Alternating series test (Wikipedia.org)
Definition of an alternating series:
A series of the form $$\sum_{k=0}^\infty (-1)^k a_k$$
where all the $a_k$ are non-negative. The emphasis here is on the non-negative.
We know the absolute value of the remainder $R_n=\sum_{k=n+1}^\infty (-1)^k a_k$ for an alternating series will be less than the next term in the series, $a_{n+1}$.
So what's the point of all this? Well, some power series might look like alternating series, but they'll only be alternating series for certain values of $x$. Thus, you can only estimate the remainder using the above estimate for values of $x$ for which the power series turns into an alternating series.
[Remark: It sort of follows from the above, that a power series that doesn't look like an alternating series, might be an alternating series for certain values of $x$]
Tuesday, December 6, 2011
Answering a student's email questions
For question 6, you should set $x=0$ and solve for $t$. You'll get two values of $t$, but only the one that gives you $y\geq 3$ will work. Then set $y=3$ and solve for $t$. Again you'll get two values of $t$, but only one gives you $x\geq 0$.
For question 7, the furthest distance from the origin is when $r$ is greatest. Thus we are trying to maximize $r$. Then from one variable calculus, just take the derivative with respect to $\theta$ to find critical points. $\frac{dr}{d\theta}=0$. There's a geometric interpretation to this, but it's much easier to explain through pictures than through words. If requested, I'll explain this.
Question 8 is asking for when $r=0$. No derivatives required. Let me know if I'm wrong on this.
For question 10, your mistake might have been that the integral should go from $0$ to $\pi$ and not from $0$ to $2\pi$.
For question 7, the furthest distance from the origin is when $r$ is greatest. Thus we are trying to maximize $r$. Then from one variable calculus, just take the derivative with respect to $\theta$ to find critical points. $\frac{dr}{d\theta}=0$. There's a geometric interpretation to this, but it's much easier to explain through pictures than through words. If requested, I'll explain this.
Question 8 is asking for when $r=0$. No derivatives required. Let me know if I'm wrong on this.
For question 10, your mistake might have been that the integral should go from $0$ to $\pi$ and not from $0$ to $2\pi$.
Taylor Remainder Term
On page 756 (of Stewart's Single Variable Calculus Early Transcendentals), the margin has the following information:
If $f^{(n+1)}$ is continuous on an interval $I$ and $x\in I$, then $$R_n(x)=\frac{1}{n!}\int_a^x(x-t)^n f^{(n+1)}(t)dt$$. This is called the integral form of the remainder term.
Lagrange's form of the remainder term:
$$R_n(x)=\frac{f^{n+1}(z)}{(n+1)!}(x-a)^{n+1})$$
for some $z$ between $x$ and $a$.
Note that Taylor's Inequality follows from Lagrange's form of the remainder term.
Let's do a similar problem to #17 on this the 2010 final exam.
$f(x)=3^x$ and $a=5$.
$f^\prime(x)=\ln(3) 3^x$
$f^{\prime\prime}(x)=\ln(3)^2 3^x$
$f^{(3)}(x)=\ln(3)^3 3^x$
Then $f$ has Taylor series $$\sum_{n=0}^\infty \frac{\ln(3)^n 3^5}{n!}(x-5)^n$$
It's remainder $R_n(x)$ will be $$R_n(x)=\frac{1}{n!}\int_5^x(x-t)^n \ln(3)^{n+1} 3^t dt$$
Note: Using this formula, I would answer the following for part (c): $$R_2(2.1)=\frac{1}{2!}\int_2^{2.1} (2.1-t)^2 \ln(2)^3 2^t dt$$
To answer part (d), just integrate the above by parts and use all the information he gave you to simplify the answer. You might get a different answer since he gave an answer I don't fully understand to part (c). See Wikipedia: Taylor's theorem > Explicit formulae for the remainder
If $f^{(n+1)}$ is continuous on an interval $I$ and $x\in I$, then $$R_n(x)=\frac{1}{n!}\int_a^x(x-t)^n f^{(n+1)}(t)dt$$. This is called the integral form of the remainder term.
Lagrange's form of the remainder term:
$$R_n(x)=\frac{f^{n+1}(z)}{(n+1)!}(x-a)^{n+1})$$
for some $z$ between $x$ and $a$.
Note that Taylor's Inequality follows from Lagrange's form of the remainder term.
Let's do a similar problem to #17 on this the 2010 final exam.
$f(x)=3^x$ and $a=5$.
$f^\prime(x)=\ln(3) 3^x$
$f^{\prime\prime}(x)=\ln(3)^2 3^x$
$f^{(3)}(x)=\ln(3)^3 3^x$
Then $f$ has Taylor series $$\sum_{n=0}^\infty \frac{\ln(3)^n 3^5}{n!}(x-5)^n$$
It's remainder $R_n(x)$ will be $$R_n(x)=\frac{1}{n!}\int_5^x(x-t)^n \ln(3)^{n+1} 3^t dt$$
Note: Using this formula, I would answer the following for part (c): $$R_2(2.1)=\frac{1}{2!}\int_2^{2.1} (2.1-t)^2 \ln(2)^3 2^t dt$$
To answer part (d), just integrate the above by parts and use all the information he gave you to simplify the answer. You might get a different answer since he gave an answer I don't fully understand to part (c). See Wikipedia: Taylor's theorem > Explicit formulae for the remainder
Monday, December 5, 2011
part of page 762
$$\frac{1}{1-x}=\sum_{n=0}^\infty x^n$$
$$e^x=\sum_{n=0}^\infty \frac{x^n}{n!}$$
$$\sin x=\sum_{n=0}^\infty (-1)^n\frac{x^{2n+1}}{(2n+1)!}$$
$$\cos x=\sum_{n=0}^\infty (-1)^n\frac{x^{2n}}{(2n)!}$$
Note: $\sin 0=0$ and $\cos 0=1$
$$\tan^{-1} x=\sum_{n=0}^\infty (-1)^n\frac{x^{2n+1}}{2n+1}$$
$$\ln(1+x)=\sum_{n=1}^\infty (-1)^{n-1}\frac{x^n}{n}$$
$$(1+x)^{k}=\sum_{n=0}^{\infty}\left(\begin{array}{c}
k\\
n\end{array}\right)x^{n}$$
$$\left(\begin{array}{c}
k\\
n\end{array}\right)=\frac{k(k-1)(k-2)\cdots(k-n+1)}{n!}$$
$$e^x=\sum_{n=0}^\infty \frac{x^n}{n!}$$
$$\sin x=\sum_{n=0}^\infty (-1)^n\frac{x^{2n+1}}{(2n+1)!}$$
$$\cos x=\sum_{n=0}^\infty (-1)^n\frac{x^{2n}}{(2n)!}$$
Note: $\sin 0=0$ and $\cos 0=1$
$$\tan^{-1} x=\sum_{n=0}^\infty (-1)^n\frac{x^{2n+1}}{2n+1}$$
$$\ln(1+x)=\sum_{n=1}^\infty (-1)^{n-1}\frac{x^n}{n}$$
$$(1+x)^{k}=\sum_{n=0}^{\infty}\left(\begin{array}{c}
k\\
n\end{array}\right)x^{n}$$
$$\left(\begin{array}{c}
k\\
n\end{array}\right)=\frac{k(k-1)(k-2)\cdots(k-n+1)}{n!}$$
Sunday, December 4, 2011
Tim's Office Status
"Funny" Example:
2011-12-04 12:00 PM Tim entered his office.
2011-12-04 12:01 PM Tim left his office to go the bathroom.
2011-12-04 12:20 PM Tim came back from the bathroom but left immediately to avoid meeting with students.
I'll try to come to the office later this evening. But if not, I'll definitely be there most of Monday and Tuesday with the exception of about two hours on Monday and two hours on Tuesday getting exercise, and an hour each day to eat.
Feel free to send me questions by e-mail. timctran at jhu dot edu. I would start my focus on doing problems which Professor Wilson has worked out in class.
2011-12-05 10:23 AM Tim entered his office.
2011-12-05 11:00 AM Krieger 205
2011-12-05 12:00 PM Krieger 300
2011-12-05 01:30 PM Ended "Review" and Left Krieger 300
Playing touch football, will be back in the office later.
2011-12-05 06:20 PM Tim entered his office. Will be around for another hour.
2011-12-05 07:20 PM Tim left the office to go do back handsprings. Might come back in two hours. Maybe you can come and study while I play games. Muahaha.
Definitely memorize the table on page 762. When allowed, using the summation notation should make the work easier.
Of course you have to be smart about finding the series. Example: Find the Maclaurin series for $e^{x^2+3}$. Some students might be tempted to just plug $x^2+3$ into the series for $e^x$, $\sum_{n=0}^\infty \frac{x^n}{n!}$. However, that wouldn't give you a Maclaurin series. Learn to manipulate your starting expression in the correct way to get what you want. In this case, it's a simple matter of writing $e^{x^2+3}$ as $e^3e^{x^2}$. Then we get $e^3 \sum_{n=0}^\infty \frac{x^2n}{n!}$.
2011-12-05 09:50 PM Tim entered his office.
2011-12-06 02:00 AM Tim left the office to go to sleep.
2011-12-06 02:00 PM Tim entered his office, but didn't update it until 02:30 PM. He'll be gone at 03:00 PM for lunch. He'll be back around 5.
2011-12-06 07:15 PM He came back around 5 and left at 7:15 for modern dance. Check out his recent post. Send him questions by e-mail. He plans to sleep around 11 or 12, so he can get to school at 8:40 AM to get the exams.
A student reminded me of an important tip. Double-check your work when it comes to parameterized curves. This includes the possibility that a curve has already completed itself.
Simple Example: $c(t)=(\cos(t),\sin(t))$ starting at $t=0$ will come back to the starting point $(1,0)$ at $t=2\pi$. $d(t)=(\cos(3t),\sin(3t))$ starting at $t=0$ will come back to the starting point $(1,0)$ at $t=2\pi/3$. If when finding the area or the length you had integrated from $0$ to $2\pi$, then you would get three times the expected answer. And unless the expected answer is 0, then you, my friend, would have the wrong answer. =D
Related to parameterized curves, areas, and lengths, keep in mind that length will always be positive. Area can be negative depending on the context. When asked to find area bounded in a region, it's usually implied that this is total area. In other words, all area is positive, regardless of it's location on the x-y plane. So where does parameterization come in? Well, depending on the direction of the parameterization, you might introduce a negative sign into your answer. So take the absolute value and smile.
Short theoretical explanation of the previous paragraph. When we parametrize a curve, we can choose to integrate from $a$ to $b$, or from $b$ to $a$. And $$\int_a^b f(x)dx=-\int_b^a f(x)dx$$.
2011-12-04 12:00 PM Tim entered his office.
2011-12-04 12:01 PM Tim left his office to go the bathroom.
2011-12-04 12:20 PM Tim came back from the bathroom but left immediately to avoid meeting with students.
I'll try to come to the office later this evening. But if not, I'll definitely be there most of Monday and Tuesday with the exception of about two hours on Monday and two hours on Tuesday getting exercise, and an hour each day to eat.
Feel free to send me questions by e-mail. timctran at jhu dot edu. I would start my focus on doing problems which Professor Wilson has worked out in class.
2011-12-05 10:23 AM Tim entered his office.
2011-12-05 11:00 AM Krieger 205
2011-12-05 12:00 PM Krieger 300
2011-12-05 01:30 PM Ended "Review" and Left Krieger 300
Playing touch football, will be back in the office later.
2011-12-05 06:20 PM Tim entered his office. Will be around for another hour.
2011-12-05 07:20 PM Tim left the office to go do back handsprings. Might come back in two hours. Maybe you can come and study while I play games. Muahaha.
Definitely memorize the table on page 762. When allowed, using the summation notation should make the work easier.
Of course you have to be smart about finding the series. Example: Find the Maclaurin series for $e^{x^2+3}$. Some students might be tempted to just plug $x^2+3$ into the series for $e^x$, $\sum_{n=0}^\infty \frac{x^n}{n!}$. However, that wouldn't give you a Maclaurin series. Learn to manipulate your starting expression in the correct way to get what you want. In this case, it's a simple matter of writing $e^{x^2+3}$ as $e^3e^{x^2}$. Then we get $e^3 \sum_{n=0}^\infty \frac{x^2n}{n!}$.
2011-12-05 09:50 PM Tim entered his office.
2011-12-06 02:00 AM Tim left the office to go to sleep.
2011-12-06 02:00 PM Tim entered his office, but didn't update it until 02:30 PM. He'll be gone at 03:00 PM for lunch. He'll be back around 5.
2011-12-06 07:15 PM He came back around 5 and left at 7:15 for modern dance. Check out his recent post. Send him questions by e-mail. He plans to sleep around 11 or 12, so he can get to school at 8:40 AM to get the exams.
A student reminded me of an important tip. Double-check your work when it comes to parameterized curves. This includes the possibility that a curve has already completed itself.
Simple Example: $c(t)=(\cos(t),\sin(t))$ starting at $t=0$ will come back to the starting point $(1,0)$ at $t=2\pi$. $d(t)=(\cos(3t),\sin(3t))$ starting at $t=0$ will come back to the starting point $(1,0)$ at $t=2\pi/3$. If when finding the area or the length you had integrated from $0$ to $2\pi$, then you would get three times the expected answer. And unless the expected answer is 0, then you, my friend, would have the wrong answer. =D
Related to parameterized curves, areas, and lengths, keep in mind that length will always be positive. Area can be negative depending on the context. When asked to find area bounded in a region, it's usually implied that this is total area. In other words, all area is positive, regardless of it's location on the x-y plane. So where does parameterization come in? Well, depending on the direction of the parameterization, you might introduce a negative sign into your answer. So take the absolute value and smile.
Short theoretical explanation of the previous paragraph. When we parametrize a curve, we can choose to integrate from $a$ to $b$, or from $b$ to $a$. And $$\int_a^b f(x)dx=-\int_b^a f(x)dx$$.
Subscribe to:
Posts (Atom)