You CAN Ace Calculus
other exams from this semester |
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exam C1 - |
Single Variable Calculus |
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Multi-Variable Calculus |
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Acceleration Vector |
Arc Length (Vector Functions) |
Arc Length Function |
Arc Length Parameter |
Conservative Vector Fields |
Cross Product |
Curl |
Curvature |
Cylindrical Coordinates |
Lagrange Multipliers |
Line Integrals |
Partial Derivatives |
Partial Integrals |
Path Integrals |
Potential Functions |
Principal Unit Normal Vector |
Differential Equations |
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Do you have a practice problem number but do not know on which page it is found? If so, enter the number below and click 'page' to go to the page on which it is found or click 'practice' to be taken to the practice problem.
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Help Keep 17Calculus Free |
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This is the final of four exams for semester C of calculus 2.
Each exam page contains a full exam with detailed solutions. Most of these are actual exams from previous semesters used in college courses. You may use these as practice problems or as practice exams. Here are some suggestions on how to use these to help you prepare for your exams.
- Set aside a chunk of full, uninterrupted time, usually an hour to two, to work each exam.
- Go to a quiet place where you will not be interrupted that duplicates your exam situation as closely as possible.
- Use the same materials that you are allowed in your exam (unless the instructions with these exams are more strict).
- Use your calculator as little as possible except for graphing and checking your calculations.
- Work the entire exam before checking any solutions.
- After checking your work, rework any problems you missed and go to the 17calculus page discussing the material to perfect your skills.
- Work as many practice exams as you have time for. This will give you practice in important techniques, experience in different types of exam problems that you may see on your own exam and help you understand the material better by showing you what you need to study.
IMPORTANT -
Exams can cover only so much material. Instructors will sometimes change exams from one semester to the next to adapt an exam to each class depending on how the class performs during the semester while they are learning the material. So just because you do well (or not) on these practice exams, does not necessarily mean you will do the same on your exam. Your instructor may ask completely different questions from these. That is why working lots of practice problems will prepare you better than working just one or two practice exams.
Calculus is not something that can be learned by reading. You have to work problems on your own and struggle through the material to really know calculus, do well on your exam and be able to use it in the future.
Exam Details | |
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Time | 2.5 hours |
Questions | 8+1EC |
Total Points | 100+10EC |
Tools | |
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Calculator | no |
Formula Sheet(s) | 1 page, 8.5x11 or A4 |
Other Tools | ruler for drawing graphs |
Instructions:
- Show all your work.
- For each problem, correct answers are worth 1 point. The remaining points are earned by showing calculations and giving reasoning that justify your conclusions.
- Correct notation counts (i.e. points will be taken off for incorrect notation).
- Give exact, simplified answers.
[10 points] Derive a numerical equation that you would plug into your calculator (if you had one) to determine \( max \abs{R_n(x)} \), an upper bound for the Taylor Series Remainder, for \(\sin(0.2)\) using \(\sin(x)\) at \(a=0\) with \(n=4\).
Problem Statement |
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[10 points] Derive a numerical equation that you would plug into your calculator (if you had one) to determine \( max \abs{R_n(x)} \), an upper bound for the Taylor Series Remainder, for \(\sin(0.2)\) using \(\sin(x)\) at \(a=0\) with \(n=4\).
Final Answer |
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\(\displaystyle{ max \abs{R_4(0.2)} = \frac{(0.2)^5}{5!} }\) |
Problem Statement |
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[10 points] Derive a numerical equation that you would plug into your calculator (if you had one) to determine \( max \abs{R_n(x)} \), an upper bound for the Taylor Series Remainder, for \(\sin(0.2)\) using \(\sin(x)\) at \(a=0\) with \(n=4\).
Solution |
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The general equation for upper bound on the remainder is
\(\displaystyle{ max \abs{R_n(x)} = \frac{max \abs{f^{(n+1)}(z)}}{(n+1)!} (x-a)^{n+1} }\).
For our problem, \(a=0\), \(n=4\) and \(x=0.2\). We will choose \(c\) once we calculate the derivatives and \(c \in [0,0.2]\). The value of \(c\) we choose will maximize \(\abs{f^{(n+1)}(c)}\). Let's build a table to find our values.
n |
n+1 |
\(f^{(n+1)}(x)\) |
0 |
1 |
\(f'(x) = \cos(x)\) |
1 |
2 |
\(f''(x) = -\sin(x)\) |
2 |
3 |
\(f^{(3)}(x) = -\cos(x)\) |
3 |
4 |
\(f^{(4)}(x) = \sin(x)\) |
4 |
5 |
\(f^{(5)}(x) = \cos(x)\) |
\(\cos(x)\) on \([0,0.2]\) |
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built with GeoGebra |
We stop at \(n=4\) since this is given in the problem. For the equation of \(max \abs{R_n(x)}\) we require the next derivative, which from our table is \(f^{(5)}(x) = \cos(x)\).
Now we need to choose \(c\). We are limited to the interval \([0,0.2]\). We need to choose the value in the interval \([0,0.2]\) that maximizes \(f^{(5)}(x) = \cos(x)\). It should be obvious that the max occurs at \(x=0\), so we choose \(c=0\) and so \(f^{(5)}(0) = \cos(0)=1\). Here is a plot to show this.
So \(max \abs{f^{(5)}(c)} = \cos(c)=\cos(0)=1\)
\(\displaystyle{ max \abs{R_4(0.2)} = \frac{max \abs{f^{(5)}(0)}}{(5)!} (0.2-0)^{5} = }\)
\(\displaystyle{ \frac{(0.2)^{5}}{(5)!} }\)
Final Answer |
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\(\displaystyle{ max \abs{R_4(0.2)} = \frac{(0.2)^5}{5!} }\) |
close solution |
[10 points] Calculate the arc length of the curve \(x=2t\), \(y=(2/3)t^{3/2}\) for \(5 \leq t \leq 12\).
Problem Statement |
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[10 points] Calculate the arc length of the curve \(x=2t\), \(y=(2/3)t^{3/2}\) for \(5 \leq t \leq 12\).
Final Answer |
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\(74/3\) |
Problem Statement |
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[10 points] Calculate the arc length of the curve \(x=2t\), \(y=(2/3)t^{3/2}\) for \(5 \leq t \leq 12\).
Solution |
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The general equation for the arc length in parametric form is
\(\displaystyle{ s = \int_{t_0}^{t_1}{ \sqrt{(dx/dt)^2+(dy/dt)^2} ~dt}}\).
In our problem,
\(x=2t \to dx/dt = 2\)
\(y=(2/3)t^{3/2} \to dy/dt = t^{1/2}\)
So our integral is
\(\displaystyle{ s = \int_{5}^{12}{ \sqrt{(2)^2+(t^{1/2})^2} ~dt} = }\)
\(\displaystyle{ \int_{5}^{12}{ \sqrt{4+t} ~dt} }\)
To solve this we use integration by substitution with \(u=4+t \to du=dt\). We choose to change the limits of integration.
lower limit: \(t=5 \to u=4+5=9\)
upper limit: \(t=12 \to u=4+12=16\)
So now we have
\(\displaystyle{ \int_{9}^{16}{u^{1/2}~du} = }\)
\(\displaystyle{ \left[ \frac{u^{3/2}}{3/2} \right]_{9}^{16} = }\)
\(\displaystyle{ \frac{2}{3} \left[ (16)^{3/2} - (9)^{3/2} \right] = 74/3 }\)
Final Answer |
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\(74/3\) |
close solution |
[15 points] Calculate the equation of the tangent line to \(r=6+3\cos\theta\) at the point \(\theta=\pi/2\). Give your answer in slope intercept form.
Problem Statement |
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[15 points] Calculate the equation of the tangent line to \(r=6+3\cos\theta\) at the point \(\theta=\pi/2\). Give your answer in slope intercept form.
Final Answer |
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\(y=x/2+6\) |
Problem Statement |
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[15 points] Calculate the equation of the tangent line to \(r=6+3\cos\theta\) at the point \(\theta=\pi/2\). Give your answer in slope intercept form.
Solution |
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The general equation for the derivative in polar coordinates is
\(\displaystyle{ \frac{dy}{dx} = \frac{r'\sin\theta+r\cos\theta}{r'\cos\theta-r\sin\theta} }\).
In our problem,
\(r=6+3\cos\theta \to r'=-3\sin\theta\)
\(\theta = \pi/2 \to r=6+3\cos(\pi/2) = 6\)
So the slope is \(\displaystyle{ \frac{dy}{dx} = \frac{-3\sin(\pi/2)+6\cos(\pi/2)}{-3\cos(\pi/2)-6\sin(\pi/2)} = \frac{1}{2} }\)
So the slope is \(1/2\). To find the point, we need to determine the x and y values associated with \((6,\pi/2)\). By inspection, we can deduce that the point is \((0,6)\). However, let's check it by using \(x=r\cos\theta\) and \(y=r\sin\theta\).
\(x=r\cos\theta = 6\cos(\pi/2) = 0\)
\(y=r\sin\theta \ 6\sin(\pi/2) = 6\)
Now we need to get the equation. We have the slope \(m=1/2\) at the point \((0,6)\).
We will use \(y=mx+b \to \) \(6 = (1/2)(0)+b \to \) \(b=6 \to y=x/2+6\)
Final Answer |
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\(y=x/2+6\) |
close solution |
[10 points] Calculate the equation of the tangent line to the curve \(x=t^2-1\), \(y=t^3+2\) at \(t=2\). Give your answer in slope intercept form.
Problem Statement |
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[10 points] Calculate the equation of the tangent line to the curve \(x=t^2-1\), \(y=t^3+2\) at \(t=2\). Give your answer in slope intercept form.
Final Answer |
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\(y=13x/4+1/4\) |
Problem Statement |
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[10 points] Calculate the equation of the tangent line to the curve \(x=t^2-1\), \(y=t^3+2\) at \(t=2\). Give your answer in slope intercept form.
Solution |
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The general equation of the slope for parametric equations is \(\displaystyle{ \frac{dy}{dx} = \frac{dy/dt}{dx/dt} }\).
For our problem
\(x=t^2-1 \to dx/dt=2t\)
\(y=t^3+t \to dy/dt = 3t^2+1\)
\(\displaystyle{ \frac{dy}{dx} = \frac{3t^2+1}{2t} }\) at \(t=2\)
\(\displaystyle{ \frac{dy}{dx} = \frac{3(2)^2+1}{4} = \frac{13}{4} }\)
So the slope of the tangent line is \(m=13/4\). Now we the point.
At \(t=2\) \(x=2^2-1=3\) and \(y=2^3+2=10\)
So \(m=13/4\) at the point \((3,10)\).
Using \(y=mx+b\), we get \(10=3(13/4)+b \to b = 10-39/4 = 1/4\)
\(y=13x/4+1/4\) is the equation of the tangent line at \(t=2\).
Final Answer |
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\(y=13x/4+1/4\) |
close solution |
[10 points] Determine the area inside \(r=\sqrt{\cos(3\theta)}\) and outside \(r=1/\sqrt{2}\) in the first quadrant.
Problem Statement |
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[10 points] Determine the area inside \(r=\sqrt{\cos(3\theta)}\) and outside \(r=1/\sqrt{2}\) in the first quadrant.
Final Answer |
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\( \displaystyle{ \frac{1}{36} [ 3\sqrt{3}-\pi ] }\) |
Problem Statement |
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[10 points] Determine the area inside \(r=\sqrt{\cos(3\theta)}\) and outside \(r=1/\sqrt{2}\) in the first quadrant.
Solution |
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The plot shows the shaded area to be calculated.
The general equation is \(\displaystyle{ R = (1/2)\int_{\alpha}^{\beta}{ r_o^2 - r_i^2~d\theta } }\). In our case, \(r_o = \sqrt{\cos(3\theta)}\) and \(r_i = 1/\sqrt{2}\). Now we need to find the angle of intersection of the two curves.
\(\begin{array}{rcl}
\sqrt{\cos(3\theta)} & = & 1/\sqrt{2} \\
\cos(3\theta) & = & 1/2 \\
3\theta = \pi/3 \\
\theta = \pi/9
\end{array}\)
\(\begin{array}{rcl}
R & = & \displaystyle{ \frac{1}{2} \int_{0}^{\pi/9}{ \cos(3\theta) - 1/2 ~d\theta } } \\
& = & \displaystyle{ \frac{1}{2} \left[ \frac{\sin(3\theta)}{3} - \frac{\theta}{2} \right]_0^{\pi/9} } \\
& = & \displaystyle{ \frac{1}{2} \left[ \frac{\sin(\pi/3)}{3} - \frac{\pi}{18} - 0 \right] } \\
& = & \displaystyle{ \frac{1}{2} \left[ \frac{3\sqrt{3}}{18} - \frac{\pi}{18} \right] } \\
& = & \displaystyle{ \frac{1}{36} [ 3\sqrt{3}-\pi ] }
\end{array}\)
Final Answer |
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\( \displaystyle{ \frac{1}{36} [ 3\sqrt{3}-\pi ] }\) |
close solution |
[25 points] Use parametric equations to describe the boundary of the region bounded by the equations \(y=x^2, x=3\) and \(y=1\). Describe the boundary with a counter-clockwise orientation. Give two sets of parameterizations. a. in parallel with each segment range \(0 \leq t \leq 1\); b. sequentially with the first segment starting at zero.
Problem Statement |
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[25 points] Use parametric equations to describe the boundary of the region bounded by the equations \(y=x^2, x=3\) and \(y=1\). Describe the boundary with a counter-clockwise orientation. Give two sets of parameterizations. a. in parallel with each segment range \(0 \leq t \leq 1\); b. sequentially with the first segment starting at zero.
Solution |
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The plot of the area is shown here. We have also shown arrows indicating counter-clockwise direction. We start our analysis at the point (1,1). Moving counter-clockwise, the first segment is designated A, the second B and the third C.
a. parallel
segment A | ||
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initial description | ||
\(x=t\) |
\(y=1\) |
\(1 \leq t \leq 3\) |
change interval to start at \(t=0\); replace \(t\) with \(t+1\) | ||
\(x=t+1\) |
\(y=1\) |
\(1 \leq t+1 \leq 3 \to \) \(0 \leq t \leq 2\) |
change interval to end at \(t=1\); replace \(t\) with \(2t\) | ||
\(x=2t+1\) |
\(y=1\) |
\(0 \leq 2t \leq 2 \to \) \(0 \leq t \leq 1\) |
segment B | ||
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initial description | ||
\(x=3\) |
\(y=t\) |
\(1 \leq t \leq 9\) |
change interval to start at \(t=0\); replace \(t\) with \(t+1\) | ||
\(x=3\) |
\(y=t+1\) |
\(1 \leq t+1 \leq 9 \to \) \(0 \leq t \leq 8\) |
change interval to end at \(t=1\); replace \(t\) with \(8t\) | ||
\(x=3\) |
\(y=8t+1\) |
\(0 \leq 8t \leq 8 \to \) \(0 \leq t \leq 1\) |
segment C | ||
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initial description | ||
\(x=t\) |
\(y=t^2\) |
\(1 \leq t \leq 3 \) |
start interval at zero; replace \(t\) with \(t+1\) | ||
\(x=t+1\) |
\(y=(t+1)^2\) |
\(1 \leq t+1 \leq 3 \to \) \(0 \leq t \leq 2 \) |
end interval at one; replace \(t\) with \(2t\) | ||
\(x=2t+1\) |
\(y=(2t+1)^2\) |
\(0 \leq 2t \leq 2 \to \) \(0 \leq t \leq 1\) |
reverse direction of the curve; | ||
\(x=2(1-t)+1 = 3-2t\) |
\(y=(2(1-t)+1)^2 = (3-2t)^2\) |
\(0 \leq t \leq 1\) |
Final Answers for part a. parallel | |
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A |
\(y=1\), \(x=2t+1\), \(0 \leq t \leq 1\) |
B |
\(x=3\), \(y=8t+1\), \(0 \leq t \leq 1\) |
C |
\(x=3-2t\), \(y=(3-2t)^2\), \(0 \leq t \leq 1\) |
b. sequential
segment A | ||
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same as part a | ||
\(x=2t+1\) |
\(y=1\) |
\(0 \leq t \leq 1\) |
segment B | ||
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from part a | ||
\(x=3\) |
\(y=8t+1\) |
\(0 \leq t \leq 1\) |
want interval to start at 1 (end of segment A); replace \(t\) with \(t-1\) | ||
\(x=3\) |
\(y=8(t-1)+1 = 8t-7\) |
\(0 \leq t-1 \leq 1 \to \) \(1 \leq t \leq 2\) |
segment C | ||
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from part a | ||
\(x=3\) |
\(y=8t+1\) |
\(0 \leq t \leq 1\) |
want interval to start at 2 (end of segment b); replace \(t\) with \(t-2\) | ||
\(x= 3-2(t-2) = 7-2t\) |
\(y=(3-2(t-2))^2 = (7-2t)^2\) |
\(2 \leq t \leq 3\) |
Final Answers for part b. sequential | |
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A |
\(y=1\), \(x=2t+1\), \(0 \leq t \leq 1\) |
B |
\(x=3\), \(y=8t-7\), \(1 \leq t \leq 2\) |
C |
\(x=7-2t\), \(y=(7-2t)^2\), \(2 \leq t \leq 3\) |
close solution |
[10 points] Use the geometric series \(\displaystyle{ \sum_{k=0}^{\infty}{ r^k } = \frac{1}{1-r} }\) for \(\abs{r} < 1 \) to find the power series for \(g(x)=\ln(1-3x)\) and determine the interval of convergence.
Problem Statement |
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[10 points] Use the geometric series \(\displaystyle{ \sum_{k=0}^{\infty}{ r^k } = \frac{1}{1-r} }\) for \(\abs{r} < 1 \) to find the power series for \(g(x)=\ln(1-3x)\) and determine the interval of convergence.
Final Answer |
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\(\displaystyle{ g(x) = - \sum_{k=1}^{\infty}{ \frac{3^k x^k}{k} } } \) on the interval \([ -1/3, 1/3 )\) |
Problem Statement |
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[10 points] Use the geometric series \(\displaystyle{ \sum_{k=0}^{\infty}{ r^k } = \frac{1}{1-r} }\) for \(\abs{r} < 1 \) to find the power series for \(g(x)=\ln(1-3x)\) and determine the interval of convergence.
Solution |
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In order to get an equation in the form \(\displaystyle{ \frac{1}{1-r} }\), we take the derivative of \(g(x)\) to get \(\displaystyle{ g'(x) = \frac{-3}{1-3x} }\).
From this we can let \(r=3x\), so our series is \(\displaystyle{ g'(x) = -3 \sum_{k=0}^{\infty}{ (3x)^k } }\) for \( \abs{3x} < 1 \to \abs{x} < 1/3 \).
In order to \(g(x)\), we need to integrate our series for \(g'(x)\).
\(\displaystyle{ g(x) = -3 \sum_{k=0}^{\infty}{ 3^k\int{x^k~dx} } = }\)
\( \displaystyle{ -3 \sum_{k=0}^{\infty}{ \frac{3^kx^{k+1}}{k+1} } + C } \) with the ratio of convergence \((-1/3,1/3)\) (since the ratio of convergence does not change with integration).
Integration Constant C |
Now we need to determine the constant \(C\). We can choose any value within the interval \((-1/3,1/3)\), so it is convenient to choose \(x=0\). So \(g(0)=\ln(1-3(0)) = 0\) and \( g(0) = \displaystyle{ -3 \sum_{k=0}^{\infty}{ \frac{3^k0^{k+1}}{k+1} } + C } \to \) \( C = 0 \).
Interval of Convergence |
Now we need to determine the interval of convergence by testing the end points. However, if we look back at the original function \(g(x)=\ln(1-3x)\), \(x=1/3\) is not in the domain, so we can say right away that it is not part of the interval of convergence. We only need to test \(x=-1/3\).
So, when we let \(x=-1/3\) in the infinite series for \(g(x)\), we get
\(\begin{array}{rcl}
g(-1/3) & = & \displaystyle{ -3 \sum_{k=0}^{\infty}{ \frac{3^k(-1/3)^{k+1}}{k+1} } } \\
& = & \displaystyle{ -3 \sum_{k=0}^{\infty}{ \frac{3^k (-1)^{k+1}}{3^{k+1}(k+1)} } } \\
& = & \displaystyle{ \sum_{k=0}^{\infty}{ \frac{-(-1)^{k+1}}{k+1} } } \\
& = & \displaystyle{ \sum_{k=0}^{\infty}{ \frac{(-1)^k}{k+1} } }
\end{array}\)
This is an alternating series, so we will use the alternating series test.
condition 1: \( \displaystyle{ \lim_{n\to\infty}{ \frac{1}{n+1} } = 0 } \) holds
condition 2: we need to show \(a_{n+1} \leq a_n\)
\(\begin{array}{rcl}
a_{n+1} & \leq & a_n \\
\frac{1}{n+2} & \leq & \frac{1}{n+1} \\
n+1 & \leq & n+2 \\
1 & \leq & 2
\end{array}\)
The last inequality, \(1 \leq 2 \), holds for all n. So the series converges by the alternating series test.
So the interval of convergence is \([ -1/3, 1/3 )\).
The sum for \(g(x)\) can be written another way if we start at \(k=1\). This will clean up the sum a bit.
\(\displaystyle{ g(x) = - \sum_{k=1}^{\infty}{ \frac{3^{k} x^k}{k} } }\)
Final Answer |
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\(\displaystyle{ g(x) = - \sum_{k=1}^{\infty}{ \frac{3^k x^k}{k} } } \) on the interval \([ -1/3, 1/3 )\) |
close solution |
[10 points] Use the series for \(\ln(x)\) at \(a=3\) to determine the Taylor Series for \(g(x)=1/x\) (also at \(a=3\)) and determine the radius of convergence for \(g(x)\).
\( \ln(x) = \ln(3) + \displaystyle{ \sum_{n=1}^{\infty}{ \frac{(-1)^{n+1}(x-3)^n}{n3^n} } } \) with interval of convergence \((0,6]\).
Problem Statement |
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[10 points] Use the series for \(\ln(x)\) at \(a=3\) to determine the Taylor Series for \(g(x)=1/x\) (also at \(a=3\)) and determine the radius of convergence for \(g(x)\).
\( \ln(x) = \ln(3) + \displaystyle{ \sum_{n=1}^{\infty}{ \frac{(-1)^{n+1}(x-3)^n}{n3^n} } } \) with interval of convergence \((0,6]\).
Final Answer |
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\( \displaystyle{ \sum_{n=1}^{\infty}{ \frac{(-1)^{n+1}(x-3)^{n-1}}{3^n} } } \) \(R=(0,6)\) |
Problem Statement |
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[10 points] Use the series for \(\ln(x)\) at \(a=3\) to determine the Taylor Series for \(g(x)=1/x\) (also at \(a=3\)) and determine the radius of convergence for \(g(x)\).
\( \ln(x) = \ln(3) + \displaystyle{ \sum_{n=1}^{\infty}{ \frac{(-1)^{n+1}(x-3)^n}{n3^n} } } \) with interval of convergence \((0,6]\).
Solution |
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We are given the Taylor Series for \(\ln(x)\). In order to get \(g(x)=1/x\), we need to take the derivative of \(\ln(x)\).
\(\displaystyle{ (\ln(x))' = \sum_{n=1}^{\infty}{ \frac{(-1)^{n+1}n(x-3)^{n-1}}{n3^n} } = }\)
\( \displaystyle{ \sum_{n=1}^{\infty}{ \frac{(-1)^{n+1}(x-3)^{n-1}}{3^n} } } \)
The radius of convergence is not changed by differentiation or integration, so it is the same as for \(\ln(x)\), which is \(R=(0,6)\).
Final Answer |
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\( \displaystyle{ \sum_{n=1}^{\infty}{ \frac{(-1)^{n+1}(x-3)^{n-1}}{3^n} } } \) \(R=(0,6)\) |
close solution |
[10 points extra credit] Determine the Taylor Series and interval of convergence for \(\ln(x)\) at \(a=3\) using Taylor Series expansion.
Problem Statement |
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[10 points extra credit] Determine the Taylor Series and interval of convergence for \(\ln(x)\) at \(a=3\) using Taylor Series expansion.
Final Answer |
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\( \displaystyle{ f(x)=\ln(x)=\ln(3)+\sum_{n=1}^{\infty}{ \frac{(-1)^{n+1}}{n3^n}(x-3)^n } } \) with interval of convergence \((0,6]\) |
Problem Statement |
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[10 points extra credit] Determine the Taylor Series and interval of convergence for \(\ln(x)\) at \(a=3\) using Taylor Series expansion.
Solution |
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The general formula for Taylor Series expansion is
\(\displaystyle{ f(x) = f(a) + f'(a)(x-a) + . . . + \frac{f^{(n)}(a)(x-a)^n}{n!} + . . . }\)
We will build a table so that we can try to see a pattern.
n |
\(f^{(n)}(x)\) |
\( f^{(n)}(a) \) |
\(0\) |
\(f(x)=\ln(x)\) |
\( f(3) = \ln(3) \) |
\(1\) |
\(f'(x)=1/x\) |
\( f'(3) = 1/3 \) |
\(2\) |
\(f'(x)=-x^{-2}\) |
\( f'(3) = -1/3^2\) |
\(3\) |
\(f'(x)=2x^{-3}\) |
\( f'(3) = 2/3^3\) |
\(4\) |
\(f'(x)=-6x^{-4}\) |
\( f'(3) = -6/3^4\) |
\(5\) |
\(f'(x)=24x^{-5}\) |
\( f'(3) = 24/3^5\) |
So the pattern from the last column looks like \( \displaystyle{ \frac{(-1)^{n+1}(n-1)!}{3^n} } \). So our Taylor Series is
\( \displaystyle{ f(x)=\ln(x)=\ln(3)+\sum_{n=1}^{\infty}{ \frac{(-1)^{n+1}(n-1)!}{3^n n!}(x-3)^n } } \). Simplifying a bit, we get
\( \displaystyle{ f(x)=\ln(x)=\ln(3)+\sum_{n=1}^{\infty}{ \frac{(-1)^{n+1}}{n3^n}(x-3)^n } } \).
Radius of Convergence |
Now we will use the Ratio Test to determine the radius of convergence.
We have
\( \displaystyle{ a_n= \frac{(-1)^{n+1}}{n3^n}(x-3)^n } \) and
\( \displaystyle{ a_{n+1}= \frac{(-1)^{n+2}}{(n+1)3^{n+1}}(x-3)^{n+1} } \).
Notice that we are ignoring the constant \(\ln(3)\) since adding or subtracting a constant from a series does not affect the convergence or divergence of the series.
In order to use the Ratio Test, we need to evaluate
\( \displaystyle{ \lim_{n\to\infty}{ \left| \frac{a_{n+1}}{a_n} \right| } } \).
\( \displaystyle{ \lim_{n\to\infty}{ \left| \frac{a_{n+1}}{a_n} \right| } } \) |
\(\displaystyle{ \lim_{n\to\infty}{ \left| \frac{(-1)^{n+2}(x-3)^{n+1}}{(n+1)3^{n+1}} \right. } }\) \(\displaystyle{ \left. \frac{n3^n}{(-1)^{n+1}(x-3)^n} \right| }\) |
\(\displaystyle{ \lim_{n\to\infty}{ \left| \frac{n}{n+1} \frac{3^n}{3^{n+1}} (x-3) \right| } }\) |
\(\displaystyle{ \left| \frac{x-3}{3} \right| < 1 }\) |
\(\abs{x-3} < 1 \to \) \( 0 \lt x \lt 6 \) |
So the radius of convergence is \(R=3\) or \(R=(0,6)\).
Interval of Convergence |
To determine the interval of convergence, we need to test each endpoint.
\(x=0\) - We can tell immediately that this value of x is not in the domain of the function \(f(x)=\ln(x)\), so it cannot be in the interval of convergence.
\(x=6\) - The first step is to substitute \(x=6\) in the series to get
\( \displaystyle{ \sum_{n=1}^{\infty}{ \frac{(-1)^{n+1}}{n3^n}(6-3)^n } = } \)
\( \displaystyle{ \sum_{n=1}^{\infty}{ \frac{(-1)^{n+1}}{n} } } \)
This series converges by the alternating series test (details shown there). Therefore the interval of convergence is \((0,6]\).
Final Answer |
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\( \displaystyle{ f(x)=\ln(x)=\ln(3)+\sum_{n=1}^{\infty}{ \frac{(-1)^{n+1}}{n3^n}(x-3)^n } } \) with interval of convergence \((0,6]\) |
close solution |