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Coherent-State-Based Twin-Field Quantum Key Distribution

Abstract

Large-scale quantum communication networks are still a huge challenge due to the rate-distance limit of quantum key distribution (QKD). Recently, twin-field (TF) QKD has been proposed to overcome this limit. Here, we prove that coherent-state-based TF-QKD is a time-reversed entanglement protocol, where the entanglement generation is realized with entanglement swapping operation via an entangled coherent state measurement. We propose a coherent-state-based TF-QKD with optimal secret key rate under symmetric and asymmetric channels by using coherent state and cat state coding. Furthermore, we show that our protocol can be converted to all recent coherent-state-based TF-QKD protocols by using our security proof. By using the entanglement purification with two-way classical communication, we improve the transmission distance of all coherent-state-based TF-QKD protocols.

Introduction

Since the first quantum key distribution (QKD) experiment with 32 cm free-space channel1, a lot of efforts have been devoted to achieving long-distance QKD. Recently, the maximum distance of point-to-point QKD has been pushed up to 421 km ultralow-loss fiber2. Several experiments show that quantum-limited measurement3,4 and QKD5 can be demonstrated by using satellite-to-ground downlink with more than 1000 km free-space channel. Furthermore, measurement-device-independent (MDI) QKD6 has been performed over 404 km ultralow-loss fiber7 by using the optimal four-intensity set8, which is immune to any attack on detection9 by exploiting the Bell state measurement. Further increasing the fiber-based transmission distance without quantum repeater is a difficult obstacle to overcome. In the literature, without the help of trusted relay or quantum repeater, people believe that the limit is approximately 500 km fiber10. The strong evidence comes from the secret key agreement capacity of repeaterless quantum channel11,12, where the optimal rate is linear scaling with the transmittance of two communication parties, known as the repeaterless bound12.

A breakthrough called twin-field (TF) QKD13 has been proposed to break this bound, resulting in many variants14,15,16,17,18,19,20,21,22 and experimental demonstrations23,24,25,26. However, each security proof of the coherent-state-based TF-QKD14,18,19,20, or called phase-matching QKD, is carefully tailored. Cat state, superposition of coherent states with two opposite phases, as an important resource, has been widely used for quantum information processing, including quantum computation27, quantum teleportation28, quantum repeater29,30, QKD31 and quantum metrology32. Importantly, cat states have been successfully generated and exploited to demonstrate various quantum tasks33,34,35,36,37.

Here, we point out the physics in coherent-state-based TF-QKD is exactly entanglement swapping operation via the entangled coherent state (ECS) measurement. The coherent-state-based TF-QKD is a time-reversed entanglement protocol by using ECS measurement, which is similar with the MDI-QKD by using the Bell state measurement. We propose a coherent-state-based TF-QKD protocol under symmetric and asymmetric channels by using coherent state and cat state coding. The entanglement purification with one-way38,39 and two-way40 classical communication techniques are used to prove the security of our protocol against coherent attacks in the asymptotic regime. The secret key rate of our protocol is larger than refs14,18,19. Furthermore, we show that our protocol can be converted to other coherent-state-based protocols14,18,19,20 by using our security proof, which means that all coherent-state-based TF-QKD can be unified under a single framework. We consider the TF-QKD with two-way classical communication, which significantly improves the transmission distance of all coherent-state-based TF-QKD protocols with large misalignment.

Results

ECS measurement

Generally, the symmetric beam splitter (BS) and single-photon detectors are used to implement the interference measurement of TF-QKD13. One can assume that the two inputs of BS are a and b modes while the two output modes are $$\tilde{a}=(a+b)/\sqrt{2}$$ and $$\tilde{b}=(a-b)/\sqrt{2}$$. The four two-mode ECS forms41,42 are $$|{\Phi }^{\pm }\rangle =(|\alpha \rangle |\alpha \rangle \pm |\,-\,\alpha \rangle |\,-\,\alpha \rangle )/\sqrt{{N}_{\pm }}$$ and $$|{\Psi }^{\pm }\rangle =(|\alpha \rangle |\,-\,\alpha \rangle \pm |\,-\,\alpha \rangle |\alpha \rangle )/\sqrt{{N}_{\pm }}$$, where $${N}_{\pm }=2(1\pm {e}^{-4\mu })$$ are the normalization factors and $$\mu =|\,\pm \,\alpha {|}^{2}$$ is the intensity of coherent states $$|\,\pm \,\alpha \rangle$$. The four ECSs are sometimes called quasi-Bell states. The quantum states $$|\,\pm \,\alpha \rangle$$ constitute the quasi-computational basis while the quantum states $$|{\xi }^{\pm }(\alpha )\rangle =(|\alpha \rangle \pm |\,-\,\alpha \rangle )/\sqrt{2}$$ constitute the quasi-dual basis. After passing through the lossless symmetric BS, the four states become

$$\begin{array}{ll}|{\Phi }^{+}{\rangle }_{ab}\,\mathop{\longrightarrow }\limits^{{\rm{BS}}}\,|{\rm{even}}{\rangle }_{\tilde{a}}|0{\rangle }_{\tilde{b}}, & |{\Phi }^{-}{\rangle }_{ab}\,\mathop{\longrightarrow }\limits^{{\rm{BS}}}\,|{\rm{odd}}{\rangle }_{\tilde{a}}|0{\rangle }_{\tilde{b}},\\ |{\Psi }^{+}{\rangle }_{ab}\,\mathop{\longrightarrow }\limits^{BS}\,|0{\rangle }_{\tilde{a}}|{\rm{even}}{\rangle }_{\tilde{b}}, & |{\Psi }^{-}{\rangle }_{ab}\,\mathop{\longrightarrow }\limits^{BS}\,|0{\rangle }_{\tilde{a}}|{\rm{odd}}{\rangle }_{\tilde{b}},\end{array}$$
(1)

where $$|{\rm{even}}{\rangle }_{\tilde{a}}$$ ($$|{\rm{odd}}{\rangle }_{\tilde{a}}$$) means that the output mode $$\tilde{a}$$ contains even (odd) photon numbers. If we consider the case of ideal photon-number-resolving detector and lossless channel, one can unambiguously discriminate the four ECSs by performing photon-number parity measurement.

For the case of lossy channel and threshold detector, one can only discriminate the case with or without detector clicks. Generally, a successful detection event in TF-QKD13 is defined that one and only one detector clicks. Therefore, we make only detector L (R) clicking represent that the result of the ECS measurement is the state $$|{\Phi }^{-}\rangle$$ $$(|{\Psi }^{-}\rangle )$$. Due to decoherence of the cat states in lossy channel, the states $$|{\Phi }^{+}\rangle$$ and $$|{\Psi }^{+}\rangle$$ will always be mistakenly measured as quantum states $$|{\Phi }^{-}\rangle$$ and $$|{\Psi }^{-}\rangle$$, respectively. However, the corresponding probabilities can be restricted to be very low when the optical intensity is low and there is no eavesdropper’s disturbance. The post-selected joint quantum states of two legitimate users have quantum correlations, which then means that coherent-state-based TF-QKD have MDI characteristic.

TF-QKD with cat state

We introduce a coherent-state-based TF-QKD with coherent state and cat state coding, as shown in Fig. 1. State preparation. Alice (Bob) randomly chooses the Z and X bases with probabilities pZ and pX. For the Z basis, Alice (Bob) randomly prepares coherent state optical pulses $$|{\alpha }_{a(b)}\rangle$$ and $$|\,-\,{\alpha }_{a(b)}\rangle$$ with equal probabilities for the logic bits 0 and 1. For the X basis, Alice (Bob) randomly prepares cat state optical pulses $$|{\xi }^{+}({\alpha }_{a(b)})\rangle$$ and $$|{\xi }^{-}({\alpha }_{a(b)})\rangle$$ with equal probabilities for the logic bits 0 and 1. Entanglement measurement. Alice and Bob send the optical pulses to the untrusted Charlie through insecure quantum channel with efficiency $${\eta }_{a}$$ and $${\eta }_{b}$$ (with detector efficiency taken into account). Charlie is supposed to perform the ECS measurement. For example, he let the two optical pulses interfere in the symmetric BS which would be detected by two threshold detectors L and R. Announcement. Charlie publicly discloses whether he has obtained a successful measurement result and which ECS is acquired. Alice and Bob only keep the data of successful measurement and discard the rest. Reconciliation. Alice and Bob announce their bases over an authenticated classical channel. They only keep the data of the same basis and discard the rest. For the Z basis, Bob flips his key bit if Charlie announces a result with $$|{\Psi }^{-}\rangle$$. For the X basis, Bob always flips his key bit. Parameter estimation. The data of the Z basis are used for constituting raw key and calculating the gain QZ and quantum bit error rate (QBER) EZ of the Z basis. The data of the X basis are all announced to calculate QBER EX of the X basis. Key distillation. Alice and Bob exploit the error correction and privacy amplification to distill secret key.

Here, we prove that coherent-state-based TF-QKD with coherent state and cat state coding is secure against coherent attacks in the asymptotic regime. Any one of two successful detections is enough for proving the security. The coherent-state-based TF-QKD can be regarded as a time-reversed entanglement protocol, where Alice and Bob prepare maximally entangled state $$|\psi {\rangle }_{a^{\prime} a}={(|+z\rangle }_{a^{\prime} }|{\alpha }_{a}{\rangle }_{a(b)}+|\,-\,z{\rangle }_{a^{\prime} }|\,-\,{\alpha }_{a}{\rangle }_{a})/\sqrt{2}$$ and $$|\psi {\rangle }_{b^{\prime} b}={(|+z\rangle }_{b^{\prime} }|{\alpha }_{b}{\rangle }_{b}+|\,-\,z{\rangle }_{b^{\prime} }|\,-\,{\alpha }_{b}{\rangle }_{b})/\sqrt{2}$$, respectively. $$|\,\pm \,z\rangle$$ are the eigenstates of Pauli’s Z operator. Alice and Bob keep the qubit and send the optical mode to Charlie, who is supposed to perform entanglement swapping via the ECS measurement. Thereby, the bipartite states between Alice and Bob have quantum correlation through Charlie’s entanglement swapping operation. One can use the entanglement purification technique38 to distill maximally entangled state and generate the secret key. If Alice and Bob measure qubits before sending optical pulses in the virtual entanglement protocol, it will become the prepare-and-measurement protocol, i.e., coherent-state-based TF-QKD. Details can be found in Supplemental Material. The efficient QKD scheme43 can be directly applied, where we let $${p}_{Z}\approx 1$$ in the asymptotic limit. The secret key rate of our coherent-state-based TF-QKD with one-way classical communication39 in the asymptotic limit is

$$R={Q}_{Z}[1-fh({E}_{Z})-h({E}_{X})],$$
(2)

where $$h(x)=-\,x\,{\log }_{2}\,x-(1-x)\,{\log }_{2}\,(1-x)$$ is the Shannon entropy and $$f=1.16$$ is the error correction efficiency. In our simulation, without Charlie’s disturbance, we have gain $${Q}_{Z}=(1-{p}_{d})[1-(1-2{p}_{d}){e}^{-2x}]$$, QBERs $${E}_{Z}=(1-{p}_{d})[{e}_{{d}_{Z}}(1-{e}^{-2x})+{p}_{d}{e}^{-2x}]$$ and $${E}_{X}=\frac{1}{2}\{1+{e}^{-2({\mu }_{a}+{\mu }_{b})}[1-(1-2{p}_{d}){e}^{2x}]/[1-(1-2{p}_{d}){e}^{-2x}]\}$$, where $$x={\mu }_{a}{\eta }_{a}={\mu }_{b}{\eta }_{b}$$, $${e}_{{d}_{Z}}$$ is the misalignment rate of the Z basis and pd is the dark count rate. The misalignment of the X basis can be neglected since Bob always flips his bit.

Here, we exploit the two-way entanglement purification40 into our coherent-state-based TF-QKD protocol to increase transmission distance. Specifically, before implementing Key distillation step, Alice randomly permutes all raw key bits and divides them into two groups, Bob does the same. Alice and Bob compute a parity on raw key of two groups and compare the parities. The second group is always discarded. If their parities are the same, they keep the bit of the first group. Otherwise, they discard it. One can repeat the above operation once for each B step. After kth B step is applied, the gain, bit and phase error rates can be given by $${Q}_{Z}^{k}=\frac{1}{2}{A}^{k-1}{Q}_{Z}^{k-1}$$, $${E}_{Z}^{k}={({E}_{Z}^{k-1})}^{2}/{A}^{k-1}$$, and $${E}_{X}^{k}=2{E}_{X}^{k-1}(1-{E}_{Z}^{k-1}-{E}_{X}^{k-1})/{A}^{k-1}$$, where we have $${A}^{k-1}={(1-{E}_{Z}^{k-1})}^{2}+{({E}_{Z}^{k-1})}^{2}$$, $${Q}_{Z}^{0}={Q}_{Z}$$, $${E}_{Z}^{0}={E}_{Z}$$ and $${E}_{X}^{0}={E}_{X}$$. The secret key rate of our coherent-state-based TF-QKD after kth B step in the asymptotic limit is

$${R}^{k}={Q}_{Z}^{k}[1-fh({E}_{Z}^{k})-h({E}_{X}^{k})].$$
(3)

Converting to other protocols

Without loss of generality, let positive-operator valued measure $${E}_{10}$$ and $${E}_{01}$$ ($${E}_{s}={E}_{10}+{E}_{01}$$) denote the successful measurement results with ECSs $$|{\Phi }^{-}\rangle$$ and $$|{\Psi }^{-}\rangle$$; let $$\hat{{\rm{P}}}(|u,v\rangle ):\,=|u\rangle \langle u|\otimes |v\rangle \langle v|$$ with $$|u,v\rangle =|u\rangle |v\rangle$$. The density matrix of the Z and X bases are $${\rho }_{Z}=\frac{1}{4}[\hat{{\rm{P}}}(|{\alpha }_{a},{\alpha }_{b}\rangle )$$$$+\hat{{\rm{P}}}(|{\alpha }_{a},-\,{\alpha }_{b}\rangle )$$ + $$\hat{{\rm{P}}}(|\,-\,{\alpha }_{a},{\alpha }_{b}\rangle )+\hat{{\rm{P}}}(|\,-\,{\alpha }_{a},-\,{\alpha }_{b}\rangle )]$$ and $${\rho }_{X}=\frac{1}{4}[\hat{{\rm{P}}}(|{\xi }^{+}({\alpha }_{a}),{\xi }^{+}({\alpha }_{b})\rangle )+$$$$\hat{{\rm{P}}}(|{\xi }^{+}({\alpha }_{a}),{\xi }^{-}({\alpha }_{b})\rangle )$$ + $$\hat{{\rm{P}}}(|{\xi }^{-}({\alpha }_{a}),{\xi }^{+}({\alpha }_{b})\rangle )+\hat{{\rm{P}}}(|{\xi }^{-}({\alpha }_{a}),{\xi }^{-}({\alpha }_{b})\rangle )]$$, where we have $${\rho }_{X}\equiv {\rho }_{Z}=\rho$$. For the cases of $$|{\xi }^{+}({\alpha }_{a}),{\xi }^{+}({\alpha }_{b})\rangle$$ and $$|{\xi }^{-}({\alpha }_{a}),{\xi }^{-}({\alpha }_{b})\rangle$$, they always generate the error gain since Bob always flips his bit, and the corresponding density matrix is $${\rho }_{X}^{E}=\frac{1}{4}[\hat{{\rm{P}}}(|{\xi }^{+}({\alpha }_{a}),{\xi }^{+}({\alpha }_{b})\rangle )+\hat{{\rm{P}}}(|{\xi }^{-}({\alpha }_{a}),{\xi }^{-}({\alpha }_{b})\rangle )]$$. Let $${Q}_{X}$$ and $${Q}_{X}^{E}$$ represent the gain and error gain of the X basis. In the case of asymptotic limit, we always have $${Q}_{X}\equiv {Q}_{Z}={\rm{Tr}}(\rho {E}_{s})$$ and $${Q}_{X}^{E}={\rm{Tr}}({\rho }_{X}^{E}{E}_{s})$$. Therefore, the QBER EX (phase error rate of the Z basis) in the asymptotic limit can be given by $${E}_{X}={Q}_{X}^{E}/{Q}_{X}={Q}_{X}^{E}/{Q}_{Z}$$. If one can acquire an upper bound of $${Q}_{X}^{E}$$, the QBER EX can be bounded.

By using the entanglement purification with one-way38,39 and two-way40 classical communication to prove security, we only require the estimation of the QBER EX. This means that we do not need to prepare cat state if we can acquire the QBER EX through alternative method. The alternative method need to ensure that the prepared state by Alice (Bob) is linearly dependent, which cannot allow Charlie to implement unambiguous-state-discrimination attack44 before performing the entanglement swapping. Here, we show that our protocol can be converted to the coherent-state-based protocols of refs14,18,19,20 by using our security proof.

For the protocol proposed in ref.20, Alice and Bob randomly prepare coherent state $$|{e}^{i{\theta }_{a}}\sqrt{{\mu }_{a}}\rangle$$ and $$|{e}^{i{\theta }_{b}}\sqrt{{\mu }_{b}}\rangle$$ if they choose the X basis, where they need phases $${\theta }_{a(b)}\in [0,2\pi )$$ and infinite intensities $${\mu }_{a(b)}$$. As pointed out in ref.20, the operator $${\rho }_{X}^{E}$$ can be approximated to arbitrary precision in the Hilbert-Schmidt norm by the discrete diagonal coherent state representation $${\rho }_{X}^{E}={\sum }_{i=1}^{\infty }\,{\lambda }_{i}\hat{{\rm{P}}}(|{\omega }_{a}^{i},{\omega }_{b}^{i}\rangle )$$, where $$|{\omega }_{a}^{i},{\omega }_{b}^{i}\rangle$$ is the tensor product of coherent state and λi is complex number. Thereby, the error gain $${Q}_{X}^{E}$$ can be precisely obtained by using coherent states with infinite intensities. Indeed, in the ideal situation with symmetric channel, $$\mu ={\mu }_{a}={\mu }_{b}$$ and $$\eta ={\eta }_{a}={\eta }_{b}$$, the secret key rate of this protocol by using our security proof with one-way classical communication is given by $$R=(1-{e}^{-\mu \eta })[1-h(\frac{1-{e}^{-4\mu +2\mu \eta }}{2})]$$, which is the same with the results of ref.20.

For the protocol proposed in refs18,19, Alice and Bob randomly prepare phase-randomized coherent state if they choose the X basis. As pointed out in refs18,19, by using the Cauchy-Schwarz inequality, we can bound the error gain $${Q}_{X}^{E}$$ with photon-number state, i.e., $${Q}_{X}^{E}={\rm{Tr}}({\rho }_{X}^{E}{E}_{s})$$$${({\sum }_{n,m=0}^{\infty }\sqrt{{P}_{2n}^{a}{P}_{2m}^{b}{Y}_{2n,2m}})}^{2}$$ + $${({\sum }_{n,m=0}^{\infty }\sqrt{{P}_{2n+1}^{a}{P}_{2m+1}^{b}{Y}_{2n+1,2m+1}})}^{2}$$. $${Y}_{n,m}$$ is the yield given that Alice and Bob send $$n$$ and $$m$$ photon states and $${P}_{n}^{a(b)}={e}^{-{\mu }_{a(b)}}{\mu }_{a(b)}^{n}/n!$$. Decoy-state method45,46,47 can be used to estimate the yield $${Y}_{n,m}$$, which has been realized with finite intensities18,48. Here, we use the three-intensity, $$0 < \omega < \nu$$, to estimate the yield, which can be found in Methods.

For the protocol proposed in ref.14, Alice and Bob always prepare the coherent state $$|{e}^{i({\theta }_{a}+{\kappa }_{a}\pi )}\sqrt{{\mu }_{a}}\rangle$$ and $$|{e}^{i({\theta }_{b}+{\kappa }_{b}\pi )}\sqrt{{\mu }_{b}}\rangle$$, where $${\kappa }_{a(b)}\in \{0,1\}$$ and $${\theta }_{a(b)}\in [0,2\pi )$$. They keep the raw key bit only if $$|{\theta }_{a}-{\theta }_{b}|=0$$ or π. As pointed out in ref.14, one can introduce a virtual trusted party who prepares a state, splits it using symmetric BS, and sends it to both Alice and Bob. We have the following observations $$\hat{{\rm{P}}}(|\alpha \rangle )+\hat{{\rm{P}}}(|\,-\,\alpha \rangle )=\hat{{\rm{P}}}(|{\xi }^{+}(\alpha )\rangle )+\hat{{\rm{P}}}(|{\xi }^{-}(\alpha )\rangle )$$ and $$\hat{{\rm{P}}}(|{\xi }^{+}(\sqrt{2}\alpha ),0\rangle )+\hat{{\rm{P}}}(|0,{\xi }^{+}(\sqrt{2}\alpha )\rangle )$$ $$\mathop{\longrightarrow }\limits^{{\rm{BS}}}$$ $$\hat{{\rm{P}}}(|{\xi }^{+}(\alpha ),{\xi }^{+}(\alpha )\rangle )+\hat{{\rm{P}}}(|{\xi }^{-}(\alpha ),{\xi }^{-}(\alpha )\rangle )$$. For the post-selected phase-matching, the error gain can be given by $${Q}_{X}^{E}={\sum }_{n=0}^{\infty }\,{e}^{-2\mu }{(2\mu )}^{2n}{Y}_{2n}/(2n)!$$, where we need to assume $${\mu }_{a}={\mu }_{b}=\mu$$. $${Y}_{n}$$ is the yield given that the total photon number sent by Alice and Bob is n. Only even photon numbers have contribution to the phase error rate in our security proof which is only the same with the results of ref.14 in the ideal situation. Different from protocols of ours and refs18,19,20, the protocol of ref.14 seems to be only suitable for symmetric channel.

Discussion

For simulation, we use the following parameters. The inherent loss of fiber is 0.16 dB/km, the efficiency and dark count rate of threshold single-photon detector are $${\eta }_{d}=85 \%$$ and $${p}_{d}={10}^{-7}$$. For simplicity, we let Protocol 1 represent our coherent state and cat state coding protocol. Let Protocol 2 represent the protocol in refs18,19 with three-intensity phase-randomized coherent state. Here, we fix the intensities of phase-randomized coherent state with $$\nu =0.1$$ and $$\omega =0.02$$. The secret key rate of our protocol is equal to that of protocol in ref.20 since cat state can be approximated to arbitrary precision in the Hilbert-Schmidt norm by the discrete diagonal coherent state representation20. The performance of Protocols 1 and 2 under symmetric channel have been shown in Fig. 2, which assumes 1 GHz system repetition rate23. Here, we do not consider the performance of protocols in ref.14, whose secret key rate has been shown lower than protocols in refs 18,19,20. The secret key rate and transmission distance of Protocol 1 are both superior to Protocol 2. The transmission distance of Protocols 1 and 2 can both be improved by using the two-way classical communication. Especially, the advantages are very clear when the system misalignment rate is large, like $${e}_{{d}_{Z}}=10 \%$$. The large system misalignment of TF-QKD is reasonable since the phase-locking and long-distance phase stabilization techniques in field are still difficult even with some experimental progresses23,24,25,26. The performance of Protocols 1 and 2 under asymmetric channel have been shown in Fig. 3. The secret key rate of Protocol 1 can still surpass the repeaterless bound when the asymmetric channel ratio is 70%. The asymmetric channel ratio is the ratio between A-C and A-B, where A-C (B) represents the distance between Alice and Charlie (Bob). Compare Fig. 2 with 3, the performance of coherent-state-based TF-QKD is the best under symmetric channel due to the single-photon-type interference.

In summary, we propose a coherent-state-based TF-QKD with optimal secret key rate by using coherent state and cat state coding. By using the ECS measurement as the entanglement swapping operation, we unify all known coherent-state-based TF-QKD protocols under a single framework. We have proved that the coherent-state-based TF-QKD is a time-reversed entanglement protocol, which means that one can use the known techniques of qubit-based QKD to further develop the coherent-state-based TF-QKD. The results show that coherent-state-based TF-QKD is suitable for building quantum communication networks within hundreds of kilometers without trusted relay or quantum repeater. We remark that cat states used for high-speed QKD will have difficulty under current experimental conditions. However, current experiments on the demonstration of various quantum tasks with cat states are very active33,34,35,36,37, implying that cat states may become a practical resource with the rapid development of technology.

Methods

Decoy-state analysis

The phase-randomized coherent state can be seen as a mixture of Fock states. Let $${Q}_{a,b}$$ represent the gain when Alice and Bob send phase-randomized coherent state with intensity $$a$$ and $$b$$, respectively. Let $${Y}_{n,m}$$ represent the yield when Alice and Bob send n-photon and m-photon. Thereby, we have $${Q}_{a,b}={\sum }_{n=0}^{\infty }\,{\sum }_{m=0}^{\infty }\,{e}^{-a-b}\frac{{a}^{n}{b}^{m}}{n!m!}{Y}_{n,m}$$. Here, we exploit the decoy-state method with three-intensity to estimate the upper bound of the yield $${Y}_{n,m}$$ with analytical method. The upper bound of $${Y}_{\mathrm{1,1}}$$, $${Y}_{\mathrm{0,2}}$$ and $${Y}_{\mathrm{2,0}}$$ can be given by

$${Y}_{1,1}\le \frac{{e}^{2\omega }{Q}_{\omega ,\omega }-{e}^{\omega }({Q}_{\omega ,0}+{Q}_{0,\omega })+{Q}_{0,0}}{{\omega }^{2}},$$
(4)
$${Y}_{0,2}\le \frac{\omega {e}^{\nu }{Q}_{0,\nu }-\nu {e}^{\omega }{Q}_{0,\omega }+(\nu -\omega ){Q}_{0,0}}{\nu \omega (\nu -\omega )/2},$$
(5)

and

$${Y}_{2,0}\le \frac{\omega {e}^{\nu }{Q}_{\nu ,0}-\nu {e}^{\omega }{Q}_{\omega ,0}+(\nu -\omega ){Q}_{0,0}}{\nu \omega (\nu -\omega )/2},$$
(6)

where we have $${Y}_{0,0}={Q}_{0,0}$$. The upper bound of $${Y}_{\mathrm{0,}n}$$ and $${Y}_{n\mathrm{,0}}$$ with $$n\ge 3$$ can be written as

$${Y}_{0,n}\le \,{\rm{\min }}\{1,\frac{\omega {e}^{\nu }{Q}_{0,\nu }-\nu {e}^{\omega }{Q}_{0,\omega }+(\nu -\omega ){Q}_{0,0}}{\nu \omega ({\nu }^{n-1}-{\omega }^{n-1})/n!}\},$$
(7)

and

$${Y}_{n,0}\le \,{\rm{\min }}\{1,\frac{\omega {e}^{\nu }{Q}_{\nu ,0}-\nu {e}^{\omega }{Q}_{\omega ,0}+(\nu -\omega ){Q}_{0,0}}{\nu \omega ({\nu }^{n-1}-{\omega }^{n-1})/n!}\}.$$
(8)

Let $${F}_{x,y}={e}^{x+y}{Q}_{x,y}-{e}^{x}{Q}_{x,0}-{e}^{y}{Q}_{0,y}+{Q}_{0,0}$$, the upper bound of $${Y}_{\mathrm{1,}n}$$ and $${Y}_{n\mathrm{,1}}$$ with $$n\ge 2$$ can be given by

$${Y}_{1,n}\le \,{\rm{\min }}\{1,\frac{\omega {F}_{\omega ,\nu }-\nu {F}_{\omega ,\omega }}{({\nu }^{n}{\omega }^{2}-\nu {\omega }^{n+1})/n!}\},$$
(9)

and

$${Y}_{n,1}\le \,{\rm{\min }}\{1,\frac{\omega {F}_{\nu ,\omega }-\nu {F}_{\omega ,\omega }}{({\nu }^{n}{\omega }^{2}-\nu {\omega }^{n+1})/n!}\}.$$
(10)

Similarly, the upper bound of $${Y}_{n,m}$$ with $$n,m\ge 2$$ can be given by

$${Y}_{n,m}\le \,{\rm{\min }}\{1,\frac{{\omega }^{2}{F}_{\nu ,\nu }-\nu \omega ({F}_{\nu ,\omega }+{F}_{\omega ,\nu })+{\nu }^{2}{F}_{\omega ,\omega }}{{\nu }^{2}{\omega }^{2}({\nu }^{n-1}-{\omega }^{n-1})({\nu }^{m-1}-{\omega }^{m-1})/n!m!}\}.$$
(11)

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Grasselli, F. & Curty, M. Practical decoy-state method for twin-field quantum key distribution. New J. Phys. 21, 073001 (2019).

Acknowledgements

We gratefully acknowledges support from the National Natural Science Foundation of China under Grant No. 61801420 and Nanjing University.

Author information

Authors

Contributions

H.-L.Y. and Z.-B.C. conceived and designed the study. H.-L.Y. performed the numerical simulation. All authors contributed extensively to the work presented in this paper.

Corresponding authors

Correspondence to Hua-Lei Yin or Zeng-Bing Chen.

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.

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Reprints and Permissions

Yin, H., Chen, Z. Coherent-State-Based Twin-Field Quantum Key Distribution. Sci Rep 9, 14918 (2019). https://doi.org/10.1038/s41598-019-50429-0

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## bbc bitesize physics

December 30, 2020 in Uncategorized

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format: plain


text:
# predict future price

Show 40 post(s) from this thread on one page
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• July 28th 2009, 02:53 AM
isaacm1
predict future price
Hello,
I need to get a prediction for 4 days ahead for an item price, I have only the last 10 days prices history of that item.

what would be the equation to solve this ?

Thanks
• July 28th 2009, 03:52 AM
pickslides
There are many different ways to predict the next 4 values. It can be dependant on many things, what were the previous 10 values you have?
• July 28th 2009, 05:03 AM
isaacm1
11.4124 , 11.4196 , 11.4289 , 11.4127, 11.4076 , 11.3988 , 11.3821 ,11.3798 , 11.3926 ,11.4025

of course most accurate method will be better but please post two if possible

Thanks
• July 28th 2009, 01:16 PM
isaacm1
• July 28th 2009, 02:12 PM
pickslides
Be patient my friend, they are many posters from all different corners of the world on this forum so everyone is on a different time zone, since my last post I have enjoyed some very deep sleep, 8 of the best you could say.

Looking at your data and considering it is daily I would suggest the next 4 values could be decreasing. There seems to be a peak at the 3rd value and then after the 8th value it increases again. 10 values is a very small set to make a prediction on. There seem to be 2 weekly increases after values 1,2,8 & 9. These could very well be weekends.

• July 28th 2009, 03:09 PM
isaacm1
sorry, the numbers are prices, it is a challenge question from someone nothing more

any reasonable prediction will do

Thanks
• July 28th 2009, 03:19 PM
pickslides
Quote:

Originally Posted by isaacm1
any reasonable prediction will do

11.4124
11.4196
11.4289
11.4127
11.4076
11.3988
11.3821
11.3798
11.3926
11.4025
11.38289
11.37689
11.37237
11.37138
• July 28th 2009, 04:17 PM
isaacm1
thanks but I need equation , he also mention to use the average of all 10 days , does that help ?
• July 28th 2009, 05:54 PM
pickslides

11.4124
11.4196
11.4289
11.4127
11.4076
11.3988
11.3821
11.3798
11.3926
11.4025

you could use a linear least squares regression, there are many options. This would give the equation $y = -0.0038x+11.425$
• July 29th 2009, 12:29 AM
isaacm1
Thanks, where x is the number of future days ?
• July 29th 2009, 01:51 PM
pickslides
yep, x=11,12,13,14 for the next 4 values. This result is linear and will tend to zero as $x \rightarrow \infty$
• July 29th 2009, 02:06 PM
isaacm1
Many Thanks,

one question please how you get the -0.0038 value ?
• July 29th 2009, 02:25 PM
pickslides
It is the gradient for y=bx+a from the least squares regression. It is a long a boring explantion but if you really want to know check out

Least-Squares Line
• July 29th 2009, 02:50 PM
isaacm1
if my numbers changed I need to know how to get the "b" value (-0.0038)
• July 29th 2009, 02:58 PM
pickslides
Quote:

Originally Posted by pickslides

How to find 'b' is in this link my firend. (Rofl)
Show 40 post(s) from this thread on one page
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text:
# Thread: Trigo graph stucked @ b

1. ## Trigo graph stucked @ b

a) Sketch the graph of $y=|2cosx|+1$ for 0 ≤ x ≤ 2 $\pi$, stating the coordinates of the turning points.

b) State the number of solutions for the equation $|2cosx|+1 = \frac{x}{2}$ for 0 ≤ x ≤ 2 $\pi$
I have already solved for part a, having problems with part b...

I am guessing that we have to equate $y=\frac{x}{2}$

2. Originally Posted by Punch
a) Sketch the graph of $y=|2cosx|+1$ for 0 ≤ x ≤ 2 $\pi$, stating the coordinates of the turning points.

b) State the number of solutions for the equation $|2cosx|+1 = \frac{x}{2}$ for 0 ≤ x ≤ 2 $\pi$
I have already solved for part a, having problems with part b...
Have you considered graphing the functions over that domain?

3. Originally Posted by Prove It
Have you considered graphing the functions over that domain?
Yes, drawing a line would solve the question. However, I do not know how to draw the line y=\frac{x}{2} on this graph...

4. Originally Posted by Punch
Yes, drawing a line would solve the question. However, I do not know how to draw the line y=\frac{x}{2} on this graph...
when x=2 , y=1

when x=4 , y=2

now you have 2 points , connect the 2 points and wherever it intersects your first function would be the solutions .

when x=2 , y=1

when x=4 , y=2

now you have 2 points , connect the 2 points and wherever it intersects your first function would be the solutions .
But on the x-axis is all in $\pi$... Do i convert $\pi$ to values?

6. Originally Posted by Punch
But on the x-axis is all in $\pi$... Do i convert $\pi$ to values?
pi is 3.142 (in radians) .


text:
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### Theory:

In the previous section, we learned about acceleration due to gravity ($$g$$) and its relation with universal gravitational constant ($$G$$). In this section, we will compute the mass of the earth.

We know that,

$\begin{array}{l}g\phantom{\rule{0.147em}{0ex}}=\phantom{\rule{0.147em}{0ex}}\frac{G×M}{{R}^{2}}\phantom{\rule{0.147em}{0ex}}\\ \\ \mathit{Where},\\ g\phantom{\rule{0.147em}{0ex}}-\phantom{\rule{0.147em}{0ex}}\mathit{Acceleration}\phantom{\rule{0.147em}{0ex}}\mathit{due}\phantom{\rule{0.147em}{0ex}}\mathit{to}\phantom{\rule{0.147em}{0ex}}\mathit{gravity}\\ \\ G\phantom{\rule{0.147em}{0ex}}-\phantom{\rule{0.147em}{0ex}}\mathit{Universal}\phantom{\rule{0.147em}{0ex}}\mathit{gravitational}\phantom{\rule{0.147em}{0ex}}\mathit{constant}\\ \\ M\phantom{\rule{0.147em}{0ex}}-\phantom{\rule{0.147em}{0ex}}\mathit{Mass}\phantom{\rule{0.147em}{0ex}}\mathit{of}\phantom{\rule{0.147em}{0ex}}\mathit{the}\phantom{\rule{0.147em}{0ex}}\mathit{Earth}\\ \\ R\phantom{\rule{0.147em}{0ex}}-\phantom{\rule{0.147em}{0ex}}\mathit{Radius}\phantom{\rule{0.147em}{0ex}}\mathit{of}\phantom{\rule{0.147em}{0ex}}\mathit{the}\phantom{\rule{0.147em}{0ex}}\mathit{Earth}\end{array}$

Rearranging the above equation, the mass of the Earth ($$M$$ is obtained as follows:

$\begin{array}{l}\mathit{Mass}\phantom{\rule{0.147em}{0ex}}\mathit{of}\phantom{\rule{0.147em}{0ex}}\mathit{the}\phantom{\rule{0.147em}{0ex}}\mathit{Earth}\phantom{\rule{0.147em}{0ex}}\left(M\right)\phantom{\rule{0.147em}{0ex}}=\phantom{\rule{0.147em}{0ex}}\frac{g×{R}^{2}}{G}\\ \\ \mathit{We}\phantom{\rule{0.147em}{0ex}}\mathit{know}\phantom{\rule{0.147em}{0ex}}\mathit{that},\\ \\ \mathit{Acceleration}\phantom{\rule{0.147em}{0ex}}\mathit{due}\phantom{\rule{0.147em}{0ex}}\mathit{to}\phantom{\rule{0.147em}{0ex}}\mathit{gravity}\phantom{\rule{0.147em}{0ex}}\left(g\right)\phantom{\rule{0.147em}{0ex}}=\phantom{\rule{0.147em}{0ex}}9.8\phantom{\rule{0.147em}{0ex}}m{s}^{-1}\\ \\ \mathit{Universal}\phantom{\rule{0.147em}{0ex}}\mathit{gravitational}\phantom{\rule{0.147em}{0ex}}\mathit{constant}\phantom{\rule{0.147em}{0ex}}\left(G\right)\phantom{\rule{0.147em}{0ex}}=\phantom{\rule{0.147em}{0ex}}6.674\phantom{\rule{0.147em}{0ex}}×\phantom{\rule{0.147em}{0ex}}{10}^{-11}\phantom{\rule{0.147em}{0ex}}N{m}^{2}{\mathit{kg}}^{-2}\\ \\ \mathit{Radius}\phantom{\rule{0.147em}{0ex}}\mathit{of}\phantom{\rule{0.147em}{0ex}}\mathit{the}\phantom{\rule{0.147em}{0ex}}\mathit{Earth}\phantom{\rule{0.147em}{0ex}}\left(R\right)\phantom{\rule{0.147em}{0ex}}=\phantom{\rule{0.147em}{0ex}}6378\phantom{\rule{0.147em}{0ex}}\mathit{km}\end{array}$

Substituting the known values of $$g$$, $$R$$ and $$G$$, you can calculate the mass of the Earth.

Variation of acceleration due to gravity ($$g$$):

We know that the acceleration due to gravity ($$g$$) is inversely proportional to the square of the geometric radius of the Earth ($g\phantom{\rule{0.147em}{0ex}}\propto \phantom{\rule{0.147em}{0ex}}\frac{1}{{R}^{2}}$); its value changes from one place to another on the surface of the Earth.

In the equatorial region, the geometric radius of the Earth is maximum, and in the polar region, it is minimum. So, the '$$g$$' value is maximum in the polar region and minimum in the equatorial region.

When you go to a higher altitude from the surface of the Earth, the value of '$$g$$' decreases. In the same way, when you go deep below the surface of the Earth, the value of '$$g$$' decreases. The value of $$g$$ is zero at the centre of the Earth.


text:
# Definition:Integral Domain/Definition 1

## Definition

An integral domain $\struct {D, +, \circ}$ is:

a commutative ring which is non-null
with a unity
in which there are no (proper) zero divisors, that is:
$\forall x, y \in D: x \circ y = 0_D \implies x = 0_D \text{ or } y = 0_D$

that is, in which all non-zero elements are cancellable.


text:
## A direction splitting approach for incompressible Brinkmann flow

• The direction splitting approach proposed earlier in [6], aiming at the efficient solution of Navier-Stokes equations, is extended and adopted here to solve the Navier-Stokes-Brinkman equations describing incompressible flows in plain and in porous media. The resulting pressure equation is a perturbation of the incompressibility constrained using a direction-wise factorized operator as proposed in [6]. We prove that this approach is unconditionally stable for the unsteady Navier-Stokes-Brinkman problem. We also provide numerical illustrations of the method's accuracy and efficiency.

### Additional Services

Author: T. Gornak, J. L. Guermond, O. Iliev, P. D. Minev urn:nbn:de:hbz:386-kluedo-33182 Berichte des Fraunhofer-Instituts für Techno- und Wirtschaftsmathematik (ITWM Report) (216) Report English 2012/10/11 2012 Fraunhofer-Institut für Techno- und Wirtschaftsmathematik 2012/10/11 16 Fraunhofer (ITWM) 5 Naturwissenschaften und Mathematik / 51 Mathematik / 510 Mathematik Standard gemäß KLUEDO-Leitlinien vom 10.09.2012

$Rev: 13581$


text:
Satellite Motion: Speed & Period Video Lessons

Concept

Problem: Satellites in near-earth orbit experience a very slight drag due to the extremely thin upper atmosphere. These satellites slowly but surely spiral inward, where they finally burn up as they reach the thicker lower levels of the atmosphere. The radius decreases so slowly that you can consider the satellite to have a circular orbit at all times.As a satellite spirals inward, does it speed up, slow down, or maintain the same speed?

FREE Expert Solution

In this problem, we need to know Kepler's law of planetary motion, which is stated as:

$\overline{)\begin{array}{rcl}{\mathbf{T}}^{\mathbf{2}}& {\mathbf{\propto }}& {\mathbf{R}}^{\mathbf{3}}\\ {\mathbf{T}}& {\mathbf{\propto }}& {\mathbf{R}}^{\frac{\mathbf{2}}{\mathbf{3}}}\end{array}}$, where T is the time period and R is the radius of the orbit.

Problem Details

Satellites in near-earth orbit experience a very slight drag due to the extremely thin upper atmosphere. These satellites slowly but surely spiral inward, where they finally burn up as they reach the thicker lower levels of the atmosphere. The radius decreases so slowly that you can consider the satellite to have a circular orbit at all times.

As a satellite spirals inward, does it speed up, slow down, or maintain the same speed?


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# How to use a pebble smartwatch as a pomodoro timer

When using the pomodoro technique, I’ve found it useful to have access to the pomodoro timer in the lowest-friction way possible. In particular, I like having the timer on my wrist. I implemented this with a Pebble watch, which I bought used for £40.

With the timer on your wrist, you can walk away from your computer during breaks, without having to go back and check how much break time you have left. During long breaks (usually 20 minutes every 4 pomos), you can enjoy a walk outside the office (pictured below). The watch will also vibrate at the beginning and end of every pomo.

The benefits of having the timer on your wrist instead of your phone may seem trivial. In practise, I have found that even a little bit of friction can cause me to fall off the pomodoro wagon.

The Pebble company is now defunct, but the watches can easily be found on eBay. The Pebble app store has also been shut down, but I still have the .pbw files of two pomodoro apps: Solanum and Simple Pomodoro. You can sideload these apps onto your Pebble through the Pebble smartphone app. The pebble smartphone app still exists on the Google and Apple app stores, and could itself be sideloaded if it ever disappears.

I prefer Simple Pomodoro, because it supports long breaks. Solanum is more beautiful.

It’s possible to use a Pebble pomo timer in tandem with any other timer. I like having a timer on my computer screen. Just start the two timers at the same time and they’ll keep going in sync.

December 28, 2018


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# 0.6 Quadratic functions and graphs

 Page 1 / 2

## Introduction

In Grade 10, you studied graphs of many different forms. In this chapter, you will learn a little more about the graphs of quadratic functions.

## Functions of the form $y=a{\left(x+p\right)}^{2}+q$

This form of the quadratic function is slightly more complex than the form studied in Grade 10, $y=a{x}^{2}+q$ . The general shape and position of the graph of the function of the form $f\left(x\right)=a{\left(x+p\right)}^{2}+q$ is shown in [link] .

## Investigation : functions of the form $y=a{\left(x+p\right)}^{2}+q$

1. On the same set of axes, plot the following graphs:
1. $a\left(x\right)={\left(x-2\right)}^{2}$
2. $b\left(x\right)={\left(x-1\right)}^{2}$
3. $c\left(x\right)={x}^{2}$
4. $d\left(x\right)={\left(x+1\right)}^{2}$
5. $e\left(x\right)={\left(x+2\right)}^{2}$
Use your results to deduce the effect of $p$ .
2. On the same set of axes, plot the following graphs:
1. $f\left(x\right)={\left(x-2\right)}^{2}+1$
2. $g\left(x\right)={\left(x-1\right)}^{2}+1$
3. $h\left(x\right)={x}^{2}+1$
4. $j\left(x\right)={\left(x+1\right)}^{2}+1$
5. $k\left(x\right)={\left(x+2\right)}^{2}+1$
Use your results to deduce the effect of $q$ .
3. Following the general method of the above activities, choose your own values of $p$ and $q$ to plot 5 different graphs (on the same set of axes) of $y=a{\left(x+p\right)}^{2}+q$ to deduce the effect of $a$ .

From your graphs, you should have found that $a$ affects whether the graph makes a smile or a frown. If $a<0$ , the graph makes a frown and if $a>0$ then the graph makes a smile. This was shown in Grade 10.

You should have also found that the value of $q$ affects whether the turning point of the graph is above the $x$ -axis ( $q<0$ ) or below the $x$ -axis ( $q>0$ ).

You should have also found that the value of $p$ affects whether the turning point is to the left of the $y$ -axis ( $p>0$ ) or to the right of the $y$ -axis ( $p<0$ ).

These different properties are summarised in [link] . The axes of symmetry for each graph is shown as a dashed line.

 $p<0$ $p>0$ $a>0$ $a<0$ $a>0$ $a<0$ $q\ge 0$ $q\le 0$

## Domain and range

For $f\left(x\right)=a{\left(x+p\right)}^{2}+q$ , the domain is $\left\{x:x\in \mathbb{R}\right\}$ because there is no value of $x\in \mathbb{R}$ for which $f\left(x\right)$ is undefined.

The range of $f\left(x\right)=a{\left(x+p\right)}^{2}+q$ depends on whether the value for $a$ is positive or negative. We will consider these two cases separately.

If $a>0$ then we have:

$\begin{array}{ccc}\hfill {\left(x+p\right)}^{2}& \ge & 0\phantom{\rule{1.em}{0ex}}\left(\mathrm{The square of an expression is always positive}\right)\hfill \\ \hfill a{\left(x+p\right)}^{2}& \ge & 0\phantom{\rule{1.em}{0ex}}\left(\mathrm{Multiplication by a positive number maintains the nature of the inequality}\right)\hfill \\ \hfill a{\left(x+p\right)}^{2}+q& \ge & q\hfill \\ \hfill f\left(x\right)& \ge & q\hfill \end{array}$

This tells us that for all values of $x$ , $f\left(x\right)$ is always greater than or equal to $q$ . Therefore if $a>0$ , the range of $f\left(x\right)=a{\left(x+p\right)}^{2}+q$ is $\left\{f\left(x\right):f\left(x\right)\in \left[q,\infty \right)\right\}$ .

Similarly, it can be shown that if $a<0$ that the range of $f\left(x\right)=a{\left(x+p\right)}^{2}+q$ is $\left\{f\left(x\right):f\left(x\right)\in \left(-\infty ,q\right]\right\}$ . This is left as an exercise.

For example, the domain of $g\left(x\right)={\left(x-1\right)}^{2}+2$ is $\left\{x:x\in \mathbb{R}\right\}$ because there is no value of $x\in \mathbb{R}$ for which $g\left(x\right)$ is undefined. The range of $g\left(x\right)$ can be calculated as follows:

$\begin{array}{ccc}\hfill {\left(x-p\right)}^{2}& \ge & 0\hfill \\ \hfill {\left(x+p\right)}^{2}+2& \ge & 2\hfill \\ \hfill g\left(x\right)& \ge & 2\hfill \end{array}$

Therefore the range is $\left\{g\left(x\right):g\left(x\right)\in \left[2,\infty \right)\right\}$ .

## Domain and range

1. Given the function $f\left(x\right)={\left(x-4\right)}^{2}-1$ . Give the range of $f\left(x\right)$ .
2. What is the domain of the equation $y=2{x}^{2}-5x-18$ ?

## Intercepts

For functions of the form, $y=a{\left(x+p\right)}^{2}+q$ , the details of calculating the intercepts with the $x$ and $y$ axes is given.

The $y$ -intercept is calculated as follows:

$\begin{array}{ccc}\hfill y& =& a{\left(x+p\right)}^{2}+q\hfill \\ \hfill {y}_{int}& =& a{\left(0+p\right)}^{2}+q\hfill \\ & =& a{p}^{2}+q\hfill \end{array}$

If $p=0$ , then ${y}_{int}=q$ .

For example, the $y$ -intercept of $g\left(x\right)={\left(x-1\right)}^{2}+2$ is given by setting $x=0$ to get:

Introduction about quantum dots in nanotechnology
what does nano mean?
nano basically means 10^(-9). nanometer is a unit to measure length.
Bharti
do you think it's worthwhile in the long term to study the effects and possibilities of nanotechnology on viral treatment?
absolutely yes
Daniel
how to know photocatalytic properties of tio2 nanoparticles...what to do now
it is a goid question and i want to know the answer as well
Maciej
Abigail
for teaching engĺish at school how nano technology help us
Anassong
Do somebody tell me a best nano engineering book for beginners?
there is no specific books for beginners but there is book called principle of nanotechnology
NANO
what is fullerene does it is used to make bukky balls
are you nano engineer ?
s.
fullerene is a bucky ball aka Carbon 60 molecule. It was name by the architect Fuller. He design the geodesic dome. it resembles a soccer ball.
Tarell
what is the actual application of fullerenes nowadays?
Damian
That is a great question Damian. best way to answer that question is to Google it. there are hundreds of applications for buck minister fullerenes, from medical to aerospace. you can also find plenty of research papers that will give you great detail on the potential applications of fullerenes.
Tarell
what is the Synthesis, properties,and applications of carbon nano chemistry
Mostly, they use nano carbon for electronics and for materials to be strengthened.
Virgil
is Bucky paper clear?
CYNTHIA
carbon nanotubes has various application in fuel cells membrane, current research on cancer drug,and in electronics MEMS and NEMS etc
NANO
so some one know about replacing silicon atom with phosphorous in semiconductors device?
Yeah, it is a pain to say the least. You basically have to heat the substarte up to around 1000 degrees celcius then pass phosphene gas over top of it, which is explosive and toxic by the way, under very low pressure.
Harper
Do you know which machine is used to that process?
s.
how to fabricate graphene ink ?
for screen printed electrodes ?
SUYASH
What is lattice structure?
of graphene you mean?
Ebrahim
or in general
Ebrahim
in general
s.
Graphene has a hexagonal structure
tahir
On having this app for quite a bit time, Haven't realised there's a chat room in it.
Cied
what is biological synthesis of nanoparticles
what's the easiest and fastest way to the synthesize AgNP?
China
Cied
types of nano material
I start with an easy one. carbon nanotubes woven into a long filament like a string
Porter
many many of nanotubes
Porter
what is the k.e before it land
Yasmin
what is the function of carbon nanotubes?
Cesar
I'm interested in nanotube
Uday
what is nanomaterials and their applications of sensors.
what is nano technology
what is system testing?
how did you get the value of 2000N.What calculations are needed to arrive at it
Privacy Information Security Software Version 1.1a
Good
Other chapter Q/A we can ask


text:
# Inner Product and projection

V is in inner product space over F and U is subspace. $p=P_u(v)$. I need to prove or disprove by an example that:

1. $||v||=||p||$

2. $<v,p>=<p,p>$

-
You need to prove/disprove, we need your attempt. – Davide Giraudo Dec 4 '12 at 21:40
What inner product spaces come to mind? – Rudy the Reindeer Dec 4 '12 at 21:41
I dont know where to start so can u please guide me? im not looking for solution – baaa12 Dec 4 '12 at 21:43
I am guiding: what inner product spaces do you know? – Rudy the Reindeer Dec 4 '12 at 21:46
can you help me in what ||p|| is?? – baaa12 Dec 4 '12 at 22:01

Hint: ad 1. What do you know about $\ker P_U$? Can a $v\in\ker P_U$ fulfill 1.?
ad 2. What does it mean for $P_U$ to be an orthogonal projection? Can you say something about $P_U(v-p)$?


text:
## Loren on the Art of MATLABTurn ideas into MATLAB

Note

Loren on the Art of MATLAB has been archived and will not be updated.

# The Winter of Our Vectorization

### Contents

Today I'd like to introduce guest blogger Matt Tearle who works on our MATLAB Product Training materials here at MathWorks. Matt is on a mission to teach the world MATLAB, but this winter is testing his resolve. Annoyed that 22" of snow forced him to reschedule a training, today he shows just how bad this winter had been.

#### Going the Whole 9 Feet

It has been a brutal winter at MathWorks' headquarters in Natick, MA (a little west of Boston). I recently read a fascinating article by MIT doctoral student Ben Letham who created a convincing visualization of just how relentless the snowfall has been this year. It was so convincing that I had to reproduce it myself (in MATLAB, of course!). In the process I had some fun with vectorization.

It's common to look at statistics like the largest amount of snowfall in a single day (23.5" on Feb 17, 2003) or the highest total for the entire winter (108.5" for 2014/15). But 6" of snow every day for a week would also cause chaos, even though it would seem mild by either of those measures. Hence, Mr. Letham's idea is to calculate the largest total of snow that fell during a range of given time periods -- three days, a week, two weeks, ... For example, 40.5" of snow fell during the worst week-long period this winter, compared to just over 31" in the worst week of 1995/96 (which had the most snow of an entire winter until this year) and only(!) 28.5" in 2002/03 (23.5" of which fell in one day).

When you plot these "worsts" for a range of time periods, you get:

Sure enough, 2014/15 had the worst week ever, the worst two weeks ever, the worst month ever, the worst of just about any time period ever! And by some distance, too. The winter of 1977/78 is included because Boston locals talk with nostalgic reverence about the blizzard of 1978 that brought the area to a standstill. 1978 was pretty bad, but nothing compared to the snowfall we've had this winter.

#### Enter MATLAB

Intrigued by Mr. Letham's analysis, I wanted to play with the numbers myself. Like any good researcher, Mr. Letham documented his sources, which turned out to be NOAA's National Climatic Data Center. This meant that I could retrieve the data for myself and start playing (and so can you!). Before doing my own thing (which turned out to be fun, but a story for another day), I wanted to check that I could reproduce Mr. Letham's results. As it turned out, this helped me remember why I love MATLAB and gain appreciation for a function I had previously overlooked.

I downloaded the data from NOAA, used the Import Tool to import it into MATLAB, and did some basic preprocessing. As I often do, at this point I saved the cleaned-up data as a MAT-file, so I could now play with impugnity.

load snowfalls

#### Time for Some Logical Indexing

To compare different winters, I need to be able to extract the snowfall data for the given time periods. This is easily achieved with a combination of the date and time variable types that were introduced in R2014b and my all-time favorite MATLAB construction: logical indexing. I used the Import Tool to read the dates as datetime variables. Now I needed to separate 2014/15 from the rest of the data. Because the Northern Hemisphere winter crosses the new year, I decided to break years by the summer solstice (June 21):

pre2015 = (t <= datetime(2014,6,21));
snowPre2015 = snow(pre2015);
snow2015 = snow(~pre2015);

Numerical comparisons work naturally with datetime variables, so pre2015 is a logical variable which I can use to extract the snowfall values. To extract the year of the legendary blizzard, I used the handy isbetween function:

blizz = isbetween(t,datetime(1977,6,22),datetime(1978,6,21));
snow1978 = snow(blizz);

#### Convoluted, Yet Simpler

So now the crux of the problem: how to calculate the largest total over an $n$-day period. The typical programming language approach would be to loop over the array, taking $n$ elements starting with the $k^{th}$ element. But thinking about this process made me realize that I was looking for something very similar to a moving average. I knew that you can calculate moving averages in MATLAB by using convolution, implemented with the conv function. In this case, I want the sum (rather than the sum divided by $n$, as I'd want for an average). Hence, I can calculate all the $n$-day totals in a vectorized way:

n = 7;
ndaytotals = conv(snow2015,ones(n,1));
max7day = max(ndaytotals)

max7day =
40.4724

#### Fun with arrayfun

Now I need to loop over $n$. I couldn't see any neat way to vectorize this, so maybe I should use a loop (remembering to preallocate):

nmax = 180;
ndaysmax = zeros(nmax,1);
for n = 1:nmax
ndaysmax(n) = max(conv(snow2015,ones(n,1)));
end

I could also use arrayfun to perform this same operation without explicitly writing a for-loop:

nmax = 180;
calcndaymax = @(n) max(conv(snow2015,ones(n,1)));
ndaysmax = arrayfun(calcndaymax,1:nmax);

But is this really achieving anything? The code is very slightly more compact, but arguably more complicated (at least to someone who isn't a hard-core MATLAB-phile). I admit that, for this very reason, I tend to look down a bit on arrayfun as "how to cheat at vectorizing". But in this case I realized I had actually stumbled upon a great reason to use arrayfun...

Having applied this clever convolution operation to the 2015 data, I now need to do the same to the pre-2015 data and the 1978 data (and, presumably, any others that might strike me as interesting). Should I copy and paste code? No! This sounds like a job for a function. The "classical" approach would be to write a function that took an array of data and a maximum value of $n$ as inputs. It would return the maximum total for each $n$ from 1 to $n_{max}$. This wouldn't be a particularly difficult function to write, given that the code is basically done already (above). But do I really need to write a function file just for this? No: anonymous functions provide a way to create functions without writing a separate function file.

The one important caveat, though, is that anonymous functions must have a one-line definition (i.e., no loops!). So here's a compelling reason to use arrayfun: I can define my operation in a single line, enabling me to create an anonymous function, which can then be applied to any data set of interest:

nmax = 180;
maxXinNdays = @(x) arrayfun(@(n) max(conv(x,ones(n,1))),1:nmax);

Now I can apply this function to the snowfall data:

ndaysmaxPre2015 = maxXinNdays(snowPre2015);
ndaysmax2015 = maxXinNdays(snow2015);
ndaysmax1978 = maxXinNdays(snow1978);

#### Putting it All Together

Now I have a fairly short script that performs the whole analysis, leveraging the power of an anonymous function vectorized with arrayfun:

load snowfalls

% Extract data for 2014/15 winter and for everything else until then
pre2015 = (t <= datetime(2014,6,21));
snowPre2015 = snow(pre2015);
snow2015 = snow(~pre2015);
% Extract data for the infamous winter of 1977/78
blizz = isbetween(t,datetime(1977,6,22),datetime(1978,6,21));
snow1978 = snow(blizz);

% Define a function that calculates the maximum total value in n
% consecutive elements of the array x, for n from 1 to nmax
nmax = 180;
maxXinNdays = @(x) arrayfun(@(n) max(conv(x,ones(n,1))),1:nmax);

% Use the function to compare snowfall totals
figure
semilogx(maxXinNdays(snowPre2015))
hold on
semilogx(maxXinNdays(snow2015))
semilogx(maxXinNdays(snow1978))
xlim([1 200])

% Annotate the graph
xlabel('Number of days')
ylabel('Total snowfall (in)')
title('Max total snowfall in an n-day period')
legend('pre-2015 record','2015','1978','Location','southeast')

#### Spring into Action!

You can download the data and code and try this out for yourself. (The data also includes temperature readings.) What other statistics or metrics can you calculate with this technique of vectorizing anonymous functions with arrayfun? Was 2015 really that bad? Let us know what you discover about the winter of our discontent here.

Published with MATLAB® R2015a


text:
# How many three-digit numbers are divisible by 7?

Question.

How many three-digit numbers are divisible by 7?

Solution:

First three-digit number that is divisible by 7 = 105

Next number = 105 + 7 = 112

Therefore, 105, 112, 119,\ldots

All are three digit numbers which are divisible by 7 and thus, all these are terms of an A.P. having first term as 105 and common difference as 7.

The maximum possible three-digit number is 999. When we divide it by 7, the remainder will be 5.
Clearly, 999 - 5 = 994 is the maximum possible three-digit number that is divisible by 7.
The series is as follows.

$105,112,119, \ldots \ldots, 994$

Let 994 be the $\mathrm{n}^{\text {th }}$ term of this A.P.

a = 105

d = 7

$a_{n}=994$

$\mathrm{n}=?$

$a_{n}=a+(n-1) d$

994 = 105 + (n – 1) 7

889 = (n – 1) 7

(n – 1) = 127

n = 128

Therefore, 128 three-digit numbers are divisible by 7.


text:
## Question about "the" before names

9

1

For example I want to discuss one movie called "A" (it's just an example name) with my friend. Do I need to use the definite article before its name. Example:

I want to talk about the "A" with you (Let's talk about the "A"). Did you like it?

or

I want to talk about "A" with you (Let's talk about "A"). Did you like it?

12

No.

If there is a definite article in the title, you say it as part of the title:

Today, I would like to discuss The Bridge on the River Kwai with you.

However, if there isn't, you don't need to add one.

Today, I would like to discuss Kind Hearts and Coronets with you.

The only time you would see an added article would be if the film title is used as an adjective phrase modifying a noun that take an article. For example, you would say:

I have the soundtrack stuck in my head.

and:

I have the soundtrack to Kind Hearts and Coronets stuck in my head.

Therefore, if you use the title as an adjective phrase, you would also say:

I have the Kind Hearts and Coronets soundtrack stuck in my head.

But this is because "soundtrack" needs the article, not because the title does.

You should also be aware that, if there is an article in the title, when you use it as an adjective you would normally not repeat the adjective. So, for example, it would be more common to hear:

I have the Bridge on the River Kwai soundtrack stuck in my head.

than:

*I have the The Bridge on the River Kwai soundtrack stuck in my head.

Thank you for your answer! May I ask you one more question: If I want to talk with my friend about my favourite music group "Queen". What is correct to say than: Let's talk about the "Queen" or Let's talk about "Queen"? – user5369 – 2014-05-12T15:51:19.653

3+1 for Guinness references. Note that if an attributive is set before the title the article is required: "Alec Guinness starred in the extraordinarily funny Kind Hearts and Coronets." – StoneyB on hiatus – 2014-05-12T15:59:17.513

5Bands follow the same rules as film titles. If you say, "Let's talk about the Queen" people will think you want to discuss Elizabeth Mountbatten-Windsor, not Freddie Mercury. There are only a few categories of proper names that regularly take articles: ship names ("the Titanic") is the only one that springs to mind. – chapka – 2014-05-12T16:10:52.873

I'd probably say: 'Today, I would like to discuss the movie "Kind Hearts and Coronets" with you.' Unless given the context it was immediately obvious that I was talking about a movie. – Cruncher – 2014-05-12T18:06:04.983

2It may be to note that if the sentence would use an article which would differ from the title, informal writing would probably favor dropping the title from the article; formal writing would virtually demand a restructuring: "I like the soundtrack from An Affair to Remember" might informally be written as "I like the Affair to Remember soundtrack". I'm not sure in what cases that would be acceptable formally--one could talk about a New York Times crossword puzzle even though the banner on the page reads The New York Times. Movie titles seem like they should keep "The", though. – supercat – 2014-05-13T04:10:06.480

2

Putting the definite article the to proper noun is not a common practice. For the movies, the definite article goes if it is in the title. For example: The Expendables. Otherwise, it's just spoken the way the title is.

I want to talk about 'The Expendables'. Did you like it?

I want to talk about 'Captain America'. Did you like it?

You can still put the this way...

I want to talk about the movie Captain America; did you like it? - it took the, it's specific, the only movie - Captain America.

Thank you for your answer! May I ask you one more question: If I want to talk with my friend about my favourite music group "Queen". What is correct to say than: Let's talk about the "Queen" or Let's talk about "Queen"? – user5369 – 2014-05-12T15:55:49.230

1

Ask yourself - does it have it in their title? If no, you cannot put. http://en.wikipedia.org/wiki/Queen_%28band%29 On the other hand, if you still want, say *Let's talk about the band Queen.* Hope it's clear.

– Maulik V – 2014-05-12T16:04:49.793

Talking about the Queen would have the British monarchy as subject rather than music, wouldn't it? – Hagen von Eitzen – 2014-05-12T20:28:37.267

Or the Danish monarchy, or Spanish, or Dutch, or whatever, @HagenvonEitzen. – TRiG – 2014-05-12T22:57:57.133

@TRIG Admittedly, this may have been my fault because I'm German and our language usage may have interfered: People often say things like "Gestern war die Queen im Fernsehen", but not "Gestern war die (englische) Königin im Fernsehen." On the other hand, all the other queens are genereally referred to as "die spanische Königin" (and not e.g. "la Reina"), "die schwedische Königin" or "Königin Silvia" etc. That's why reading "the Queen" (especially capitalized) locked my brain on Britain. – Hagen von Eitzen – 2014-05-13T17:55:34.427