Use this page to ask anything related to MA 123. If you are registered and logged in, your posts will display your user name. If you would like to post anonymously, make sure you are logged out. Anyone is allowed to post.
Use this page to ask anything related to MA 123. If you are registered and logged in, your posts will display your user name. If you would like to post anonymously, make sure you are logged out. Anyone is allowed to post.
For #3 in the homework, find the maclaurin series for $$f(x) = \cos^2 x$$,
can I just use the maclaurin series of \(\cos x\) and square that?
You can, but keep in mind series don’t square like you would like them to. That is,
$$
\left( \sum_{n =0}^{\infty} a_n \right)^2 \not = \sum_{n =0}^{\infty} a_n^2
$$
To square a series, it’s best to write the series in expanded form and multiply using the distributive property.
For example,
$$
\begin{aligned}
\left(\sum_{n = 0}^{\infty} x^n \right)^2 & = (1 + x + x^2 + x^3 + \cdots) \cdot (1+x+x^2+x^3 + \cdots ) \\
& = 1 + 2x + 3x^2 + 4 x^3 + \cdots \\
& = \sum_{n = 1}^{\infty} nx^{n-1}
\end{aligned}
$$
There is a more reasonable way to get the square of this series. Hint: what is the common function that has the power series inside the square.
Use the hint: \( \cos^2 x = \frac{1+ \cos 2x}{2} \)
$$
\begin{aligned}
\cos^2 x & = \frac{1+ \cos 2x }{2} = \frac{1}{2}+ \frac{1}{2} \cos 2x \\
& = \frac{1}{2} + \frac{1}{2} \sum_{n =0}^{\infty} \frac{(-1)^n (2x)^{2n}}{(2n)!} \\
& = \frac{1}{2} + \frac{1}{2} \sum_{n =0}^{\infty} \frac{(-1)^n 2^{2n} x^{2n}}{(2n)!}
\end{aligned}
$$
I need help finding the interval of convergence for this webassign problem. I tried doing the ratio test and I thought the interval of convergence was (-1/6, 1/6) but I got it wrong.
$$
\sum_{n=3}^{\infty} ( -1 )^n \frac{n^3x^n}{6^n}
$$
Apply the ratio test:
$$
\require{cancel}
\begin{aligned}
\lim_{n \rightarrow \infty} & \left| \frac{a_{n+1}}{a_n} \right| =
\lim_{n \rightarrow \infty} \left| \frac{ ( -1 )^{n+1} \frac{(n+1)^3x^{n+1}}{6^{n+1}} }{ ( -1 )^n \frac{n^3x^n}{6^n}} \right| \\
& = \lim_{n \rightarrow \infty} \left| \frac{(n+1)^3x^{n+1}}{6^{n+1}} \frac{6^n}{n^3x^n} \right| \\
& = \lim_{n \rightarrow \infty} \left| \frac{(n+1)^3 \bcancel{x^n} x}{\bcancel{6^n} 6} \frac{\bcancel{6^n}}{n^3\bcancel{x^n}} \right|\\
& = \lim_{n \rightarrow \infty} \left| \frac{(n+1)^3 x}{6 n^3} \right| \\
& = \left|\frac{x}{6}\right| \lim_{n \rightarrow \infty} \left| \frac{(n+1)^3}{n^3} \right| \\
& = \left|\frac{x}{6}\right| <1.
\end{aligned}
$$
That is, for any \(x\) in \((-6,6)\) the series converges. Now we test the endpoints.
For \(x = 6\):
$$
\sum_{n=3}^{\infty} ( -1 )^n \frac{n^3x^n}{6^n} =\sum_{n=3}^{\infty} ( -1 )^n n^3
$$
which diverges by the divergence test.
For \(x = -6\):
$$
\sum_{n=3}^{\infty} ( -1 )^n \frac{n^3x^n}{6^n} =\sum_{n=3}^{\infty} n^3
$$
which also diverges by the divergence test.
So, the interval of convergence is \((-6,6)\)
Hello, I’m trying to do this problem in WebAssign:
$$ \sum_{n=1}^\infty \frac{x^n}{1\cdot 3\cdot 5\cdot \cdot \cdot (2n-1)} $$
I need to find the interval of convergence and its radius. I am thinking that the denominator is:
$$ 1\cdot 3\cdot 5\cdot \cdot \cdot (2n-1) = (2(n!)-1) $$
Using the ratio test and simplifying as much as I could, I got
$$
\left | x \right | \cdot \lim_{n \to \infty} \frac{2(n!)-1}{2(n+1)!-1}
$$
At this point, I don’t know how to simplify the limit because of the terms. What do you think?
Thanks,
BV
Apply the ratio test:
$$
\require{cancel}
\begin{aligned}
\lim_{n \rightarrow \infty} & \left| \frac{a_{n+1}}{a_n} \right| =
\lim_{n \rightarrow \infty} \left| \frac{\frac{x^{n+1}}{1 \cdot 3 \cdot 5 \cdots (2(n+1)-1) } }{\frac{x^n}{1 \cdot 3 \cdot 5 \cdots (2n-1) }} \right| \\
& = \lim_{n \rightarrow \infty} \left| \frac{x^{n+1}}{1 \cdot 3 \cdot 5 \cdots (2(n+1)-1) } \frac{1 \cdot 3 \cdot 5 \cdots (2n-1) }{x^n} \right|\\
& = \lim_{n \rightarrow \infty} \left| \frac{\bcancel{x^n} \cdot x}{\bcancel{1 \cdot 3 \cdot 5 \cdots (2n-1)}(2n+1) } \frac{\bcancel{1 \cdot 3 \cdot 5 \cdots (2n-1)} }{\bcancel{x^n}} \right| \\
& = \lim_{n \rightarrow \infty} \left| \frac{x}{2n+1} \right| \\
& = |x| \lim_{n \rightarrow \infty} \left| \frac{1}{2n+1} \right| \\
& = |x| \cdot 0 = 0 <1 \text{ regardless of the value of } x.
\end{aligned}
$$
Thus, the interval of convergence is \((-\infty, \infty)\) and the radius of convergence is \(\infty\).
To help see why the terms above cancel like they do, set \(n = 10, 11, 12\) and see what cancels in the ratio.
I cant figure out this problem on webassign: Show that the series is convergent. How many terms of the series do we need to add in order to find the sum to the indicated accuracy?
The summation from n=1 to infinity of
$$\frac{\left ( -1 \right )^n}{n10^n}$$
\(|error| \leq 0.0001\)
The Alternating Series Test is applicable. The series is alternating, the terms go to zero, and the sequence \(b_n\) is decreasing. So the series converges.
You can use the Alternating Series Estimation Theorem which gives a bound on the error.
In particular, we have that
$$
|error| \leq b_{n+1}
$$
and since we want the error to be less than \(0.0001\), it suffices to find \(n\) such that
$$
b_{n+1} < 0.0001 $$ That is, solve: $$ \frac{1}{10^n} < 0.0001 $$ which is equivalent to $$ 10000 < 10^n $$ So \(n >4 \).
That is, we need to choose \(n\) at least equal to 5.
Hi can I have some help for this webassign problem:
Express the following as a ratio of integers
$$4.470\overline{20}$$
I can’t find a way to set up a series that infinitely adds 0.00020.
Thanks
$$
\begin{aligned}
4.470 \overline{20}
& = 4 + \frac{470}{1000} + \frac{2}{100^2} + \frac{2}{100^3} + \frac{2}{100^4}+\frac{2}{100^5} + \cdots \\
& = 4 + \frac{470}{1000} +2 \sum_{n = 2}^{\infty} \left(\frac{1}{100} \right)^n \\
& = 4 + \frac{470}{1000} + \frac{2}{100} \sum_{n = 2}^{\infty} \left(\frac{1}{100} \right)^{n-1} \\
& = 4 + \frac {470} {1000} + \frac {2} {100}\left ( \frac {a} {1 – r} – a_ 1 \right) \\
& = 4 + \frac {470} {1000} + \frac {2} {100}\left ( \frac {1} {1 – \frac {1} {100}} – 1 \right) \\
&=\frac{8851}{1980}
\end{aligned}
$$
If the problem does not require you to use series, you can do the following:
Let \( x = 4.470 \overline{20} \). Then \( 1000 x = 4470.20 \overline{20} \) and \(100000 x = 447020.20 \overline{20} \). So,
$$
100000 x – 1000x = 447020.20 \overline{20} – 4470.20 \overline{20}
$$
That is,
$$
99000 x = 442550
$$
Solving for \(x\), we get:
$$
x = \frac{442550}{99000} \text{ or } \frac{8851}{1980}
$$
# 8 on WebAssign gave
$$ \sum_{n=1}^{\infty} \frac{x^n}{5^n}$$
the Watch It (video tutorial) says that I can immediately equate this series to $$ \frac{\frac{x}{5}}{1-\frac{x}{5}} $$
even though it is not in the form of
$$ \sum_{n=1}^{\infty}ar^{n-1} $$
Am I allowed to to this even though it doesn’t follow the form \(ar^{n-1}\)?
You should definitely write it in the appropriate form before applying the formula. The video may have skipped this step, but it doesn’t me that we should. Here are the requisite steps:
$$
\begin{aligned}
\sum_{n=1}^{\infty} \frac{x^n}{5^n}
& = \sum_{n=1}^{\infty} \left( \frac{x}{5} \right)^n \\
& = \frac{x}{5} \sum_{n=1}^{\infty} \left( \frac{x}{5}\right)^{n-1} \\
& = \frac{x}{5} \frac{1}{1- \frac{x}{5}} \\
& = \frac{x}{5-x}
\end{aligned}
$$
Hello, just a question in webassign;
$$ a_n = 3 – (0.3)^n $$
This supposedly converges to 3, but when I try it with the following method, I get something else:
$$ \lim_{n \to \infty} 3 – (0.3)^n $$
$$\lim_{n \to \infty} 3 – 0.3 \left(\lim_{n \to \infty} (0.3)^{n-1}\right)$$
$$ 3 – 0.3 \left(\lim_{n \to \infty} (0.3)^{n-1}\right)$$
Then the part inside the big parenthesis is equal to:
$$ \frac{a_1}{1-r} $$
$$ 3 – 0.3 \left(\frac{2.7}{0.7} \right) \ne 3 $$
Anything I’m missing or am I using the wrong method?
Remember that the formula
$$
\frac{a}{1-r}
$$
is the sum for a geometric series
$$
\sum_{n = 1}^{\infty} a r^{n-1}
$$
when \(|r|<1\) but we are taking the limit of a sequence.
To find the limit of the given sequence do the following:
$$ \lim_{n \to \infty} (3 – (0.3)^n) = 3 -\lim_{n \to \infty} (0.3)^n = 3 - 0 = 3 $$
To see why \(\displaystyle \lim_{n \to \infty} (0.3)^n = 0 \), recall that a positive fraction raised to a power tends to zero as that power grows large.
Sorry, I thought the code would translate into the equation.
You just need to wrap your code:
$$ your code here $$ or \( your code here \)
Hi, this is a webassign problem, could you please help me:
The sum from n=1 to infinity of
$$
\left ( \frac{7}{e^n}+\frac{6}{n(n+1)} \right )
$$
I know this converges, so I need to find the sum of it.
I know I can split it up into two sums, and can solve the 7/e^n but the second one is confusing me.
$$
\begin{aligned}
\sum_{n=1}^{\infty} \left( \frac{7}{e^n} + \frac{6}{n(n+1)}\right) & =
\sum_{n=1}^{\infty} \frac{7}{e^n} +\sum_{n=1}^{\infty} \frac{6}{n(n+1)}\\
& =
\frac{7}{e}\sum_{n=1}^{\infty} \frac{1}{e^{n-1}} +6\sum_{n=1}^{\infty} \frac{1}{n(n+1)}\\
& =
\frac{7}{e}\sum_{n=1}^{\infty} \left( \frac{1}{e} \right)^{n-1} +6\sum_{n=1}^{\infty} \frac{1}{n(n+1)}\\
& =
\frac{7}{e}\sum_{n=1}^{\infty} \left( \frac{1}{e} \right)^{n-1} +6\sum_{n=1}^{\infty}\left( \frac{1}{n}-\frac{1}{n+1} \right)\\
& = \frac{7}{e} \frac{1}{1-\frac{1}{e}} + 6 \\
& = \frac{7}{e-1} + 6
\end{aligned}
$$
where,
Hello,
Can you help me set up the geometric series for number 3 in the homework? I got
$$
\frac{3}{4}[(a_{n-1})+8]
$$
but that isn’t geometric. I assume that the common ratio, r is > 1 and thus divergent.
Thanks
Problem 3 in Homework 1 asks: Once a day, eight tons of pollutants are dumped into a bay. Of this, 25% is removed by natural processes each day. Model the long-run quantity of pollutants right after a dump by a geometric series. Does the geometric series have a limit? What does this fact tell you about the quantity of pollutants in the bay over time?
Let’s calculate the total amount of pollution in the bay day-by-day.
Day 1: we dump 8 tons of pollution. Current total is 8 tons
Day 2: we start with 3/4ths of 8 tons from day 1 and we dump 8 tons. Current total is
$$
8 + 8 \frac{3}{4}
$$
Day 3: we start with 3/4ths from the 8 tons dumped in day 2 plus 3/4ths from the 8 tons dumped on day 1 plus we dump 8 tons. Current total is
$$
8 + 8 \frac{3}{4} + 8 \left(\frac{3}{4} \right)^2
$$
Day n: Current total would be
$$
\sum_{k = 1}^n 8 \left( \frac{3}{4}\right)^{k-1}
$$
You should check that this is correct by evaluating the sum for \(n = 1, 2, 3\) and see if it matches what we have for Days 1,2, and 3, respectively.
Finally the geometric series is obtained by letting \(n\) tend to infinity (this is what we mean by “in the long run”).
$$
\sum_{k =1}^{ \color{red} \infty} 8 \left( \frac{3}{4} \right)^{k-1}
$$
Side note: You can start with a recursive relation that calculates the total amount of pollution in the bay based on the total pollution in the bay the day before. That is, if \(q_n\) is the total pollution in the bay on day \(n\), then
$$
q_n = \frac{3}{4} q_{n-1} + 8.
$$
(This is close to what you have)
Then, this recursive relation has to be solved. To find the solution you can google:
“non-homogeneous recursive relations of first degree”
Compare your results. Also note that this method does not involve ever writing down a geometric series and so not a valid solution, as the problem explicitly states “Model the long-run quantity of pollutants right after a dump by a geometric series.”