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# Research: Looking for a sequence that produce variation's of Pascal's triangle

Prologue

I am an undergraduate so if my terminology or approach seem inappropriate/confusing please explain in the comments.

I created a notation where

$$F(0 \rightarrow n,x) = [\hspace{1mm}F(0 ,x),\hspace{2mm}F(1, x),\hspace{2mm}F(2, x), \ldots,\hspace{2mm} F(n-2,x),\hspace{2mm}F(n-1,x),\hspace{2mm}F( n,x)\hspace{1mm}]$$

Last semester in Calculus 1 I created a function noted as $\mu(e,f)$ where $f$,$e$ are just positive integers including 0 and if c is any numerical value then $\mu(c,c)=0$ and $\frac{d \mu(c,c)}{dx}=0$. The interesting part was $$\frac{d \mu(e,f)}{dx}=\mu(e+1,f)+\mu(e,f+1)$$ I then figured out by playing around with the function that if $f>e + c$ for any real c then: $$\frac{d^\lambda \mu(e,f)}{(dx)^\lambda}=\sum_{i=0}^{\lambda}\binom{\lambda}{i}\mu(e+i,f+\lambda-i)$$

I figured this out by putting the Function into Pascal's Triangle, where each row represents the derivative of the row before it. I will call this triangle Pascal's Triangle for Functions (PTF). For clarity I made some stuff that makes the patterns easier to see. In the equation $\frac{d \mu(e,f)}{dx}=\mu(e+1,f)+\mu(e,f+1)$, The $\mu(e+1,f)$ will be presented as $\swarrow$ and $\mu(e,f+1)$ will be presented as $\searrow$. Let $G(j,k) = \mu(e+j,f+k)$ therefore $\frac{d G(j,k)}{dx}=G(j+1,k)+G(j,k+1)$.

$$G(0,0)\\ \swarrow\searrow \\ G(1,0)+G(0,1)\\ \swarrow\searrow\hspace{12mm}\swarrow\searrow\\ G(2,0)+2G(1,1)+G(0,2)\\ \swarrow\searrow\hspace{12mm}\swarrow\searrow\hspace{12mm}\swarrow\searrow \\ G(3,0)+3G(2,1)+3G(1,2)+G(0,3)\\ \swarrow\searrow\hspace{12mm}\swarrow\searrow\hspace{12mm}\swarrow\searrow\hspace{12mm}\swarrow\searrow \\ G(4,0)+4G(3,1)+6G(2,2)+4G(1,3)+G(0,4)\\ \swarrow\searrow\hspace{12mm}\swarrow\searrow\hspace{12mm}\swarrow\searrow\hspace{12mm}\swarrow\searrow\hspace{12mm}\swarrow\searrow \\ G(5,0)+5G(4,1)+10G(3,2)+10G(2,3)+5G(1,4)+G(0,5)\\ \ldots$$

Anyways knowing $\mu(c,c)=0$ and $e$,$f$ are finite then let $\alpha = e-f$ so $\mu(f+\alpha,f)$. I as the writer am going to ignorantly for the sake of clarity assume if a function ever becomes 0 that it no longer exists and therefore loses it's significant. Therefore 0 will be replaced with white space for these illustrations.

If $\alpha=1$ and $e =0$

$$\hspace{5mm}\mu(0,1)\\ \hspace{5mm}\swarrow\searrow\\ [\hspace{1mm}0=\mu(1,1)\hspace{1mm}]+\mu(0,2)\\ \hspace{26mm}\swarrow\searrow\\ \hspace{25mm}\mu(1,2)+\mu(0,3)\\ \hspace{28mm}\swarrow\searrow\hspace{12mm}\swarrow\searrow\\ \hspace{15mm}[\hspace{1mm}0=\mu(2,2)\hspace{1mm}]+2\mu(1,3)+\mu(0,4)\\ \hspace{50mm}\swarrow\searrow\hspace{12mm}\swarrow\searrow\\ \hspace{51mm}2\mu(2,3)+3\mu(1,4)+\mu(0,5)\\ \hspace{58mm}\swarrow\searrow\hspace{12mm}\swarrow\searrow\hspace{10mm}\swarrow\searrow\\ \hspace{45mm}2[\hspace{1mm}0=\mu(3,3)\hspace{1mm}]+2\mu(2,4)+3\mu(1,5)+\mu(0,6)\\ \hspace{78mm}\swarrow\searrow\hspace{12mm}\swarrow\searrow\hspace{10mm}\swarrow\searrow\\ \hspace{75mm}2\mu(3,4)+5\mu(2,5)+4\mu(1,6)+\mu(0,7)\\ \ldots$$

if $\alpha=2,e=0$: $$\hspace{5mm}\mu(0,2)\\ \hspace{5mm}\swarrow\hspace{5mm}\searrow\\ \hspace{5mm}\mu(1,2)+\mu(0,3)\\ \hspace{8mm}\swarrow\hspace{5mm}\searrow\hspace{2mm}\swarrow\hspace{5mm}\searrow\\ \hspace{0mm}[\hspace{1mm}0=\mu(2,2)\hspace{1mm}]+2\mu(1,3)+\mu(0,4)\\ \hspace{28mm}\swarrow\hspace{5mm}\searrow\hspace{2mm}\swarrow\hspace{5mm}\searrow\\ \hspace{30mm}2\mu(2,3)+3\mu(2,4)+\mu(0,5)\\ \hspace{30mm}\swarrow\hspace{5mm}\searrow\hspace{3mm}\swarrow\hspace{5mm}\searrow\hspace{3mm}\swarrow\hspace{5mm}\searrow\\ \hspace{15mm}2[\hspace{1mm}0=\mu(3,3)\hspace{1mm}]+5\mu(2,4)+4\mu(1,5)+\mu(0,6)\\ \hspace{50mm}\swarrow\hspace{5mm}\searrow\hspace{3mm}\swarrow\hspace{5mm}\searrow\hspace{3mm}\swarrow\hspace{5mm}\searrow\\ \hspace{50mm}5\mu(3,4)+9\mu(2,5)+5\mu(1,6)+\mu(0,7)\\ \hspace{55mm}\swarrow\hspace{5mm}\searrow\hspace{6mm}\swarrow\hspace{5mm}\searrow\hspace{6mm}\swarrow\hspace{5mm}\searrow\hspace{6mm}\swarrow\hspace{5mm}\searrow\\ \hspace{40mm}5[\hspace{1mm}0=\mu(4,4)\hspace{1mm}]+14\mu(3,5)+14\mu(2,6)+6\mu(1,7)+\mu(0,8)\\ \ldots$$

I created a function noted as $\Upsilon_\alpha(\beta,R)$, where $\beta=0,1,2...$ represents the row number, $\alpha=1,2...$ represents variation, $R$ represents the column of a row. using the illastrations above I created some examples but I noted $\Upsilon_\alpha(\beta,R)$as $\binom{\beta}{R}_\alpha$ and I did not include the arrows.

If $\alpha=1$ and $e =0$

$$\hspace{5mm}\binom{0}{0}_1\mu(0,1)\\ [\hspace{1mm}0=\mu(1,1)\hspace{1mm}]+\binom{0}{0}_1\mu(0,2)\\ \hspace{25mm}\binom{1}{0}_1\mu(1,2)+\binom{1}{1}_1\mu(0,3)\\ \hspace{15mm}[\hspace{1mm}0=\mu(2,2)\hspace{1mm}]+\binom{1}{0}_1\mu(1,3)+\binom{1}{1}_1\mu(0,4)\\ \hspace{51mm}\binom{2}{0}_1\mu(2,3)+\binom{2}{1}_1\mu(1,4)+\binom{2}{2}_1\mu(0,5)\\ \hspace{45mm}2[\hspace{1mm}0=\mu(3,3)\hspace{1mm}]+\binom{2}{0}_1\mu(2,4)+\binom{2}{1}_1\mu(1,5)+\binom{2}{2}_1\mu(0,6)\\ \hspace{65mm}\binom{3}{0}_1\mu(3,4)+\binom{3}{1}_1\mu(2,5)+\binom{3}{2}_1\mu(1,6)+\binom{3}{3}_1\mu(0,7)\\ \ldots$$

if $\alpha=2,e=0$: $$\hspace{5mm}\binom{0}{0}_2\mu(0,2)\\ \hspace{5mm}\binom{1}{0}_2\mu(1,2)+\binom{1}{1}_2\mu(0,3)\\ \hspace{0mm}[\hspace{1mm}0=\mu(2,2)\hspace{1mm}]+\binom{1}{0}_2\mu(1,3)+\binom{1}{1}_2\mu(0,4)\\ \hspace{30mm}\binom{2}{0}_2\mu(2,3)+\binom{2}{1}_2\mu(2,4)+\binom{2}{2}_2\mu(0,5)\\ \hspace{15mm}[\hspace{1mm}0=\mu(3,3)\hspace{1mm}]+\binom{2}{0}_2\mu(2,4)+\binom{2}{1}_2\mu(1,5)+\binom{2}{2}_2\mu(0,6)\\ \hspace{50mm}\binom{3}{0}_2\mu(3,4)+\binom{3}{1}_2\mu(2,5)+\binom{3}{2}_2\mu(1,6)+\binom{3}{3}_2\mu(0,7)\\ \hspace{30mm}[\hspace{1mm}0=\mu(4,4)\hspace{1mm}]+\binom{3}{0}_2\mu(3,5)+\binom{3}{1}_2\mu(2,6)+\binom{3}{2}_2\mu(1,7)+\binom{3}{3}_2\mu(0,8)\\ \ldots$$

if $\alpha=c,e=0$: $$\binom{0}{0}_c\mu(0,c)\\ \binom{1}{0}_c\mu(1,c)+\binom{1}{1}_c\mu(0,c+1)\\ \ldots\\ \binom{c-1}{0}_c\mu(c-1,c)+\binom{c-1}{1}_c\mu(c-2,c+1)+\ldots+\binom{c-1}{c-2}_c\mu(1,2c-2)+\binom{c-1}{c-1}_c\mu(0,2c-1)\\ [\mu(c,c)=0]+\binom{c-1}{0}_c\mu(c-1,c+1)+\ldots+\binom{c-1}{c-2}_c\mu(1,2c-1)+\binom{c-1}{c-1}_c\mu(0,2c)\\$$

I then realized that if I allow.

$$R_{max}=\left\{ \begin{array}{@{}ll@{}} \frac{\beta+\alpha+mod_2(\beta+\alpha)}{2}, & \text{if }\beta\geq\alpha \\ \beta, & \text{if }\beta<\alpha \end{array}\right.$$

AND

$$\Gamma=\left\{ \begin{array}{@{}ll@{}} \frac{\beta-\alpha+mod_2(\beta-\alpha)}{2}, & \text{if }\beta\geq\alpha \\ 0, & \text{if }\beta<\alpha \end{array}\right.$$

Then the sum can be summarized by:

$$\frac{d^\beta \mu(e,e+\alpha)}{(dx)^\beta}=\sum_{i=0}^{R_{max}}\Upsilon_\alpha(\beta,R_{max}-i)\mu(e+i+\Gamma,e+\alpha+\beta-i)$$

And since I've always called $\binom{i}{R}$ the Pascal's Triangle notation I will call $\Upsilon_\alpha(\beta,R)$ the Variant of Pascal's Triangle notation (VPT)

another notable behavior is When $\beta<\alpha$ then $R_{max}=\beta, \Gamma=0$, remembering $f = e+\alpha$ then $$\frac{d^\beta \mu(e,e+\alpha)}{(dx)^\beta}=\frac{d^\beta \mu(e,f)}{(dx)^\beta}$$ $$\sum_{i=0}^{\beta}\Upsilon_\alpha(i,\beta-i)\mu(e+i,f+\beta-i)=\sum_{i=0}^{\beta}\binom{\beta}{\beta-i}\mu(e+i,f+\beta-i)$$ therefore $\Upsilon_\alpha(\beta,R_{max}-i)=\binom{\beta}{\beta-i}$ for $\beta<\alpha$

BUT If $\beta\geq\alpha$ then all of the Pascal's rules apply for all $R$'s in $\beta$ EXCEPT when $\beta=\alpha+1$ a the horizontal column of blank space called the "You Shall Not Pass Line", YSNPL, prevents any new functions from becoming created at the beyond or on the line. This is because of the property $\mu(c,c)=0$ and $\frac{d \mu(c,c)}{dx}=0$ prevent any function from going beyond or on that line of 0's.

examples of $\Upsilon_\alpha(\beta,R)$

In order to understand what I just wrote I created 3 examples classified as functions called Pascal's Triangle Numerical Version (PTN$(\alpha)$), Variant of Pascal's Triangle Numerical Version (VPTN$(\alpha)$), and Variant of Pascal's Triangle $\Upsilon_\alpha$ form (VPTY$(\alpha)$).

Both PTN$(\alpha)$ and VPTN$(\alpha)$ will produce 6 rows of their given sequence.

PTN$(\alpha)$ will produce Pascal's Triangle with $\alpha$ number circles around the left most 1 on a given row, starting with the top row. The circles represent that part of Pascal's Triangle will not change after it encounter the given VPT.

VPTN$(\alpha)$ will show the VPT in numerical form.

VPTY$(\alpha)$ will produce the VPT in $\Upsilon_\alpha$ form

$\hspace{10mm}\text{PTN(0):}\hspace{54mm}\text{VPTN(0):}\hspace{30mm}\text{VPTY(0):}\\ \hspace{12mm}(1)\hspace{70mm}1\hspace{40mm}\Upsilon_0(0 \rightarrow 0,0)\\ \hspace{12mm}1,1\hspace{72mm}1\hspace{38mm}\Upsilon_0(0 \rightarrow 0,1)\\ \hspace{10mm}1,2,1\hspace{68mm}1,1\hspace{36mm}\Upsilon_0(0 \rightarrow 1,2)\\ \hspace{8mm}1,3,3,1\hspace{68mm}2,1\hspace{34mm}\Upsilon_0(0 \rightarrow 1,3)\\ \hspace{6mm}1,4,6,4,1\hspace{64mm}2,3,1\hspace{32mm}\Upsilon_0(0 \rightarrow 2,4)\\ \hspace{2mm}1,5,10,10,5,1\hspace{62mm}5,4,1\hspace{30mm}\Upsilon_0(0 \rightarrow 2,5)$

$\hspace{10mm}\text{PTN(1):}\hspace{55mm}\text{VPTN(1):}\hspace{30mm}\text{VPTY(1):}\\ \hspace{12mm}(1)\hspace{70mm} 1\hspace{40mm}\Upsilon_1(0 \rightarrow 0,0)\\ \hspace{10mm}(1),1\hspace{66mm}1,1\hspace{38mm}\Upsilon_1(0 \rightarrow 1,1)\\ \hspace{10mm}1,2,1\hspace{68mm}2,1\hspace{35mm}\Upsilon_1(0 \rightarrow 1,2)\\ \hspace{8mm}1,3,3,1\hspace{64mm}2,3,1\hspace{33mm}\Upsilon_1(0 \rightarrow 2,3)\\ \hspace{6mm}1,4,6,4,1\hspace{64mm}5,4,1\hspace{31mm}\Upsilon_1(0 \rightarrow 2,4)\\ \hspace{2mm}1,5,10,10,5,1\hspace{58mm}5,9,5,1\hspace{28mm}\Upsilon_1(0 \rightarrow 3,5)\\$

$\hspace{10mm}\text{PTN(2):}\hspace{55mm}\text{VPTN(2):}\hspace{30mm}\text{VPTY(2):}\\ \hspace{12mm}(1)\hspace{70mm}1\hspace{40mm}\Upsilon_2(0 \rightarrow 0,0)\\ \hspace{10mm}(1),1\hspace{66mm}1,1\hspace{38mm}\Upsilon_2(0 \rightarrow 1,1)\\ \hspace{8mm}(1),2,1\hspace{62mm}1,2,1\hspace{36mm}\Upsilon_2(0 \rightarrow 2,2)\\ \hspace{8mm}1,3,3,1\hspace{64mm}3,3,1\hspace{33mm}\Upsilon_2(0 \rightarrow 2,3)\\ \hspace{6mm}1,4,6,4,1\hspace{60mm}3,6,4,1\hspace{31mm}\Upsilon_2(0 \rightarrow 3,4)\\ \hspace{2mm}1,5,10,10,5,1\hspace{58mm}9,10,5,1\hspace{26mm}\Upsilon_2(0 \rightarrow 3,5)\\$

$\hspace{10mm}\text{PTN(3):}\hspace{55mm}\text{VPTN(3):}\hspace{30mm}\text{VPTY(3):}\\ \hspace{12mm}(1)\hspace{70mm}1\hspace{40mm}\Upsilon_3(0 \rightarrow 0,0)\\ \hspace{10mm}(1),1\hspace{66mm}1,1\hspace{38mm}\Upsilon_3(0 \rightarrow 1,1)\\ \hspace{8mm}(1),2,1\hspace{62mm}1,2,1\hspace{36mm}\Upsilon_3(0 \rightarrow 2,2)\\ \hspace{6mm}(1),3,3,1\hspace{58mm}1,3,3,1\hspace{34mm}\Upsilon_3(0 \rightarrow 3,3)\\ \hspace{6mm}1,4,6,4,1\hspace{60mm}4,6,4,1\hspace{31mm}\Upsilon_3(0 \rightarrow 3,4)\\ \hspace{2mm}1,5,10,10,5,1\hspace{52mm}4,10,10,5,1\hspace{26mm}\Upsilon_3(0 \rightarrow 3,5)\\$

Remember $\alpha$ represents a variation of Pascal's Triangle, $\beta$ represents a row in $\alpha$ and $R$ represents a column in $\beta$. One should have noticed for each $\alpha$ there is a invisible line where no new number exists on or to the left of that line, YSNPL. One should have noticed for every column in $\beta$ the top right column and the top left column sum together in the row underneath, $\beta+1$, to create a new column in between both of the original columns.

While researching this phenomena I produced some results for $\alpha=1,2,\infty$

$$\Upsilon_2(\beta+1,R)=\Upsilon_1(\beta,R) \text{ can be found in }\left\{ \begin{array}{@{}ll@{}} \frac{R+1}{\beta+1}\binom{\beta+1}{(\frac{\beta-R}{2})}, & \text{if }R<\beta \text{ and } mod_2(\beta-R)=0\\ , & \text{otherwise} \end{array}\right.$$

let $\eta=2R-mod_2(\beta+\alpha-1)$

if maximuim $R$ = $R_{max}$

$$\Upsilon_1(R,\beta)=\Upsilon_2(R,\beta+1)=\frac{\eta+1}{\beta+1}\binom{\beta+1}{\frac{\beta-\eta}{2}}$$

I found this by creating my own variant from the formula given in A008313

$\Upsilon_3$ in A026009 EQ UNKNOWN

$\Upsilon_4$ in A026022 EQ UNKNOWN

Since YSNPL only exists if $\beta=\alpha+1$ and $\beta\neq\infty$ if $\beta$ is finite. Therefore the following statement must be true.$$\Upsilon_\infty(R,\beta) = \binom{R}{\beta}$$ My Question: Is this a known pattern? Are there other known $\alpha$'s? This code can find any triangle with a given $\alpha,\beta$

def tri(alpha,beta) : if alpha > 0 : D = [0]*((beta)//2+alpha); D[0] = 1 b = False; h = 1 for i in range(beta) : print([D[z] for z in range(0,h,1)]) if i < alpha-1 : for R in range(h-1, -1, -1) : D[R+1] += D[R] h += 1 else : if b : for R in range(h, -1, -1) : D[R] += D[R-1] h += 1 else : for R in range(0, h-1, 1) : D[R] += D[R+1] b = not b tri(1,10)

• Where you wrote $\alpha{\leq}\beta{\leq}i$ I changed it to $\alpha\leq\beta\leq i$ simply by changing the code from \alpha{\leq}\beta{\leq}i to \alpha\leq\beta\leq i. Why do you make that extra effort to put in those {curly braces} whose only effect is to prevent proper spacing? $\qquad$ – Michael Hardy Jun 14 '16 at 3:42
• @MichaelHardy I thought I always had to use the brackets to separate known words from characters and numbers which I believed included known words. But thank's you for the assistance. – Jon Jun 14 '16 at 3:47
• To write $\int_0^{12}$ you write \int_0^{12}, and if you omit the braces you see $\int_0^12$. To write $\frac 1 {12}$ you can write \frac 1 {12} and if you omit the braces you see $\frac 1 12$. For $\int_0^\infty$ you can write \int_0^\infty but for $\int_0^{+\infty}$ you write \int_0^{+\infty} and if you omit the braces you see $\int_0^+\infty$. Those are some examples of what braces do. $\qquad$ – Michael Hardy Jun 14 '16 at 4:25
• Binary relation symbols and binary operators have some space before and after them, as in $5+3=8$, coded as 5+3=8, and if you write 5{+}3{=}8 then there's nothing before or after the "+" or the "=" so that space does not appear and you see $5{+}3{=}8$ instead of $5+3=8$. $\qquad$ – Michael Hardy Jun 14 '16 at 4:27
• I don't see where you actually define $\Upsilon_\alpha(\beta,\gamma)$. – joriki Jun 14 '16 at 4:33

$$\Upsilon_\alpha(\beta,R)=\left\{ \begin{array}{@{}ll@{}} \binom{\beta}{R}, & \text{if }\beta\leq\alpha \\ \binom{\beta}{R}, & \text{if }\beta\leq R+\alpha \\ \binom{\beta}{R}-\binom{\beta-\alpha}{R}, & \text{otherwise} \end{array}\right.$$


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# How do you find the greatest common factor for the list of terms 70x^2, 30x^6, 60x^9?

Feb 12, 2017

See the entire solution process below:

#### Explanation:

To find the Greatest Common Factor of these three terms we must first factor each term separately:

$70 {x}^{2} = 2 \times 5 \times 7 \times x \times x$

$30 {x}^{6} = 2 \times 5 \times 3 \times x \times x \times x \times x \times x \times x$

$90 {x}^{9} = 2 \times 5 \times 3 \times 3 \times x \times x \times x \times x \times x \times x \times x \times x \times x$

The Common Factors are:

$70 {x}^{2} = \textcolor{red}{2 \times 5} \times 7 \times \textcolor{red}{x \times x}$

$30 {x}^{6} = \textcolor{red}{2 \times 5} \times 3 \times \textcolor{red}{x \times x} \times x \times x \times x \times x$

$90 {x}^{9} = \textcolor{red}{2 \times 5} \times 3 \times 3 \times \textcolor{red}{x \times x} \times x \times x \times x \times x \times x \times x \times x$

Therefore the GCG (Greatest Common Factor) is:

$\textcolor{red}{2 \times 5 \times} \textcolor{red}{x \times x} = 10 {x}^{2}$

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# Holder regularity for the heat potentials

First I apologize for my bad English and for any error: this is my first question.

I need some regularity results for the single and double layer heat potentials. If $\Gamma(t,x)$ is the fundamental solution of the heat equation, the single and double layer potentials are defined as follows:

(SLP) $u(t,x)= \int_{0}^{t}\int_{\partial \Omega}\Gamma(t-\tau,x-y)\varphi(\tau,y)d\sigma(y) d\tau$

(DLP) $v(t,x)= \int_{0}^{t}\int_{\partial \Omega}\frac{\partial\Gamma(t-\tau,x-y)}{\partial\nu(y)}\varphi(\tau,y)d\sigma(y) d\tau$.

In the cylinder $Q=\Omega\times[0,T]$, $\Omega\subset\mathbb{R}^n$, $\Omega$ bounded domain regular enough and $\nu$ in the unit outer normal to $\partial \Omega$

My questions are:

1) It is true that if $\varphi\in C^{1+\alpha/2,2+\alpha}(Q)$ then $v$ extends with continuity to $\bar{\Omega}\times[0,T]$ and this extension stay in $C^{1+\alpha/2,2+\alpha}(\bar{Q})$?

2)It is true that if $\varphi\in C^{(1+\alpha)/2,1+\alpha}(Q)$ then $u$ extends with continuity to $\bar{\Omega}\times[0,T]$ and this extension stay in $C^{1+\alpha/2,2+\alpha}(\bar{Q})$?

Where the functions spaces $C^{1+\alpha/2,2+\alpha},C^{(1+\alpha)/2,1+\alpha}$ are the parabolic holder spaces for example defined in "Lectures on Elliptic and Parabolic Equations in Holder Spaces, N.V. Krylov, pg 117,118"

(Is possible that the parameters of the Holder spaces are different.)

These questions are well known for the armonic potentials and I want to know if similar results hold true for the heat potentials.

• You take $\varphi\in C^{1+\alpha/2,2+\alpha}(Q)$ but densities should be defined on $\partial\Omega\times[0,T]$. Does it means that the trace of $\varphi$ is taken? Sep 23 '14 at 12:33
• You're right, i think that the density $\varphi$ must stay in $C^{1+\alpha/2,2+\alpha}(\partial\Omega\times[0,T])$ Sep 23 '14 at 12:49

The answer is yes for the smooth enough boundary, provided $\varphi|_{t=0}=0$ and for 2) additionally $\partial_t\varphi|_{t=0}=0$. Say, for the SLP from $C^{1+\alpha/2,2+\alpha}(\bar{Q})$ it should hold $u|_{t=0}=\partial_tu|_{t=0}=0$. From here the mentioned conditions on $\varphi$ at $t=0$ follow.

• Thank you.Can you give me some references for these results? Sep 23 '14 at 12:54
• When $\Omega$ is a half-space it is written in O.A. Ladyzhenskaya, V.A. Solonnikov, N.N. Uraltseva, "Linear and Quasilinear Equations of Parabolic Type". For general case I havn't a reference handy, but smoothness of the heat potentials was studied in the works of Kamynin. Sep 23 '14 at 13:10
• Thank you. I think that the works of Kamynin are what I was looking for. Sep 24 '14 at 13:08
• In the Kamynin's works I think that the SLP and the DLP don't reach the regularity $C^{1+\alpha/2,2+\alpha}$ but $C^{1+\alpha'/2,2+\alpha'}$ for all $\alpha'<\alpha$. If you recall some references for the result with $\alpha$ with holder coefficient, came you give me? Thank you in advance Sep 25 '14 at 11:06
• @foo90 If you can read Russian and would provide some e-mail address, I can send you my article (in Russian) for SLP of a parabolic operator with variable coefficients, which, as a special case, have the required result. Sep 26 '14 at 14:03


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# GRE®

Free Version
Easy

GRE-GGSE9J

Given that $f(x) = 3x+6$ and $g(x) = 5-2x$, find the value of $f(2) + g(-2)$.

A

$13$

B

$12$

C

$108$

D

$21$


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# Why is the Lunar Node in a daily motion?

I've read a lot about the Lunar Nodes online (on Wikipedia, astro.com and a bunch of other websites), but there is one thing I cannot figure out and the sources don't seem to clearly explain: why are the nodes moving on a daily basis? As the node is a point of intersection of Moon's orbit and the Ecliptic I understand that this point does not fall in the same place in space every time, but I would expect that once it's passed by the Moon it stays where the crossing occurred and its position gets updated on the next crossing of the Ecliptic, so in ephemeris I would expect to see the same values of the coords between the crossings, but instead I can see they're changing on the daily basis (ie. this ephemeris has 2 columns showing true and mean nodes' positions: http://www.astro.com/swisseph/ae/2000/ae_2016.pdf). I'm sure the ephemeris cannot be wrong, so I must be missing some details. I thought that finding the Moon's nodes would be as easy as finding the dates/times/longitudes when the Moon's latitude equals 0, but apparently this way I can only find the times and static points of the crossings of the Ecliptic, not the moving nodes as they're shown in ephemeris tables.

The nodes are the two points on the orbit of the moon at which the plane of the orbit crosses the elliptic. The position of the nodes change in a predictable way due to the perturbation by the sun (and the other planets). Since the orbit is constantly changing, the position of the longitude. The node is not the position where the crossing last occurred. Rather it is the intersection of two imaginary lines in the sky, the ecliptic and the orbit of the moon.

The position of the nodes are measured as an angle, clockwise from the first point of Aries, thus it is a longitude, or right ascension. On the first of September 2016 the longitude of the ascending node of the moon is 162.484 degrees, and decreasing slowly.

The linked pdf indicates a value of 12♍41, apparently this means 12 degrees 41 minutes within Virgo. The notation is not used in astronomy. If you are interested in the scientific study of the sky you should be aware that astro.com is an astrology website. The presentation of ephemeris may not represent scientific practice. I consider modern astrology to be pseudoscience and fraudulent.

• Thanks, James. I'm using pyephem for astro calculations. Do you think it would be a good method of tracing Moon's nodes: 1. get dates where Moon is at 0 lat; 2. find longitudes of the crossing points on those dates; 3. calculate average daily rate of motion using longitudes of two consecutive crossing points, ie. (lon2 - lon1)/days. ??? Aug 27 '16 at 6:06
• pyephem knows ascending node. m=ephem.Moon();m.compute('2016/9/1');print(m._raan) Aug 27 '16 at 7:09
• Yes, I knew about _raan, but unfortunately it doesn't work for me for some reason giving AttributeError: 'ephem.Moon' object has no attribute '_raan', but that's a separate issue I'm about to investigate (upgrading pyephem package didn't help...). Aug 27 '16 at 9:07
• In pyephem _raan is the attribute of EarthSatelites, but not the Moon, see github.com/brandon-rhodes/pyephem/issues/110 . Aug 27 '16 at 19:34
• @mac13k Did my comment on that github thread help?
– user21
Aug 30 '16 at 12:54


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# Correct statistics for reporting speedup results

Say I have slow and fast versions of some code, and want to report a speedup number comparing the two. I run the slow version $n$ times and the fast version $m$ times, producing times $(s_1, \ldots, s_n)$ and $(f_1, \ldots, f_m)$. The simplest way to produce a speedup is to average the means: $$\frac{\bar{s}}{\bar{f}} = \frac{m \sum_{i<n} s_i}{n \sum_{j<m} f_j}$$ However, this does not take outliers into account.

Question: What is the best statistic to use when reporting speedup numbers?

• How big is the standard deviation in comparison to the mean? Whatever you do, you should report what you did and probably put error bars if they are large. If they are really large, you should investigate the source. Most computer code should run pretty deterministically in time unless there's a random component to the program itself or you are sharing computer resources with others (this could be networking or disk, not just cluster nodes). If competition for disk resources is the problem, you might consider reporting performance with I/O disabled (quite common)--just be sure to note it. – Bill Barth Feb 23 '14 at 21:36
• On Edison (a Cray supercomputer), I have a 2% difference between two samples. On my laptop I see a 6-8% standard deviation measured over 10 samples. Both are for compute kernel only, no I/O. – Geoffrey Irving Feb 23 '14 at 21:50
• To clarify why I am mentioning outliers if the variances are already reasonably low: this is a sufficiently fundamental statistical quantity that I would like to know the ideal way to report it, even I nonideal ways are fine in this particular case. – Geoffrey Irving Feb 23 '14 at 21:52
• The question is what are you trying to communicate, and formula would communicate that best? I don't think I've ever seen a paper that reports the run-to-run variability in speedup unless the cause was central to the paper. Given that we posit a linear relationship between run time and processor/task/thread count, you're probably fine to use the ratio of means, but then error bar that with the ratio of max-to-min and min-to-max if you think showing the range is important. Also, you should probably look at your frequency scaling and task pinning options to cut down on your variability. :) – Bill Barth Feb 23 '14 at 23:06
• There can be a lot of trickery in eliminating IO. Between compiler optimizations to "Copy On Write" tricks there can be really non-obvious ties downward. I usually follow the prototype of d1 = loadData(); d2 = copy(d1); r1=algo(d2); r2=algo(d1), and only consider the time of the second run. – meawoppl Feb 26 '14 at 1:29

In addition to all that Bill Barth has already said above, let me mention that people often report the fastest of several runs. The rationale is that the actual run time is the ideal run time plus any number of slow downs resulting from other processes running, OS delays, network delays, etc. Since these are all noise we are not interested in, using the fastest run time comes closest to the one we really want to know.

• Unfortunately, this principle doesn't help when reporting a speedup between two algorithms. – Geoffrey Irving Feb 24 '14 at 6:02
• @GeoffreyIrving, why not? Both algorithms have a theoretical performance expectation vs. problem size (or processor count or other non-statistical parameter) with low-order and parameter-independent terms ignored. Using the fastest time (and noting this fact) is simply helping you to ignore these additional terms. Which seems like a fine strategy. Unless you tell us differently, it seems like you're trying to figure out how to communicate the difference between to algorithms most effectively, and Wolfgang's suggestion is conventional and expected so it might convey that information best. – Bill Barth Feb 24 '14 at 13:34
• Oops, yep, you're right. I happily withdraw my statement. – Geoffrey Irving Feb 24 '14 at 17:45
• (+1) A side-question: I complete see your point about non-symmetric noise distribution, etc. Let's say though that I make an implementation A, and an implementation B and I benchmark them and after a reasonable amount of runs, the 25-th quantile and the median and mean are ~4.5x faster in A than B while the 0% quantile is ~3x. When comparing implementation A to B, despite the fact that: yes A is theoretically only ~3x faster, isn't a ~3x speed-up unrepresentative of the speed-up on might expect when using implementation A instead of B? (This is a real-life example by the way) – usεr11852 Apr 19 '15 at 7:59
• @usεr11852: It all depends on the system you're on. If your median or 25th quantile are so far apart as to distort the statistics in the way you hypothesize here, then you're likely on a system that has a lot of noise. For example, it may be used by others at the same time, etc. That may not be representative of the systems others have for their repeat experiments, and it would sound to me like you're overselling your results in that case. So, I still suggest to report the best runs. Whatever you do, you should report in the paper what statistics you use. – Wolfgang Bangerth Apr 19 '15 at 18:41

I suggest you use the median to give a statistical estimate. Unlike the mean, the median is not corrupted by outliers.

• For data where all noise is positive (i.e., with a non-symmetric noise distribution), the median is as bad as any other statistic. For run-times, this is indeed the case, see my answer above. – Wolfgang Bangerth Feb 27 '14 at 13:23

If the standard deviation is not negligible, you could use two box plots side-by-side, constructed each with the timing of one of the algorithms. They are by all means not standard in numerical analysis, but they do a great job in displaying this kind of information.


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# Logistics of a hovering watercraft in a fantasy setting

So in my fantasy world, there exists a substance that (when applied to the surface of an object) repels water in the same way a very strong magnet would. This allows boats to hover above the surface of the water.

The people that use this substance are a small tropical island culture, and use their boats to hunt large creatures that can manipulate the water to defend themselves.

My question is this: what would be the optimal design for a boat that uses this substance in terms of balance, propulsion, and handling large waves? Assume pre-industrial technology, but any materials you’d like, since I haven’t nailed down that part of the world yet anyway.

• Can we get a bit more information on this prey? How large are they, what kind of attacks do they utilize? – Mormacil Mar 4 at 22:53
• "This allows boats to hover above the surface of the water": no it doesn't. Water is a fluid. The craft will float just like any other boat, with the difference that the volume of the hull will be somewhat larger corresponding to the strength of the water-repelling field. – AlexP Mar 4 at 23:07
• @AlexP Either that, or it'd just immediately sink right down to the bottom as it repels the water underneath it, sinks a little bit, repels the new water underneath it, sinks a little more, ad infinitum down to the bottom of the ocean. – nick012000 Mar 5 at 4:06
• I'm with AlexP here. If you repel the water, the water repels you. So it will mostly work the same way than a real boat, with the exception of being unafected by the currents. A wave will still be a wave and hit the boat, just from farther away. The only clear advantage is that you'll stay dry, assuming the boat didn't flip. @nick012000 The water you repel will be blocked by the density of the ocean and will push back on the boat, the same way a boat already does with gravity only (Archimedes principle and all that). – Echox Mar 5 at 9:46
• Since we have Aron the sailing instructor in the game, I thought I should offer the kayaker's perspective on this. A frictionless hull would be an uncontrollable nightmare. Everything you do is a balance of forces between paddle and hull resistance, taking away one of those factors would make it hell to control. – Separatrix Mar 5 at 14:28

Strangely enough, your boats aren't going to be THAT different from the boats we already have in terms of propulsion and wave handling. Balance is a completely different matter, but let's deal with each of these criteria one at a time.

Propulsion
You really have two options with your level of technology and they're the same options that everyone else had; oars and sails. Depending on how far above the water you sit though, oars are problematic because they have to be longer to get into the water meaning you need to be stronger to pull the oar. For open sea journeys this is an issue because putting the rowers close enough to the water to make oars effective is counter productive to being able to handle large waves, where you want high watertight 'walls' on the side of your boat (more on that later) so I'd stick with sails.

Sails could be managed by the people you describe technologically, especially as pre-industrial doesn't mean pre-science. There are plenty of examples in history of pre-industrial sailors who used sophisticated means to get their boats from one place to another in terms of both navigation and wind management so this is the best option. Just bear in mind, these kinds of large sailing vessels were a massive expenditure prior to industrialisation and they would be for your world as well.

Large Waves
Ultimately the best defence against high waves is high walls. On conventional ships, they sat very tall in the water (with massive ballast reserves in the hull to keep them upright) so as to survive high seas. Your floating boat will need the same walls, so your boat will still have a number of decks on it with walls on the side to stave off wave strike.

Balance (and Navigation because they're related)
These boats will have flat bottoms. It's that simple. You don't need to keep the boat from drifting in current because your boat floats above it in the first place, so there's no need for deep hulls. You'll need keels (and rudders) though, because the sails are only part of the propulsion equation in that boats actually rely on some resistance against the hull to change direction. Rudders for instance need contact with the water to reorient the boat. So, your boat probably has a large rectangular square bottom to maximise the repulsion area against the water, hence maximising the balance of the boat. BUT, it also has a series of long keels that dip into the water (not coated with the repulsive material) that help with steering, and at least some of these will be on swivels that can be controlled from within the boat for steering as a rudder. Ideally, these would be on the outer edges of the flat surface to preserve stability, like a catamaran.

As Aron points out in comments, this may not work. Certainly, the resistance that a keel can generate is minimal by comparison to a hull, so the idea that you could successfully do anything other than use a rudder like control surface is in doubt and as such, should be taken as speculative.

So, your boat would look like a very large floating bathtub, with keels and sails below and above it respectively. That will allow it to balance, survive high seas, and move & navigate on the open sea.

• Comments are not for extended discussion; this conversation has been moved to chat. – L.Dutch - Reinstate Monica Mar 6 at 3:22
• It seems I accidentally clicked on the "downvote" button and now it's stuck there until the answer is edited. Sorry about that. :( – Vilx- Mar 6 at 9:05

Alright I think a flat hull will be by far the most stable design. You want a constant repelling force and a flat wide hull gives you stability against waves. At the same time you do want a somewhat high front against waves.

## A

A tall front to steer against waves. This also gives you a vantage point to engage your prey with if they're large enough. Obviously this comes equipped with a railing, in fact your whole boat will be.

## B

Back/front view. A wide flat hull for stability with a high railing on the main deck.

## C

I was considering using your water repellent for some sort of ingenious propulsion mechanic before I realized you need to push that down in the water with more force then the boat weights. But maybe it will inspire you or someone else to make it work with some force multipliers.

## D

An upgrade could be extra hulls not unlike a catamaran. The idea being you place those far and wide to give addition stability against waves from the sides. This will require some very strong water repellent to work. Extra strength by joining them with the mast against the central line of the ship.

## E

Sails will be your best propulsion system. With no contact with water you should be able to get some frightening speeds.

Now for more specific details I turn you to regular ship design. You might want more storage space then a single space between your hull and the main deck. You probably want regular rudders in the back to aid with steering your vessel, regular rudders should work, uncoated. You might want multiple sails, again regular sources on sail design should have you covered.

• Ex sailing instructor here, sailing a hovercraft is literally impossible. A hover craft would not be able to generate lateral resistance needed for a sail to push against to generate lift. – Aron Mar 5 at 1:40
• In regards to D, "very strong repellant" shouldn't be a necessity for the extra hulls. The pressure on the hulls will be distributed equally according to surface area. As long as the repellant exerts a particular force in relation to the surface area it is painted on, the repellant will work the same (in regards to flotation) for all sizes of hulls. – Flater Mar 5 at 9:30
• @Aron: Just to specify for us landlubbers; when you say "sailing", you mean steered sailing as opposed to simple propulsion in the direction of the wind; right? As far as I understand the mechanics of sailing, having a keel that still drags in the water (by not coating it in the repellant) would solve that issue, no? – Flater Mar 5 at 9:33
• @Aron, Steering might be possible with rudders and centerboard/keel that are not water repelling. – ShadoCat Mar 5 at 18:18
• @Aron: Sure, but contrast that to the Polynesians who used their knowledge of water/wind currents to plan their routes. The ship could have other (more expensive) means of steering/propulsion but the ability to sail/drift (which would be more preferable) when possible. Even then, you're starting from the basis of a ship with a fixed destination, which is based on the expectation of the ship taking you where you want to go. But when you have a ship that is not possible to be steered, you don't generally expect it to go to a fixed destination and adapt accordingly. – Flater Mar 7 at 7:43

I'd think about it completely differently than a boat.

It's true that sailing would be impossible since it's been pointed out that sailing relies on the resistance of the hull in the water to navigate effectively. But that's not true for aviating like you might do in a balloon, which is essentially what you have. You have a flying craft that simply flies very close to the ground.

You haven't specified how far above the surface the craft would hover, but I'm assuming that it's at least a meter (~3ft) so that it can get over most of the mundane waves and swells. Your technology level doesn't really allow for using fans like a zeppelin or blimp would use to steer, but you could still navigate just like a hot-air balloonist or glider pilot would.

Like the Polynesians, your civilization would be familiar with ocean and wind currents, but with a bias towards the wind aspect as that's the strongest form of propulsion you'll use. Knowledge of the ocean currents would mostly only be useful for tracking prey.

As far as the design of the craft, it should definitely get wider as it goes down into a large flat base for the repelling material. Water would just stream off it down the sides. You can use sails or even kites, balloons or parachutes to grab the wind and pull you along. Control surfaces like wings and tails on airplanes help you steer. Remember, you're in the air not on the water. As long as you can keep your means of moving and navigating intact, your craft should survive just about anything.

As other answers suggest, the wider base will provide stability and since you have basically a sink proof material lining the bottom it won't really matter if the craft is totally swallowed by a wave. As long as the bottom is heavier than the top, the craft will rise back up from underwater. Maybe some kind of air bladder/attic area(s) sealed into the upper/highest parts of the ship. As long as your decks drained you could temporarily be fully submerged and probably come out ok.

• as long as the bottom is heavier than the top, the craft will rise back up from underwater That is not a given. Consider that overturned ships don't always naturally flip back up, even if their (intended) bottom is heavier than their (intended) top. Imagine a really wide but very low ship, with a heavy bottom in the middle. Now put that ship upside down on the water. It's never going to flip itself upright because it's much too wide for it to flip naturally. This is why catamarans widen their base so they become flip-resistant. But a flipped catamaran will also be harder to flip back. – Flater Mar 5 at 9:37
• Something that is just repels water is not going to float if entirely submerged, unless it is strong enough to bring some air with it. If the air bubble is lost due to turbulence then the magic substance is only going to have natural buoyancy. – Paul Johnson Mar 5 at 14:04
• Despite the inventiveness, since the keel could also be coated and would have a larger impact on the water below without touching it, you could steer it just fine with a keel. – Demigan Mar 5 at 22:14

For the sake of concision, I'm going to refer to the anti-water magnetic coating as AWMC. I'm assuming that AWMC is polar, that it can be turned off or directed, or that the repulsion is significantly stronger on one side, since otherwise the crew wouldn't be able to go near once it was applied (humans being mostly water). I'm also assuming that the AWMC resistance force is quite powerful.

The optimal hull design is to stick to regular hull designs just with a thin layer of AWMC that functions as frictionless coating + waterproofing.

The real genius is using AWMC for propulsion, since it's been shown by our resident sailors and kayakers that traditional forms of naval propulsion require friction/resistance. Introducing... the WHIFFLE MOTOR (patent pending)!! In its simplest form, the whiffle motor is a board with holes poked in it and AWMC applied to the back side. When it's first dropped into the water, the board generates thrust by resisting the water, but this creates a vacuum/pressure gradient in the water, most strongly right at the transition point where water comes in through the intake holes. Water is pulled in by the gradient and then pushed backwards by the AWMC coating, and the cycle begins. Better designs streamline the overall setup so that the most water possible is pushed backwards the most evenly, thus generating the most even and powerful thrust. In my unprofessional opinion, the best designs probably would probably look like a jet engine or turbofan engine, making it so that water only comes in from the intake and all the water is ejected in a smooth stream. Steering is now a question of having some lesser boards that can be angled more freely.

Basically, you now have spacecraft in a roughly 2D plane. Speed is limited by the direct resistance that water provides to being pushed apart, so you don't get to go stupidly fast just with a basic WHIFFLE MOTOR (patent pending).

• Very much like a water powered pulse jet. – MongoTheGeek Mar 5 at 18:40
• But then, where is the energy coming from? Or is it a free energy machine? – Eth Mar 8 at 10:22
• @Eth it's a free energy machine only if you think magnets are free energy - presumably the AWMC field weakens over time just like a magnet does. – Jeutnarg Mar 8 at 16:35

Super surfboard.

In answers so far the water repulsion has been considered a sort of ultrafloatation. Really, though, this tech would be super useful for propulsion. Imagine a surfboard. Only the back is treated - possible at an angle, not parallel to the water surface. The treated back repels water and so rides higher. One would have exactly the same effect with a regular surfboard if the back were physically higher than the front which is the case where you are on a wave. The net effect of the vectors is to push the board forward.

With this water repulsion tech, the surfboards will move forward even on still water. Because the mid to front of the board is in the water, one can still use friction of the board against the water to steer just as with a regular board. A skeg or keel could still be on the board to improve maneuverability - just don't treat that with repulsion tech.

The main benefit of this application of the water repulsion tech is awesomeness. It will not be like a bunch of dudes hunched in a Teflon whaleboat. It will be more like Mongols on horseback - but faster. The party surfing out from shore at sunrise will be cool as can be. If there are waves (maybe created by the water creature?) they can do some regular surfing too. They will hang ten with their harpoon lassos. It might even be possible to surf right up onto the sea creatures.

• Take a mundane surfboard. Add a trivial weight transparent bulg hanging down from its back. This "repels water". Why doesn't your technique work? How is this magic technique fundamentally different? – Yakk Mar 5 at 16:35
• You describe buoyancy. The OP makes this akin to magnetism. You can float a magnet atop another magnet. You can push a magnet along a surface with another magnet. I thought this magic magnetlike repulsion was like this pushing / floating phenomenon. It would be interesting to make a mockup with a flat surface made of magnets S side up taped together and a HotWheels car with an S side down magnet under its back end. If the magnet sheet were deformable like water this would be a fine simulation. – Willk Mar 5 at 17:26
• It shouldn't matter. A wedge doesn't slide forward towards it's pointed edge. You are essentially making a wedge with the back lifted by magic. – MongoTheGeek Mar 5 at 18:36
• @MongoTheGeek: Suppose I lift the back of my surfboard by applying a force to the water. I am applying the force with a powerful diesel leafblower. I apply the force at 45 degrees from vertical. This force vector can be considered half in the plane parallel to the surfboard and half perpendicular. The perpendicular component is responsible for the lift. Would my surfboard move forward because of the opposite reaction from the parallel component? – Willk Mar 5 at 18:51
• If you apply the force at an angle it will push forward. @Jeutnarg's answer builds on this. – MongoTheGeek Mar 5 at 23:14

Another way to use this magic substance for propulsion comes to mind, extrapolating on Willk's answer: A stone or piece of wood, with a hole bored through, and some of the substance smeared around the opening on one side would make a handy no-fuel water rocket, especially if it was put coaxially inside a tube of bamboo or something to direct the water flow.

You'd start the torpedo by sticking the front end into the water and pushing it down until a jet of water starts shooting out the back, and then lower the whole shebang into the water.

Basically I'm thinking of a ramjet-type operating principle, the water comes in pressurised to an extent by the intake shape and is vaporized by the cavitation effect after coming out of the stone nozzle, since under water the ring of well-placed magic substance (and it's toroid-shaped anti-water field) creates a no-water area right after the nozzle, leaving a small "hole" in the middle of the nozzle where the field is the weakest. There's more room in the exhaust section of the tube for the expanded vapor to escape, and less pressure, so the vapor escapes that way, creating suction for bringing in new water through the nozzle.

I don't know if this is really feasible give only a water-repelling magic substance, but sounds plausible enough for a magic-using world.

As a bonus, the question absolutely brings to mind the hydrophobe wizards of Krull from The Colour of Magic :)

Depending on the sea conditions, you might be able to do something interesting with waves. Ignoring wind and air resistance for a moment, a craft like this will tend to slide down a wave, so if you are going the same way as the waves then propulsion is very simple. If you combine this with a sail then you could, if you are clever enough, use the difference in wave and wind movement to control your direction.

The Wandering Albatross does something like this to glide without flapping its wings. Your islanders are not going to do exactly that, but they will probably have related techniques. The simplest thing would be to use a sail to move sideways along a wave, using the trough of the wave a bit like a keel, thereby going to the left or right of the wind and wave direction. I don't think you could tack like this though.

A couple of notes on my approach to this before we begin the main design specs.

1. I'm assuming the coating only repels free water, not the water in the bodies of the people in the boat.

2. A mass of water equal to the mass of the vessel will be displaced from below the vessel. Whether the vessel is in contact with the water or not doesn't change this. In practice exactly the same rules of buoyancy apply as to a normal boat, if you fail to displace sufficient volume of water for the mass of your vessel, it will still sink. It will sink in a little bubble of repelled water, but it will sink.

Start with a dragonboat, they're beautifully designed fast boats which take a lot of people. We're going to run this like a normal boat, hull in the water. Lightly coat the gunwales and inside of the hull with your hydrophobic substance. The most important factor in any boat is keeping the people inside and the water outside, this use of your magic coating will serve particularly well for this. No need to worry about waves washing over the side or enemies squirting water at you risking swamping or sinking your boat. It'll also make it significantly more comfortable inside for your (dry) paddlers.

Add outriggers, these are going to make full use of your magic coating. Full length, wide enough for a warrior to stand on and fully coated in your magic substance. These are going to act both for stability and as a fighting/hunting platform. They'll have a stabilising effect significantly greater than their mass and volume would normally allow for. What's going to surprise you here is that the greatest risk is making the outriggers too buoyant. To a certain extent they need to be overwashed as they hit waves independently of the main hull, otherwise the forces on them transmit negatively back to the hull, making the whole thing a lot more uncomfortable than it needs to be along with risking high stresses on their struts and breaking the whole thing up.

Your magic substance is used for stability, improved versatility and comfort, but not as a critical element. Controlling a boat is already a fine art, having the hull in the water makes the boat controllable, whether you choose to sail it, row it, or paddle it like a big canoe, boat control is about balance of forces. Once you start taking some of those forces out of the system you get an unbalanced equation and control becomes particularly difficult.

Hull design isn't a simple game, whether you choose hard or soft chines, a "V" hull or flat bottom, carvel or clinker, daggerboard, centreboard or leeboard, fixed keel or lift keel, bilge keel or wing keel, all these things are significant, in most cases they're not compromises but deliberate design decisions fundamentally changing the boat for different purposes. If you fully coat the hull in your substance you take away all these options and leave yourself with at best a vague shapeless hull. Keep the boat in the water.

Being able to aim your craft while sailing relies on balancing the forces of the wind with the forces you apply on the water.

The boat is long and narrow, and has much much more resistance to movement perpendicular to its hull than parallel. When you add a force from the sails, it lets you "lever" that wind force into force going the direction you want. In extreme "tacking" cases you are even sailing into the wind using its force to pull you towards it!

This means that naive designs won't work. A flat-bottomed boat will act more like a balloon than a ship, and will be tossed around by the wind in the direction the wind goes.

While keels can help, that keel needs to be reinforced against the forces and large enough to provide the force you need all alone; in effect you end up with a hull.

But not all is lost. Under newton's law, all forces generate equal and opposite reactions.

Make the ship long and narrow. Have it force a line into the water, actually shaping the water and forming an impression.

Going forward a relatively narrow section needs to be moved, while going sideways a much larger section of water.

So you'll want the ship to be shaped much like a hull; or, more accurately, so that the "virtual" hull around it caused by the repulsion field is shaped like a conventional hull.

Defence against waves acts similarly, except the shape we care about is the virtual hull instead of the hull. Note that "holes" in the virtual hull may be harder than holes in a hull if the effect radiates out in all directions.

If the "virtual" hull can be shaped without shaping the hull things can get interesting. Does the effect end when you leave line of sight? Then you could do interesting things with "covers" to turn off/on the effect.

Imagine a tube picking up water. It fills a chamber. Then the entry tube is closed, the repulsion field is "turned on", firing the water out the rear. Then it is turned off, feeding more water into the chamber.

What more, the "hull" of the ship need not be solid. It could be a net of treated material held rigid in a frame. The virtual hull would be far smoother than the net, and would have no gaps.

Rowing could be done out holes in this "hull", extending through the virtual hull unopposed, and pushing against the water. The surface area of rowing would be larger than on a conventional ship, who is mostly limited by the water-air barrier around the ship.

The oars themselves could be treated with this substance and never themselves get wet. If you could cover up the effect, "oars" that permit that could be used for propulsion, generating an effect similar to the "jet" above.

# Steam Propulsion

We normally think of steam as being an industrial age discovery. However the first "steam engine" was constructed in Roman times by Hero of Alexandra.

This should serve as the prototype of a simple steam based propulsion system, whereby a large steam boiler is vented directly into the water behind the vessel. Giving a good turn off speed, but limited range and manoeuvrability.

After a few centuries of trial, error and explosions, people will discover the...

# Valveless Pulse Jet

This is much lighter than the Steam "rocket" engine and much smaller and cheaper to produce. This allows it to be mounted similar to an outboard engine.

The disadvantages of it include loud noise (all users of this tech would be deaf, 140dB range), being prone to exploding and an extremely bumpy ride.

https://en.m.wikipedia.org/wiki/Pop_pop_boat

• Venting steam into the water might be risky around "water manipulating prey" (even more so for the pop pop boat design, which also draws water into it's boiler as part of the cycle). This leaves the ship with an "Achilles heel": direct access into its steam engine boiler from the water... – G0BLiN Mar 6 at 9:37

If this water propulsion is absolute (the substance cannot be made to come into contact with water at all), then wide craft would be the best bet to prevent tipping. If this substance is expensive, you can make a catamaran or trimaran hull.

For propulsion, you could use sail with keel and rudders in the water. Steering will still be a bit dicey (think putting a sail in a Frisbee (round plastic toy for those who aren't old enough to have seen one) on a flat icy lake. You'll spend almost as much time going sideways as going forward.

Another option is to dip a non-repulsing scoop into the water and have it fall into a tube with a sloghtly angled piece of the substance capping off the front end. that will drive the water out the back end of the tube. Add tubes until you have the propulsion you need. Note that you need to have some tubs of water on deck to start the ship moving. If you go this route, have the tubs created by some guy named Bussard.

I think that if your material is not a coating, but a kind of magnet (so that a ship is kept in suspension by an array of at least 4 of them), the hull of the ships should be studied much more like a building rather than a ship (think that it would have pillars where the magnets are located), and anyway it should be much lighter that a ship.
This is because a surface floating on the water could support much more weight than one suspended in the air (think a large raft with a heavy weight in the center: would it be more likely to break if it is floating on water or if it is kept raised by four pillars at its vertices?). So, it would be built so that all the weight is somehow distributed onto the magnets, because if the weight is loaded among them, they would risk to break down.
This would probably limit the size and total weight of the ship, even in consideration that the more massive the ship, the more difficult it will be to manoeuver it (because of the lack of lateral friction with water, you will need more force to counteract its inertia).

About the propulsion, as other answers point, in some conditions the magnets could be used to move and steer the ship. If it is possible to shield the water-repulsive effect, I think that placing a shielding on the sides of some of the magnets, and orienting them in diagonal, the resulting force would have components both normal and tangential with respect to the surface of water, allowing to move and not only to float.
Of course, I don't know how it could be explained in terms of energy conservation (maybe the magnets discharge with the use?)

I do have an idea, however it does not exactly fit the description of pre-inductrial tech, or at least I do not think it does.

The ship could be able to instead of using a sail using a wind turbine to harness the power of the wind at a much higher efficiency to work on a geared shaft that turns a propellor. The turbine itself would be a vertical axis wind turbine similar to this:

Now what this would do is create a vessel capable of sailing in any direction, because it can capture the wind from all angles. Simply because the blades face in all directions. It would also allow you to sail directly into the wind and in fact it would be strongest when sailing up or downwind because of it's omnidirectional (on a 2D plane) nature. It also would be able to better harness gusty winds than a horizontal bladed turbine. It has an advantage over saild in the fact that it isn't adversely affected by storm conditions and doesnt get damaged as easily since it's not a fabric that can tear, its most likely metal or wood.

You could use it to power some sort of catamaran for maximum speed and stability. And as the technology of your society advances you can create some forms of electric storage.


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# Quaternion Camera Implementation

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5 replies to this topic

### #1Coldon  Members

Posted 16 October 2007 - 07:55 PM

I'm trying to get a quaternion camera working but i'm struggling a bit, strangely enough not with the maths behind the quaternion but rather how to implement it in opengl. The tutorials i've found all use different methods. Some directly modify the modelview matrix while others make using of transformations via glRotate/glTranslate's and finally some use the gluLookAt() function (i know this is a shortcut to the transforms). My quaternion maths is right and my class can produce anything needed for any of the above methods. can you give me a brief overview of where i need to start and how to implement the camera? like should i use heading and pitch variables (kinda like euler angles), should i use a heading vector and work from that? Whats the difference between these methods? Is there a prefered solution? here are my vector, quaternion and camera classes, i've implemented a basic rotate method (from the NEHE tutorial) which doesnt work as intended, it seems to rotate spirally ?!? Point and Vector Classes :
class point3f
{
public:

//members
//------------------------------------------------------------------------------

float x, y, z;

//constructors
//------------------------------------------------------------------------------

point3f() : x(0),y(0),z(0) {}
point3f(float ix, float iy, float iz) : x(ix), y(iy), z(iz) {}

//methods
//------------------------------------------------------------------------------

point3f operator +(point3f p)
{
return point3f(x + p.x, y + p.y, z + p.z);
}

//subtraction
point3f operator -(point3f p)
{
return point3f(x -p.x, y - p.y, z - p.z);
}

//convert to array
void toArray(float* a)
{
a[0] = x;
a[1] = y;
a[2] = z;
}
};

//vector class
class vector3f
{

public:

//members
//------------------------------------------------------------------------------

float x, y, z;

//constructors
//------------------------------------------------------------------------------

//default constructor
vector3f() : x(0),y(0),z(0) {}

//create vector from point
vector3f(point3f p) : x(p.x), y(p.y), z(p.z) {}

vector3f operator =(point3f p)
{
return vector3f(p);
}

//create vector with co-ordinates
vector3f(float ix, float iy, float iz) : x(ix), y(iy), z(iz) {}

//methods
//------------------------------------------------------------------------------

//cross product
vector3f crossProduct(const vector3f &v)
{
vector3f cp;

//cross product
cp.x =	y * v.z - z * v.y;
cp.y =	-( x * v.z - z * v.x );
cp.z =	x * v.y - y * v.x;

return cp;
}

//normalize
void normalize()
{
float sumSquares = x*x + y*y + z*z;

//dont normalise if vector is close enough
if ( abs(sumSquares - 1.0) > VECTOR_TOLERANCE )
{
float L = sqrt(sumSquares);

x /= L;
y /= L;
z /= L;
}
}

//return normalize copy of current vector
vector3f getNormalized()
{
//copy current quarternion and return conjugate
vector3f v(*this);
v.normalize();

return v;
}

//dot product
float dotProduct(const vector3f &v)
{
return x * v.x + y * v.y + z * v.z;
}

//shortcut for cross product
vector3f operator *(vector3f &v)
{
return crossProduct(v);
}

vector3f operator +(vector3f &v)
{
return vector3f(x + v.x, y + v.y, z + v.z);
}

//subtraction
vector3f operator -(vector3f &v)
{
return vector3f(x - v.x, y - v.y, z - v.z);
}
};

Quaternion Class:
class quaternion
{
public:

//members
//------------------------------------------------------------------------------

float x,y,z,w;

public:

//constructors
//------------------------------------------------------------------------------

//empty constructor
quaternion() : x(0), y(0), z(0), w(0) {}

//specific constructor
quaternion(float xi, float yi, float zi, float wi) : x(xi), y(yi), z(zi), w(wi) {}

//from vector
quaternion(vector3f v) : x(v.x), y(v.y), z(v.z), w(0) {}

//create from an angle and an arbitrary axis
quaternion(float degrees, vector3f axis)
{
float theta = (degrees / 180.0f) * PI;
float sinTheta = sin( theta / 2.0f );
float cosTheta = sin( theta / 2.0f );

//calculate quaternion values
x = axis.x * sinTheta;
y = axis.y * sinTheta;
z = axis.z * sinTheta;
w = cosTheta;
}

//methods
//------------------------------------------------------------------------------

//define quaternion multiplication operator
quaternion operator *(const quaternion &b)
{
quaternion result;

result.x = w * b.x + x * b.w + y * b.z - z * b.y;
result.y = w * b.y - x * b.z + y * b.w + z * b.x;
result.z = w * b.z + x * b.y - y * b.x + z * b.w;
result.w = w * b.w - x * b.x - y * b.y - z * b.z;

//return new quaterion
return(result);
}

//define vector multiplication operator
vector3f operator *(const vector3f &vi)
{
//create quaternion from normalized vector
vector3f vn(vi);
vn.normalize();
quaternion v(vn), result;

//apply multiplications
result = v * getConjugate();
result = *this * result;

return vector3f(result.x, result.y, result.z);
}

//normalise quaterion
void normalize()
{
float sumSquares = x * x + y * y + z * z + w * w;

if ( abs(sumSquares - 1.0) > QUATERNION_TOLERANCE )
{
float L = sqrt(sumSquares);

x /= L;
y /= L;
z /= L;
w /= L;
}
}

//return normalized copy of current quaternion
quaternion getNormalized()
{
//copy current quarternion and return conjugate
quaternion q(*this);
q.normalize();

return q;
}

//conjugate quaterion
void conjugate()
{
x = -x;
y = -y;
z = -z;
}

//return conjugate copy of current quaternion
quaternion getConjugate()
{
//copy current quarternion and return conjugate
quaternion q(*this);
q.conjugate();

return q;
}

//to vector
vector3f toVector()
{
return vector3f(x, y, z);
}

//to matrix (homogenous 4x4 matrix)
void toMatrix(float* matrix)
{
// First row
matrix[0]	= 1.0f - 2.0f * ( y * y + z * z );
matrix[1]	= 2.0f * (x * y + z * w);
matrix[2]	= 2.0f * (x * z - y * w);
matrix[3]	= 0.0f;

// Second row
matrix[4]	= 2.0f * ( x * y - z * w );
matrix[5]	= 1.0f - 2.0f * ( x * x + z * z );
matrix[6]	= 2.0f * (z * y + x * w );
matrix[7]	= 0.0f;

// Third row
matrix[8]	= 2.0f * ( x * z + y * w );
matrix[9]	= 2.0f * ( y * z - x * w );
matrix[10]	= 1.0f - 2.0f * ( x * x + y * y );
matrix[11]	= 0.0f;

// Fourth row
matrix[12]	= 0;
matrix[13]	= 0;
matrix[14]	= 0;
matrix[15]	= 1.0f;
}

};

Camera Class:
class camera
{
public:

//constructors
//------------------------------------------------------------------------------

camera(vector3f p, vector3f v, vector3f u) : position(p), view(v), up(u) {}

//methods
//------------------------------------------------------------------------------

void rotate(float h, float p)
{
GLfloat m[16];

// Make the Quaternions that will represent our rotations
quaternion qh(h,vector3f(1,0,0));
quaternion qp(p,vector3f(0,1,0));

// Combine the pitch and heading rotations and store the results in q
quaternion result = qp * qh;
result.toMatrix(m);

// Let OpenGL set our new prespective on the world!
glMultMatrixf(m);
}

void changePosition()
{

}
};

[Edited by - Coldon on October 17, 2007 3:32:14 AM]

### #2nicmenz  Members

Posted 17 October 2007 - 03:35 AM

use Cos if you write Cos ;)

float cosTheta = sin( theta / 2.0f ); // SHOULD BE COS

I don't think that's your only problem here :/ your methods look well implemented, though. could you describe the artifacts?

### #3nicmenz  Members

Posted 17 October 2007 - 03:42 AM

again it's me.

I never implemented a quat-based camera, but like the trackball it can be imagined as the task to rotate one starting point P to a target point P' on the unit sphere. to compute the quaternion you are looking for, you have to
a) compute the rotation axis (P.cross(P'))
b) the rotation angle (acos(P.dot(P'))

you can then use your constructor "quaternion(float degrees, vector3f axis)" to build the rotation Q that moves P into P'. last step should be to convert Q into a 4x4 matrix (I use glLoadMatrix()).

I don't know if it's necessary for a camera-implementation, but for the trackball you have to store the last computed quaternion as starting point for the next rotation...but not sure here.

greetx,
me.

### #4Coldon  Members

Posted 17 October 2007 - 07:14 AM

hey, lol - that was a dumb mistake with the cos function, dangers of cutting and pasting. I'll try what you suggest and get back to you tomorrow.

### #5dpoon  Members

Posted 17 October 2007 - 12:34 PM

I have a quaternion camera implementation in opengl and direct3d over on my site that might be of help to you.

### #6Coldon  Members

Posted 20 October 2007 - 10:33 PM

@dpoon:

your camera examples are excellent! I really like your coding style as its really similar to my own and its neat clear and concise!

That stupid typo was causing the problems, once i fixed it the quaternion rotations work as intended, i'm busy getting movement working but its proving to be a little tricky but i'll get it (eventually)...

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text:
Question

# Which compound is used in making LED?

Open in App
Solution

## L.E.D:L.E.D stands for the light emitting diode.It emits light when the current flows through a semiconductor.Compounds/Materials in L.E.D:Some metals that are used in L.E.D are arsenic, indium as well as gallium.The compound that is used for making LED is Gallium arsenide phosphide $\left({\mathrm{GaAs}}_{1-\mathrm{x}}{\mathrm{P}}_{\mathrm{x}}\right)$.Gallium arsenide phosphide: It is an alloy of gallium arsenide and gallium phosphide.Gallium arsenide phosphide is used for manufacturing different colored LEDs such as orange, red, or yellow.

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text:
# Prove or disprove integration proof

• May 12th 2009, 01:31 AM
nankor
Prove or disprove integration proof
May I plz get some assistance on the following question;

If f and g are integrable on a closed set and f(x)<=h(x)<=g(x) for all x in in that set then h (a function) is integrable on that set.
• May 12th 2009, 04:09 AM
Plato
Quote:

Originally Posted by nankor
If f and g are integrable on a closed set and f(x)<=h(x)<=g(x) for all x in in that set then h (a function) is integrable on that set.

What do you mean by "f and g are integrable"?
• May 12th 2009, 06:03 AM
putnam120
The statement is false.

Take $f(x)=0$ and $g(x)=1$ on the interval $[0,1]$ and $h(x)=1$ if $x\in\mathbb{Q}$ and 0 other wise. This shows that the statement is false for the Riemann integral.

For Lebesgue integrable take $h(x)=1$ on a set $\mu\subset[0,1]$ and $\mu$ is not Lebesgue measurable, and 0 else where. Then $h(x)$ is not Lebesgue integrable.


text:
# Repeated references with multibib

I am trying to cite multiple references, sorted by reference type (journal vs. conference). Some of the journal references cite the corresponding conference paper where the results originally appeared.

However, when I cite the conference paper in the journal reference, the conference paper bibliographic details re-appear in the journal bibliography, like so:

[J1] Blah blah. Journal paper. Extended version of [C1].

[J2] Blah blah. Conference paper.

[C1] Blah blah. Conference paper.

I only want the citation (eg. [C1]) and not the repeated reference (eg. [J2]).

Below is a MWE of my document code.

.tex file

% !TEX TS-program = pdflatexmk
\documentclass{article}

\usepackage[resetlabels, labeled]{multibib}

\newcites{%
J,%
C%
}%
{%
Journal Articles,%
Conference Articles%
}

\begin{document}

\nociteJ{DoeKnuthJournal}
\nociteC{DoeKnuthConference}

\bibliographystyleJ{unsrt}
\bibliographystyleC{unsrt}

\bibliographyJ{./Publications}
\bibliographyC{./Publications}

\end{document}

.bib file

@article{
DoeKnuthJournal,
author  =   "Doe, John and Knuth, Donald",
title   =   "A really good paper",
journal =   "Journal of Good Papers",
note    =   "Extended version of GP 2000 article \cite{DoeKnuthConference}",
}

@inproceedings{
DoeKnuthConference,
author  =   "Doe, John and Knuth, Donald",
title   =   "A really good paper",
booktitle = "Proceedings of Good Papers (GP 2000)",
}

Try splitting the bib file into two files, one for journals and one for conferences:

.tex file:

\documentclass{article}

\usepackage[resetlabels, labeled]{multibib}

\newcites{%
J,%
C%
}%
{%
Journal Articles,%
Conference Articles%
}

\begin{document}

\nociteJ{DoeKnuthJournal}
\nociteC{DoeKnuthConference}

\bibliographystyleJ{unsrt}
\bibliographystyleC{unsrt}

\bibliographyJ{PublicationsJ}
\bibliographyC{PublicationsC}

\end{document}

PublicationsJ.bib:

@article{
DoeKnuthJournal,
author  =   "Doe, John and Knuth, Donald",
title   =   "A really good paper",
journal =   "Journal of Good Papers",
note    =   "Extended version of GP 2000 article \cite{DoeKnuthConference}",
}

PublicationsC.bib:

@inproceedings{
DoeKnuthConference,
author  =   "Doe, John and Knuth, Donald",
title   =   "A really good paper",
booktitle = "Proceedings of Good Papers (GP 2000)",
}

This is somewhat unsatisfying, because it would be nice if you can leave all the references in the same bib file, but this seems to be a known limitation of multibib; see Section 1.3 of the documentation.


text:
# First difference or forward demeaning: Implications for the method of moments estimators

Econometric Reviews, 2017, vol. 36, issue 6-9, 883-897

Abstract: In this paper, we consider the method of moment estimation for dynamic panel models based on either forward demeaning (FOD) or first difference (FD) transformations to eliminate the individual-specific effects, using either all lags or one lag as instruments. We show that the Arellano–Bond-type generalized method of moment (GMM) based on FD is asymptotically biased of order $$$$\sqrt{c}$$$$c using all lags or one lag as instruments where $$c={{T} \over {N}}\lte{}\infty$$c=TN<∞ as N,T→∞. For GMM based on FOD, it is asymptotically biased of order $$$$\sqrt{c}$$$$c when using all lags, but it is asymptotically unbiased when using only fixed number of lags as instruments. We also discuss these findings in light of the simple IV estimator. Monte Carlo simulations confirm our findings in this paper.

Date: 2017
References: Add references at CitEc
Citations: View citations in EconPapers (5) Track citations by RSS feed

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http://hdl.handle.net/10.1080/07474938.2017.1307594 (text/html)
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Handle: RePEc:taf:emetrv:v:36:y:2017:i:6-9:p:883-897


text:
# Can a functional language be homoiconic?

In a homoiconic language the primary representation of programs is also a data structure in a primitive type of the language itself.

I was wondering if the lambda calculus was homoiconic (I assume not, because applications are not data (which, for the calculus, would mean variable or abstraction?)), and if not, what it would take to make it homoiconic (wrap all applications in an abstraction with a useless variable, a la \a.b c?).

But the more important question is this: If the only primitive data type in the language is the function, is it possible to achieve homoiconicity? (Does homoiconicity even make sense if all you have are functions?) If not, then is there some formal definition of what data-properties must be present in order to achieve homoiconicity?

The definition implies not only that programs are represented by a "primitive" datatype, but presumably also that it is possible to inspect the elements of this type, i.e., one can actually get at the code (otherwise I can just declare a primitive abstract type code and never let you look at it).

In general, the construct you are looking for is known as quote. It takes a piece of code (in $\lambda$-calculus or anywhere else) and returns its representation, either as an abstract syntax tree, a string, or just an integer (because integers can encode syntax trees). Of course, in the case of $\lambda$-calculus such trees, strings or numbers would be encoded in the usual way that we encode datastructures in $\lambda$-calculus.

It is known that having quote in general increases the expressive power of the language. For example, in $\lambda$-calculus we cannot implement the statement "all functions are continuous" but we can if we also have quote at our disposal (this is the essence of Kreisel-Lacombe-Shoenfield-Tseitin theorem).

The opposite of quote is eval, which many languages have. However, eval does not add to the expressive power of a language because it can be implemented within the language, assuming the language is general enough (eval is essentially an interpreter for the language).

It is worth mentioning that quote is an abstract way of making a programming language be a lot more like the Turing machine model (TM) of computation. In the TM model a piece of code is a Turing machine, and it is represented on the tape as a sequence of symbols, which is the only datatype that Turing machines know about.

Note: it is common knowledge that "all models of computation are equivalent". What this means is that they compute the same class of functions from numbers to numbers $\mathtt{nat} \to \mathtt{nat}$, but they are not equivalent when it comes to higher-order functions such as $(\mathtt{nat} \to \mathtt{nat}) \to \mathtt{nat}$.

• Say LC had a quote method... What would it return? We can construct sequences and use K and I to access its members... If I ran quote (\a.a b) would it return the sequence \a.a, b? What if I ran quote \a.a? Would it return the sequence a, a? How can a functional language "inspect" functions? (I'm an idiot, thank you for humoring me.) Oct 17 '15 at 8:36
• And regarding the last paragraph, I thought LC and TM were computationally equivalent... How can TM be capable of doing something LC can't? Oct 17 '15 at 8:38
• Marking as accepted, because it explains what I was looking for, but I'm afraid I still don't quite understand. LC doesn't have quote... but it can create data structures that can represent programs, and it can implement its own eval and pass those data structures to it, so what expressive power does quote add? I can see finding myself in a situation where I wish I could quote an expression, but can't I fix that by rewriting my code so that I'm dealing with those syntax trees and passing them to eval instead writing the code directly? It sounds like quote is just convenience. Oct 20 '15 at 0:58
• Nope, it is not just convenience. There is no $\lambda$-term $q$ such that, for every $\lambda$-term $t$ we would have that $q \, t$ is the representation of the syntax of $t$. Rewriting code does not count because you are not a $\lambda$-term. Oct 20 '15 at 21:25


text:
# The Roots of a Polynomial

We can learn a lot about the roots, or the zeros, of a polynomial without knowing what the roots are. The coefficients of a polynomial tell us a lot about the roots. We will define the polynomial that we are speaking of before we begin. The polynomials we will be talking about are ones that have real coefficients and powers that are integers. Let $P(x)={{a}_{n}}{{x}^{n}}+{{a}_{n-1}}{{x}^{n-1}}+{{a}_{n-2}}{{x}^{n-2}}+{{a}_{n-3}}{{x}^{n-3}}+\text{ }...\text{ }+\text{ }{{a}_{0}}$. Then, an, an–1, an–2, an–3, ..., a0, are all real numbers.

1. A polynomial of degree n has n number or roots or less. The roots can be real and/or non-real. For example, a polynomial of degree 5 has at most 5 roots. How can a polynomial of degree n have less than n number of roots? Consider the polynomial $P(x)={{x}^{3}}+3{{x}^{2}}-9x+5$. This polynomial has only 2 real roots, 1 and -5, and no non-real roots. When we factor the polynomial, we see why: $P(x)=(x-1)(x-1)(x+5)$. The root 1 occurs twice.
2. Non-real roots of a polynomial are always conjugate pairs. If y + zi is a root, then If y - zi is also a root.
3. Polynomials that have an odd degree have at least 1 real root. This can easily be observed since the domain of odd functions is (–∞, +∞) and 0 falls within the domain. The function crosses the x-axis at least once, which is the root of the polynomial.

## Sums of Roots

We now focus on the sums of the roots and how they are related to the coefficients. Let the polynomial $P(x)={{a}_{n}}{{x}^{n}}+{{a}_{n-1}}{{x}^{n-1}}+{{a}_{n-2}}{{x}^{n-2}}+{{a}_{n-3}}{{x}^{n-3}}+\text{ }...\text{ }+\text{ }{{a}_{0}}$ be a polynomial with roots r1, r2, r3, ..., rn. Then it is possible to find $r_{1}^{k}+r_{2}^{k}+r_{3}^{k}+r_{4}^{k}+\text{ }...\text{ }+\text{ }r_{n}^{k}$, for any integer k, in terms of the coefficients of the polynomial, an, an–1, an–2, an–3, ..., a0. We are going to let an = 1 to make calculations easier. If it is not, we can always divide the whole polynomial with an to make it 1.

If we work backward to find the polynomial coefficients, we will be able to find the coefficients in terms of the roots.

(i) $P(x)=(x-{{r}_{1}})(x-{{r}_{2}})(x-{{r}_{3}})(x-{{r}_{4}})\times \ldots\times (x-{{r}_{n}})$

(ii) )$P(x)={{x}^{n}}-({{r}_{1}}+{{r}_{2}}+{{r}_{3}}+{{r}_{4}}++{{r}_{n}}){{x}^{n-1}}+({{r}_{1}}{{r}_{2}}+{{r}_{1}}{{r}_{3}}+{{r}_{1}}{{r}_{4}}++{{r}_{2}}{{r}_{3}}+{{r}_{2}}{{r}_{4}}++{{r}_{n-1}}{{r}_{n}}){{x}^{n-2}}+({{r}_{1}}{{r}_{2}}{{r}_{3}}+{{r}_{1}}{{r}_{2}}{{r}_{4}}+{{r}_{1}}{{r}_{2}}{{r}_{5}}++{{r}_{2}}{{r}_{3}}{{r}_{4}}+{{r}_{2}}{{r}_{3}}{{r}_{5}}++{{r}_{n-2}}{{r}_{n-1}}{{r}_{n}}){{x}^{n-1}}+...\pm{{r}_{1}}{{r}_{2}}{{r}_{3}}\times\ldots\times{{r}_{n}}$

The coefficients are:

• $a_{n}=1$
• ${{a}_{n-1}}=-({{r}_{\text{1}}}+{{r}_{\text{2}}}+{{r}_{\text{3}}}+{{r}_{\text{4}}}+\text{ }\ldots \text{ }+{{r}_{n}})$ = sum of product of roots taken 1 at a time, or simply the sum
• ${{a}_{n-2}}=\text{ }({{r}_{\text{1}}}{{r}_{\text{2}}}+{{r}_{\text{1}}}{{r}_{\text{3}}}+{{r}_{\text{1}}}{{r}_{\text{4}}}+\text{ }\ldots {{r}_{\text{1}}}{{r}_{n}})\text{ }+\text{ }({{r}_{\text{2}}}{{r}_{\text{3}}}+{{r}_{\text{2}}}{{r}_{\text{4}}}+{{r}_{\text{2}}}{{r}_{\text{5}}}+\text{ }\ldots \text{ }+{{r}_{\text{2}}}{{r}_{n}})\text{ }+\text{ }({{r}_{\text{3}}}{{r}_{\text{4}}}+{{r}_{\text{3}}}{{r}_{\text{5}}}+{{r}_{\text{3}}}{{r}_{\text{6}}}+\text{ }\ldots .{{r}_{\text{3}}}{{r}_{n}})\text{ }+\text{ }\ldots$ = sum of product of roots taken two at a time
• ${{a}_{n-3}}=-[({{r}_{\text{1}}}{{r}_{\text{2}}}{{r}_{\text{3}}}+{{r}_{\text{1}}}{{r}_{\text{3}}}+{{r}_{\text{1}}}{{r}_{\text{4}}}+\text{ }\ldots {{r}_{\text{1}}}{{r}_{n}})\text{ }+\text{ }({{r}_{\text{2}}}{{r}_{\text{3}}}+{{r}_{\text{2}}}{{r}_{\text{4}}}+{{r}_{\text{2}}}{{r}_{\text{5}}}+\text{ }\ldots \text{ }+{{r}_{\text{2}}}{{r}_{n}})\text{ }+\text{ }({{r}_{\text{3}}}{{r}_{\text{4}}}+{{r}_{\text{3}}}{{r}_{\text{5}}}+{{r}_{\text{3}}}{{r}_{\text{6}}}+\text{ }\ldots .{{r}_{\text{3}}}{{r}_{n}})\text{ }+\text{ }\ldots]$ = sum of product of roots taken 2 at a time
• ${{a}_{n-4}}=({{r}_{1}}{{r}_{2}}{{r}_{3}}+{{r}_{1}}{{r}_{2}}{{r}_{4}}+...+{{r}_{1}}{{r}_{2}}{{r}_{n}})+({{r}_{1}}{{r}_{3}}{{r}_{4}}+{{r}_{1}}{{r}_{3}}{{r}_{5}}+...+{{r}_{1}}{{r}_{3}}{{r}_{6}})+...+({{r}_{2}}{{r}_{3}}{{r}_{4}}+{{r}_{2}}{{r}_{3}}{{r}_{5}}+...+{{r}_{2}}{{r}_{3}}{{r}_{n}})+...$ = sum of product of roots taken 3 at a time
• ....
• ${{a}_{0}}=\pm ({{r}_{\text{1}}}\times {{r}_{\text{1}}}\times {{r}_{\text{1}}}\times {{r}_{\text{1}}}{{\times }_{\ldots }}\times {{r}_{n}})$ = product of the roots. a0 is the constant.

The sign of the last term, the constant, depends on whether the polynomial is odd or even. Because the signs alternate, if the polynomial is even, then the product will have a positive sign. If it's odd, then the sign is negative.

For a polynomial of any degree, the sum of the roots is given by: $\underset{i}{\mathop{\sum }}\,{{r}_{i}}=-\frac{{{a}_{n-1}}}{{{a}_{n}}}$.

### Example 1

Let’s try it with a real polynomial. Let $P(x)=(x-1)(x+2)(x-3)$, which has roots 1, –2, and 3. Multiplying out, we have: $P(x)=({{x}^{2}}+x-2)(x-3)={{x}^{3}}-2{{x}^{2}}-5x+6$. So, a3 = 1, a2 = –2, a1 = –5, and a0 = 6.

The sum of the roots is
1 – 2 + 3 = 2 = –a2.
The sum of the products of the roots taken 2 at a time is
(1 × –2) + (1 × 3) + (–2 × 3) = –2 + 3 – 6 = –5 = a1.
The product is 1 × –2 × 3 = –6 = –a0.
Remember the polynomial is odd and the sign of the last term is negative. We take the negative of a0 for the product.

The math checks correctly.

### Example 2

The following example involves complex roots and real roots. Let $P(x)={{x}^{4}}-5{{x}^{3}}+11{{x}^{2}}+11x-78$. There’s really no “sure-thing” method of solving for the roots for higher degrees. We can use Newton’s method as the next best thing. However, just for illustration, I derived this polynomial working forward from $P(x)=(x-2+3\mathsf{i})(x-2+3\mathsf{i})(x+2)(x-3)$. The roots are 2 + 3i, 2 – 3i, –2, and 3.

The sum of the roots is
2 + 3i + 2 – 3i – 2 + 3 = 5.
Using the formula, the sum is –a3 = 5.
The product of the roots is
(2 + 3i) × (2 – 3i) × –2 × 3 = (4 + 9)(–6) = –78, which equals a0.
Polynomial is even so we keep the sign as is.

For the sum of the roots, only an and an-1 are relevant. How do the other coefficients play a role? We will see their role in the sum of the square of the roots.

## Sum of Square of Roots

To find the sum of the squares, we first square the sum.

(iii) ${{({{r}_{\text{1}}}+{{r}_{\text{2}}}+{{r}_{\text{3}}}+{{r}_{\text{4}}}+\text{ }\ldots \text{ }+{{r}_{n}})}^{2}}=(r_{1}^{2}+r_{2}^{2}+r_{3}^{2}+r_{4}^{2}+\text{ }\ldots \text{ }+r_{n}^{2})+2({{r}_{\text{1}}}{{r}_{\text{2}}}+{{r}_{\text{1}}}{{r}_{\text{3}}}+{{r}_{\text{1}}}{{r}_{\text{4}}}+\text{ }\ldots {{r}_{\text{1}}}{{r}_{n}})$

(iv) ${{\left( \frac{{{a}_{n-1}}}{{{a}_{n}}} \right)}^{2}}=(r_{1}^{2}+r_{2}^{2}+r_{3}^{2}+r_{4}^{2}+\text{ }\ldots \text{ }+r_{n}^{2})+2\cdot \frac{{{a}_{n-2}}}{{{a}_{n}}}$

Solving for the squares, we have the sum of the squares of the roots:

The sum of the square of the roots is given by: $\underset{i}{\mathop{\sum }}\,r_{i}^{2}={{\left( \frac{{{a}_{n-1}}}{{{a}_{n}}} \right)}^{2}}-\frac{2{{a}_{n-2}}}{{{a}_{n}}}$.

### Example 3

Using the polynomial in Example 2$P(x)={{x}^{4}}-5{{x}^{3}}+11{{x}^{2}}+11x-78$, let’s find the sum of the squares manually and using the formula. The roots are 2 + 3i, 2 – 3i, –2, and 3.

Manually: (2 + 3i)2 + (2 – 3i)2 + (–2)2 + 32 =
4 + 12i – 9 + 4 – 12i – 9 + 4 + 9 = 3
Using the formula: $\frac{{{(-5)}^{2}}}{1^2} -2\left( \frac{11}{1} \right)=25-22=3$

The formula checks.

## Sum of Higher Powers

You can find the sum of higher powers of roots on your own. The formulas for the cubes and the fourth power are below. With each higher power, it gets more and more complex.

The sum of the cube of the roots is: $\underset{i}{\mathop{\sum }}\,r_{i}^{3}=-{{\left( \frac{{{a}_{n-1}}}{{{a}_{n}}} \right)}^{3}}+\frac{3{{a}_{n-1}}{{a}_{n-2}}}{a_{n}^{2}}-\frac{3{{a}_{n-3}}}{{{a}_{n}}}$.

### Example 4

We'll use the same polynomial again for the sum of the cubes of the roots manually and with the formula: $P(x)={{x}^{4}}-5{{x}^{3}}+11{{x}^{2}}+11x-78$. The roots are 2 + 3i, 2 – 3i, –2, and 3.

Manually: (2 + 3i)3 + (2 – 3i)3 + (–2)3 + 33 =
–46 + 9i – 46 – 9i – 8 + 27 = –73
Using the formula:
$-{{ \frac{(-5)}{1^3} }^{3}}+\frac{3\cdot -5\cdot 11}{1}-\frac{3\cdot 11}{1}=125-165-33=-73$

The formula checks.

The sum of the roots to the fourth power: $\underset{i}{\mathop{\sum }}\,r_{i}^{4}={{\left( \frac{{{a}_{n-1}}}{{{a}_{n}}} \right)}^{4}}-\frac{4a_{n-1}^{2}{{a}_{n-2}}}{a_{n}^{3}}+\frac{4{{a}_{n-1}}{{a}_{n-3}}+2a_{n-2}^{2}}{a_{n}^{2}}-\frac{4{{a}_{n-4}}}{{{a}_{n}}}$.

## Sum of Inverse of Roots

Now, what about the sum of the inverse of the roots? Here we use a clever tactic: divide the polynomial by the constant.

(v) $P(x)=\frac{1}{{{r}_{1}}{{r}_{2}}{{r}_{3}}\times ...\times {{r}_{n}}}{{x}^{n}}-\left( \frac{1}{{{r}_{1}}{{r}_{2}}}+\frac{1}{{{r}_{1}}{{r}_{2}}}+\frac{1}{{{r}_{1}}{{r}_{3}}}+\frac{1}{{{r}_{1}}{{r}_{4}}}+...+\frac{1}{{{r}_{n-1}}{{r}_{n}}} \right){{x}^{n-1}}+...\mp \left( \frac{1}{{{r}_{1}}}+\frac{1}{{{r}_{2}}}+\frac{1}{{{r}_{3}}}+\frac{1}{{{r}_{4}}}+...+\frac{1}{{{r}_{n}}} \right)x\pm 1$.

You can see that the coefficient of the x1 term is the sum of the reciprocal of the roots. By dividing the polynomial with the product, we have essentially converted the sum of the products to the sum of the reciprocal of the products. Hence,

The sum of the reciprocal of the roots is: $\sum\limits_{i}{\frac{1}{{{r}_{i}}}}=-\frac{{{a}_{1}}}{{{a}_{0}}}$.

Note that we don’t have to worry about the sign. It will always be negative because if the polynomial is even and the constant is positive, the x term preceding the constant will have a negative sign. If the polynomial is odd and the constant is negative, then the x term preceding the constant will have a positive sign. Hence, in both cases, we have a negative sign.

### Example 5

Let’s try it with an example. Let $P(x)=(x-2)(x-5)(x+1)(x+3)(x+2)={{x}^{5}}-{{x}^{4}}-21{{x}^{3}}-11{{x}^{2}}+68x+60$. The roots are 2, 5, –1, –3, and –2. The sum is 1. Since a4 = –1, this checks. The product of the roots is –60. Since P(x) is odd, we take the negative of the constant. Hence –a0 = –60. This also checks.

Now, let’s find the sum of the reciprocals manually and with the formula.

Manually:$\frac{1}{2}+\frac{1}{5}-1-\frac{1}{3}-\frac{1}{2}=\frac{15}{30}+\frac{6}{30}-\frac{30}{30}-\frac{10}{30}-\frac{15}{30}=-\frac{17}{15}$
Formula: $-\frac{{{a}_{1}}}{{{a}_{0}}}=-\frac{68}{60}=-\frac{17}{15}$

$\frac{{{\pi }^{2}}}{6}=\frac{1}{{{1}^{2}}}+\frac{1}{{{2}^{2}}}+\frac{1}{{{3}^{2}}}+\frac{1}{{{4}^{2}}}+\frac{1}{{{5}^{2}}}+\frac{1}{{{6}^{2}}}+\text{....}=\sum\limits_{n=1}^{\infty }{\frac{1}{{{n}^{2}}}}$.
He assumed that the formula for the sum of the reciprocal of the roots works for an infinite series as well. He manipulated the Taylor series of sine to arrive with a polynomial of the form $1+{{a}_{1}}x+{{a}_{2}}{{x}^{2}}+{{a}_{3}}{{x}^{3}}+{{a}_{4}}{{x}^{4}}+{{a}_{5}}{{x}^{5}}+...$. I extended Euler's method and found the convergence of series of the form: $\sum\limits_{n=1}^{\infty }{\frac{1}{{n^2}-{{z}^{2}}}}$ and $\sum\limits_{n=1}^{\infty }{\frac{1}{{n^2}+{{z}^{2}}}}$. You can read about this topic here: Euler's Series of π.


text:
# Challenge: Answer this gedanken (PFP - Perpendicularly Fired Photon)

I'm challenging anyone who can answer the following question objectively:

As usual, imagine a railway station and trains which are equipped with single photon sources (one each in the platform and inside the trains) and a narrow strip of detector which is positioned right above the photon sources. The photons are fired vertically upwards, perpendicular to the train’s direction of travel. In addition, let us assume that apart from the photo detectors on the train and the platform, the roof of the train is open, and that the ceiling of the station above the train also has photo detectors, corresponding to the position of the photo detectors on the platform. Now, the photon sources in the station and the trains are set to trigger at the same moment – i.e. when the train’s photon source passes in front of Alice. Now, Bob is sitting in front of the photon source in the train which is travelling at approximately half the speed of light, towards the station.

When the train in which Bob travels crosses Alice, both photon sources emit a photon (on platform and on the train). Hence, one photon is emitted towards the roof of the railway station by the photon source stationed in the platform; another is emitted inside the train above which is the photo detector on the train (moving w.r.t. Alice) and the photo detector on the station’s roof (stationary w.r.t. Alice) above the train. Which means, Alice will (? might) record two photons hitting the photo detectors on the ceiling of the station (the photon from the platform and the photon from the train). The photon emanating from inside the train should hit the station’s ceiling since the train’s detector would have moved away by the time the photon travels upwards. Now from Bob’s view point, the photon emitted inside the train will be detected by the sensor IN the train; the photon emitted by the emitter on the platform will miss the detector on top of it (since the station is moving relative to Bob’s train). Either way, SR would not work for both observers at the same time.

In case you are thinking that there is some trickery with which this question can be answered in a manner similar to the barn door paradox, let Charlie travel in the track parallel to Bob, in the opposite direction (both Bob and Charlie are equidistant from Alice’s view point and are approaching with the same speed). Now we will have three different observers with different predictions; but due to law of conservation of energy, the photons can be detected by only one of the sensors, which should settle the confusion (or if they do separate, we would be able to tell because of the lesser energy liberated by the photon). In general, the experiment suggests that depending on the relative speed of the train to the station, there are an infinite number of positions in which the photon can be found, which could either support the many worlds theory, or is at least at odds with the Special relativity.

I have been digging around this for almost 13 months; still no solution. Any ideas? Would theory of General relativity help?

Should I rephrase my question -or- add details?

-
I want to add that this seems like a trivial question - and probably it is, since this is similar to the Michelson-Morley experiment, only with a slight modification. But for the heck of it, I cannot figure what is wrong in my PFP gedanken! – Vignesh Mar 12 '14 at 19:17
The suspense is killing me - can some body tell me if I'm right or wrong - I REALLY need an answer... – Vignesh Mar 12 '14 at 20:19

can some body tell me if I'm right or wrong - I REALLY need an answer

Your premise that Alice and Bob observe both photon trajectories as vertical is false.

The locus of events that form the worldline of either photon are absolute but the coordinates Alice and Bob assign to these events are not.

If Alice observes the station emitted photon to have a constant $x$ coordinate $x = 0$ then, assuming the photon is emitted at $t = t' = 0$, the Lorentz transformation to Bob's coordinate is

$$x' = -\gamma vt$$

In other words, Bob does not observe the photon to have a vertical trajectory but, rather, to have a horizontal component too.

Similarly, Alice observes the train emitted photon to have x coordinate

$$x = \gamma vt$$

Are you not familiar with the well-known light clock thought experiment?

If the Galilean transformation were true, Alice would observe Bob's photon to have a vertical speed of $v_z = c$ and a horizontal speed of $v_x = v$ and, thus, the photon would have a total speed of $u = \sqrt{c^2 + v^2}$

According to the Lorentz transformation, Alice observes Bob's photon to have a total speed $u = c$ and a horizontal speed of $v_x = v$ thus, the vertical speed of the photon is reduced, according to Alice, to $v_z = \sqrt{c^2 - v^2}$.

-
I think I'm missing elementary things - I will read throug the answers thoroughly (I think all answers point in the same direction, that one of my assumptions is wrong - will correct myself). Thanks. – Vignesh Mar 14 '14 at 9:48
Many thanks for your detailed answer - this clears it! – Vignesh Mar 17 '14 at 18:09

No issue with SR and no many worlds - when Bob is firing his photon upwards towards the train's detector, Alice as well sees the photon of Bob moving to the train's detector, but for her the photon's direction is not upwards but diagonal. For her the photon is taking more time than it takes for Bob because it is going diagonally - this phenomenon is called time dilation.

-
Hmmm.. That doesn't answer my question - Why would stationary Alice see a diagonal path for the photon? – Vignesh Mar 12 '14 at 21:06
Why would stationary Alice see a diagonal path for the photon? For the exact same reason Bob observes a diagonal path for the station emitted photon. This is elementary. The events that constitute the worldline of the photon are absolute - the coordinates Alice and Bob assign to those events are not. If Alice observes that the station emitted photon has constant x coordinate of zero, then by the Lorentz transformation, Bob observes the photon to have non-constant x' coordinate: $x' = -\gamma vt$ – Alfred Centauri Mar 12 '14 at 22:35
Sorry, did not get it the first time around. I stand corected. – Vignesh Mar 17 '14 at 18:03

The point you are confused about is the very fundamental point that the laws of physics are the same in all inertial frames. Let's look at what Bob sees. When the photon is emitted, bob sees the photon go straight up and hit the detector on the train, becuase photons move in straight lines in inertial frames, and the train is an intertial frame. Since the photon is going straight up in Bob's frame, and Bob's frame is moving with respect to Alice's frame, Alice sees the photon going up diagonally. Since the photon was emitted directly underneath the detector on the station, and the photon travels diagonally, Alice will not observe the photon from the train hitting the detector in the station. Thus both people will observe the station photon hitting the station detector, and the train photon hitting the train detector.

Notice that the only thing used here was the laws of physics being the same in all reference frames. Thus we can do the same thought experiment in newtonian dynamics with a ball being thrown into the air (ignore gravity). Then you would have said that the ball thrown upward from inside the train "should hit the station’s ceiling since the train’s detector would have moved away by the time the ball travels upwards." But this is not true because we know that the train is an inertial frame and so in the frame of the train the ball must move straight up and so Alice will see the ball moving diagonally, so the ball will hit the train's detector and miss the station's detector.

-
Thanks for the answer, I think I got it this time. I have one more question regarding this - 'Do photons have a preferential frame?' - I thinks that is another question, so I will frame a new query... – Vignesh Mar 17 '14 at 18:06

When the photons are emmited, Alice will see that Bob's photon has a finite horizontal momentum component. That is, it's not travelling vertically towards the roof, but diagonally towards the place where Bob's sensor will be some time later (how much time depends on Bob's speed relative to Alice, which determines the horizontal component of Bob's photon momentum). Both Alice and Bob will agree that Bob detects his own photon. Think of it as if Bob threw a ball upwards inside the train (an ultrarelativistic, massless ball traveling at exactly the speed of light).

By the way, you may want to reformulate your question. Saying that the photon sources trigger "at the same moment" does not make sense: the same moment respect to which reference frame? Simultaneity notions change when you go from one frame to another in SR.

-
Got it! But instead of reformulating the silly question, I will ask another question which might be more relevant! Thanks for your answer. – Vignesh Mar 17 '14 at 18:08


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26.11.2019 | Focus | Ausgabe 4/2020 Open Access

# A new uncertain DEA model and application to scientific research personnel

Zeitschrift:
Soft Computing > Ausgabe 4/2020
Autoren:
Meilin Wen, Xue Yu, Fei Wang
Wichtige Hinweise
Communicated by Y. Ni.
The original article has been revised: Due to open choice cancellation.

## Publisher's Note

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## Abstract

Data envelopment analysis (DEA), which has been widely used since it was introduced by Charnes et al. (J Econom 30:91–107, 1985), is an effective method for evaluating the relative efficiency of decision-making units. DEA models require accurate input data and output data; however, if no sample is available to estimate accurate data, then uncertain DEA is introduced. This paper reports on several new studies on uncertain DEA using the Hurwicz criterion, which attempts to find the intermediate area between extremes. Some uncertain DEA models, as well as their crisp equivalent models, are presented. Then, the Hurwicz ranking method is proposed based on these models, which can give an evaluation to all the decision-making units. By varying the parameter $$\beta$$ in the Hurwicz criterion, which reflects the optimism of the decision maker, the new ranking method can exhibit various forms. Finally, an application to scientific personnel is provided to prove the advantage of the proposed method.
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