diff --git a/ptx/review-exercises-limits.ptx b/ptx/review-exercises-limits.ptx
index 8c14ddf4f..09dd87313 100644
--- a/ptx/review-exercises-limits.ptx
+++ b/ptx/review-exercises-limits.ptx
@@ -272,7 +272,7 @@
For a numerical approximation, make a table:
-
+ x
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
Adjust the initial condition in this interactive figure to answer the following.
@@ -1073,7 +1072,7 @@
Slope field for the logistic differential equation \yp = y(1-y) from
- Graph of slope field for the logistic differential equation y’=y(1-y) from the example.
+ Graph of slope field for the logistic differential equation y'=y(1-y) from the example.
@@ -1161,7 +1160,7 @@
with a few representative solution curves
- Graph of slope field for the logistic differential equation y’=y(1-y) with representative solution curves.
+ Graph of slope field for the logistic differential equation y'=y(1-y) with representative solution curves.
Finding the extreme values of f(x)= 2x^3+3x^2-12x in
-
+ xf(x)
@@ -879,7 +879,7 @@
Finding the extreme values of a piecewise-defined function in
-
+ xf(x)
@@ -983,7 +983,7 @@
Finding the extrema of f(x)= \cos\mathopen{}\left(x^2\right)\mathclose{} in
-
+ xf(x)
@@ -1094,7 +1094,7 @@
Finding the extrema of the half-circle in
-
+ xf(x)
diff --git a/ptx/sec_hyperbolic.ptx b/ptx/sec_hyperbolic.ptx
index e6c1ef962..cfd78f0ff 100644
--- a/ptx/sec_hyperbolic.ptx
+++ b/ptx/sec_hyperbolic.ptx
@@ -781,7 +781,7 @@
Domains and ranges of the hyperbolic and inverse hyperbolic functions
-
+ FunctionDomainRange
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 @@
-
Iterations of the Bisection Method of Root Finding
-
+ Iteration #IntervalMidpoint Sign
@@ -3274,7 +3274,7 @@
If f(m)\gt 0 at the midpoint m, put + for the midpoint sign. If f(m)\lt 0, put - for the midpoint sign.
-
+ IterationIntervalMidpoint Sign
@@ -3364,7 +3364,7 @@
If f(m)\gt 0 at the midpoint m, put + for the midpoint sign. If f(m)\lt 0, put - for the midpoint sign.
-
+ IterationIntervalMidpoint Sign
@@ -3454,7 +3454,7 @@
If f(m)\gt 0 at the midpoint m, put + for the midpoint sign. If f(m)\lt 0, put - for the midpoint sign.
-
+ IterationIntervalMidpoint Sign
@@ -3544,7 +3544,7 @@
If f(m)\gt 0 at the midpoint m, put + for the midpoint sign. If f(m)\lt 0, put - for the midpoint sign.
-
+ IterationIntervalMidpoint Sign
diff --git a/ptx/sec_limit_infty.ptx b/ptx/sec_limit_infty.ptx
index 55dc94d27..6fb8b198b 100644
--- a/ptx/sec_limit_infty.ptx
+++ b/ptx/sec_limit_infty.ptx
@@ -759,7 +759,7 @@
Observing that f(x)=\sin(1/x) has no limit as x\to0 in
-
+ x\sin(1/x)
@@ -1434,7 +1434,7 @@
The difference quotient evaluated at values of h near 0
-
+ h\frac{f(1+h)-f(1)}{h}
@@ -1994,7 +1994,7 @@
For a numerical approximation, make a table:
-
+ x
@@ -2087,7 +2087,7 @@
For a numerical approximation, make a table:
-
+ x
@@ -2183,7 +2183,7 @@
For a numerical approximation, make a table:
-
+ x
@@ -2294,7 +2294,7 @@
For a numerical approximation, make a table:
-
+ x
@@ -2405,7 +2405,7 @@
For a numerical approximation, make a table:
-
+ x
@@ -2512,7 +2512,7 @@
For a numerical approximation, make a table:
-
+ x
@@ -2619,7 +2619,7 @@
For a numerical approximation, make a table:
-
+ x
@@ -2724,7 +2724,7 @@
For a numerical approximation, make a table:
-
+ x
@@ -2829,7 +2829,7 @@
For a numerical approximation, make a table:
-
+ x
@@ -2937,7 +2937,7 @@
For a numerical approximation, make a table:
-
+ x
@@ -3028,7 +3028,7 @@
For a numerical approximation, make a table:
-
+ x
@@ -3117,7 +3117,7 @@
For a numerical approximation, make a table:
-
+ x
@@ -3218,7 +3218,7 @@
For a numerical approximation, make a table:
-
+ x\big\lfloor\lvert x\rvert\big\rfloor !
@@ -3333,7 +3333,7 @@
For a numerical approximation, make a table:
-
+ x\big\lfloor\lvert x\rvert\big\rfloor !
@@ -3482,7 +3482,7 @@
All quantities are functions of time, t (min), defined as follows:
-
+ V(t)Volume of water in a tank (\text{m}^3) at time, t
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 @@
-
+
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.
-
-
-
-
-
-
- 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.
-
-
-
-
-
- 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.
-
-
-
-
-
-
- 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.
-
-
-
-
-
-
-
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 @@
-
+
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 .
-
-
-
-
- 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 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 @@
-
+
We start by using algebra to break the series apart:
@@ -1398,31 +1393,10 @@
.
This is illustrated in .
-
-
-
-
- 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 @@
-
+
+
-
-
-
-
+
+
+ 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 @@
-
-
-
-
+
+
-
+
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_shell_method.ptx b/ptx/sec_shell_method.ptx
index 2185229f5..d7df5a394 100644
--- a/ptx/sec_shell_method.ptx
+++ b/ptx/sec_shell_method.ptx
@@ -1447,7 +1447,7 @@