University of Washington Department of Chemistry Chemistry 453 Winter Quarter 2012

Homework Assignment 1 Due at 5 p.m. on 1/09/12 in Catalyst Drop-In Box..

This homework is worth a total of 10 points. This homework set is intended as a quick and relatively painless way for you to brush up on your math skills, and for me to understand your math background. Questions of these types will appear in lectures and exams.

QUESTION SET A: The Simple Functions of Applied Math Properties of exponential functions, factorial functions, and logarithms are widely exploited in Chemistry 453. Product and summation symbols are also widely used in chapters on quantum and statistical mechanics.

1A) Using only the following facts, and not the logarithm keys on your calculator, determine ln(60603/2) ~ ? ln(2) ~0.693; ln(3) ~1.099; ln(5) ~1.609; ln(1±x)~±x if x<<1. Show your reasoning.

33 ln() 60603/2 ==×× ln() 6060 ln() 1.01 6 1000 22

333 =++=+++⎣⎦⎡⎤ln1.01 ln 6 ln10[] ln1.01 ln 2 ln 3 3ln10 22 33 =++++=+++[]ln1.01 ln 2 ln 3 3ln 2 3ln 5[] ln1.01 4ln 2 ln 3 3ln 5 22 3 =+⎡⎤0.01 4() 0.693 ++ 1.099 3 () 1.609 = 13.062 2 ⎣⎦

2A) Using only the following facts, and not the logarithm keys on your calculator, determine ln(9901/2) ~ ? ln(2) ~0.693; ln(3) ~1.099; ln(5) ~1.609; ln(1±x)~±x if x<<1. Show your reasoning

11 11 ln() 9901/2 ==×=+ ln() 990 ln() .990 1000 ln() .990 ln() 103 22 22 1 3 0.01 3 =−+ln() 1 0.01 ln() 10 =−+× ln() 2 5 2222 33 3 3 =−0.005 + ln 2 + ln 5 =− 0.005 +() 0.693 +() 1.609 = 3.448 22 2 2 3A) Using only the following facts, and not the exponential key on your calculator, determine e-8010/1000~ ? e3 ~20.086; e5~148.413; ln(1±x)~±x if x<<1. Show your reasoning,

11 e−−8010/1000==ee 8.01−+()8 .01 = e −++ ( 3 5 0.01 ) = eee −−− 3 5 0.01 ≈ × ×−() 1 0.01 20.086 148.413 ==()()()0.0498 0.00674 0.99 0.000332

4A) The factorial function nnnn!12321=−−( )( )ii increases so rapidly, that evaluating 100! with your calculator is impossible because of memory limitations (try it). There is an expression called Stirling’s Approximation for evaluating factorial n ⎛⎞n expressions: nn!2≈ ⎜⎟ π . Use Stirling’s Approximation to evaluate ln(100!). ⎝⎠e

n ⎡⎤⎛⎞n 1 lnnnnnnn !≈=−++ ln⎢⎥⎜⎟ 2ππ ln() ln 2 ln ⎣⎦⎢⎥⎝⎠e 2 1 ∴ln100!=−++ 100ln100 100() ln 2π ln100 2 1 =−+460.5 100() 1.837 + 4.605 =+= 360.5 3.221 363.7 2

5 5 ⎛⎞1 1 1 1 1 111 6 4 3 13 5A) ∏⎜⎟= /2· /3· /4 6A) ∑ = ++= + + = x=3 ⎝⎠x −1 x=3 x −1 2 3 4 12 12 12 12

QUESTION SET B: Vectors, Matrices, and Determinants Linear algebra is also used in chemistry 453 so familiarity with vectors, matrices, and determinants is advisable. Many physical properties including forces, particle velocities, electric fields and dipole moments are represented mathematically as vectors. Vectors can be multiplied together to obtain a scalar result (i.e. inner or dot product) or a vector result (i.e. cross or outer product). For example, energy is a scalar and the energy of interaction between a dipole moment and a field is obtained by taking an inner product. Torque is an outer product of two vectors which can be evaluated with a determinant. The reorientation of a vector in space is mathematically represented by multiplying a vector with a matrix. Representing a vector both in Cartesian and spherical coordinate systems is also a necessary skill.

ˆˆˆ 1B) A vector in three dimensional space is designated Vvivjvkvvv=++xy z =() xyz,, where iˆ , ˆj , and kˆ are unit vectors in the x, y, and z directions, respectively. Let A = ()1, 2, 3 and B =−()4, 4,1 . Calculate the scalar or “dot” product ABi = ? Solution: ABii=−=+−+=−()()()()1, 2, 3 4, 4,1 1 4( 2)( 4) ( 3)( 1) 1

2B) Let A = ()1, 2, 3 and B =−()2, 4, 8 . Determine AB+=++3 (1 6,2 12,3 − 24) = (7,14, − 21) .

3B) Let A = ()3, 4, 3 . Calculate the magnitude (i.e. length) of this vector. Calculate also the angles θ and φ that define the orientation of this vector in a spherical coordinate system. 222 222 Solution: A=++=++= aaaxyz 343 34

⎛⎞a 32π −−11⎜⎟z ⎛⎞ θπ===×==cos cos⎜⎟ 59 0.33 1.04 radians ⎜⎟aaa222++ ⎝⎠34 360 ⎝⎠xyz

−−11⎛⎞ay ⎛⎞42 π ϕ ===×=tan⎜⎟ tan⎜⎟ 53.1 0.93 radians ⎝⎠ax ⎝⎠3 360

⎛⎞⎛⎞13− 4 3 ⎜⎟⎜⎟ 4B) Determine the vector that results from the operation:⎜⎟⎜⎟822−= 2 ? ⎜⎟⎜⎟ ⎝⎠⎝⎠−62 1 1

⎛⎞⎛⎞⎛⎞⎛⎞13−−−− 4 3 364 7 ⎜⎟⎜⎟⎜⎟⎜⎟ ⎜⎟⎜⎟⎜⎟⎜⎟8 2 2−= 2 24 −+= 4 2 22 ⎜⎟⎜⎟⎜⎟⎜⎟ ⎝⎠⎝⎠⎝⎠⎝⎠−−−+−6 2 1 1 18 4 1 21

⎛⎞⎛⎞−12 6 9 5B) Determine the matrix that results from the product:⎜⎟⎜⎟= ? ⎝⎠⎝⎠4521−−−

⎛⎞⎛⎞⎛−−−−−−−1 2 6 9 64 92 ⎞⎛⎞ 1011 ⎜⎟⎜⎟⎜== ⎟⎜⎟ ⎝⎠⎝⎠⎝4−−− 5 2 1 24 + 10 36 + 5 ⎠⎝⎠ 34 41

⎛⎞13− 4 ⎜⎟ 6B) Evaluate the determinant of the matrix ⎜⎟822 ⎜⎟ ⎝⎠−62 1

13− 4 22 82 82 822=−() 1() 3 +−() 4 21−− 61 62 −62 1 =−−+−+=−−−=−()(1 2 4 ) ()( 3 8 12 ) ( 4 )( 16 12 ) 2 60 112 174

102−−x 7B) Solve the determinant for x: 000−x = −−204x

102−−x 00014225−==−−−−−−−=−+x (xx )()()()()() x x x32 x

−−204x −−=⇒=xx2 ()50 x 0,5

8B) The outer or cross product of Aaaa= ( xyz,,) and B = (bbbxyz,,) is AB× ., and is evaluated using the determinant expression: ijkˆˆˆ AB×= axyz a a. Determine the outer product of A = (4,3, 2) and B =−()1, 2, − 3 by

bbbxyz evaluating the appropriate determinant.

ijkˆˆˆ ˆˆˆ A×= B ax a yz a = i() ab yzzy − ab − j() ab xzzx − ab + k() ab xyyx − ab

bbbxyz =−−−ijˆˆ()33() 22 () () 43() −−−+ 21 () kˆ() 4231() −− ( ) . =−13ijkˆˆ + 10 + 11ˆ

QUESTION SET C: Derivatives and their Applications The derivative is the calculus operation used most frequently in this course. Derivatives are used generally throughout mechanics. Derivatives are used in Taylor expansions to approximate values of functions and are used to find the minima and maxima of functions. Derivatives are fundamental components of differential equations. Differential equations are used to solve mechanical problems.

df f ()θθθ==2cos sin ? dθ

df⎛⎞ dcosθθ d sin 22 1C) ∴ =+=−+=2⎜⎟ sinθ cosθθθθ 2() sin cosor 2cos 2 dddθθθ⎝⎠

2C) df fx() == ()1 – x4 / x ? dx df d d =+xxxx−−1441()1 – () 1 – () dx dx() dx 34 4 −−13 4 2 41xxx− 51− =−+xx()()()41 – xx −=−−22 =− x xx 3C) 1− edf− x 2 fx()==? 2xdx2 df2 ddf dd1− e−x == dx2 dx dx dx dx2 x

dd⎡⎤1 d−1 =−+−()()11eex−−xx() 2 dx⎣⎦⎢⎥2 x dx dx dd1 =−−=−−⎡⎤⎡⎤221−−111exxx − −() e − x − 2 ex − xx − 1() 1 e − x − 2 dx⎣⎦⎣⎦2 dx −x 11⎡⎤−−−xxxdded11⎡⎤ − −− 22 d − =+−−−−⎢⎥exx⎢⎥()11 e xx() e 22⎣⎦dx dx⎣⎦ dx dx 1 1 =−⎡⎤x−21exe−−−xx−−−−−⎡⎤21() exxe −−−− x 32 x 2 ⎣⎦2 ⎣⎦ ee−−xx1− =− + 2x x3

∂ ⎛⎞463 36− 3 ∂ ⎛⎞463 45 4C) ⎜⎟286x yxyx+=2 − ⎜⎟212x yxy+=2 ∂ xx⎝⎠ ∂ yx⎝⎠

5C) Using a Taylor expansion show that close to 1, ln(x) ≈ x -1. Use a Taylor expansion to approximate ln(1.05) = ? to first order.

Taylor Series: f(x) = f(a) + (x-a)f’(a) +(1/2!)(x-a)2f’’(a)… To first order, ln(x) = ln(a) + (x-a)(1/a) for x ≈ 1, set a=1 so that ln x ≈ (x – 1) ln(1.05) = 0 + (1.05 – 1)(1/1) = 0.05

6C) Find the local maximum and minimum of the equation f(x) = x3 – 27x (ignore any max or min at ± ∞)

df d2f min occurs at /dx = 0 and /dx2 > 0 x = 3 is min, df d2f max occurs at /dx = 0 and /dx2 < 0 x = -3 is max

df d2 f =−=⇒=±3270x2 xx 3; = 6 dx dx2

QUESTION SET D: Integrals and their Applications Integrals and derivatives go hand-in-hand in mechanics courses. In Chemistry 453 integrals are commonly used to calculate averages of various types. Although we expect you to be able to perform simple integrations from scratch, it is far more often the case that you are required to evaluate an integral that has standard form. Standard form means the integral has a form with a generally recognized answer, which you can find in books or on the Web. In Chemistry 453 very often the challenge is to correctly frame a physical problem in terms of an integral with standard form.

3 3 22x3 52 1D) ∫ 2271xdx2 ==−=() x=1 331 3

V f 11V f 1 1Vf 2D) dv==−=ln x() ln Vfi ln V ln ∫ 33vVVi 3 3 Vi i

2/3L 2 2 ⎛⎞π x 3D) Evaluate the integral; ∫ sin ⎜⎟dx where L is a constant, using the standard LLL/3 ⎝⎠ x 1 form sin2 axdx=− sin 2 ax + C , where a and C are constants. ∫ 24a 2/3L 2/3L 2222422 ⎛⎞ππxxLx⎛⎞⎛ ⎛ ⎞⎛⎞ LL ⎛⎞ ππ ⎛⎞ ⎞ sin⎜⎟dx =− sin ⎜ ⎟ =−⎜⎟ sin ⎜⎟ − sin ⎜⎟ LLL∫ ⎜⎟⎜24ππ LL 64 3 3 ⎟ L/3 ⎝⎠⎝⎠⎝ ⎝ ⎠L/3 ⎝⎠ ⎝⎠ ⎝⎠ ⎠ +∞ +∞ 2 2 4D) Evaluate the integrals ∫ edx−κ xkT/2 and ∫ edp− pmkT/2 using the standard form −∞ −∞ ∞ 1/2 −cx2 ⎛⎞π integral: ∫ edx= ⎜⎟. Note κ, k, T, and m are constants. In the standard form c 0 ⎝⎠4c indicates a constant. Note the function being integrated is an even function: f(x)=f(-x). +∞ +∞ 2222ππkT kT edxedx−−κκxkT/2===22 xkT /2 ∫∫ 4κκ −∞ 0 +∞ +∞ 222π mkT ∫∫edpedp−−pmkT/2===222 pmkT /2 π mkT −∞ 0 4 In each case the limits of the integral are changed from -∞ to ∞ to 0 to ∞, and the integralis doubled. This is a result of the integrand being an even function.

5D) The average kinetic energy of a very small spherical particle moving in one +∞ 2 1 p 2 dimension is: Eedp= ∫ − pmkT/2 , where m, k, T are constants. Calculate 2π mkT −∞ 2m ∞ 2 135⋅⋅( 2n − 1) π E using the standard form: xe2nax− dx= , where n is a positive ∫ nn+1 0 2 aa integer and a is a constant greater than zero. Note the function being integrated is even.

+∞2 +∞ 111p 22 E== e−−p/2 mkT dp p 2 e p /2 mkT dp 22ππmkT∫∫2mm mkT −∞ 0 12mkT kT ==2π mkT 2π mkT 42m

STATISTICS QUESTIONS: Statistical techniques are indispensible tools used to calculate mechanical properties averaged over a very large number of systems. In Chemistry 453 you have to know how to handle simple probability calculations and how to handle very large numbers. Here are some examples:

1E) The factorial function is defined as nnnn!1221= ( −−)( ) ⋅. For example 5!54321120=××××= . There is a formula called Stirling’s approximation which states that ln(N!)=NlnN-N, where N is a very large number. Most calculators have factorial keys. Calculate ln(N!) for N=5 and N=1,000,000 using your calculator by employing first the factorial key and then the logarithm key. Describe and explain any problems that you encounter. Then use Stirling’s approximation to calculate ln(N!) for N=5 and N=1,000,000. Explain the utility and limitations of Stirling’s approximation.

Using the calculator: 5!=120. But 1,000,000! Results in an error. Using Stirling’s Approximation: ln(5!)=5ln5-5=3.05. Therefore 5!=e 3.05=21.1, so Stirling’s approximation does not work well for factorials of small numbers. But: Ln(1000000)!=1000000ln1000000-1000000=12,815,510. Then 1000000!=e12,815,510 There is not enough memory in your calculator to evaluate such a large exponential.