Mathematics 303(2010), Final exam Answers, page 1 of * First Name: ANSWERS Last Name: Student Numer: MATHEMATICS 303 AY CLASS Instructor: Z.V. Kovarik URATION OF EXAMINATION: 3 HOURS MCMASTER UNIVERSITY FINAL EXAMINATION April, 2010 THIS EXAMINATION PAPER INCLUES 8 PAGES AN QUESTIONS. YOU ARE RESPONSIBLE FOR ENSURING THAT YOUR COPY OF THE PAPER IS COMPLETE. BRING ANY ISCREPANCY TO THE ATTENTION OF YOUR INVIGILATOR. Special Instructions: Use of Casio FX-991 calculator only is allowed. No notes or ooks. Last two pages are the Formula Sheet and the Normal istriution tale. The four pages efore that are for rough work. If you must use these pages for overflow work, please make clear reference to this fact. The maximum marks for parts of the questions are indicated on the left margin. Maximum points: 78 Question Max Score 1 14 2 3 12 4 12 5 12 6 14 Total 78
Mathematics 303(2010), Final exam Answers, page 2 of * 1. Consider 0a Þ [4] (a) Find the residues of 0 at all its poles. Hint: One of the poles is 3Þ All poles: They are the fourth roots of a : 3, 3ß 3ß 3 We can use LHospitals Rule (applicale to analytic functions) 3 At 3Àlim a3 lim a Ä3 Ä3 Ä3 a w 3 a a 3 lim a)3 3 At 3 À 3 At 3À 3 At 3À 3 [4] () For V ß let GßV e the semicircle kk Vß e, GßV the segment cvßvd and GV GßV GßV, the closed path traced positively. ( e denotes the imaginary part.) Find ) GV 0a.Þ The poles inside G V are 3ß 3with residues 3, 3 8 ) GV 0a. 13 Res 0a 4 4 1 1 ) 13ˆ ˆ 3 ˆ 3 [6] (c) Find y showing that as 0aB.B G 0a.B Ä V Ä Þ ßV Remark. There was a typo: 0a.B should e 0a., ut oth GßV GßV integrals tend to y the same type of inequalities, although the one with the typo does not lead to the conclusion aout 0aB.B. On GßVß kk Vß so y triangle inequality (rememerig that V Ñ V V k0ak k k Ÿ V ¹ GßV 0a.B¹ Ÿ V 1 V V lengthagßv V 1 V and y comparison test, since lim also VÄ V ß lim VÄ GßV 0a. (typo corrected) Conclusion: As limit for VÄß 1 0aB.B
Mathematics 303(2010), Final exam Answers, page 3 of * 2. Consider 0a / Þ [4] (a) Find all singularities of 0aÞ (/ has no singularities in the finite plane, ut has roots) The singularities od 0a are the roots of /, i,e, solutions of / Þ 31513 Solved: / / ß 5 integer ln a5 13ß 5 integer [4] () For Taylor expansion of 0a around, find the radius of convergence. The radius of convergence is the distance from the centre to the nearest singularity Such a singularity is any of the ln 13ß ln 13Þ istance from 0 is radius, equal kln 13k Éaln 1 [6] (c) Find the first three terms of this Taylor expansion. (1) By long division: / á Truncated quotient: * & á ˆ á l á á * á ) á (2) By undetermined coefficients: á - - - á Cross-multiply: ˆ á a- - - á á Collect and compare: - - - - - - * & Solve: - - - Expansion again: 0a á * & w / / a ww / a/ / a w 0 a ww 0 a x x * & ÐÑ Using derivatives: 0 a ß 0 a 0a 0 a á á
Mathematics 303(2010), Final exam Answers, page 4 of * 3. Consider the conformal mapping A0a Þ [ ] (a) Find the inverse function 1aAÞ irectly aaß AA aaß 1A a A A [5] () For the vertical line P À d ß find the image 0aPÞ (You can parametrize P y 3>, and study kak). 3> A0a3> 3> k k 3> È > A 3> È, so the image is the unit circle > (the image of is A ) (Not to e penalized for omitting this): Conversely, if / 3) is on the unit circle and ) 1 then >cot 3> 3 corresponds to /. 3> [4] (c) Find the image of the right half-plane d under 0. Interior point test: The oundary is the unit circle, we need to test an interior point: Pick ß 0a, falls inside the circle. The image is the (open) unit disc kak Þ Or: Tracing the oriented oundary: The right half-plane has oundary P oriented positively when it is traced downwards (from positive imaginary part to negative). 3> The images of 3> are 3>. We can determine the orientation of the images y picking three points in this order: A5 0a5ß c ß ß d c3ß ß 3d caßaßa d c3ß ß 3d and the unit circle is traced counterclockwise, corresponding to the positive orientation relative to the interior of the circle.
Mathematics 303(2010), Final exam Answers, page 5 of * 4. (Laplace Transform question) Consider the differential equation ww C C 0 > where 0 > Ÿ > 1 a a > 1 w with initial conditions C a ß C a [4] (a) Find the Laplace Transform J a of 0 a > Þ 0 a > L a>1 e0 fa / 1 1 1 > 1 Or from definition: > / / /.> ¹ Þ > [6] () Find the Laplace Transform ] a of C a > ec fa ] ww w ec fa ] C a C a ] 1 ww / ec C fa a ] 1 / ] a a y partial fractions ] ˆ 1 / ˆ [4] (c) Find the solution C> ain the form one function of > for Ÿ > 1 C> a another function of > for > 1 C e] f a cosa> a cosa a>1l a>1 a cosa> a cosa> L a> 1 a > a L a > > Ÿ> C> a a cos a 1 > 1
Mathematics 303(2010), Final exam Answers, page 6 of * B / B 5. Consider a random variale \ with proaility density : ab - - B [3] (a) Find the (Fourier style) characteristic function of \. ÐTo e a proailty density, - ) 9a / : a B.B / -/.B a - 3 - B - a 3 - B /.B / - -3 3B 3B -B 3 - [6] () Given a random variale ] independent of \ ut with the same proaility density : ab, find the proaility density : ab of Y\]Þ Convolution: For Bß: ab For B ß : ab a: : ab : ab>: a>.> B B> > - -a - - B B - - / /.> /.> - B/ B Summary: B B/ B : ab - - B [3] (c) Find the characteristic function of Y found in part (). Use convolution property: - / : a B.B 9a 9a 3B a-3 or directly, if we do not recall convolution: the antiderivative is otained through integration y parts, - B/ B / 3B.B a-3 B - - B/ a-3 B - / -3 a-3 3B -B 3B / : ab.b - B/ /.B a-3 B a-3 B - B/ - / Š -3 ¹ a-3 - again. a-3
Mathematics 303(2010), Final exam Answers, page 7 of * 6. The averaged test scores of students were recorded, with a sample mean ( points, and from past records, the standard deviation was known to e 2 points. [4] (a) etermine the value of + such that the standard normal variale falls into the interval c+ß+ dwith proaility 0.95. (A tale is attached) B Î Let :B a mean the standard normal proaility density / ÞBy symmetry, + + + È 1 :B.B a :B.BÞ*& a for :B.B a 0.475 + From the tale: + Þ* [6] () Calculate the half-width - of the confidence interval for the mean.. From the formula page, + 5 + 5 Confš B È Ÿ. ŸB 8 È8 means that the half-width is + 5 È for + Þ*ß 5 ß 8 ß 8 + 5 È8 aþ* a * Þ) [4] (c) etermine the 95 confidence interval for the actual mean. (of the much larger population). Same formula, Confe( Þ) Ÿ. Ÿ ( Þ) f ConfeÞ* Ÿ. Ÿ (Þ) f
Mathematics 303(2010), Final exam Answers, page 8 of * Formula Page cosa ) cosa ) a+ a, a a+, a+, ) a+ a, a a+, a+,.@ cos @ @ a+ a, a a+, a+, 3 3 / / ß sin 3 3 a 3a/ / 3 Š È 1 ß 8 x x 8x log lnk k arg 8 sina+ cosa, asina+, sina+, cos ) cos sin sin sin sin cos cos cos cos tan sin sin cos cos cos cos ln sin cos / á 3 (many-valued) Residue Theorem: 0 analytic on a simple closed curve G, with finitely many 8 isolated singularities ßáß inside GÀ) 0. a 3 Res 0 a Residue at a simple pole : 1 4 0a a -0a 8 G 4 - Res - limä- 3B Fourier Transform (not scaled): Yc0aB da / 0aB.B w erivative rules: Yc0 ab da 3 Yc0aB da. YcB0aB da 3. Yc0aB da Laplace Transform: e fa > 0 / 0 a>.>, I a, - >, ß ecosa-> f - ß esina-> f - Ÿ>+ + L a>+ ß e0 a>+ L a>+ f / e0f > + Expected value of \ with proaility density :B a ÀI\ a B:B.B a Expected value of 0 a\ : I0\ a a 0B:B.B a a Variance: Vara\ I ˆ a\ I a\ I a\ ai a\ Sum of independent r.v.s: : : : : w w w : abb : ab.b \] \ ] \ ] 8 8 8 4 8 4 4 4 Sample mean and sample variance: B Bß ab B Confidence interval for. (with known 5 ): confidence level (, standard normal + 5 + 5 È8 È8 Confš B Ÿ. ŸB ßwhere Proc+ŸŸ+ d ( > Gamma Function: I a > /.>ß recursion Ia Ia some values Ia for integer, Iˆ 8 8x 8 È1
Mathematics 303(2010), Final Exam, page 9 of * Standard Normal istriution z 0.00 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.0.00000.00399.00798.01197.01595.01994.02392.02790.03188.03586 0.1.03983.04380.04776.05172.05567.05962.06356.06749.07142.07535 0.2.07926.08317.08706.09095.09483.09871.10257.10642.11026.11409 0.3.11791.12172.12552.12930.13307.13683.14058.14431.14803.15173 0.4.15542.15910.16276.16640.17003.17364.17724.18082.18439.18793 0.5.19146.19497.19847.20194.20540.20884.21226.21566.21904.22240 0.6.22575.22907.23237.23565.23891.24215.24537.24857.25175.25490 0.7.25804.26115.26424.26730.27035.27337.27637.27935.28230.28524 0.8.28814.29103.29389.29673.29955.30234.30511.30785.31057.31327 0.9.31594.31859.32121.32381.32639.32894.33147.33398.33646.33891 1.0.34134.34375.34614.34849.35083.35314.35543.35769.35993.36214 1.1.36433.36650.36864.37076.37286.37493.37698.37900.38100.38298 1.2.38493.38686.38877.39065.39251.39435.39617.39796.39973.40147 1.3.40320.40490.40658.40824.40988.41149.41309.41466.41621.41774 1.4.41924.42073.42220.42364.42507.42647.42785.42922.43056.43189 1.5.43319.43448.43574.43699.43822.43943.44062.44179.44295.44408 1.6.44520.44630.44738.44845.44950.45053.45154.45254.45352.45449 1.7.45543.45637.45728.45818.45907.45994.46080.46164.46246.46327 1.8.46407.46485.46562.46638.46712.46784.46856.46926.46995.47062 1.9.47128.47193.47257.47320.47381.47441.47500.47558.47615.47670 2.0.47725.47778.47831.47882.47932.47982.48030.48077.48124.48169 2.1.48214.48257.48300.48341.48382.48422.48461.48500.48537.48574 2.2.48610.48645.48679.48713.48745.48778.48809.48840.48870.48899 2.3.48928.48956.48983.49010.49036.49061.49086.49111.49134.49158 2.4.49180.49202.49224.49245.49266.49286.49305.49324.49343.49361 2.5.49379.49396.49413.49430.49446.49461.49477.49492.49506.49520 2.6.49534.49547.49560.49573.49585.49598.49609.49621.49632.49643 2.7.49653.49664.49674.49683.49693.49702.49711.49720.49728.49736 2.8.49744.49752.49760.49767.49774.49781.49788.49795.49801.49807 2.9.49813.49819.49825.49831.49836.49841.49846.49851.49856.49861 3.0.49865.49869.49874.49878.49882.49886.49889.49893.49896.49900 3.1.49903.49906.49910.49913.49916.49918.49921.49924.49926.49929 3.2.49931.49934.49936.49938.49940.49942.49944.49946.49948.49950 3.3.49952.49953.49955.49957.49958.49960.49961.49962.49964.49965 3.4.49966.49968.49969.49970.49971.49972.49973.49974.49975.49976 The value corresponding to a given is the proaility T a Ÿ ^ Ÿ. For example, T a Ÿ ^ Ÿ Þ» Þ*( EN