124 - Answers to even (and some other) assigned problems
5.5:
5.7:
- 16) a) A/x + B/x^2 + C/(x + 1) b) A/x + (Bx + C)/(x^2 + 1)
Appendix G:
- 16) 3/2 + ln(3/2) (NOTE: ln 3 - ln 2 = ln(3/2))
- 34) ln|x| + 1/(x^2 + 1) + C
5.10:
- 2) a) Not improper, only discontinuity is at 1/2 which is not in range
of integration.
b) Improper, not continuous at 1/2, which is in range of integration.
c) Improper since +/- infinity.
d)Improper since undefined at x = 1.
7.4:
- 4) a) 120 , b) 120e^((ln 125)/6)t) = 120 e^((ln 5)/2)t) = 120*5^(t/2) ,
c) 6,708 , d) 5,398.2 , e) 9.2 hours
- 10) a) 12.25 years , b) 28.45 years
- 12) 5e^(2x)
- 20) a) 11.6 years , b) 6.2 %
7.5:
- 6) Assume K = 4 billion = 4000 million.
- a) With t = 0 being 1990, units of t in years and units of P(t) in millions
get P(t) = 4000/(1 + 15*e^(-kt))
- b) k = -(1/10)*ln(149/165) or approx. equal .0102
- c) approx. 679.8 million in 2100 , approx. 1.45 billion in 2200.
- d) The population will be greater than 300 million in 2010.
- 8)
- a) Using k is approx. 1.19 , P(t) = 10000/(1 + 24*e^(-1.19t))
- b) approx. 2.68 years.
8.1
- 16) Diverges
- 18) Converges to 1
- 20) Converges to pi/2
- 22) Diverges
- 28) Converges to 0
- 38) a) L b) 1, .5, .66667 ,
.6, .625, .61538 , .61905, .61765, .61818, .61798 c) (-1 + \sqrt(5))/2
8.2
- 14) Diverges
- 20) Diverges
- 22)
Diverges
- 26) Converges to (cos 1)/( 1 - cos 1)
- 30) Diverges
8.3:
- 4) a) Diverges b) Nothing
- 8) Converges
- 14) Diverges
- 20) Converges
- 22) Diverges
- 26) Converges
- 28)
- a) S_10 = 1.0820366... using calculator. R_10
< 1/3000 < .00034 < .0004 (rounding up to one significant digit)
- b)
Aprroximating S by the midpoint of the interval [S_10 + 1/3993 , S_10 +
1/3000] gives an approximate value of S of 1.0823285... . This gives
|error| < 4.2 x 10^(-5) < 5 x 10^(-5) = .00005 (rounding up to one
significant digit). Since this is < .0001 we would say accurate to three
decimal places and thus we can write S is approximately 1.082... .
- c)
If use |R_n| < \int_n^\infty 1/x^4 dx to get |R_n| < .00001 we need 1/(3n^3)
< .00001 , which gives n > 32.1 , so n = 33 will work. If we compute S_33
since we have it is accurate to four decimal places we can write S is
approximately 1.0823... .
- 30) n = 4 is the smallest n that will
work. So S is approximately S_4 = 1.036
- 31) Need n = 40,000 to get R_n
< .01 , but using the midpoint approximation just need n with (1/sqrt(n) -
1/sqrt(n+1)) < .01 . This is difficult to solve for n, but calculating
values will show the smallest n that works is n = 14. So calculating S_14
shows S = 2.60... .
- 32) e^(100) .
8.4:
- 2) a) Diverges
b) Converges c) Nothing
- 8) Converges, but not absolutely.
- 15) Need a value of n with 1/(2n+1)! < .00001 , i.e. (2n+1)! > 100,000 .
One can't solve for n here, but experimenting with values of n shows n = 4
is the smallest n that works. Since S_4 = .84146... we can say with
certainty that the first four decimals of S are .8414 . Note however that
S_4 < S (the last term of S_4 is negative) so by rounding UP the fifth
digit 6 in S_4 one would have S is closer to .8415 then to .8414.
- 16) Need a value of n with 1/(2n+2)! < .00001 , i.e. (2n+2)! > 100,000.
Experimenting with values of n shows n=4 is the smallest n that will work.
We can S is approximately \sum_{n=0}^4 (-1)^n/(2n)! (note this is actually
S_5 = sum of first FIVE terms since series begins with n = 0).
Since S_5 = .54030... we can say S = .5403 (accurate to four decimal places).
- 17) Need a value of n with (n+1)^2/10^n < .00001 . We can't solve for
n, but experimenting with values of n shows the samllest n that works is
n=6. Note S_6 < S (since the last term of S_6 is negative). Since S_6 -
.067614 we can say S is approximately .0676 (to four decimal places).
- 19) Does not converge
absolutely. Diverges (by n-th term test).
- 20) Converges absolutely.
- 21) Converges absolutely.
- 22) Does not converge absolutely, but
converges.
- 23) Does not converge absolutely but converges.
- 24)
Does not converge absolutely. Diverges.
- 25) Converges absolutely.
- 26) Converges absolutely.
- 27) Converges absolutely.
- 28)
Does not converge absolutely. Diverges.
8.6
- 2) It converges for |x| < 2 because it is the integral of the given
series and integrating a series does not change the radius of convergence.
- 12 a) \sum_{n=0}^\infty (-1)^n x^{n+1}/(n + 1) for |x| < 1 b)
\sum_{n=0}^\infty (-1)^n x^{n+2}/(n + 1) for |x| < 1 c) \sum_{n=0}^\infty
(-1)^n x^(2n+2)/(n+1) for |x| < 1.
8.7 - 2
a) If it were the series for f(x) at 1, then the degree one coefficient
would be f'(1) = -0.8. However the graph of the function shows the slope at
x = 1 , which is equal to f'(1), satisfies f'(1) > 0.
- 2b) If it were
the series for f(x) at 2, then the degree two coefficient would be f"(2)/2 =
1.5, which would imply f"(2) = 3. However the graph of the function is
concave down at x = 2, which would imply f"(2) < 0.
8.9
- 11) With M = (3/8)*4^(-5/2) = 3/256 get |R_2(x)| < 1.6 x 10^-5.
- 13) With M = (56/81)*(.8)^(-10/3) < 1.5 get |R_3(x)| < 1 x 10^-4 .
- 15) With M = (16(.1)^4 + 48(.1)^2 + 12)*3^(.01) < 12.7 get |R_3(x)| <
5.3 x 10^-5 .
- 17) With M = 6 get |R_4(x)| < .05 .
Chapter 8 Review
- 2) Converges to zero.
- 18) Diverges.
- 20) cos(sqrt(pi)/3)
- 22) e^(-e)
- 25) We need a value of n with 1/(n+1)^5 < .00001 = 10^(-5),
i.e. (n+1)^5 > 10^5, so n+1 > 10 or n > 9. Taking n = 10 and computing
S_10 gives S_10 = .97211... , so we can say S = .9721 (accurate to four
decimal places).