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SCIENCE CHINA Information Sciences, Volume 61, Issue 2: 022302(2018) https://doi.org/10.1007/s11432-016-9042-8

## Fractional full duplex cellular network: a stochastic geometry approach

Wenping BI1,2,3, Limin XIAO1,2,3, Xin SU1,2,3, Shidong ZHOU1,2,3,*
• AcceptedFeb 28, 2017
• PublishedJul 28, 2017
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### Abstract

In-band full-duplex (FD) communication has been considered as a promising technology to enhance the spectral efficiency (SE) for the next generation of wireless system. However, the severe interference in FD cellular network may largely degrade the system performance, especially for cell edge users. In this paper, multi-cell cellular network consisting of BSs with FD capability and legacy half duplex (HD) users is studied. In order to relieve the received interference and enhance the system SE, a new resource allocation strategy named fractional FD (FFD) is proposed and its main idea and procedure are described as follows.Firstly, the frequency and time resources are partitioned into FD resource blocks (RBs) and HD RBs based on the network topology. Then all the users are classified into two groups named cell center users (CCUs) and cell edge users (CEUs) based on their channel conditions. At last, in each cell, each FD RB is allocated to a pair of CCUs with one in uplink and the other in downlink transmission direction, while only one CEU in the uplink (or downlink) transmission direction is scheduled over each HD RB. Tractable results of both the coverage probabilities and the ergodic rates of FFD, FD and HD systems are derived using stochastic geometry method. Numerical results show that, FFD can significantly improve the coverage probability of all users especially for CEUs compared with FD cellular system, and higher system SE is obtained compared with FD and HD cellular network. With proper design of the classification criterion and under the simulation settings of this paper, the SE of FFD system outperforms FD and HD system by $1.25$ and $1.38$ times, respectively.

### Acknowledgment

This work was supported by National Basic Research Program of China (Grant No. 2012CB316002), National Natural Science Foundation of China (Grant No. 61631013), National High Technology Research and Development Program of China (863 Program) (Grant No. 2015AA01A706), National Natural Science Foundation of China (Grant No. 61321061), Tsinghua University Initiative Scientific Research Program (Grant No. 2015Z02-3), National S&T Major Project (Grant No. 2014ZX03001011), Key Project of International Science and Technology Innovation Cooperation Between the Government (Grant No. 2016YFE0122900), and Huawei Technologies.

### Supplement

Appendix

Proof of Lemma sect. 3.1

Starting with the definition of laplace transform, we can get \begin{aligned} {L_{{I_{{\rm h}q}}}}\left( s \right) \mathop = \limits^{\left( {\rm a} \right)} {{{\rm E}}_{{\Phi _{{\rm h}q}}}}\left\{ {\prod\limits_{z \in {\Phi _{{\rm h}q}}\backslash c_o} {{{{\rm E}}_{{h_{z,d_o}}}}\exp \left( { - s{h_{z,d_o}}D_{z,d_o}^{ - {\alpha _{\rm bu}}}} \right)} } \right\}\mathop = \limits^{\left( {\rm b} \right)} \exp \left\{ { - 2\pi {\lambda _{\rm s}}\int\nolimits_R^\infty {\left\{ {1 - {{{\rm E}}_{{\rm h}}}\left[ {{{\mathop{\rm e}\nolimits} ^{\left( { - s{h}v^{ - {\alpha _{\rm bu}}}} \right)}}} \right]} \right\}v{\rm d}v} } \right\}, \end{aligned} \tag{23} where $c_o$ and $d_o$ are $b_o$ ($u_o$) and $u_o$ ($b_o$) if $q=d(u)$, respectively. (a) follows the fact that $\Phi~_{{\rm~h}q}$ and $h_{z,d_o}$ are independent from each other. Therefore, the expectation order can be exchanged. (b) follows from the probability generating function of PPP. Carrying on the proof in (23), the followings can be obtained \begin{align} &\exp \left\{ { - 2\pi {\lambda _{\rm s}}\int\nolimits_R^\infty {\left\{ {1 - {{{\rm E}}_{\rm h}}\left[ {\exp \left( { - sh{v^{ - {\alpha _{\rm bu}}}}} \right)} \right]} \right\}v{\rm d}v} } \right\} \\ &\mathop = \limits^{\left( {\rm c} \right)} \exp \left\{ - \frac{2}{{{\alpha _{\rm bu}}}}\pi {\lambda _{\rm s}}{{{\rm E}}_{\rm h}}\left\{ - {{\left( {sh} \right)}^{\frac{2}{{{\alpha _{\rm bu}}}}}}\Gamma \left( { - \frac{2}{{{\alpha _{\rm bu}}}}} \right)- \frac{{{\alpha _{\rm bu}}}}{2}{R^2} + {{\left( {hs} \right)}^{\frac{2}{{{\alpha _{\rm bu}}}}}}\Gamma \left( { - \frac{2}{{{\alpha _{\rm bu}}}},hs{R^{ - {\alpha _{\rm bu}}}}} \right) \right\} \right\}, \tag{24} \end{align} where (c) is proofed in [17]. Based on the Rayleigh fading assumption, $h\sim~\exp(1)$, we can obtain the expectations as \begin{eqnarray}&\displaystyle{{{\rm E}}_{\rm h}}\left\{ {{{ h }^{\frac{2}{{{\alpha _{\rm bu}}}}}}} \right\} = \int\nolimits_0^\infty {{{ x }^{\frac{2}{{{\alpha _{\rm bu}}}}}}{{\rm e}^{ - x}} {\rm d}x} \mathop = \limits^{\left( {\rm d} \right)} \Gamma \left( {\frac{2}{{{\alpha _{\rm bu}}}} + 1} \right) , \tag{25} \\ &\displaystyle{ {\rm E}_{\rm h}}\left\{ {{{ h}^{\frac{2}{{{\alpha _{\rm bu}}}}}}\Gamma \left( { - \frac{2}{{{\alpha _{\rm bu}}}},hs{R^{ - {\alpha _{\rm bu}}}}} \right)} \right\}= \int\nolimits_0^\infty {{x^{\frac{2}{{{\alpha _{\rm bu}}}}}}\Gamma \left( { - \frac{2}{{{\alpha _{\rm bu}}}},xs{R^{ - {\alpha _{\rm bu}}}}} \right){{\rm e}^{ - x}}{\rm d}x}\mathop = \limits^{\left( {\rm e} \right)} J\left( {\frac{2}{{{\alpha _{\rm bu}}}} + 1,1, - \frac{2}{{{\alpha _{\rm bu}}}},s{R^{ - {\alpha _{\rm bu}}}}} \right), \tag{26} \end{eqnarray} where $J(\mu~,\beta~,\nu~,\gamma~)$ is defined in (11) and proofed in [17]. (d) and (e) can be obtained by Eqs. (3.478) and (3.381) 3), respectively. Substitute (25) and (26) into (24) and we can get the result of Lemma sect. 3.1.

Gradshteyn I, Ryzhik I. Table of Integrals, Series, and Products. Manhattan: Academic Press, 2014. 346–370.

Proof of Lemma 3.2

Starting with the definition of laplace transform, we can get \begin{align} &{L_{{I_{{\rm f}q}}}}\left( s \right) \mathop = \limits^{\left({\rm f}\right)}{{{\rm E}}_{D,h,{\Phi _{\rm fd}}}}\left\{ {\prod\limits_{z \in {\Phi _{{\rm fd}\backslash b_o}}} {\exp \left( { - s{P_{\rm fd}}{h_{z,d_o}}{{k}_{z,d_o}}D_{z,d_o}^{ - {\alpha _{\rm d}}}} \right)} } \right\}{{{\rm E}}_{D,h,{\Phi _{\rm fu}}}}\left\{ {\prod\limits_{x \in {\Phi _{\rm fu}}\backslash c_o} {\exp \left( { - s{P_{\rm fu}}{h_{x,d_o}}{{k}_{x,d_o}}D_{x,d_o}^{ - {\alpha _{\rm u}}}} \right)} } \right\} \\ & \mathop = \limits^{\left({\rm g}\right)} \exp \left\{ { - 2\pi {\lambda _{\rm s}}{{\rm E}_{\rm h}}\int\nolimits_{R_{\rm d}}^\infty {\left\{ {1 - \left[ {\exp \left( { - {s_{\rm d}}h{v^{ - {\alpha _{\rm d}}}}} \right)} \right]} \right\}v{\rm d}v} } \right\} \exp \left\{ { - 2\pi {\lambda _{\rm s}}{{\rm E}_{\rm h}}\int\nolimits_{R_{\rm u}} ^\infty {\left\{ {1 - \left[ {\exp \left( { - {s_{\rm u}}h{v^{ - {\alpha _{\rm u}}}}} \right)} \right]} \right\}v{\rm d}v} } \right\}, \tag{27} \end{align} where $c_o$ and $d_o$ are $\emptyset$ ($u_o$) and $u_o$ ($b_o$) if $q=d({\rm~u})$, respectively. (f) follows the fact that $\Phi~_{\rm~fu}$ and $\Phi~_{\rm~fd}$ are assumed to be independent. In Step (g), when we focus on the downlink, the inter-user interference distance can be arbitrarily close, therefore the integration lower bound $R_{\rm~u}$ is $\eta~_{\rm~uu}R$ when taking the interference cancelation strategy into consideration. Based on the association principle, the interfering distance between focused downlink user and neighbor BSs is at least $R$, so $R_{\rm~d}~=~R$. While when considering uplink transmission, the inter-BS interference distance between focused BS and the neighbor BSs can also be arbitrarily close, as a consequence $R_{\rm~d}$ is $\eta~_{\rm~bb}R$ when SIC is applied. In the same way, the distance from the neighbor uplink users to the focused BS is at least $R$, which gives $R_{\rm~u}=~R$. Then following the proof in sect. 1, we can get the result in (13).

Proof of Lemma 3.3

Before solving joint LT of $I_{\rm~d}$ and $\hat~I_{\rm~d}$, some properties of the sum of two exponential distribution random variables are analysed.

We define $G$ as the sum of two exponential distribution random variables as follows: $$G = {s_2}{h_1} + {s_2}{h_2}, \tag{28}$$ where both $h_1$ and $h_2$ follow exponential distribution, i.e, $h_1~~\text{and}~~h_2\sim~{\rm~exp}(1)$. Then, when $s_1=s_2=s$, $G$ is the sum of two exponential random variables with same rate. As presented in [17], $G$ follows Erlang distribution. While, if $s_1~\ne~s_2$, $G$ follows the hypo-exponential distribution.

Based on the PDF of $G$, we can have {{{\rm E}}_G}\left( {{G^\delta }} \right) = \left \{ \begin{aligned} &{s^\delta }\Gamma \left( {\delta + 2} \right), \text{if s_1=s_2=s}, \\ &\frac{1}{{{s_2} - {s_1}}}\Gamma \left( {\delta + 1} \right)\left( {{s_2}^{\delta + 1} - {s_1}^{\delta + 1}} \right), \text{otherwise}, \end{aligned} \right. \tag{29} where (29) can be obtained by Eq. (3.478)$^{3)}$. Furthermore, according to Eq. (3.381)$^{3)}$, we can also obtain $$\begin{array}{l} {{\rm E}_G}\left( {{G^{\mu - 1}}\Gamma \left( {\nu ,\gamma G} \right)} \right) = \left\{ \begin{array}{l} {s}^2J\left( {\mu {\rm{ + }}1,{s}^{ - 1},\nu ,\gamma } \right), {\rm{ if }} {{\rm{s}}_{\rm{1}}}{\rm{ = }}{{\rm{s}}_{\rm{2}}}{\rm{ = s }}, \\ \frac{1}{{{s_2} - {s_1}}}\left( {J\left( {\mu ,\frac{1}{{{s_2}}},\nu ,\gamma } \right) - J\left( {\mu ,\frac{1}{{{s_1}}},\nu ,\gamma } \right)} \right), {\rm{ otherwise}}, \end{array} \right. \end{array} \tag{30}$$ where $J(~{\cdot,~\cdot,~\cdot,~\cdot~})$ is defined in Eq. (11). We also start with the definition of joint LT. \begin{align}&{\mathcal{L}_{{{\hat I}_{\rm d}},{I_{\rm d}}}}\left( {{s_{\rm h}},{s_{\rm f}}} \right) = {{{\rm E}}_{{{\hat I}_{\rm d}},{I_{\rm d}}}}\left( {{{\rm e}^{ - {s_{\rm h}}{{\hat I}_{\rm d}} - {s_{\rm f}}{I_{\rm d}}}}} \right) \\ &\mathop = \limits^{\left(k \right)}{\rm E}\left( \exp \left( - {s_{\rm hd}}\left( {\sum\limits_{z \in {\Phi _{\rm hd}}\backslash b_o} {{h_{z,u_o}}{D_{z,u_o}}^{ - {\alpha _{\rm bu}}}} } \right)- \left( {\sum\limits_{z \in {\Phi _{{\rm fd}\backslash b_o}}} {{s_{\rm fd}}{h_{z,u_o}}{{ {{D_{z,u_o}}} }^{ - {\alpha _{\rm bu}}}}} + \sum\limits_{x \in {\Phi _{\rm fu}}} {{s_{\rm fu}}{h_{x,u_o}}{{ {{D_{x,u_o}}} }^{ - {\alpha _{\rm uu}}}}} } \right) \right) \right) \\ &\mathop = \limits^{\left( l \right)} {\rm E}\left( \exp \left( - \left( {\sum\limits _{z \in {\Phi _{\rm fd}}\backslash b_o} {\left( {{s_{\rm hd}}{h_{z,u_o}} + {s_{\rm fd}}{h_{z,u_o}}} \right){D_{z,u_o}}^{ - {\alpha _{\rm bu}}}} } \right) - {s_{\rm fu}}\left( {\sum\limits_{x \in {\Phi _{\rm fu}}} {{h_{x,u_o}}{{\left( {{D_{x,u_o}}} \right)}^{ - {\alpha _{\rm uu}}}}} } \right) \right) \right) \\ &\mathop = \limits^{\left( m \right)} \exp \left\{ { - 2\pi {\lambda _{\rm s}}{\rm E}\int\nolimits_R^\infty {1 - {{\mathop{\rm e}\nolimits} ^{\left( { - \left( {{{s}_{\rm fd}}{h_{\rm fd}} + {{s}_{\rm hd}}{h_{\rm hd}}} \right){v^{ - {\alpha _{\rm bu}}}}} \right)}}v{\rm d}v} } \right\}\exp \left\{ { - 2\pi {\lambda _{\rm s}}{\rm E}\int\nolimits_{\eta _{\rm uu}R} ^\infty {1 - {{\mathop{\rm e}\nolimits} ^{\left( { - {{s}_{\rm fu}}{h_{\rm fu}}{v^{ - {\alpha _{\rm uu}}}}} \right)}}v{\rm d}v} } \right\} \\ &\mathop = \limits^{\left( n \right)}\exp \left\{ - 2\pi {\lambda _{\rm s}}{{\rm E}_{{G_{\rm u}}}}\left\{ - \frac{1}{{{\alpha _{\rm bu}}}}{G_{\rm u}}^{\frac{2}{{{\alpha _{\rm bu}}}}}\Gamma \left( - \frac{2}{{{\alpha _{\rm bu}}}}\right) - \frac{1}{2}{R^2}+ \frac{1}{{{\alpha _{\rm bu}}}}{G_{\rm u}}^{\frac{2}{{{\alpha _{\rm bu}}}}}\Gamma \left( - \frac{2}{{{\alpha _{\rm bu}}}},\frac{{{G_{\rm u}}}}{{{R^\alpha }}}\right) \right\} \right\}\omega \left( {{\alpha _{\rm uu}},s_{\rm fu}\eta _{\rm uu}R } \right), \tag{31} \end{align} where ${s}_{\rm~fd}={{s_{\rm~f}}}{{P_{\rm~fd}}}{{k}_{\rm~fd}},~~{s}_{\rm~hd}={{s_{\rm~h}}}{{P_{\rm~hd}}}{{k}_{\rm~hd}}$ and${s_{\rm~fu}}={{s_{\rm~f}}}{{P_{\rm~fu}}}{{k}_{\rm~fu}}$ in Step (k). In Step (l), the interference BS sets in HD RBs and FD RBs are the same, therefore, the interfering distances between the focused downlink user and the BSs after classifying the user as CEU is the same as the distances when the focused user is regarded as cell CCU. Hence, we can write them together as the first sum item of Step (l). $G_{\rm~u}={{{s}_{\rm~fd}}{h_{\rm~fd}}~+~{{s}_{\rm~hd}}{h_{\rm~hd}}}$ which is the same as (28) and its properties have already been given by (29) and (30). Then the conclusion of Lemma 3.3 is obtained.

Proof of Theorem sect. 4.1

The proof begins with the definition of the CEU and coverage probability, then we can obtain \begin{align} &F_{{\rm edge},q}\left( {T} \right) = \mathbb{P}\left( {{{{\rm SINR}}} > T|{\rm SINR} < {{\gamma}_q}} \right) \\ &\mathop = \limits^{\left({\rm h}\right)} \frac{{{\rm E}\left( {\mathbb{P}\left( {{{\rm SINR}} > T,{\rm SINR} < {{\gamma}_q}|R,{I_{{\rm f}q}},{I_{{\rm h}q}}} \right)} \right)}}{{{\rm E}\left( {\mathbb{P}\left( {{\rm SINR} < {{\gamma}_q}|R,{I_{{\rm f}q}}} \right)} \right)}}\mathop = \limits^{\left(I\right)} \frac{{{\rm E}\left( {{{\mathop{\rm e}\nolimits} ^{\left( { - {s_{\rm h}}\left( {{\sigma ^2} + {{\hat I}_{\rm e}}} \right)} \right)}}\left( {1 - {{\mathop{\rm e}\nolimits} ^{\left( { - {s_{\rm f}}\left( {{\sigma ^2} + {I_{\rm e}} + {\delta }{I_{\rm SI}}} \right)} \right)}}} \right)} \right)}}{{{\rm E}\left( {\left( {1 - {{\mathop{\rm e}\nolimits} ^{\left( { - {s_{\rm f}}\left( {{\sigma ^2} + {I_{\rm e}} + {\delta }{I_{\rm SI}}} \right)} \right)}}} \right)} \right)}}, \tag{32} \end{align} where SINR and $I_{\rm~e}$ denote the received SINR and interference when the user is regarded as CCU. While SINR$~and~$hat I_rm e$~represent~the~received~SINR~and~interference~after~the~user~is~classified~as~CEU.~SINR~is~defined~as~(\ref{equation2}).~Step~(h)~follows~Bayes~theorem.~Step~(I)~is~obtained~based~on~two~facts.~one~is~the~independency~of~the~small~scale~fading~over~FD~and~HD~RBs,~the~other~fact~is~that~the~small~scale~fading~is~assumed~to~be~Rayleigh,~i.e,~$h∼ rm exp1)$,~so~$mathbbPh<t)=1-exp (-t).~Then~carry~on~the~proof~in~Step~(I)~and~we~have \begin{align} \label{equation_coverage_edge} &F_{{\rm edge},q}\left( {T} \right) = \mathbb{P}\left( {{{\rm SINR}} > T|{\rm SINR} < {{\gamma}_q}} \right)= \frac{{\int\nolimits_{R = 0}^\infty {{\rm E}\left( {{{\mathop{\rm e}\nolimits} ^{\left( { - {s_{\rm h}}{\sigma ^2}} \right)}}\left( {{{\mathop{\rm e}\nolimits} ^{ - {s_{\rm h}}{{\hat I}_{\rm e}}}} - {{\mathop{\rm e}\nolimits} ^{\left( { - {s_{\rm f}} {{{\sigma }^2} + {s_{\rm h}}{\hat I_{\rm e}} + {s_{\rm f}}{I_{\rm e}}} } \right)}}} \right)} \right){f_R}\left( R \right){\rm d}R} }}{{\int\nolimits_{R = 0}^\infty {{\rm E}\left( {1 - {{\mathop{\rm e}\nolimits} ^{\left( { - {s_{\rm f}}\left( {{{\sigma }^2} + {I_{\rm e}}} \right)} \right)}}} \right){f_R}\left( R \right){\rm d}r} }}\nonumber\\ &\mathop = \limits^{\left({\rm j}\right)} \frac{{\int\nolimits_{R = 0}^\infty {\left( {{{\mathop{\rm e}\nolimits} ^{ - {s_{\rm h}}{\sigma ^2}}}\left( {{L_{{I_{{\rm h}q}}}}\left( {{s_{\rm h}}} \right) - {{\mathop{\rm e}\nolimits} ^{ - {s_{\rm f}}\sigma {^2}}}{L_U}\left( {{s_{\rm h}},{s_{\rm f}}} \right)} \right)} \right){f_R}\left( R \right){\rm d}R} }}{{\int\nolimits_{R = 0}^\infty {\left( {1 - {{\mathop{\rm e}\nolimits} ^{ - {s_{\rm f}}\sigma {^2}}}{L_{{I_{{\rm f}q}}}}\left( {{s_{\rm f}}} \right)} \right){f_R}\left( R \right){\rm d}R} }}, \end{align} where~f_R(R)$~is~defined~in~(\ref{distribution_r}).~We~can~obtain~Step~(j)~according~to~the~definition~of~LT.~$L_I_m$~is~the~LT~of~interference~$I_m$~($m ∈ łeft rm hq,rm fq right$~)~and~they~are~defined~in~(\ref{laplace_hd})~and~(\ref{laplace_fq}).~The~joint~LT~$L_U$~is~different~in~the~downlink~and~uplink~transmission.~In~downlink~transmission,~the~interfering~BSs~are~the~same~before~and~after~allocating~HD~RBs~to~the~CEUs,~which~results~in~$hat I_rm e$~and~$I_rm e$~are~correlated~with~each~other.~As~discussed~in~Lemma~\ref{Theorm_laplace_jiont},~$L_U$~is~defined~in~\ref{L_U}.~While~in~the~uplink,~the~uplink~interfering~user~sets~have~already~changed~when~the~serving~RBs~of~CEU~switch~from~FD~RBs~to~HD~RBs.~Therefore,~in~uplink~$L_U$~can~be~written~as~$L_rm hułeft(s_rm hright)L_rm fułeft(s_rm fright). Then the conclusion of Theorem sect. 4.1 is obtained.

### References

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• Figure 1

An exemplary interference scenario of HD/FD cellular network. (a) Synchronous TDD cellular network; protectłinebreak (b) FD cellular network.

• Figure 2

Resource blocks allocation.

• Figure 9

SE vs. successive interference cancellation capability.

• Table 1   Simulation parameters
 RB band width Number of RB Thermal noise $P_b$ $P_{\rm~u}$ $\eta~$ $\lambda_{\rm~s}$ $\lambda_{\rm~u}$ $\gamma_{\rm~d}$ ${\left|~{{\varepsilon~}}~\right|^2}$ Path loss [25] 1 MHz 128 $-$174 dBm$\cdot$Hz$^{-1}$ 0.1 W 0.1 W 0 ${10^{{\rm{~-~}}3}}$ m$^{-2}$ ${1}$ m$^{-2}$ 0 dB $-110$ dB 140.7 + 36.7lg$R$ ($R$ in km)
• #### 0

Citations

• Altmetric

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# Torsion and angle of twist

1. Oct 16, 2008

### aznkid310

1. The problem statement, all variables and given/known data

Unfortunately, I dont have a picture to upload, so I'll describe it the best that I can.

A prismatic bar AB of length L and solid circular cross section (diameter d) is loaded by a distributed torque of constant intensity t per unit distance. Determine the angle of twist W between the ends of the bar.

2. Relevant equations

3. The attempt at a solution

d(Torque) = tdx --> Torque T = integral (from 0 to L) [tdx] = tL

W = int(0 to L) [T(x)dx/GI(x)] , where G = shear modulus, I = polar moment of inertia

Is my T(x) equal to T = tL ?

How do I find I? I realize that I = int over the area [x^2 dA], where x is the distance from the center to dA

Can i just use the formula I = [(pi)r^4]/4 for a solid bar?

2. Oct 16, 2008

### Pyrrhus

if the bar is prismatic it means its cross section is constant along its length.

For this case a differential twist angle will be

$$d \phi = \frac{T(x)dx}{GI_{polar}}$$

No, T = f(x). It's not constant for this case.

Yes, but that will just make your work longer.

That formula is wrong. That's the area moment of inertia of the section. Look up the polar moment of inertia.

Last edited: Oct 16, 2008
3. Oct 17, 2008

### aznkid310

Sorry i typed it in wrong. It should be I = (pi/2)(r^4)

Could you get me started on finding T(x)?

4. Oct 17, 2008

### Pyrrhus

What do you understand by "... a distributed torque of constant intensity t per unit distance..." ?

5. Oct 17, 2008

### aznkid310

that the torque varies linearly with distance?

6. Oct 17, 2008

### Pyrrhus

so T(x) = ?

7. Oct 18, 2008

### aznkid310

T(x) = tx?

W = (t/GI)int(0 to L) [ xdx]

= (tL^2)/2GI

= [16tL^2]/[(pi)Gd^4]

Last edited: Oct 18, 2008
8. Feb 25, 2010

### diyana

hi...how do i get the angle of twist (in radian) if the question give the revolutions number?

for more understanding, here is the question :

the ship at the surface,A has just started to drill for oil on the ocean floor at a depth of 1500m. knowing that the top of the 200-mm-diameter steel drill pipe (G=77.2GPa) rotates two complete revolutions before the drill bit at the bottom,B starts to operate, determine the maximum shear stress caused in the pipe by torsion.

i know that angle of twisT,@ = TL/JG where J= (pi/2)(c^4) .....c=outermost radius.

and when i got the T (torsion), i can use the equation ; shear stress = Tc/J.

am i correct here? i just got confuse how do i get the T..

format: plain


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### Home > APCALC > Chapter 5 > Lesson 5.2.5 > Problem5-106

5-106.

Use your derivative tools to write the second derivative, $\large\frac { d ^ { 2 } y } { d x ^ { 2 } }$, of each function below. Homework Help ✎

1. $y=\frac{\sin(x)}{x}$

2. $y=\csc^2\left(x\right)-\cot^2\left(x\right)$

3. $y=\sqrt{\frac{1}{x}}$

You could use the quotient rule AND the chain rule...Or you could rewrite y as with exponents and quickly use the Power Rule!

4. $y=|x-2|$

Rewrite y as a piecewise function. The derivative will be a piecewise function as well. So will the second derivative.


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4 out of 5 dentists recommend this WordPress.com site

### Continuous linear operators are bounded.: decoding the proof, and how the mathematician chances upon it

Here we try to prove that a linear operator, if continuous, is bounded.

Continuity implies: for any $\epsilon>0, \|Tx-Tx_0\|<\epsilon$ for $\|x-x_0\|<\delta$

We want the following result: $\frac{\|Ty\|}{\|y\|}\leq c$, where $c$ is a constant, and $y$ is any vector in $X$.

What constants can be construed from $\epsilon$ and $\delta$, knowing that they are prone to change? As $T$ is a linear operator, $\frac{\epsilon}{\delta}$ is constant. We need to use this knowledge.

We want $\frac{\|Ty\|}{\|y\|}\leq \frac{\epsilon}{\delta}$, or $\delta\frac{\|Ty\|}{\|y\|}\leq {\epsilon}$.

We have $\|Tx-Tx_0\|=\|T(x-x_0)\|<\epsilon$.

Hence, $x-x_0=\delta.\frac{y}{\|y\|}$.

$\|T(\delta.\frac{y}{\|y\|})\|=\frac{\delta}{\|y\|}\|Ty\|$.

We have just deconstructed the proof given on pg.97of Kreyszig’s book on Functional Analysis. The substitution $x-x_0=\delta.\frac{y}{\|y\|}$ did not just occur by magic to him. It was the result of thorough analysis. And probaby such investigation.

But hey! Let’s investigate this. $\frac{\delta}{\epsilon}$ is also constant! Let us assume $\epsilon\frac{\|Ty\|}{\|y\|}\leq \delta$. Multiplying on both sides by $\frac{\epsilon}{\delta}$, we get $\frac{\epsilon^2}{\delta}\frac{\|Ty\|}{\|y\|}\leq \epsilon$. This shows $x-x_0=\frac{\epsilon^2}{\delta}\frac{y}{\|y\|}$. Does this substitution also prove boundedness?

We have to show $\|x-x_0\|<\delta$. $\frac{\epsilon^2}{\delta}<\delta$ only if $\epsilon<\delta$. Hence, this is conditionally true.

Similar investigations taking $(\frac{\epsilon}{\delta})^n$ to be constant can also be conducted.

### Linear operators mapping finite dimensional vector spaces are bounded,

Theorem: Every linear operator $T:V\to W$, where $V$ is finite dimensional, is bounded.

Proof $\frac{\|Tx\|}{\|x\|}=\frac{\|T(a_1e_1+a_1e_2+\dots+a_ne_n)\|}{\|a_1e_1+a_1e_2+\dots+a_ne_n\|}\leq \frac{\|T(a_1e_1+a_1e_2+\dots+a_ne_n)\|}{c(|a_1|+|a_2|+\dots+|a_n|)}\leq \frac{\|T(e_i)\|}{c}$

where $\|T(e_i)\|=\max\{\|T(e_1)\|,\|T(e_2)\|,\dots\}$.

What we learn from here is

$\|e_i\|(|a_1|+|a_2|+\dots+|a_n|)\geq\|a_1e_1+a_1e_2+\dots+a_ne_n\|\geq c(|a_1|+|a_2|+\dots+|a_n|)$

where

$\|e_i\|=\max\{\|e_1\|,\|e_2\|,\dots,\|e_n\|\}$.

Similarly,

$\|e_i\|(|a_1|+|a_2|+\dots+|a_n|)\geq\|a_1e_1+a_1e_2+\dots+a_ne_n\|\geq \|e_k\|(|a_1|+|a_2|+\dots+|a_n|)$

where

$\|e_k\|=\min\{\|e_1\|,\|e_2\|,\dots,\|e_n\|\}$

Another proof of the assertion is

$\frac{\|Tx\|}{\|x\|}=\frac{\|T(a_1e_1+a_1e_2+\dots+a_ne_n)\|}{\|a_1e_1+a_1e_2+\dots+a_ne_n\|}\leq \frac{\|T(a_1e_1+a_1e_2+\dots+a_ne_n)\|}{\|e_k\|(|a_1|+|a_2|+\dots+|a_n|)}\leq \frac{\|T(e_i)\|}{\|e_k\|}$

which is a constant.

Note: why does this not work in infinite dimensional spaces? Because the max and min of $\|e_r\|$ and $\|Te_r\|$ might not exist.

### Riesz’s lemma decoded

This is a rant on Riesz’s lemma.

Riesz’s lemma- Let there be a vector space $Z$ and a closed proper subspace $Y\subset Z$. Then $\forall y\in Y$, there exists a $z\in Z$ such that $|z-y|\geq \theta$, where $\theta\in (0,1)$, and $|z|=1$.

A proof is commonly available. What we will discuss here is the thought behind the proof.

For any random $z\in Z\setminus Y$ and $y\in Y$, write $\|z-y\|$. Let $a_{y\in Y}=\inf\|z-y\|$. Then $\|z-y\|\geq a$. Also, there exists a $y_0\in Y$ such that $\|z-y_0\|\leq\frac{a}{\theta}$. Then $\left\|\frac{z}{\|z-y_0\|}-\frac{y}{\|z-y_0\|}\right\|\geq\theta$. Because the vector space $Z$ is closed under scalar multiplication, we have effectively proved $\|z-y\|\geq\theta$ for any $\theta\in (0,1)$ and $y\in Y$.

If there is some other vector $v$ such that $\|v-v_0\|\leq\frac{a}{\theta}$, then $\|\frac{z}{\|v-v_0\|}-\frac{y}{\|v-v_0\|}\|\geq\theta$.

Hence, one part of Riesz’s lemma, that of exceeding $\theta$ is satisfied by every vector $z\in Z\setminus Y$. The thoughts to take away from this is dividing by $\theta$ or a number less than $1$ increases everything, even a small increase from the infimum exceeds terms of a sequence converging to the infimum, and every arbitrary term in the sequence is greater than the infimum. When we say $\theta$ can be any number in the interval $(0,1)$, we know we’re skirting with boundaries. We could aso have thought of a proof in this direction: let $b=\sup_{y\in Y} \|z-y\|$. Then $b\theta\leq\|z-y_0\|\leq b$. However, for an arbitrary $y\in Y$, $\left\|\frac{z}{\|z-y_0\|}-\frac{y}{\|z-y_0\|}\right\|\leq\frac{1}{\theta}$.

Hence, for every $\theta\in (0,1)$, $\theta\leq \|z-y\|\leq\frac{1}{\theta}$.

Now what about $\|z\|=1$? This condition is satisfied only when $z=z-y_0$ in the expression $\left\|\frac{z}{\|z-y_0\|}-\frac{y}{\|z-y_0\|}\right\|\geq\theta$.

Hence, over in all, for every vector $z\in Z-Y$, there are infinite vectors which satisfy the condition of Riesz’s lemma. Also, for every such $z$, there is AT LEAST one unit vector which satisfies Riesz’s lemma (there can be more than one). Hence, to think there can be only one unit vector in $Z-Y$ which satisfies Riesz’s lemma would be erroneous.


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# Taylor approximation fails in my model

Hi,
I’m working on a model with money, in which I take the ‘money in utility’ form to construct my model. (U=…+gammalog(M/P)+…)
It naturally gives a F.O.C. in the form of:
gamma/(M/P)=C^(-signa)
(i/(1+i))

Then during the process of log-linearization, the right side of the equation would be transformed into an expression with log(i/rho) (where rho is the steady-state interest rate).
(I define I=log((1+i)/(1+rho)) (in the traditional way), so approximately I=i-rho.)
Then using log-linearization, we can write down:
log(i/rho)=log((I+ rho)/rho)=log(1+I/rho)

We can fairly say that log(1+x)=x when x is small (|x|<0.1), but in this situation, I/rho would be in the interval [0,3] in the normal shock size, which absolutely does not suit for Taylor Approximation. (eg. log(1+3)=1.39<<3)
However, it seems that we have to use Taylor Approximation in Dynare, which would lead to bad accuracy. Even if you use higher order, it only makes the approximation worse, because using the Taylor Approximation
log(1+x)=x-x^2/2+x^3/3+…+(-x)^n/n+…
when |x|>1, this series just explodes…

I was wondering if you can help me solve this problem…

I guess that the 3 in your discussion above means an upper bound for rho of 3%. Then the upper bound of your interval is 1+0.03 and not 1+3

Sorry, but in my model rho=0.02, and the absolute value of i can be [0,0.06] if I set the initial shock value to be 0.1.
Temporarily I reset shock to 0.01 to avoid this problem, but I’m very curious about the experiment if the shock is bigger.

I’m not sure to understand why do you feel necessary to reset the value of the shock, but if you are dealing with big shocks in a small model, you should consider solving it via value function iterations or global methods. None of these approaches are provided by Dynare.

I am also puzzled by the statement on higher order approximations. Obviously the log is also going to infinity. So higher order approximations are usually more and not less accurate.

As I have said before, using Taylor expansion we can get
log(1+x)=x-x^2/2+x^3/3+…+(-1)^(n-1)*(x^n)/n+…
So if |x|<1, we can get as accurate as we want by just increasing the order of Taylor approximation, because the series above is convergent on interval (-1,1]. However, if the shock size is so big that |x|>1, the series above will diverge, then Taylor approximation will fail.

Now I get it. If you move outside of the convergence radius, you will of course get problems. That is a case where the local solution techniques available in Dynare do not work accurately. As @MichelJuillard pointed out, you would go for global methods in this case. Take projection methods. Here you could use Chebyshev polynomials that do not suffer from the same problem as normal polynomials.


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# A nanodroplet cell processing platform facilitating drug synergy evaluations for anti-cancer treatments

## Abstract

Therapeutic drug synergism intervened in cancer treatments has been demonstrated to be more effective than using a single effector. However, it remains inherently challenging, with a limited cell count from tumor samples, to achieve potent personalized drug cocktails. To address the issue above, we herein present a nanodroplet cell processing platform. The platform incorporates an automatic nanodroplet dispenser with cell array ParaStamp chips, which were fabricated by a new wax stamping approach derived from laser direct writing. Such approach enables not only the on-demand de-wetting with hydrophobic wax films on substrates but also the mask-less fabrication of non-planar microstructures (i.e. no photolithography process). The ParaStamp chip was pre-occupied with anti-cancer drugs and their associate mixtures, enabling for the spatially addressable screening of optimal drug combinations simultaneously. Each droplet with a critical volume of 200 nl containing with 100 cells was utilized. Results revealed that the optimal combination reduces approximate 28-folds of conducted doses compared with single drugs. Tumor inhibition with the optimally selected drug combination was further confirmed by using PC-3 tumor-bearing mouse models. Together, the nanodroplet cell processing platform could therefore offer new opportunities to power the personalized cancer medicine at early-stage drug screening and discovery.

## Introduction

Synergistic combination of two or more drugs has been a major avenue targeting cancers1,2. This regimen not only improves the therapeutic efficacy by triggering synthetic lethality in target cells but also minimizes the side effects by reducing doses of each drug3,4. Therefore, the identification of optimal combination of various possible concentrations from a set of drugs presents a substantial challenge.

Several approaches to optimize the selection regime have been demonstrated, in terms of large scale simulations5,6 and stochastic search algorithms7,8. The power of feedback system control (FSC) methodology can facilitate the screenings down to 10–20 iterative tests out of million possible combinations; however, challenges still remain9. For example, the time for cell preparation adopted among the total independent iterations would last for weeks. In addition, the usage of conventional multi-well plate assays would counter the feasibility for personalized medicine, which is inherently subject to a limited cell count from tumor samples10. Although the multilayered culture technique has been demonstrated to better predict the in vivo efficacy targeting tumor microenvironments, the time, required cell amount and complexity of experimental setup are not addressed as well11. As such, a new technique needs to be addressed against the shortcomings described above.

In the past decade, cell-based microarray platforms have been demonstrated to address the issues of critical drug volume, cell-source limitation, or high-throughput and high-content screening12,13,14,15,16,17. Notably, a recently developed approach involved cell microarray with cancer stem cells (CSCs) to potentially address the tumor heterogeneity in vivo18. Likewise, other researches revealed that cells cultured in microscales could contribute distinct drug responses, attributing to the higher cell to volume ratio compared with macroscale cultures19,20. Although the translation from in vitro to in vivo optimal drug combinations has been successfully performed with standard multi-well plates, relatively little effort has been directed toward using cell arrays as biosensor tools for pre-clinical in vivo assays8,11,21.

To address the technology gap described above, we herein present the nanodroplet cell processing platform for high-throughput screenings of optimal drug combinations. The platform incorporates a programmable nanodroplet dispenser with ParaStamp cell array chips. The chips were fabricated following a new mask-less approach developed to improve the hydrophobic wax stamping with well arrays with a uniform size (i.e. without photolithography process). We demonstrated that approximate 500-fold miniaturization does not impact the in vivo outcome (81 test spots per 22 × 22mm2; 100 cells in 200 nl per spot). Taken together, these findings highlight our newly developed nanodroplet cell processing platform could become a cost-effective, purpose-tailored and high-throughput toolkit for improving pre-clinical drug screening efficacy.

## Results and Discussion

### Characteristics of polydimethylsiloxane (PDMS) ParaStamp

We incorporated a mask-less approach with a laser direct-writing availability in four steps to fabricate the PDMS stamp (Fig. 1): programmable engraving on adhesive tapes with a CO2 laser; stamping a hydrophobic paraffin wax with corresponding patterns from the engraved tapes on glass substrates [a representative image shown in Fig. 2(a) and the water contact angles are 105.4° and 41.7° for patterned substrate and bare glass, respectively, shown in Fig. 2(b)]; placing the patterned glasses in a water bath to enable the sessile droplets to spontaneously reflect the underlying patterns [a representative image shown in Fig. 2(c)]; and casting the concave microstructures by PDMS prepolymers to present the ParaStamp with repeatable stamping ability [a representative image shown in Fig. 2(d)]. We demonstrated that the PDMS ParaStamp could be achieved with various characteristic sizes [Fig. 2(e)], in which the sizes behave in a linear relationship depending on the designed parameters for laser micromachining [Fig. 2(f)]. In addition, the aspect ratio of the PDMS ParaStamp was obtained with an approximate 0.13 of height to width [Fig. 2(g)].

CO2 laser micromachining has been widely adopted in biomedical applications, owing to its rapid and mask-less prototyping potentials22,23,24. In this paper, we first and successfully demonstrate that such the micromachining can be applied to facilitate the ParaStamp technique25. It could achieve the critical size of PDMS ParaStamp fabricated down to 120 μm [Fig. 2(f)], which is parallel to the typical trench width with 140 μm engraved in poly(methyl methacrylate) (PMMA)22. Most importantly, the PDMS ParaStamp could be re-operated up to 50 times without destroying the patterns (data not shown here). These highlight the power of using PDMS ParaStamp for a wide range of applications, e.g. cell patterning and rewritable droplet storage as well as cost-effective drug synergy screening25.

### High-throughput drug synergy screening via the nanodroplet cell processing platform

It has been demonstrated with a potential to improve the therapeutic relevant selectivity by synergistic drug combinations. Our developed nanodroplet platform also facilitates the drug synergy screening from the proof-of-conceptual experiment, as illustrated in Figs 3 and 4 (see also Supplementary Fig. S1). We applied the PDMS ParaStamp to fabricate 9 × 9 well arrays on a glass substrate (i.e. ParaStamp chip), in which the average diameter of patterned wells is 1151 μm with a coefficient of variation (CV) to be 3.4%. The ParaStamp chip contains arrayed dehydrolyzed drugs, in which each drug droplet (200 nl) is first dispensed with a single drug or a mixture of drugs with different concentrations and then dried. A PDMS-glass gasket was applied to prevent the evaporation of nanodroplets during experiments (Supplementary Fig. S2). To direct the advantage of using such chip, we first treated prostate PC-3 cancer cells with different concentrations of four commonly used chemotherapeutic drugs – cisplatin, paclitaxel, 5-FU and doxorubicin for a 1-day single drug treatment (Fig. 4). We observed that the IC50 values derived from the chip were 62.9 ± 5.0, 8.0 ± 0.1, 8324.2 ± 2475.2 and 13.4 ± 2.2 μg/ml for cisplatin, paclitaxel, 5-FU and doxorubicin, respectively.

For demonstration the drug synergy screening, we treated PC-3 cells with three different concentrations of the four anti-cancer drugs described above and their associated mixtures [Fig. 5(a)]. It therefore yielded totally 81 drug combinations, including control, 5 and 10 μg/ml of cisplatin, 0.6 and 1.2 μg/ml of paclitaxel, 500 and 1000 μg/ml of 5-FU, 10 and 20 μg/ml of doxorubicin, and the rest 72 associate mixtures. Based on the first screening with regression coefficient, we further determined cisplatin, paclitaxel, and doxorubicin as the potent effectors to inhibit PC-3 cell proliferation [Fig. 5(b)]. The optimal drug combination was selected to be 10 μg/ml cisplatin, 0.6 μg/ml paclitaxel and 20 μg/ml doxorubicin, which is depending on the second screening with combinational synergy from the 3-drug cocktails [Fig. 5(c)]. This optimum was demonstrated to be more efficient than single drug treatment [Fig. 5(d)]. Moreover, it could reduce the single drug doses of cisplatin, paclitaxel, and doxorubicin down to 45.4, 34.4 and 5.3 folds, respectively, under the 82% optimal cell inhibition. It therefore contributed an average 28.4-fold reduction among the three drugs (combination A in Supplementary Table S1).

In this paper, we conducted a drug pre-coating approach to facilitate the drug screening with the chip, thus excluding the issue of droplet evaporation during the sequenced dispensing with drugs. Where it may cause the distinction while the aqueous drugs turn into dried solids, possibly lacking their pharmaceutical activity. We therefore performed an experiment to compare our approach used and the conventional adding-with-drug in 96-well plates. Results revealed that no significant difference is detected between the two approaches (Supplementary Fig. S3). It therefore highlights that our proposed approach could be applied for a potential live-cell biosensor that is pre-occupied with drugs.

### Partially successful translation from in vitro drug combination to in vivo drug administration

To demonstrate the feasibility of using the nanodroplet platform to predict the in vivo efficacy, we first employed PC-3 tumor-bearing mouse models to compare the three different drug combinations screened from in vitro assays (combination A in Supplementary Table S1). Based on an empirical correlation, one could convert the drug doses to be administered between animals and humans. However, there is relatively less proper formula to systematically translate the in vitro drug doses to in vivo sets8,11,21,26. We therefore developed a translated algorithm, as exhibited in Eqs (2) and (3), to be applied into the in vivo models (details can be referred in Methods). The translated doses were listed in Supplementary Table S1.

Results of the mouse model assays showed that the screened combination A presents a tumor inhibition (p < 0.05) compared with the control, whereas no difference is observed among the single drugs administered [Fig. 6(a,b)]. For the requirement of power = 0.8, the total sample size needed was calculated to be 25 mice, which is larger than 18 mice as power = 0.65 from our experimental data at Day 15 (see also the section of Power analysis in the Supplementary Information). In addition, there was no significant weight change of mice among the individual drug treatments except for the cisplatin treatment at Day15 [Fig. 6(c)].

We further compared the in vivo prediction of drug responses derived from our platform and the standard 96-well plate assay (Supplementary Fig. S4). Both of them presented the same in vitro drug combinations, similar responses in cell viability (18.2% from platform and 20.0% from plate) and similar synergistic CI values (0.2 from platform and 0.5 from plate), as listed in Supplementary Table S1. The drug doses translated from the platform (combination A) to mice were determined to be 1.1 mg/kg cisplatin, 4.2 mg/kg paclitaxel and 1.2 mg/kg doxorubicin, whereas the plate (combination B) was obtained with 4.7 mg/kg cisplatin, 17.9 mg/kg paclitaxel and 5.3 mg/kg doxorubicin. Although the three in vivo drug doses in combination B were particularly 3-folds larger than that in A, the efficacy of tumor inhibition from A was significantly higher than B (Supplementary Fig. S4). Indeed, the single drug treatment with a higher concentration would lead a slower tumor growth (Supplementary Fig. S5). In other words, it indicates that the nanodroplet platform could predict the in vivo drug efficacy more efficiently than the 96-well plate assay.

Previous studies have revealed that the synergistic drug combinations, derived from well-plate assays, can benefit the therapeutic outcomes2,8,9. In this paper, we furthermore demonstrate that our nanodroplet cell processing platform could not only streamline the screening of potent drug combinations within one day, but also properly predict the in vivo efficacy. In addition, we reveal that approximate 500-fold miniaturization (200 nl of volume in platform versus 100 μl in 96-well plate per test) does not impact the in vivo outcome (Fig. 6 and Supplementary Fig. S4). Its critical volume needed per test (200 nl) is significantly smaller than 1.5 μl of recommended working volume for Corning 1536 well plates. Moreover, the optimal drug combination derived from our platform could reduce approximately 28-folds of conducted doses compared with single drugs used. Our findings herein imply that our developed platform could further ignite new applications for drug screening on rare cells, e.g. CSCs or circulating tumor cells (CTCs)27.

## Conclusions

In conclusion, we have presented the nanodroplet cell processing platform – automatic nanodroplet dispensing and ParaStamp cell array chips) – in which the chips were fabricated by a new methodology by using laser direct writing-derived wax stamping. The new approach has been successfully demonstrated to facilitate not only the on-demand surface patterning with hydrophobic wax films but also the mask-less fabrication of concave microstructures. In addition, we successfully demonstrate that the nanodroplet platform can be applied for the high-throughput drug synergy screening in vitro and in vivo. Beyond the above concerns, we anticipate that the new wax patterning technique and its resulted bio-assay cell array chip will direct a facile and effective avenue to achieve the high-throughput requirement for future biomedical applications.

## Methods

### Preparation of PDMS ParaStamp and ParaStamp chip

The PDMS ParaStamp was prepared as the schematic shown in Fig. 1. First, a thermally conductive adhesive tape (8805, 3MTM) was patterned by design and ablated by a CO2 laser engraver (Mercury II, LaserPro Inc. Grand Prairie, TX). Second, a layer of hydrophobic/de-wetting paraffin wax was patterned on a plain glass slide or a cover glass (with a 22 × 22 mm2 of total area), following the below approach. A commercial Parafilm “M” was tailored and sandwiched by a foil and the patterned adhesive prior to heating at 70 °C for 30 s. The patterned adhesive was peeled off and stamped on a glass substrate followed by cooling for another 30 seconds at room temperature. After peeling off the adhesive stamp, the wax patterns transferred was achieved. Third, the aqueous mold with patterns was simultaneously performed by dipping and then slowly pulled with the wax-patterned glass substrate from a water bath. Afterwards, the corresponding microstructures with concave features were cast by PDMS prepolymers at 70 °C for 1 h, leading the formation of a PDMS ParaStamp and its resulted ParaStamp chips. The weight ratio of PDMS base to curing agent was 10:1. The patterned adhesive can only be utilized to transfer the corresponding wax patterns for one time, due to the thermal deformation of itself. Instead, the PDMS ParaStamp can be repeatedly used for various substrates on demand because of its thermal resistance.

### Cell line

Human prostate cancer cell line PC-3 (CRL-1435, ATCC, Manassas, VA) was maintained in RPMI-1640 medium (SH30027.02, HyClone), supplemented with 10% fetal bovine serum (FBS; 10437028, Gibco) and 1% penicillin/streptomycin (P/S; GT-SPS100, GeneTeks Bioscience), and cultured in a humidified 5% CO2 incubator at 37 °C.

### High-throughput drug synergy screening

To demonstrate the feasibility of drug synergy screening via the nanodroplet platform, we fabricated a 9 × 9 ParaStamp chip following the procedure as shown in Fig. 1. The array chip contained total 81 paraffin wax-made wells patterned on a 22 × 22 mm2 cover glass, in which the diameter of the wells (bare glass surfaces for cell culturing) is 1151 μm and the center-to-center distance is 2.2 mm (Fig. 3). The chip was sterilized by UV for 30 min prior to use.

For the high-throughput drug synergy screening, we followed the procedure shown in Fig. 3. Prostate PC-3 cells were conducted under the single or synergistic effects of four chemotherapeutic drugs – cisplatin (Fresenius kabi, Solan, Himachal Pradesh, India), paclitaxel (Phyxol; Sinphar Pharmaceutical, Taiwan), 5-FU (Haupt Pharma, Wolfratshausen, Germany), and doxorubicin (Adriamycin, Pfizer, New York, NY). First, cisplatin droplets with different concentrations were dispensed onto the corresponding wells by using a programmable nanodroplet dispensing machine (Versa 10 spotter, Aurora Instruments Ltd. Vancouver, CA). Drugs were diluted in deionized (DI) water and the volume of each droplet was 200 nl. The array chip was then incubated at room temperature for 10 minutes, allowing the droplets to evaporate and therefore to convert into drug powders accordingly. Subsequently, paclitaxel, 5-FU, and doxorubicin drugs were dispensed following the approach described above, respectively. Afterwards, cell droplets with a volume of 200 nl (100 cells per droplet) were dispensed onto the array chip, followed by assembling with a PDMS-glass gasket cover, and then incubated for one day. After the 1-day drug treatment, cell viability was determined by staining live cells with Calcein AM (4 μM; Invitrogen, Carlsbad, CA). The fluorescence intensities of live cells were captured by a charge-coupled device (CCD) camera mounted on an upright microscope, and analyzed by a Fiji imaging macro software. The live cell percentage, according to the detected fluorescence intensity, was normalized against the untreated cells. The IC50 values were fitted and evaluated by a four-parameter logistic equation presented28.

For drug screening performed in 96-well plates, media containing PC-3 cells with a density of 3000 cells per well were first conducted. The concentrated drug solutions were then added into the corresponding wells, leading the same drug concentrations compared with that used in nanodroplet platform and a final volume with 100 μl per well. The cell viability was evaluated from a plate reader (TCX-LS07, NTU) based on MTT (3-[4,5-dimethylthiazol-2-y1]-2,5-diphenyl tetrazolium bromide) assay.

### Linear regression model

The stepwise model was based on a linear regression equation:

$$y={b}_{0}+\sum _{i=1}^{k}{b}_{i}{x}_{i}+\sum _{i=1}^{k}{b}_{ii}{{x}_{i}}^{2}+\sum _{i=1}^{k}\sum _{j=1}^{k}{b}_{ij}{x}_{i}{x}_{j}+c$$
(1)

where y is the cell viability. b0, bi, bii and bij are the intercept, single drug linear interaction, quadratic and bilinear 2-drug interaction terms, respectively. x represents the drug dose. c is the error term with a mean equaling to zero2.

### Combination index (CI)

To determine the effect of drug combination synergy, the CI values evaluated from the CompuSYN software were adopted to represent the synergistic interaction (CI < 0.8), the additive effect (0.8 ≤ CI ≤ 1) or the antagonism (CI > 1).

### Translating of selected in vitro drug combinations to in vivo drug administrations

The translation from in vitro to in vivo drug doses was based on the formula we utilized in this work:

$${D}_{x,invivo}=(\frac{MT{D}_{x}}{DR{I}_{ave}})\times n$$
(2)

where x is the specific drug selected from the in vitro combinations, Dx,in vivo is the drug dose evaluated for the in vivo drug x administration, MTDx is the maximum tolerated dose of drug x used for the in vivo assay, and n is the total number of drug administrations during the period of the assay. DRIave is the average of dose reduction indexes (DRIs), which is calculated from

$$DR{I}_{ave}=\frac{\sum _{x=1}^{m}DR{I}_{x}}{n}$$
(3)

where DRIx is the DRI value of drug x, which is determined from the CompuSYN software based on the cell viability data (Fig. 4), and m is the number of total drugs x used.

### In vivo subcutaneous xenograft model assay

Male athymic nu/nu (nude) mice aged 4–6 weeks were obtained from Lasco Biotech Industry (weight 20–25 g). Each mouse was injected subcutaneously with a total number of 3 × 106 Matrigel-encapsulated PC-3 cells. Tangible tumors were visible within 5–7 days, at which time the drug administration by intraperitoneal (IP) injection was initialized. Each drug combination or single drug was administered by multiple injections of the translated in vivo dose at days 0, 3, 5, 7, 9, 11, and 13 to mice (i.e. totally seven time points), at which time the tumors were scored. Each experimental group was applied with four mice. The tumor volume V was evaluated by the equation V = 4 × π × (l/2)3/3, in which the mean diameter l = (a × b)1/2 (a and b denote the two orthogonal diameters of each tumor). We compared two different drug combinations A and B to determine the effective tumor inhibition under the selected drug optimizations (Supplementary Table S1). The drug combination A was derived from the nanodroplet platform. In contrast, the combination B was from the 96-well plate assay. The average DRIs (DRIave) were evaluated from the equation (3), which results in (45.4 + 34.4 + 5.3)/3 = 28.4 and (5.1 + 18.6 + 4.0)/3 = 9.2 for the combinations A and B, respectively. The MTDs of nude mice treated with cisplatin, paclitaxel or doxorubicin were typically at the range of 4.0–6.6 mg/kg29,30, 15–36 mg/kg31,32 and 2–10 mg/kg30,31 per assay, respectively. Based on the Eq. (2), it leaded the Dcisplatin,in vivo, Dpaclitaxel,in vivo and Ddoxorubicin,in vivo from the combination A to be 1.0–1.6 mg/kg, 3.7–8.9 mg/kg and 0.5–2.5 mg/kg, respectively. Similarly, the translated doses from the combination B were 3.0–5.0 mg/kg (cisplatin), 11.4–27.3 mg/kg (paclitaxel) and 1.5–7.6 mg/kg (doxorubicin). We therefore determined the in vivo drug doses based on the selecting margin described above (Supplementary Table S1). Note that, the selecting criterion of the translated dose should not exceed the MTD of each drug. In addition, the drug combinations in A and B expressed the same doses in vitro but not in vivo.

### Statistical analysis

To compare data from two groups, Student’s t test was used. For multiple groups comparison, one-way ANOVA test was used. Two-way ANOVA test was adopted to compare data in subcutaneous xenograft model assay. A statistical significance was considered as p < 0.05.

### Compliance with ethic requirements

All institutional and national regulations for the care and use of laboratory mice were followed. All relevant experiments were performed according to the experimental protocols approved by the Animal Care and Use Committee of National Taiwan University, Taiwan.

## Data Availability

Data in this paper will be available after a kind request from the corresponding authors (C.T.K., J.H.L. and H.L.).

## References

1. 1.

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## Acknowledgements

This work was supported by Grant 104-2911-I-002 -578 from the Ministry of Science and Technology (MOST) Taiwan, Grant 106-S3399 and 107-S3777 from the National Taiwan University Hospital (NTUH), Grant 103-2314-B-002-072 and 104-2314-B-002-168-MY3 from the Ministry of Science and Technology (MOST) Taiwan.

## Author information

C.T.K. and J.Y.W. developed the nanodroplet platform technique, performed all experiments, analyzed all data and wrote the manuscript. S.R.L. and Y.S.L. performed the experiment of cell culturing and drug screening. H.H.C. and J.T.H. provided drugs and suggestions. A.M.W. supplied the CO2 laser engraver. C.T.K., B.P.C., J.H.L. and H.L. provided overall guidance and critically reviewed the manuscript.

Correspondence to Ching-Te Kuo or Jen-Her Lu or Hsinyu Lee.

## Ethics declarations

### Competing Interests

The authors declare no competing interests.

Publisher’s note: Springer Nature remains neutral with regard to jurisdictional claims in published maps and institutional affiliations.


text:
# Find $\lim_{n\to\infty}P_n$ where $P_n =\big (1+\frac{1}{a_1}\big)\big(1+\frac{1}{a_2}\big)\big(1+\frac{1}{a_3}\big)\cdots\big(1+\frac{1}{a_n}\big)$

Let $a_1=1$ and $a_n=n(a_{n-1}+1)$ for $n=2,3,\ldots$

Find $$\lim_{n\to\infty}P_n\\ P_n = \left(1+\frac{1}{a_1}\right)\left(1+\frac{1}{a_2}\right)\left(1+\frac{1}{a_3}\right)\cdots\left(1+\frac{1}{a_n}\right); \ n=1,2,\ldots$$

My approach:

$$P_n = \left(\frac{a_1+1}{a_1}\right)\left(\frac{a_2+1}{a_2}\right)\cdots\left(\frac{a_n+1}{a_n}\right)$$

or $$P_n = \left(\frac{a_2}{2a_1}\right)\left(\frac{a_3}{3a_2}\right)\left(\frac{a_4}{4a_3}\right)\cdots\left(\frac{a_{n+1}}{(n+1)a_n}\right)$$

or $$P_n = \left(\frac{a_{n+1}}{a_1(n+1)!}\right)$$

or $$\lim_{n\to\infty} P_n =\lim_{n\to\infty} \left(\frac{a_{n+1}}{(n+1)!}\right)$$

How to find $\lim_\limits{n\to\infty}a_{n+1}$? I am stuck here.

Hint. From $$a_n=n(a_{n-1}+1),$$ by dividing by $n!$ one gets $$\frac{a_n}{n!}=\frac{a_{n-1}}{(n-1)!}+\frac{1}{(n-1)!}$$or $$\frac{a_n}{n!}-\frac{a_{n-1}}{(n-1)!}=\frac{1}{(n-1)!}$$ then by summing from $n=2$ to $n=N$ terms telescope giving $$\frac{a_N}{N!}-\frac{a_{1}}{1!}=\sum_{n=2}^N\frac{1}{(n-1)!}$$ that is $$\frac{a_N}{N!}=\sum_{n=0}^{N-1}\frac{1}{n!},\quad N\ge1.$$ One may recall that
$$e^x=\sum_{n=0}^{\infty}\frac{x^n}{n!},\quad x\in \mathbb{R}.$$


text:
# Edge Detection using Laplacian Filter

#### Machine Learning (ML)

Laplacian Filter (also known as Laplacian over Gaussian Filter (LoG)), in Machine Learning, is a convolution filter used in the convolution layer to detect edges in input.

Ever thought how the computer extracts a particular object from the scenery. How exactly we can differentiate between the object of interest and background.
Well a simple baseline answer could be: Edge Detection.

Before diving into the mechanics of it, let’s understand why specifically edge detection.

Try and look around you and find any object of your choice. It may be a book, apple or anything within your sight. Now, see the shape of that object.

Ask yourself, how did you figure out the shape of that particular object? Well, internally you are doing a subtraction. Subtraction of the object from its surroundings. That’s how you can say that, this is the shape. Not only that, if you were not able to distinguish that then you won’t even know that there is an object or not. What happens when you place a black book in front of black box? You won’t know there is anything, especially if they resonate the same colors.

Hence, in identifying the object, the first step is to separate it from the background. And to put it into technical terms: to detect the edge of the object.
Now, that you have an incentive to learn the topic, let’s start.

Laplacian filters are derivative filters used to extract the vertical as well as horizontal edges from an image. This is how they separate themselves from the usual sobel filters.

Sobel filters are single derivative filters, that means that they can only find edges in a single dimension.

# Why are they called derivative? And how exactly are we going to get the edges?

Wait, we will explain it all. The thing is that how can you detect and edge exactly? You find a difference between the borders and hence you say that this is one object and that is another. For example, if you have a green book placed on a blue table, then the color change from blue to green denotes a change of object. Hence, draws a line separating book from the table.

To put it technically we are checking for a change in pixel intensities, as we always work on grayscale images and the best way to see different colors on a grayscale is to check the pixel intensities.

So, if there is a bump in the pixel intensity, we can say that there is an edge.
Now to detect an edge we need to do it mathematically. To do so we can take a derivative of intensity and hence can find a bump in intensity wherever it exists and that is how we find the answer.

This is how sobel filters work. They take one derivative and find an edge in either of the one dimension (x or y).
But with Laplacian filter, we can get edges in both dimensions, hence we take double derivative of the intensities. And what happens when we do double derivation, the graph points to zero. So we will check those pixels which lead to zero and then mark them as edge points. The reference graph (Credit: OpenCV.org) is as follows:

Some of the most common filters used to create the laplacian are:

0 1 0
1 -4 1
0 1 0

#### Second

-1 -1 -1
-1 8 -1
-1 -1 -1

Both of these are created by the following equation. As we discussed we need double derviation of every pixel, so that we can check the pixel intensities.

Now as we are clear with the theory, let’s look at the actual steps.

One thing to note is that the Laplacian filter is a bit too sensitive. So, it will work badly if there is noise in the image. Hence we apply something known as a Gaussian Blur to smooth the image and make the Laplacian filter more effective.

Note: Due to this addition of the gaussian filter, the overall filter is always in a pair. And in normal dialogues you may hear Laplacian over the Gaussian Filter (LoG). Which shouldn’t be confused as another filter, but just that it comes with the Gaussian blur, nothing else.

Now, let's break down the Laplacian as well as the Gaussian blur functions and implement our own functions.

## Steps:

• We will define a function for the filter
• Then we will make the mask
• Then we will define function to iterate that filter over the image(mask)
• We will make a function for checking the zeros as explained
• And finally a function to bind all this together

One thing to keep in mind is that to create a combination of the filters, i.e. Laplacian over gaussian filter (LoG), then we can use the following formula to combine both of them.

Following is the function to define the above equation:

def l_o_g(x, y, sigma):
# Formatted this way for readability
nom = ( (y**2)+(x**2)-2*(sigma**2) )
denom = ( (2*math.pi*(sigma**6) ))
expo = math.exp( -((x**2)+(y**2))/(2*(sigma**2)) )
return nom*expo/denom

Note: Don't be confused by the equation and the code being abit different. But the code has been written for the simplified equation (which can be found by compounding the inner subtraction as well as multiplication of outer denominator).

As we are breaking down the filter, we would also need some additional functions. Those are:

• A function to take the filter and convolve around the image.
• A function for creating the actual mask.
• And finally a function to check the zeros.

#### Function(s) logic:

Create_log fucntion:

1. Calculate w by using the size of filter and sigma value
2. Check if it is even of not, if it even make it odd, by adding 1
3. Iterate through the pixels and apply log filter and then append those changed pixels into a new array, and then reshape the array.

Convolve function:

1. Extract the heights and width of the image
2. Now make a range of pixels which are covered by the mask (output of filter)
3. Make an array of zeros (res_image)
4. Now iterate through the image and append the values of product of mask and the original image and append it to the res_image array.

Zeros function (To mark the edges):

1. Make a zeros array of same dimensions of the log_image (zc_image)
2. Now iterate over the image and check two values, if they are equal to zero or not. If they are not equal to zero then check if values are positive or negative.
If the pixel value is zero, that means there an edge exist.
If none of the pixels, are zero, that means there is no edge.
3. Now whichever pixels were zero, append those coordinates to zc_image array.
4. Make all the pixels in zc_image as 1, meaning white. That will show the edges.

Now that we understand the logic of coding the functions, let us see how we can implement them in Python.

#### Creating the filter image:

def create_log(sigma, size = 7):
w = math.ceil(float(size)*float(sigma))

if(w%2 == 0):
w = w + 1

w_range = int(math.floor(w/2))
print("Going from " + str(-w_range) + " to " + str(w_range))
for i in range_inc(-w_range, w_range):
for j in range_inc(-w_range, w_range):

#### Convolving the filter over the original image

def convolve(image, mask):
width = image.shape[1]
height = image.shape[0]

res_image = np.zeros((height, width))

# Iterate over every pixel that can be covered by the mask
for i in range(w_range,width-w_range):
for j in range(w_range,height-w_range):
# Then convolute with the mask
for k in range_inc(-w_range,w_range):
for h in range_inc(-w_range,w_range):
return res_image

#### Checking the zeros to mark the edges

def z_c_test(l_o_g_image):
z_c_image = np.zeros(l_o_g_image.shape)

# Check the sign (negative or positive) of all the pixels around each pixel
for i in range(1,l_o_g_image.shape[0]-1):
for j in range(1,l_o_g_image.shape[1]-1):
neg_count = 0
pos_count = 0
for a in range_inc(-1, 1):
for b in range_inc(-1,1):
if(a != 0 and b != 0):
if(l_o_g_image[i+a,j+b] < 0):
neg_count += 1
elif(l_o_g_image[i+a,j+b] > 0):
pos_count += 1

# If all the signs around the pixel are the same and they're not all zero, then it's not a zero crossing and not an edge.
# Otherwise, copy it to the edge map.
z_c = ( (neg_count > 0) and (pos_count > 0) )
if(z_c):
z_c_image[i,j] = 1

return z_c_image

And this is how we put it all together, to finally generate the image. Do note, we need to pass a grayscale image in the final function, not the original one.
So we will make one final function to take in the image and given the sigma value, generate the masked image.

def run_l_o_g(bin_image, sigma_val, size_val):

# Smooth the image by convolving with the LoG mask
print("smoothing")

# Display the smoothed imgage
blurred.imshow(l_o_g_image, cmap=cm.gray)

# Find the zero crossings
print("finding zero crossings")
z_c_image = z_c_test(l_o_g_image)
print(z_c_image)

#Display the zero crossings
edges.imshow(z_c_image, cmap=cm.gray)
pylab.show()

And finally you may experiment with other values of sigma, and see the changes in intensities.

If this seems like too much work, we can also implement it using the OpenCv library and the in-built functions. This is how we can implement it in Python.

### OpenCV Implementation

#### Steps:

• Remove the noise by applying the Gaussian Blur.
• Convert the image into grayscale.
• Apply Laplacian Filter.
• See the output.

We will use the OpenCV library to code this in. We will also implement the filters from scratch. The code for the same is followed after the base code.

#### Code:

Import the libraries:

import cv2
import matplotlib.pyplot as plt

source = cv2.imread('WallpaperStudio10-62073.jpg', cv2.IMREAD_COLOR)

Here the first argument is to define the path of the image and second defines how you want to read the image. Here we want to read the image in color hence we use IMREAD_COLOR.

Remove noise by applying Gaussian Blur:

source = cv2.GaussianBlur(source, (3, 3), 0)

Here, first argument is the image. Second defines the kernel size.

Convert to grayscale:

source_gray = cv2.cvtColor(source, cv2.COLOR_BGR2GRAY)

Apply the Laplacian Filter:

dest = cv2.Laplacian(source_gray, cv2.CV_16S, ksize=3)
abs_dest = cv2.convertScaleAbs(dest)

Show the output:

plt.imshow(abs_dst, cmap="gray")

Now you can see the Laplacian filter gets the mug edges clearly and also takes in the internal text on the mug.

Pretty decent right? This is how you can apply an edge detector by using Laplacian or Laplacian over Gaussian filter. With this article at OpenGenus, you must have the complete idea of Laplacian Filter. Enjoy.


text:
# shear , viscosity and expansion of universe

what is the meaning of expansion ,shear and viscosity in context of universe? how can we conclude a result after getting a numerical value of above terms?

-
The second question is better than the first here, in a sense. The problem is that "viscosity" is a term for fluids, shear applies to materials and expansion applies to anything with length. So what really is that first question? – Roy Simpson Mar 21 '11 at 11:14
@Roy, "shear" applies to fluids as well, without shear you cant measure viscosity. And materials can be liquid, solid or gaseous. – Georg Mar 21 '11 at 11:55
Indeed, so this makes the question extremely general, apparently about all of physics. – Roy Simpson Mar 21 '11 at 11:57
The Raychaudhuri equation (en.wikipedia.org/wiki/Raychaudhuri_equation) is often useful in modeling the expanding Universe, when we want to consider the possibility of inhomogeneous and/or anisotropic expansion. Shear and expansion are central there. – Ted Bunn Mar 21 '11 at 13:12
This theorem mentions "vorticity" not "viscosity" though - but it could be the question here. – Roy Simpson Mar 21 '11 at 16:48


text:
# Ahhh Double Star Question AGAIN

1. Jan 16, 2005

### NotaPhysicsMan

Ahhh Double Star Question...AGAIN!!

I think my prof really doesn't like us or maybe he simply has an over expectation of our abilities...

As the drawing shows (attached), a pond has the shape of an inverted cone with the top sliced off and has a depth of 5.00m. The atmospheric pressure is 1.01 x 10^5 PA. The circular top surface (radius=R2) and circular bottom surface (radius=R1) of the pond are both parallel to the ground. The magnitude of the force acting on the top surface is the same as the manitude of the force acting on the bottom surface. Obtain a)R2 and b)R1.

Okay, this looks like the question involving two equations of two unknowns. First I know that F1=F2... So P1A=P2A. The other I believe might have to do with the volume of this pond. A normal cone would have a volume of 1/3pie x r^2 x h. But since this is a cut-off cone, I'm lost here. Also, that 60 degrees is quite confusing. I think it may have to do with the area of the middle section when unfolded, a trapezoid shape perhaps? Any Ideas appreciated!
Thanks.

#### Attached Files:

• ###### cylinderprof.JPG
File size:
4.5 KB
Views:
66
2. Jan 16, 2005

### NotaPhysicsMan

*bump* Anyone?

3. Jan 16, 2005

### dextercioby

1.Compute the 2 forces and set them equal.Show your work.

Daniel.

4. Jan 16, 2005

### NotaPhysicsMan

ok, so P2 is simply atmospheric pressure which is
1.01 x 10^5. P1 can be found by P2=P1+pgh. That turns out to be pretty much the same as P2, so 1.01 x 10^5. Ok so the pressures P2=P1. Area is pie x r1^2 and pie x r2^2. Ok in terms of force:

F=PA
so P2 x (pie x r2^2)= P1 x (pie x r1^2). Since pie and Pressures are virtually the same, I can cancel them out. So I have r1^2=r2^2 which doesn't make any sense.

5. Jan 16, 2005

### dextercioby

Of course it doesn't.The pressures are DIFFERENT.U can't take them as equal.This is a trick in the problem...

Daniel.

6. Jan 16, 2005

### NotaPhysicsMan

P2 x (pie x r2^2)= P1 x (pie x r1^2). <---oK is this at least right? P1 has to be found using this formula, P2=P1+pgh. The pressure at the bottom will definitely be bigger than the top, but only by a factor of pgh, which is 49050. So P1= 1.50 x 10^5 Pa. Ok, now I have the two, now solve for one of the r's. ERR, what's the other equation! I think it may be the volume, some help.

7. Jan 16, 2005

### dextercioby

U don't need the volume.You need to apply the definition of 'tangent'.Think of the rightangle triangle in that trapese.And use the fact that u know ine side of the triangle (namely the height of the trapese) and one angle in the triangle (which is found by using that 60° which the problem gives you).

Daniel.

8. Jan 16, 2005

### NotaPhysicsMan

Yes, I thought that the 60 degrees could be used when you slice the cone in half you get a trapezoid. Maybe I could find the length of one the top or bottom and that would be the circumference. and C=2pie x r. r =C/2pie. I can see how the 60 degrees can be used to find the hypoteneuse and the base but I still can't get the either lengths no matter how many little triangles I cut the thing into...

9. Jan 16, 2005

### NotaPhysicsMan

I just can't seem to get it. HELP

10. Jan 16, 2005

### NotaPhysicsMan

BUMP!!! I'm still getting no where...

11. Jan 16, 2005

### dextercioby

You don't know geometry,if u can't prove this formula
$$\tan 30°=\frac{R_{2}-R_{1}}{h}$$

Then if u don't know geometry,how do u expect solving physics problems?? :uhh:

Daniel.

12. Jan 16, 2005

### NotaPhysicsMan

I did that from the start, but didn't realize that I could simply go R2-R1 and use that as my second equation. I thought I could find an actual number from the 60 degrees..Ah well.