diff --git a/ptx/apex.ptx b/ptx/apex.ptx index 18062bbf5..dd9a54d63 100644 --- a/ptx/apex.ptx +++ b/ptx/apex.ptx @@ -83,7 +83,7 @@ - + diff --git a/ptx/sec_FTC.ptx b/ptx/sec_FTC.ptx index 189840c64..79d218056 100644 --- a/ptx/sec_FTC.ptx +++ b/ptx/sec_FTC.ptx @@ -129,7 +129,7 @@ x \geq 1 is given by A(x)=\frac12 (x)(2x)-\frac12 (1)(2)=x^2-1.

-
+
The area of the shaded region is F(x) = \int_1^x 2t\, dt diff --git a/ptx/sec_Modeling.ptx b/ptx/sec_Modeling.ptx index 29391f743..654bed141 100644 --- a/ptx/sec_Modeling.ptx +++ b/ptx/sec_Modeling.ptx @@ -42,7 +42,7 @@ \draw [firstcolor] (-1.5,0) node [text width=60pt,align=center] (a) { \centering The rate of change of the population}; \draw [firstcolor,->] (a) -- (.3,.7); - \draw [firstcolor] (2,.25) node [text width=60pt,align=center] (b) { \centering the population.}; + \draw [firstcolor] (2,0) node [text width=60pt,align=center] (b) { \centering the\\ population.}; \draw [firstcolor,->] (b) -- (1.6,.8); \draw [firstcolor] (.5,2) node [text width=32pt,align=center] (c) { \centering is}; diff --git a/ptx/sec_int_comp_tests.ptx b/ptx/sec_int_comp_tests.ptx index 5dc0ab996..af755b346 100644 --- a/ptx/sec_int_comp_tests.ptx +++ b/ptx/sec_int_comp_tests.ptx @@ -513,6 +513,18 @@ Since the limit on the left side diverges to \infty, we can say that \lim\limits_{n \to \infty}\sum_{i=N}^n b_i also diverges to \infty.

+

@@ -520,18 +532,7 @@ -
Video presentation of diff --git a/ptx/sec_lhopitals_rule.ptx b/ptx/sec_lhopitals_rule.ptx index 495c466c8..fda68e0b7 100644 --- a/ptx/sec_lhopitals_rule.ptx +++ b/ptx/sec_lhopitals_rule.ptx @@ -814,7 +814,7 @@

- \lim\limits_{x\to 1} \frac{ }{} + \lim\limits_{x\to 1} \frac{ }{}

diff --git a/ptx/sec_riemann.ptx b/ptx/sec_riemann.ptx index 521884db5..cfc8f5356 100644 --- a/ptx/sec_riemann.ptx +++ b/ptx/sec_riemann.ptx @@ -794,13 +794,13 @@ \draw (1.5,1) node {$\displaystyle x_i = x_0 + i\Delta x$}; \draw [firstcolor] (1,0) node [text width=32pt,align=center] (a) { \centering starting\\[-5pt] value}; - \draw [firstcolor,->] (a) -- (.95,.75); + \draw [firstcolor,->] (a) -- (1.25,.75); \draw [firstcolor] (2,2.5) node [text width=120pt,align=center] (b) { \centering number of subintervals\\[-5pt] between $x_0$ and $x_i$}; - \draw [firstcolor,->] (b) -- (1.95,1.25); + \draw [firstcolor,->] (b) -- (2.05,1.25); \draw [firstcolor] (3,0) node [text width=60pt,align=center] (c) { \centering subinterval\\[-5pt] size}; - \draw [firstcolor,->] (c) -- (2.75,.75); + \draw [firstcolor,->] (c) -- (2.45,.75); \end{tikzpicture} diff --git a/ptx/sec_sequences.ptx b/ptx/sec_sequences.ptx index fdd2bb488..25f3fec70 100644 --- a/ptx/sec_sequences.ptx +++ b/ptx/sec_sequences.ptx @@ -733,7 +733,7 @@

- Scatter plot for the sequence in + Scatter plot for the sequence in of A scatter plot showing a representative sample of points from the third sequence in this example. @@ -1514,7 +1514,7 @@

    -
  1. +

  2. a_{n+1}-a_n \amp = \frac{n+2}{n+1} - \frac{n+1}{n} @@ -1523,84 +1523,13 @@ \amp \lt 0 \text{ for all \(n\). } Since a_{n+1}-a_n\lt 0 for all n, - we conclude that the sequence is decreasing. -

    -
    3. - -
  3. -

    - - a_{n+1}-a_n \amp = \frac{(n+1)^2+1}{n+2} - \frac{n^2+1}{n+1} - \amp = \frac{\big((n+1)^2+1\big)(n+1)- (n^2+1)(n+2)}{(n+1)(n+2)} - \amp = \frac{n^2+3n}{(n+1)(n+2)} - \amp \gt 0 \text{ for all \(n\). } - - Since a_{n+1}-a_n\gt 0 for all n, - we conclude the sequence is increasing. -

    -
    4. - -
  4. -

    - We can clearly see in , - where the sequence is plotted, that it is not monotonic. - However, it does seem that after the first 4 terms it is decreasing. - To understand why, perform the same analysis as done before: - - - a_{n+1}-a_n \amp = \frac{(n+1)^2-9}{(n+1)^2-10(n+1)+26} - \frac{n^2-9}{n^2-10n+26} - \amp = \frac{n^2+2n-8}{n^2-8n+17}-\frac{n^2-9}{n^2-10n+26} - \amp = \frac{(n^2+2n-8)(n^2-10n+26)-(n^2-9)(n^2-8n+17)}{(n^2-8n+17)(n^2-10n+26)} - \amp = \frac{-10n^2+60n-55}{(n^2-8n+17)(n^2-10n+26)} - . + we conclude that the sequence is decreasing, illustrated in .

    -

    - We want to know when this is greater than, or less than, 0. - The denominator is always positive, - therefore we are only concerned with the numerator. - For small values of n, - the numerator is positive. - As n grows large, - the numerator is dominated by -10n^2, - meaning the entire fraction will be negative; - , for large enough n, a_{n+1}-a_n \lt 0. - Using the quadratic formula we can determine that the numerator is negative for n\geq 5. - In short, the sequence is simply not monotonic, - though it is useful to note that for n\geq 5, - the sequence is monotonically decreasing. -

    -
    5. - -
  5. - -

    - Again, the plot in - shows that the sequence is not monotonic, - but it suggests that it is monotonically decreasing after the first term. - We perform the usual analysis to confirm this. - - a_{n+1}-a_n \amp = \frac{(n+1)^2}{(n+1)!} - \frac{n^2}{n!} - \amp = \frac{(n+1)^2-n^2(n+1)}{(n+1)!} - \amp = \frac{-n^3+2n+1}{(n+1)!} - - When n=1, the above expression is \gt 0; - for n\geq 2, the above expression is \lt 0. - Thus this sequence is not monotonic, - but it is monotonically decreasing after the first term. -

    -
    6. -
-

- -
- Plots of sequences in - - -
- +
+ Plot of the sequence in of - + Plot of the first sequence in this example. It is decreasing and bounded below.

@@ -1631,11 +1560,24 @@

+ + +
  • +

    + + a_{n+1}-a_n \amp = \frac{(n+1)^2+1}{n+2} - \frac{n^2+1}{n+1} + \amp = \frac{\big((n+1)^2+1\big)(n+1)- (n^2+1)(n+2)}{(n+1)(n+2)} + \amp = \frac{n^2+3n}{(n+1)(n+2)} + \amp \gt 0 \text{ for all \(n\). } + + Since a_{n+1}-a_n\gt 0 for all n, + we conclude the sequence is increasing, illustrated in . +

    -
    - +
    + Plot of the sequence in of - + Scatter plot for the second sequence in this example. It is increasing but not bounded.

    @@ -1667,13 +1609,43 @@

    - +
    • + +
  • +

    + We can clearly see in , + where the sequence is plotted, that it is not monotonic. + However, it does seem that after the first 4 terms it is decreasing. + To understand why, perform the same analysis as done before: + + + a_{n+1}-a_n \amp = \frac{(n+1)^2-9}{(n+1)^2-10(n+1)+26} - \frac{n^2-9}{n^2-10n+26} + \amp = \frac{n^2+2n-8}{n^2-8n+17}-\frac{n^2-9}{n^2-10n+26} + \amp = \frac{(n^2+2n-8)(n^2-10n+26)-(n^2-9)(n^2-8n+17)}{(n^2-8n+17)(n^2-10n+26)} + \amp = \frac{-10n^2+60n-55}{(n^2-8n+17)(n^2-10n+26)} + . +

    - -
    - +

    + We want to know when this is greater than, or less than, 0. + The denominator is always positive, + therefore we are only concerned with the numerator. + For small values of n, + the numerator is positive. + As n grows large, + the numerator is dominated by -10n^2, + meaning the entire fraction will be negative; + , for large enough n, a_{n+1}-a_n \lt 0. + Using the quadratic formula we can determine that the numerator is negative for n\geq 5. + In short, the sequence is simply not monotonic, + though it is useful to note that for n\geq 5, + the sequence is monotonically decreasing. +

    + +
    + Plot of the sequence in of - + Scatter plot for the third sequence in this example. It is not monotonic.

    @@ -1710,12 +1682,32 @@

    -
    - +
    • + +
  • + +

    + Again, the plot in of + shows that the sequence is not monotonic, + but it suggests that it is monotonically decreasing after the first term. + We perform the usual analysis to confirm this. + + a_{n+1}-a_n \amp = \frac{(n+1)^2}{(n+1)!} - \frac{n^2}{n!} + \amp = \frac{(n+1)^2-n^2(n+1)}{(n+1)!} + \amp = \frac{-n^3+2n+1}{(n+1)!} + + When n=1, the above expression is \gt 0; + for n\geq 2, the above expression is \lt 0. + Thus this sequence is not monotonic, + but it is monotonically decreasing after the first term. +

    + +
    + Plot of the sequence in of - + Scatter plot for the last sequence in this example. It is not monotonic.

    @@ -1754,9 +1746,17 @@

    - - -
    • + + + +

    + + + + + + + Video solution diff --git a/ptx/sec_series.ptx b/ptx/sec_series.ptx index 68de229a4..7fcb7c391 100644 --- a/ptx/sec_series.ptx +++ b/ptx/sec_series.ptx @@ -983,7 +983,7 @@ Partial sums of the series are plotted in .

    -
    +
    Scatter plots relating to the series of @@ -1087,7 +1087,7 @@

      -
    1. +

    2. We can decompose the fraction 2/(n^2+2n) as @@ -1121,47 +1121,11 @@ so \infser \frac1{n^2+2n} = \frac32. This is illustrated in .

      -
      3. - -
    3. -

      - We begin by writing the first few partial sums of the series: - - S_1 \amp = \ln\left(2\right) - S_2 \amp = \ln\left(2\right)+\ln\left(\frac32\right) - S_3 \amp = \ln\left(2\right)+\ln\left(\frac32\right)+\ln\left(\frac43\right) - S_4 \amp = \ln\left(2\right)+\ln\left(\frac32\right)+\ln\left(\frac43\right)+\ln\left(\frac54\right) - - At first, this does not seem helpful, - but recall the logarithmic identity: - \ln(x) +\ln(y) = \ln(xy). - Applying this to S_4 gives: - - S_4 \amp = \ln\left(2\right)+\ln\left(\frac32\right)+\ln\left(\frac43\right)+\ln\left(\frac54\right) - \amp = \ln\left(\frac21\cdot\frac32\cdot\frac43\cdot\frac54\right) = \ln\left(5\right) - . - We can conclude that \{S_n\} = \big\{\ln(n+1)\big\}. - This sequence does not converge, - as \lim\limits_{n\to\infty}S_n=\infty. - Therefore \ds\infser \ln\left(\frac{n+1}{n}\right)=\infty; - the series diverges. - Note in how the sequence of partial sums grows slowly; - after 100 terms, it is not yet over 5. - Graphically we may be fooled into thinking the series converges, - but our analysis above shows that it does not. -

      -
      4. -
    -

    - -
    - Scatter plots relating to the series in - -
    - - - +
    + Scatter plots of the sequence and corresponding partial sums in of + + Scatter plots of the sequence, and corresponding partial sums, for the first part of this example.

    @@ -1204,13 +1168,42 @@ - -

    + +
    + -
    - - - +
  • +

    + We begin by writing the first few partial sums of the series: + + S_1 \amp = \ln\left(2\right) + S_2 \amp = \ln\left(2\right)+\ln\left(\frac32\right) + S_3 \amp = \ln\left(2\right)+\ln\left(\frac32\right)+\ln\left(\frac43\right) + S_4 \amp = \ln\left(2\right)+\ln\left(\frac32\right)+\ln\left(\frac43\right)+\ln\left(\frac54\right) + + At first, this does not seem helpful, + but recall the logarithmic identity: + \ln(x) +\ln(y) = \ln(xy). + Applying this to S_4 gives: + + S_4 \amp = \ln\left(2\right)+\ln\left(\frac32\right)+\ln\left(\frac43\right)+\ln\left(\frac54\right) + \amp = \ln\left(\frac21\cdot\frac32\cdot\frac43\cdot\frac54\right) = \ln\left(5\right) + . + We can conclude that \{S_n\} = \big\{\ln(n+1)\big\}. + This sequence does not converge, + as \lim\limits_{n\to\infty}S_n=\infty. + Therefore \ds\infser \ln\left(\frac{n+1}{n}\right)=\infty; + the series diverges. + Note in how the sequence of partial sums grows slowly; + after 100 terms, it is not yet over 5. + Graphically we may be fooled into thinking the series converges, + but our analysis above shows that it does not. +

    + +
    + Scatter plots of the sequence and corresponding partial sums in of + + Scatter plots of the sequence, and corresponding partial sums, for the second part of this example.

    @@ -1255,8 +1248,10 @@

    - -
    • + + +

    + Video solution @@ -1388,7 +1383,7 @@

      -
    1. +

    2. We start by using algebra to break the series apart: @@ -1398,31 +1393,10 @@ . This is illustrated in .

      -
      3. - -
    3. -

      - This looks very similar to the series that involves e in . - Note, however, - that the series given in this example starts with n=1 and not n=0. - The first term of the series in the Key Idea is 1/0! = 1, - so we will subtract this from our result below: - - \infser \frac{1000}{n!} \amp = 1000\cdot\infser \frac{1}{n!} - \amp = 1000\cdot (e-1) \approx 1718.28 - . - This is illustrated in . - The graph shows how this particular series converges very rapidly. -

      - -
      - Scatter plots relating to the series in - - -
      - - - +
      + Scatter plots of the sequence and corresponding partial sums in of + + Scatter plots of the sequence, and corresponding partial sums, for the first part of this example.

      @@ -1468,13 +1442,28 @@ - +

      +
      4. -
      - - - +
    4. +

      + This looks very similar to the series that involves e in . + Note, however, + that the series given in this example starts with n=1 and not n=0. + The first term of the series in the Key Idea is 1/0! = 1, + so we will subtract this from our result below: + + \infser \frac{1000}{n!} \amp = 1000\cdot\infser \frac{1}{n!} + \amp = 1000\cdot (e-1) \approx 1718.28 + . + This is illustrated in . + The graph shows how this particular series converges very rapidly. +

      +
      + Scatter plots of the sequence and corresponding partial sums in of + + Scatter plots of the sequence, and corresponding partial sums, for the first part of this example.

      @@ -1514,13 +1503,11 @@ - -

      - -
      5. + +
    -
  • +

  • The denominators in each term are perfect squares; we are adding \ds \sum_{n=4}^\infty \frac{1}{n^2} diff --git a/ptx/sec_taylor_poly.ptx b/ptx/sec_taylor_poly.ptx index ba993e962..9e6424885 100644 --- a/ptx/sec_taylor_poly.ptx +++ b/ptx/sec_taylor_poly.ptx @@ -988,7 +988,7 @@ -