Mark Bly's Math 142 Webpage


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Office Hours
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Math Tutorial Center

Ayres G012
Mon: 9am-5pm
Tue: 9am-5pm
Wed: 9am-5pm
Thu: 9am-5pm
Fri: 9am-3pm

Hodges North Commons
Sun: 5-10pm
Mon: 5-10pm
Tue: 5-10pm
Wed: 5-10pm
Thu: 5-10pm


Internet Videos

Test 1
(5.3)
The antiderivative
Power rule for integrals
Integral of sin(x), cos(x) & e^x
Integral of x^-1
A more complicated integral expression
Initial value problem with pos/vel/acc
(5.1, 5.2)
Approximating area under curve (N rectangles)
Intro to Riemann Sum
Area under curve using graph
(5.4)
The fundamental theorem
Evaluating a definite integral with FTC
Breaking up an integral inverval
Swapping the bounds on an integral
(5.5)
Derivative of an integral
Area under a rate function
(5.6)
Integral and pos/vel/acc
Water in tub problem
(5.7)
Integration by u-substitution
An example using u-substitution
Another example using u-substitution
Integral of tan(x) via u-substitution
Area under curve with u-substitution
Test 2
(6.1)
Area between two curves
Area between three curves
(6.2)
Volume by cross-sections
Another volume by cross-sections
Volume of a sphere by cross-sections
Average value of a function
Average value example
(6.3)
Volumes of revolution - Disks
Volumes of revolution - Between 2 disks
(6.4)
Volume by shells
Shell Method Example
Shells with two functions of x
Shells with two functions of y
(6.5)
Work - Raising Weight
Work - Pumping Water (Triangular Prism)
Work - Pumping Water (Inverted Cone)
Work - Pumping Water (Cylinder)
(7.1)
Integration by parts (one iteration)
Integration by parts (two iterations)
Integration by parts (manipulating equations)
Integral of ln(x) (by parts)
(7.2)
Odd power of sinx or cosx example
Another odd power of sinx or cosx example
Even power of sinx and cosx example
Integral of tan(x) via u-substitution
(7.3)
Integral using trig substitution
Another integral using trig substitution
A trig substitution w/ an even power of cosx
Part two of the video above
Test 3
(7.5)
Integration using partial fractions
Integration using partial fractions (w/long division)
(7.7)
A convergent improper integral
A divergent improper integral
Polynomial Long Division Explanation
Polynomial Long Division Example
(7.9)
Trapezoids to approximate area under curve
An example using trapezoids
Explanation of Simpson's Rule formula
Complete justification for Simpson's Rule
(8.1, 11.2)
Arc length formula
Arc length example
(11.4)
Area between f(theta) and origin
Example with area between f(theta) and origin
(8.3)
Center of mass basics
Test 4
(10.1)
Convergence of a sequence
A convergent sequence example
(10.2)
Partial sums
Infinite series are a limit of partial sums
Finite geometric series formula
Infinite geometric series formula
Infinite bouncing ball and geometric series
(10.3)
Comparison test
Comparison test example
Integral test
Integral test example
Limit comparison test
(10.4)
Alternating series test
Absolute vs conditional convergence
(10.5)
Ratio test
(10.6)
Power series & radius of convergence
Using ratio test to find radius of convergence
Representing a function as a power series
(10.7)
Maclauren/Taylor series intro
Series for e^x
Series for sinx
Series for cosx


HW Help

Test 1
(5.3) 1-15, 40-44

[Q1]

(9) If taking the integral of the given functions f(x) is proving difficult, you might try taking the derivative of the given functions F(x) and work backwards.

(40, 41) Like the advice given in 9, it might help to work backwards. Rather that looking at the graph of f and deciding which is a graph of its antiderviative, you might look at the possible antiderivative graphs and ask which of them has f for the graph of its derivative. By process of elimination, you can find the solution.

Email Q&A
(5.1) 8-16, 22-30

[Q1]

(8-16, 21, 22) To find R_N, you must divide the interval [a,b] into N equal subregions. Then, you build rectangles as tall as the function is tall at the right endpoint of each of those equal subregions. Finally, you add up the areas of those N rectangles, and you will have R_N computed. For L_N, you will build rectangles as tall as the function is tall from the left endpoint of each subregion. For M_N, you will build rectangles as tall as the function is tall from the midpoint of each subregion.

(23-28) You want to look at the form for each sum given and notice what values that make up each term are changing. The values that are changing need to use j to represent them in summation notation. The values that are not changing will not require the use of j.

(5.2) 2, 4, 8, 14, 34, 36, 44, 46, 50, 54

[Q1]

(2, 4, 8) Once you have the graph of each function drawn, your task will be to compute the area between f(x) and the x-axis on the interval [a,b] given. Regions of area below the x-axis will be accounted for negatively. Regions of area above the x-axis, positively.

(14) The function |f(x)| will take every region of f(x) that is below the x-axis and reflect it to be above the x-axis, hence making every function value that was negative into a positive of equal magnitude.

(46) If you distribute the integral across the sum, you can then compute the integral of the function x on [0,5] by drawing a graph and computing the area geometrically.

(5.4) 2, 3, 5-10, 25-30, 43, 44, 49, 50, 55, 56, 59, 60

[Q2]

(27-29) Since you are integrating a trig function with something inside other than just x or theta, you need to properly account for the chain rule when you find the antiderivative function for each. For example, note that the trig function with (1/3)*theta inside would result in a multipled factor of (1/3) when taking the derivative. Hence, we are going to need a multipled factor of 3 when taking the antiderivative.

(43-44) Where the expression inside the absolute value is negative, the absolute value will flip that negative to a positive. For example, if we are looking at f(x)=|3-x|, this function is just 3-x when x<3. But, when x>3, the function |3-x| is the opposite of the negative value inside...so the function acts like x-3 for those x>3 values.

(55-56) The area between the function and the x-axis on the interval [-2,3] is the same as the area on the interval [-2,2] plus the area between [2,3]. This is simply a matter of a whole being the sum of its parts.

(59-60) The F(1)=3 bit can help you find the C associated with the general form of F when you integrate. Once you find what C needs to be, finding F(4) is purely a matter of computation.

(5.5) 4-5, 7-10, 17-18, 21-24, 28-30, 42

[Q2]

(17-18) You don't actually have to integrate the functions. The integral symbol will be in your final expression for F. The initial condition merely helps you with bounds.

(28) Note the top bound is not just x but x^2. The chain rule will need to be applied to x^2 when we take the derivative to find G'(x).

(29-30) Similar to 28. The top bound is not just x. We need to apply the chain rule to that top bound. (Note: Even in the case where the top bound is just x, we can still apply the chain rule if we wish...that will just multiply the expression we get by 1...which is exactly what you usually get in those cases.)

(42) The integral on [a,b] calculates an error free estimate for the area between f(x) and the x-axis. Note that when the function is below the x-axis, the function values will be negative, which will cause the regions of area below the x-axis to be accounted negatively by the integral calculation. This function f(x)=x^2-9 only has so much negative area. Isolate that negative area with a careful choice of a and b and you will account for all the negative area and none the positive area by this integral, hence achieving a minimum for the value of this integral.

(5.6) 2-4, 8-10
(8-10) To compute distance traveled, you have to consider both area above the x-axis and area below the x-axis as being positive. This will demand you first determine at which points the function crosses the x-axis (ie. where the function takes a value of zero). At those points, you will need to break your integral area computation into pieces. If any of the area calculations are negative, you will need to use the absolute value of those value in your computation. Ultimately, you will be just adding up the positive values of the areas for the intergrals you computed.
Email Q&A
(5.7) 7-10, 20-22, 29-43 odd, 80-86
(29-43 odd) When you have two pieces of your function that are polynomials differing by a degree of 1, substitution is a method that should come to mind. If you set the higher power of those polynomials to u, then du will be of one degree less. You will then need to manipulate the expression for du to be able to be substituted into the original integral.

(80-86) Note the functions given are all in terms of x. That is to say, all of these functions live in the xy-plane...the bounds given for the area regions given along the x-axis. Once you change variable to u in order to integrate, your function will now be living in the uy-plane. You will need to change the bounds on your area regions accordingly (using the relation between x and u you used to substitute originally) so that the bounds given along the x-axis now are bounds along the u-axis.

Email Q&A
Test 2
(5.8) 7-9, 20-22

(5.9) 5, 6

[Q3]

(5.8 #8) Note tan^-1(x) is not necessary. This integral can be computed via substitution.

(5.8 #20-21) Using u=x^2 will get these functions starting to look like the standard forms of tan^-1(x) and sec^-1(x).

(5.8 #22) You know facts like (2+3)/7 = (2/7)+(3/7). If you apply this kind of idea here, you'll be able to break this single integral into a pair of integrals, each of which will be just like problems you saw previously in 7-9.

(5.9 #5-6) k=ln(2)/(doubling time) will give you the constant you need to put into N(t)=(N_0)*e^(kt).

(6.1) 2-4, 9, 19, 35

(6.2) 4, 8, 9

[Q3]

(6.1 #3-4) You need to find the intersection points to determine the bounds of the enclosed region.

(6.1 #19, 35) When encountering three functions, you can observe that moving left to right within your region, the region is always only dependent upon two functions. Breaking into proper subregions accordingly, you can compute the area of the desired region. See Image.

(6.2 #8) Drawing a line from the top vertex of an equilateral triangle to its base, you can use 30-60-90 triangles to show that the area of an equilateral triangle is (sqrt(3)/4)*s^2, where s is the length of a side.

Email Q&A
(6.2) 17, 19

(6.3) 1-2, 5-6

[Q4]

(6.2 #17) You will need to use similar triangles to solve for each the length of the minor axis and length of the major axis at a generic height. Using the major axis, you will get something like (12/6) = ((12-y)/b). The minor axis is similar but using the side length of 4.

(6.2 #19) This is a tough problem. Drawing pictures and keeping the letters straight will be a major concern. To find (a), you can use similar traingles. Modifying the picture used to analyze the similar triangles in (a), you can find (b). If you've found (a) and (b), then (c) is merely a matter of integration. Recall all letters except x will be treating as if they were constants (ie. some number).

(6.3 #1-2, 5-6) When revolving a function of x around the x-axis, chop up the x-axis interval and draw a slice. The slice will be a thin round disk.

(6.3) 15-16, 27-28, 39-40

(6.4) 7-10, 21-22, 35-36

[Q4]

(6.3 #15-16, 27-28) The more explicit you are at describing the top function and the bottom function, the easier it is to apply this information when considering the volume of a representative slice of these regions.

(6.3 #39-40) The same hint as for 15-16, 27-28 applies. Additionally here, we must consider not just how far the top and bottom functions are from the x-axis but how far they are from the axis of rotation.

(6.4 #7-10) Find intersection points to have bounds for your region of interest.

(6.4 #21-22) Even thoough there are three functions given, your expression for the area will not need to depend on all three. Even if done roughly, a sketch can be really helpful when trying to visualize the region being revolved.

(6.4 #35-36) Be as explicit as possible in your description of the functions that define the boundaries of this region. This will allow you to practically plug and chug when expressing the volume of a slice.

(6.2) 39-42

(6.5) 4, 6, 12, 13

[Q4]

(6.2 #39-42) These are direct applications of the average value formula. You simply need to find the area between f(x) and the x-axis on the intervals and then divide that number by the width of the interval.

(6.5 #12-13) Work problems involve determining a general expression for the work on a given slice of the figure of interest. In particular, if the column or cone has height of 4, you could think of these objects existing between y=0 and y=4. The slice you take would have to be at a generic y between 0 and 4. You can use known methods from sections 6.3 or 6.4 to find the volume of tye slice. Multiply by density to find mass. Multiply by g to find weight. Multiply by the height of the lift (namely y) to find work. Integrate along the region of interest (y=0 to y=4), and you'll have it.

(7.1) 4, 8, 12, 14, 20, 24, 50, 52

[Q5]

(7.1 #8) If you distribute what is in the parentheses, you'll get one integral of 2x*e^x (which requires parts) and another of just e^x (which is cookie cutter).

(7.1 #8, 12, 14) For integrals of the form x^n times cosine, sine, or e, you need to set u equal to x^n and dv equal to everything else. The power of x will indicate how many times you will need to use parts before you will have fully evaluated this kind of integral.

(7.1 #20, 24, 50) For integrals that involve ln(x), you can set u equal to ln(x), or (ln(x))^2, and let dv be everything else.

(7.1 #52) This is a ticky problem. You may want to use substitution and let u=x^3 since the denominator has an expression of degree 2 within it. However, if you try this out, you will notice it does not work. To exploit substitution, you can make dv equal (x/sqrt(x^2_9))*dx...you will be able to utilize substitution on that when you find v. Hence, u will have to be everything else...namely x^2. If you now run through parts, you will be able to make the integration work. Tough problem!

(7.1) 60, 61, 82, 85

[Q5]

(7.1 #60-61) See hint above from 7.1 #8, 12, 14.

(7.1 #82) See hint aboce from 7.1 #20, 24, 50

(7.1 #85) Treat b^x almost like e^x. Just known that the derivative of b^x is not simply b^x, but b^x*ln(b). With this and the hint from 7.1 #8, 12, 14 in mind, you can solve this problem without too much trouble.

(7.2) 1, 9, 10

(7.3) 1-5, 7, 8

(7.2 #1) Integrals with only powers of sine and cosine being multiplied can be attacked with cos^2x+sin^x=1 when you have an odd power. Use that identity to reduce the odd power to a power of 1 and then utilize substitution.

(7.2 #9, 10) Integrals involving powers and sine and cosine that do not have an odd power demand the half angle formulas of cos^2x=(1/2)+(1/2)*cos2x and sin^2x=(1/2)-(1/2)*cos2x. Repeated use of these formulas will allow integrals of sine and cosine without odd powers to be solved.

(7.2) 55, 57

(7.3) 15, 17, 19

(7.2 #55) To integrate tangent, you can express this in terms of sine over cosine and utilize substitution. This is a classic Calc II problem!

(7.2 #57) Taking cos^2x+sin^2x=1 and dividing both sides by cos^2x yields 1+tan^2x=sec^2x. Applying this once to the expression sec^4x should give you a pair of integrals to work with, each of which can be dealt with with (at worst) substitution.

(7.3 #15, 17, 19) Integrals involving sqrt(a+x^2) need the substituion x=sqrt(a)*tan(theta). Integrals involving sqrt(x^2-a) need the substitution x=sqrt(a)*sec(theta). Integrals involving sqrt(a-x^2) need the substitution x=sqrt(a)*sin(theta). These three facts should be memorized, and it will probably be worth your time to become familiar with integrals of these types.

Test 3
(7.5) 5-8, 9-10, 15-18

[Q6]

(7.5 #5-8) Since numerator degree is not less than denominator degree, we will need to perform long division before using partical fraction decomposition. Fortunately, all four denominators are already fully factored with respect to the real numbers, so integrating the remainder fraction will remain somewhat manageable.

(7.5 #16) Note that x^2+x is not fully factored yet. You'll need to factor it fully before you attempt to decompose into partial fractions.

(7.5 #17-18) The repeated factors in the denominator of these expressions need to be dealt with by including a fraction in the partial fraction decomposition for each power of the repeated factor (up to the power present).

(7.5) 27-30

[Q6]

(7.5 #27) Since the factor in the denominator is a repeat factor of degree three, you will need three fractions in your partial fraction decomposition. One with the factor to the first power in the denominator, one with the factor to the second power, and one with the factor to the third.

(7.5 #28-29) Similar to the hint immediately above, but with two fractions in your decomposition for each of the repeated factors of degree 2.

(7.5 #30) Note the numerator degree is not less than the denominator degree, so we will need to perform long division before using partial fraction decomposition.

(7.7) 5-10, 19-24, 38, 40, 45-47

[Q7]

(7.7 #9-10) Note the undefined nature of the function at x=0. Even though the bounds of this integral look unintimidating on the surface, this undefined nature means a limit will need to be used on the integral bound that is zero to properly account for this infinitely tall region of area.

(7.7 #22-23) Similar to above. The bounds on these integrals appear on the surface to present no issue. Looking further reveals an undefined nature at x=1 for #22 and at x=3 for #23. The bounds corresponding to these asymptotes will need to be accounted for with a limit.

(7.7 #24) Here, we also have bounds that appear harmless on the surface. Moreover, neither x=-4 nor x=0 are locations of an asymptote for this function. Be careful though. There is an asymptote between x=-4 and x=0...in particular, at x=-2. To properly account for this, we'll need to split the integral into two regions: one on [-4,-2], and another on [-2,0]. For each integral, we will need to take a limit for the bound at x=-2.

(7.7 #38) ln(1)=0.

(7.7 #45-47) To deal with the doubly improper integral region, we will have to break this into a pair of singly improper integrals. Pick any value along the number line to split the integral. For simplicity, let's say x=0. This will give us two regions: one on (-inf,0], and another on [0,inf). Dealing with each infinity via a limit, we can properly account for the area between this curve and the x-axis.

(7.7) 54-56, 58, 59, 60-64

[Q7]

(7.7 #59) Since sin(x) ranges between -1 and 1, the numerator of this expression ranges between 0 and 2. This means, no matter the x value, the function inside the integral is less than or equal to 2/x^2. Now, you can use comparison to show convergence.

(7.7 #61-62) In 61, since we have an x^5 expression square rooted, the effective power of x is 5/2. You can use this for comparison. #62 is similar, but with an effective power of 3/2.

(7.7 #63) The effective power here is 1/2. Knowing what we know about the p-test for integrals, we should expect divergence and work by comparison to show this.

(7.7 #64) Note we are dealing with a vertical asymptote at x=0. When x gets small (in absolute value), the smaller powers of x will dominate the expression. Comparison with x^(1/3) will allow you to show convergence.

(7.9) 1, 4, 5, 10, 12

[Q7]

Note that M_N and T_N are just ways to estimate the area under a curve. Both involve chopping the interval of interest into N subintervals. M_N involves sampling function values at the midpoint of each subinterval. T_N involves averaging the function values at the left and right endpoints of each subinterval (effectively an average of R_N and L_N).
(7.9) 13, 18, 27

[Q8]

(7.9 #27) Use the data points given as your function values. You can think of these data points as the height of the function at each point. Then, it is simply a matter of applying the formula for Simpson's Rule.
(8.1) 3-8, 11, 13

(11.2) 1-5

[Q8]

(8.1 #4) Note that ((1/4)x^3 + x^-3)^2 is equal to 1+(f')^2.

(8.1 #7) You can use u-substitution, letting u equal what is inside the sqrt.

(8.1 #8) Note that ((1/2)x^(1/2) + (1/2)x^(-1/2))^2 is equal to 1+(f')^2.

(8.1 #11, #13) Let your function be the arc length formula, sqrt(1+(f')^2). This, use the trapezoidal rule or simpsons rule on that formula (on the interval given).

(11.2 #4) Similar to 8.1 #7.

(11.2 #5) If you factor out the sqrt(t^2)...or just t, you can use u-substitution letting u equal the degree 2 expression remaining inside the sqrt. The t your factored out will work well with du.

(11.4) 5, 6, 8, 10, 25, 28

[Q8]

(8.1) 48, 49

[Q8]

(8.3) 1-10, 11, 20, 21

[Q8]

After Test 3
(10.1) 1, 2, 3, 5, 14
(14) Use limit rules like lim(a_n + b_n) = lim(a_n) + lim(b_n), lim(a_n * b_n) = lim(a_n) * lim(b_n), and lim(c * a_n) = c* lim(a_n).
(10.2) 2-4, 11, 12, 23-26, 40
(11) Note: many of the terms in each partial sum will cancel. You will be able to express S_N in terms of just two terms. Moreover, one of those two terms will go to zero as N-->infinity.

(12) Decompose 1/[n(n+1)] into partial fractions. Then, many of the terms will cancel just like in (11) above.

(25) Negative exponents can be expressed as positive exponents by bringing numerator terms to denominator (and vice versa). Hence, (a/b)^(-n) is equal to (b/a)^n.

(40) Note that .217217217... is equal to .217 + .000217 + .000000217 + ..., and this is equal to 217/1000 + 217/(1000)^2 + 217/(1000)^3 + ... . Factoring out the first term, you get something of the form c*(r^0 + r^1 + r^2 + ...).

(10.3) 1, 4, 9, 16, 19, 20, 24
(4, 9, 16, 20, 24) You can compare highest degree terms in the numerator and denominator to use limit comparison. A couple examples follow. In #4, the highest degree terms take the form of 1/[n^(1/2)]. Let this be your b_n in the limit comparison. In #16, the highest degree terms take the form of 1/[sqrt(n^2)], or just 1/n. This can be your b_n in limit comparison.

(19) You know 1/(2^n) is the same as (1/2)^n. A series of the form r^n is geometric, and can be easily determined to be convergent or divergent based on the absolute value of r. You can use the (1/2)^n as comparison in this case.

(10.3) 39-41, 49-53

(10.4) 1-6, 17-20

(10.3 #39, 40, 41, 49, 51, 53) Similar to 10.3 #4, 9, 16, 20, 24. For example, in #41, highest degree terms take the form of n/[sqrt(n^3)], which is n/[n^(3/2)], or simply 1/[n^(1/2)]. Use this for limit comparison.

(10.3 #50, #62) Recall that sin(Any Angle) and cos(Any Angle) is between -1 and 1. In #50, you encounter a cos^2 term, which must always be less than 1...so cos^2(-)/n^2 must be less than 1/n^2. Compare. In #62, the cos(n) being between -1 and 1 will not dictate the limit as n --> infinity. The n^2 term in the denominator will.

(10.4 #4) The terms are of the form n^4/n^3 and will not approach zero.

(10.4 #6) Recalling that sin(Any Angle) is between -1 and 1, then this sum takes a value between that of -1/n^2 and 1/n^2. And you know the value of those two sums (convergent, by p-test).

(10.4 #17) Similar to 10.2 #25. In particular, 7^(-n) is equal to (1/7)^n.

(10.4 #19) Observe that 5^n-3^n is greater than 2^n when n>1. This will allow you to compare.

(10.4 #20) Highest degree terms take the form 1/n. Limit compare.

(10.5) 1-4, 9-12

(10.6) 9, 10

(10.5 #3) Note that (n+1)/n > 1. Hence, [(n+1)/n]^n (some number greater than 1 times itself n times) is greater than 1 as well. Finally, note the expression [(n+1)^(n+1)]/[n^n] is equal to (n+1)*[(n+1)/n]^n. Using the fact that latter tem in [--] is greater than 1, we can conclude something about the limit of a_(n+1)/a_n.

(10.5 #4) This expression is not conducive to the ratio test (it lacks an n! and an exponent of n). Use limit comparison to get a conclusive result.

(10.5 #11) Similar to 10.5 #3, but with a factor of e as well. Just recall that e is a fixed number. The n+1 term though grows boundlessly as a limit when n --> infinity.

(10.5 #12) Note that [n^40/(n+1)^40] expands to be of the form [n^40/(n^40+...)], where the dot dot dot is terms of degree lower than 40. Since the numerator and denominator are of equal degree, the limit as n --> infinity will be the ratio of the leading coefficients...namely 1.

(10.6 #9, 10) Since power series necessarily have an exponent of n, the ratio test is going to be the test of choice with any power series radius of convergence problem. The |a_(n+1)/a_n|>1 and <1 cases are conclusive. Since the ratio test does not yield a conclusive result when |a_(n+1)/a_n|=1, we have to be sure to test a_(n+1)/a_n=-1 and a_(n+1)/a_n=1 cases separately.

Series Test User Guide
(10.6) 7, 16, 31, 32

(10.7) 3, 8, 29, 30, 42, 53, 55

(10.6 #7, 16, 31, 32) Similar to 10.6 #9, 10. Power series, since they have n in the exponent, work well with the ratio test. In #16, note that regardless of the fixed choice of x, the expression x/(n+1) will limit to zero as n --> infinity. In #31, note that [x^(n+1)+...]/[x^n+...] is approximately equal to x^(n+1)/x^n as n gets large. Like in the case of #16, regardless of the choice of x, note that -5/(n+1) will limit to zero as n --> infinity. In #32, note that only a *very* small value of x will allow |(n+1)*x| to limit to be less than 1.

(10.7 #3) Note that 1/(1-2x) is of the form 1/(1-r), where r=2x. This series takes the form r^0 + r^1 + r^2 + ...

(10.7 #8) We need to use Taylor's formula and find a trend in the terms f^(n)(0)/n!. Note that f'(x)=4e^(4x) via the chain rule. Hence f^(n)(x)=4^n*e^(4x).

(10.7 #29, #30) Each coefficient of a_0 + a_1(x-c) + a_2(x-c)^2 + ... is of the form a_n = f^(n)(x)/n!. Compute the first few terms of this form to see the trend in the expression of the coeffcients.

(10.7 #42) Similar to 10.7 #3. 1/(1-x) is of the form 1/(1-r), where r=x.

(10.7 #53) Use cos(t) = 1 - t^2/2! + t^4/4! - t^6/6! + ..., then subtract 1 and divide out the t.

(10.7 #55) Use ln(1+t) = t - t^2/2! + t^3/3! - t^4/4! + ...