(x_0 + \Delta x, y_0 + \Delta y) - u(x_0, y_0 + \Delta y) = u_x(x*, y_0 + \Delta y) \Delta x$ ... ...

Peter

[ NOTE ... ... In Wendell Fleming's book: "Functions of Several Variables" (Second Edition) the Mean Value Theorem reads as follows:

Fleming - Mean Value Theorem ... .png

... ... ]

Consider the following metrices in $\mathbb{R}^2$. For $x,y\in \mathbb{R}^2$ let \begin{align*}&d_1(x,y)=|x_1-y_1|+|x_2-y_2| \\ &d_2(x,y)=\sqrt{(x_1-y_1)^2+(x_2-y_2)^2} \\ &d_{\infty}(x,y)=\max \{|x_1-y_1|,|x_2-y_2|\}\end{align*}

Draw the unit ball $B_i(0,1)=\{y\in X\mid d_i(0,y)<1\}$ in each metric.

Show that the metrices $d_1, d_2, d_{\infty}$ are strongly equivalent.

Could you give me a hint for the first part?

As for the second part:

Without loss of generality, we assume that $\max \{|x_1-y_1|,|x_2-y_2|\}=|x_2-y_2|$.

\begin{align*}d_{\infty}(x,y)=\max \{|x_1-y_1|,|x_2-y_2|\}=|x_2-y_2|\leq |x_1-y_1|+|x_2-y_2|=d_1(x,y)\end{align*}

\begin{align*}d_1(x,y)&=|x_1-y_1|+|x_2-y_2|\leq \max \{|x_1-y_1|,|x_2-y_2|\}+\max \{|x_1-y_1|,|x_2-y_2|\}\\ & =2\max \{|x_1-y_1|,|x_2-y_2|\}=2d_{\infty}(x,y)\end{align*}

So we get \begin{equation*}d_{\infty}(x,y)\leq d_1(x,y)\leq 2d_{\infty}(x,y)\end{equation*} Does this mean that $d_{\infty}(x,y)$ and $d_1(x,y)$ are strongly equivalent? (Wondering)

\begin{align*}d_{\infty}(x,y)&=\max \{|x_1-y_1|,|x_2-y_2|\} =|x_2-y_2|=\sqrt{|x_2-y_2|^2}=\sqrt{(x_2-y_2)^2}\\ & \leq \sqrt{(x_1-y_1)^2+(x_2-y_2)^2}=d_2(x,y)\end{align*}

\begin{align*}d_2(x,y)&=\sqrt{(x_1-y_1)^2+(x_2-y_2)^2}\leq \sqrt{(x_1-y_1)^2}+\sqrt{(x_2-y_2)^2}\\ & =|x_1-y_1|+|x_2-y_2| \leq \max \{|x_1-y_1|,|x_2-y_2|\}+\max \{|x_1-y_1|,|x_2-y_2|\}\\ & =2\max \{|x_1-y_1|,|x_2-y_2|\} = 2d_{\infty}(x,y)\end{align*}

So we get \begin{equation*}d_{\infty}(x,y)\leq d_2(x,y)\leq 2d_{\infty}(x,y)\end{equation*} Does this mean that $d_{\infty}(x,y)$ and $d_2(x,y)$ are strongly equivalent? (Wondering)

Do we have to show that also for $d_1$ and $d_2$ ? Or does this follow from the above? (Wondering) ]]>

I know that it means $ \displaystyle \frac1{1+2}+\frac1{1+2+3}+\frac1{1+2+3+4}+\frac1{1+2+3+4+5}+\frac1{1+2+3+4+5+6}+...+\frac1{231}$, but how do I answer? It's from a student worksheet for 7th graders, so they haven't learnt about sequence and series yet. Granted, the book also says that the question was from a middle school-level math contest. ]]>

ok not sure of the next step but

$\dfrac{dy}{dx}=\dfrac{2x-y}{x+2y}$ ]]>

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The positive reals $x,\,y$ satisfy $x^2+y^3\ge x^3+y^4$. Show that $x^3+y^3\le 2$.

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Remember to read the POTW submission guidelines to find out how to submit your answers! ]]>

of the series.

$$\sum_{n=1}^{\infty}\dfrac{(-1)^n x^n}{\sqrt[3]{n}}$$

(1)

$$a_n=\dfrac{(-1)^n x^n}{\sqrt[3]{n}}$$

(2)

$$\left|\dfrac{a_{a+1}}{a_n}\right|

=\left|\dfrac{(-1)^{n+1} x^{n+1}}{\sqrt[3]{n+1}}

\cdot\dfrac{\sqrt[3]{n}}{(-1)^n x^n}\right|

=-\frac{\sqrt[3]{n}x\left(n+1\right)^{\frac{2}{3}}}{n+1}$$

(3) W|A Convergence Interval is

$$-1\le \:x\le \:1$$

ok on (2) I was expecting a different result to take $\infty$ to

on (3) the example I was trying to follow was ??? on how W|A got this interval ]]>

ok again I used an image since there are macros and image

I know this is a very common problem in calculus but think most still stumble over it

inserted the graph of v(t) and v'(t) and think for v'(t) when the graph is below the x-axis that participle is moving to the left

the integral has a - interval but I think the total is an absolute value...

my take on some of it.

finally this will be my last AP calculus exam question for a while

I was surprized how many views these got... must be a big concern for many