chapter7_ex
数据挖掘第七章__聚类分析
Chapter 7. 聚类分析
• 聚类分析概述 • 聚类分析的数据类型
• 主要聚类分析方法分类
划分方法(Partitioning Methods)
分层方法
基于密度的方法
基于网格的方法
基于模型(Model-Based)的聚类方法
火龙果 整理
• 差异度矩阵
– (one mode)
0 d(2,1) 0 d(3,1 ) d ( 3, 2 ) : : d ( n,1) d ( n,2)
0 : ... ... 0
火龙果 整理
1.数据矩阵 数据矩阵是一个对象—属性结构。它是n个对象组
6.3 聚类分析中的数据类型
假设一个要进行聚类分析的数据集包含 n
个对象,这些对象可以是人、房屋、文件等。
聚类算法通常都采用以下两种数据结构:
火龙果 整理
两种数据结构
• 数据矩阵
– (two modes)
x11 ... x i1 ... x n1 ... x1f ... ... ... xif ... ... ... xnf ... x1p ... ... ... xip ... ... ... xnp
• 保险: 对购买了汽车保险的客户,标识那些有较高平均赔偿 成本的客户;
• 城市规划: 根据类型、价格、地理位置等来划分不同类型的 住宅; • 地震研究: 根据地质断层的特点把已观察到的地震中心分成 不同的类;
火龙果 整理
生物方面,聚类分析可以用来对动物或植物分类,或 根据基因功能对其进行分类以获得对人群中所固有的
(6.2)
火龙果 整理
汉英翻译(chapter seven)
首字母缩写。
P&G
宝洁,它是Procter and Gamble(William Procter and James Gamble )
的首字母缩写。
WST
世界卫星终端公司,它是World Satellite Terminal的首字母缩写。琼州学院外国语学院 Nhomakorabea11
三、商标翻译
商标是商品的门面,各国厂商历来对它十分重视,都希望 在高质量产品上有个译得美的商标,给厂家带来可观的经 济效益,这里的所谓“美”,起码要做到译出的商标词所 含的文化信息能够符合译入语民族的文化心理。如要使商 标名字译得美,就要针对不同的商标,采用不同的译法。 进口药品的译名,不少都有“康”、“定”、“福”、 “平”、“安”、“宁”、“灵”等。如:脑复康 (Piracetam),咪康唑(Miconazole),优福定(UFT),利福 定(Rifandin),尼群地平(Nitrendipine),非洛地平 (Felodipine),普尔安(Propanidid),癫健安(Valpromide), 喘特宁(Volmex),脉宁平(Prazosin),克霉灵(Montricin), 美可灵(Micoren)。这是汉民族求安、向善、祈福心理决 定的。
Value (有益) + derma (希腊语:皮肤)——益肤药皂;英国
5、多外来词
Shampoo
香波;(印度语:按摩、推拿)
LUX
“力士”香皂,(英国Unilevier (联合利华);拉丁文;阳光)使人联想到Luck,
Luxury等词语,A其广告词“super rich shine”更是让使用者不禁由阳光联想到健 康亮泽的肌肤。 的、大型全球性护肤品与身体护理品品牌。
国际财务课后习题答案chapter7
CHAPTER 7 FUTURES AND OPTIONS ON FOREIGN EXCHANGESUGGESTED ANSWERS AND SCX.UTIONS TO END-OF-CIIAPTERQUESTIONS AND PROBLEMSQUESTIONS1 ・ Expl ain the basi c dificrcnces between the operation of a ciurcncy ibrward market and a fiitures market. Answer: Tlie forward mark巩is an OTC market where the forward contract for purchase or sale of foreign currency is tailor-made between the client and its international bank・ No money cliangess hands miti l tlie maturity d^:e of the contract when deliveiy and receipt are typically made. A fijtiues contract is ail exchaiige-tiaded instniment with standaidized featuies specifying contiact size and deliveiy date. Futiues coohacts aie m aiked-to-iiiarket daily to reflect chaiig es in tlie settlement pii ce. Delivery i s seldom made in a futures market. Rather a l^versi ng trade is made to close out a long or short position.2.In order for a derivatives market to fimetion most cfFio cntly, two types of economic agetits are needed: hedgers atid speculdtors・ Explain・Answer: Two types of niarket paiticipants arc ncccssaiy for the 巴尸五ci ent operat on of a derivatives market:speculators and hedgers. A speculator attempts to profit fiom a ijiange in the futures price・ To do this, the speculator wi 11 take a long or sliort position iti a futures contract depending upon his expectations of iuture price movern ent. A hedger, on- tlie-otlier- han d, desir es to avoid price var iation by locking in a purchase pri ce of die wide dying asset thiougli a long position in a liitures ccutract ci a sales price through a short positi on. hi effect, the hedger passes off the iisk of price vai iation to the speculator who is better able, or al I ea?t more willing, to bear this risk.3.Wliy are most fiitiu es positions cl os ed out th rough a reversing trade rath er than held to deliveiy? Answer: In forward markets,approximately 90 percent of all cont「勺cts that 前e initially establishucl result in the short making deliveiy to the long of the asset underlying the contract. This is natural because tlie term s of forward contracts ai e tailor — ni ade between th c long and s hort ・ By contrast, only about one percent of CIUTCIICV fiitiwcs cotit roots result in del i very ・ While futures contracts arc useful for speculation and hedging, their standardized delivery dates make them unlikely to correspond to the actual future dates wiicti forogii cxdiange transactions will occur. Tlius, they are genet'al closed out in arcs z ei^ing trade. In fact, the commission that buyers mid sell ers pay to transact hi the futures maket is s singl c amount that covei^ the round-trip tiTais actio ns of initiating and closing out the position・4.How can the FX firtut es maiket be used for pH cc discovery?Answer: To the extent tliat FX forward pri oes are ail unbiased predictor of fiiture spot exchange rates, tlie market anticipates whether one cuiyenoy will appreciate or depreciate versus another・ Because FX Eitiires conti-acts trade in an expi ration cycle, cliflerent c onto acts expire at diHeient peiiDclic dates into tlie iiiture ・Tlie patteiri of the prices of these contacts provides information as to tlie mai kefs ciin ent belief about th巴rd ative fiiture value of one ciui ency versus anothei at the scheduled expiration dates of the contracts. One will genially see a steadily appreci ating or depreciating pattern; howevei; it may be mixed at times. TTius, the fiitures market is usefill for price dscovery. i.e., obtaining the market's forecast of the spot exchange rate at different future:d“tES・5・ What is tlie major di fFerenoe in the obligation of one with a long positi on in a fijturcs (or fonvaid) contract in comparison to an options confraot?Answer: A futures (or forwaid) contract is a vehicle for buying or sei ing a staled amount of foreign exchange at a stated price per unit at a specified time in tlie fiitiue・ If the long holds the contract to the delivery date, he pays tlie effective contractual fiitures (or foiwat'd) price, regardless; of whether i t is an advantageous pti ce in oompaiiscn to tlie spot price at the delkeiy date. By oontiasf, ail option is a contiact giving the loris tlie riglit to buy ci sell a given quantity of an asset at a specified price at some time in the fiihn e? but not enforcing any obligation on him if the spot pti ce is more favorable tliaii the exercise prize. Because th巴opti on ownei* does not have to exercise t he option if it i s to his disadvaiit^e, the option has a price, or premium, whereas no price is p已id at inception to enter into a futures (orforward) contract6. Wliat is nieait by the terminology that an option is in-, at-, or oiit・of— the-money?Answer: A call (put.) option with St >E (E> S^) is refetred to as trading in-thc-monej r・ If E tlie option is hading at-tlic-money. If Sf< E (E<§) the call (put) option is trading out-ofithe- JYiOYie^ ・7 List the arguments (variables) of which an FX call or put oj^tion model price is a function ・ How does tiic call and put preininm change with respect to s change in the ai'guments?Answer: Botii call mid put options are fiinctions of only six variables:£, E, r讣讣 T an de When all else remains tlic same、the price of aEuicpcaii FX call (put) option w 11 incressc:1・ tlic lai'gei'(smaller) isS,2. the small曰(larged is E、3・ tlie smaller (larger) is r n4・ tlie i aiger (smalleij is r t>5.the laiger (smaller) is relative to r f, arid6.the gj eater is aWhen and are not too niucli different in size, a European FX call and put will increase in price when the option term-to-matiirity increases・ Howe、曰;when 飞 is very mu ch laigei than a European FX cal I will increase in price, but the put premium wil I decrease, whe厂i the option tenn-to-m increascs. The opposite i s tme when i s vety much greater than r$. ForAmerican 二X opti oils tlic analysis is I傑s complicated Since a longer tenn American option can be exercised on any date that a shorter tenn opti on can be exercised or □ some later date, it follows tliat tlie all else remaining the sarne. tlie longer tenn Americen opti on will sell at a price at least as laige as tlie shorter tenn option.PROBLEMS1. Assiunc toda>r,s settlement price on a CME EUR futures contract is S1.314O/EUR. Yon have a short position in one contract. Your performaiicc bond accoimt ciurcntly has a balance of $L 700. The next tlii'cc days, scttleincnt prices ETC $1.3126, $1.3133, arid S1.3049. Calculate the changes in tlic perfonnaiicc bond account from d已ily marking-to-market andthe balance of tlie perfotTnance bond accoiuit after the third day. Solution: $1, 700 +[〔$1.314 O・ S1.3126) + ($1.3126 -Si. 3133)+ (Sl.3133 - SI.3049)]XEUR125,000= $2,837.50,where EUR125, 000 is the contractual size of one EUR contract.2- Do problem 1 again assuming you have a long position in the futures conti act・Solution: $1,700+ [($1.3126 ・ $1.3140)+($ 1 ・ 31 33 ・S1・3126)十($L3(Mg • $1 .3133)] xEUR125,0OO= $562.50,where EUR125, OOO istlie contrachial sizeuf one EUR contract.With only $562・ 50 in your petfonnancc bond account, you would experience a tnargiti call requesting that additional fijnds be added to youipeiionnance bond account to bring tlic balance back up to tlie initial petdonnaiice bond level・3・ Using the quotations in Exlubit 7.3、cal cul ate the face value of the open interest in the June 2005 Swiss franc fiitures contiact ・Solution: 2401 contracts x SF125Q00 二SF262, 625JD00.vvhei'e SF125, 0C0 is tlie couti actnal size of one SF contract ・ing tlie quotations in Exliibit 7. 3, note that the June 2005 Mexican peso Mur es contract has a price of SO. 08845. You believe tlie spot piice in June wil be $ 0. 09500. WhM speculative position would you enter into to attempt to profit frotn your beliefs ? Calculate your anticipated profits, assuming you taP;e aposition in tlwee contracts ・ Wliat is the size of your profit (loss) if the fhtures price is indeed an unbiased predictor of the fiitii re spot price and this price materializes?Solution: If you expect the Mexi can peso to li se from SO.08845 to SO. 09500, you would take a long position in fiitiucs since the fiitiires price of $ 0.08845 is less than your expected spot price.Your anticipated profit from a I ong position in tiirec contracts is: 3 x ($0.09 500 -$0・ 08845)xMP500.0C'0= $9, 825.00. where MP500.00C1 isthe contractual size of one MP contrast.If the fiitures price is sn unbiased predictor of the expected spot price, the expected spot price is tlic iutca cs price of $0.08845///MP・ If tliis spot price materializes, you will not hsrs r e any profits or losses from your short position in three futures contracts: 3 x ($O・ 08845 -$0.08845) XMP500.000 =0.5.Do problem 4 again assuming you believe the Jiuie 2005 spot price will be $0.08500. Solution: If you expect tlie Mexi can psso to depieci ate fi-oni $ 0.08&15 to $ 0.07500, you wou d take a short position in fiitures since the futures price of $0.08845 is greater tliaii your expected spot price・Yciu anticipated p io fit from a sh or t pos ition in three contract s is: 3 x i, $ 0 ・ 08845 ・ $0.07500) xXlP500,000= $20,175.00 ・If tlie fiitiues price is an unbiased predictor of the Future spot price and this price materializes? you will not profit or lose from your long futwes pzisiti on.6.George Johnson is considering a possible ax-motith SI 00 million LJBClR-bascd, floating-ratebank loan to Hind a project 址terms shown in the tabic below. Johnson fears a possible ti ss in the LIBOR rate by December mid wants to use the December Eurodollar fiitures contrast to hedge thi s risk・ Tlie contract expires December 20< 1999. has a US$ 1 mi Ilion contract size,and a discount yield of 7. 3 pei cent・Joints on will ignore the cash flow implications of marking to market、i nitial margin requirements, and any timing niisinatch between ex change-ti'aded fiitures contiact cash flows and tlie interest payments due in March. Loan TermsaLoan First loan payment (9%) Second paynie nto initiated and fiitures contract expires and principal•••5 9/20/99 町2/20/933/20/00a・ Fonnulatc Jolmsotrs September 20 floating-to-f xed-ratc sti ategy using the Eurodollai futui c contracts discussed in tlie text above. Showthat tliis strategy would result in a fixed-rate Icaih assiuning ail increase in tlieLIBOR rate to 7・ 8 percent by December20, which remains at 7.8 peicent tbrougli March 2O・ Show all calculations.Johnson is considering s 1 2 — moutli loan as aii alternatiue・Ihi $ approach wi II result in two additional unceilain cash flows, as follows:I.oaricFii sbSecond Tliii(UFoin1li pa^nnento initiated payment (9%) s>payin ent payment a and principal9/20/99 12/20/99 码/20/00 6/20/009/20/00b. Describe tlie strip hedge that Johnson could use and explain how it liedges the 12-month loan (spec 迅'number of contracts). No calculatious are needed.CFA Guidel ine Answer孔Tlie basis point value (BPV) of a Eiu odollai' fiihu es cxDiiti act can be found by substituting the contract specifications into the following money m aiket relationship:a BPV FUI = Ciiange in Value = (face value) x (days to maturity / 360)x (change in yield)a q尸$(1 milion) x (90 / 360)x (.0001 )$25Tlie nimibcr of contract, N. can be found by:N = (BPV spot) / (BPV fiitiires)x($2,500)/($25)3 = 100aORo N = (value of spot position) f (face value of each Futures contract)尸($ 100 million) / (SI million)a =1CO(value of spot position) / (value of iutiucs position)b □ S(1 OO, 000, 000) / ($ 981,750)where value of fiitiires position = $1,000,000 x [1-(0.073/4)]« 102 contractsTlicreforc on September 20, Johnson would sell ICO (or 102) December Eurodollar futures contracts at the 7.3 percent yield. The imp: iedL1BOR rate in December is 7・3 percent as indicated by the December Eiuofiitiu'es discount yield of 7.3 percent・ fhus a boniowing rate of 9・3 percent (7.3 percent + 200 basis points) can be locked in if tlie hedge is cciTcctly implemented.A rise in the rate to 7.8 percent represents a 50 basis point (bp) increase over tlie implied LIBOR rate. For s 50 basis point increase in LIBOR, the cash flow on the short futures position is:o = ($25 per basis point per contract) x 50 bp x 100 contractsx$125,000.However, the cash flow on die floating rate liabi lity is:x -0.098x ($100,000,000/4)=・ S2,45O, 000.Combining the cash flow fiom tlie hedge with the cash flcwfi-om the loan results in a net outflow of S2?325,GOO, which translates into an annual rate of 9.3 percent:=($2,325,000x4) / $100,000,030 = 0.093This is precisely the implied bor rowii^ rate that Johnson locked in on September 2(). Regardless of the LIBOR rate on December 20. the net outflow will be $ 2,325,000, which translates into ail annualized rate of 9.3 percent. Consequently, tlie floating rate liability 1ms been converted to a fixed rate liabil ity i n the sense tliat tlie interest rate imcertaintv associated with tlie March 20 payine nt (using tlie December 20 contiact) has been removed as of Sepzember 20・ b・【1】a strip hedge, Johnson would sell 100 December futiues (for the March payment), 100 March fiitiires (lor the June payment)、and 100 June firtiu'es (for the September payment)・ The objective is to hedge each interest rate psynient sepaiately using tlie appropriate muiiber of contiacts. The probl em is the same as in Pai! A except here tlii ee cash flows sie subject to rising rat es and a strip of fiitu res is used to hedge this interest rate risk. Tliis pi obi em is simplified somewhat because the cash flow mismatch behveen the fiitiires and the loan payment is ignored ・ Therefore, in ord er to hedge each cash flow, Johnson simply sells 100 contracts far each payment・The strip hedge traiisfbrrns the floating rate loan into a strip of fixed rWc payments・As was done in Part A、the f xed rates are found by adding 200 basis points to tlie imp I icd Foiwar d LIBOR rate indicated by tile dis count yield of the tlirce diiFcrcnt Eiu^odollar fiitiires contracts・ Tlic fixed pajments will be equal wlicn the LIBOR temi structure is flat for the first year ・7.Jacob Bower has a liability that:•has a pnncipal balance of S1 DO million on June 30,1998,•accrues interest quarterly stalling on June 30. 1998.•pf^s interest quarterly、•has a one-yeai' tenn to maturity, end•calculates interest due based on 90-day LIBOR (die London Intel bank Offeredo Rate.)Bower wishes to hedge hi s remaining i nterest payments against changes in interest rates・Bower has coircctly cal cul ated that he needs to sell (short) 300 Eurodollar fhturcs contracts to accomplish the hedge ・ He is considei*ing the altemative hedging strategies out I inedin the following table.Initial Position(6/30/98) in90 Day LlBOR Eurodollai- Contracts曰Explanwhy strategy B is a more effective hedge than strategy A when the yield curve undergoes em instant aiieous iionparallel shift.b・ Discuss ail interest rate scenario in which strategy A would be superior to strates^/ B・CFA Guide! ine Answei*a.^Strategy, B's SuperiorityStrategy B is a strip hedge that is constructed by selling (shoiiiiig) 1 OO Bjtures contracts m aturiiig in each of the next three quailers. With tlie strip liedge in place, each qiiaiter of the coming year is hedged against shifts in interest rates for th at qnailei*. The r eason Strategy B will be a more effective hedge than Strategy A for Jacob Bower is that Strategy B is likely to work well whether a parallel shift or a nonpai'allcl shift occurs over th ㊁onc-yeat' term of Bow er 7s liability. That is, regardless of what happens to the term structiwc, Strategy B structures the fiihires hedge sc that the rates reflected by the Einodollar fiihwcs cash price match the applicable fates forthe undciiying liability-tlic 90day LIBOR-based rate on Bower's liability. The same is net true forStrategy A. Because Jacob Bowers liability cemcs a floating interest rate that resets quaitcrly ・ he needs a sti ategy that provides a series of th rec-month hedges ・ Strategy A will need to be re^triictm'ed when tlie three -montii September contract expires. In particular, if the yield curve twists upward (fijtures yields rise more for distant expirations than for neai' expirstious), Strategy A will produce iiife ioi hedge results・b. Scenaiio in Which Stiategy A is SuperiorStrategy A is a stack hedge stiategy that initially involves selling (sliortirig) 300 September contracts・ Strategy A is raiely better than Stiategy B as a hedging orrisk-nediiction strategy. Only from the perspecti ve of faxorable cnsh flows is Strategj r A better than Strategy B. Such cash flows occur only in certain interest rate scenarbs・For example Strategy A wil 1 work aswclI ss Strategy B for Bowct^s liability if interc^z rates (inst antatieously) change in parallel fashion. Another interest rate scenario where Sfratcgy A outpctioniis Strategy B is one in which tlie yield ciuve rises but witli ahvist so that futures yields rise more for neai' expi rations than for distant expirations. Upon expiration of the September co厂tract. Bower will have to rol 1 out his hedge by selling 200 December contracts to hedge the remaining interest payments. Tliis action will have the effect tliat tlie cash flow from Stiat 已gy A wi 11 be larger than the cash flow from Strategy B b©cause tlie appreciation ou the 300 slioi! September fiitures contracts will be larger tliaii the cumulative appreciation in the 300 contracts shorted in Strategy B (i.e., 100 Septem ber, 100 Deceinber, and 100 Mauch). Consequently, the cash flow fi-oni Strategy A will more thai offset the increase in the intei est payment on the liability, whereas the cash ilowfi-om Strategy B wil I exactly offset the increase in the interest payment on the I lability・e the quotations in Exliibit 7.7 to calcinate the intrinsic value and the time value of the 97 September Japanese yen Amet iceii call arid put options.Solution: Premium- Intrinsic Value = Time Value97 Sep Call 2.08 -Max [95.80 -97.00= - 1.20. o] =2.08 cents per 100 yen97Sep Put 2.47 - Ma>c[97.C0 - 95.80 =1. 20, 0] = 1.27 cents per 1 OOyen9.Assume spot Swiss franc is $ 0.7000 and the six-month fbrwaid rate is $0.6950. What is tlie minitnuni price that a six-month Ametican call option with a striking price of SO.6800 should sellFor in a rational market f Assume the aimualizcdsix-niontii Ewodollar rate is 3 % percent・ Solution:Note to Instnictor: A complete solution to this problem relies on the boiindaiy expressions presented in footnote 3 of the text 济Chapter 7.C a>Mzx[(70 — 68)、(6950 - 68)/(1.0175), O]>Zl4zx[ 2. 1.47. 0] = 2 cents10・ Do problem 9 again assimiing ail American put option instead of a call option・Solution:心必4(68 -70), (68-69. 50)/(1.0175), 0]-2, -1.47. 0] = 0 cents1 1 ・ Use tlie European option-pii ci ng models developed in tlie chapter to value the cal 1 of probl an 9 and tlie put of problem 1 0・ Assume the aimualized volatility of the Swiss fi*anc is 14.2 percent. This problem can be solved using tlie FXOPM.xJ s spreadsheet・Solution:^ = [/n(69.50/68)+.5(. 142)2(.50)]/(.142)心O=.2675<4= £・.142*50 =・ 2765 ・.1004 = .1671N(di) = .6055N(d^ =・ 5664M呦= .3945N(-d^) = .4336Q 二[69.50(.6055)・ 68(・5664)]e"3%j°)= 3.51 centsP. - [68(.4336)-69.50(. 3945张心珈刘=2.03 cents12. Use the binomial option-pricing model developed in the chapter to value the call of problem 9・ Tlie volatility of tlie Swissiiaiic is 14.2 percent・Solut ion: Tlie spot rate at T will be either 77.390 = 70・00c(l・ 1056) or 63.32 0 = 70.00^(.9045), where u = &*灼=1. 1056 and a? = 7血=・©045. At the exerci se price of E =6& the option wi II only be exercised at time T if the Swi ss franc appreciates; its exercise value would be C u f= 9.390 = 77.39^ - 68. If tlie Swiss franc depreciates it would not be rational to exercise the opti on ; its value woul cl be C dT = O.TVie hedge ratio is% = (9.39 一0)/(7739-63 J2)=・ 6674・Thus, the call premium is:=?k^{[69.50(.6674)-68((70/681 (. 6674 - 1)+])]/(1.0175), 70 -68}= Max[l. 64, 2] = 2 cctits per SF.国际财务管理课后习题答案chapter 711/11GMINI CASE : THE OPTIONS SPECULATORA speciil at or is ccnsid ering the purchase of five three - month Japanese yen call options with a strikingprice of 96 cents per 100 yeti. Tlie premium is 1.35 cents per 100 yen ・ Tlie spot price is 95.28 cents per 100 yen and tlie 90・day forward rate is 95.71 cents. The speculator believes tlie yen will ^Jpreciate to $ 1.00 per 1 00 yen ovei the next du es months. As tlie speculator's assistant, you liavebeen asked to prepare the following :1 ・ Graph the call option cash flow schedule.2. Det 已 rmine the speculator's profit if the yen appreciates to $1. 00/100 yen.3. Det 曰 Triine the speculators profit if the yen only appreciates to the fonvaid rate.4. Determine the fiitiu c spot price at which the speculator will only break wen.Suggested Solution to tlie Options Speculator:2. (5 x¥6,250000) x [(1 00 - 96)- 1. 35]/1 0000 = $&281・25・3. Sin c e the oprj on expi res out — of-the — money, the -s p ec u lator will let the opt ion expi leworthless ・ He uvill only lose tlie option pi emium ・4. = E +C=96 + 1.35 = 97.35 ceiitsper 100 yen.。
25614294_Chapter_7_The_call_of_the_wild_第七章野性的呼唤
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Chapter-07
• forms a “web of trust”
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PGP Operation – Email Compatibility
• when using PGP will have binary data to send (encrypted message etc)
system
Email Security Enhancements
• confidentiality
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– protection from modification
• also use key ID in signatures
PGP Message Format
PGP Key Rings
• each PGP user has a pair of keyrings:
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Chapter7-6(电磁波的散射和吸收介质的色散)解析
平均散射能流为
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4 2
cE r
2 0 0
2 2
e
2
r
sin
入射波强度I0定义为平 0c 2 I 0 S0 E0 均入射能流
2
散射波能流可写为
re S 2 sin2 I 0 r
2
2
2 0
1 1 sin cos d 1 cos2 2
2 2
得对非偏振入射波的平均散射能流
re 1 2 S 2 1 cos I 0 r 2
单位立体角的散射功率与入 射波强度I0之比称为微分散 射截面,记为d/d, 得汤 姆孙散射微分截面
2
d re 1 cos2 d 2
0 , 则在入射波电场E0e-it作用下的振子运动方程为
.. . e i t 2 x x 0 x E0e m
以 x = x0e-it代入得这方程的稳态解
it e e 1 x E0 e 2 2 m 0 i m
1
§7.6 电磁波的散射和吸收
介质的色散
一、自由电子对电磁波的散射 二、束缚电子的散射 三、电磁波的吸收 四、介质的色散 五、经典电动力学的局限性
1
一、自由电子对电磁波的散射
当一定频率的外来电磁波投射到电子上时,电磁波的振 荡电场作用到电子上,使电子以相同频率作强迫振动.振 动着的电子向外辐射出电磁波,把原来入射波的部分能量 辐射出去,这种现象称为电磁波的散射.
eE0 it x e 2 m
大学物理chapter-7
q + F p 。 。 -q
E
0, M 0
稳定平衡
π, M 0
非稳定平衡
返回
退出
-
F
F
+
π 0 2 p -
-q 。 。 +q F
F
E
π π 2
+ F
E
p
F
E
在非均匀外电场中 电偶极子所受合力不为零, 力矩不为零。
返回
退出
sin 2 sin 1 Ex 4π 0 a
讨论
cos1 cos 2 Ey 4π 0 a
1. 无限长带电直线: 1 =0 ,2 =
P
Ex 0
E Ey 2 π 0 a
4 π 0 a
返回
退出
2、半无限长带电直线: 1 = 0 ,2 = /2 Ex E y
r a / sin
x a cot
dx a csc 2 d
cos 2 Ex a csc d 2 2 4π 0 a csc
2 1
Ex (sin 2 sin 1 ) 4π 0 a
(cos1 cos 2 ) 同理 E y 4π 0 a
第七章 静止电荷的电场
§7-1 物质的电结构 库仑定律 §7-2 静电场 电场强度
§7-3 静电场的高斯定理
§7-4 静电场的环路定理 电势 §7-5 电场强度与电势梯度的关系 §7-6 静电场中的导体 §7-7 电容器的电容 §7-8 静电场中的电介质 §7-9 有电介质时的高斯定理 电位移 §7-10 静电场的能量
11
尸体派对
《尸体派对:驭血》白金攻略心得》首先是全WRONG END的收集顺便这里要说一句的是结局列表里的黑星星的点亮顺序,有些时候跟你正常打流程时是有出入的。
拿第九章举个例子吧。
第二个黑星星是碰黑烟死,第四个是被墙壁碾死,所以我一开始玩游戏时认为第三个WROND END应该就在这两段剧情中间。
结果卡了一个多小时章节结局进入方法CH1被肉酱拍死~CH2最后不阻止犬丸开柜子CH31.在女厕所被淹死2.后期在2楼西侧教室里触发剧情(有张CG)被幽灵拍死~CH41.黄毛登场时被人体模型刻命君抡死2.收集本馆和别馆的链接通道的六鬼门时,不提醒班长把手拿出来3.别馆2F东侧,美术室在大树处收集六鬼门是时存心上去被幽灵摸一下(这里必须自己上前作死,待在原地不动一会就发生剧情了。
)4.本馆1楼左上角体育馆(嘛,本作新加?的场景,其实也就是2U的那个。
)拿完六鬼门下便便螺旋时不动,就会被淹没而死了5.本馆1楼左下泳池收集六鬼门时,水里不动就行6.流程最后是男人(班长来着?)就撑60S环节,被斧头拍死CH51.玄关黄毛被碧池掐住的时候不挣扎2.依旧是玄关,被碧池和刻命怕死3.在1楼中间教室(这个是本作新增的教室呢)发生剧情时,选择不出门CH6别馆1楼右上侧教室内收集六鬼门镜子处CH⑨1.班长在本馆2楼西侧最下面的教室,选择钻狗洞去隔壁教室2.班长拿斧头砍国王3.持田在地下防空洞还没将光之水晶和暗之水晶放在台座上,就去地图右上角的黑烟处4.救出酋长后被墙壁碾死5.本章最后炮姐啥都不干,或者选错道具CH101.有一处发生剧情时,门里会闪现出犬丸的身影,这时候立刻从这门追出去,选择杀,或者啥都不选。
(这里选救的话,有段不错的剧情可以看哦~)2.最终BOSS烧鸡3.最终BOSS惜败4.4个灵魂全部用完击倒最终BOSS主要的写完了,下面是写零碎的东西~1 无伤和不用手电打通一关可以在第一章完成2 不受到精神污染过关第一章完成不了,这章应该还没开启黑化系统,我建议和《不殺》这个杯子一起做了(第九章不进化一个敌人过关)3 名札的话,要说的就是第九章的隐藏的一个。
概率论和数理统计(李慧斌)复习大纲-第7章-置信区间-Confidence-Intervals
概率论与数理统计(李慧斌)复习大纲Chapter 7 Confidence Intervals置信区间7.1 Sampling Distribution 抽样分布统计量的分布称为抽样分布。
在本节中,我们将从正态分布推导出随机样本的样本方差分布,以及样本均值和样本方差的各种函数的分布。
复习:Thm 5.5.2若X1, X2,…, X n独立且满足,i= 1,2,…,n,若C1, C2,…, C n不全为零,则Corollary 5.5.2 设随机变量X1, X2,…, X n组成随机样本,满足正态分布,其中均值μ和方差σ2,则7.2 χ2Distribution卡方分布定义:若随机变量X1, X2,…, X n独立同分布且其中每个随机变量都满足标准正态分布,所以有着以n阶自由度卡方分布(χ2distribution with n degrees of freedom),记作,n来源于独立随机变量中以n阶自由度的χ2分布的概率密度函数其中欧拉函数定义为χ2分布的性质:定理1定理2 (χ2分布的可加性)若X ~χ2 (n) , Y ~χ2(m),X, Y独立,则X+Y ~ χ2 (n+m)例:设X1, X2,…, X n是正态分布的随机样本,证明Thm 7.3.1 设X1, X2,…, X n是正态分布的随机样本,则:(1)与独立;(2)注:,虽然基于n个,但是它们之和为0,所以指定数量的n-1确定剩余值。
因此有n-1阶自由度。
结果表明,只有从正态分布中抽取随机样本,样本均值和样本方差才是独立的。
证明如下:的联合概率分布函数为其中A为正交矩阵(orthogonal matrix),且的联合概率分布函数为因此独立且⇒与独立且7.4 The t Distribution t分布定义:设X ~ N(0, 1), Y ~χ2 (n)且X和Y独立,则随机变量所满足的分布称为n阶自由度t分布,记作,其中的概率密度函数为t分布的性质:(1)f(x)图像呈钟型,且中心为0;(2)它的一般形状类似于平均分布0的正态分布的概率密度函数。
ESD系统 黑马ESD系统资料 Chapter 7,8,9_ch
3.3 PES H41q-HRS
························································································· 17
3.3.1 系统结构 ··································································································· 17 3.3.2 中央模件 F8652A ······················································································ 17 3.3.3 协处理器模件 F8621A ··············································································· 19 3.3.4 通讯模件 ··································································································· 19 3.3.5 机械结构与设计························································································· 19 3.3.6 24 伏直流供电 ··························································································· 19 3.3.7 5 伏直流供电 ····························································································· 20 3.3.8 I/O 总线 ····································································································· 20 3.3.9 安全停车(Safety Shutdown) ································································· 20 3.4 输入/输出级 ········································································································· 21 3.4.1 24 伏直流供电 ··························································································· 21 3.4.2 I/O 模件 ····································································································· 21 3.4.3 (Ex)i 模件 ··································································································· 21
托马斯微积分课件7
Rule, and Improper Integrals
7.1 Basic Integration Formulas 7.2 Integration by Parts 7.3 Partial Fractions 7.4 Trigonometric Substitutions 7.5 Integral Tables, CAS, and Monte Carlo
Integration 7.6 L’ Hospital Rule 7.7 Improper Integrals
目录 上页 下页 返回 结束
7.2
Integration by Parts
(分部积分法)
目录 上页 下页 返回 结束
Suppose that Fu f u
f g x g xdx f g xdg x
uv uv uv.
Integrating both sides with respect to x, we have
uvdx uv uv dx uv dx uv dx uv dx uvdx uv dx u dv uv v du
目录 上页 下页 返回 结束
Yet, what shall we do with the integral f xdx?
therefore, we have
x
sin
xdx
1 2
x sin x x2 cos xdx
.
目录 上页 下页 返回 结束
Example 1. Evaluate the indefinite intln xdx
ln
x
x2 2
dx
x2
《管理学原理》英语教学课件CHAPTER7-Managing Change
Stress (1)
An individual is confronted with an opportunity, constraint, or demand related what desired. Outcome is perceived to be both uncertain & important.
Authority relationships Coordinating mechanisms Job redesign Spans of control
People
Attitudes Expectations Perceptions Behavior
+ Technology
Work Process Work method Equipment
Coercion
When a powerful group’s endorsement is needed
Inexpensive, easy May be illegal; way to gain may undermine change agent’s support credibility
Organization Development
Differentiate between creativity and innovation.
Explain how organizations can stimulate innovation.
2019/4/14 郭志文 2001 Copyright 3
What Is Change?
Structure
商务英语翻译 chapter 7 & 8
2. Syntactic features 1) simple sentences KISS principle :Keep it short and sweet. e.g. Coca-cola is it. (还是可口可乐好) A diamond lasts forever. (diamond ring) 2)elliptical sentences (space and cost ) e.g. It is a moment you planned for. Reached for. Struggled for. A long-awaited moment of success. Omega, for this and all your significant moments. (Omega watch) 3) parallel structures E.g. Designed with a computer. Silenced by a laser built by a robot. (Volvo automobile ) 3. Rhetorical features 1) Figure of speech a) Simile E.g. Some people are as reliable as sunrise…Theses are Amway people. Cool as a mountain stream… Cool as fresh Consulate. (Consulate is a kind of American cigarettes.) b) Metaphor E.g. High efficiency. Our big bird can be fed even at night. (French Air Cargo Carriers ) We’re rolling out the red carpet for Asia’s elite travelers. (Dragonaire Airlines ) 2)pun (the artistic use of homonyms and homophones) a)Homophonic pun E.g. OIC (OIC glasses ) More sun and air for your son and your heir. (seaside bathing spot) b) Homographic pun E.g. Try our sweet corn. You’ll smile from ear to ear. Money doesn’t grow on trees. But it blossoms at our branches. (British Lloyd Bank) III. Linguistic features of business ads 3) parody (this word came from Greek word ‘paroidia’ which means satirical poem) a) words E.g. My goodness! My Guinness! (爱尔兰最畅销啤酒—健力士 Guinness) b)idioms or proverbs Like son, like father. Like daughter, like mother. (talcum powder 爽身粉) c)maxims or aphorisms Not all cars are created equal. (mitsubishi) d) passage Pepsi cola hits the spot; twelve full ounces, that’s a lot. Twice as much for a nickel, too; Pepsi Cola is the drink for you. (adapted from a British folk music) 4) Rhyming (including alliteration and rhyme Hi-Fi, Hi-fun, Hi-fashion, only from Sony. (Sony products) Be specific---Go Union Pacific.) 5) Personification Oscar de La Renta knows what makes a woman beautiful. (Oscar Cosmetics) 6) Exaggeration
大一上册微积分课件chapter7
s in n 1 x c o s x ( n 1) (1 s in 2 x ) s in n 2 x d x s in n 1 x c o s x ( n 1) s in n 2 x d x
(n 1) sin n xdx
T h erefo re
n s in n x d x c o s x s in n 1 x ( n 1) s in n 2 x d x
3x
2x 2 ln x 33
2 3
xdx
3
2x 2 ln x 4 x 2 C
3
9
Example6 x arctan xdx
arctan
xd
x2 2
arctan x x2 2
x2 2
1
1 x2
dx
x2 2
arctan
x
1 2
1
1
1 x
2
dx
x2 arctan x 1 x arctan x C
b
(
f
(x)g(x)
f
( x) g ( x))dx
a
a
f (x)g(x)]ba
b f (x)g(x)dx
a
b f (x)g(x)dx
a
b f (x)g(x)dx a
f (x)g(x)]ba
b f (x)g(x)dx
a
or
b a
udv
uv]ba
b
vdu
a
Exampe
1
arctan xdx.
Thus
s in n x d x 1 c o s x s in n 1 x n 1 s in n 2 x d x
n
n
s in n x d x s in n 1 x d c o s x s in n 1 x c o s x c o s x d s in n 1 x s in n 1 x c o s x ( n 1) c o s 2 x s in n 2 x d x s in n 1 x c o s x ( n 1) (1 s in 2 x ) s in n 2 x d x
化工热力学_Chapter6-7_习题与解答_III_201406
化工热力学chapter67习题与解答iii201406热力学第一定律习题化工热力学习题工程热力学习题化工热力学习题精解基础会计习题与解答数值分析习题解答固体物理学习题解答合同法习题集及解答和声学教程习题解答
化工热力学习题课-III (第六章、第七章) June 13, 2014
第六章 化工过程的能量分析 作业题(1,2)_(May 16 & 23, 2014)
取整个装置为体系,忽略各设备散热损失,且冷凝器出口的冷却水所 携带的㶲通常难以利用,可忽略,则总㶲损:
DK总 Ex5 -Ex6 Ws 4.438 106 0.9731106 1.2152 106 2.250 106 kJ
对废热锅炉作㶲衡算,忽略热损失,又无功交换,Ws=0,其㶲损:
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1Chapter 7 Problem SetChapter 7PROBLEMS1.[M, None, 7.4] Figure 1 shows a practical implementation of a pulse register. Clock Clk isideal with 50% duty cycle.Figure 0.1Pulse register.Data : V DD = 2.5V, t p,inv= 200ps, node capacitances are C Clkd = 10fF, C x= 10fF, both true and complementary outputs node capacitances are 20fF.a.Draw the waveforms at nodes Clk, Clkd, X and Q for two clock cycles, with D=0 in onecycle and D = 1 in the other.b.What is the approximate value of setup and hold times for this circuit?c.c)If the probability that D will change its logic value in one clock cycle is α, with equalprobability of being 0 or 1, what is the power consumption of this circuit? (exclude thepower consumption in the clock line) f clk=100 MHz.2Chapter 7 Problem Set2.[M, None, 7.4] Figure 2 shows a register that attempts to statistically reduce power consump-tion using a data-transition look-ahead technique.Figure 0.2Pulse register.a.Briefly describe the operation of the circuit.b.If all the NMOS transistors are of the same size, and all of the PMOS transistors are of thesame size, two times wider than the NMOS, roughly determine the input switching proba-bility under which this flip-flop reduces power, compared to an equivalent flip-flop with-out data-transition look-ahead circuitry.3.[E, None, 7.6] Shown in Figure 3 is a novel design of a Schmitt trigger. Determine the W/Lratio of transistor M1 such that V M+=3V Tn. V DD = 2.5V. The W/L ratios of other transistors areshown in figure. You may ignore the body effect in this question. The other transistor param-eters are as given in Chapter 3.NMOS: V Tn = 0.4V, k n’ = 115µA/V2, V DSAT = 0.6V, λ = 0, γ = 0V1/2Digital Integrated Circuits - 2nd Ed 3PMOS: V Tp = -0.4V , k p ’ = -30µA/V 2, V DSAT = -1V , λ = 0, γ = -0V 1/24.[M, None, 7.6] Consider the circuit in Figure 4. The inverter is ideal, with VM =V DD /2 andinfinite slope. The transistors have V T _=0.4V , k n ’=120 µA/V 2 and k p = 40 µA/V 2. M 1 has(W/L)1=1. Ignore all other parasitic effects in the transistors.a.As V IN goes from 0 to V DD and back to 0 explain the sequence of events which makes thiscircuit operate as a Schmitt Trigger.b.Find the value of (W/L)2 such that when V IN increases from 0 to VDD the output willswitch at V in = 0.8V .c.Find the value of (W/L)3 such that when V in decreases from V DD to 0 the output willswitch at V in = 0.4V . If you don’t trust your value from b., you may use (W/L)2=5.Figure 0.3Schmitt trigger.Figure 0.4Schmitt trigger.4Chapter 7 Problem Set5.[M, None, 7.6] Figure 5 shows an astable multivibrator. Calculate and draw voltage wave-forms at the capacitor V C and at the output V out. What is the oscillation frequency of the mul-tivibrator?Figure 0.5Astable multivibrator.Assume that the amplifier is ideal, with symmetric supplies (V out max = V DD, V out min = -V SS)R1 = 1kΩ, R2 = 3kΩ, R3 = R4 = 4kΩ, C = 1nF, V DD = -V SS == 5V, diode voltage V D = 0.6V (ideal diode), V out (t = 0-) = -V SS.6.[E, None,7.6] An oscillator is shown in Figure 6. Draw the signal waveforms for this circuitat nodes X, Y, Z, A, and B. Determine the oscillation frequency. You may assume that thedelay of the inverters, the resistances of the MOS transistors, and all internal capacitors can beignored. The inverter switch point is set at 1.25V. Assume that nodes Y and Z are initially at0V and 2.5V, respectively.Figure 0.6Oscillator.Digital Integrated Circuits - 2nd Ed57.[E, None, 7.6] Consider the oscillator in Figure7. Assume that the “n” switches turn “on” forvoltages above V DD/2, and the “p” switches turn “on” for voltages below V DD/2. Assume thatthe current sources stop when the node voltage charges to either V DD or ground.Figure 0.7Oscillator.a.Find the oscillation period for V DD = 3V.b.Draw the waveforms at nodes X, Y, and Z for two periods.c.Find the oscillation period.8.[M, None, 7.6] The circuit in Figure8 operates at a supply voltage of 3V and uses twoSchmitt triggers with the following threshold voltages: V M+ = 2V, V M- = 1V.Figure 0.8 A circuit composed of Schmitt triggers.a.Identify whether the circuit is monostable, bistable, or astable?b.Draw the waveforms at nodes X, Y, and A. Mark all important voltage levels.c.Calculate the key timing parameter for this circuit (propagation delay for bistable, pulsewidth for monostable, and time period for astable) in terms of R and C. You can assumethat gate delays are negligible compared to the delay of the RC network.。