- 9 | Sequences and Series
- 10 | Approximating Functions Using Series
- 12 | Functions of Several Variables
- 13 | A Fundamental Tool: Vectors
- MA 123: Homework
- MA 123: Workshop
- List of Maclaurin Series of Some Common Functions
- Series Convergence Tests
- Review Problems

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Hello Aori,

I have a question with domain of multivariables \(\ln(1-x^2-y^2).\) Is the inequality \(1-x^2-y^2 > 0\) correct? Also, what does the graph of \(1-x^2-y^2 > 0\) look like? Is it a circle with radius 1 centered at origin?

Thanks.

The domain of a logarithmic function is the set of all inputs for which the argument to the logarithmic function is positive. That is, the domain of \(\ln u\) is the set of all \(u > 0.\) Consequently, the domain of the multivariable function \(\ln(1-x^2-y^2)\) is all real ordered pairs \((x,y)\) such that \(1-x^2-y^2 > 0. \) We can represent the domain in set builder notation as

$$

\left \lbrace (x,y) \; | \; 1-x^2 – y^2 >0 \right \rbrace.

$$

We can also represent the domain graphically.

The shaded portions and solid curves of the graph represent the domain of the function \(\ln(1-x^2-y^2).\)

Can you explain how to find the area of the parallelogram in Workshop 6 MA 123 and explain question 2 in the worksheet too please.

There isn’t a question in workshop 6 asking for the area of a parallelogram. Also, solutions to workshop 6 are available on moodle. Did you mean homework 4?

I am sorry. It asks for the area of Q. It is workshop 6 from feb 26,2013( MA 123). I need help with both questions 1 and 2.

P1. The four points \(A = (1,-2,1), B = (3,1,2), C= (0,2,1),\) and \(D = (-2,-1,0)\) form the corners of a quadrilateral \(Q\).

Let

$$

\begin{aligned}

\vec{AB} &= \langle 3,1,2 \rangle – \langle 1,-2,1\rangle = \langle 2,3,1\rangle \\

\vec{CD} &= \langle -2,-1,0 \rangle – \langle 0,2,1 \rangle = \langle -2,-3,-1\rangle \\

\vec{BC} &= \langle 0,2,1 \rangle – \langle 3,1,2 \rangle = \langle -3,1,-1\rangle \\

\vec{AD} &= \langle -2,-1,0 \rangle – \langle 1,-2,1 \rangle = \langle -3,1,-1\rangle \\

\end{aligned}

$$

Then

A ball rests on a straight track aligned in the direction \(\langle 2, 1, 0 \rangle.\) The ball will remain stationary unless it experiences a force of at least \(10\) N along the direction of the track. If a wind is blowing in the direction \(\langle -1,-2, 6 \rangle, \) what is the magnitude of the wind’s force required to move the ball, and in which direction will the ball move?

Let \(\vec{a} = \langle 2,1,0 \rangle \) and \(\vec{b} = \langle -1,-2, 6 \rangle. \) The amount of wind blowing in the direction of the track, \(\vec{a}\), is given by

$$

\text{proj}_{\vec{a}} \vec{b} = \frac{\vec{a} \cdot \vec{b}}{\vec{a} \cdot \vec{a}} \vec{a} = \left \langle \frac{-8}{5}, \frac{-4}{5}, 0 \right \rangle

$$

We need that the magnitude of this vector be 10 N. That is, \(|c\text{proj}_{\vec{a}} \vec{b}| = \frac{4c}{\sqrt{5}} = 10.\) So, \(c = \sqrt{5}/2.\)

The ball will move in the direction \(\vec{a} + c\text{proj}_{\vec{a}} \vec{b} = \left \langle 2-\frac{4}{\sqrt{5}},1-\frac{2}{\sqrt{5}},0 \right \rangle.\) This is the same direction as \(\langle 2,1,0 \rangle. \)

Hi,

For the problem with the wind, you answered c = sqrt 5/2. Is this the wind force ( final answer for part a ) or do we have to work from there.

The force necessary to move the ball is \(c\) times \(proj_{\vec{a}} \vec{b}.\)

Hello,

I need help with homework 3, question 1. I know that the series is ln(1+x) and I also did the ratio test, but I am not sure about how to use those values to find x. Basically, I got like convergence to -1 but they gave us big values and ask us to find x. How to do it?

Here is a similar problem.

For \(k = 1,-10\), determine whether or not there exists a number \(x\) such that

$$

1 + x + \frac{x^2}{2} + \frac{x^3}{3!} + \cdots = k

$$

If \(x\) exists, find it.

Note that

$$

1 + x + \frac{x^2}{2} + \frac{x^3}{3!} + \cdots = e^x

$$

for all \(x\) in the interval \((-\infty, \infty)\). Thus, we are looking for solutions to

$$

e^x = k

$$

for \(k = 1\) and \(k = -1\).

When \(k = 1\), \(e^x = 1 \Rightarrow x = 0\). When \(k = -1\), \(e^x = -1\) which has no solutions.

Be aware interval on which the series is allowed to be replaced by the corresponding elementary function, plays an important role. In the homework, the series is allowed to be be replaced by the elementary function \(\ln(1+x)\) for \(x\) in the interval \((-1,1]\). Therefore, any solutions you obtain after replacing the series with its corresponding elementary function must be also lie within this interval. If not, then it is not a solution to the original equality. If so, then it is a solution to the original equality.

Thank You. Can you explain the paragraph you typed in detail please.

Here is another example,

For \(k = -1, -1/2, 5000\) determine whether or not there exists a number \(x\) such that

$$

1 + 2(x-1) + 2^2(x-1)^2 + 2^3(x-1)^3 + \cdots = k

$$

If \(x\), exists find it.

Note that

$$

1 + 2(x-1) + 2^2(x-1)^2 + 2^3(x-1)^3 + \cdots = \frac{1}{1-2x}

$$

for all \(x\) in the interval \((1/2,3/2)\). The interval of convergence can be determined by applying the ratio test and testing the endpoints.

Thus, we are looking for solutions to

$$

\frac{1}{1-2x} = k

$$

for \(k = -1, -1/2, 5000\).

If \(k = -1\), then \(1/(1-2x) = -1 \Rightarrow x = 1\). As \(x = 1\) is in the interval of convergence, \(x = 1\) solves the original equation where \(k = -1\).

If \(k = -1/2\), then \(1/(1-2x) = -1/2 \Rightarrow x = 3/2\). As \(x = 3/2\) is not in the interval of convergence, \(x = 3/2\) does not solve the original equation where \(k = -1/2\).

If \(k = 5000\), then \(1/(1-2x) = 5000 \Rightarrow x = .4999\). As \(x = .4999\) is not in the interval of convergence, \(x = .4999\) does not solve the original equation where \(k = 5000\).

I have a WebAssign problem that I’m unsure why I am getting the wrong answer. We haven’t learned how to find the length of a curve yet, but I looked in the book to figure out how to solve it. My problem reads:

Find the length of the curve.

$$

r(t)= \left \langle 4t,t^2, \frac{1}{6}t^3 \right \rangle, 0\leq t\leq 4

$$

I took the derivative, squared it, square rooted that, and took the integral from 0 to 4 and got 128/3. I’m not exactly sure what I’m doing wrong.Thanks!

The steps you’ve outlined are the right ones. A bit of terminology is wrong, but the idea is correct. In particular, you mentioned taking the derivative and squaring it. What you mean to say is that you added the square of each component of \(\mathbf{r}'(t)\) and then took the square root. I’ve worked out the solution to the problem and got an answer different from 128/3. There is probably a computational error somewhere.

Hello Aori, I don’t know how to approach #2 in Homework 5.

A ball rests on a straight track aligned in the direction (2,1,0). The ball will remain stationary unless it experiences a force of at least 10 N along the direction of the track. If a wind is blowing in the direction (-1,-2,6), what is the magnitude of the wind’s force required to move the ball, and in which direction will the ball move?

Can you give me a sort of hint on how to go about solving this?

Let \(B = \langle 2,1,0 \rangle \) and \(W = \langle -1, -2, 6 \rangle \). The effect of \(W\) on \(B\) is determined by

$$

\text{comp}_\mathbf{B} \mathbf{W}

$$

However, just computing this for the given values of \(W\) and \(B\) isn’t enough. We need to factor in that the wind can blow harder. This can be accomplished by scaling the vector \(W\) by some nonzero constant, say \(c\).

Working with the vectors \(cW\) and \(B\) one can find the appropriate value of \(c\) and consequently the vector \(cW\). Once this is done, the magnitude of the force will be \(\left| cW \right|\). The direction can be computed using the formula

$$

\text{proj}_{\mathbf{B}} c\mathbf{W}

$$

Hi, I was wondering if you can help me with this problem. How can I find the

$$

\sum_{n=1}^{\infty} n(n+1) (1/2)^{n-1}

$$

from problem 4 on Homework 3?

Observe that,

$$

\begin{aligned}

\frac{d^2}{dx^2}\left[ \frac{1}{1-x}\right] = \frac{2}{(1-x)^3}

\end{aligned}

$$

and that,

$$

\begin{aligned}

\frac{d^2}{dx^2}\left[ \frac{1}{1-x}\right]

& = \frac{d^2}{dx^2}\left[ \sum_{n=0}^{\infty} x^n \right] \\

& = \frac{d}{dx}\left[ \sum_{n=1}^{\infty} \frac{d}{dx}[x^n] \right] \\

& = \frac{d}{dx}\left[ \sum_{n=1}^{\infty}n x^{n-1} \right] \\

& = \sum_{n=2}^{\infty}\frac{d}{dx}[n x^{n-1}] \\

& = \sum_{n=2}^{\infty}n(n-1) x^{n-2} \\

& = \sum_{n=1}^{\infty}(n+1)n x^{n-1}

\end{aligned}

$$

So that

$$

\frac{2}{(1-x)^3} = \sum_{n=1}^{\infty}(n+1)n x^{n-1}

$$

which holds true for \(|x|<1\).

Substituting in \(x = 1/2\) we obtain the series we want to find the sum of on the right and the sum of that series on the left.

Some things to note in the above calculuations:

Hello Aori, I’m a bit confused with getting the next term, such as in the ratio test or estimating sums if its an alternating series.

For example:

$$ \sum_{n=0}^{\infty} \frac{x^{2n}}{n^{2n+1}} $$

If I were to add 1 to n would it be

$$ \sum_{n=0}^{\infty} \frac{x^{2(n+1)}}{n^{2(n+1)+1}} $$

or

$$ \sum_{n=0}^{\infty} \frac{x^{2n+1}}{n^{2n+2}} $$

Thank you.

First note that the \(n\)-th term in the sum

$$ \sum_{n=1}^{\infty} \frac{x^{2n}}{n^{2n+1}} $$

is denoted \(a_n\) and is defined by

$$

a_n = \frac{x^{2n}}{n^{2n+1}}

$$

(By the way, the sum here must start at \(n=1\) since \(n=0\) makes the denominator zero).

Also note that when you are looking at the \(n\)-th term you must drop the summation symbol.

Now,

$$

\begin{aligned}

a_{n+1} &= \frac{x^{2(n+1)}}{n^{2(n+1)+1}}\\

& = \frac{x^{2n+2}}{(n+1)^{2n+2}}

\end{aligned}

$$

You simply replace where ever you see \(n\) by \(n+1\) and simplify to obtain the form of the \(n+1\)-th term.

However, these computations are done inline within the ratio test starting with

$$

\lim_{n \rightarrow \infty} \left| \frac{a_{n+1}}{a_n}\right|

$$

Here is how it would look:

$$

\require{cancel}

\begin{aligned}

\lim_{n \rightarrow \infty} \left| \frac{a_{n+1}}{a_n}\right| &= \lim_{n \rightarrow \infty} \left| \frac{\frac{x^{2n+2}}{(n+1)^{2n+2}}}{\frac{x^{2n}}{n^{2n+1}}}\right| \\

& = \lim_{n \rightarrow \infty}\left| \frac{x^{2n+2}}{(n+1)^{2n+2}} \cdot \frac{n^{2n+1}}{x^{2n}} \right| \\

& = \lim_{n \rightarrow \infty}\left| \frac{\bcancel{x^{2n}}x^2}{(n+1)^{2n+2}} \cdot \frac{n^{2n+1}}{\bcancel{x^{2n}}} \right| \\

& = x^2 \cdot \lim_{n \rightarrow \infty} \left| \frac{n^{2n+1}}{(n+1)^{2n+2}} \right| \\

& = x^2 \cdot \lim_{n \rightarrow \infty} \frac{1}{n} = x^2 \cdot 0 = 0

\end{aligned}

$$

Since the limit of the ratio is zero regardless of the value of \(x\), the series converges for all values of \(x\). That is the radius of convergence is \(\infty\) and the interval of convergence is \((-\infty, \infty)\).

As for using the Alternating Series Estimation Theorem, please choose an example and I will go through a detailed solution. The question you’re asking is too broad.

Hello. I have a question regarding this WebAssign problem

$$

\cos(x) \approx 1 – \frac{x^2}{2} + \frac{x^4}{24},\; (|error| < 0.05 )

$$

It is asking for the range of values of x for which the given approximation is accurate to within the stated error. I get the the range of values of (-1.047,1.047), but that is not the correct answer.

Here is an example: Use the Alternating Series Estimation Theorem to estimate the range of values of x for which the given approximation is accurate to within the stated error.

$$

\arctan(x) \approx x – \frac{x^3}{3} + \frac{x^5}{5}, \; (|error| < 0.0005 ) $$

The term that plays the role of \(b_{n+1}\) in the Alternating Series Estimation Theorem is the term in the Maclaurin series for \(\arctan x\) after \(\frac{x^5}{5}\) in absolute value, which is \( \left| \frac{x^7}{7} \right| \).

This term bounds the error and so it suffices to set

$$

\left| \frac{x^7}{7} \right| < 0.0005 $$ which has as its solution set the interval \((-\sqrt[7]{.0035}, \sqrt[7]{.0035}) \). Note: I tried this problem on WebAssign; it seems to have issues with rounding. If three decimal places doesn't work, do four.

Hello, I have a question about the maclaurin series of: $$ f(x) = (1+x)^n = \sum_{n=0}^\infty\binom{k}{n} x^n $$

What exactly does the array of k and n mean? How does it expand to many terms?

Thank you.

By definition,

$$

{ k \choose n } = \frac{ k(k-1)(k-2) \cdots (k-n + 1)}{n!}

$$

Hello Aori,

I have a question about the binomial series.

I understand that if $$f(x) = (1+x)^{k}$$,

$$\frac{d^{n}}{dx^n} = kPn * (1+x)^{k-n}$$,

and therefore the McLauren series for $$f(x) = \sum_{n=0}^{\infty }[ kCn * x^n ].$$

However, I don’t understand how to compute \(kCn\) when k<n.

From what I understand, $$kCn = \frac{kPn}{ n!} = \frac{k!}{ n! (k-n)! }$$

For example, if \(k = 2\), and \(n = 5\),

$$kCn = 2C5 = \frac{2!}{5! (-3)!} $$

Or when k is a negative number, such as \(k = -2\),

$$kCn = \frac{(-2)!}{ n! (-2-n)!}$$

I looked up negative factorials on wikipedia, but it said that factorials are not defined for negative integers, even when the factorial function is extended by the gamma function.

If this is the case, how does the binomial series work?

The problem seems to be with the identity

$$

\frac{kPn}{n!} = \frac{k!}{n! (k-n)!}.

$$

For your first example, note that

$$

2 C5 = \frac{2(2-1)(2-2)(2-3)(2-4)}{5!} = 0

$$

For your second example, with \(n = 5\) note that

$$

\begin{aligned}

(-2) C5 &= \frac{-2(-2-1)(-2-2)(-2-3)(-2-4)(-2-4)}{5!} \\

& = \frac{(-2)(-3)(-4)(-5)(-6)}{5!}

\end{aligned}

$$

which equals \(-\frac{2 \cdot 3 \cdot 4 \cdot 5 \cdot 6}{5!}\).

When \(k\) is not a non-negative integer we compute, \(kCn\) using the identity

$$

kCn = \frac{kPn}{n!}

$$