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# Tom and John traveled in the same direction along the equal route at

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Tom and John traveled in the same direction along the equal route at  [#permalink]

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14 May 2016, 02:59
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Tom and John traveled in the same direction along the equal route at their constant speed rates of 15 km per hour and 10 km per hour, respectively. After 15 minutes Tom passed John, Tom reaches a certain Gas station, how many minutes it takes John to reach the station?
A. 5 min
B. 6 min
C. 7 and 1/2 min
D. 8 min
E. 10 min

* The answer will be posted in two days.

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"Only $79 for 1 month Online Course" "Free Resources-30 day online access & Diagnostic Test" "Unlimited Access to over 120 free video lessons - try it yourself" ##### Most Helpful Expert Reply Math Expert Joined: 02 Aug 2009 Posts: 7978 Re: Tom and John traveled in the same direction along the equal route at [#permalink] ### Show Tags 14 May 2016, 10:19 MathRevolution wrote: Tom and John traveled in the same direction along the equal route at their constant speed rates of 15 km per hour and 10 km per hour, respectively. After 15 minutes Tom passed John, Tom reaches a certain Gas station, how many minutes it takes John to reach the station? A. 5 min B. 6 min C. 7 and 1/2 min D. 8 min E. 10 min * The answer will be posted in two days. Firstly the Q should be " how many MORE minutes it takes John to reach the station?" TOM's speed = 15kmph, ........... JOHN's speed = 10KMPH.... so in 15 minutes, tom takes a lead of $$\frac{(15-10)}{60} *15 = \frac{5}{60}* 15 = \frac{5}{4}km$$.. John has to cover this$$\frac{5}{4}$$ km.. With speed of 10kmph, he will cover this distance in $$\frac{5}{4}*\frac{60}{10}= \frac{15}{2} = 7 and \frac{1}{2}..min$$ C _________________ ##### Most Helpful Community Reply Senior SC Moderator Joined: 14 Nov 2016 Posts: 1348 Location: Malaysia Tom and John both ride a bicycle in the same direction on an equal rou [#permalink] ### Show Tags 22 Feb 2017, 18:16 2 5 Tom and John both ride a bicycle in the same direction on an equal route at their constant speed rates of 20 km per hour and 12 km per hour, respectively. After 10 minutes Tom passes John, he reaches a gas station. How many minutes does it take John to reach the gas station? A. 5 min B. 6 min C. 6 and $$\frac{2}{3}$$ min D. 10 min E. 15 min _________________ "Be challenged at EVERY MOMENT." “Strength doesn’t come from what you can do. It comes from overcoming the things you once thought you couldn’t.” "Each stage of the journey is crucial to attaining new heights of knowledge." Rules for posting in verbal forum | Please DO NOT post short answer in your post! Advanced Search : https://gmatclub.com/forum/advanced-search/ ##### General Discussion Marshall & McDonough Moderator Joined: 13 Apr 2015 Posts: 1684 Location: India Re: Tom and John traveled in the same direction along the equal route at [#permalink] ### Show Tags 14 May 2016, 10:14 1 Relative speed = 10 - 5 = 5 Distance traveled in 15 mins or 1/4 hrs = 5/4 kms John travels 10 kms in 1 hour John travels 5/4 kms in 5/(4*10) = 1/8 hrs = 7.5 mins Answer: C Math Revolution GMAT Instructor Joined: 16 Aug 2015 Posts: 8017 GMAT 1: 760 Q51 V42 GPA: 3.82 Re: Tom and John traveled in the same direction along the equal route at [#permalink] ### Show Tags 16 May 2016, 19:56 Since the question states “after 15 minutes”, we can say Tom traveled 15/4km for 15 minutes as he can travel 15km per hour. Hence, using the same logic, we can say John traveled 10/4km as he travels 10km per hour. So, John has to travel (15/4)-(10/4)km=5/4km more. Since John’s speed is 10km/hour, which means 1km/6minutes. As he has to travel 5/4km more, it is going to take him 6(5/4) minutes. Hence, 6(5/4)=15/2 minutes. The correct answer is C. _________________ MathRevolution: Finish GMAT Quant Section with 10 minutes to spare The one-and-only World’s First Variable Approach for DS and IVY Approach for PS with ease, speed and accuracy. "Only$79 for 1 month Online Course"
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Tom and John both ride a bicycle in the same direction on an equal rou  [#permalink]

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22 Feb 2017, 19:27
AustinKL wrote:
Tom and John both ride a bicycle in the same direction on an equal route at their constant speed rates of 20 km per hour and 12 km per hour, respectively. After 10 minutes Tom passes John, he reaches a gas station. How many minutes does it take John to reach the gas station?

A. 5 min

B. 6 min

C. 6 and $$\frac{2}{3}$$ min

D. 10 min

E. 15 min

since passing John, Tom has gone 10 min*1/3 kpm=10/3 k
in the same ten minutes, John has gone 10 min*1/5 kpm=2 k
John needs 10/3-2=4/3 k more to reach gas station
(4/3 k)/(1/5 kpm)=6 2/3 min for John to reach gas station
C
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Re: Tom and John traveled in the same direction along the equal route at  [#permalink]

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20 Mar 2017, 09:09
chetan2u wrote:

Firstly the Q should be " how many MORE minutes it takes John to reach the station?"

That makes such a huge difference!!! I was not able to solve this problem because the question was not worded correctly
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Re: Tom and John traveled in the same direction along the equal route at  [#permalink]

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20 Mar 2017, 14:31
2
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Math Revolution, if you want people to buy your product, you should ensure the prompts use clear and concise English.
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Re: Tom and John both ride a bicycle in the same direction on an equal rou  [#permalink]

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11 May 2017, 00:27
gracie wrote:
AustinKL wrote:
Tom and John both ride a bicycle in the same direction on an equal route at their constant speed rates of 20 km per hour and 12 km per hour, respectively. After 10 minutes Tom passes John, he reaches a gas station. How many minutes does it take John to reach the gas station?

A. 5 min

B. 6 min

C. 6 and $$\frac{2}{3}$$ min

D. 10 min

E. 15 min

since passing John, Tom has gone 10 min*1/3 kpm=10/3 k
in the same ten minutes, John has gone 10 min*1/5 kpm=2 k
John needs 10/3-2=4/3 k more to reach gas station
(4/3 k)/(1/5 kpm)=6 2/3 min for John to reach gas station
C

if both are travelling in same direction...then it means as per question john was 10 mins ahead of TOM...so as relative speed of 8kmph it took 4/3 km more distance that tom had to cover and john covered only 2km as (12 kmph for 10mins)...so john travelled 2km to reach gas station....so it took him 10 mins
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Re: Tom and John both ride a bicycle in the same direction on an equal rou  [#permalink]

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11 May 2017, 02:08
ziyuen wrote:
Tom and John both ride a bicycle in the same direction on an equal route at their constant speed rates of 20 km per hour and 12 km per hour, respectively. After 10 minutes Tom passes John, he reaches a gas station. How many minutes does it take John to reach the gas station?

A. 5 min

B. 6 min

C. 6 and $$\frac{2}{3}$$ min

D. 10 min

E. 15 min

$$20km/h = \frac{20km}{1h}=\frac{20km}{60min}=\frac{1km}{3min}=\frac{1}{3}km/min$$
$$12km/h = \frac{12km}{1h}=\frac{12km}{60min}=\frac{1km}{5min}=\frac{1}{5}km/min$$

It took Tom 10 mins to travel to Gas station. The distance is: $$\frac{1}{3} \times 10 = \frac{10}{3} km$$

To reach gas station, John needs to travel in: $$\frac{10}{3}:\frac{1}{5}=\frac{50}{3}=16\frac{2}{3} min$$ after Tom passes John.

Hence, since Tom reached gas station, John needs to travel $$16\frac{2}{3}-10=6\frac{2}{3} min$$ to reach gas station.

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Re: Tom and John both ride a bicycle in the same direction on an equal rou  [#permalink]

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15 May 2017, 17:29
ziyuen wrote:
Tom and John both ride a bicycle in the same direction on an equal route at their constant speed rates of 20 km per hour and 12 km per hour, respectively. After 10 minutes Tom passes John, he reaches a gas station. How many minutes does it take John to reach the gas station?

A. 5 min

B. 6 min

C. 6 and $$\frac{2}{3}$$ min

D. 10 min

E. 15 min

We are given that Tom rides at 20 km per hour and John rides at 12 km per hour.

If Tom reaches a gas station in 10 minutes, or 1/6 hour, after passing John, then he has ridden 20 x 1/6 = 10/3 kilometers.

Since time = distance/rate, it will take John (10/3)/12 = 10/36 = 5/18 hour to reach the gas station.

5/18 hour is 5/18 x 60 = 50/3 = 16 2/3 minutes.

Thus, John arrives at the gas station 16 2/3 - 10 = 6 2/3 minutes after Tom reaches it.

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Re: Tom and John both ride a bicycle in the same direction on an equal rou  [#permalink]

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20 May 2017, 02:44
1
hazelnut wrote:
Tom and John both ride a bicycle in the same direction on an equal route at their constant speed rates of 20 km per hour and 12 km per hour, respectively. After 10 minutes Tom passes John, he reaches a gas station. How many minutes does it take John to reach the gas station?

A. 5 min

B. 6 min

C. 6 and $$\frac{2}{3}$$ min

D. 10 min

E. 15 min

Please write correct English when posting questions. The question is very easy but very badly worded. Please correct it.

Posted from my mobile device
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Re: Tom and John traveled in the same direction along the equal route at  [#permalink]

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06 Sep 2017, 05:25
MathRevolution wrote:
Tom and John traveled in the same direction along the equal route at their constant speed rates of 15 km per hour and 10 km per hour, respectively. After 15 minutes Tom passed John, Tom reaches a certain Gas station, how many minutes it takes John to reach the station?
A. 5 min
B. 6 min
C. 7 and 1/2 min
D. 8 min
E. 10 min

* The answer will be posted in two days.

Using relative speed , the distance travelled = 15*5/60 = 5/4 km.

Time taken for J = 5/4*1/10 hr= 1/8 hrs = 7.5 mins.

Ans:C
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Re: Tom and John both ride a bicycle in the same direction on an equal rou  [#permalink]

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06 Sep 2017, 05:39
hazelnut wrote:
Tom and John both ride a bicycle in the same direction on an equal route at their constant speed rates of 20 km per hour and 12 km per hour, respectively. After 10 minutes Tom passes John, he reaches a gas station. How many minutes does it take John to reach the gas station?

A. 5 min

B. 6 min

C. 6 and $$\frac{2}{3}$$ min

D. 10 min

E. 15 min

Using relative speed distance= 10 mins* 8km/60 = 4/3 kms.

Now John needs = 4/3* 60/12= 20 /3 mins

Ans:C
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Re: Tom and John traveled in the same direction along the equal route at  [#permalink]

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11 Sep 2017, 10:55
MathRevolution wrote:
Tom and John traveled in the same direction along the equal route at their constant speed rates of 15 km per hour and 10 km per hour, respectively. After 15 minutes Tom passed John, Tom reaches a certain Gas station, how many minutes it takes John to reach the station?
A. 5 min
B. 6 min
C. 7 and 1/2 min
D. 8 min
E. 10 min

In 15 minutes, or 1/4 hour, Tom travels 15 x 1/4 = 15/4 km and John travels 10 x 1/4 = 10/4 km. Thus, John has to travel 15/4 - 10/4 = 5/4 km to catch up to Tom.

Therefore, it takes John (5/4)/10 = 5/40 = 1/8 hours, or 7.5 minutes, to catch up to Tom.

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Tom and John both ride a bicycle in the same direction on an equal rou  [#permalink]

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12 Sep 2017, 08:41
Tom gains 8 km/h on John, or (4/3)km per 10 minutes. Now we have to find out in how many minutes is John able to cover that distance.

Since average speed = distance/time:

$$Time = (4/3)/12 = (1/9) hours$$

(1/9) hours * 60 minutes = (20/3) minutes -> 6 minutes and 40 seconds, or 6 minutes and (2/3) minutes.
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Re: Tom and John traveled in the same direction along the equal route at  [#permalink]

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01 Oct 2017, 08:22
i felt the question is not a gmat level question as it is not having correct explanation of the time asked for starts from when.

same question given with different data set in following link... with different difficulty level.
https://gmatclub.com/forum/tom-and-john ... 34625.html

MathRevolution wrote:
Tom and John traveled in the same direction along the equal route at their constant speed rates of 15 km per hour and 10 km per hour, respectively. After 15 minutes Tom passed John, Tom reaches a certain Gas station, how many minutes it takes John to reach the station?
A. 5 min
B. 6 min
C. 7 and 1/2 min
D. 8 min
E. 10 min

* The answer will be posted in two days.
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Posts: 5017
Location: India
Concentration: Sustainability, Marketing
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Re: Tom and John traveled in the same direction along the equal route at  [#permalink]

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03 Oct 2019, 18:43
MathRevolution wrote:
Tom and John traveled in the same direction along the equal route at their constant speed rates of 15 km per hour and 10 km per hour, respectively. After 15 minutes Tom passed John, Tom reaches a certain Gas station, how many minutes it takes John to reach the station?
A. 5 min
B. 6 min
C. 7 and 1/2 min
D. 8 min
E. 10 min

* The answer will be posted in two days.

total distance covered ; 15/60 * 5 ; 5/4 km
so Tom will take ; 5/4 * 60/10 ; 7.5 mins
IMO C
Re: Tom and John traveled in the same direction along the equal route at   [#permalink] 03 Oct 2019, 18:43
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My Math Forum Limit generalizing

 Calculus Calculus Math Forum

 October 23rd, 2016, 01:35 PM #1 Newbie   Joined: Oct 2016 From: Czech Republic Posts: 1 Thanks: 0 Limit generalizing I'm confused with this: I need to generalize the formula $\lim_{n\to\infty }\frac{a_n}{b_n}$ = $\frac{\lim_{n\to\infty} a_n}{\lim_{n\to\infty } b_n}$ so that it can be applied even in exceptional cases. I see that the exceptional case is when $b_n=0$. I appreciate all of you help
 October 23rd, 2016, 02:02 PM #2 Math Team   Joined: Dec 2013 From: Colombia Posts: 6,936 Thanks: 2265 Math Focus: Mainly analysis and algebra Can you give some context to this problem?
 October 24th, 2016, 11:13 AM #3 Math Team   Joined: Jan 2015 From: Alabama Posts: 2,633 Thanks: 678 What do you mean by "generalize"? The statement you give should be "As long as $lim_{n\to \infty} a_n$ and $\lim_{n\to\infy} b_n$ exists and the latter limit is not 0, $\lim_{n\to\infty}\frac{a_n}{b_n}= \frac{\lim_{n\to\infty} a_n}{\lim_{n\to\infty} b_n}$."

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Research Papers: Contact Mechanics

# Experimental Investigations on the Coefficient of Restitution for Sphere–Thin Plate Elastoplastic Impact

[+] Author and Article Information
Deepak Patil

Mechanical Engineering Department,
Carnegie Mellon University,
Pittsburgh, PA 15213

C. Fred Higgs, III

Mechanical Engineering Department,
Carnegie Mellon University,
Pittsburgh, PA 15213
e-mail: higgs@rice.edu

1Corresponding author.

Contributed by the Tribology Division of ASME for publication in the JOURNAL OF TRIBOLOGY. Manuscript received March 11, 2017; final manuscript received June 29, 2017; published online August 22, 2017. Assoc. Editor: Sinan Muftu.

J. Tribol 140(1), 011406 (Aug 22, 2017) (13 pages) Paper No: TRIB-17-1081; doi: 10.1115/1.4037212 History: Received March 11, 2017; Revised June 29, 2017

## Abstract

In multiparticle simulations of industrial granular systems such as hoppers, tumblers, and mixers, the particle energy dissipation is governed by an important input parameter called the coefficient of restitution (COR). Oftentimes, the wall thickness in these systems is on the order of a particles diameter or less. However, the COR value implemented in event-driven simulations is either constant or a monotonically decreasing function of the impact velocity. The present work experimentally investigates the effect of wall thickness on the COR through sphere–thin plate elastoplastic impacts and elucidates the underlying impact phenomena. Experiments were performed on 0.635 cm and 0.476 cm diameter (d) spheres of various materials impacting aluminum 6061 plates of different thicknesses (t) with the low impact velocities up to 3.1 m/s. Besides COR, indentation measurements and numerical simulations are performed to gain a detailed understanding of the contact process and energy dissipation mechanism. As the “t/d” ratio decreases, a considerable amount of energy is dissipated into flexural vibrations leading to a significantly lower COR value. Based on the results, it can be concluded that using a constant COR input value in particle simulations may not always be an appropriate choice, especially, in the case of thin plates. However, these new COR results validate that when the wall thickness is more than twice the sphere diameter (i.e., t/d > 2), a constant COR value obtained for an impact with semi-infinite plate can be reasonably used.

<>

## Figures

Fig. 1

Drop test apparatus (COR rig): (a) laboratory photograph and (b) schematic side view (A) hollow metal support, (B) plate, (C) sphere, (D) sphere holder, (E) Plexiglass casing, (F) suction device, (G) hose, and (H) air pump

Fig. 2

Plate installation on the hollow metal support: (a) supporting metal frame, (b) plate mounted on the frame, and (c) clamped plate

Fig. 3

COR as a function of distance from plate support for tungsten carbide sphere (d = 0.635 cm) impacting aluminum 6061 plates at Vi = 3.1 m/s

Fig. 4

Coefficient of restitution results for brass spheres: (a) d = 0.476 cm and (b) d = 0.635 cm

Fig. 5

COR as a function t/d ratio for various materials, results plotted for impact velocity Vi = 1.1 m/s

Fig. 6

COR as a function t/d ratio for various materials, results plotted for impact velocity Vi = 2.3 m/s

Fig. 7

COR as a function t/d ratio for various materials, results plotted for impact velocity Vi = 3.1 m/s

Fig. 8

Comparison between analytical solutions and the current work's experimental data of S2 steel sphere for impact velocity Vi = 2.3 m/s

Fig. 9

Plate indentation measurements: (a) contour plot, (b) cross section profile, and (c) crater image

Fig. 10

Indentation depth on various plates after impact with brass sphere (d = 0.635 cm)

Fig. 11

Indentation depth on various plates after impact with tungsten carbide sphere (d = 0.635 cm)

Fig. 12

Indentation depth on various plates after impact with S2 tool steel sphere (d = 0.635 cm)

Fig. 13

Finite element model mesh layout

Fig. 14

Comparison of experimental and simulation COR results for various materials

Fig. 15

Contact forces developed during the impact for various plates, brass sphere (d = 0.635 cm) impacting at Vi = 2.3 m/s

Fig. 16

Sphere–plate energy transfer during the impact process, brass sphere (d = 0.635 cm) impacting at Vi = 2.3 m/s on plate t = 0.476 cm

Fig. 17

Stress (von Mises equivalent) and plastic strain contours at the instance of maximum contact force for brass sphere (d = 0.635 cm) impacting at Vi = 2.3 m/s on thickest and thinnest plate: (a) stress contours, t = 0.160 cm, (b) stress contours, t = 1.270 cm, (c) plastic strain contours, t = 0.160 cm, and (d) plastic strain contours, t = 1.270 cm

Fig. 18

Relative energy loss between plastic deformation and flexural vibrations during the impact process, brass sphere (d = 0.635 cm) impacting at Vi = 2.3 m/s

Fig. 19

Experimental versus simulation results for indentation, brass sphere (d = 0.635 cm) impacting at Vi = 2.3 m/s

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text:
# How do you find the center and radius of the circle x^2 −2x + y^2 − 6y = 26?

May 1, 2016

Center is $\left(1 , 3\right)$ and radius is $6$

#### Explanation:

General equation of a circle is of the form

${x}^{2} + {y}^{2} + 2 g x + 2 f y + c = 0$, whose center is $\left(- g , - f\right)$ and radius is $\sqrt{{g}^{2} + {f}^{2} - c}$

Hence in the equation of circle ${x}^{2} - 2 x + {y}^{2} - 6 y = 26$ is of the form

${x}^{2} + {y}^{2} - 2 x - 6 y - 26 = 0$ and hence

$g = - 1$, $f = - 3$ and $c = - 26$

Hence, center of circle is $\left(- \left(- 1\right) , - \left(- 3\right)\right)$ or $\left(1 , 3\right)$

and radius is $\sqrt{{\left(- 1\right)}^{2} + {\left(- 3\right)}^{2} - \left(- 26\right)} = \sqrt{1 + 9 + 26} = \sqrt{36} = 6$

graph{x^2+y^2-2x-6y-26=0 [-10, 10, -5, 5]}


text:
# If a one-sided limit of a function doesn't exist, how can a function

#### Eclair_de_XII

Homework Statement
"Let $f(x)=\sqrt{x}$. Prove or disprove $\lim_{x\rightarrow0} \sqrt{x}=0$."
Homework Equations
A function $f$ has a limit $L$ at $a$ iff $\lim_{x\rightarrow a^-}=L=\lim_{x\rightarrow a^+}$.
Instinct tells me to just plug in the number, say the limit is zero, and be done with it. But at the same time, while reading the statement from the "Relevant equations" section of this post, I cannot feel but feel some doubt as to whether or not this is the right approach. I mean, only the right-sided limit is defined, while the left-sided limit isn't. So am I to conclude that since the root function doesn't even have a left-handed limit, that I don't need to worry about the statement in the "Relevant equations" section?

Sorry; it's been many years since I last took Calculus I, so I am sorry for being dense when it comes to simple concepts like this.

Related Calculus and Beyond Homework Help News on Phys.org

#### Mark44

Mentor
Problem Statement: "Let $f(x)=\sqrt{x}$. Prove or disprove $\lim_{x\rightarrow0} \sqrt{x}=0$."
Relevant Equations: A function $f$ has a limit $L$ at $a$ iff $\lim_{x\rightarrow a^-}=L=\lim_{x\rightarrow a^+}$.

Instinct tells me to just plug in the number, say the limit is zero, and be done with it. But at the same time, while reading the statement from the "Relevant equations" section of this post, I cannot feel but feel some doubt as to whether or not this is the right approach.
It isn't. What's usually meant when you're asked to prove a limit is to do so using a $\delta - \epsilon$ argument. And "plugging in a number" works only for the simplest of functions. For example, $\lim_{x \to 1}\frac {x^2 - 1}{x - 1}$ exists, but you can't just "plug in 1" for x.
Eclair_de_XII said:
I mean, only the right-sided limit is defined, while the left-sided limit isn't. So am I to conclude that since the root function doesn't even have a left-handed limit, that I don't need to worry about the statement in the "Relevant equations" section?
On the contrary, your relevant definition is very important here. If the right-side limit exists, but the left-side limit doesn't exist, what does that say about the existence of the two-sided limit?
Eclair_de_XII said:
Sorry; it's been many years since I last took Calculus I, so I am sorry for being dense when it comes to simple concepts like this.

#### Eclair_de_XII

what does that say about the existence of the two-sided limit?
So the two-sided limit does not exist? But doesn't that contradict the continuity of $\sqrt{x}$ at $x=0$?

Last edited:

#### Orodruin

Staff Emeritus
Homework Helper
Gold Member
2018 Award
Why do you say that the left side limit does not exist? It is certainly the case if you restrict yourself to functions taking real values, but the limit does exist regardless if you consider complex valued functions.

If you consider real valued functions only then $\sqrt x$ is not defined for $x < 0$ and it should come as no surprise that it is therefore not continuous on $\mathbb R$.

#### Eclair_de_XII

it is therefore not continuous on $\mathbb{R}$.
Oh! If I restrict the domain of $f$ to non-negative real numbers, will the condition stated in the "Relative equations" section then be a non-issue?

Last edited:

#### Orodruin

Staff Emeritus
Homework Helper
Gold Member
2018 Award
Oh! If I restrict the domain of $f$ to non-negative real numbers, will the condition stated in the "Relative equations" section then be a non-issue?
If you also restrict the codomain.

The concept of continuity on $\mathbb R$ is a special case of the general topologigal definition.

#### Mark44

Mentor
Why do you say that the left side limit does not exist? It is certainly the case if you restrict yourself to functions taking real values
I assumed that the OP was talking about the real-valued square root function.

#### Eclair_de_XII

If you also restrict the codomain.
Okay, so it's continuous if I restrict the codomain and domain to the non-negative real numbers. Wait, but if I restrict the domain to the non-negative real numbers, won't the codomain also be restricted as a result?

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#### MidgetDwarf

So the two-sided limit does not exist? But doesn't that contradict the continuity of $\sqrt{x}$ at $x=0$?
continuity and limit of a function are not equivalent. Recall that the definition of limits of a function requires that the number x is approaching be a limit point of the domain of the function. Just because a number is a limit point of a set, it does not imply that the limit point be in the set (think of open sets for the trivial case).

On the other hand, the definition of continuity states to consider if a function is continuous at a point, the minimum is that the point must be the domain of the function.

#### MidgetDwarf

Why do you say that the left side limit does not exist? It is certainly the case if you restrict yourself to functions taking real values, but the limit does exist regardless if you consider complex valued functions.

If you consider real valued functions only then $\sqrt x$ is not defined for $x < 0$ and it should come as no surprise that it is therefore not continuous on $\mathbb R$.
However, it is defined if we restrict the domain to all real number greater than or equal to x. It is very trivial to show this.

#### MidgetDwarf

I think you should review the concept of limit points, open, closed, compact set, limit of a function, definition of continuity.

#### WWGD

Gold Member
.....but if I restrict the domain to the non-negative real numbers, won't the codomain also be restricted as a result?
No, notice $f(x)=f(-x)$

#### Mark44

Mentor
... but if I restrict the domain to the non-negative real numbers, won't the codomain also be restricted as a result?
No, notice $f(x)=f(-x)$
How does this make any sense with $f(x) = \sqrt x$, assuming the real-valued square root function?

#### WWGD

Gold Member
How does this make any sense with $f(x) = \sqrt x$, assuming the real-valued square root function?
Instead of considering both the positive and negative root, just ignore the negative root, that's all.

#### PeroK

Homework Helper
Gold Member
2018 Award
Why do you say that the left side limit does not exist? It is certainly the case if you restrict yourself to functions taking real values, but the limit does exist regardless if you consider complex valued functions.

If you consider real valued functions only then $\sqrt x$ is not defined for $x < 0$ and it should come as no surprise that it is therefore not continuous on $\mathbb R$.
This is very much the physicist's view. In real analysis, however, the mathematician cannot extend the concept of a number to the complex field. A physicist is, of course, free to adduce the complex numbers in any case.

Instead, the key point missed by the OP is that the formal definition of continuity restricts any points under consideration to the domain of the function.

In this case the "left-sided" limit is not valid at the point $x =0$. And, in fact, the limit and the right-sided limit are equivalent at this point.

#### Mark44

Mentor
Mark44 said:
How does this make any sense with $f(x) = \sqrt x$, assuming the real-valued square root function?
Instead of considering both the positive and negative root, just ignore the negative root, that's all.
This still makes no sense. The real-valued square root function doesn't have negative roots.

#### WWGD

Gold Member
This still makes no sense. The real-valued square root function doesn't have negative roots.
Yes, I phrased it poorly , yet many people consider the possibility of two roots notwithstanding the fact that an expression with two outputs cannot be a function. I thought this is what poster was referencing.

#### Mark44

Mentor
yet many people consider the possibility of two roots

#### WWGD

Gold Member
Maybe we can match the misconception with Mr conception, Dave or otherwise ( I have a Venezuela connection so I know of some of the famous players) :).

#### Math_QED

Homework Helper
On the contrary, your relevant definition is very important here. If the right-side limit exists, but the left-side limit doesn't exist, what does that say about the existence of the two-sided limit?
I'm sorry but this response is as wrong as it can get.

The only definition of limit is the $\epsilon-\delta$ definition or something equivalent with it. The "definition" the OP is using is a theorem that holds under certain conditions. Here, these conditions are not satisfied. The notion of left limit simply doesn't make sense for the function $[0,\infty[\to \mathbb{R}: x\mapsto \sqrt{x}$.

If you would apply the $\epsilon-\delta$ definition of limit to this function, you can show that the limit exists and equals $0$.

It is certainly wrong to say the limit does not exist, as you seem to be suggesting.

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#### Mark44

Mentor
I'm sorry but this response is as wrong as it can get.

The only definition of limit is the $\epsilon-\delta$ definition or something equivalent with it. The "definition" the OP is using is a theorem that holds under certain conditions. Here, these conditions are not satisfied. The notion of left limit simply doesn't make sense for the function $[0,\infty[\to \mathbb{R}: x\mapsto \sqrt{x}$.

If you would apply the $\epsilon-\delta$ definition of limit to this function, you can show that the limit exists and equals $0$.

It is certainly wrong to say the limit does not exist, as you seem to be suggesting.
What I'm saying is that the two-sided limit, $\lim_{x \to 0} \sqrt x$, doesn't exist. Clearly $\lim_{x \to 0^-} \sqrt x$ does not exist, so the two-sided limit can't exist.

#### Math_QED

Homework Helper
What I'm saying is that the two-sided limit, $\lim_{x \to 0} \sqrt x$, doesn't exist. Clearly $\lim_{x \to 0^-} \sqrt x$ does not exist, so the two-sided limit can't exist.

The limit $\lim_{x\to 0}\sqrt{x}$ does exist.

With this, I mean that for every $\epsilon>0$, there is a $\delta >0$ such that if $x\geq 0$ satisfies $0 <|x| <\delta$, then $|\sqrt{x}|<\epsilon$.

#### Mark44

Mentor
I'm sure you know what it is, and is in terms of $\delta$ and $\epsilon$, in which $|x - a| < \delta$, where x is in some open interval around a. There's a theorem about when a two-sided limit exists, involving the two one-sided limits. That is, $\lim_{x \to a}f(x)$ exists if an only if $\lim_{x \to a^-}f(x)$ and $\lim_{x \to a^+}f(x)$ exist and are equal.

#### PeroK

Homework Helper
Gold Member
2018 Award
I'm sure you know what it is, and is in terms of $\delta$ and $\epsilon$, in which $|x - a| < \delta$, where x is in some open interval around a. There's a theorem about when a two-sided limit exists, involving the two one-sided limits. That is, $\lim_{x \to a}f(x)$ exists if an only if $\lim_{x \to a^-}f(x)$ and $\lim_{x \to a^+}f(x)$ exist and are equal.
Yes, but that doesn't apply to an end point on a closed interval. The left sided limit if you allow it to be valid is vacuously any number you want. There are simply no qualifying values of $x <0$ in the domain of the function.

#### Math_QED

Homework Helper
I'm sure you know what it is, and is in terms of $\delta$ and $\epsilon$, in which $|x - a| < \delta$, where x is in some open interval around a. There's a theorem about when a two-sided limit exists, involving the two one-sided limits. That is, $\lim_{x \to a}f(x)$ exists if an only if $\lim_{x \to a^-}f(x)$ and $\lim_{x \to a^+}f(x)$ exist and are equal.
I think you will be interested in reading this. It explains why we can't apply this theorem here. We need the point at which the limit is taken to be an interior point of the domain.

"If a one-sided limit of a function doesn't exist, how can a function"

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# Commands

These commands examine or modify the general model state. These are either settings that are saved/restored/reset along with the model, or general model controls that do not pertain to a specific module (like zone, structure elements, or ball).

Model Commands


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Lesson 4, Topic 5
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Longitude & Time

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Recall:

→   The Earth rotates about its axis to make complete rotation in 24hours a whole day.

→  This implies that 1-hour interval is equivalent to 15° interval between two meridians

e.g. Ghana and Cameroon

→   That is 4 minutes is equivalent 1° interval.

→   Places to the East gain time, while places to the West loses time.

Example 1

X(0o, 35oE) and Y(0o, xoE) are two places on the surface of the Earth such that Y is due  East of X. If the difference in the local times at X and Y is 2 hours 40 minutes, find xo.

Solution:

Recall 2hours 40minutes = 160 minutes

i.e. 160 minutes ≡y°

4 minutes ≡1°

y° = 1 × $$\frac{160}{4}$$

y° = 40°

Since Y is due East of x, i.e. Y gains time rather x.

i.e. Longitude of Y = 35 + 40 =75oE

i.e. xo = 75oE

Example 2

Points A and B are two places on Latitude 57oN. The distance between them is 6,671Km. If B is due East of A, find the:

(i) Local time at A, if the local time at B is 6:30am

(ii) Local time at B if the local time at A is 10:45pm.

Solution

(i) Distance $$\scriptsize AB = 6671 = \normalsize \frac{θ}{360} \scriptsize \; \times \; 2 \; \times \; \normalsize \frac{22}{7} \scriptsize \; \times \; 6400$$

= θ × 60.8612

i.e.    θ = 109.609

θ = 109.6°

109.6o ≡ $$\frac{109.6}{15}$$hours

i.e. 109.6 ≡ 7.3 hours

If local time is 6:30am at B, then A will lose time for 7 hours 18 minutes.

i.e. at A = 11:12pm of previous day.

(ii)  Local time at A = 10:45pm, then B will gain time for 7hours 18minutes.

i.e. local time at B = 6:03am of the next day.

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## Getting the integrated equalizer to work, from Debian Jessie, KDE 4.

I happen to have an older laptop, which I name ‘Klystron’, that is running Debian 8 / Jessie, with KDE 4 as its desktop manager. Don’t ask me why, but I tend to leave older builds of Linux running on some of my computers, just because they seem to be running without any deficiencies.

That laptop has lousy speakers. I decided a few days ago, that it would benefit, if I could get the 14-band graphical equalizer to work, that is generally available on Linux computers which, like that laptop, use the ‘PulseAudio’ sound server. However, on this old version of Linux, achieving that was a bit harder than it’s supposed to be. Yet, because I succeeded, I felt that I should share with the community, what the steps were, needed to succeed.

First of all, this is what the equalizer looks like, which I can now open on that laptop:

And it works!

In order to get this sort of equalizer working with PulseAudio, eventually, the following two modules need to be loaded:

module-equalizer-sink

module-dbus-protocol

And, if I gave the command ‘load-module…’ naively from the command-line, as user, because under my builds of Linux, PulseAudio runs in user mode, both these modules seem to load fine, without my having to install any additional packages.

On more recent builds of Linux, one needs to install the package ‘pulseaudio-equalizer’ to obtain this feature, or some similarly-named package. But, because these two modules just loaded fine under Debian / Jessie, I guess the functionality once came integrated with PulseAudio.

But I soon started to run in to trouble with these modules, and discovered why, then, the equalizer function was not set up to run out-of-the-box…

(Updated 6/26/2020, 10h30… )

## Getting Pulseaudio to schedule real-time threads under Debian / Stretch and beyond.

On the computer which I name ‘Phosphene’, I have Debian 9 / Stretch installed, and, going the popular route, the Debian maintainers have decided that ‘Pulseaudio’ should be the main sound server, while in certain cases, it can be made to coexist with the ‘Jack’ / ‘Jack v2′ daemon, the latter for more demanding needs. In my case, starting Jack suspends Pulse.

In addition, Debian systems I’m familiar with run ‘pulseaudio’ as user, and not as root.

But, the way my Pulse daemon is configured, includes a ‘loopback’ module, that sends audio that has been input and sampled from the analog input of my (“Creative X-Fi”) sound card, back to its analog output, so that I can listen to an old-fashioned-but-modern radio receiver, over my speakers, without having to connect anything special to do so.

The problem that I was experiencing was, buffer underruns, and sound drop-outs, due to the way this default sound server was operating the sound card. And, surely enough, it had something to do with a by-now famous issue, of allowing Pulse to schedule certain threads to run with real-time priority. This can be enabled in the config file ‘/etc/pulse/daemon.conf‘, but fails to become actual when just enabled so.

Trying to find an answer to why this attempt was failing, sent me on a trip to many Web-sites and BB-postings, most of which had as disadvantage, the fact that with Pulse, there is an old way and a new way to accomplish this, and the fact that most of the BB-postings assumed the old way, being dated to the year 2016 and earlier.

Rest assured, under Debian 9 / Stretch, authorizing Pulse to schedule some threads with RT priority, is managed according to ‘the new way’. Just to recap how ‘the old way’ worked:

The file ‘/usr/bin/pulseaudio‘ needed to be set with ‘SETUID root’, and when run this way, if the sound server detected that the real user (not the effective user) belonged to the group ‘pulse-rt‘, would schedule the threads it needed to run with RT, and would then drop privileges, so that according to the process-monitor ‘top’, it would never even give a hint that anything had in fact been scheduled with RT.

This is how ‘the new way’ is supposed to work:

If the user needs for Pulse specifically to run with RT, he must actually install the package ‘rtkit‘, which runs as a system user, and which sets certain threads to have RT priority in a curated way, that’s meant to be ‘safer’, than it still is for any one user to belong to a group, that simply allows him or her to run arbitrary processes with RT. In fact, ‘the old way’ was also meant as a similar safeguard.

I had installed the specified package, rebooted my system – with Pulse given its new startup parameters, and found alas, that the file ‘/var/log/syslog‘, which was benefiting from an elevated logging level set by me, revealed numerous messages that I did not understand.

Trying to figure out what was happening resulted in more than one day’s frustration, and then finally, to some peace of mind…

(Updated 4/16/2020, 13h15 … )

## PulseAudio Restart Bug – Solved

I enjoy my Linux computers, and one reason is the fact that many technical issues can be resolved, without having to reboot endlessly. However, in my past usage, there has been an irritating exception to this pattern. It’s common under Linux, that we can simply restart the PulseAudio Server from the command-line, using one out of several methods, and that the subject should be done with. But alas, every time I have ever restarted PulseAudio in this way, or, if the server restarted on its own, afterwards, when looking up the Plasma 5 -generated status display (which is actually referred to as “Phonon”), I’d be missing a Devices List, like so:

This type of display can be interpreted to mean several things, such as, that the PulseAudio server did restart, but that perhaps, it simply failed to connect to the inter-process, session-unique, message-bus. Therefore, in the past, whenever I had such a display, I eventually did what I thought I had to do, which was, to log out and back in again, or, to reboot. On my system, PulseAudio is configured such, that it runs as one user-name, and not as root.

In fact, a peculiar side effect of this bug was, that the list of available output devices was still being displayed, within ‘pavucontrol‘.

But this ordeal has now become even more inconvenient than it ever was because on the computer which I name ‘Phosphene’, the need may recur more frequently, ‘just to restart the PulseAudio server’.

However, I have finally found the true cause for this malfunction, which was, that when PulseAudio is restarted from within an existing session, it simply fails to load one module, which is also the module that it needs, in order to be able to list the available devices:

module-device-manager

In fact, there exists a script in ‘/usr/bin‘, that loads a series of X11-related modules.

Therefore, after a restart of this service, I can simply give the following command now:

/usr/bin/start-pulseaudio-x11

And Eureka! I can now obtain a list of available devices, without ever having to log out and back in, or, without ever having to reboot:

In fact, I have created a shell-script, which I can click on as user, and which carries out this task, safely.

I’ve also discovered that the ‘ProjectM’ music visualization application still works, and detects the beat of the playing music as before. What this means is that theoretically, after ‘ProjectM’ was installed, instead of rebooting, I could have just restarted the PulseAudio server as described here, to get that working.

( Edited 2019/10/29, 23h35 … )

I know that there exists an unrelated problem, that just happens to give the same appearance within ‘Phonon’, but that cannot be resolved in this way…

## Technical Impediment In Getting Sound From My Linux Tablet (Solved)

One of the facts which I’ve blogged about is, that I have a Linux Guest System installed on the Android Tablet, that’s a Google Pixel C.

Another fact which I blogged about a long time ago was, that I am able to share the PulseAudio Sound Server that resides on the computer now named ‘Phosphene’, for use by the client computer I name ‘Klexel’.

A basic limitation to my Linux Tablet remains, that it isn’t suited to play back audio streams, or video streams that have audio, because inherently, I’m just running its Linux Guest as a VNC Session. And so a logical thought on that would be:

‘Why not specify the Sound-Providing Server, as the place that the Linux Guest System streams its sound to, at least as long as I am on my own LAN?’

And while in theory this sounds like a good idea, in practice the implementation is still some distance away.

The main problem? While ‘Klexel’ is connected to this Sound Server, it ties up the only TCP Port, which is therefore unable to accept new connections, say from my Linux Tablet. Now, I can tell ‘Klexel’ to relinquish its session on the Sound Server, but doing so has an unexpected consequence. This corrupts the module on the PulseAudio Server, that was listening for remote connections. I need to unload the module, and reload it with the same parameters as before, just so that ‘Klexel’ can reconnect.

The long-term effect of this will be, that the Linux Tablet may be able to obtain one session on ‘Phosphene’ for sound, but that every time this tablet disconnects, again, that module on Phosphene’s PulseAudio Server will go into a corrupted state.

Hence, I have not yet worked this into a practical solution. But if I ever did, I’d be able to expand the applications of the Linux Guest System – on the Tablet – into audiovisual applications.

Update:

I am now one step closer to permitting Linux audiovisual applications on my tablet to access the sound server on my LAN. What I have discovered is, that the module in question can be loaded more than once on the PulseAudio Server, as long as each instance of it listens on a different port number. I.e., the second instance can be configured to listen on port 4318 instead of the default, port 4317. The configuration lines which accomplish this are as follows:

The command ‘pactl list modules‘ confirms that both instances are loaded and stable. Further, when the video-player ‘xine’ is finished with its connection to the server, it closes the TCP Port in a way that does not corrupt the module, so that ‘xine’ can be started a second time and will cause sound to play for the second time.


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Browse Questions

# Find the principal and general solution of the following equation $\tan x=\sqrt 3$

$\begin{array}{1 1}(A)\;\large\frac{\pi}{3},\frac{4\pi}{3},\normalsize x=n\pi+\large\frac{\pi}{3}\\(B)\;\large\frac{7\pi}{3},\frac{2\pi}{3},\normalsize x=n\pi+\large\frac{\pi}{2}\\(C)\;\large\frac{\pi}{5},\frac{4\pi}{5},\normalsize x=n\pi+\large\frac{\pi}{5}\\(D)\;\large\frac{\pi}{7},\frac{4\pi}{7},\normalsize x=n\pi+\large\frac{\pi}{7}\end{array}$

Toolbox:
• $\tan \large\frac{\pi}{3}$$=\sqrt 3 • \tan x=\tan y when y is the principal value of x • x=n\pi+y We know that \tan \large\frac{\pi}{3}$$=\sqrt 3$
$\tan (\pi+\large\frac{\pi}{3})=$$\tan \large\frac{\pi}{3}$$=\sqrt 3$
$\Rightarrow \tan \large\frac{4\pi}{3}$
$\therefore \tan \large\frac{\pi}{3}=$$\tan \large\frac{4\pi}{3}$$=\sqrt 3$
$\therefore$ Principal solutions are $\large\frac{\pi}{3},\frac{4\pi}{3}$
If $\tan x=\tan \large\frac{\pi}{3}$
$\therefore$ General solution: $x=n\pi+\large\frac{\pi}{3}$where $n \in z$
Hence (A) is the correct answer.


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# Statistics for Environmental Engineers

Скачать в pdf «Statistics for Environmental Engineers»

If we accept this model, the intervention estimation problem becomes how to determine the weighting factor в. Fortunately, an alternate formulation of the model makes this reasonably simple. The white noise-random walk model is equivalent to an ARIMA (0,1,1) model (Box et al., 1994):

yt = yt_ 1 + a, _ ва,_!

where 0 < в < 1, and a, = independent random noise distributed as N(0,o<2).

The white noise and the random walk error components are related to o2 as follows:

o2 = во2

o2 = (1 _ в)2о2

The equivalence of the two forms of the model and the derivations are given in Pallesen et al. (1985).

A recursive iteration is done separately for each section of data before and after the intervention to estimate в. The method is to:

1. Choose a starting value for в and use at= yt— yt-1+ ва,-1 to recursively calculate the residuals at t = 2, 3,…, T. Assume at=0= 0 to start the calculations.

2.    Calculate the residual sum of squares, RSS(e) = Y*a] for each section. Add these to get the total RSS for the entire series. If a gap has been used to account for a transition period, data in the gap are omitted from these calculations.

3.    Search over a range of в to minimize RSS(e).

4.    Use the minimum RSS to estimate al = RSS(e), where n is the total number of residuals

n


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how to import Agilent CGH custom design xml file into limma
1
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Entering edit mode
Jarek Bryk ▴ 110
@jarek-bryk-3457
Last seen 8.0 years ago
Hello, I am trying to analyse my two-color mouse aCGH data on Agilent platform. I have custom-designed 1M probe CGH arrays for which I have a xml file that defines the arrays' design (layout etc.). My goal (for now) is to identify CNVs in the samples. I am trying to use snapCGH to analyse the samples, but I stumble upon some basic preprocessing steps. I can read in the Feature Extraction files with no problem (following almost literally snapCGH user guide, just for single sample): > RG1<-read.maimages(file="sample1.txt",source="agilent") Read sample1.txt > str(RG1) Formal class 'RGList' [package "limma"] with 1 slots ..@ .Data:List of 8 .. ..$: num [1:974016, 1] 297 33.6 35.2 145.7 333.7 ... .. .. ..- attr(*, "dimnames")=List of 2 .. .. .. ..$ : NULL .. .. .. ..$: chr "sample1" .. ..$ : num [1:974016, 1] 25 25 25 25 25 25 25 25 25 25 ... .. .. ..- attr(*, "dimnames")=List of 2 .. .. .. ..$: NULL .. .. .. ..$ : chr "sample1" .. ..$: num [1:974016, 1] 333.8 78.2 54.7 111.8 170.5 ... .. .. ..- attr(*, "dimnames")=List of 2 .. .. .. ..$ : NULL .. .. .. ..$: chr "sample1" .. ..$ : num [1:974016, 1] 52 52 52 53 53 53 53 53 53 52 ... .. .. ..- attr(*, "dimnames")=List of 2 .. .. .. ..$: NULL .. .. .. ..$ : chr "sample1" .. ..$:'data.frame': 1 obs. of 1 variable: .. .. ..$ FileName: chr "sample1" .. ..$:'data.frame': 974016 obs. of 9 variables: .. .. ..$ Row : int [1:974016] 1 1 1 1 1 1 1 1 1 1 ... .. .. ..$Col : int [1:974016] 1 2 3 4 5 6 7 8 9 10 ... .. .. ..$ Start : int [1:974016] 0 0 0 0 0 0 0 0 0 0 ... .. .. ..$Sequence : chr [1:974016] "" "" "" "CCCATTCACTAATCTACATATTATCTCCATCCAACAAAAATTTCTTTCAGTAAGGTGTGG" ... .. .. ..$ ProbeUID : int [1:974016] 0 1 1 3 5 7 9 11 13 15 ... .. .. ..$ControlType : int [1:974016] 1 1 1 0 0 0 0 0 0 0 ... .. .. ..$ ProbeName : chr [1:974016] "MmCGHBrightCorner" "DarkCorner" "DarkCorner" "A_67_P21917383" ... .. .. ..$GeneName : chr [1:974016] "MmCGHBrightCorner" "DarkCorner" "DarkCorner" "Msn" ... .. .. ..$ SystematicName: chr [1:974016] "MmCGHBrightCorner" "DarkCorner" "DarkCorner" "chrX:93323379-93323438" ... .. ..$: chr "agilent" .. ..$ :List of 4 .. .. ..$ngrid.r: num 1 .. .. ..$ ngrid.c: num 1 .. .. ..$nspot.r: int 1068 .. .. ..$ nspot.c: int 912 > RG1$design<- -1 # I have flipped data where red is the reference > RG2<-backgroundCorrect(RG1,method="normexp") Array 1 corrected Array 1 corrected > MA<-normalizeWithinArrays(RG2,method="loess") And after this command I think I need to supply the design file to limma/snapCGH so that it knows what to do with processCGH or genomePlot, because this happens during the next steps: > MA2<-processCGH(MA, method.of.averaging=mean,ID="ProbeUID") Error in order(MA$genes$Chr, MA$genes$Position) : argument 1 is not a vector > genomePlot(MA,array=1) Error in order(datainfo$Chr, datainfo$Position) : argument 1 is not a vector I am shooting in the dark with the ProbeUID field here, as this should be the unique probe identifier, but it's not where the problem is. I need position of genes in the genome, and this is in the design file. I think I should use the readPositionalInfo to relay this information to limma/snapCGH, but it only accepts objects of class RGList or MAList, and my design is an xml file - I don't know how to parse the file... I would be very grateful for any clarification on how to proceed with the analysis. cheers jarek > sessionInfo() R version 2.12.0 (2010-10-15) Platform: i386-apple-darwin9.8.0/i386 (32-bit) locale: [1] en_US.UTF-8/en_US.UTF-8/C/C/en_US.UTF-8/en_US.UTF-8 attached base packages: [1] splines stats graphics grDevices utils datasets methods base other attached packages: [1] snapCGH_1.20.0 DNAcopy_1.24.0 limma_3.6.0 aCGH_1.28.0 multtest_2.6.0 Biobase_2.10.0 [7] survival_2.35-8 cluster_1.13.1 GLAD_2.12.0 loaded via a namespace (and not attached): [1] affy_1.28.0 affyio_1.18.0 annotate_1.28.0 AnnotationDbi_1.12.0 [5] DBI_0.2-5 genefilter_1.32.0 grid_2.12.0 lattice_0.19-13 [9] MASS_7.3-8 preprocessCore_1.12.0 RColorBrewer_1.0-2 RSQLite_0.9-2 [13] strucchange_1.4-2 tilingArray_1.28.0 tools_2.12.0 vsn_3.18.0 [17] xtable_1.5-6 -- Jarek Bryk | www.evolbio.mpg.de/~bryk Max Planck Institute for Evolutionary Biology August Thienemann Str. 2 | 24306 Pl?n, Germany tel. +49 4522 763 287 | bryk at evolbio.mpg.de aCGH CGH probe aCGH snapCGH aCGH CGH probe aCGH snapCGH • 1.3k views ADD COMMENT 0 Entering edit mode @sean-davis-490 Last seen 3 months ago United States On Wed, Oct 20, 2010 at 11:26 AM, Jarek Bryk <bryk@evolbio.mpg.de> wrote: > Hello, > > I am trying to analyse my two-color mouse aCGH data on Agilent platform. I > have custom-designed 1M probe CGH arrays for which I have a xml file that > defines the arrays' design (layout etc.). My goal (for now) is to identify > CNVs in the samples. I am trying to use snapCGH to analyse the samples, but > I stumble upon some basic preprocessing steps. > > I can read in the Feature Extraction files with no problem (following > almost literally snapCGH user guide, just for single sample): > > > RG1<-read.maimages(file="sample1.txt",source="agilent") > Read sample1.txt > > str(RG1) > Formal class 'RGList' [package "limma"] with 1 slots > ..@ .Data:List of 8 > .. ..$ : num [1:974016, 1] 297 33.6 35.2 145.7 333.7 ... > .. .. ..- attr(*, "dimnames")=List of 2 > .. .. .. ..$: NULL > .. .. .. ..$ : chr "sample1" > .. ..$: num [1:974016, 1] 25 25 25 25 25 25 25 25 25 25 ... > .. .. ..- attr(*, "dimnames")=List of 2 > .. .. .. ..$ : NULL > .. .. .. ..$: chr "sample1" > .. ..$ : num [1:974016, 1] 333.8 78.2 54.7 111.8 170.5 ... > .. .. ..- attr(*, "dimnames")=List of 2 > .. .. .. ..$: NULL > .. .. .. ..$ : chr "sample1" > .. ..$: num [1:974016, 1] 52 52 52 53 53 53 53 53 53 52 ... > .. .. ..- attr(*, "dimnames")=List of 2 > .. .. .. ..$ : NULL > .. .. .. ..$: chr "sample1" > .. ..$ :'data.frame': 1 obs. of 1 variable: > .. .. ..$FileName: chr "sample1" > .. ..$ :'data.frame': 974016 obs. of 9 variables: > .. .. ..$Row : int [1:974016] 1 1 1 1 1 1 1 1 1 1 ... > .. .. ..$ Col : int [1:974016] 1 2 3 4 5 6 7 8 9 10 ... > .. .. ..$Start : int [1:974016] 0 0 0 0 0 0 0 0 0 0 ... > .. .. ..$ Sequence : chr [1:974016] "" "" "" > "CCCATTCACTAATCTACATATTATCTCCATCCAACAAAAATTTCTTTCAGTAAGGTGTGG" ... > .. .. ..$ProbeUID : int [1:974016] 0 1 1 3 5 7 9 11 13 15 ... > .. .. ..$ ControlType : int [1:974016] 1 1 1 0 0 0 0 0 0 0 ... > .. .. ..$ProbeName : chr [1:974016] "MmCGHBrightCorner" "DarkCorner" > "DarkCorner" "A_67_P21917383" ... > .. .. ..$ GeneName : chr [1:974016] "MmCGHBrightCorner" "DarkCorner" > "DarkCorner" "Msn" ... > .. .. ..$SystematicName: chr [1:974016] "MmCGHBrightCorner" "DarkCorner" > "DarkCorner" "chrX:93323379-93323438" ... > .. ..$ : chr "agilent" > .. ..$:List of 4 > .. .. ..$ ngrid.r: num 1 > .. .. ..$ngrid.c: num 1 > .. .. ..$ nspot.r: int 1068 > .. .. ..$nspot.c: int 912 > > RG1$design<- -1 # I have flipped data where red is the reference > > RG2<-backgroundCorrect(RG1,method="normexp") > Array 1 corrected > Array 1 corrected > > MA<-normalizeWithinArrays(RG2,method="loess") > > Hi, Jarek. You definitely DO NOT want to be loess-normalizing CGH data. There is an expected positive correlation between average intensity and log-ratio; loess-normalization will effectively eliminate that signal. > And after this command I think I need to supply the design file to > limma/snapCGH so that it knows what to do with processCGH or genomePlot, > because this happens during the next steps: > > > MA2<-processCGH(MA, method.of.averaging=mean,ID="ProbeUID") > Error in order(MA$genes$Chr, MA$genes$Position) : > argument 1 is not a vector > You may need something like: MA$design=rep(1,ncol(MA)) This assumes that there are no dye swaps. If there are, you may need to put some -1 in there. > genomePlot(MA,array=1) > Error in order(datainfo$Chr, datainfo$Position) : > argument 1 is not a vector > > I am shooting in the dark with the ProbeUID field here, as this should be > the unique probe identifier, but it's not where the problem is. I need > position of genes in the genome, and this is in the design file. I think I > should use the readPositionalInfo to relay this information to > limma/snapCGH, but it only accepts objects of class RGList or MAList, and my > design is an xml file - I don't know how to parse the file... > > This information is stored in the MA$Genes\$SystematicName above. You will need to split out the pieces (chromosome, position) and then use those for the locations. You will also probably need to remove control probes. Hope that helps. Sean > I would be very grateful for any clarification on how to proceed with the > analysis. > > cheers > jarek > > > sessionInfo() > R version 2.12.0 (2010-10-15) > Platform: i386-apple-darwin9.8.0/i386 (32-bit) > > locale: > [1] en_US.UTF-8/en_US.UTF-8/C/C/en_US.UTF-8/en_US.UTF-8 > > attached base packages: > [1] splines stats graphics grDevices utils datasets methods > base > > other attached packages: > [1] snapCGH_1.20.0 DNAcopy_1.24.0 limma_3.6.0 aCGH_1.28.0 > multtest_2.6.0 Biobase_2.10.0 > [7] survival_2.35-8 cluster_1.13.1 GLAD_2.12.0 > > loaded via a namespace (and not attached): > [1] affy_1.28.0 affyio_1.18.0 annotate_1.28.0 > AnnotationDbi_1.12.0 > [5] DBI_0.2-5 genefilter_1.32.0 grid_2.12.0 > lattice_0.19-13 > [9] MASS_7.3-8 preprocessCore_1.12.0 RColorBrewer_1.0-2 > RSQLite_0.9-2 > [13] strucchange_1.4-2 tilingArray_1.28.0 tools_2.12.0 > vsn_3.18.0 > [17] xtable_1.5-6 > > -- > Jarek Bryk | www.evolbio.mpg.de/~bryk > Max Planck Institute for Evolutionary Biology > August Thienemann Str. 2 | 24306 PlÃ¶n, Germany > tel. +49 4522 763 287 | bryk@evolbio.mpg.de > > _______________________________________________ > Bioconductor mailing list > Bioconductor@stat.math.ethz.ch > https://stat.ethz.ch/mailman/listinfo/bioconductor > Search the archives: > http://news.gmane.org/gmane.science.biology.informatics.conductor > [[alternative HTML version deleted]]


text:
# deformations of Lie algebroids

In the paper "Deformations of Lie brackets"- by I. Moerdijk and M. Crainic, they define deformations of a Lie algebroid as follows:

Let $$A$$ be a fixed vector bundle, and $$I\subset \mathbb{R}$$ and interval.

1. A family of Lie algebroids over $$I$$ is a collection $$(A_t)_{t \in I}$$ of Lie algebroids $$A_t= (A, [.,.]_t, \rho_t)$$ varying smoothly with respect to $$t$$. (Here $$\rho_t$$ denotes the anchor map.)\
2. A deformation of a Lie algebroid $$(A, [.,], \rho)$$ is a family $$(A_t)_{t \in I}$$ of Lie algebroids over an interval containing the origin with $$A_0=A$$.

My question is: what is meant by "varying smoothly with respect to $$t$$" in this context?

A Lie algebroid over a manifold $$M$$ is

• a vector bundle $$\pi:E\rightarrow M$$

• a Lie bracket $$[\cdot,\cdot]:\Gamma(E)\times \Gamma(E)\rightarrow \Gamma(E)$$

• a map $$\rho:E\rightarrow TM$$ (called the anchor map)

such that $$[X,fY]=f[X,Y]+\rho(X)(f).Y$$ for all $$X,Y\in\Gamma(E)$$ and $$f\in C^{\infty}(M)$$.

• I am trying to understand your question... You want to define what it means by a family of Lie algebroids over an interval.. Right? – Praphulla Koushik Dec 26 '18 at 9:36


text:
## The Integrating Factor Method

The integrating factor method is a method of solving first order differential equations:
It can be used for equations of the form
$f(x)\frac{dy}{dx}+yg(x)=h(x)$

Divide by
$f(x)$
to get
$\frac{dy}{dx}+ \frac{yg(x)}{f(x)}= \frac{h(x)}{f(x)}$

Multiply by a factor
$e^{\int \frac{g(x)}{f(x)}dx}$
so the equation becomes
$\frac{dy}{dx}e^{\int \frac{g(x)}{f(x)}dx}+ye^{\int \frac{g(x)}{f(x)}dx} \frac{g(x)}{f(x)}=h(x)e^{\int \frac{g(x)}{f(x)}dx}$

We can write this as
$\frac{d}{dx}(ye^{\int \frac{g(x)}{f(x)}dx})=h(x)e^{\int \frac{g(x)}{f(x)}dx}$

Then
$ye^{\int \frac{g(x)}{f(x)}dx}= \int h(x)e^{\int \frac{g(x)}{f(x)}dx} \rightarrow y= \frac{\int h(x)e^{\int \frac{g(x)}{f(x)}dx}}{e^{\int \frac{g(x)}{f(x)}dx}}$
.
Example:
$x \frac{dy}{dx}+(x+1)y=x^2$
,
Divide by
$x$
.
$\frac{dy}{dx}+(1+ \frac{1}{x})y=x$
,
The integrating factor is
$e^{\int (1+ \frac{1}{x})dx}=e^{x+lnx}=e^xe^{lnx}=e^xx=xe^x$
.
The equation becomes
$xe^x \frac{dy}{dx}+ (1+ \frac{1}{x}) xe^x y= x^2e^x+$

We can write this as
$\frac{d}{dx}(yxe^x)=x^2e^x$

Integrating both sides
$xe^xy=x^2e^x-2xe^x+2e^x+A$

Then
$y=x-2+ \frac{2}{x} + \frac{A}{x}e^{-x}$


text:
# Two vectors A and B have equal magnitudes. The magnitude of (A + B) is ‘n’ times the magnitude of (A – B). The angle between A and B is

This question was previously asked in
JEE Mains Previous Paper 1 (Held On: 10 Jan 2019 Shift 2)
View all JEE Main Papers >
1. $${\rm{si}}{{\rm{n}}^{ - 1}}\left( {\frac{{{n^2} - 1}}{{{n^2} + 1}}} \right)$$
2. $${\rm{si}}{{\rm{n}}^{ - 1}}\left( {\frac{{n - 1}}{{n + 1}}} \right)$$
3. $${\rm{co}}{{\rm{s}}^{ - 1}}\left( {\frac{{{n^2} - 1}}{{{n^2} + 1}}} \right)$$
4. $${\rm{co}}{{\rm{s}}^{ - 1}}\left( {\frac{{n - 1}}{{n + 1}}} \right)$$

Option 3 : $${\rm{co}}{{\rm{s}}^{ - 1}}\left( {\frac{{{n^2} - 1}}{{{n^2} + 1}}} \right)$$
Free
JEE Mains Previous Paper 1 (Held On: 12 Apr 2019 Shift 2)
8346
90 Questions 360 Marks 180 Mins

## Detailed Solution

Concept:

Given,

Magnitude of A = Magnitude of B

|A| = |B|

Or A = B      ----(1)

Let magnitude of (A + B) is R and for (A – B) is R'.

Given, magnitude of R is n times the magnitude of R'

Thus, R = nR'

Now,

R = A + B

R2 = A2 + B2 + 2AB cos θ

[∵using equation (1)]

R2 = A2 + A2 + 2AA cos θ

R2 = 2A2 + 2A2 cos θ      ----(2)

Again,

R' = A – B

$${\left( {R'} \right)^2} = {A^2} + {B^2} - 2AB\cos \theta$$

[∵ using equation (1)]

$${\left( {R'} \right)^2} = {A^2} + {A^2} - 2AA\cos \theta$$

$${\left( {R'} \right)^2} = 2{A^2} - 2{A^2}\cos \theta$$       ----(3)

Calculation:

Given

R = nR'

$${\left( {\frac{R}{{R'}}} \right)^2} = {n^2}$$

Dividing equation (ii) with equation (iii), we get

$${n^2} = \frac{{2{A^2} + 2{A^2}\cos \theta }}{{2{A^2} - 2{A^2}\cos \theta }}$$

$${n^2} = \frac{{2{A^2}\left( {1 + \cos \theta } \right)}}{{2{A^2}\left( {1 - \cos \theta } \right)}}$$

$$\frac{{{n^2}}}{1} = \frac{{1 + \cos \theta }}{{1 - \cos \theta }}$$

Thus,

$${n^2} = \frac{{1 + \cos \theta }}{{1 - \cos \theta }}$$

The value of $$\frac{{{n^2} - 1}}{{{n^2} + 1}}$$ will be,

$$\frac{{{n^2} - 1}}{{{n^2} + 1}} = \frac{{\left( {\frac{{1 + \cos \theta }}{{1 - \cos \theta }}} \right) - 1}}{{\left( {\frac{{1 + \cos \theta }}{{1 - \cos \theta }}} \right) + 1\;}}$$

$$\frac{{{n^2} - 1}}{{{n^2} + 1}} = \frac{{\left( {\frac{{1 + \cos \theta - \left( {1 - \cos \theta } \right)}}{{1 - \cos \theta }}} \right)}}{{\left( {\frac{{1 + \cos \theta + \left( {1 - \cos \theta } \right)}}{{1 - \cos \theta }}} \right)\;}}$$

$$\frac{{{n^2} - 1}}{{{n^2} + 1}} = \frac{{\left( {1 + {\rm{cos}}\theta } \right) - \left( {1 - {\rm{cos}}\theta } \right)}}{{\left( {1 + {\rm{cos}}\theta } \right) + \left( {1 - {\rm{cos}}\theta } \right)}}$$

$$\frac{{{n^2} - 1}}{{{n^2} + 1}} = \frac{{2{\rm{cos}}\theta }}{2} = {\rm{cos}}\theta$$

$$\theta = {\cos ^{ - 1}}\left( {\frac{{{n^2} - 1}}{{{n^2} + 1}}} \right)\;$$

The angle between A and B is $${\cos ^{ - 1}}\left( {\frac{{{n^2} - 1}}{{{n^2} + 1}}} \right)\;$$


text:
# Backpropagation for noobs

I am trying to understand neural networks and how they work, by programming my own one from scratch in nodejs. Currently, i managed to build a network, that has weights, layers and neurons. I also understood what an activation function is and i am using the sigmoid function.

I've now come to the point of back propagation with the gradient descent algorithm. My Problem is, that i only know 10th grade Math and every tutorial/explanation i managed to find, uses complex functions and math which i cannot manage to understand.

I would really like to finish this project and get back propagation to work, so if someone could explain me how to use the back propagation without using too complex math (e, derivatives, functions, etc.) or explaining the complex math it would be greatly appreciated.

My source Code: https://gitlab.com/milan44/node (Training is happening in Network.prototype.train)

• (+1) without 'complex' math backprop cannot be implemented. You will have to get a handle on the complex maths somehow. – naive Feb 26 '19 at 11:17


text:
There are two variants. The default one is based on 1406.1078v3 and has reset gate applied to hidden state before matrix multiplication. The other one is based on original 1406.1078v1 and has the order reversed.

layer_gru(
object,
units,
activation = "tanh",
recurrent_activation = "sigmoid",
use_bias = TRUE,
return_sequences = FALSE,
return_state = FALSE,
go_backwards = FALSE,
stateful = FALSE,
unroll = FALSE,
time_major = FALSE,
reset_after = TRUE,
kernel_initializer = "glorot_uniform",
recurrent_initializer = "orthogonal",
bias_initializer = "zeros",
kernel_regularizer = NULL,
recurrent_regularizer = NULL,
bias_regularizer = NULL,
activity_regularizer = NULL,
kernel_constraint = NULL,
recurrent_constraint = NULL,
bias_constraint = NULL,
dropout = 0,
recurrent_dropout = 0,
...
)

## Arguments

object What to compose the new Layer instance with. Typically a Sequential model or a Tensor (e.g., as returned by layer_input()). The return value depends on object. If object is: missing or NULL, the Layer instance is returned. a Sequential model, the model with an additional layer is returned. a Tensor, the output tensor from layer_instance(object) is returned. Positive integer, dimensionality of the output space. Activation function to use. Default: hyperbolic tangent (tanh). If you pass NULL, no activation is applied (ie. "linear" activation: a(x) = x). Activation function to use for the recurrent step. Boolean, whether the layer uses a bias vector. Boolean. Whether to return the last output in the output sequence, or the full sequence. Boolean (default FALSE). Whether to return the last state in addition to the output. Boolean (default FALSE). If TRUE, process the input sequence backwards and return the reversed sequence. Boolean (default FALSE). If TRUE, the last state for each sample at index i in a batch will be used as initial state for the sample of index i in the following batch. Boolean (default FALSE). If TRUE, the network will be unrolled, else a symbolic loop will be used. Unrolling can speed-up a RNN, although it tends to be more memory-intensive. Unrolling is only suitable for short sequences. If True, the inputs and outputs will be in shape [timesteps, batch, feature], whereas in the False case, it will be [batch, timesteps, feature]. Using time_major = TRUE is a bit more efficient because it avoids transposes at the beginning and end of the RNN calculation. However, most TensorFlow data is batch-major, so by default this function accepts input and emits output in batch-major form. GRU convention (whether to apply reset gate after or before matrix multiplication). FALSE = "before" (default), TRUE = "after" (CuDNN compatible). Initializer for the kernel weights matrix, used for the linear transformation of the inputs. Initializer for the recurrent_kernel weights matrix, used for the linear transformation of the recurrent state. Initializer for the bias vector. Regularizer function applied to the kernel weights matrix. Regularizer function applied to the recurrent_kernel weights matrix. Regularizer function applied to the bias vector. Regularizer function applied to the output of the layer (its "activation").. Constraint function applied to the kernel weights matrix. Constraint function applied to the recurrent_kernel weights matrix. Constraint function applied to the bias vector. Float between 0 and 1. Fraction of the units to drop for the linear transformation of the inputs. Float between 0 and 1. Fraction of the units to drop for the linear transformation of the recurrent state. Standard Layer args.

## Details

The second variant is compatible with CuDNNGRU (GPU-only) and allows inference on CPU. Thus it has separate biases for kernel and recurrent_kernel. Use reset_after = TRUE and recurrent_activation = "sigmoid".

## Input shapes

N-D tensor with shape (batch_size, timesteps, ...), or (timesteps, batch_size, ...) when time_major = TRUE.

## Output shape

• if return_state: a list of tensors. The first tensor is the output. The remaining tensors are the last states, each with shape (batch_size, state_size), where state_size could be a high dimension tensor shape.

• if return_sequences: N-D tensor with shape [batch_size, timesteps, output_size], where output_size could be a high dimension tensor shape, or [timesteps, batch_size, output_size] when time_major is TRUE

• else, N-D tensor with shape [batch_size, output_size], where output_size could be a high dimension tensor shape.

This layer supports masking for input data with a variable number of timesteps. To introduce masks to your data, use layer_embedding() with the mask_zero parameter set to TRUE.

## Statefulness in RNNs

You can set RNN layers to be 'stateful', which means that the states computed for the samples in one batch will be reused as initial states for the samples in the next batch. This assumes a one-to-one mapping between samples in different successive batches.

For intuition behind statefulness, there is a helpful blog post here: https://philipperemy.github.io/keras-stateful-lstm/

To enable statefulness:

• Specify stateful = TRUE in the layer constructor.

• Specify a fixed batch size for your model. For sequential models, pass batch_input_shape = list(...) to the first layer in your model. For functional models with 1 or more Input layers, pass batch_shape = list(...) to all the first layers in your model. This is the expected shape of your inputs including the batch size. It should be a list of integers, e.g. list(32, 10, 100). For dimensions which can vary (are not known ahead of time), use NULL in place of an integer, e.g. list(32, NULL, NULL).

• Specify shuffle = FALSE when calling fit().

## References

Other recurrent layers: layer_cudnn_gru(), layer_cudnn_lstm(), layer_lstm(), layer_rnn(), layer_simple_rnn()


text:
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text:
# SOHCAHTOA: Find the sides of a right triangle

#### How does SOHCAHTOA help us find side lengths?

After you are comfortable writing sine,cosine, tangent ratios, you might want to use sohcahtoa to find the sides of a right triangle. That is exactly the topic that this page focuses on.

### Example of finding Side Length

How to use sine,cosine,tangent to determine side X in the triangle below

Directions: Use sohcahtoa to find the given side length.

Step 1

Based on your givens and unknowns, determine which sohcahtoa ratio to use

Since we know the 67 angle, its adjacent side length and we want to know length of the opposite side, we should use tangent

Step 2

Set up an equation based on the ratio you chose in the step 1

$tan(67) = \frac{opposite}{adjacent} \\ tan(67) = \frac{x}{14}$

Step 3

Cross multiply and solve the equation for the side length. (Let's round to the nearest hundredth)

$tan(67) = \frac{x}{14} \\ 14\times tan(67) = x \\ x \approx 32.98$

### Practice Problems

Step 1

Based on your givens and unknowns, determine which sohcahtoa ratio to use

Since we know the 53 angle, the hypotenuse,and we want to find length of the opposite side, we should use sine

Step 2

Set up an equation based on the ratio you chose in the step 1

$sin(53) = \frac{opposite}{hypotenuse} \\ sin(53) = \frac{x}{15}$

Step 3

Cross multiply and solve the equation for the side length. (round to the nearest hundredth)

$15\times sin(53) = x \\ x \approx 11.98$

Step 1

Based on your givens and unknowns, determine which sohcahtoa ratio to use

Since we know the 53 angle, the adjacent side ,and we want to find length of the hypotenuse, we should use cosine

Step 2

Set up an equation based on the ratio you chose in the step 1

$cos(53) = \frac{adjacent}{hypotenuse} \\ cos(53) = \frac{45}{x}$

Step 3

Cross multiply and solve the equation for the side length. (round to the nearest hundredth)

$x=\frac{45}{cos(53)} \\ x \approx 74.8$

$cos(63) = \frac{adjacent}{hypotenuse} \\ cos(63) = \frac{3}{x} \\ x = \frac{3}{cos63} \approx 6.6$

Step 1

Use SOHCAHTOA to find  Y in the triangle on the left.

$tan(55) = \frac{y}{22} \\ y = 22 \times tan(55) \approx 31.4$

Step 2

How long is NM?

$NM^2 = 22^2+ 31.4^2 \approx 1471.1 \\ NM = \sqrt{1471.1} \approx 38.3$

Method 1

Use SOHCAHTOA  and set up a ratio such as sin(16) = 14/x  (From here solve for X) By the way, you could also use cosine.

Method 2

Set up the following equation using the Pythagorean theorem: x2=482+142(From here solve for X)

### Ultimate Math Solver (Free)

Free Algebra Solver ... type anything in there!


text:
# 0. How to I get my computer set up for today’s class?

# Install Packages
if(!require(dplyr)) {install.packages("sjlabelled", repos='https://cran.csiro.au/', dependencies=TRUE)}
if(!require(sjlabelled)) {install.packages("sjlabelled", repos='https://cran.csiro.au/', dependencies=TRUE)}
if(!require(sjmisc)) {install.packages("sjmisc", repos='https://cran.csiro.au/', dependencies=TRUE)}
if(!require(sjstats)) {install.packages("sjstats", repos='https://cran.csiro.au/', dependencies=TRUE)}
if(!require(sjPlot)) {install.packages("sjlabelled", repos='https://cran.csiro.au/', dependencies=TRUE)}
if(!require(summarytools)) {install.packages("summarytools", repos='https://cran.csiro.au/', dependencies=TRUE)}
if(!require(ggplot2)) {install.packages("ggplot2", repos='https://cran.csiro.au/', dependencies= TRUE)}
if(!require(ggthemes)) {install.packages("ggthemes", repos='https://cran.csiro.au/', dependencies= TRUE)}

# Load packages into memory
library(dplyr)
library(sjlabelled)
library(sjmisc)
library(sjstats)
library(sjPlot)
library(summarytools)
library(ggplot2)
library(ggthemes)

# Turn off scientific notation
options(digits=5, scipen=15)

# Stop View from overloading memory with a large datasets
RStudioView <- View
View <- function(x) {
if ("data.frame" %in% class(x)) { RStudioView(x[1:500,]) } else { RStudioView(x) }
}

elect_2013 <- read.csv(url("https://methods101.com/data/elect_2013.csv"))

# 4. How do I make beautiful correlation matricies in R? sjPlot for beautiful tables and plots

The command I have taught you above - cor.test() - is fine for testing the relationship between just two variables. However, often you want to be able to look at the correlations between a large set of variables all the at the one time.

In addition, most journals expect you to produce a correlation matrix of all your variables - both so that the bivariate relationships between variables are clear to the reader, and as a prelude to conducting multivariate analysis.

# Why we recommend ‘sj’ packages

We (Hang Young Lee and myself) recommend using the various functions in ‘sj’ packages (sjlabelled, sjPlot, sjstats, sjmisc), combined with functionality from ggplot2, and also cascading style sheets (CSS), to do the formating of your publication quality tables and plots.

The ‘sj’ packages have several important advantages over other approaches:

1. Labelled data + SPSS/Stata compatability: they support labelled data, similar to that used in SPSS and Stata, which means you can easily import data with labelled values in SPSS or Stata format, and then use those labels in figures, tables, and analysis easily;
2. Easy data manipulation: they have a consistent set of data manipulation functions, which allow easy, intuitive transformation of data;
3. Very large range of functions: they have a very large range of functions for plotting figures, drawing tables, and running analysis
4. Commands are more intuitive than standard R: many of the commands for generating figures, tables, and analysis are simpler and more intuitive than the standard R commands (so many of the plotting commands are simpler than ggplot2; and the data manipulation is simpler than using base R);
5. Compatible with tidyverse: they work seamlessly with ‘tidyverse’, meaning you can use %>% piping, and all the range of data manipuation in this core set of R packages, particularly (dyplr and ggplot2);
6. Outputs can be modified with advanced R and HTML code: in almost all cases, the commands are written so that the figures, tables, and analysis they return can be further modified with very advanced commands (e.g. virtually all ggplot2 options work with any ‘sj’ plot, and virtually all html tables generated by ‘sj’ can be modified with cascading style sheets (CSS)).

## 4.1 How to I make a correlation table (which I can paste into Excel, for example? sjt.corr() - correlation matrix as a HTML table

First we are going to learn how to make a HTML table of the correlation matrix.

What does it do?

• It makes a correlation matrix
• Sends it to your browser as a webpage
• From here you can either screenshot it, or cut and paste it into Excel for reformating.

The advantages of HTML tables are:

• They keep the output as text (not pixels/a picture), so you can actually change fonts, cut and paste into Excel, or move to another program.
• It is a nice simple black and white layout.

The disadvantages of HTML tables are:

• They are a little simple
• They don’t include all the amazing functionality of ggplots, such as colours and charts and shapes.

The command for correlation tables in sjPlots is sjt.corr(). Note that the only difference in the name of the command for tables and plots is that the plot command (below) is called sjp.corr(). But remember, the arguments (settings send to the function), and the output (HTML table vs picture) are completely different.

lga <- readRDS(url("https://www.methods101.com/data/nsw-lga-crime-clean.RDS"))
first <- which( colnames(lga)=="astdomviol" )
last <- which(colnames(lga)=="transport")
crimes <- lga[, first:last ]
sjt.corr(crimes,              # the dataset
na.deletion = c("pairwise"),  # pairwise deletion keeps more data than listwise
corr.method = c("pearson"),   # pearsons r is the standard way to calculate correlation coefficients
title = NULL,                 # You can put a title for your table here
var.labels = NULL,            # This is for variable lables if they aren't in your data already
wrap.labels = 100,            # stops long variable labels wrapping
show.p = TRUE,                # show's p-values as stars (*) or (**) or (***)
p.numeric = FALSE,            # show's p-values as numbers
fade.ns = TRUE,               # non-significant results appear in light grey
val.rm = NULL,                #
digits = 2,                   # number of digits to round correlation coefficient
triangle = "both",            # Above and below the diagonal in a matrix are the same. Do you want both?
string.diag = NULL,           #
CSS = NULL,                   # This is for putting css code to make tables pretty
encoding = NULL,              #
file = NULL,                  #
use.viewer = TRUE,            # Sends output to browser
remove.spaces = TRUE)         #

## 4.2 How do I make a colourful correlation plot/figure? sjp.corr() - correlation matrix as a figure/plot.

lga <- readRDS(url("https://www.methods101.com/data/nsw-lga-crime-clean.RDS"))
first <- which( colnames(lga)=="astdomviol" )
last <- which(colnames(lga)=="transport")
crimes <- lga[, first:last ]

sjp.corr(crimes,                # the dataset
show.legend  = TRUE,          # shows the legend that appear on the right
corr.method = "pearson",      # chooses pearsons r as the correlation method
na.deletion = "pairwise",     # pairwise deletion, to save data
sort.corr = FALSE             # if this was set to TRUE, it would sort so similar variables are
)                             # next to each other. Makes it easier to spot patterns in data.
## Computing correlation using pearson-method with pairwise-deletion...

set_theme(                      # function in sjPlot to set various themes of tables
base = theme_blank()          # this is a simple default theme, which makes text in cells
)                               # and labels the same size, and does a few other neat things.
sjp.corr(crimes,
decimals = 1,         # sets number of decimals for correlations
wrap.labels =100,     # stops variable labels wrapping
show.p=TRUE,          # shows stars for p-values
show.values=TRUE,     # shows correlation coefficients in figure
show.legend = TRUE,   # shows the ledgent on the right hand side of figure
sort.corr = TRUE,     # sorts the rows and columns by correlations
# - this makes it easier to see patterns
geom.colors = "RdBu",       # colour scale for cells. This is Red-Blue.
corr.method = "pearson",    # calculates pearson correlation. Default is spearman correlation.
na.deletion = "pairwise") + # uses pairwise deletion. Default is listwise deletion.
theme(axis.text.x=element_text(angle=90,
hjust=0.95,vjust=0.2)) +  # rotates x-axis labels and aligns them to the right
labs(fill="Pearson's r") +  # put a title on the legend
theme(legend.title = element_text(hjust = 0.5)) + # centres the legend title
guides(fill = guide_colourbar(ticks = FALSE,barwidth = 2, barheight = 15))  # removes the white ticks
## Computing correlation using pearson-method with pairwise-deletion...

                                                                              # in the legend; also makes
# colour bar wider and higher.
sjp.corr(crimes,
decimals = 1,         # sets number of decimals for correlations
wrap.labels =100,     # stops variable labels wrapping
show.p=TRUE,          # shows stars for p-values
show.values=TRUE,     # shows correlation coefficients in figure
show.legend = TRUE,   # shows the ledgent on the right hand side of figure
sort.corr = TRUE,     # sorts the rows and columns by correlations
# - this makes it easier to see patterns
geom.colors = "Spectral",       # colour scale for cells. This is Spectral,
# which is Blue, green, yellow, orange, red.
corr.method = "pearson",    # calculates pearson correlation. Default is spearman correlation.
na.deletion = "pairwise") + # uses pairwise deletion. Default is listwise deletion.
theme(axis.text.x=element_text(angle=90,
hjust=0.95,vjust=0.2)) +  # rotates x-axis labels and aligns them to the right
labs(fill="Pearson's r") +  # put a title on the legend
theme(legend.title = element_text(hjust = 0.5)) + # centres the legend title
guides(fill = guide_colourbar(ticks = FALSE,barwidth = 2, barheight = 15))  # removes the white ticks
## Computing correlation using pearson-method with pairwise-deletion...

                                                                              # in the legend; also makes
# colour bar wider and higher.
Last updated on 26 August, 2019 by Dr Nicholas Harrigan (nicholas.harrigan@mq.edu.au)


text:
PSY 470 Week 6 Article Critique 3

# PSY 470 Week 6 Article Critique 3

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PSY 470 Week 6 Article Critique 3

Topic: Severe Mental Disorders

After reading Schizophrenia: A Sibling’s Tale, write a paper of 500-750 words in which you address the following questions:

1. What did you think of the article?

2. How did the article relate to topics presented in the textbook?

3. What interesting questions did the article raise for you?

4. Identify the positive and negative symptoms of schizophrenia and give an example of each.

5. Describe the types of therapy discussed in Schizophrenia: A Sibling’s Tale.

Use the GCU e-Library databases, include a minimum of two outside references or other resources to highlight and support your viewpoint.

Prepare this assignment according to the APA guidelines found in the APA Style Guide, located in the Student Success Center. An abstract is not required.

This assignment uses a rubric. Please review the rubric prior to beginning the assignment to become familiar with the expectations for successful completion.

You are required to submit this assignment to Turnitin. Please refer to the directions in the Student Success Center.

PSY 470
ajotatxe

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0 points

#### Oh Snap! This Answer is Locked

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Filename: psy-470-week-6-article-critique-3-43.docx

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Print Length: 5 Pages/Slides

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Surround your text in *italics* or **bold**, to write a math equation use, for example, $x^2+2x+1=0$ or $$\beta^2-1=0$$

Use LaTeX to type formulas and markdown to format text. See example.


text:
# Calculate [Cr^3+] in equilibrium with Cr(OH4)- when 0.010 mol of Cr(NO3)3 is dissolved in a liter of solution buffered at pH 10.0

I tried setting up a RICE chart with the initial conditions being:

$$\begin{array}{c|c} \hline \text{Species} & \text{Concentration / M} \\ \hline \ce{Cr^3+} & 0.010 \\ \ce{OH-} & 10^{-4.0} \\ \ce{Cr(OH)4-} & 0 \\ \hline \end{array}$$

I'm assuming I should use the given $K_\mathrm{f}$ value for $\ce{Cr(OH)4-}$ ($8 \times 10^{29}$).

However, when I solve for the change I end up with a negative value for the concentration of $\ce{Cr^{3+}}$? The correct answer is supposed to be $10^{-16}$.

• You probably mean $\ce{Cr^{3+}}$ and not $\ce{Cr3+}$. Here is a link on how to properly typeset chemical equations and other formulae using MathJax. Also, please write down the balanced reaction and we can proceed from there. – Todd Minehardt Jan 31 '17 at 2:07
• I forgot to mention that you do not have nitrate in your query. You'll need to account for that in the overall reaction. – Todd Minehardt Jan 31 '17 at 2:13
• You don't have to worry about the nitrate. The answer is the same as if you threw some $\ce{Cr(OH)3}$ into a solution buffered at pH 10. Until you know what you are doing don't ski[p steps by doing some of this in your head. Write out reaction, then the equilibrium equation, solve the equilibrium equation for $\ce{[Cr^{3+}]}$, and then do the RICE chart. – MaxW Jan 31 '17 at 4:25
• I've written out the full RICE chart and tried setting it up multiple ways, with no luck. – khajiit Feb 9 '17 at 22:30
• Could you please add the source of the question? (Book title, page number or question number) – orthocresol Mar 26 '17 at 22:57

I know this problem and the answer the book gives, and unfortunately the book is in error. Let me explain. Adding 0.010 mol of Cr(NO3)3 will also give you 0.010 mol of $\ce{Cr^3+}$, as we can assume the $\ce{Cr^3+}$ will completely dissociate from the $\ce{NO^3-}$. The $\ce{OH^-}$ concentration can be found from the pH, and it is 0.0001 $\frac{moles}{L}$.

The "before reaction starts" gets set up accordingly:

$\ce{Cr^3+ + 4OH- <=> Cr(OH)4-}$

0.010 M .0001 M 0 M

Notice, however, that there is a 4:1 ratio of $\ce{OH^-}$ to $\ce{Cr^3+}$. It takes 4 $\ce{OH^-}$ for every 1 $\ce{Cr^3+}$, and $\ce{OH^-}$ is quite obviously the limiting reactant. So after the reaction, what is the concentration of $\ce{Cr^3+}$? Let's make it 1:1, then.

$\frac{.0001 M}{4}$ = $\ce{2.5 x 10^{-5}}$. And this is essentially the [$\ce{Cr^3+}$] and [$\ce{Cr(OH)4-}$]. However, this is not what the book says. They meant to put the concentration of $\ce{Cr(OH)4-}$ as .010 M, not $\ce{Cr^3+}$. Let's try it that way, and see what we get!

Cr(OH)4- <=> Cr3+ + 4OH-

Here, in the equation, the concentrations are:

$\ce{Cr(OH)4-}$ = .010 M

$\ce{Cr^3+}$ = 0

4$\ce{OH^-}$ = .0001 M

-x +x +4x

.010 - x +x .0001 + 4x

Because .010 - x and .0001 + 4x are so small, let .010 - x = .010 and .0001 + 4x = .0001.

The $\ce{K_f}$ = 8 x $10^{29}$, but that is for the equation when it looks like:

$$\ce{Cr(OH)4- <=> Cr3+ + 4OH-}$$

We switched it around to find out how much $\ce{Cr(OH)4-}$ would dissolve to make $\ce{Cr^3+}$, so we have to use $\frac{1}{Kf}$, which = 1.25 x $10^{-30}$.

$\ce{K_f}$ = $\frac{[{Cr^3+}][{OH-}]^4}{[\ce{Cr(OH)4-}]}$

$\ce{K_f}$ = $\frac{(x)(.0001)^4}{(.010)}$

1.25 x $10^{-30}$ = $\frac{1 x 10^{-16}x}{.010}$

You can take it from there. The answer will be 1.25 x $10^{-16}$ M.


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# Calculating the expected value of free tickets in a paytable?

Lets says I have a lottery game where the ticket costs $1 and has the following probability/prize distribution: • 0.3 -> \$1
• 0.2 -> X
• 0.5 -> \$0 If X = \$1, then the expected value is: 0.3(\$1) + 0.2(\$1) = \$0.50 If X = FREE_TICKET, then I've calculated (and confirmed) via sampling that the EV is either: • 0.3(\$1) / (0.3+0.5) = 0.375
• 0.3(\$1) + 0.2 * \$1 * 0.3/(0.3+0.5) = 0.375

Obviously, a FREE_TICKET isn't the same as \$1, the price of a ticket. However, I'm stumped when calculating the EV of the table for the following two cases: • X = 2 FREE_TICKETS • X = \$5 + FREE_TICKET

How do I calculate them?

-
You have the expected value as $Y = 0.3\cdot 1 + 0.2\cdot X + 0.5\cdot 0$. If, in addition, $X = 2Y$, or $X = 5 + Y$, you can solve for $X$ and $Y$ simultaneously. – mjqxxxx Nov 28 '12 at 18:24
Thanks, you're right, it's just basic algebra. – AlexD Nov 29 '12 at 14:31