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n / 2 n , n 2

bounded, decreasing for n 1

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sin n

bounded, not monotone

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n 1 / n , n 3

bounded, decreasing

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tan n

not monotone, not bounded

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Determine whether the sequence defined as follows has a limit. If it does, find the limit.

a 1 = 2 , a 2 = 2 2 , a 3 = 2 2 2 etc.

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Determine whether the sequence defined as follows has a limit. If it does, find the limit.

a 1 = 3 , a n = 2 a n 1 , n = 2 , 3 ,… .

a n is decreasing and bounded below by 2 . The limit a must satisfy a = 2 a so a = 2 , independent of the initial value.

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Use the Squeeze Theorem to find the limit of each of the following sequences.

cos ( 1 / n ) 1 1 / n

0

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a n = sin n sin ( 1 / n )

0 : | sin x | | x | and | sin x | 1 so 1 n a n 1 n ) .

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For the following sequences, plot the first 25 terms of the sequence and state whether the graphical evidence suggests that the sequence converges or diverges.

[T] a n = cos n

Graph oscillates and suggests no limit.
This is a graph that oscillates between 1 and -1 from 0 to 25 on the x axis. There appears to be no limit.

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Determine the limit of the sequence or show that the sequence diverges. If it converges, find its limit.

a n = ( 2 n ) 1 / n n 1 / n

n 1 / n 1 and 2 1 / n 1 , so a n 0

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a n = ln ( n 2 ) ln ( 2 n )

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a n = ( 1 2 n ) n

Since ( 1 + 1 / n ) n e , one has ( 1 2 / n ) n ( 1 + k ) −2 k e −2 as k .

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a n = ln ( n + 2 n 2 3 )

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a n = 2 n + 3 n 4 n

2 n + 3 n 2 · 3 n and 3 n / 4 n 0 as n , so a n 0 as n .

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a n = ( n ! ) 2 ( 2 n ) !

a n + 1 a n = n ! / ( n + 1 ) ( n + 2 ) ( 2 n ) = 1 · 2 · 3 n ( n + 1 ) ( n + 2 ) ( 2 n ) < 1 / 2 n . In particular, a n + 1 / a n 1 / 2 , so a n 0 as n .

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Newton’s method seeks to approximate a solution f ( x ) = 0 that starts with an initial approximation x 0 and successively defines a sequence x n + 1 = x n f ( x n ) f ( x n ) . For the given choice of f and x 0 , write out the formula for x n + 1 . If the sequence appears to converge, give an exact formula for the solution x , then identify the limit x accurate to four decimal places and the smallest n such that x n agrees with x up to four decimal places.

[T] f ( x ) = x 2 2 , x 0 = 1

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[T] f ( x ) = ( x 1 ) 2 2 , x 0 = 2

x n + 1 = x n ( ( x n 1 ) 2 2 ) / 2 ( x n 1 ) ; x = 1 + 2 , x 2.4142 , n = 5

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[T] f ( x ) = e x 2 , x 0 = 1

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[T] f ( x ) = ln x 1 , x 0 = 2

x n + 1 = x n x n ( ln ( x n ) 1 ) ; x = e , x 2.7183 , n = 5

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[T] Suppose you start with one liter of vinegar and repeatedly remove 0.1 L, replace with water, mix, and repeat.

  1. Find a formula for the concentration after n steps.
  2. After how many steps does the mixture contain less than 10 % vinegar?
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[T] A lake initially contains 2000 fish. Suppose that in the absence of predators or other causes of removal, the fish population increases by 6 % each month. However, factoring in all causes, 150 fish are lost each month.

  1. Explain why the fish population after n months is modeled by P n = 1.06 P n 1 150 with P 0 = 2000 .
  2. How many fish will be in the pond after one year?

a. Without losses, the population would obey P n = 1.06 P n 1 . The subtraction of 150 accounts for fish losses. b. After 12 months, we have P 12 1494 .

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[T] A bank account earns 5 % interest compounded monthly. Suppose that $ 1000 is initially deposited into the account, but that $ 10 is withdrawn each month.

  1. Show that the amount in the account after n months is A n = ( 1 + .05 / 12 ) A n 1 10 ; A 0 = 1000 .
  2. How much money will be in the account after 1 year?
  3. Is the amount increasing or decreasing?
  4. Suppose that instead of $ 10 , a fixed amount d dollars is withdrawn each month. Find a value of d such that the amount in the account after each month remains $ 1000 .
  5. What happens if d is greater than this amount?
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[T] A student takes out a college loan of $ 10,000 at an annual percentage rate of 6 % , compounded monthly.

  1. If the student makes payments of $ 100 per month, how much does the student owe after 12 months?
  2. After how many months will the loan be paid off?

a. The student owes $ 9383 after 12 months. b. The loan will be paid in full after 139 months or eleven and a half years.

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[T] Consider a series combining geometric growth and arithmetic decrease. Let a 1 = 1 . Fix a > 1 and 0 < b < a . Set a n + 1 = a . a n b . Find a formula for a n + 1 in terms of a n , a , and b and a relationship between a and b such that a n converges.

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[T] The binary representation x = 0 . b 1 b 2 b 3 ... of a number x between 0 and 1 can be defined as follows. Let b 1 = 0 if x < 1 / 2 and b 1 = 1 if 1 / 2 x < 1 . Let x 1 = 2 x b 1 . Let b 2 = 0 if x 1 < 1 / 2 and b 2 = 1 if 1 / 2 x < 1 . Let x 2 = 2 x 1 b 2 and in general, x n = 2 x n 1 b n and b n 1 = 0 if x n < 1 / 2 and b n 1 = 1 if 1 / 2 x n < 1 . Find the binary expansion of 1 / 3 .

b 1 = 0 , x 1 = 2 / 3 , b 2 = 1 , x 2 = 4 / 3 1 = 1 / 3 , so the pattern repeats, and 1 / 3 = 0.010101 .

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[T] To find an approximation for π , set a 0 = 2 + 1 , a 1 = 2 + a 0 , and, in general, a n + 1 = 2 + a n . Finally, set p n = 3.2 n 2 a n . Find the first ten terms of p n and compare the values to π .

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For the following two exercises, assume that you have access to a computer program or Internet source that can generate a list of zeros and ones of any desired length. Pseudorandom number generators (PRNGs) play an important role in simulating random noise in physical systems by creating sequences of zeros and ones that appear like the result of flipping a coin repeatedly. One of the simplest types of PRNGs recursively defines a random-looking sequence of N integers a 1 , a 2 ,… , a N by fixing two special integers K and M and letting a n + 1 be the remainder after dividing K . a n into M , then creates a bit sequence of zeros and ones whose n th term b n is equal to one if a n is odd and equal to zero if a n is even. If the bits b n are pseudorandom, then the behavior of their average ( b 1 + b 2 + + b N ) / N should be similar to behavior of averages of truly randomly generated bits.

[T] Starting with K = 16,807 and M = 2,147,483,647 , using ten different starting values of a 1 , compute sequences of bits b n up to n = 1000 , and compare their averages to ten such sequences generated by a random bit generator.

For the starting values a 1 = 1 , a 2 = 2 ,… , a 1 = 10 , the corresponding bit averages calculated by the method indicated are 0.5220 , 0.5000 , 0.4960 , 0.4870 , 0.4860 , 0.4680 , 0.5130 , 0.5210 , 0.5040 , and 0.4840 . Here is an example of ten corresponding averages of strings of 1000 bits generated by a random number generator: 0.4880 , 0.4870 , 0.5150 , 0.5490 , 0.5130 , 0.5180 , 0.4860 , 0.5030 , 0.5050 , 0.4980 . There is no real pattern in either type of average. The random-number-generated averages range between 0.4860 and 0.5490 , a range of 0.0630 , whereas the calculated PRNG bit averages range between 0.4680 and 0.5220 , a range of 0.0540 .

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[T] Find the first 1000 digits of π using either a computer program or Internet resource. Create a bit sequence b n by letting b n = 1 if the n th digit of π is odd and b n = 0 if the n th digit of π is even. Compute the average value of b n and the average value of d n = | b n + 1 b n | , n = 1 ,... , 999 . Does the sequence b n appear random? Do the differences between successive elements of b n appear random?

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Source:  OpenStax, Calculus volume 2. OpenStax CNX. Feb 05, 2016 Download for free at http://cnx.org/content/col11965/1.2
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