text:
# lyx bibliography does not display math symbols

I am using the IEEEtran template. I have to cite one reference, this reference has several special math characters, when I tried to convert it to pdf file, LyX gave many errors about it,

Errors: six "undefined control sequence."

Description:
...ncertainty measures for {$\ell^{2}(\mathbb {R}^{n})$},
The control sequence at the end of the top line
of your error message was never \def'ed. If you have
misspelled it (e.g., \hobx'), type I' and the correct
spelling (e.g., I\hbox'). Otherwise just continue,
and I'll forget about whatever was undefined.

From the error and description, it should be the title, and the title of reference is :

title="Entropy-Based Uncertainty Measures for {$\ell^{2}(\mathbb{R}^{n})$},
{$\ell^{2}(\mathbb{Z})$}, and {$\ell^{2}(\mathbb{Z}/N\mathbb{Z})$}
with a Hirschman Optimal transform for {$\ell^{2}(\mathbb{Z}/N\mathbb{Z})$}"

However, when I add one special relation symbol \triangleq or complicated binary operators, similar symbols in the body of the LyX file anywhere (not Greek letters), LyX can convert into pdf file smoothly, but leave unwanted symbols in the pdf file.

I don't have anything in the preamble. and the bibliography works with the same author with different paper title, so it is not a problem with the author name but with the math symbols.

-

You need to load the amssymb` package.


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# 31) Thomas Corporation produces stopwatches. According to company standards, it should take 1 hour..

31) Thomas Corporation produces stopwatches.  According to company standards, it should take 1 hour of direct labor to produce a stopwatch. Thomas' standard labor cost is $18 per hour. During June, Thomas produced 5,000 stopwatches and used 5,150 hours of direct labor at a total cost of$102,500.  What is Thomas' direct labor price variance for June?

A) $9,800 favorable B)$9,515 unfavorable

C) $9,515 favorable D)$9,800 unfavorable

32) The actual cost of direct materials is $51.30 per pound. The standard cost per pound is$55.00. During the current period, 7,240 pounds were used in production. The standard quantity for actual units produced is 7,000 pounds. How much is the direct materials price variance?

A) $25,900 favorable B)$25,900 unfavorable

C) $26,788 favorable D)$26,788 unfavorable

33) The following information describes a company's usage of direct labor in a recent period:

 Actual direct labor hours used 32,500 Actual rate per hour $18.00 Standard rate per hour$16.50 Standard hours for units produced 32,000

How much is the direct labor price variance?

A) $8,250 favorable B)$8,250 favorable

C) $9,000 unfavorable D)$9,000 unfavorable

34) The following information describes a company's usage of direct labor in a recent period:

 Actual direct labor hours used 32,500 Actual rate per hour $18.00 Standard rate per hour$16.50 Standard hours for units produced 32,000

How much is the direct labor price variance?

A) $48,000 favorable B)$48,750 favorable

C) $48,000 unfavorable D)$48,750 unfavorable

35) The actual cost of direct materials is $11.70 per pound. The standard cost per pound is$11.50. During the current period, 8,200 pounds of direct materials were used in production and 18,300 pounds were purchased. The standard quantity of direct materials for actual units produced is 18,100 pounds. How much is the direct materials efficiency variance?

A) $113,850 favorable B)$115,830 favorable

C) $115,830 unfavorable D)$113,850 unfavorable

36) The actual cost of direct labor per hour is $15.20 and the standard cost of direct labor per hour is$14.00. One and a half budgeted direct labor hours are allowed per finished unit. During the current period, 500 finished goods were produced using 2,050 direct labor hours. How much is the direct labor price variance?

A) $600 favorable B)$22,460 favorable

C) $2,460 unfavorable D)$600 unfavorable

37) The actual cost of direct labor per hour is $15.20 and the standard cost of direct labor per hour is$14.00. One and a half standard direct labor hours are allowed per finished unit. During the current period, 500 finished goods were produced using 2,050 direct labor hours. How much is the direct labor efficiency variance?

A) $18,200 unfavorable B)$19,760 favorable

C) $18,200 favorable D)$19,760 unfavorable

38) The actual cost of direct labor per hour is $13.00. Two standard direct labor hours are allowed per finished good. During the current period, 2,000 finished goods were produced using 3,800 direct labor hours. The direct labor efficiency variance is$1,300, unfavorable. Calculate the standard direct labor rate per hour.

A) $26.00 B)$6.50

C) $13.00 D)$10.00

39) The standard cost of direct labor per hour is $10.00. Two standard direct labor hours are allowed per finished good. During the current period, 512 finished goods were produced using 1,000 direct labor hours. The direct labor price variance is$100, unfavorable. Calculate the actual cost of direct labor per hour.

A) $2.40 B)$10.00

C) $10.10 D)$20.00

40) Cateye Corporation Company budgeted 600 pounds of direct materials costing $27.00 per pound to make 6,000 units of product. The company actually used 585 pounds of direct materials costing$28.00 per pound to make the 6,000 units. What is the direct materials efficiency variance?

A) $405 favorable B)$420 favorable

C) $405 unfavorable D)$420 unfavorable

format: plain


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# Day 21

Thursday, April 20, 2017

# Sounds and spectra

(from last time in case needed)

This code plots the spectrum of a sound in a wav file: plot_spectrum.py

Example wav files: guitar low E, piano low F, brass bar

Here is the spectrum of the brass bar:

And here is the spectrum of the low piano note:

(Notice that the "fundamental" (the lowest frequency) is almost not present.)

Online tone generator in case it's useful

## Synthesis

This code produces a sound with components we specify: replicate_sound.py

Let us see if we can synthesize the recorded brass bar sound above using frequencies and amplitudes read from the spectrum of the recording:

synthesized_brass.wav

# Drum (circular stretched membrane)

Let's record the sound a drum, and analyze and synthesize it.

[In case of difficulties, I prepared these in advance: small tom closed, small tom open]

# Solving the wave equation on a disk

Let's solve this problem by separation of variables, find the spectrum, and see if it matches the recording of the real drum.

Relationship between F and f:

Code that made the picture above: circular_membrane_modes.py

# Synthesis of drum sound from solution of wave equation on disk

# enter frequency components here
freqs = 120.*array([1,1.593,2.136,2.295,2.653,2.917])
amps  = array([1.,1,1,1,1,1])

synthesized_drum.wav (first effort, just guessing frequency of fundamental)

actual drum recording


text:
# Special case of confirmatory factor analysis

I remember that I read an article about factor analysis where they assume that each item loads on one and only one latent factor. For example, in the following factor loading matrix $$\begin{pmatrix}\lambda_1&0\\\lambda_2&0\\\lambda_3&0\\0&\lambda_4\\0&\lambda_4\\0&\lambda_5\\0&\lambda_6\\0&\lambda_7\\\end{pmatrix},$$ the first three items load on the first latent factor while the last four load on the second factor.

In that article, the author calls this kind of the factor analysis exercise by a special name. However, I do not remember the title nor the name of this kind.

Do you have any idea?

• Are you referring to "simple structure"? – jsakaluk Apr 12 '18 at 6:01
• @jsakaluk: You help me a lot with this term. I am looking for a paper taking about it to take it as a reference for my paper. – TDT Apr 12 '18 at 8:47
• confirmatory factor analysis? – Jeremy Miles Apr 21 '18 at 20:31


text:
# Refractive index and critical angle

1. Nov 22, 2004

### Cheman

The equation relating these is :

sine(critical angle) = refractive index between 2 meterials.

Could someone please supply a proof for this?

Also, as a side question , is there a proof for the sin i / sin r = v1/ v2 , not including the one in which you draw triangles and substitute d1 for v1t, etc?

Thanks.

2. Nov 22, 2004

### ZapperZ

Staff Emeritus
It's hard to know why, for example, you can't derive this yourself. May I suggest that you first find the definition of what is meant by the "critical angle"? This is because, once you understand what it means, then you'll realize that all that is involved in finding the "proof" of this is simply setting one of the appropriate angle in Snell's Law to be 90 degrees (upon which, the sine of that angle "magically" becomes 1). This doesn't involve a rigorous proof. It does, however, involve the understanding of the meaning of the variables that are being used.

Zz.

3. Nov 22, 2004

### Cheman

I am not stupid so please try to avoid patronising me. I do know what all of these terms mean, but I believe that if you actually carry out the proof you are saying it will not work out.

Let me show you:

sin i / sin r = mu ( refractive index)

sin i = sin c
sin r = sin 90 = 1

Therefore, sin c = mu; which cannot be true - most refractive indices are above one and you cannot inverse sine something larger than one - there is no solution.

Now I have checked up what the resulting formula should be and apparently it should be sin c = 1/ mu. Now do you understand why I am asking the question - I can't get the proof to work.
I would really appreciate if someone could please give me a hand - cause I'm stumped.

Thanks.

4. Nov 22, 2004

### ZapperZ

Staff Emeritus
This is why I said it is HARD to understand why you are having a problem doing this, because you NEGLECTED to describe in detail WHERE you are stuck! I'm sorry, but contrary to popular belief, I am not a mind-reader, so I have no idea WHERE exactly you are stuck and what you actually know! (Oy vey!)

Secondly, "mu" is the index of refraction? Are you sure? Or is "mu" simply the RATIO of the index of refractions of the two medium, and that it is in particular a ratio when compared to vacuum where n=1? Start with the Snell's Law. There are TWO different index of refractions involved! It has to, the light is going from one medium to another, so it makes no sense to only have it as a function of one index. However, the RATIO of the two indexes would be relevant. ONLY when one of the medium is a vacuum (or even air as the first approximation) would this ratio would be simply the index of refraction of one of the medium.

I assume that I do not need to work out the detail of this since I hate to be accused of "patronizing" you with baby algebra that you already know.

Zz.

5. Nov 22, 2004

### Staff: Mentor

Start with Snell's law: $n_1 sin\theta_1 = n_2sin\theta_2$
If the second medium is air ($n_2 = 1$) and the angle of refraction is 90 degrees ($sin\theta_2 = 1$), then you can solve for the critical angle.

This is what ZapperZ was saying. But you've got to start with the correct Snell's law.

6. Nov 22, 2004

### Cheman

But 1mu2 is defined as the refractive index. ie - it equals v1/ v2. Therefore, can we not start by using this?

7. Nov 22, 2004

### Staff: Mentor

The definition of index of refraction is $n_1 = c/v_1$, where c is the speed of light in vacuum and v is the speed of light in the medium of interest. Thus, $v_1/v_2 = n_2/n_1$.

8. Nov 23, 2004

### Cheman

Could someone please prove the equation using sin i / sin r = v1 / v2 = 1mu2? After all, if the other equation is true, and this one is then we should be able to derive one from the other.

Thanks.

9. Nov 23, 2004

### Staff: Mentor

What equation are you trying to prove? What is "1mu2"?

$sin\theta_i/sin\theta_r = v_1/v_2$ is equivalent to Snell's law given the definition of index of refraction ($n_1 = c/v_1$).

10. Nov 23, 2004

### Cheman

Trying to proove sin c = 1mu2, from sin i / sin r = v1 / v2 = 1mu2. 1mu2 is the refractive index between the two materials.

Thanks.

11. Nov 23, 2004

### Staff: Mentor

I don't know what refractive index "between the two materials" means; v1/v2 equals the inverse ratio of the indices of refraction. (Of course, if the second material is vacuum or air, then $n_2 = 1$.)

Combine: $sin\theta_i/sin\theta_r = v_1/v_2$
With: $n_1 = c/v_1$
Gives: $sin\theta_i/sin\theta_r = n_2/n_1$

For the critical angle $sin\theta_r = 1$ and let $n_2 = 1$,
then: $sin\theta_c = 1/n_1$

12. Nov 23, 2004

### tumor

When diamond is immersed in the water ray of light is not reflected back but passes thru it, how come?


text:
# Simple Swapping Integral Question (A Proof in Univalent Functions by Pommerenke)

This is really a simple question involving swapping the order of integration. The step I'm confused about comes from a proof in Pommerenke's "Univalent Functions," and for those of you with a reference, it is Theorem 1.2.

Let $C$ be a (closed) piecewise smooth curve homologous to zero with respect to the domain $H\subset \mathbb{C}$. We want to show that

$$\int_C \overline{h(w)} h'(w) dw = \dfrac{1}{\pi} \int_{H} |h'(w)|^2 \left( \int_C \dfrac{dz}{z-w} \right) d\Omega_w$$

where $d\Omega_w$ is the Lebesgue measure.

The first step in the proof is to swap the order of integration, but I'm not really sure why one could do that. I'm guessing it has something to do with a remark in the text about changing $H$ slightly to assume that $h$ is analytic on $\overline{H}$ rather than just on $H$.

Several things...

1. I think I can assume that $H$ is a bounded domain.
2. Assuming that $H$ is a bounded domain, with the additional assumption that $h$ is analytic on $\overline{H}$, I can assume that $h$ is bounded on $H$.

I'm guessing I have to use something like Fubini, but if I want to use Fubini, I'll have to show that

$$\int_C \int_H \dfrac{1}{|z-w|} d\Omega_w |dz|$$

is convergent.

I can see that the inner integral is convergent for all $z$ in $H$. But then I'm not sure why the iterated integral is convergent.

Is it true that

$$\int_H \dfrac{1}{|z-w|} d\Omega_w$$

is a continuous function in $z$? I think it's true, but I'm not sure how to show this...

-
As for "changing $H$ slightly" the point must be that your original $H$ is open. Let $D$ be the image of the zero homotopy of $C$; then $D$ is a closed subset of $H$. Now, $\mathbb C\setminus H$ is also closed, and since $\mathbb C$ is a $T_4$ space these two disjoint closed sets can be separated by disjoint open neighborhoods, and the closure of the open neighborhood of $D$ will be entirely contained in $H$, so $h$ is analytic there (and all points of interest are still within its interior). –  Henning Makholm Jan 2 '12 at 0:49
Errr, sorry for not clarifying: I'm using the convention that "domain" means open and connected. –  Braindead Jan 2 '12 at 1:08
Yes, I got that. Sorry if what I told was something you already knew; it was not entirely clear from the question whether this was something you were unsure about or just scene-setting for the real question. –  Henning Makholm Jan 2 '12 at 1:12


text:
Power in Circuits Video Lessons

Concept

# Problem: Two light bulbs have resistances of 400 Ω and 800 ΩPart AThe two light bulbs are connected in series across a 120- V line. Find the current through each bulb.Enter your answers numerically separated by a comma.I400, I800 =______APart BFind the power dissipated in each bulb.P400, P800 =________WPart CFind the total power dissipated in both bulbs.P =______W

###### FREE Expert Solution

Part A

Current:

$\overline{){\mathbf{i}}{\mathbf{=}}\frac{\mathbf{V}}{{\mathbf{R}}_{\mathbf{e}\mathbf{q}}}}$

For resistors in series:

$\begin{array}{rcl}{\mathbf{R}}_{\mathbf{e}\mathbf{q}}& \mathbf{=}& {\mathbf{R}}_{\mathbf{1}}\mathbf{+}{\mathbf{R}}_{\mathbf{2}}\\ & \mathbf{=}& \mathbf{400}\mathbf{+}\mathbf{800}\end{array}$

Req = 1200Ω

83% (458 ratings)
###### Problem Details

Two light bulbs have resistances of 400 Ω and 800 Ω

Part A

The two light bulbs are connected in series across a 120- V line. Find the current through each bulb.
I400, I800 =______A

Part B
Find the power dissipated in each bulb.
P400, P800 =________W

Part C
Find the total power dissipated in both bulbs.
P =______W


text:
## electrode reaction rate constants

Also contains definitions of: conditional rate constant of an electrode reaction, standard rate constant of an electrode reaction
https://doi.org/10.1351/goldbook.E01961
The electrode reaction rate constants are related to the partial currents by $k_{\mathrm{ox}}=\frac{l_{\text{a}}}{n\ F\ A\ \Pi\: c_{i}^{\nu _{i}}}$ $k_{\mathrm{red}}=\frac{l_{\text{c}}}{n\ F\ A\ \Pi\:c_{i}^{\nu _{i}}}$ where kox and kred are the rate constants for the oxidizing (anodic) and reducing (cathodic) reactions respectively, n is the charge number of the cell reaction, F is the Faraday constant and A is the geometric area of the electrode, the product Π c i ν i includes all the species i which take part in the partial reaction, ci is the volume concentration of species i and ν i is the order of the reaction with respect to species i. The conditional rate constant of an electrode reaction is the value of the electrode reaction rate constant at the conditional (formal) potential of the electrode reaction. When α the transfer coefficient is independent of potential, $k_{\text{c}} = \frac{k_{\text{ox}}}{\text{e}^{\frac{\alpha _{\text{a}}\ (E- E_{\text{c}}^{0'})\ n\ F}{\nu \ R\ T}}} = \frac{k_{\text{red}}}{\text{e}^{\frac{-\alpha _{\text{c}}\ (E- E_{\text{c}}^{0'})\ n\ F}{\nu \ R\ T}}}$ where αa and αc are the anodic and cathodic transfer coefficients respectively, E is the electric potential difference, E c is the conditional (formal) potential, ν is the stoichiometric number, R is the gas constant and T is the thermodynamic temperature. Similar rate constants can be defined using activities in place of concentrations in the first two equations, and the standard electrode potential in place of the conditional potential in the latter two equations. This type of rate constant is called the standard rate constant of the electrode reaction. The observable electrode rate constant is the constant of proportionality expressing the dependence of the rate of the electrode reaction on the interfacial concentration of the chemical species involved in the reaction.
Sources:
PAC, 1974, 37, 499. 'Electrochemical nomenclature' on page 515 (https://doi.org/10.1351/pac197437040499)
PAC, 1980, 52, 233. 'Electrode Reaction Orders, Transfer Coefficients and Rate Constants. Amplification of Definitions and Recommendations for Publication of Parameters' on page 236 (https://doi.org/10.1351/pac198052010233)


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# How to visually or intuitively understand single element multi-dimensional tensors?

Consider the following code in PyTorch

>>>torch.tensor([8]).shape
torch.Size([1])
>>>torch.tensor([[8]]).shape
torch.Size([1, 1])
>>>torch.tensor([[[8]]]).shape
torch.Size([1, 1, 1])

We can notice that we want to store only a single element $$8$$ in a tensor. But it is possible in tensors to store $$8$$ in any n-dimensional tensor where $$n \in \mathbb{N}$$. In strict case $$\mathbb{N}$$ may be replaced by $$\mathbb{W}$$.

But, I am facing difficulty in understanding this fact of a single element contributing to all dimensions. If the element is present in all dimensions, then I am assuming that it has to be present multiple times, which is not the case. I can't understand how a single element is contributing any number of dimensions without repeating itself multiple times.

How to understand this phenomenon? How should I interpret or visualize this fact intuitively?

How should I interpret or visualize this fact intuitively?

You could visualize it as a point in a geometrical space:

• 8 is just a number
• [8] is just a number in a line
• [[8]] is a number in a plane
• [[[8]]] is a number in a space

The object (number 8) won't change. The space around it changes.

• You can never represent a complex object (3d-cube) in a simpler shape (2d-plane).
• But you can always represent a simpler object (2d-square) in a higher dimensional shape (3d-space).

A number is a simplest possible object, and therefore it "fits inside" (can be represented in) any dimension.

The number is not repeated if it is the only element of some high dimensional space.

An $$n$$-dimensional (vector) space is a set of objects (known as vectors, although this term is more commonly used to refer to the objects in $$1$$-dimensional spaces), which are ordered/organized/shaped in specific ways, but composed of scalars/numbers, and to which you can also apply certain operations.

It may be a good idea to think that the shape/organization of each object in an $$n$$-d space is determined by some "container", which has a certain shape, and that the container of the objects of higher-dimensional spaces generalizes (or it can include) the containers of lower-dimensional spaces.

• In the case of a $$0$$-dimensional space (e.g. $$\mathbb{R}$$), there's no container because you don't need it, as every object is composed of only one number (this is always the case!).

• In the case of $$1$$-dimensional spaces, an object could be composed of more than one number (but this does not have to be always the case: it depends on the specific $$1$$-d space, so there are many $$1$$-d spaces), so you need a way to organized these numbers. You organize them in a sequence that follows some direction. So, the objects $$[0] \in \mathbb{R}^1$$, $$[1, 2] \in \mathbb{R}^2$$, $$[2, 3, 1] \in \mathbb{R}^3$$, $$[0, 4, 2, 2] \in \mathbb{R}^4$$, etc., are all $$1$$-d objects because each of them is composed of one or more numbers, which are organized in sequence. $$\mathbb{R}^1$$, $$\mathbb{R}^2$$, $$\mathbb{R}^3$$, etc., are all $$1$$-d spaces, because they organize each of their objects in a sequence: their only difference is the number of numbers/scalars in each object.

• In the case of $$2$$-d spaces, you organize the objects not in a line, but in a rectangle. The rectangles can have different shapes, so they are not necessarily just squares. So, $$[0] \in \mathbb{R}^{1 \times 1}$$, $$\begin{bmatrix} 1 & 1\end{bmatrix} \in \mathbb{R}^{1 \times 2}$$, $$\begin{bmatrix} 1 & 1 \\ 0 & 2\end{bmatrix} \in \mathbb{R}^{1 \times 2}$$.

• In the case of $$3$$-d spaces, you organize each object into cuboids. This does not mean that these cuboids contain more than one number. In general, they could, but you also have cuboids that contain only one number, i.e. $$[0] \in \mathbb{R}^{1 \times 1 \times 1}$$, which, in PyTorch, would be printed as [[[0]]] to give you the idea that the number 0 is inside a container [0], which is inside another container [ ], which can contain only one small container that can contain only one number (e.g. [0], but you could also have had [1] or [10]), which is inside another container [ ], which can contain only one container, which can contain only another container, which, in turn, can contain only one number.

To give you an analogy, think of having many boxes of different sizes and also have balls that you can put inside the smallest boxes. These boxes are the "containers" and the balls are the numbers. You can put smaller boxes inside the bigger ones (such that the boxes cannot slide around) and you can put the balls inside only the smallest boxes (so that they do not move).


text:
# American Institute of Mathematical Sciences

June  2018, 38(6): 2731-2762. doi: 10.3934/dcds.2018115

## Asymptotic properties of various stochastic Cucker-Smale dynamics

 Institut de Mathématiques de Toulouse, 118, route de Narbonne, F-31062 Toulouse Cedex 9, France

Received  March 2017 Revised  December 2017 Published  April 2018

Starting from the stochastic Cucker-Smale model introduced in [14], we look into its asymptotic behaviours for different kinds of interaction. First in term of ergodicity, when $t$ goes to infinity, seeking invariant probability measures and using Lyapunov functionals. Second, when the number $N$ of particles becomes large, leading to results about propagation of chaos.

Citation: Laure Pédèches. Asymptotic properties of various stochastic Cucker-Smale dynamics. Discrete & Continuous Dynamical Systems - A, 2018, 38 (6) : 2731-2762. doi: 10.3934/dcds.2018115
##### References:

show all references

##### References:
Case $\gamma = 0.2$, with $D = 0.1$, $\lambda = 5$, $\beta = 2$ and $K_{\gamma} = 8.3$
Case $\gamma = 2$, with $D = 0.1$, $\lambda = 5$, $\beta = 2$ and $K_{\gamma} = 17$
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text:
# Is a short one easier?

I came up with this short riddle while washing up:

The right words get rid of little old me,

Otherwise baffled with puzzlement be.

Hint:

There is a clue or more on every line, take time to reflect

• Welcome to Puzzling.SE! This looks like a nice neat first riddle! If you haven't already, feel free to take the tour and visit the help center to learn more about the site, and I hope you stick around! – F1Krazy Jul 4 '18 at 12:57

A riddle?

The right words get rid of little old me

The puzzle will be solved if you come up with the solution

Otherwise baffled with puzzlement be.

Riddles can indeed be baffling and puzzling.

Hint

As well as each line having a clue, the riddle itself is a clue... because it's a riddle.

Title

This is indeed a very short riddle.

• I was one minute too late. I should have just checked the newly answered post (namely, yours) before continuing mine $\stackrel{\bullet\bullet}{\smile}$ $(+1)$ – Mr Pie Jul 4 '18 at 13:55
• Rightly guessed. Main clues were "rid" and "puzzlement"; the right words get rid of the riddle which are the answer; which is its name. "Little old me" because it is a little riddle; "old" being a familiar term to refer to itself, which of course is the riddle. – Sam Liddicott Jul 6 '18 at 14:17

Is it?

Dishwasher

I came up with this short riddle while washing up:

Normally you need to get down to arrange and remove the contents in a dishwasher. Hence the title. The short ones like kids need not bend.

The right words get rid of little old me:

Setting the right mode and timer makes the contents inside new.

Otherwise baffled with puzzlement be.

Not knowing about the controls make you see it in confusion.

There is a clue or more on every line, take time to reflect

After the cleaning time, the contents are as new that you can see your own reflection on it.

I think you are:

An encrypted message

The right words get rid of little old me,

The right words are the deciphering key. Once applied it replaces the original message (the little old you) with a new, readable one.

Otherwise baffled with puzzlement be.

If the right deciphering key is not applied, you only get gibberish that will baffle you.

Germs?

The right words get rid of little old me

The "right words" could mean the right-most words in your opening sentence - "washing up". Washing gets rid of germs, which are certainly "little" (microscopic) and "old" (microorganisms are generally believed to be the oldest living things).

"There is a clue or more on every line"

This is what makes me believe that your opening words about washing up are not superfluous but part of the puzzle.

I think you are:

A crossword

The right words get rid of little old me,

filling in a crossword with the right words completes the crossword

Otherwise baffled with puzzlement be.

not being able complete a crossword may leave you baffled with puzzlement


text:
# 14-9 ALTERNATIVE DIVIDEND POLICIES In 2008, Keenan Company paid dividends totaling $3,600,000 on... 1 answer below » 14-9 ALTERNATIVE DIVIDEND POLICIES In 2008, Keenan Company paid dividends totaling$3,600,000 on net income of $10.8 million. Note that 2008 was a normal year and that for the past 10 years, earnings have grown at a constant rate of 10%. However, in 2009, earnings are expected to jump to$14.4 million, and the firm expects to have profitable investment opportunities of $8.4 million. It is predicted that Keenan will not be able to maintain the 2009 level of earnings growth—the high 2009 earnings level is attributable to an exceptionally profitable new product line introduced that year—and the company will return to its previous 10% growth rate. Keenan’s target capital structure is 40% debt and 60% equity. a. Calculate Keenan’s total dividends for 2009 assuming that it follows each of the following policies: (1) Its 2009 dividend payment is set to force dividends to grow at the long-run growth rate in earnings. (2) It continues the 2008 dividend payout ratio. (3) It uses a pure residual dividend policy (40% of the$8.4 million investment is financed with debt and 60% with common equity).

(4)     It employs a regular-dividend-plus-extras policy, with the regular dividend being based on the long-run growth rate and the extra dividend being set according to the residual policy.

b.       Which of the preceding policies would you recommend? Restrict your choices to the ones listed but justify your answer.

c.        Assume that investors expect Keenan to pay total dividends of $9,000,000 in 2009 and to have the dividend grow at 10% after 2009. The stock’s total market value is$180 million. What is the company’s cost of equity?

d.       What is Keenan’s long-run average return on equity? [Hint: g = Retention rate X

ROE = (1.0 - Payout rate)(ROE).]

e.        Does a 2009 dividend of \$9,000,000 seem reasonable in view of your answers to Parts c and d? If not, should the dividend be higher or lower? Explain your answer.

b)Financial institution


text:
# Bank PO Questions

Q:

If 6 years are subtracted from the present age of Arun and the remainder is divided by 18, then the present age of his grandson Gokul is obtained. If Gokul is 2 years younger to Madan whose age is 5 years, then what is the age of Arun ?

 A) 72 years B) 54 years C) 60 years D) 47 years

Explanation:

Let the Present age of arun is x
Gokul's age = 5-2 = 3
x-6/18 = 3
x-6 = 54
x = 60
Arun's present age = 60 years

2 1313
Q:

A person lends Rs. 300 to A at 4% and Rs.540 to B at 6% for the same time. if the total amount that he gets at the end is Rs. 1032. What is the time?

Money lent to A+B = (300 + 540) = Rs. 840

Interest received = Rs. 1032 - Rs.840 = Rs. 192

Interest on Rs. 300 lent to A for 1 year = $\frac{300×1×4}{100}$ = Rs. 12

Interest on Rs.540 lent to B for 1 year = $\frac{540×6×1}{100}$ = Rs. 36

Total yearly interest = Rs.(12+36) = Rs.48

If interest is Rs. 48, time = 1 year

If interest is Rs. 192, time =  192/48 = 4 years

1313
Q:

_ po _ mp _ n _ _ onmpo_.

 A) opmpnm B) mponpo C) mnompn D) mnopnm

Explanation:

The pattern 'mpon' is repeated.

Filed Under: Series - Verbal Reasoning - Mental Ability
Exam Prep: AIEEE , Bank Exams , CAT , GRE , TOEFL
Job Role: Bank Clerk , Bank PO

6 1311
Q:

Derrick's dad bought new tires for \$900 using a credit card. His card has an interest rate of 19%. If he has no other charges on his card and does not pay off his balance at the end of the month, how much money will he owe after one month?

 A) 11.25 B) 13.25 C) 14.25 D) 15.25

Explanation:

I=prt

0 1310
Q:

Tanjore kingdom was founded by Raja Odeyar. True or False ?

 A) TRUE B) FALSE

Explanation:

Mysore kingdom was founded by Raja Odeyar.

Subject: Indian History
Exam Prep: AIEEE , Bank Exams , CAT
Job Role: Bank Clerk , Bank PO

7 1309
Q:

When Ram was born, his father was 32 years older than his brother and his mother was 25 years older than his sister. If Ram's brother is 6 years older than Ram and his mother is 3 years younger than his father, how old was Ram's sister when Ram was born ?

 A) 6 years B) 8 years C) 10 years D) 12 years

Explanation:

Ram age when he was born = 0 years
=> His brother's age = 6 year
=> His father's age = brother age + 32years = 6+32 = 38
=> His mother's age = father's age - 3 = 35
So sister's age = 35-25 = 10years.

1 1309
Q:

Ravi, Anand and Pranay are running a business firm in partnership. What is Anand's share in the profit earned by them?

a. Ravi, Anand and Pranay invested the amounts in the ratio of 2 : 4 : 7.

b. Pranay's share in the profit is Rs. 8750.

 A) Only a is sufficient B) Neither a nor b is sufficient C) Only b is sufficient D) Both a and b sufficient

Explanation:

Given Pranay's profit in the share is Rs. 8750

But given that their investments are in the ratio 2:4:7

$\frac{7}{13}×P$ = 8750

P = 16250

Now Anand's share = $\frac{4}{13}×16250$ = Rs.5000

Thus both staements a and b are sufficient to answer this question.

2 1309
Q:

What is the difference between the local values of 3 in the number 43403 ?

 A) 0 B) 3030 C) 3003 D) 2997


text:
# Math Help - Proof of No Local Max/Min

1. ## Proof of No Local Max/Min

Okay.. so I've been doing calculus for about 4 hours at this point. And I haven't the slightest clue how to go about this proof... and my brain feels like jello.

Prove that the function

$f(x)=x^{101}+x^{51}+x+1$

has neither a local maximum nor a local minimum.

I took the derivative and now do not have any clue of what I need to do... how can I find critical points for it? Please help.

$f'(x)=101x^{100}+51x^{50}+1$

2. Originally Posted by banshee.beat
Okay.. so I've been doing calculus for about 4 hours at this point. And I haven't the slightest clue how to go about this proof... and my brain feels like jello.

Prove that the function

$f(x)=x^{101}+x^{51}+x+1$

has neither a local maximum nor a local minimum.

I took the derivative and now do not have any clue of what I need to do... how can I find critical points for it? Please help.

$f'(x)=101x^{100}+51x^{50}+1$
notice that this is actually a quadratic in disguise. it is quadratic in $x^{50}$. if you can't see it, let $a = x^{50}$, then you have $101a^2 + 51a + 1$

now can you find the critical points? set the function to zero and see what happens


text:
## Is anti-Semitism a homogeneous construct?

2009
##### Series
Diskussionsbeiträge der Projektgruppe Friedensforschung; 63
##### Publication type
Working Paper/Technical Report
##### Abstract
The present paper investigates how the distinction of different facets of anti-Semitism is to be understood from a methodological point of view. Does it mean that there is a unique (quantitative) dimension of anti-Semitism on which the participants can be ordered due to a variety of expressions of anti-Semitic prejudice, or does it imply that these facets are independent dimensions of anti-Semitism, which combine themselves into (qualitatively) different syndromes of anti-Semitism?
After discussing various methodological approaches to the analysis of homogeneity vs. inhomogeneity of tests or questionnaires, the paper applies these methodologies to data from an anti-Semitism survey by Petzold (2004) and comes to the conclusion that the facets are no independent constructs. All of them are indicators of one and the same anti-Semitic attitude whose expression is partly modified during the strengthening of anti-Semitic prejudice, however.
##### Summary in another language
The present paper investigates how the distinction of different facets of anti-Semitism is to be understood from a methodological point of view. Does it mean that there is a unique (quantitative) dimension of anti-Semitism on which the participants can be ordered due to a variety of expressions of anti-Semitic prejudice, or does it imply that these facets are independent dimensions of anti-Semitism, which combine themselves into (qualitatively) different syndromes of anti-Semitism?
After discussing various methodological approaches to the analysis of homogeneity vs. inhomogeneity of tests or questionnaires, the paper applies these methodologies to data from an anti-Semitism survey by Petzold (2004) and comes to the conclusion that the facets are no independent constructs. All of them are indicators of one and the same anti-Semitic attitude whose expression is partly modified during the strengthening of anti-Semitic prejudice, however.
150 Psychology
##### Keywords
Antisemitismusforschung,Item-Response-Modelle,Anti-Semitism Research,Item-Response-Models,Attitude Measurement
##### Cite This
ISO 690KEMPF, Wilhelm, 2009. Is anti-Semitism a homogeneous construct?
BibTex
@techreport{Kempf2009antiS-10306,
year={2009},
series={Diskussionsbeiträge der Projektgruppe Friedensforschung},
title={Is anti-Semitism a homogeneous construct?},
number={63},
author={Kempf, Wilhelm}
}

RDF
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<dcterms:abstract xml:lang="eng">The present paper investigates how the distinction of different facets of anti-Semitism is to be understood from a methodological point of view. Does it mean that there is a unique (quantitative) dimension of anti-Semitism on which the participants can be ordered due to a variety of expressions of anti-Semitic prejudice, or does it imply that these facets are independent dimensions of anti-Semitism, which combine themselves into (qualitatively) different syndromes of anti-Semitism?<br />After discussing various methodological approaches to the analysis of homogeneity vs. inhomogeneity of tests or questionnaires, the paper applies these methodologies to data from an anti-Semitism survey by Petzold (2004) and comes to the conclusion that the facets are no independent constructs. All of them are indicators of one and the same anti-Semitic attitude whose expression is partly modified during the strengthening of anti-Semitic prejudice, however.</dcterms:abstract>
<dcterms:issued>2009</dcterms:issued>
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Yes


text:
# Comic book database scripts

As promised in yesterday’s post , this one will detail the scripts and macros I used to build my database of comic books. The focus will be on generating CSV files that can be imported into almost any database system.

Let’s start with what I had: a box with over 500 comics. I hadn’t gone through the box in decades other than to pick out a few issues here and there, but I believed it was mostly sorted in alphabetical order by series title and then numerically by issue number within each series. My plan was to go through the box one series at a time, figure out which issues I had for that series, and use the Grand Comics Database to pull out all the relevant information for each issue.

The key to getting information on a particular issue from the GCD is to know the ID number the GCD has assigned to that issue. That ID is used to construct URLs that access the publication and contents information for the issue. The fastest way to get all the issue IDs for a comic book series is to go to the web page for that series and scrape it.

Here’s an example of a series page, the page for the 1963–1996 run of The Avengers.

In the Index Status section of the page (the green part in the screenshot) is a table with links to all the individual issues, ordered by issue number.1 The links include within their URLs the issue IDs we want. The goal is to build a list of the issue IDs of every comic book I own and then use that list to collect the information from the GCD to put into my database.

The process for getting all the issue IDs is this:

1. Navigate to the page for a series. This has to be done “by hand” using the search features of the GCD website.
2. Invoke a Keyboard Maestro macro and enter the list of issue numbers I own for that series. Here’s what that looks like:

The list is defined by a combination of numbers, commas, and hyphens in the manner you would expect. The macro then scrapes the page, extracts the links for those issues from the Index Status section, and adds a line with the issue ID for each issue to a file named GCD.csv.

3. Repeat these steps for every series I own.

The Keyboard Maestro macro invoked in Step 2 is this one:

The first two actions define variables, one for the URL of the series page (the current page showing in Safari) and one for the string that defines list of issue numbers. The third action is a Python script that does all the work:

python:
1:  #!/usr/bin/env -S -P${HOME}/anaconda/bin python 2: 3: import os 4: import requests 5: from bs4 import BeautifulSoup 6: import csv 7: import sys 8: 9: # Initialize 10: seriesURL = os.environ['KMVAR_seriesURL'] 11: csvPath = os.environ['HOME'] + '/Dropbox/comics/GCD.csv' 12: issueListStr = os.environ['KMVAR_issueList'] 13: 14: # Function for parsing the list of issues 15: def makeIssueList(listStr): 16: '''Make a Python list of integers from a formatted string. 17: 18: The string will look like "1-5,7, 10-13, 15,17"''' 19: 20: issues = [] 21: # Eliminate all spaces 22: s = listStr.replace(' ', '') 23: 24: # Split on commas if there are any 25: if ',' in s: 26: sublists = s.split(',') 27: sublists = [ x.strip() for x in sublists ] 28: else: 29: sublists = [s.strip()] 30: 31: # Loop through sublists 32: for r in sublists: 33: if '-' in r: 34: first, last = [ int(x.strip()) for x in r.split('-') ] 35: issues += range(first, last+1) 36: else: 37: issues.append(int(r.strip())) 38: 39: # Clean up duplicates and sort 40: issues = sorted(list(set(issues))) 41: 42: return issues 43: 44: # Convert the issue list string into a proper Python list 45: issueList = makeIssueList(issueListStr) 46: 47: # Get the page HTML 48: req = requests.get(seriesURL) 49: 50: # Parse the HTML 51: soup = BeautifulSoup(req.text, "html5lib") 52: 53: # Get the series title 54: seriesTitle = soup.title.text.split('::')[-1].strip() 55: 56: # Get the section with the list of issues 57: issueSection = soup.find('div', class_='status_grid') 58: 59: # Create a list of issues. Each item in the list will be 60: # a sublist of seriesTitle, issueNum, issueID 61: issues = [] 62: issueTags = issueSection.find_all('div', class_='status_grid_flex') 63: for i in issueTags: 64: if int(i.a.text.split()[0]) in issueList: 65: issueNum = i.a.text 66: issueID = i.a.get('href').split('/')[-2] 67: issues.append([seriesTitle, issueNum, issueID]) 68: 69: # Write out the results in CSV format 70: with open(csvPath, 'a') as f: 71: writer = csv.writer(f) 72: writer.writerows(issues)  The weird shebang line at the top is there because I wanted Keyboard Maestro to run the script through the version of Python 3 I installed via Anaconda. Lines 10 and 12 extract the values from the KM variables defined in the first two steps. Line 11 defines the file where we’re collecting the issue IDs. Lines 15–42 define the makeIssueList function, which does what it says: takes the string entered in the dialog box from Action 2 of the macro and builds a Python list of integers from it. I think the comments are reasonably good at explaining what it does. One part that is worth noting is Line 40. It converts the list into a set and then back into a list, which might seem stupid, but it’s an easy way to eliminate any duplicates. The rest of the script is web scraping, aided by the BeautifulSoup library. The series title is in the <title> of the page, before the double colon; Line 54 gets that. The Index Status section, where the table of issue links is, is defined by a <div> with class status_grid, so we extract that portion of the page in Line 57. The table of links is, oddly enough, not an HTML <table>, but a set of <div>s with class status_grid_flex. We collect all of these in Line 62 and then loop through them in Lines 63–67, pulling out the anchor text (which is the issue number) and the portion of the URL that corresponds to the issue ID. When the loop is done, we have a list of lists. Lines 70–72 write those out to the GCD.csv file defined back in Line 11. The Avengers portion of that file looks like this: The Avengers,131,28139 The Avengers,132,28301 The Avengers,134,28513 The Avengers,135,28606 The Avengers,136,28719 The Avengers,137,28835 The Avengers,138,28941 The Avengers,139,29052 The Avengers,140,29102 The Avengers,141,29262 The Avengers,142,29362 The Avengers,143,29458 The Avengers,144,29543 The Avengers,145,29641 The Avengers,146,29735 The Avengers,147,29828 The Avengers,148,29910 The Avengers,149,30007  Strictly speaking, I don’t really need the series title or issue number, but I put them there to make debugging easier. These scripts don’t just spring out of my head fully formed like Athena from the head of Zeus. With this macro built, it was time for the no-fun part of the job: pulling all the comics out of their box and going through them to get the issue lists for every series. But a Christmas Miracle occurred. Near the top of the box, hidden between a couple of issues of FOOM was a printed list of everything in the box. I knew I had made this list in the 80s, but I thought it had been lost when it wasn’t at the top of the box. I can’t remember what software I used to make it, and I have no idea why it was printed using the ImageWriter’s internal font instead of a real Mac font, but there it was. Two copies, even. So instead of having to sort through 500+ comic books, I could just sit down at my iMac and work through this list. It took about an hour or two, much less time than I had been expecting. With the semi-automated creation of the GCD.csv file out of the way, I could start the fully automated building of the CSV files for issues, covers, and stories. As I mentioned in yesterday’s post, you can download information for a particular issue in CSV form. This FAQ explains how to get the CSV through a link on the issue page. It didn’t take much effort to figure out that the CSV could be downloaded directly using a URL like https://www.comics.org/issue/28139/export_issue_csv/  where the numeric part in the middle is the issue ID. The first line of the CSV is the information for the issue, and the remaining lines are the information for each “story” in the issue. I put “story” in quotes because the GCD categorizes lots of things that aren’t really stories as “stories”: covers, letters pages, ads, tables of contents—basically everything in an issue that isn’t the issue itself. The fields for the issue line and the “story” lines are described here, but the description is incomplete. Instead of 17 fields for both issue and “story” records, there are 17 fields for issues and 18 for “stories.” The first 17 fields of “stories” match the documentation, but I could never figure out what the 18th was supposed to be—it was empty in all the records I looked at. I wanted my database to consist of three tables: issues, covers, and stories, where my definition of a story was limited to stuff inside the comic that most people would think of as a narrative or something related to a narrative. Tables of contents and one-off illustrations were fine, but no ads and no letters pages. I was able to do this filtering (and separate out the covers) by looking at the “Type” field. We’ll see this shortly when we go through the data collection script. One more thing before we look at the script. The CSV information is available only to registered GCD account holders. The link above won’t work unless you send a cookie with it to indicate you’re logged in to the site. There are probably many ways to figure out the necessary cookie. I logged in using Firefox and used the “Copy as cURL” command from the Network section of its Developer Tools to read all the headers that were sent when I accessed a GCD page. It looked like this: curl 'https://www.comics.org/series/1571/' -H 'User-Agent: Mozilla/5.0 (Macintosh; Intel Mac OS X 10.13; rv:64.0) Gecko/20100101 Firefox/64.0' -H 'Accept: text/html,application/xhtml+xml,application/xml;q=0.9,*/*;q=0.8' -H 'Accept-Language: en-US,en;q=0.5' --compressed -H 'Referer: https://www.comics.org/searchNew/?q=avengers' -H 'Connection: keep-alive' -H 'Cookie: csrftoken=areallylongstringwithabunchofparts' -H 'Upgrade-Insecure-Requests: 1' -H 'Cache-Control: max-age=0'  I copied the Cookie portion into the script to be sent as a header whenever I made a CSV request. Here’s the Python script that reads the GCD.csv file and writes out the issues.csv, covers.csv, and stories.csv files. It’s named comics.py. python: 1: import csv 2: import requests 3: import sys 4: 5: # Parameters 6: csvURL = 'https://www.comics.org/issue/{}/export_issue_csv/' 7: imageName = '{}.jpg' 8: cookies = dict(csrftoken='areallylongstringwithabunchofparts') 9: junkWords = 'advertisement promo letters ownership'.split() 10: skeletonCSV = sys.argv[1] 11: issueCSV = 'issues.csv' 12: coverCSV = 'covers.csv' 13: storyCSV = 'stories.csv' 14: issueInputHeaders = '''Number Volume Publisher Brand Publication_Date 15: Date Frequency Price Pages Editor Isbn Notes Barcode On_Sale_Date 16: Issue_Title Reprint_Links Keywords'''.split() 17: issueOutputHeaders = '''ID Title Number Date Price Pages 18: Publisher Frequency Editor Link'''.split() 19: contentInputHeaders = '''Title Type Feature Pages Scripter Penciller Inker 20: Colorist Letterer Editor Genre Characters Job_Number 21: Reprint_Info Synopsis Notes Keywords Other'''.split() 22: coverOutputHeaders = '''ID Title Image ImageName Penciller Inker 23: Colorist Letterer Characters Issue'''.split() 24: storyOutputHeaders = '''ID Title Type Scripter Penciller Inker 25: Colorist Letterer Pages Characters Synopsis Issue'''.split() 26: 27: # Initialize issue and story lists 28: issues = [] 29: covers = [] 30: stories = [] 31: currentIssueID = 0 32: currentCoverID = 0 33: currentStoryID = 0 34: 35: # Open and loop through the skeleton info file 36: with open(skeletonCSV, newline='', encoding='utf-8') as idCSV: 37: idReader = csv.DictReader(idCSV) 38: for book in idReader: 39: # Show progress via series and issue number 40: print('{} #{}'.format(book['series'], book['number'])) 41: req = requests.get(csvURL.format(book['id']), cookies=cookies) 42: # req.encoding = 'utf-8' 43: reader = csv.reader(req.text.strip().split('\r\n')) 44: for row in reader: 45: if len(row) == len(issueInputHeaders): 46: currentIssueID += 1 47: issue = dict(zip(issueInputHeaders, row)) 48: # Dates have '00' for day; change to '01' 49: # If month is '00', change to blank 50: year, month, day = [ int(x) for x in issue['Date'].split('-') ] 51: if month == 0: 52: issue['Date'] = '' 53: else: 54: day = 1 55: issue['Date'] = '{:02d}/{:02d}/{:4d}'.format(month, day, year) 56: # Get rid of ' USD' from Price 57: issue['Price'] = issue['Price'].replace(' USD', '') 58: # Get rid of floating point part of Pages 59: issue['Pages'] = issue['Pages'].replace('.000', '') 60: issue['Title'] = book['series'] 61: issue['ID'] = '{:04d}'.format(currentIssueID) 62: issue['Link'] = 'https://www.comics.org/issue/{}/'.format(book['id']) 63: issues.append({k: issue[k] for k in issueOutputHeaders}) 64: else: 65: content = dict(zip(contentInputHeaders, row)) 66: if content['Type'] == 'cover': 67: currentCoverID += 1 68: content['ID'] = '{:04d}'.format(currentCoverID) 69: content['Image'] = '' 70: content['ImageName'] = imageName.format(book['id']) 71: content['Issue'] = '{:04d}'.format(currentIssueID) 72: covers.append({k: content[k] for k in coverOutputHeaders}) 73: elif any(x in content['Type'] for x in junkWords): 74: continue 75: else: 76: currentStoryID += 1 77: # Get rid of floating point part of Pages 78: content['Pages'] = content['Pages'].replace('.000', '') 79: content['ID'] = '{:04d}'.format(currentStoryID) 80: content['Issue'] = '{:04d}'.format(currentIssueID) 81: stories.append({k: content[k] for k in storyOutputHeaders}) 82: 83: with open(issueCSV, 'w', newline='', encoding='utf-8') as f: 84: writer = csv.DictWriter(f, fieldnames=issueOutputHeaders) 85: writer.writeheader() 86: writer.writerows(issues) 87: 88: with open(coverCSV, 'w', newline='', encoding='utf-8') as f: 89: writer = csv.DictWriter(f, fieldnames=coverOutputHeaders) 90: writer.writeheader() 91: writer.writerows(covers) 92: 93: with open(storyCSV, 'w', newline='', encoding='utf-8') as f: 94: writer = csv.DictWriter(f, fieldnames=storyOutputHeaders) 95: writer.writeheader() 96: writer.writerows(stories)  The parameters section in Lines 6–25 defines a bunch of things we’ll use later in the script: • Templates for the CSV URLs and the cover image files (which we’ll talk about later). • A dictionary for the GCD cookie discussed above. • A list of words in the Type field that indicate records I don’t want in my database. • The input and output files. The input is given as an argument to the script so I could debug it using a smaller input file, one with only 5 lines instead 500. • The input fields for the issue information and the content (or “story”) information. These came from the documentation discussed above, with an 18th field (“Other”) added for the content. • The output fields for the issue, cover, and story tables. Lines 28–30 initialize the lists we’ll use to hold the information as we collect it. Lines 31–33 initialize serial numbers that will be used as the primary keys for our three tables. I could have used the GCD issue ID as the primary key for the issues table but decided to create my own, since that’s what I was doing for the other tables. Lines 36–37 then opens and parses the input file (GCD.csv) into a list of dictionaries, one for each issue, and Line 38 starts looping through the list. For each issue, Line 40 prints out a progress line to the terminal to let us know where we are, and Line 41 uses requests and the cookie defined in Line 8 to download the CSV. We then look at each line of the CSV and decide what kind of line it is. We use the length of the dictionary to decide if it’s for an issue or content (Line 45). If it’s for an issue, we • Generate a new serial number for that issue and add it to the dictionary. • Put the Date field in mm/dd/yyyy format so FileMaker can understand it. GCD saves the date as either yyyy-mm-00 or yyyy-00-00 depending on how much it knows about date. I decided, somewhat arbitrarily, to set the dates with known months to the first of the month. I need a day figure for FileMaker (or any database) to understand that it’s a date, and the first is as good as any. Dates with unknown months (there are only a few of these) are set to blank. I suspect I can fill in the blanks later by looking in the small print at the bottom of the title pages of these issues. • Change the Price to just a floating point number with no “USD” suffix. All of my comic books have prices in US dollars. • Change the Pages field from floating point to integer. There are no fractional pages. • Add a link to the GCD page for that issue. • Add a field for the series title. • Add the dictionary to the issues list, but just the fields we’re interested in, as defined by outputIssueFields. If the dictionary is for content, we use its Type field to determine whether its a cover or not (Line 66). If it’s a cover, we • Generate a new serial number for the cover and add it to the dictionary. • Add a blank Image field. We’ll fill this later with the cover image. • Add a field for the name of the cover image file. The image file will be a JPEG named for the issue ID, and we’ll discuss the script that downloads all the cover images when we’re done with this one. • Add the issue serial number so we can link the tables. • Add the dictionary to the covers list, but just the fields we’re interested in, as defined by outputCoverFields. If the dictionary doesn’t have a Type of cover, check to see if it’s stuff I don’t want as defined by the list of junkWords (Line 73) . If not, we’ll consider it a legitimate story and • Generate a new serial number for the story and add it to the dictionary. • Change the Pages field from floating point to integer. • Add the issue serial number so we can link the tables. • Add the dictionary to the stories list, but just the fields we’re interested in, as defined by outputStoryFields. When all the comic books have been looped through, Lines 83–96 write out the CSV files for the three tables. These are in a format that can be imported by pretty much any database program. This script, run via python comics.py GCD.csv  takes from six to eight minutes to complete; the variance, I assume, is due to response time from the GCD. We’re in the home stretch now. As I mentioned yesterday, artwork is what makes a comic book a comic book, and I couldn’t pass up the opportunity to add cover images to the database. GCD has images at for all of my comics, so I wrote a script, covers.py, to download them. Here it is: python: 1: import csv 2: import requests 3: from bs4 import BeautifulSoup 4: import sys 5: import shutil 6: 7: # Parameters 8: coverURL = 'https://www.comics.org/issue/{}/cover/4/' 9: cookies = dict(csrftoken='areallylongstringwithabunchofparts') 10: skeletonCSV = sys.argv[1] 11: 12: # Open and loop through the skeleton info file 13: with open(skeletonCSV, newline='', encoding='utf-8') as idCSV: 14: idReader = csv.DictReader(idCSV) 15: for book in idReader: 16: # Show progress via series and issue number 17: print('{} #{}'.format(book['series'], book['number'])) 18: 19: # Get the cover page 20: req = requests.get(coverURL.format(book['id']), cookies=cookies) 21: 22: # Parse the HTML 23: soup = BeautifulSoup(req.text, "html5lib") 24: 25: # Get the URL of the cover image 26: imageDiv = soup.find('div', class_='issue_covers') 27: imageURL = imageDiv.find('a', href='/issue/{}/'.format(book['id'])).img['src'] 28: img = requests.get(imageURL, stream=True) 29: if img.status_code==200: 30: with open('covers/{}.jpg'.format(book['id']), 'wb') as f: 31: img.raw.decode_content = True 32: shutil.copyfileobj(img.raw, f)  Once again, we’ll use the GCD.csv file as input to get the issue IDs for all my comic books. Line 8 defines a template for the URL of a page with the highest resolution cover image GCD has. We loop through all the issues, get the page with the image via requests and the same cookie we used before, and parse the page with BeautifulSoup. The image we want is in a <div> of class issue_covers (Line 26). We search that <div> for an anchor with an href attribute that includes our issue ID (Line 27). The URL of the image is in the src attribute of the <img> within that anchor. Then we fire up requests again to download that image URL (Line 28) and (if the download worked) treat the download as a set of raw bytes and save them to a JPEG file named for the issue ID (Lines 30–32). After running this script via python covers.py GCD.csv  which took 15–20 minutes, I had a folder of over 500 JPEGs. The names of the files correspond to the entries in the ImageName field of the Covers table. Getting the images into the database depends on the database you’re using. For FileMaker, after I had set up all the tables and imported the three CSV files, I wrote this script to loop through the Covers table, read the ImageName field, and insert that image into the Image field. Sorry about the small type size. FileMaker won’t let me change it, perhaps because you don’t actually type FileMaker scripts, perhaps because FileMaker’s programmers are all 25-year-olds who don’t think beyond themselves. Here’s what it says Go to Record/Request/Page [First] Loop Set Variable [$image ; Value: Covers::ImageName ]
Go to Field [ Covers::Image ]
Insert Picture [ "\$image" ]
Go to Record/Request/Page [ Next ; Exit after last: On ]
End Loop

You might be wondering how the script knows where to look for the image files, as there’s no obvious folder name in the script. The full path is hidden in the Insert Picture step. If you click on it, you can bring up the Specify Picture File window, where the full path is written out.

Here’s what an entry in the database looks like in FileMaker Go on my phone.

Because the tables are linked, I can show the cover and story information on the page for an issue.

And there you have it. Two years of comics, forty years of neglect, a few days of programming. What will I do with this information now that I have it in my pocket? I can’t tell you, but I know it’s mine.

1. To be clear about the terminology, what I’m calling the issue number is the number printed on the cover of the comic. It’s the number used by most people when referring to a comic, e.g. Spider-man debuted in Amazing Fantasy #15. What I’m calling the issue ID is a number internal to the GCD, unique for every individual issue in their database. As far as I know, its value cannot be determined by any rational formula or algorithm; it is assigned by the database itself when an issue is entered.


text:
# What to do after this last step?

I am solving a question from book in which I have to use summation. It is as follows:

$$\frac{1}{10}\sum_{n=0}^{9} e^{-jk\omega_0n}$$ The value of $$\omega_0$$ is $$\frac{2\pi}{10}$$. What I have done so far is:

$$\frac{1}{10} \left(\frac{1-e^{-9jk\frac{2\pi}{10}}.e^{jk\frac{2\pi}{10}}} {1-e^{-jk\frac{2\pi}{10}}}\right)$$

After this step I am stuck and can not figure out the next step?

• I believe the solution can be found by considering the fact that the exponential expressions represent complex numbers which can be intuitively added in the complex plane. Hint: what is the sum of $e^{0} + e^{-j\pi}$? – applesoup Nov 13 '18 at 13:35

This sum appears quite often in DSP.

\begin{align} \sum_{n=0}^{N-1} \exp(-j\alpha n) &\stackrel{(a)}{=} \frac{1- \exp(-j\alpha N)}{1 - \exp(-j\alpha )}\\ &= \frac{e^{-j\alpha N/2}(e^{+j\alpha N/2}- e^{-j\alpha N/2})}{e^{-j\alpha /2}(e^{+j\alpha /2}- e^{-j\alpha /2})}\\ &\stackrel{(b)}{=} e^{-j\frac{\alpha}{2}(N-1)} \frac{\sin(N\alpha/2)}{\sin(\alpha/2)} \end{align} where the last step $$(b)$$ comes from identity $$e^{+j\theta}-e^{-j\theta} = 2j\sin(\theta)$$ and be careful at $$(a)$$ as Matt's comment.

Substitute $$\alpha=k\omega_0=k\frac{2\pi}{N}$$, $$\sin(N\alpha/2)=\sin(k\pi)=0$$ if $$k\in\mathbb{Z}$$ hence so is the sum.

## Update for $$\alpha=0$$

It is trivial that $$\exp(-j\alpha n) \stackrel{\alpha=0}{=} 1$$ then the sum equals $$N$$.

It can be thought as the case of the final result after $$(b)$$ when $$\alpha \to 0$$ then $$e^{-j\frac{\alpha}{2}(N-1)} \frac{\sin(N\alpha/2)}{\sin(\alpha/2)} \stackrel{\alpha \to 0}{\to} N$$ where the limit is evaluated by L'Hôpital's rule.

Further interesting discussion can be found in Laurent Duval's answer.

• Note that technically speaking your first equation is not correct for $\alpha =0$ (i.e., $k=0$ in the OP's example), because you're dividing by zero; luckily, in the final formula, $\alpha=0$ leads to the correct result if taken as a limit. That's also why most people never realize / don't care. – Matt L. Nov 13 '18 at 14:09
• it's true @MattL. – AlexTP Nov 13 '18 at 14:11
• @MattL. Is it really luckily or will it always be the same when taken as a limit ? Are there any such summation formulas (in the context of DSP) that you know which do not equate to their L'Hopital limits at the 0/0 condition ? – Fat32 Nov 13 '18 at 21:31
• I would argue that the OP did not claim $k$ was integer. – AlexTP Nov 14 '18 at 12:18
• @AlexTP: Engineers using $k$ as a continuous (real) variable shouldn't be trusted ;) And, what is more, that formula looks very much like a DFT of a quite simple sequence, in which case $k$ would indeed be integer. But, you're right, who am I to assume anything that is not explicitly stated in the question!? – Matt L. Nov 14 '18 at 18:15

As a good answer has been given by @AlexTP, I'll go from the detail to a larger picture.

Although the "De L'Hospital's rule" seems attractive, it should be handled with a lot of care (see my answer to Why does L'Hopital's rule fail in this case?), because it requires somehow strong conditions!

I do recommend to use weaker conditions instead, when one can. And with sines of positive arguments around zero $$(0\le x\le\epsilon)$$, luckily, one can the standard simple bounds:

$$x(1-x^2/6) \le \sin(x) \le x$$

which suffice to bound the sine ratio $$\frac{\sin(N\alpha/2)}{\sin(\alpha/2)}$$ as $$\alpha \to 0$$ with $$N$$ given. More precisely, for both $$0\le \alpha/2\le\epsilon$$ and $$0\le N\alpha/2\le\epsilon$$, we have

$$N\alpha/2(1-(N\alpha/2)^2/6) \le \sin(N\alpha/2) \le N\alpha/2$$ and $$\frac{1}{\alpha/2} \le \frac{1}{\sin(\alpha/2)} \le \frac{1}{\alpha/2(1-(\alpha/2)^2/6)}$$

thus

$$N(1-(N\alpha/2)^2/6) \le \frac{\sin(N\alpha/2)}{\sin(\alpha/2)} \le \frac{N}{(1-(\alpha/2)^2/6)}\,,$$ whose central term thus tends to $$N$$ as $$\alpha \to 0_+$$. And you can plug it into AlexTP solution.

What is important here is that the underlying function of $$k$$ (just suppose $$k \in \mathbb{R}$$) is obviously continuous, as it is a finite sum of ($$N=10$$) continuous functions $$f_n(k)=\exp(-jk w_0 n)$$ (this could not work with infinite sums, as an infinite sum of continuous functions might be discontinuous, especially if the convergence is not uniform).

So, you can conveniently use the geometric sum formula anywhere it holds (with a non-zero denominator), and just use the limit (as the function IS continuous) otherwise.

This sum of an exponential geometric series is important: it is the discrete equivalent to the Fourier transform of a continuous rectangular window. The former yields a sinc, a cardinal sine. Your formula, for a discrete rectangular window, yield a (phased) discrete sinc called aliased sinc (asinc) or periodized sinc (psync), with details on the JOS website signal processing site.

• [off topic] adding an infinitely many continuous functions can produce a discontinuous resultant ? Such as the ideal LPF frequency response. But according to mathematical induction, it should be continuous too? So why is mathmeatical induction failing here? Do you know conditions on which mathematical induction can be applied or not? According to m.induction, if sum of two continuous functions are continuous -which is-, then the sum of N cont.func. will be also continuous for all N. So why is the contradiction ? – Fat32 Nov 14 '18 at 0:03
• From my forgotten past, if the convergence is uniform, then the resulting function is continuous as well. The set of natural number $\mathbb{N}$ does not contains infinity. Another example: a finite sum of rationals is rational. But $1-\frac{1}{3}+ \frac{1}{5} - \cdots \to \pi/4$. – Laurent Duval Nov 14 '18 at 7:43
• Second thought: it is more about the notion of limit. Continuity does not always hold under limits. Infinite sums are special cases of limits – Laurent Duval Nov 14 '18 at 7:46
• @Fat32, This is the very question that spawned Real Analysis (my least favorite math courses). There is not a failure of induction. The limit case is not part of the sequence. The limit of a sequence of continuous functions can be a discontinuous functions, but the limit is never reached it is only approached. It's one of those epsilon and delta arguments. – Cedron Dawg Nov 14 '18 at 13:17
• Infinity can be treated, as in Non-standard analysis, but it usually requires an additional set of axioms – Laurent Duval Nov 14 '18 at 16:01


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## Section6.3Multiplying Polynomials

###### ObjectivesPCC Course Content and Outcome Guide

Previously, we have learned to multiply monomials in Section 6.1 (such as $(4xy)\left(3x^2\right)$) and to add and subtract polynomials in Section 6.2 (such as $(4x^2-3x)+(5x^2+x-2)$). In this section, we will learn how to multiply polynomials.

###### Example6.3.2Revenue

A local organic jam company currently sells about $1500$ jars a month at a price of $\13$ per jar. They have also realized that for every $25$-cent increase in the selling price of a jar of jam, they will sell $50$ fewer jars of jam each month.

In general, this company's revenue can be calculated by multiplying the cost per jar by the total number of jars of jam sold.

If we let $x$ represent the number of $25$-cent increases in the price, then the price per jar will be the current price of thirteen dollars/jar plus $x$ times $0.25$ dollars/jar, or $13+0.25x\text{.}$

Continuing with $x$ representing the number of $25$-cent increases in the price, we know the company will sell $50$ fewer jars each time the price increases by $25$ cents. The number of jars the company will sell will be the $1500$ they currently sell each month, minus $50$ jars times $x\text{,}$ the number of price increases. This gives us the expression $1500-50x$ to represent how many jars the company will sell after $x$ $25$-cent price increases.

Combining this, we can now write a formula for our revenue model:

\begin{align*} \text{revenue} \amp= \left(\text{price per item}\right)\left(\text{number of items sold}\right)\\ R \amp= \left(13+0.25x\right)\left(1500-50x\right) \end{align*}

To simplify the expression $\left(13+0.25x\right)\left(1500-50x\right)\text{,}$ we'll need to multiply $13+0.25x$ by $1500-50x\text{.}$ In this section, we'll learn how to multiply these two expressions that each have multiple terms.

### Subsection6.3.1Review of the Distributive Property

The first step in almost every polynomial multiplication exercise will be a step of distribution. Let's quickly review the distributive property from Section 2.8, which states that $a(b+c)=ab+ac$ where $a, b\text{,}$ and $c$ are real numbers or variable expressions.

When we multiply a monomial with a binomial, we apply this property by distributing the monomial to each term in the binomial. For example,

\begin{align*} \highlight{-4x}(3x^2+5) \amp= \multiplyleft{(-4x)}(3x^2)+\multiplyleft{(-4x)}(5) \\ \amp=-12x^3-20x \end{align*}

A visual approach to the distributive property is to treat the product as finding a rectangle's area. Such rectangles are referred to as generic rectangles and they can be used to model polynomial multiplication.

The big rectangle consists of two smaller rectangles. The big rectangle's area is $2x(3x+4)\text{,}$ and the sum of those two smaller rectangles is $2x\cdot3x+2x\cdot4\text{.}$ Since the sum of the areas of those two smaller rectangles is the same as the bigger rectangle's area, we have:

\begin{align*} 2x(3x+4) \amp= 2x\cdot3x+2x\cdot4\\ \amp= 6x^2+8x \end{align*}

Generic rectangles are frequently used to visualize the distributive property.

Multiplying a monomial with a polynomial involves two steps: distribution and monomial multiplication. We also need to rely on the rules of exponents 6.1.15 when simplifying.

###### Remark6.3.7

We can use the distributive property when multiplying on either the left or the right. This means that we can state $a(b+c)=ab+ac\text{,}$ or that $(b+c)a=ba+ca\text{,}$ which is equivalent to $ab+ac\text{.}$ As an example,

\begin{align*} (3x^2+5)\highlight{(-4x)} \amp= (3x^2)\multiplyright{(-4x)}+(5)\multiplyright{(-4x)} \\ \amp=-12x^3-20x \end{align*}

### Subsection6.3.2Approaches to Multiplying Binomials

##### Multiplying Binomials Using Distribution

Whether we're multiplying a monomial with a polynomial or two larger polynomials together, the first step to carrying out the multiplication is a step of distribution. We'll start with multiplying binomials and then move to working with larger polynomials.

We know we can distribute the $3$ in $(x+2)3$ to obtain $(x+2)\multiplyright{3}=x\multiplyright{3}+2\multiplyright{3}\text{.}$ We can actually distribute anything across $(x+2)\text{.}$ For example:

\begin{equation*} (x+2)\cat=x\cdot \cat + 2\cdot \cat \end{equation*}

With this in mind, we can begin multiplying $(x+2)(x+3)$ by distributing the $(x+3)$ across $(x+2)\text{:}$

\begin{equation*} (x+2)\highlight{(x+3)} = x\highlight{(x+3)} + 2\highlight{(x+3)} \end{equation*}

To finish multiplying, we'll continue by distributing again, but this time across $(x+3)\text{:}$

\begin{align*} (x+2)\highlight{(x+3)} \amp= x\highlight{(x+3)} + 2\highlight{(x+3)}\\ \amp= x \cdot x + x \cdot 3 + 2 \cdot x + 2 \cdot 3\\ \amp=x^2+3x+2x+6\\ \amp=x^2+5x+6 \end{align*}

To multiply a binomial by another binomial, we simply had to repeat the step of distribution and simplify the resulting terms. In fact, multiplying any two polynomials will rely upon these same steps.

##### Multiplying Binomials Using FOIL

While multiplying two binomials requires two applications of the distributive property, people often remember this distribution process using the mnemonic FOIL. FOIL refers to the pairs of terms from each binomial that end up distributed to each other.

If we take another look at the example we just completed, $(x+2)(x+3)\text{,}$ we can highlight how the FOIL process works. FOIL is the acronym for "First, Outer, Inner, Last".

\begin{align*} (x+2)(x+3)\amp= (\overbrace{{x} \stackrel{}{\cdot} {x}}^{\text{F}}) + (\overbrace{{3} \stackrel{}{\cdot} {x}}^{\text{O}}) + (\overbrace{{2} \stackrel{}{\cdot} {x}}^{\text{I}}) + (\overbrace{{2} \stackrel{}{\cdot} {3}}^{\text{L}})\\ \amp=x^2+3x+2x+6\\ \amp=x^2+5x+6 \end{align*}
F: $x^2$

The $x^2$ term was the result of the product of first terms from each binomial.

O: $3x$

The $3x$ was the result of the product of the outer terms from each binomial. This was from the $x$ in the front of the first binomial and the $3$ in the back of the second binomial.

I: $2x$

The $2x$ was the result of the product of the inner terms from each binomial. This was from the $2$ in the back of the first binomial and the $x$ in the front of the second binomial.

L: $6$

The constant term $6$ was the result of the product of the last terms of each binomial.

##### Multiplying Binomials Using Generic Rectangles

We can also approach this same example using the generic rectangle method. To use generic rectangles, we treat $x+2$ as the base of a rectangle, and $x+3$ as the height. Their product, $(x+2)(x+3)\text{,}$ represents the rectangle's area. The next diagram shows how to set up generic rectangles to multiply $(x+2)(x+3)\text{.}$

The big rectangle consists of four smaller rectangles. We will find each small rectangle's area in the next diagram by the formula $\text{area}=\text{base}\cdot\text{height}\text{.}$

To finish finding this product, we need to add the areas of the four smaller rectangles:

\begin{align*} (x+2)(x+3)\amp=x^2+3x+2x+6\\ \amp=x^2+5x+6 \end{align*}

Notice that the areas of the four smaller rectangles are exactly the same as the four terms we obtained using distribution, which are also the same four terms that came from the FOIL method. Both the FOIL method and generic rectangles approach are different ways to represent the distribution that is occurring.

###### Example6.3.11

Multiply $(2x-3y)(4x-5y)$ using distribution.

Solution

To use the distributive property to multiply those two binomials, we'll first distribute the second binomial across $(2x-3y)\text{.}$ Then we'll distribute again, and simplify the terms that result.

\begin{align*} (2x-3y)\highlight{(4x-5y)}\amp=2x\highlight{(4x-5y)}-3y\highlight{(4x-5y)}\\ \amp=8x^2-10xy-12xy+15y^2\\ \amp=8x^2-22xy+15y^2 \end{align*}
###### Example6.3.12

Multiply $(2x-3y)(4x-5y)$ using FOIL.

Solution

First, Outer, Inner, Last: Either with arrows on paper or mentally in our heads, we'll pair up the four pairs of monomials and multiply those pairs together.

\begin{align*} (2x-3y)(4x-5y)\amp= (\overbrace{{\stackrel{}{2x}}\cdot{4x}}^{\large\text{F}})+ (\overbrace{{\stackrel{}{2x}}\cdot{(-5y)}}^{\large\text{O}})+ (\overbrace{{\stackrel{}{-3y}}\cdot{4x}}^{\large\text{I}})+ (\overbrace{{\stackrel{}{-3y}}\cdot{(-5y}}^{\large\text{L}})\\ \amp=8x^2-10xy-12xy+15y^2\\ \amp=8x^2-22xy+15y^2 \end{align*}
###### Example6.3.13

Multiply $(2x-3y)(4x-5y)$ using generic rectangles.

Solution

We begin by drawing four rectangles and marking their bases and heights with terms in the given binomials:

Next, we calculate each rectangle's area by multiplying its base with its height:

Finally, we add up all rectangles' area to find the product:

\begin{align*} (2xy-3)(4xy-5)\amp=8x^2y^2-10xy-12xy+15\\ \amp=8x^2y^2-22xy+15 \end{align*}

### Subsection6.3.3More Examples of Multiplying Binomials

When multiplying binomials, all of the approaches shown in Subsection 6.3.2 will have the same result. The FOIL method is the most direct and will be used in the examples that follow.

###### Example6.3.16

Multiply and simplify the formula for the jam company's revenue, $R$ (in dollars), from Example 6.3.2 where $R= (13+0.25x)(1500-50x)$ and $x$ represents the number of 25-cent price increases to the selling price of a jar of jam.

Solution

To multiply this, we'll use FOIL:

\begin{align*} R \amp= \left(13+0.25x\right)\left(1500-50x\right)\\ \amp= \left(13\cdot1500\right) + \left(13 \cdot (-50x) \right) + \left( 0.25x \cdot 1500 \right) + \left( 0.25x \cdot (-50x) \right) \\ \amp= 19500 - 650x + 375x - 12.5x^2 \\ \amp= -12.5x^2 - 275x + 19500 \end{align*}
###### Example6.3.17

An artist sells his paintings at $$10.00$ per piece. Currently, he can sell $100$ paintings per year. Thus, his annual income from paintings is $10\cdot100=1000$ dollars. He plans to raise the price. However, for each$$2.00$ of price increase per painting, his customers would buy $5$ fewer paintings annually.

Assume the artist would raise the price of his painting $x$ times, each time by $$2.00\text{.}$ Use an expanded polynomial to represent his new income per year. Solution Currently, each painting costs$$10.00\text{.}$ After raising the price $x$ times, each time by $2.00\text{,}$ each painting’s new price would be $10+2x$ dollars. Currently, the artist sells $100$ paintings per year. After raising the price $x$ times, each time selling $5$ fewer paintings, he would sell $100-5x$ paintings per year. His annual income can be calculated by multiplying each painting’s price with the number of paintings he would sell: \begin{align*} \text{annual income}\amp=(10+2x)(100-5x)\\ \amp=100(10+2x)-5x(10+2x)\\ \amp=1000+200x-50x-10x^2\\ \amp=-10x^2+150x+1000 \end{align*} After raising the price $x$ times, each time by$2.00\text{,}$ the artist’s annual income from paintings would be $-10x^2+150x+1000$ dollars.

### Subsection6.3.4Multiplying Polynomials Larger Than Binomials

The foundation for multiplying any pair of polynomials is distribution and monomial multiplication. Whether we are working with binomials, trinomials, or larger polynomials, the process is fundamentally the same.

###### Example6.3.20

Multiply $\left( x+5 \right)\left( x^2-4x+6 \right)\text{.}$

We can approach this product using either distribution generic rectangles. We cannot directly use the FOIL method, although it can be helpful to draw arrows to the six pairs of products that will occur.

Using the distributive property, we begin by distributing across $\left( x^2-4x+6 \right)\text{,}$ perform a second step of distribution, and then combine like terms.

\begin{align*} \left(x+5\right)\highlight{\left( x^2-4x+6 \right)} \amp= x\highlight{\left( x^2-4x+6 \right)}+5\highlight{\left( x^2-4x+6 \right)}\\ \amp= x\cdot x^2 - x\cdot 4x +x\cdot 6 +5\cdot x^2 - 5\cdot 4x +5\cdot 6\\ \amp= x^3 -4x^2 +6x +5x^2 -20x +30 \\ \amp= x^3+x^2-14x+30 \end{align*}

With the foundation of monomial multiplication and understanding how distribution applies in this context, we are able to find the product of any two polynomials.

### Subsection6.3.5Exercises

Multiplying Monomials with Binomials

###### 1

Find the product of the monomial and the binomial.

${6x}\left({x-6}\right)=$

###### 2

Find the product of the monomial and the binomial.

${9x}\left({x+4}\right)=$

###### 3

Find the product of the monomial and the binomial.

${2x}\left({3x+6}\right)=$

###### 4

Find the product of the monomial and the binomial.

${-3x}\left({-6x-10}\right)=$

###### 5

Find the product of the monomial and the binomial.

${-5x^{2}}\left({x+4}\right)=$

###### 6

Find the product of the monomial and the binomial.

${-3x^{2}}\left({x-8}\right)=$

###### 7

Find the product of the monomial and the binomial.

${5y^{2}}\left({-6y^{2}-9y}\right)=$

###### 8

Find the product of the monomial and the binomial.

${-2r^{2}}\left({-3r^{2}-5r}\right)=$

###### 9

Find the product of the monomial and the trinomial.

${8r^{2}}\left({-10r^{2}-9r-6}\right)=$

###### 10

Find the product of the monomial and the trinomial.

${-5t^{2}}\left({-7t^{2}+5t-10}\right)=$

###### 11

Find the product

$\displaystyle{ ({-10x^{13}y^{8}})({-6x^{13}+5y^{14}}) = }$

###### 12

Find the product

$\displaystyle{ ({2x^{15}y^{16}})({-9x^{19}-5y^{4}}) = }$

###### 13

Find the product

$\displaystyle{ ({-3a^{17}b^{5}})({4a^{6}b^{13}+5a^{18}b^{4}}) = }$

###### 14

Find the product

$\displaystyle{ ({-4a^{18}b^{12}})({8a^{12}b^{3}-4a^{12}b^{4}}) = }$

###### 15

Find the product

$\displaystyle{ ({5a^{9}})({-3a^{6}+4a^{4}b^{3}+10b^{5}}) = }$

###### 16

Find the product

$\displaystyle{ ({-6a^{9}})({-4a^{3}-2a^{4}b^{6}+3b^{5}}) = }$

Applications of Multiplying Monomials with Binomials

###### 17

A rectangle’s length is $6$ feet shorter than $5 \text{ times of}$ its width. If we use $w$ to represent the rectangle’s width, use a polynomial to represent the rectangle’s area in expanded form.

$\displaystyle{ \text{area}=}$ square feet

###### 18

A rectangle’s length is $7$ feet shorter than $4 \text{ times of}$ its width. If we use $w$ to represent the rectangle’s width, use a polynomial to represent the rectangle’s area in expanded form.

$\displaystyle{ \text{area}=}$ square feet

###### 19

A triangle’s height is $8$ feet longer than $\text{twice}$ its base. If we use $b$ to represent the triangle’s base, use a polynomial to represent the triangle’s area in expanded form. A triangle’s area can be calculated by $A=\frac{1}{2}bh\text{,}$ where $b$ stands for base, and $h$ stands for height.

$\displaystyle{ \text{area}=}$ square feet

###### 20

A triangle’s height is $10$ feet longer than $6 \text{ times of}$ its base. If we use $b$ to represent the triangle’s base, use a polynomial to represent the triangle’s area in expanded form. A triangle’s area can be calculated by $A=\frac{1}{2}bh\text{,}$ where $b$ stands for base, and $h$ stands for height.

$\displaystyle{ \text{area}=}$ square feet

###### 21

A trapezoid’s top base is $3$ feet longer than its height, and its bottom base is $9$ feet longer than its height. If we use $h$ to represent the trapezoid’s height, use a polynomial to represent the trapezoid’s area in expanded form. A trapezoid’s area can be calculated by $A=\frac{1}{2}(a+b)h\text{,}$ where $a$ stands for the top base, $b$ stands for the bottom base, and $h$ stands for height.

$\displaystyle{ \text{area}=}$ square feet

###### 22

A trapezoid’s top base is $1$ feet longer than its height, and its bottom base is $3$ feet longer than its height. If we use $h$ to represent the trapezoid’s height, use a polynomial to represent the trapezoid’s area in expanded form. A trapezoid’s area can be calculated by $A=\frac{1}{2}(a+b)h\text{,}$ where $a$ stands for the top base, $b$ stands for the bottom base, and $h$ stands for height.

$\displaystyle{ \text{area}=}$ square feet

Multiplying Binomials

###### 23

Find the product of the two binomials.

$\left({x+9}\right)\left({x+5}\right)=$

###### 24

Find the product of the two binomials.

$\left({y+6}\right)\left({y+9}\right)=$

###### 25

Find the product of the two binomials.

$\left({3y+4}\right)\left({y+5}\right)=$

###### 26

Find the product of the two binomials.

$\left({9r+8}\right)\left({r+2}\right)=$

###### 27

Find the product of the two binomials.

$\left({r+5}\right)\left({r-9}\right)=$

###### 28

Find the product of the two binomials.

$\left({t+2}\right)\left({t-5}\right)=$

###### 29

Find the product of the two binomials.

$\left({t-3}\right)\left({t-1}\right)=$

###### 30

Find the product of the two binomials.

$\left({x-6}\right)\left({x-7}\right)=$

###### 31

Find the product of the two binomials.

$\left({2x+9}\right)\left({5x+2}\right)=$

###### 32

Find the product of the two binomials.

$\left({5x+3}\right)\left({4x+1}\right)=$

###### 33

Find the product of the two binomials.

$\left({3y-4}\right)\left({2y-10}\right)=$

###### 34

Find the product of the two binomials.

$\left({2y-10}\right)\left({6y-10}\right)=$

###### 35

Find the product of the two binomials.

$\left({8r-6}\right)\left({r-4}\right)=$

###### 36

Find the product of the two binomials.

$\left({5r-2}\right)\left({r-6}\right)=$

###### 37

Find the product of the two binomials.

$\left({2t-7}\right)\left({t+2}\right)=$

###### 38

Find the product of the two binomials.

$\left({7t-3}\right)\left({t+9}\right)=$

###### 39

Find the product of the two binomials.

$\left({3x-9}\right)\left({4x^{2}-1}\right)=$

###### 40

Find the product of the two binomials.

$\left({6x-5}\right)\left({2x^{2}-1}\right)=$

###### 41

Find the product of the two binomials.

$\left({7x^{3}+10}\right)\left({x^{2}+1}\right)=$

###### 42

Find the product of the two binomials.

$\left({4y^{3}+8}\right)\left({y^{2}+10}\right)=$

###### 43

Find the product of the two binomials.

$\left({6y^{2}-2}\right)\left({3y^{2}-1}\right)=$

###### 44

Find the product of the two binomials.

$\left({4r^{2}-8}\right)\left({2r^{2}-1}\right)=$

###### 45

Expand the following polynomial in factored form.

${3\!\left(x+2\right)\!\left(x+3\right)}=$

###### 46

Expand the following polynomial in factored form.

${-3\!\left(x+2\right)\!\left(x+3\right)}=$

###### 47

Expand the following polynomial in factored form.

${x\!\left(x-2\right)\!\left(x+2\right)}=$

###### 48

Expand the following polynomial in factored form.

${-x\!\left(x+2\right)\!\left(x+3\right)}=$

###### 49

Expand the following polynomial in factored form.

${-\left(4x+1\right)\!\left(x+4\right)}=$

###### 50

Find the product of the two binomials.

$\displaystyle{ ({a-4b})({a+8b}) = }$

###### 51

Find the product of the two binomials.

$\displaystyle{ ({a-5b})({a-3b}) = }$

###### 52

Find the product of the two binomials.

$\displaystyle{ ({a+9b})({6a+10b}) = }$

###### 53

Find the product of the two binomials.

$\displaystyle{ ({a+5b})({7a+8b}) = }$

###### 54

Find the product of the two binomials.

$\displaystyle{ ({8a-5b})({2a-9b}) = }$

###### 55

Find the product of the two binomials.

$\displaystyle{ ({9a-9b})({8a+9b}) = }$

###### 56

Find the product of the two binomials.

$\displaystyle{ ({10ab+4})({5ab-8}) = }$

###### 57

Find the product of the two binomials.

$\displaystyle{ ({2ab+7})({2ab+8}) = }$

Applications of Multiplying Binomials

###### 58

A rectangle’s base can be modeled by ${x+3}$ meters, and its height can be modeled by ${x-2}$ meters. Use a polynomial to represent the rectangle’s area in expanded form.

$\displaystyle{ \text{area}=}$ square meters

###### 59

A rectangle’s base can be modeled by ${x-4}$ meters, and its height can be modeled by ${x-6}$ meters. Use a polynomial to represent the rectangle’s area in expanded form.

$\displaystyle{ \text{area}=}$ square meters

###### 60

An artist sells his paintings at ${\13.00}$ per piece. Currently, he can sell $100$ paintings per year. Thus, his annual income from paintings is $13\cdot100=1300$ dollars. He plans to raise the price. However, for each ${\5.00}$ of price increase per painting, his customers would buy $9$ fewer paintings annually.

Assume the artist would raise the price of his painting $x$ times, each time by ${\5.00}\text{.}$ Use an expanded polynomial to represent his new income per year.

$\displaystyle{ \text{new annual income}=}$

###### 61

An artist sells his paintings at ${\14.00}$ per piece. Currently, he can sell $130$ paintings per year. Thus, his annual income from paintings is $14\cdot130=1820$ dollars. He plans to raise the price. However, for each ${\3.00}$ of price increase per painting, his customers would buy $7$ fewer paintings annually.

Assume the artist would raise the price of his painting $x$ times, each time by ${\3.00}\text{.}$ Use an expanded polynomial to represent his new income per year.

$\displaystyle{ \text{new annual income}=}$

Multiplying Larger Polynomials

###### 62

Find the product of the binomials with the trinomial

$\left({-4x-4}\right)\left({x^{2}-4x-2}\right)=$

###### 63

Find the product of the binomials with the trinomial

$\left({4x+2}\right)\left({x^{2}+4x-2}\right)=$

###### 64

Find the product of the two polynomials.

$\left({5x+4}\right)\left({-4x^{3}+3x^{2}+4x+2}\right)=$

###### 65

Find the product of the two polynomials.

$\left({-5x+4}\right)\left({-3x^{3}-2x^{2}+4x-5}\right)=$

###### 66

Find the product of the trinomial and the trinomial.

$\left({x^{2}+2x+3}\right)\left({x^{2}+4x+5}\right)=$

###### 67

Find the product of the trinomial and the trinomial.

$\left({x^{2}+2x+3}\right)\left({x^{2}+4x+5}\right)=$

###### 68

Find the product

$\displaystyle{ ({a-3b})({a^{2}+4ab-7b^{2}}) = }$

###### 69

Find the product

$\displaystyle{ ({a+4b})({a^{2}-7ab+7b^{2}}) = }$

###### 70

Find the product

$\displaystyle{ ({a+b+5})({a+b-5}) = }$

###### 71

Find the product

$\displaystyle{ ({a+b-6})({a+b+6}) = }$


text:
Relative Sizes of Convergents of Simple Continued Fraction

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## MathSofiya 2 years ago Intro to Statistics: $\mu=\sum xP(x)$ $\mu$ is the expected value of x So would I add the x values and P(x) values and then multiply them to get $\mu$? x: 0 1 2 3 4 5 P(x): 0.237 0.396 0.264 0.088 0.015 0.001

1. hartnn

multiply corresponding x and p(x) values, like 0*0.237+1*0.396 .......

2. slaaibak

you would multiply 0 by 0.237 add it to 1 times .396 add it to 2 times 0.264 etc... the sum would be u

3. MathSofiya

oh ok. It seemed too simple

4. MathSofiya

Thanks

5. hartnn

welcome ^_^

6. MathSofiya

for the standard deviation I have the equation $\sigma=\sqrt{\sum(x-\mu)^2P(x)}$ So, here I do: $0-\mu=-1.253$ $1-\mu=1-1.253=-.253$ $2-\mu=2-1.253=.747$ $3-\mu=3-1.253=1.747$ $4-\mu=4-1.253=2.747$ $5-\mu=5-1.253=3.747$ square each solution then multiply to P(x), then add those solutions up

7. hartnn

yes! finally take square root then.

8. MathSofiya

oh yes of course. Thank you!

9. MathSofiya

how do I make a table in latex? I have to submit my work electronically and a table would make it easier.

10. MathSofiya

I couldn't find it here http://en.wikipedia.org/wiki/Help:Displaying_a_formula#Diagrams_in_TeX

11. hartnn

let me search.

12. hartnn
13. hartnn

maybe write in matrix form .....of nX2 i'll look up something specifically for table

14. MathSofiya

oh ok, thanks

15. hartnn
16. MathSofiya

\hlines what does that do?

17. MathSofiya

I think I get it. I'll just use \\ \hlines before i move on to the next line. Makes sense

18. hartnn
19. MathSofiya

oh perfect! There's the explanations for \hlines. Thanks! =D