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The set of real numbers $x$ for which

$\dfrac{1}{x-2009}+\dfrac{1}{x-2010}+\dfrac{1}{x-2011}\ge 1$

is the union of intervals of the form $a<x\le b$.

Find the sum of the lengths of these two intervals.

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

The generalized meaning of this is if we are given a divergent integral, say $ \displaystyle \int _0^{\infty} f(x) ~ dx$, we can (supposedly) change the problem to $ \displaystyle \lim_{ \Lambda \to \infty } \int_0^{ \Lambda } R( \Lambda ) f(x) ~ dx$ such that the integral will become convergent so long as $ \displaystyle \lim_{ \Lambda \to \infty } R (\Lambda ) = 1$. The cut-off value of $ \displaystyle \Lambda$ is taken as $ \displaystyle x << \Lambda$. (I know not just any function R will work and I don't know how to find one.)

But that changes the nature of the integral! I don't see how this is supposed to work. My Physics text says that certain properties of the integral don't change so we can use the regularized integral for those. It doesn't directly say what those properties are that are kept, just that presumably finite properties, like mass, can be split up into finite and infinite parts. I can give you more on the Physics but I can't seem to find a good source on this lower level. My QFT text starts talking about the renormalization group and Ward identities and suddenly the problem goes over my head. I am just working Mathematically on the cut-off technique for specific integrals for right now.

Specifically I am working on the integral

$ \displaystyle \int _0^{ \infty } \left ( \dfrac{1}{q^2 - r^2} \right ) \left ( \dfrac{1}{(q - p)^2 - s^2} \right ) ~ q^3 ~ dq$

where p, r, and s are constants. This integral is logarithmically divergent.

The regularization used in my text is

$ \displaystyle \lim_{ \Lambda \to \infty} \int _0^{ \Lambda } \left ( \dfrac{- \Lambda ^2 }{ q^2 - \Lambda ^2} \right ) \left ( \dfrac{1}{q^2 - r^2} \right ) \left ( \dfrac{1}{(q - p)^2 - s^2} \right ) ~ q^3 ~ dq$

which does converge.

-Dan ]]>

We have the below maps:

- $f_1:\mathbb{R}^2\rightarrow \mathbb{R}^2, \ \ \begin{pmatrix}x \\ y\end{pmatrix}\mapsto \begin{pmatrix}-x \\ -y\end{pmatrix}$
- $f_2:\mathbb{R}^3\rightarrow \mathbb{R}^3, \ \ \begin{pmatrix}x \\ y\\ z\end{pmatrix}\mapsto \begin{pmatrix}x \\ -y\\ z\end{pmatrix}$
- $f_3:\mathbb{R}^3\rightarrow \mathbb{R}^3, \ \ \begin{pmatrix}x \\ y\\ z\end{pmatrix}\mapsto \begin{pmatrix}x+2 \\ y-3 \\ z+1\end{pmatrix}$
- $f_4:\mathbb{R}^3\rightarrow \mathbb{R}^3, \ \ \begin{pmatrix}x \\ y\\ z\end{pmatrix}\mapsto \begin{pmatrix}x-z \\ y-z \\ 0\end{pmatrix}$

I want to give the geometric interpretation of these maps.

I have done the following:

- We have that $\begin{pmatrix}-x \\ -y\end{pmatrix}=\begin{pmatrix}-1 & 0 \\ 0 & -1\end{pmatrix}\begin{pmatrix}x \\ y\end{pmatrix}$ which means that this is a rotation of $180^{\circ}$.

- Since $x$ and $z$ are unchanged and $y$ changes the sign, it is a reflection at the $xz$-plane.

- Since we add to $\begin{pmatrix}x \\ y \\ z\end{pmatrix}$ the vector $\begin{pmatrix}2 \\ -3 \\ 1\end{pmatrix}$ it is a translation by the vector $\begin{pmatrix}2 \\ -3 \\ 1\end{pmatrix}$.

- We have that $\begin{pmatrix}x \\ y\\ z\end{pmatrix}\mapsto \begin{pmatrix}x-z \\ y-z \\ 0\end{pmatrix}=\begin{pmatrix}1 & 0 & -1 \\ 0 & 1 & -1 \\ 0 & 0 & 0\end{pmatrix}\begin{pmatrix}x \\ y\\ z\end{pmatrix}=\begin{pmatrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0\end{pmatrix}\begin{pmatrix}1 & 0 & -1 \\ 0 & 1 & -1 \\ 0 & 0 & 1\end{pmatrix}\begin{pmatrix}x \\ y\\ z\end{pmatrix}$. The first matrix describes the projection to the $xy$-plane and the second one the shear mapping parallel to the $xy$-plane by the value $-1$ in the $x$ direction and by the value $-1$ in the $y$ direction.

Is everything correct? (Wondering)

I want to check also if $f_4$ is injective and surjective.

For that let $a,b,c,x,y,z\in \mathbb{R}$ with $f\begin{pmatrix}a \\ b \\ c\end{pmatrix}=f\begin{pmatrix}x \\ y \\ z\end{pmatrix} \Rightarrow \begin{pmatrix}a-c \\ b-c \\ 0\end{pmatrix}=\begin{pmatrix}x-z \\ y-z \\ 0\end{pmatrix}$. From that we get $a-c=x-z$ and $b-c=y-z$. Subtracting these two relations we get $a-b=x-y$. This equality is also satisfied with $x=-b$ and $y=-a$ and this means that $f_4$ is not injective.

The vector $\begin{pmatrix}1 \\ 1\\ 1\end{pmatrix}$ has no preimage under $f_4$ and so $f_4$ is not surjective.

Are both justifivations and results correct? (Wondering)

I want to check also if the reflection to the $z$-axis is injective and surjective.

For that does it hold that each point of the plane is the image of exactly one point and so this transformationis bijective? (Wondering) ]]>

totally lost and confused with this question:

A machine is subject to two vibrations at the same time.

one vibration has the form: 2cosωt and the other vibration has the form: 3 cos(ωt+0.785). (0.785 is actually expressed as pi/4)

determine the resulting vibration and express it in the general form of: n cos(ωtąα) ]]>

Let $v_1:\begin{pmatrix}1 \\ 1\\ 1\end{pmatrix}, \ \ v_2:\begin{pmatrix}1 \\ 0\\ 1\end{pmatrix}\in \mathbb{R}^3$.

- Let $w=\begin{pmatrix}1 \\ 0 \\2\end{pmatrix}\in \mathbb{R}^3$. If possible, give a linear map $\phi:\mathbb{R}^3\rightarrow \mathbb{R}^2$ such that $\phi (v_1)=\begin{pmatrix}1 \\ 0\end{pmatrix}, \ \ \phi (v_2)=\begin{pmatrix}0 \\ 1\end{pmatrix}, \ \ \phi (w)=\begin{pmatrix}0 \\ 0\end{pmatrix}$.
- Let $w'=\begin{pmatrix}0 \\ 1 \\0\end{pmatrix}\in \mathbb{R}^3$. If possible, give a linear map $\phi:\mathbb{R}^3\rightarrow \mathbb{R}^2$ such that $\phi (v_1)=\begin{pmatrix}1 \\ 0\end{pmatrix}, \ \ \phi (v_2)=\begin{pmatrix}0 \\ 1\end{pmatrix}, \ \ \phi (w')=\begin{pmatrix}0 \\ 0\end{pmatrix}$.

I have done the following:

- The three vectors $v_1,v_2, w$ are linearly independent and so they form a basis of $\mathbb{R}^3$.

We consider an arbitrary vector in $\mathbb{R}^3$ and we write it as a linear combination of the vectors of the basis.

\begin{equation*}\begin{pmatrix}x_1 \\ x_2\\ x_3\end{pmatrix}=c_1\begin{pmatrix}1 \\ 1\\ 1\end{pmatrix}+c_2\begin{pmatrix}1 \\ 0\\ 1\end{pmatrix}+c_3\begin{pmatrix}1 \\ 0 \\2\end{pmatrix}=\begin{pmatrix}1 & 1 & 1\\ 1 & 0 & 0 \\ 1 & 1 & 2\end{pmatrix}\begin{pmatrix}c_1\\ c_2 \\ c_3\end{pmatrix}\end{equation*}

We multiply by the inverse matrix and we get

\begin{equation*}\begin{pmatrix}c_1\\ c_2 \\ c_3\end{pmatrix}=P^{-1}\begin{pmatrix}x_1 \\ x_2\\ x_3\end{pmatrix}=\begin{pmatrix}0 & 1 & 0\\ 2 & -1 & -1 \\ -1 & 0 & 1\end{pmatrix}\begin{pmatrix}x_1 \\ x_2\\ x_3\end{pmatrix}=\begin{pmatrix}x_2 \\ 2x_1-x_2-x_3\\ -x_1+x_3\end{pmatrix}\end{equation*}

So

\begin{equation*}\begin{pmatrix}x_1 \\ x_2\\ x_3\end{pmatrix}=x_2\begin{pmatrix}1 \\ 1\\ 1\end{pmatrix}+(2x_1-x_2-x_3)\begin{pmatrix}1 \\ 0\\ 1\end{pmatrix}+( -x_1+x_3)\begin{pmatrix}1 \\ 0 \\2\end{pmatrix}\end{equation*}

We calculate $\phi \begin{pmatrix}x_1 \\ x_2\\ x_3\end{pmatrix}$ using the linearity of $\phi$:

\begin{align*}\phi \begin{pmatrix}x_1 \\ x_2\\ x_3\end{pmatrix}&=\phi \left (x_2\begin{pmatrix}1 \\ 1\\ 1\end{pmatrix}+(2x_1-x_2-x_3)\begin{pmatrix}1 \\ 0\\ 1\end{pmatrix}+( -x_1+x_3)\begin{pmatrix}1 \\ 0 \\2\end{pmatrix}\right )\\ & =x_2\cdot \phi \begin{pmatrix}1 \\ 1\\ 1\end{pmatrix}+(2x_1-x_2-x_3)\cdot \phi \begin{pmatrix}1 \\ 0\\ 1\end{pmatrix}+( -x_1+x_3)\cdot \phi \begin{pmatrix}1 \\ 0 \\2\end{pmatrix}\\ & =x_2 \begin{pmatrix}1 \\ 0\end{pmatrix}+(2x_1-x_2-x_3) \begin{pmatrix}0\\ 1\end{pmatrix}+( -x_1+x_3) \begin{pmatrix}0 \\ 0 \end{pmatrix}\\ & = \begin{pmatrix}x_2 \\ 2x_1-x_2-x_3 \end{pmatrix}\end{align*}

- The vectors $v_1, v_2, w'$ are linearly dependent, since \begin{equation*}\begin{pmatrix}0 \\ 1\\ 0\end{pmatrix}=\begin{pmatrix}1 \\ 1\\ 1\end{pmatrix}-\begin{pmatrix}1 \\ 0\\ 1\end{pmatrix}\Rightarrow w'=v_1-v_2\end{equation*}

We apply $\phi$ at this equation and we get: \begin{equation*}\phi (w')=\phi (v_1-v_2)=\phi (v_1)-\phi (v_2)\Rightarrow \begin{pmatrix}0 \\ 0\end{pmatrix}=\begin{pmatrix}1 \\ 0\end{pmatrix}-\begin{pmatrix}0 \\ 1\end{pmatrix}\Rightarrow \begin{pmatrix}0 \\ 0\end{pmatrix}=\begin{pmatrix}1 \\ -1\end{pmatrix}\end{equation*}

Therefore there is no linear map $\phi:\mathbb{R}^3\rightarrow \mathbb{R}^2$ such that$\phi (v_1)=\begin{pmatrix}1 \\ 0\end{pmatrix}$, $\phi (v_2)=\begin{pmatrix}0 \\ 1\end{pmatrix}$ und $\phi (w')=\begin{pmatrix}0 \\ 0\end{pmatrix}$.

Is everything correct? (Wondering) ]]>

Let $a\in \mathbb{R}$. We define the map $\text{cost}_a:\mathbb{R}\rightarrow \mathbb{R}$, $x\mapsto a$. We define also $-f:=(-1)f$ for a map $f:\mathbb{R}\rightarrow \mathbb{R}$.

Let $f:\mathbb{R}\rightarrow\mathbb{R}$ be a map and $\lambda\in \mathbb{R}$.

Show that:

- for $a,b\in \mathbb{R}$ it holds that $\text{cost}_a+\text{cost}_b=\text{cost}_{a+b}$.
- for $a\in \mathbb{R}$ it holds that $\lambda\text{cost}_a=\text{cost}_{\lambda a}$.
- $-(f+g)=(-f)+(-g)$.
- $f+f=2f$.
- $f+(-f)=\text{cost}_0$.

Could you give me a hint how we could these? Aren't all of these trivial? (Wondering) ]]>

I am currently reading Chapter 3: Continuous Functions on Intervals and am currently focused on Section 3.1 Limits and Continuity ... ...

I need some help with the proof of Corollary 3.13 ...

Corollary 3.13 reads as follows:

Browder - Corollary 3.13 ... .png

Can someone help me to prove that if $ \displaystyle f$ is continuous then $ \displaystyle f^+ = \text{max} (f, 0)$ is continuous ...

My thoughts are as follows:

If $ \displaystyle c$ belongs to an interval where $ \displaystyle f$ is positive then $ \displaystyle f^+$ is continuous since $ \displaystyle f$ is continuous ... further, if $ \displaystyle c$ belongs to an interval where $ \displaystyle f$ is negative then $ \displaystyle f^+$ is continuous since $ \displaystyle g(x) = 0$ is continuous ... but how do we construct a proof for those points where $ \displaystyle f(x)$ crosses the $ \displaystyle x$-axis ... ..

Help will be much appreciated ...

Peter

Answer:-

How to answer this question using geometry or calculus or by using both techniques. ]]>

I am currently reading Chapter 3: Continuous Functions on Intervals and am currently focused on Section 3.1 Limits and Continuity ... ...

I need some help in understanding the proof of Proposition 3.12 ...

Proposition 3.12 and its proof read as follows:

Browder - Proposition 3.12 ... .png

In the above proof by Browder we read the following:

" ... ... Since $ \displaystyle f(I) \subset J$, $ \displaystyle f^{ -1 } ( g^{ -1 }(V) ) = f^{ -1 } (U) \cap f^{ -1 } (J) = f^{ -1 } (U)$ ... ... "

My question is as follows:

Can someone please explain exactly why/how $ \displaystyle f^{ -1 } (U) \cap f^{ -1 } (J) = f^{ -1 } (U)$ ... ...

Help will be much appreciated ...

Peter

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Let $a>0$ and $P(x)$ be a polynomial with integer coefficients such that

$P(1)=P(3)=P(5)=P(7)=a$ and

$P(2)=P(4)=P(6)=P(8)=-a$.

What is the smallest possible value of $a$?

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

I want to prove the following properties:

- $\left (e^x\right )^y=e^{xy}$
- $\ln (1)=0$
- $\ln \left (x^y\right )=y\ln (x)$
- $a^x\cdot a^y=a^{x+y}$ and $\frac{a^x}{a^y}=a^{x-y}$
- $a^x\cdot b^x=\left (ab\right )^x$ and $\frac{a^x}{b^x}=\left (\frac{a}{b}\right )^x$
- $\left (a^x\right )^y=a^{xy}$

I have done the following:

- We have that $\displaystyle{\left (e^x\right )^y=e^{\ln \left (e^x\right )^y}=e^{y\cdot \ln e^x}=e^{y\cdot x}=e^{xy}}$

We used the rule $\displaystyle{\log \left (x^a\right )=a\cdot \log x}$. - We have that $e^0=1$. We apply the logarithm and we get $\displaystyle{\ln \left (e^0\right )=\ln (1)\Rightarrow 0\cdot \ln (e)=\ln (1)\Rightarrow 0=\ln (1)}$.
- We have that $e^{\ln x}=x$. We raise the equation to $n$ and we get $\left (e^{\ln x}\right )^y=x^y \Rightarrow e^{y\cdot \ln x}=x^y$. We take the logarithm of the equation and we get $\ln \left (e^{y\cdot \ln x}\right )=\ln \left (x^y\right ) \Rightarrow y\cdot \ln x=\ln \left (x^y\right )$.

At 1. I used this rule, although I proved that here. Would there be also an other way to prove the property 1.? (Wondering)

Is everything correct so far? (Wondering)

Could you give me a hint for the remaining $3$ properties? Do we use the previous properties? (Wondering) ]]>

I look at the problem with hilbert's hotel and the busses. First one bus with infinitely many guests, then four busses with infinitely many guests and then infinitely many busses with infinitely many guests.

At each case we move all the guests that are already in the hotel to the odd room numbers.

So the even room numbers are free for the new guests.

In the first with one bus we use the formula $n = 2(i-1)+2$ where $i\in \mathbb{N}\setminus\{0\}$ is the number of place in the bus and $n$ is the even room number that this guest will get.

In the second case with the four busses we use the following:

\begin{equation*}\begin{matrix} & 1. \text{ Bus } & 2. \text{ Bus } & 3. \text{ Bus } & 4. \text{ Bus } \\ 1. \text{ Person } & \text{ Room } 2 & \text{ Room } 4 & \text{ Room } 6 & \text{ Room } 8 \\ 2. \text{ Person } & \text{ Room } 10 & \text{ Room } 12 & \text{ Room } 14 & \text{ Room } 16 \\ 3. \text{ Person } & \text{ Room } 18 & \text{ Room } 20 & \text{ Room } 22 & \text{ Room } 24 \\ \ldots & \ldots & \ldots & \ldots & \ldots \\ i. \text{ Person } & \text{ Room } 8(i-1)+2 & \text{ Room } 8(i-1)+4 & \text{ Room } 8(i-1)+6 & \text{ Room } 8(i-1)+8\end{matrix}\end{equation*}

In the last case with the infinitely many busses we use the formula $2^{i}\left (2(j-1)+1\right )$ where $i$ is the place in the bus and $j$ is the number of the bus.

Using this we get \begin{equation*}\begin{matrix} \text{Place}/\text{Bus} & 1. \text{ Bus} & 2. \text{ Bus} & 3. \text{ Bus} & 4. \text{ Bus} & 5. \text{ Bus} & \ldots \\ 1 & 2 & 6 & 10 & 14 & 18 & \ldots \\ 2 & 4 & 12 & 20 & 28 & 36 & \ldots \\ 3 & 8 & 24 & 40 & 56 & 72 & \ldots \\ \ldots & \ldots & \ldots & \ldots & \ldots & \ldots & \ldots \end{matrix}\end{equation*}

I want to describe the injectivity and the surjectivity of the maps, and if the maps for the old and the new guests are injective and surjective.

The map for the old guests is injective, since it means that the new room thet they get either had no previous guest or one.

The same holds also for the map for the new guests. Their new rooms either had no previous guests or one.

As for the surjectivity. As for the rooms of the old guests, only the odd room umbers have a preimage and so the map is not surjective. As for the rooms of the new guests, only the even room umbers have a preimage and so the map is not surjective.

Is that correct? (Wondering) ]]>

I made a large circular tortilla.

Ate half of it. Then decided to put the rest into the fridge on a smaller plate.I raised the knife to cut the remaining semi-circle in two, and then went : "Hmmmmmmmm...".

Anyway, it's in the fridge now with an approximate solution, but I'm wondering if anyone knows the mathematical one?

I'm wondering if there are 3 different optimal solutions:

1) In which the straight-line cut necessarily is from the center of the original circle to the edge. I.e. a radius.

2) In which the straight-line cut may divide the semi-circle in any possible way.

3) In which the cut is not necessarily a straight line.

Thanks in advance if anyone can offer a demonstration.Capture.JPG