text:
DOC HOME SITE MAP MAN PAGES GNU INFO SEARCH

# (make.info.gz) Wildcard Pitfall

Info Catalog (make.info.gz) Wildcard Examples (make.info.gz) Wildcards (make.info.gz) Wildcard Function

4.4.2 Pitfalls of Using Wildcards
---------------------------------

Now here is an example of a naive way of using wildcard expansion, that
does not do what you would intend.  Suppose you would like to say that
the executable file foo' is made from all the object files in the
directory, and you write this:

objects = *.o

foo : $(objects) cc -o foo$(CFLAGS) \$(objects)

The value of objects' is the actual string *.o'.  Wildcard expansion
happens in the rule for foo', so that each _existing_ .o' file
becomes a prerequisite of foo' and will be recompiled if necessary.

But what if you delete all the .o' files?  When a wildcard matches
no files, it is left as it is, so then foo' will depend on the
oddly-named file *.o'.  Since no such file is likely to exist, make'
will give you an error saying it cannot figure out how to make *.o'.
This is not what you want!

Actually it is possible to obtain the desired result with wildcard
expansion, but you need more sophisticated techniques, including the
wildcard' function and string substitution.   The Function
wildcard' Wildcard Function.

Microsoft operating systems (MS-DOS and MS-Windows) use backslashes
to separate directories in pathnames, like so:

c:\foo\bar\baz.c

This is equivalent to the Unix-style c:/foo/bar/baz.c' (the c:'
part is the so-called drive letter).  When make' runs on these
systems, it supports backslashes as well as the Unix-style forward
slashes in pathnames.  However, this support does _not_ include the
wildcard expansion, where backslash is a quote character.  Therefore,
you _must_ use Unix-style slashes in these cases.

Info Catalog (make.info.gz) Wildcard Examples (make.info.gz) Wildcards (make.info.gz) Wildcard Function
automatically generated byinfo2html


text:
# Consider an asset that costs $130 today. You are going to hold it for 1 year and then sell it... 1 answer below » Consider an asset that costs$130 today. You are going to hold it for 1 year and then sell it. Suppose that there is a 25 percent chance that it will be worth $100 in a year, a 25 percent chance that it will be worth$120 in a year, and a 50 percent chance that it will be worth $160 in a year. Instructions: In part a, round your answer to 3 decimal places. In parts b and d, round your answers to 2 decimal places. If you are entering any negative numbers be sure to include a negative sign (-) in front of those numbers. a. What is its average expected rate of return? b. Next, figure out what the investment’s average expected rate of return would be if its current price were$140 today.

c. Does the increase in the current price increase or decrease the asset’s average expected rate of return?

(Click to select) Decrease Increase

d. At what price would the asset have a zero average expected rate of return?

This is the last question in the assignment. To submit, use Alt + S. To access other questions, proceed to the question map button.Ne

## 1 Approved Answer

VIJAYAKUMAR G
5 Ratings, (9 Votes)

The solution is attached...

## Recent Questions in Macro Economics

Submit Your Questions Here !
Copy and paste your question here...
Attach Files

format: plain


text:
Find all School-related info fast with the new School-Specific MBA Forum

 It is currently 10 Feb 2016, 02:00

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

# GMAT Data Sufficiency (DS)

 Question banks Downloads My Bookmarks Reviews Important topics Go to page Previous    1  ...  6   7   8   9   10   11  ...  148    Next Search for:
Topics Author Replies   Views Last post

Announcements

65
 150 Hardest and easiest questions for DS   Tags:
Bunuel

3

8810

07 Dec 2015, 08:09

348
 DS Question Directory by Topic and Difficulty
bb

0

105018

07 Mar 2012, 07:58

Topics

44
 Lines n and p lie in the xy-plane. Is the slope of line n
BANON

18

9001

06 Jan 2016, 12:43

7
 What is the area of the parallelogram inscribed in a circle?
ritula

18

2614

22 Jul 2015, 23:09

30
 Machines A and B each produce tablets at their respective co
tk1tez7777

18

4411

05 Jan 2016, 21:32

8
 Is the measure of one of the interior angles of
coffeeloverfreak

18

2780

11 Sep 2013, 01:48

26
 Mike recently won a contest in which he will have the opport
ctrlaltdel

18

5600

27 Aug 2015, 00:06

18
 If a and n are integers and n > 1, is n odd?
conty911

18

6667

19 Dec 2015, 04:28

5
 If ABD is a triangle, is triangle ABC a right triangle
joylive

18

3371

18 Jun 2014, 06:38

7
 Is |a| = b - c ?
gmat1011

18

4417

03 Feb 2015, 06:43

15
 Is x^2 < x - |y|?
summer101

18

1397

14 Jan 2014, 21:52

36
 If x and y are positive integers is y odd?
devinawilliam83

18

7631

09 Mar 2015, 10:56

56
 If p and n are positive integers and p > n, what is the
BANON

18

16503

17 Mar 2015, 10:32

29
 A certain one-day seminar consisted of a morning session and

18

10253

06 Jan 2016, 13:12

4
 Is (x^7)(y^2)(z^3)>0?
Barkatis

18

2069

20 Apr 2013, 03:54

2
 If x, y, and z are positive numbers, is x > y > z ?
amansingla4

18

5575

09 Mar 2014, 23:29

26
 In the xy-coordinate plane, line l and line k intersect at
M8

18

4906

13 Nov 2015, 00:46

21
 When the positive number a is rounded to the nearest tenth
frankiegar

18

6068

02 Jul 2015, 18:38

32
 Is 1/(a-b) < b-a ? (1) a
likar

18

9372

08 May 2015, 02:49

2
 A total of $60,000 was invested for one year. Part of this ong 18 2545 24 Jul 2014, 15:59  If x and y are positive, is 4x > 3y? udaymathapati 18 2598 21 May 2014, 19:24 4  A number of people each wrote down one of the first 30 Walkabout 18 4200 29 Apr 2015, 10:03 6  If a and b are integers, and |a| > |b|, is a |b| < a yangsta8 18 3647 10 Feb 2015, 01:18 25  Each employee of a certain task force is either a manager or Accountant 18 8743 01 Nov 2013, 00:15 41  If u(u+v) <> 0 and u > 0 is 1/(u + v) < 1/u + v ? PrashantPonde 18 4827 16 Dec 2015, 06:03 3  Given a sequence: a_1, a_2, a_3, ... a_{14}, a_{15} In the Sam Kana 18 3840 18 Sep 2010, 08:31 4  What is the units digit of integer s? (1) The units digit of Tags: Poor Quality dreambeliever 18 1547 19 Jun 2015, 11:33 35  A newer machine, working alone at its constant rate gmacforjyoab 18 4115 21 Jan 2016, 03:56 26  In isosceles triangle RST what is the measure of angle R? gmatgg 18 20705 12 May 2015, 08:06 6  If n is not equal to 0, is |n| < 4 ? lbsgmat 18 5027 17 Jul 2014, 00:30 66  If a rectangular region has perimeter P inches and area A sq olifurlong 18 11625 03 Feb 2016, 12:34 54  Beginning in January of last year, Carl made deposits of mydreammba 18 7453 22 Jan 2016, 08:55 13  Lines a and b have different y-intercepts. When line a is bhandariavi 18 3847 21 Aug 2014, 20:49 29  Juan bought some paperback books that cost$8 each and some
neeraj.kaushal

18

5969

11 Oct 2015, 02:12

19
 If x is different from 0, is |x| < 1 ?
carcass

18

3795

23 Sep 2014, 07:04

41
 The average of 5 distinct single digit integers is 5. If
qweert

18

7581

15 Dec 2015, 19:53

7
 k, n, 12, 6, 17 What is the value of n in the list above?
sidbidus

18

4500

01 Mar 2015, 03:28

17
 Of the 60 families in a certain neighborhood, 38 have a cat.
calreg11

18

4744

02 Jan 2016, 22:22

24
 A group of men and women gathered to compete in a marathon.
enigma123

18

6009

24 Oct 2015, 05:11

 Sequence A is defined by the equation An = 3n + 7, where n
somethingbetter

18

1683

04 Nov 2014, 03:48

13
 If r and s are positive integers, is r/s an integer?
chicagocubsrule

18

3919

05 Nov 2015, 05:47

10
 A box of doughnuts contains glazed, cream-filled, and jelly
Nsentra

18

4680

20 Nov 2014, 07:43

24
 What is the value of |x| ?
Bunuel

18

8877

07 May 2015, 21:52

8
 Is |x-1| < 1?
devinawilliam83

18

2447

24 Jul 2014, 17:43

7
 Pat is reading a book that has a total of 15 chapters. Has
gmat620

18

1855

29 Dec 2014, 04:37

17
 Does the decimal equivalent of P/Q, where P and Q are
vaivish1723

18

9554

25 Dec 2014, 11:47

4
 Is x^2 greater than x?
jscott319

18

7845

29 Dec 2015, 22:03

5
 If Pat saved \$600 of his earnings last month, how much did
Bunuel

18

6625

25 Jul 2015, 13:40

6
 On the coordinate plane, is point (u,v) closer to point
gmat620

18

2254

24 Aug 2014, 05:23

1
 If m is integer, how many +ve integers are factors of m ? 1.
asmit123

18

2768

05 Dec 2011, 04:52

14
 In the rectangular co-ordinate system shown does the line k
gmatcraze

18

4357

28 Sep 2015, 04:08

8
 If x and y are integers, is x^y
dreambeliever

18

2648

09 Jul 2015, 09:29

 Question banks Downloads My Bookmarks Reviews Important topics Go to page Previous    1  ...  6   7   8   9   10   11  ...  148    Next Search for:
 Who is online In total there is 1 user online :: 0 registered, 0 hidden and 1 guest (based on users active over the past 15 minutes) Users browsing this forum: No registered users and 1 guest Statistics Total posts 1447242 | Total topics 176199 | Active members 432130 | Our newest member cnbcravi

 Powered by phpBB © phpBB Group and phpBB SEO Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.


text:
Session 56 - Frontiers of Ultraviolet Astrophysics.
Display session, Wednesday, June 10
Atlas Ballroom,

## [56.03] ORFEUS Observations of the Contact Binary 44i Boo

N. S. Brickhouse, A. K. Dupree (Smithsonian Astrophysical Observatory)

We report ORFEUS observations of the contact binary 44i Boo (HD 133640) at two orbital phases (0.91 and 0.40). The contact binaries provide the exception to the rule of increasing magnetic activity with rotation rate among cool stars, as the radiation reaches a constant maximum level for the fastest rotators. The ratio of emission lines C III \lambda977; \lambda1176 indicates relatively high electron density, N_e = 2 \times 10^10 cm^-3 at T_e \sim 6 \times 10^4 K, with a pressure five orders of magnitude less than the coronal pressure at T_e = 6 \times 10^6 K determined from EUVE (Brickhouse amp; Dupree 1998, Ap J, 502, in press). The shape of the emission measure distribution for the system from T_e \sim 3 \times 10^4 to the limit temperature T_e \sim 2 \times 10^7 K is determined from IUE, ORFEUS, EUVE, and ASCA, with ORFEUS providing the critical O VI \lambda1032 (T_e \sim 3 \times 10^5 K), linking the upper transition region to the coolest observed EUV Fe ion (Fe IX at T_e \sim 6 \times 10^5 K). The minimum occurs near 6 \times 10^5 K, reminiscent of that of Capella.

Emission lines shifted by the relative orbital velocity (\sim 200 km/s) between the two stars would be separable at ORFEUS resolution at both phases; however, the line profiles of Si III \lambda1206, C III \lambda977, and O VI \lambda1032, sampling a decade of temperatures (from 3 \times 10^4 to 3 \times 10^5 K), are not consistent with such a simple model. Moreover, these three line profiles show significant differences from each other. Since IUE light curves of lower transition region lines (C II, C IV; see Vilhu, Neff, amp; Rahunen 1989, Aamp; A, 208, 201) are consistent with the partial eclipses observed in the photosphere, these new line profile measurements present an intriguing clue to the puzzle of coronal heating and confinement.


text:
GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 19 Jan 2019, 13:00

GMAT Club Daily Prep

Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

Events & Promotions

Events & Promotions in January
PrevNext
SuMoTuWeThFrSa
303112345
6789101112
13141516171819
20212223242526
272829303112
Open Detailed Calendar
• FREE Quant Workshop by e-GMAT!

January 20, 2019

January 20, 2019

07:00 AM PST

07:00 AM PST

Get personalized insights on how to achieve your Target Quant Score.
• Free GMAT Strategy Webinar

January 19, 2019

January 19, 2019

07:00 AM PST

09:00 AM PST

Aiming to score 760+? Attend this FREE session to learn how to Define your GMAT Strategy, Create your Study Plan and Master the Core Skills to excel on the GMAT.

If a motorist had driven 1 hour longer on a certain day and

Author Message
TAGS:

Hide Tags

Senior Manager
Joined: 23 Oct 2010
Posts: 350
Location: Azerbaijan
Concentration: Finance
Schools: HEC '15 (A)
GMAT 1: 690 Q47 V38
Re: If a motorist had driven 1 hour longer on a certain day and  [#permalink]

Show Tags

01 Jul 2012, 03:50
1)
(V+5)(T+1)-VT=70
(V+10)(T+2)-VT=X
============
2)
5T+V=65
10T+2V=X-20
2*65=X-20 =>X=150
_________________

Happy are those who dream dreams and are ready to pay the price to make them come true

I am still on all gmat forums. msg me if you want to ask me smth

Manager
Joined: 14 Nov 2011
Posts: 122
Location: United States
Concentration: General Management, Entrepreneurship
GPA: 3.61
WE: Consulting (Manufacturing)
Re: If a motorist had driven 1 hour longer on a certain day and  [#permalink]

Show Tags

03 Jun 2013, 08:03
Bunuel wrote:
20. If a motorist had driven 1 hour longer on a certain day and at an average rate of 5 miles per hour faster, he would have covered 70 more miles than he actually did. How many more miles would he have covered than he actually did if he had driven 2 hours longer and at an average rate of 10 miles per hour faster on that day?
(A) 100 (B) 120 (C) 140
(D) 150 (E) 160

Let $$t$$ be the actual time and $$r$$ be the actual rate.

"If a motorist had driven 1 hour longer on a certain day and at an average rate of 5 miles per hour faster, he would have covered 70 more miles than he actually did" --> $$(t+1)(r+5)-70=tr$$ --> $$tr+5t+r+5-70=tr$$ --> $$5t+r=65$$;

"How many more miles would he have covered than he actually did if he had driven 2 hours longer and at an average rate of 10 miles per hour faster on that day?" --> $$(t+2)(r+10)-x=tr$$ --> $$tr+10t+2r+20-x=tr$$ --> $$2(5t+r)+20=x$$ --> as from above $$5t+r=65$$, then $$2(5t+r)+20=2*65+20=150=x$$ --> so $$x=150$$.

OR another way:

70 miles of surplus in distance is composed of driving at 5 miles per hour faster for $$t$$ hours plus driving for $$r+5$$ miles per hour for additional 1 hour --> $$70=5t+(r+5)*1$$ --> $$5t+r=65$$;

With the same logic, surplus in distance generated by driving at 10 miles per hour faster for 2 hours longer will be composed of driving at 10 miles per hour faster for $$t$$ hours plus driving for $$r+10$$ miles per hour for additional 2 hour --> $$surplus=x=10t+(r+10)*2$$ --> $$x=2(5t+r)+20$$ --> as from above $$5t+r=65$$, then $$x=2(5t+r)+20=150$$.

Note that the solutions proposed by dushver and dimitri92 are not correct (though correct answer was obtained). For this question we can not calculate neither $$t$$ not $$r$$ of the motorist.

Hope it helps.

Exactly, this is what I wanted to ask. Thanks again.
Their solution means that they were traveling at speed S for t hours and only in the last hour they traveled by S+5 speed, which is wrong.
Senior Manager
Joined: 13 May 2013
Posts: 425
Re: If a motorist had driven 1 hour longer on a certain day and  [#permalink]

Show Tags

10 Aug 2013, 11:15
If a motorist had driven 1 hour longer on a certain day and at an average rate of 5 miles per hour faster, he would have covered 70 more miles than he actually did. How many more miles would he have covered than he actually did if he had driven 2 hours longer and at an average rate of 10 miles per hour faster on that day?

d=r*t
d+70 = (r+5) * (t+1)

d = (r+10) * (t+2)

(r+10) * (t+2) + 70 = (r+5) * (t+1)
rt+2r+10t+20 +70 = rt + r + 5t + 5
r + 10t + 90 = 5t + 5

As you can see, I solved for d [(r+10) * (t+2)]then plugged in d for d+70 = (r+5) * (t+1). Can someone please explain why this is an incorrect approach?

(A) 100
(B) 120
(C) 140
(D) 150
(E) 160
Director
Joined: 14 Dec 2012
Posts: 748
Location: India
Concentration: General Management, Operations
GMAT 1: 700 Q50 V34
GPA: 3.6
Re: If a motorist had driven 1 hour longer on a certain day and  [#permalink]

Show Tags

10 Aug 2013, 11:32
1
WholeLottaLove wrote:
If a motorist had driven 1 hour longer on a certain day and at an average rate of 5 miles per hour faster, he would have covered 70 more miles than he actually did. How many more miles would he have covered than he actually did if he had driven 2 hours longer and at an average rate of 10 miles per hour faster on that day?

d=r*t
d+70 = (r+5) * (t+1)

d = (r+10) * (t+2)

(r+10) * (t+2) + 70 = (r+5) * (t+1)
rt+2r+10t+20 +70 = rt + r + 5t + 5
r + 10t + 90 = 5t + 5

As you can see, I solved for d [(r+10) * (t+2)]then plugged in d for d+70 = (r+5) * (t+1). Can someone please explain why this is an incorrect approach?

(A) 100
(B) 120
(C) 140
(D) 150
(E) 160

the error is in highlited part:
that should be:
$$X = (r+10) * (t+2)$$ (d is the actual distance)....X==>distance coveres when travelling with 10 miles more average speed and 2 hours more
now you have to calculate X-d
hope its clear
_________________

When you want to succeed as bad as you want to breathe ...then you will be successfull....

GIVE VALUE TO OFFICIAL QUESTIONS...

learn AWA writing techniques while watching video : http://www.gmatprepnow.com/module/gmat-analytical-writing-assessment

Senior Manager
Joined: 10 Jul 2013
Posts: 314
Re: If a motorist had driven 1 hour longer on a certain day and  [#permalink]

Show Tags

10 Aug 2013, 12:54
el1981 wrote:
If a motorist had driven 1 hour longer on a certain day and at an average rate of 5 miles per hour faster, he would have covered 70 more miles than he actually did. How many more miles would he have covered than he actually did if he had driven 2 hours longer and at an average rate of 10 miles per hour faster on that day?

(A) 100
(B) 120
(C) 140
(D) 150
(E) 160

....
used 3times s=vt and input 65=v+5t ,then got 150miles
_________________

Asif vai.....

Senior Manager
Joined: 13 May 2013
Posts: 425
Re: If a motorist had driven 1 hour longer on a certain day and  [#permalink]

Show Tags

13 Aug 2013, 10:33
blueseas wrote:
WholeLottaLove wrote:
If a motorist had driven 1 hour longer on a certain day and at an average rate of 5 miles per hour faster, he would have covered 70 more miles than he actually did. How many more miles would he have covered than he actually did if he had driven 2 hours longer and at an average rate of 10 miles per hour faster on that day?

d=r*t
d+70 = (r+5) * (t+1)

d = (r+10) * (t+2)

(r+10) * (t+2) + 70 = (r+5) * (t+1)
rt+2r+10t+20 +70 = rt + r + 5t + 5
r + 10t + 90 = 5t + 5

As you can see, I solved for d [(r+10) * (t+2)]then plugged in d for d+70 = (r+5) * (t+1). Can someone please explain why this is an incorrect approach?

(A) 100
(B) 120
(C) 140
(D) 150
(E) 160

the error is in highlited part:
that should be:
$$X = (r+10) * (t+2)$$ (d is the actual distance)....X==>distance coveres when travelling with 10 miles more average speed and 2 hours more
now you have to calculate X-d
hope its clear

Thanks!

"How many more miles would he have covered than he actually did" doesn't refer to his actual distance but hypothetical miles. For example, if his actual distance (d) equaled 200 miles then his hypothetical distance (x) may = 50 miles which means his hypothetical total would be 200+50 = 250 miles.

We have two equations:
d+70 = (r+5) * (t+1)
d+70 = rt+r+5t+5
d+65 = rt+r+5t
d = rt-r-5t-65

x = (r+10)*(t+2)
x = rt+2r+10t+20
x - 2r - 10t - 20 = rt

Because we are looking for the difference between the hypothetical distance minus the actual distance we have to find x-d.
x-d
(rt+2r+10t+20) - (rt-r-5t-65)
(rt+2r+10t+20) - rt+r+5t+65
3r +15t + 85

But from here...I am stuck!
Intern
Status: Bring it on.....!
Joined: 11 Jul 2013
Posts: 26
Location: India
Concentration: Technology
GMAT 1: 660 Q49 V31
Re: If a motorist had driven 1 hour longer on a certain day and  [#permalink]

Show Tags

16 Oct 2013, 18:08
orignal speed = s ;

if you increase s by 5 m/hr, distance travelled is 70 miles in that extra 1 hour, therefore :

==> (s+5)*1=70 ; s=65

Now, in 2 more hours with an increased 10m/hr over original speed, distance traveled :

d=(65+10)*2 ; d=150. Ans (D)
_________________

Live as if your were to die tomorrow. Learn as if you were to live forever - Gandhi

Intern
Joined: 21 Dec 2014
Posts: 6
Re: If a motorist had driven 1 hour longer on a certain day and  [#permalink]

Show Tags

24 Jan 2015, 14:18
I see thias as that, think is easier:
1) traveled 70 km in one additional hour, at an extra 5 m/h
2) so, in that un hour at that extra speed, advance 70 miles. s2 * t2 = d2 ------ s2 =(s1 + 5)-----s1=x ----t2=1;
Therefore: (x+5) * 1 = 70 ---- x =65
3) now 65 plus 10 m/h, new speed is 75 m/h; then, in 2 additional hours, he adnvaced 2 *75 = 150 miles.

Nelson
EMPOWERgmat Instructor
Status: GMAT Assassin/Co-Founder
Affiliations: EMPOWERgmat
Joined: 19 Dec 2014
Posts: 13346
Location: United States (CA)
GMAT 1: 800 Q51 V49
GRE 1: Q170 V170
Re: If a motorist had driven 1 hour longer on a certain day and  [#permalink]

Show Tags

24 Jan 2015, 17:58
Hi All,

This question can be solved by TESTing VALUES (as some of the explanations have noted). You can keep the values really small though and save some time on the calculations:

We're told that driving 1 extra hour AND at a speed that was 5 miles/hour faster (for the entire trip) would have increased TOTAL distance by 70 miles.

IF.....
We originally drove for 1 hour at 60 miles/hour, then we would have traveled 60 miles.

Adding 1 extra hour and increasing speed by 5 miles/hour would give us.....
2 hours at 65 miles/hour, which gives us a total distance of 130 miles (which is 70 miles MORE than originally traveled).

So now we we've established the starting time, speed and distance, so we can answer the given question:

How many MORE miles would be traveled if the original time was increased by 2 hours AND the original speed was increased by 10 miles/hour?

1+2 = 3 hours
60 + 10 = 70 miles/hour
3 hours at 70 miles/hour = 210 miles

Since we originally traveled 60 miles, the 210 total miles is 150 miles MORE than originally traveled.

GMAT assassins aren't born, they're made,
Rich
_________________

760+: Learn What GMAT Assassins Do to Score at the Highest Levels
Contact Rich at: Rich.C@empowergmat.com

Rich Cohen

Co-Founder & GMAT Assassin

Special Offer: Save \$75 + GMAT Club Tests Free
Official GMAT Exam Packs + 70 Pt. Improvement Guarantee
www.empowergmat.com/

*****Select EMPOWERgmat Courses now include ALL 6 Official GMAC CATs!*****

VP
Joined: 07 Dec 2014
Posts: 1152
Re: If a motorist had driven 1 hour longer on a certain day and  [#permalink]

Show Tags

18 Feb 2016, 15:24
let t=original time
r=original rate
equation 1: 5t+r=65
equation 2: 10t+2r=m-20
multiply equation 1 by 2 and subtract equation 2
m=150 miles
Intern
Joined: 01 Apr 2015
Posts: 20
Location: United States
Re: If a motorist had driven 1 hour longer on a certain day and  [#permalink]

Show Tags

04 Mar 2017, 11:43
Since he traveled an extra 70 miles at 5mph more plus one hour of travel, at 10mph more he would have covered double that amount (140 miles) in the same time, plus one more extra hour of travel (10 miles), for a total of 150 miles. Solved in seconds without any computations... is this logic correct?
Manager
Joined: 12 Feb 2017
Posts: 70
Re: If a motorist had driven 1 hour longer on a certain day and  [#permalink]

Show Tags

02 Sep 2017, 12:09
Let's assume original distance as x, speed as s and time as t

distance= speed *time
therefore,
x= st -----(1)
According to first scenario,
x+70= (s+5) * (t+1)
x+70= st+s+5t+5
x+70= x+s+5t+5 ---from 1

s+5t=65 -----(2)

Now consider the second scenario,
It is given that the speed would be 10 Miles/hr faster and total time would be 2 hrs longer
therefore ,
(s+10) * (t+2) will give us the total distance.
distance= st+2s+10t+20
distance= x+2(s+5t)+20 ---- from (1)
distance= x+2(65)+20 ------from (2)
distance= x+150.

hence in given scenario, extra 150 miles will be covered. Hence Answer is D

Kudos if it helps.
Manager
Joined: 14 Jul 2014
Posts: 96
Location: India
Concentration: Social Entrepreneurship, Strategy
GMAT 1: 620 Q41 V34
WE: Information Technology (Computer Software)
Re: If a motorist had driven 1 hour longer on a certain day and  [#permalink]

Show Tags

23 Sep 2018, 11:35
Let’s start with the basic D=RT equation. From there, the first theoretical trip can be represented as D + 70 = (R + 5)(T + 1), which expands to D + 70 = RT + R + 5T +5 and can be simplified to 65 = R + 5T.

The second theoretical trip can be represented as (R+10)(T+2), which expands to RT + 2R + 10T + 20 (note that you only have an expression since you don’t know what the distance is). The two middle terms (2R + 10T) can be factored to 2(R+5T), which allows us to use the second equation. RT + 2(R+5T) + 20 = RT + 2(65) + 20 = RT + 150. Since the original distance was RT, the additional distance is 150, or choice D.

Additionally, this problem can be solved with number picking. Imagine that the original speed was 50mph - then you know that the one extra hour would have resulted in 55 extra miles so there must have been 3hrs at the additional 5mph to create the extra 15 miles to get to 70. Now apply those same numbers in the second case: with an extra two hours you get 2(60mph) or 120 extra miles and then the 3hrs at the additional 10mph to get 150. This will yield 150 no matter what numbers you try for the original case!
Re: If a motorist had driven 1 hour longer on a certain day and &nbs [#permalink] 23 Sep 2018, 11:35

Go to page   Previous    1   2   [ 33 posts ]

Display posts from previous: Sort by


text:
# Peclet number

basdomle The Peclet number is defined as

$Pe \equiv \frac{U_\infty L}{\alpha}$

Also

$Pe \equiv Re Pr$

where

• $U_\infty$ is the freestream velocity
• $L$ is the characteristic dimension of the problem
• $\alpha$ is the thermal diffusion coefficient


text:
# Normal distribution problem

## Homework Statement

A student sits a test and is told that the marks follow a normal distribution with mean 100. The student receives a mark of 124 and is told that he is at the 68th percentile. Calculate the variance of the distribution.

## The Attempt at a Solution

I replaced 0.68 for f (x) and 100 for μ and 124 for x. I tried to rearange to solve for σ but was unsuccessful only getting as far as,
0.553 = -288/σ^2 - lnσ
I think my approach might be completely wrong, or is there a way of doing this with the aid of a graphic calculator?

Ray Vickson
Homework Helper
Dearly Missed

## Homework Statement

A student sits a test and is told that the marks follow a normal distribution with mean 100. The student receives a mark of 124 and is told that he is at the 68th percentile. Calculate the variance of the distribution.

## The Attempt at a Solution

I replaced 0.68 for f (x) and 100 for μ and 124 for x. I tried to rearange to solve for σ but was unsuccessful only getting as far as,
0.553 = -288/σ^2 - lnσ
I think my approach might be completely wrong, or is there a way of doing this with the aid of a graphic calculator?

Your ##f##-formula is incorrect: is should be
$$f(x) = \frac{1}{\sqrt{2 \pi} \delta} e^{- \displaystyle \frac{(x-\mu)^2}{2 \delta^2}}$$
Anyway, that is not relevant for this problem: a percentile is a point on the cumulative distribution, not on the probability density function.

There is no closed-form formula for the cumulative distribution, so you need to use tables of the cumulative, or use appropriate software. Some scientific calculators even have a "normal cumulative" button.

Last edited:
NihalRi
SteamKing
Staff Emeritus
Homework Helper

## Homework Statement

A student sits a test and is told that the marks follow a normal distribution with mean 100. The student receives a mark of 124 and is told that he is at the 68th percentile. Calculate the variance of the distribution.

## The Attempt at a Solution

I replaced 0.68 for f (x) and 100 for μ and 124 for x. I tried to rearange to solve for σ but was unsuccessful only getting as far as,
0.553 = -288/σ^2 - lnσ
I think my approach might be completely wrong, or is there a way of doing this with the aid of a graphic calculator?

Your formula shows the PDF for the normal distribution with mean μ and SD δ. The percentile of the student's score represents the area under the PDF distribution curve from -∞ to x. For example, the 50th percentile means that x = μ. Like Ray Vickson posted, you'll have to use standard normal tables to find out the SD.

NihalRi
The percentile of the student's score represents the area under the PDF distributio

∫ f (x)dx=0.68

Where f (x) is the PDF

Will this work as a definite integral from 0 to 124 and I try to solve again for σ ?
Well this seemed like it would work until I actually tried to integrate it:)

Last edited:
SteamKing
Staff Emeritus
Homework Helper
∫ f (x)dx=0.68

Where f (x) is the PDF

Will this work as a definite integral from 0 to 124 and I try to solve again for σ ?
No, that's not how the standard normal tables are constructed.

The standard normal distribution has μ = 0 and δ = 1.0. In order to accommodate normal distributions with other values of μ and δ, a transformation formula is used:

##Z = \frac{x-μ}{δ}##

The total area under the standard normal distribution curve is

##\int_{-∞}^{+∞} f(Z)dZ = 1.0##,

but since the curve is symmetrical about μ = 0

##\int_{-∞}^0 f(Z)dZ = 0.5##

Knowing the value of the cumulative distribution, you want to find Z which corresponds to that amount from a standard normal table, like this one:

https://www.mathsisfun.com/data/standard-normal-distribution-table.html

Once you find the correct Z value, then using the student's score and the mean of the distribution, you should be able to work back and solve for the S.D., δ.

So I set it up so that
P (X≤124) = 0.68
and
P (Z≤(124-100)/σ)
Using invNorm on my calculator for the standard (z) distribution I found
(124-100)/σ≈0.468
σ≈51.3
Variance = 10051
Nice :)

Thank you

SteamKing
Staff Emeritus
Homework Helper
So I set it up so that
P (X≤124) = 0.68
and
P (Z≤(124-100)/σ)
Using invNorm on my calculator for the standard (z) distribution I found
(124-100)/σ≈0.468
σ≈51.3
Variance = 10051
Nice :)

Thank you
Everything looks OK, except the variance should be equal to σ2. IDK how you got 10051.

NihalRi
Everything looks OK, except the variance should be equal to σ2. IDK how you got 10051.
Your right, I must have put the wrong number in the calculator.


text:
6.2: Orbitals

Hartree Description

The energies and wave functions within the most commonly used theories of atomic structure are assumed to arise as solutions of a Schrödinger equation whose Hamiltonian $$h_e(r)$$ possess three kinds of energies:

1. Kinetic energy, whose average value is computed by taking the expectation value of the kinetic energy operator $$-\dfrac{\hbar^2}{2m} \nabla^2$$ with respect to any particular solution $$\phi_j(r)$$ to the Schrödinger equation: $KE = \langle\phi_j| -\dfrac{\hbar^2}{2m} \nabla^2 |\phi_j\rangle$
2. Coulombic attraction energy with the nucleus of charge $$Z$$: $\langle\phi_j| -\dfrac{Z_e^2}{r} |\phi_j\rangle$
3. Coulomb repulsion energies with all of the $$N-1$$ other electrons, which are assumed to occupy other atomic orbitals (AOs) denoted $$\phi_K$$, with this energy computed as

$\sum_K \langle\phi_j(r) \phi_K(r’) |\frac{e^2}{|r-r’|} | \phi_j(r) \phi_K(r’)\rangle.\label{6.1.2}$

The Dirac notation $$\langle\phi_j(r) \phi_K(r’) |\dfrac{e^2}{|r-r’|} | \phi_j(r) \phi_K(r’)\rangle$$ is used to represent the six-dimensional Coulomb integral

$J_{J,K} = \int |\phi_j(r)|^2 |\phi_K(r’)|^2 \dfrac{e^2}{r-r'} dr dr’ \label{6.1.3}$

that describes the Coulomb repulsion between the charge density $$|\phi_j(r)|^2$$ for the electron in $$\phi_j$$ and the charge density $$|\phi_K(r’)|^2$$ for the electron in $$\phi_K$$. Of course, the sum over $$K$$ must be limited to exclude $$K=J$$ to avoid counting a “self-interaction” of the electron in orbital $$\phi_j$$ with itself.

The total energy $$\epsilon_J$$ of the orbital $$\phi_j$$, is the sum of the above three contributions:

$\epsilon_J = \langle\phi_j| - \frac{\hbar^2}{2m} \nabla^2 |\phi_j\rangle + \langle\phi_j| -\frac{Z_e^2}{r} |\phi_j\rangle + \sum_K \langle\phi_j(r) \phi_K(r’) |\frac{e^2}{|r-r’|} | \phi_j(r) \phi_K(r’)\rangle.\label{6.1.4}$

This treatment of the electrons and their orbitals is referred to as the Hartree-level of theory. As stated above, when screened hydrogenic AOs are used to approximate the $$\phi_j$$ and $$\phi_K$$ orbitals, the resultant $$\epsilon_J$$ values do not produce accurate predictions. For example, the negative of $$\epsilon_J$$ should approximate the ionization energy for removal of an electron from the AO $$\phi_j$$. Such ionization potentials (IP s) can be measured, and the measured values do not agree well with the theoretical values when a crude screening approximation is made for the AO s.

LCAO-Expansion

To improve upon the use of screened hydrogenic AOs, it is most common to approximate each of the Hartree AOs {$$\phi_K$$} as a linear combination of so-called basis AOs {$$\chi_\mu$$}:

$\phi_J = \sum_\mu C_{J,\mu} \chi_\mu.\label{6.1.5}$

using what is termed the linear-combination-of-atomic-orbitals (LCAO) expansion. In this equation, the expansion coefficients {$$C_{J,\mu}$$} are the variables that are to be determined by solving the Schrödinger equation

$h_e \phi_J = \epsilon_J \phi_J. \label{6.1.6}$

After substituting the LCAO expansion for $$\phi_J$$ into this Schrödinger equation, multiplying on the left by one of the basis AOs $$\chi_\nu$$, and then integrating over the coordinates of the electron in $$\phi_J$$, one obtains

$\sum_\mu \langle\chi_\nu| h_e| \chi_\mu\rangle C_{J,\mu} = \epsilon_J \sum_\mu \langle\chi_\nu| \chi_\mu\rangle C_{J,\mu} . \label{6.1.7}$

This is a matrix eigenvalue equation in which the $$\epsilon_J$$ and {$$C_{J,\mu}$$} appear as eigenvalues and eigenvectors. The matrices $$\langle\chi_\nu| h_e| \chi_\mu\rangle$$ and $$\langle\chi_\nu| \chi_\mu\rangle$$ are called the Hamiltonian and overlap matrices, respectively. An explicit expression for the former is obtained by introducing the earlier definition of he:

$\langle\chi_\nu| h_e| \chi_\mu\rangle = \langle\chi_\nu| - \frac{\hbar^2}{2m} \nabla^2 |\chi_\mu\rangle + \langle\chi_\nu| -\frac{Ze^2}{r} |\chi_\mu\rangle \label{6.1.8}$

$+ \sum_{\eta,\gamma} \sum_K C_{K,\eta} C_{K,\gamma} \langle\chi_\nu(r) \chi_\eta(r’) |\frac{e^2}{|r-r’|} | \chi_\mu(r) \chi_\gamma(r’)\rangle. \label{6.1.9}$

An important thing to notice about the form of the matrix Hartree equations is that to compute the Hamiltonian matrix, one must know the LCAO coefficients {$$C_{K,\gamma}$$} of the orbitals which the electrons occupy. On the other hand, these LCAO coefficients are supposed to be found by solving the Hartree matrix eigenvalue equations. This paradox leads to the need to solve these equations iteratively in a so-called self-consistent field (SCF) technique. In the SCF process, one inputs an initial approximation to the {$$C_{K,\gamma}$$} coefficients. This then allows one to form the Hamiltonian matrix defined above. The Hartree matrix equations

$\sum_\mu \langle\chi_\nu| h_e| \chi_\mu\rangle C_{J,m} = \epsilon_J \sum_\mu \langle\chi_\nu| \chi_\mu\rangle C_{J,\mu} \label{6.1.10}$

are then solved for new {$$C_{K,\gamma}$$} coefficients and for the orbital energies {$$\epsilon_K$$}. The new LCAO coefficients of those orbitals that are occupied are then used to form a new Hamiltonian matrix, after which the Hartree equations are again solved for another generation of LCAO coefficients and orbital energies. This process is continued until the orbital energies and LCAO coefficients obtained in successive iterations do not differ appreciably. Upon such convergence, one says that a self-consistent field has been realized because the {$$C_{K,\gamma}$$} coefficients are used to form a Coulomb field potential that details the electron-electron interactions.

Basis Sets

Slater-type orbitals and Gaussian-type orbitals

As noted above, it is possible to use the screened hydrogenic orbitals as the {$$\chi_\mu$$}. However, much effort has been expended at developing alternative sets of functions to use as basis orbitals. The result of this effort has been to produce two kinds of functions that currently are widely used. The basis orbitals commonly used in the LCAO process fall into two primary classes:

1. Slater-type orbitals (STOs) $c_{n,l,m} (r,\theta,\phi) = N_{n,l,m,z} Y_{l,m} (\theta,\phi) r_{n-1} e^{-zr}$ are characterized by quantum numbers $$n$$, $$l$$, and $$m$$ and exponents (which characterize the orbital’s radial size) $$z$$. The symbol $$N_{n,l,m,z}$$ denotes the normalization constant.
2. Cartesian Gaussian-type orbitals (GTOs) $c_{a,b,c} (r,\theta,\phi) = N'_{a,b,c,a} x_a y_b z_c e^{-ar^2}$ are characterized by quantum numbers $$a$$, $$b$$, and $$c$$, which detail the angular shape and direction of the orbital, and exponents $$a$$ which govern the radial size.

For both types of AOs, the coordinates $$r$$, $$\theta$$, and $$\phi$$ refer to the position of the electron relative to a set of axes attached to the nucleus on which the basis orbital is located. Note that Slater-type orbitals (STO's) are similar to hydrogenic orbitals in the region close to the nucleus. Specifically, they have a non-zero slope near the nucleus

$\dfrac{d}{dr}(e^{-zr})_{r=0} = -z.$

In contrast, GTOs, have zero slope near $$r=0$$ because

$\dfrac{d}{dr}(e^{-ar^2})_{r=0} = 0.$

We say that STOs display a cusp at $$r=0$$ that is characteristic of the hydrogenic solutions, whereas GTOs do not. Although STOs have the proper cusp behavior near nuclei, they are used primarily for atomic and linear-molecule calculations because the multi-center integrals

$\langle\chi_\mu(1) \chi_\kappa(2)|\dfrac{e^2}{|r_1-r_2|}| \chi_\nu(1) \chi_\gamma(2)\rangle\label{6.1.11}$

which arise in polyatomic-molecule calculations (we will discuss these integrals later in this Chapter) cannot efficiently be evaluated when STOs are employed. In contrast, such integrals can routinely be computed when GTOs are used. This fundamental advantage of GTOs has lead to the dominance of these functions in molecular quantum chemistry.

To overcome the primary weakness of GTO functions (i.e., their radial derivatives vanish at the nucleus), it is common to combine two, three, or more GTOs, with combination coefficients which are fixed and not treated as LCAO parameters, into new functions called contracted GTOs (CGTOs). Typically, a series of radially tight, medium, and loose GTOs are multiplied by contraction coefficients and summed to produce a CGTO that approximates the proper cusp at the nuclear center (although no such combination of GTOs can exactly produce such a cusp because each GTO has zero slope at $$r = 0$$.

Although most calculations on molecules are now performed using Gaussian orbitals, it should be noted that other basis sets can be used as long as they span enough of the regions of space (radial and angular) where significant electron density resides. In fact, it is possible to use plane wave orbitals of the form

$\chi(r,\theta,\phi) = N\exp[i(k_x r \sin{\theta} \cos{\phi} + k_y r \sin_{\theta} \sin{\phi} + k_z r \cos{\theta})],\label{6.1.12}$

where $$N$$ is a normalization constant and $$k_x$$, $$k_y$$, and $$k_z$$ are quantum numbers detailing the momenta or wavelength of the orbital along the $$x$$, $$y$$, and $$z$$ Cartesian directions. The advantage to using such simple orbitals is that the integrals one must perform are much easier to handle with such functions. The disadvantage is that one must use many such functions to accurately describe sharply peaked charge distributions of, for example, inner-shell core orbitals while still retaining enough flexibility to also describe the much smoother electron density in the valence regions. Much effort has been devoted to developing and tabulating in widely available locations sets of STO or GTO basis orbitals for main-group elements and transition metals. This ongoing effort is aimed at providing standard basis set libraries which:

1. Yield predictable chemical accuracy in the resultant energies.
2. Are cost effective to use in practical calculations.
3. Are relatively transferable so that a given atom's basis is flexible enough to be used for that atom in various bonding environments (e.g., hybridization and degree of ionization).

Fundamental Core and Valence Basis

In constructing an atomic orbital basis, one can choose from among several classes of functions. First, the size and nature of the primary core and valence basis must be specified. Within this category, the following choices are common:

1. A minimal basis in which the number of CGTO orbitals is equal to the number of core and valence atomic orbitals in the atom.
2. A double-zeta (DZ) basis in which twice as many CGTOs are used as there are core and valence atomic orbitals. The use of more basis functions is motivated by a desire to provide additional variational flexibility so the LCAO process can generate molecular orbitals of variable diffuseness as the local electronegativity of the atom varies. A valence double-zeta (VDZ) basis has only one CGTO to represent the inner-shell orbitals, but uses two sets of CGTOs to describe the valence orbitals.
3. A triple-zeta (TZ) basis in which three times as many CGTOs are used as the number of core and valence atomic orbitals (of course, there are quadruple-zeta and higher-zeta bases also). Moreover, there are VTZ bases that treat the inner-shell orbitals with one CGTO and the valence orbitals with three CGTOs.

Optimization of the orbital exponents (z’s or a's) and the GTO-to-CGTO contraction coefficients for the kind of bases described above has undergone considerable growth in recent years. The theory group at the Pacific Northwest National Labs (PNNL) offer a world wide web site from which one can find (and even download in a form prepared for input to any of several commonly used electronic structure codes) a wide variety of Gaussian atomic basis sets. This site can be accessed here. Professor Kirk Peterson at Washington State University is involved in the PNNL basis set development project, but he also hosts his own basis set site.

Polarization Functions

One usually enhances any core and valence basis set with a set of so-called polarization functions. They are functions of one higher angular momentum than appears in the atom's valence orbital space (e.g., $$d$$-functions for C, N, and O and $$p$$-functions for H), and they have exponents ($$z$$ or $$a$$) which cause their radial sizes to be similar to the sizes of the valence orbitals ( i.e., the polarization $$p$$ orbitals of the H atom are similar in size to the $$1s$$ orbital rather than to the $$2s$$ valence orbital of hydrogen). Thus, they are not orbitals which describe the atom's valence orbital with one higher l-value; such higher-l valence orbitals would be radially more diffuse.

A primary purpose of polarization functions is to give additional angular flexibility to the LCAO process in forming bonding orbitals between pairs of valence atomic orbitals. This is illustrated in Figure 6.1.2 where polarization dp orbitals on C and O are seen to contribute to formation of the bonding $$p$$ orbital of a carbonyl group by allowing polarization of the carbon atom's $$p_\pi$$ orbital toward the right and of the oxygen atom's $$p_\pi$$ orbital toward the left.

Polarization functions are essential in strained ring compounds such as cyclopropane because they provide the angular flexibility needed to direct the electron density into regions between bonded atoms, but they are also important in unstrained compounds when high accuracy is required.

Diffuse Functions

When dealing with anions or Rydberg states, one must further augment the AO basis set by adding so-called diffuse basis orbitals. The valence and polarization functions described above do not provide enough radial flexibility to adequately describe either of these cases. The PNNL web site data base cited above offers a good source for obtaining diffuse functions appropriate to a variety of atoms as does the site of Prof. Kirk Peterson.

Once one has specified an atomic orbital basis for each atom in the molecule, the LCAO-MO procedure can be used to determine the $$\chi_{\mu,i}$$ coefficients that describe the occupied and virtual (i.e., unoccupied) orbitals. It is important to keep in mind that the basis orbitals are not themselves the SCF orbitals of the isolated atoms; even the proper atomic orbitals are combinations (with atomic values for the $$\chi_{\mu,i}$$ coefficients) of the basis functions. The LCAO-MO-SCF process itself determines the magnitudes and signs of the $$\chi_{\nu,i}$$. In particular, it is alternations in the signs of these coefficients allow radial nodes to form.


text:
# Suggestion for a better way to organize data to generate frequent item-sets?

I have a data of a bag of words in a document. The data has 3 columns: {document number, word number, count of the word in the number}. I am supposed to generate frequent item-sets of a particular size.

I thought that I would make list of all words that appear in a document, create a table of this list, and then generate frequent item-sets using Mlxtend or Orange . However, this approach does not seem to be efficient.

If the size is reasonable (i.e. not too many documents and not too many words in a document), you could try to build a map for each possible itemset, for instance like this:

// Assuming data is an array of size N containing all the documents
// clusters is a map associating each itemset with a set of documents
for i=0 to N-1
for j=i+1 to N-1
group = overlap(data[i], data[j])
add data[i] to the set clusters[group]
add data[j] to the set clusters[group]

An alternative version if the number of different values and size of the sets allow it and/or if it's possible to precompute the itemsets of interest:

for i=0 to N-1
for every subset S of data[i]
add data[i] to the set clusters[S]


text:
Are extensions of linear algebraic groups (over a field) themselves linear algebraic? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-23T06:59:34Z http://mathoverflow.net/feeds/question/109456 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/109456/are-extensions-of-linear-algebraic-groups-over-a-field-themselves-linear-algebr Are extensions of linear algebraic groups (over a field) themselves linear algebraic? Michael Thaddeus 2012-10-12T12:22:34Z 2012-10-12T16:26:19Z
The title says it all.

A very similar question was asked and answered about linear groups, but none of the counterexamples are algebraic: http://mathoverflow.net/questions/22814/are-extensions-of-linear-groups-linear

If $A$, $B$ are affine and there is a rational section of $C \to A$ in $1 \to B \to C \to A \to 1$, then $C \to A$ is affine, so $C$ is affine. But if not?
 http://mathoverflow.net/questions/109456/are-extensions-of-linear-algebraic-groups-over-a-field-themselves-linear-algebr/109462#109462 Answer by Angelo for Are extensions of linear algebraic groups (over a field) themselves linear algebraic? Angelo 2012-10-12T15:07:34Z 2012-10-12T16:26:19Z
Yes. The point is that $C$ is a $B$-torsor over $A$. Since being affine is a local property in the fpqc topology, $C$ is affine over $A$.

: Sorry I had not noticed grp's comment, or I wouldn't have posted an answer.

At to why there are local section, well, to me that's by the definition of an extension. Alternatively, assuming you are on a field, the injectivity of $A \to B$ means, I suppose, that $A$ is an embedding of algebraic groups. This defines a free action of $A$ on $C$; take the quotient $B/A$ (as an fppf sheaf, or étale, if $A$ is smooth); the projection $B \to B/A$ has local sections, by construction. It's a basic result that $B/A$ is represented by a group scheme. Then the exactness of the sequence should meant that $B \to C$ induces an isomorphism of $B/A$ with $C$.

If you are over an algebraically closed of characteristic 0, exactness of the sequence can be checked, in fact, at the level of closed points.


text:
# Mirror of local Calabi-Yau

What is the mirror manifold of the total space of the bundle $O(-1)\oplus O(-1)$ over $P^1$? I have tried to find the answer on the web but failed. Is there a good reference for this? Thanks.

• Try googling "resolved conifold" and "deformed conifold". Mar 18, 2010 at 18:27

The physicists (see e.g. this paper of Aganagic and Vafa) will write the mirror as a threefold $X$ which is an affine conic bundle over the holomorphic symplectic surface $\mathbb{C}^{\times}\times \mathbb{C}^{\times}$ with discriminant a Seiberg-Witten curve $\Sigma \subset \mathbb{C}^{\times}\times \mathbb{C}^{\times}$. In terms of the affine coordinates $(u,v)$ on $\mathbb{C}^{\times}\times \mathbb{C}^{\times}$, the curve $\Sigma$ is given by the equation $$\Sigma : \ u + v + a uv^{-1} + 1 = 0,$$ and so $X$ is the hypersurface in $\mathbb{C}^{\times}\times \mathbb{C}^{\times} \times \mathbb{C}^2$ given by the equation $$X : \ xy = u + v + a uv^{-1} + 1.$$
From geometric point of view it may be more natural to think of the mirror not as an affine conic fibration over a surface but as an affine fibration by two dimensional quadrics over a curve. The idea will be to start with the Landau-Ginzburg mirror of $\mathbb{P}^{1}$, which is $\mathbb{C}^{\times}$ equipped with the superpotential $w = s + as^{-1}$ and to consider a bundle of affine two dimensional quadrics on $\mathbb{C}^{\times}$ which degenerates along a smooth fiber of the superpotential, e.g. the fiber $w^{-1}(0)$. In this setting the mirror will be a hypersurface in $\mathbb{C}^{\times}\times \mathbb{C}^{3}$ given by the equation $$xy - z^2 = s + as^{-1}.$$ Up to change of variables this is equivalent to the previous picture but it also makes sense in non-toric situations. Presumably one can obtain this way the mirror of a Calabi-Yau which is the total space of a rank two (semistable) vector bundle of canonical determinant on a curve of higher genus.
In some sense the answer is $T^*S^3$. Look at http://arxiv.org/pdf/hep-th/0211098v1


text:
# High contrast stacked distribution plots

A while ago a colleague showed me an image of a beautiful way of plotting several distributions to make them easily comparable. I have since failed to find these plots or managed to figure out what they are called. But the concept was to plot densities for several variables in such a way that they appeared behind each other in depth.

Let me first illustrate the effect before we get to an implementation.

%pylab inline
import pandas as pd
import seaborn as sns
sns.set_style('white')

xx = np.linspace(0, 4 * np.pi, 512)

We just make some random sinus curves for illustration

df = pd.DataFrame()
for i in range(1, 32):
df[i] = np.sin((np.random.rand() + 0.5) * xx) + 1

figsize(16, 16)
for i, ind in enumerate(df):
offset = 1.1 * i
plt.fill_between(xx, df[ind] + offset, 0 * df[ind] + offset,
zorder=-i,
facecolor='k',
edgecolor='w',
lw=3)

sns.despine(left=True, bottom=True)
plt.xticks([])
plt.yticks([]);

plt.savefig('plots/1.png');

There are just a few small tricks to the effect. The zorder=-i parameter makes the curves stack in the correct order. Making the edge colors the same color as the background gives the effect of depth. Personally I like the very high contrast style of using black fills. (Another colleague who saw these plots called them “Moomin plots” )

You can also make some pretty, regular, patterns with this look.

df = pd.DataFrame()
for i in range(1, 16):
df[i] = np.sin(i * xx) + 1

figsize(14, 8)
for i, ind in enumerate(df):
offset = 1.1 * i
plt.fill_between(xx, df[ind] + offset, 0 * df[ind] + offset,
zorder=-i,
facecolor='k',
edgecolor='w',
lw=3)

sns.despine(left=True, bottom=True)
plt.xticks([])
plt.yticks([]);

plt.savefig('plots/2.png');

So how can we plot distribution for comparison with this technique? First let us create some simulated data. In this case I’m randomly mixing unimodal and bimodal normally distributed data, this is what a lot of the data we study in my research group looks like. Letting the mean be between 0 and 16 is arbitrary of course, and I just picked the variance to be 2 because I thought it looked good.

def unimodal():
m = random.randint(0, 16)
return random.randn(256) * 2 + m

def bimodal():
m1 = random.randint(0, 16)
s1 = random.randn(128) * 2 + m1
m2 = random.randint(0, 16)
s2 = random.randn(128) * 2 + m2
return np.hstack((s1, s2))

df = pd.DataFrame()
distributions = [unimodal, bimodal]
for i in range(16):
distribution = random.choice(distributions)
df[i] = distribution()

We’ll use the Gaussian KDE from scipy for this.

from scipy.stats import kde

Now we pick an x-axis which should cover the entire range of the data in all the columns of the DataFrame we keep the data in.

We handle the y-offsets of the data by picking half the height of the previous density plot. This can be tweaked and give slightly different effect dependong on what fraction of the height is used. I found half height to be most generally good looking though.

To make it easy to relate a label to a distribution we plot a line the density can “rest on”. If it is thick enough it will give the effect of the density growing up from it.

figsize(6, 8)
xx = np.linspace(df.min().min(), df.max().max(), 256)
yy = None
offset = 0
offs = [offset]
for i in df:
if yy is not None:
offset += 0.5 * yy.max()
offs.append(offset)

s = df[i]
f = kde.gaussian_kde(s)
yy = f(xx)

plt.fill_between(xx, yy + offset, 0 * yy + offset,
zorder=-i,
facecolor='k',
edgecolor='w',
lw=3)

plt.axhline(y=offset,
zorder=-i,
linestyle='-',
color='k',
lw=4)

sns.despine(left=True, bottom=True)
plt.yticks(offs, ['Sample {}'.format(s + 1) for s in df.columns]);

plt.tight_layout()
plt.savefig('plots/3.png');

I like these a lot, though it looks a bit messy. Of course, with real data the ordering of the variables on the y-axis would be informed by some prior knowledge so that trends could be easily spotted.

But even so, if we don’t know how we should order these for easy comparison between the distributions, we can use hierarchical clustering to arrange the distribution so they become more comparable.

import fastcluster as fc
from scipy.cluster.hierarchy import dendrogram

figsize(6, 8)

dend['leaves']

xx = np.linspace(df.min().min(), df.max().max(), 256)
yy = None
offset = 0
offs = [offset]
for i, ind in enumerate(df[dend['leaves']]):
if yy is not None:
offset += 0.5 * yy.max()
offs.append(offset)

s = df[ind]
f = kde.gaussian_kde(s)
yy = f(xx)

plt.fill_between(xx, yy + offset, 0 * yy + offset,
zorder=-i,
facecolor='k',
edgecolor='w',
lw=3)

plt.axhline(y=offset,
zorder=-i,
linestyle='-',
color='k',
lw=4)

sns.despine(left=True, bottom=True)
plt.yticks(offs, ['Sample {}'.format(s + 1) for s in df[dend['leaves']].columns]);

plt.tight_layout()
plt.savefig('plots/4.png');


text:
# Cyclohexene conformation in fused ring

The picture (a) is from a organic chemistry book by Clayden, Chapter 32. That means drawing (a) is right expression of the molecule.

What I wonder is that is it possible for (a) to ring-flip? If so, is (b) correct as a ring flip version of (a)? If it's not possible to ring flip, why?

Second question: Is drawing (c) also valid to draw this molecule? If it's so, is it same with (a)? or is it ring flip of (a)? If it's not, what's wrong with it?

Personally, I thought (c) was the right drawing for the molecule. Below (c), I drew the conformation of cyclohexene. The first conformation has a pseudoaxial hydrogen and the inner side of carbon faces downward. This is why I drew (c) with methyl at pseudoaxial, and inner carbon facing downward. But the book only has a drawing of (a).

• You can also just tell me where I can find those kind of conformation, in book, website, anything because my four organic chemistry books are needless now. – Zillai Mar 28 '18 at 15:54
• I noticed that you had typed your question in proper paragraphs to insert a line break use double space or <br>. – Avnish Kabaj Mar 29 '18 at 0:09
• I suggest you build a model (simple molymod is fine) - it will all become clear (and much easier than any 2D representation that I could draw for you) – NotEvans. Mar 30 '18 at 11:42
• @NotEvans thanks maybe I will buy one later. I don't have one now. Do you know any website that I can see the ringflip molecular form? Chemspider shows me a good 3d image but no function to change conformation. – Zillai Mar 30 '18 at 16:09
• You can ring-flip the enone ring but the methyl must be axial to both rings. If the methyl were equatorial, the double bond would be too twisted. In other words in this example, if a double bond is exocyclic to ring B (steroid lettering), there aren't enough carbons in ring A to attached axially to ring B. Models would be helpful. – user55119 Apr 5 '18 at 20:11


text:
Home ActiveRecord joining data from 2 tables with one query without looping
 I have a companies table and a leads table. I am pulling a bunch of leads and need to combine each lead with 2 ID columns from the companies table. I want to keep the companies table as the single source of truth for these ID's (and hence not adding these columns to the leads table where it could get out of sync). Right now I pull all the leads and loop through everyone to add this data. I am assuming there's a more efficient way to do this but I'm not sure how. Can someone help? leads = @user.find_leads(start_date: start_date, end_date: end_date).leads('true').includes(:company).select(fields).limit(limit).offset(offset).order("created_at DESC") leads.each do |l| company = l.company.present? ? l.company.company_external_id : "" distributor = l.company.present? ? l.company.mfg_distributor_id : "" lead_data = l.attributes.merge({external_company_id: company, external_distributor_id: distributor}) final_lead = lead_data.each_with_object({}) { |(k,v),g| g[k] = (Hash === v) ? denilize(v) : v ? v : "" } all_leads << final_lead.stringify_keys end


text:
ACFCI - ACF Confidence Interval Function

Calculates the confidence interval limits (upper/lower) for the autocorrelation function.

Syntax

ACFCI(X, Order, K, Method, alpha, upper)

X is the univariate time series data (a one dimensional array of cells (e.g. rows or columns)).

Order is the time order in the data series (i.e. the first data point's corresponding date (earliest date=1 (default), latest date=0)).

Order Description
1 ascending (the first data point corresponds to the earliest date) (default)
0 descending (the first data point corresponds to the latest date)

K is the lag order (e.g. k=0 (no lag), k=1 (1st lag), etc.). If missing, the default of k=1 is assumed.

Method is the calculation method for estimating the autocorrelation (0= Sample Autocorrelation (Default), 1=Periodogram-based estimate , 2=Cross Correlation).

Value Method
0 Sample autocorrelation method.(default)
1 Periodogram-based estimate.
2 Cross-correlation method

alpha is the statistical significance level. If missing, the default of 5% is assumed.

upper If true, returns the upper confidence interval limit. Otherwise, returns the lower limit.

Value Description
0 return lower limit
1 return upper limit

Remarks

1. The time series is homogeneous or equally spaced.
2. The time series may include missing values (e.g. #N/A) at either end.
3. The lag order (k) must be less than the time series size, or else an error value (#VALUE!) is returned.
4. The ACFCI function calculates the confidence limits as: $$\hat\rho_k - Z_{\alpha/2}\times \sigma_{\rho_k} \leq \rho_k \leq \hat\rho_k+ Z_{\alpha/2}\times \sigma_{\rho_k}$$ Where:
• $\rho_k$ is the population autocorrelation function.
• $\sigma_{\rho_k}$ is the standard error of the sample autocorrelation.
• $\hat{\rho_{k}}$ is the sample autocorrelation function for lag k.
• $Z\sim N(0,1)$
• $P(\left|Z\right|\geq Z_{\alpha/2}) = \alpha$
5. For the case in which the underlying population distribution is normal, the sample autocorrelation also has a normal distribution:

$\hat \rho_k \sim N(\rho_k,\sigma_{\rho_k}^2)$

Where:
• $\hat \rho_k$ is the sample autocorrelation for lag k.
• $\rho_k$ is the population autocorrelation for lag k.
• $\sigma_{\rho_k}$ is the standard error of the sample autocorrelation for lag k.
6. Bartlett proved that the variance of the sample autocorrelation of a stationary normal stochastic process (i.e. independent, identically normal distributed errors) can be formulated as: $$\sigma_{\rho_k}^2 = \frac{\sum_{j=-\infty}^{\infty}\rho_j^2+\rho_{j+k}\rho_{j-k}-4\rho_j\rho_k\rho_{i-k}+2\rho_j^2\rho_k^2}{T}$$
7. Furthermore, the variance of the sample autocorrelation is reformulated: $$\sigma_{\rho_k}^2 = \frac{1+\sum_{j=1}^{k-1}\hat\rho_j^2}{T}$$ Where:
• $\sigma_{\rho_k}$ is the standard error of the sample autocorrelation for lag k.
• $T$ is the sample data size.
• $\hat\rho_j$ is the sample autocorrelation function for lag j.
• $k$ is the lag order.

Examples

Example 1:

A B
1 Date Data
2 1/1/2008 #N/A
3 1/2/2008 -1.28
4 1/3/2008 0.24
5 1/4/2008 1.28
6 1/5/2008 1.20
7 1/6/2008 1.73
8 1/7/2008 -2.18
9 1/8/2008 -0.23
10 1/9/2008 1.10
11 1/10/2008 -1.09
12 1/11/2008 -0.69
13 1/12/2008 -1.69
14 1/13/2008 -1.85
15 1/14/2008 -0.98
16 1/15/2008 -0.77
17 1/16/2008 -0.30
18 1/17/2008 -1.28
19 1/18/2008 0.24
20 1/19/2008 1.28
21 1/20/2008 1.20
22 1/21/2008 1.73
23 1/22/2008 -2.18
24 1/23/2008 -0.23
25 1/24/2008 1.10
26 1/25/2008 -1.09
27 1/26/2008 -0.69
28 1/27/2008 -1.69
29 1/28/2008 -1.85
30 1/29/2008 -0.98

Formula Description (Result)
=ACF(\$B\$2:\$B\$30,1,1) Autocorrelation of order 1 (0.235)
=ACFCI(\$B\$2:\$B\$30,1,1,5%,1) Upper confidence interval for ACF of order 1 (0.37)
=ACFCI(\$B\$2:\$B\$30,1,1,5%,0) lower confidence interval for ACF of order 2 (-0.37)


text:
# 2000

## On the rate of convergence of a positive approximation process

Abstract In this paper we are dealing with a class of summation integral operators on unbounded interval generated by a…

## Application of Popoviciu’s high convexity to the study of some sequences properties

Abstract Starting by the notion of convex funcitons of $$n$$-order introduced by Tiberiu Popoviciu, we aim to record those theorems…

## More than a summing up about Meyer-Konig and Zeller operators

Abstract AuthorsOctavian Agratini Department of Mathematics, Babes-Bolyai University, Cluj-Napoca, Romania Keywords? Paper coordinatesO. Agratini, More than a summing up about…

## On a certain class of approximation operators

Abstract The paper is devoted to the study of an approximation process $$K_{n}^{H}$$ representing an integral form in Kantorovich sense…

## Fixed point theorems for set-valued maps and existence principles for integral inclusions

AbstractNew fixed point theorems of Mönch type are presented for set-valued maps. These theorems are then used to establish general…

## A Monch type generalization of the Eilenberg-Montgomery fixed point theorem

Abstract We establish a Monch type generalization of the Eilenberg-Montgomery fixed point theorem for multi-valued maps with acyclic values. AuthorsRadu…

## Discrete continuation methods for boundary value problems on bounded sets in Banach spaces

AbstractThe paper deals with existence, uniqueness and iterative approximation of solutions to boundary value problems for second-order differential equations on…

## Discrete continuation method for nonlinear integral equations in Banach spaces

AbstractThe continuation theorem for contractive mappings on spaces endowed with two metrics is used to obtain existence, uniqueness and iterative…

## Programmed motion for a class of families of planar orbits

AbstractAuthorsKeywordsReferencesPDFScanned paper. Latex version of the paper. Cite this paper as:Anisiu MC., Bozis G., Programmed motion for a class of families…

## An implicit numerical spline method for systems for ODEs

Abstract An implicit numerical method is introduced that gives piecewise polynomial spline approximations for the solution of an initial value…


text:
# Chapter 26 - The Electric Field - Exercises and Problems - Page 775: 7

The electric field strength 5.0 cm from the wire is 4000 N/C

#### Work Step by Step

The strength of the electric field at a distance $r$ from a charged wire is $E = \frac{\lambda}{2\pi ~\epsilon_0~r}$ Let $r_1 = 10.0~cm$. We can write an expression for the electric field strength at a distance of $r_1$: $E_1 = \frac{\lambda}{2\pi ~\epsilon_0~r_1}$ Let $r_2 = 5.0~cm$. We can write an expression for the electric field strength at a distance of $r_2$: $E_2 = \frac{\lambda}{2\pi ~\epsilon_0~r_2}$ We can divide $E_1$ by $E_2$: $\frac{E_1}{E_2} = \frac{\frac{\lambda}{2\pi ~\epsilon_0~r_1}}{\frac{\lambda}{2\pi ~\epsilon_0~r_2}}$ $E_2 = \frac{r_1~E_1}{r_2}$ $E_2 = \frac{(10.0~cm)(2000~N/C)}{5.0~cm}$ $E_2 = 4000~N/C$ The electric field strength 5.0 cm from the wire is 4000 N/C.

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.


text:
# Strontium Valence Electrons

### Learning Objectives

1. Anaturally occurring amino acid with the formula c9h11no3 gives the following proton nmr spectrum when determined in deuterium oxide solvent with dcl added. The amino, carboxyl, and hydroxyl protons merge into a single peak at 5.1 ppm (4 h) in d2o. Determine the struc- ture of this amino acid and explain it.
2. Valence electrons - electrons in outermost shell, for Sr this is energy level 6 Strontium has 6 energy levels and 4 valence electrons. Note: If you are in a grade 12 class or later then you probably need to use the more complex (but correct) energy level model, but this is the simple Bohr model you use through most of high school: 2 electrons.
3. Strontium S Element 38 Mass Number: 88 Atomic weight: 87.62 g/mol Atomic number (Z): 38 Electrons: 38 Protons: 38 Neutrons: 50 Period: 5 Group: 2 Block: s.
4. Comprehensive information for the element Strontium - Sr is provided by this page including scores of properties, element names in many languages, most known nuclides and technical terms are linked to their definitions.

Determining Valence Electrons 1. Give the correct number of valence electrons for the element strontium, Sr, atomic #38. Which of the following electron dot notations is correct for the element calcium, Ca, atomic #20? Which of the noble gases does not have eight electrons in its outermost shell?

By the end of this section, you will be able to:

• Write Lewis symbols for neutral atoms and ions

## Lewis Symbols of Monoatomic Elements

In almost all cases, chemical bonds are formed by interactions of valence electrons in atoms. To facilitate our understanding of how valence electrons interact, a simple way of representing those valence electrons would be useful.

A Lewis electron dot diagram (or electron dot diagram or a Lewis diagram or a Lewis structure) is a representation of the valence electrons of an atom that uses dots around the symbol of the element. The number of dots equals the number of valence electrons in the atom. These dots are arranged to the right and left and above and below the symbol, with no more than two dots on a side. (It does not matter what order the positions are used.)

For example, the Lewis electron dot diagram for calcium is simply

Figure 1 shows the Lewis symbols for the elements of the third period of the periodic table.

Figure 1. Lewis symbols illustrating the number of valence electrons for each element in the third period of the periodic table.

Lewis symbols can also be used to illustrate the formation of cations from atoms, as shown here for sodium and calcium:Likewise, they can be used to show the formation of anions from atoms, as shown below for chlorine and sulfur:Figure 2 demonstrates the use of Lewis symbols to show the transfer of electrons during the formation of ionic compounds.

Figure 2. Cations are formed when atoms lose electrons, represented by fewer Lewis dots, whereas anions are formed by atoms gaining electrons. The total number of electrons does not change.

### Example 1: Writing Lewis DoT SYmbols of Elements

What is the Lewis electron dot diagram for each element?

1. aluminum
2. selenium
1. The valence electron configuration for aluminum is 3s23p1. So it would have three dots around the symbol for aluminum, two of them paired to represent the 3s electrons (or three single dots around the atom):

2. The valence electron configuration for selenium is 4s24p4. In the highest-numbered shell, the n = 4 shell, there are six electrons. Its electron dot diagram is as follows:

#### Check Your Learning

What is the Lewis electron dot diagram for each element?

1. phosphorus
2. argon

### Example 2: Writing Lewis DoT SYmbols of Ions

What is the Lewis electron dot diagram for each ion?

1. Ca2+
2. O2−
1. Having lost its two original valence electrons, the Lewis electron dot diagram is just Ca2+.

Ca2+

2. The O2− ion has gained two electrons in its valence shell, so its Lewis electron dot diagram is as follows:

#### Check Your Learning

The valence electron configuration of thallium, whose symbol is Tl, is 6s25d106p1. What is the Lewis electron dot diagram for the Tl+ ion?

### Key Takeaways

• Lewis electron dot diagrams use dots to represent valence electrons around an atomic symbol.
• Lewis electron dot diagrams for ions have fewer (for cations) or more (for anions) dots than the corresponding atom.

### Exercises

1. Explain why the first two dots in a Lewis electron dot diagram are drawn on the same side of the atomic symbol.

2. Is it necessary for the first dot around an atomic symbol to go on a particular side of the atomic symbol?

3. What column of the periodic table has Lewis electron dot diagrams with two electrons?

4. What column of the periodic table has Lewis electron dot diagrams that have six electrons in them?

5. Draw the Lewis electron dot diagram for each element.

a) strontium

b) silicon

6. Draw the Lewis electron dot diagram for each element.

a) krypton

b) sulfur

7. Draw the Lewis electron dot diagram for each element.

a) titanium

b) phosphorus

8. Draw the Lewis electron dot diagram for each element.

a) bromine

b) gallium

9. Draw the Lewis electron dot diagram for each ion.

a) Mg2+

b) S2−

10. Draw the Lewis electron dot diagram for each ion.

a) In+

b) Br

11. Draw the Lewis electron dot diagram for each ion.

a) Fe2+

b) N3−Movies about knives.

12. Draw the Lewis electron dot diagram for each ion.

a) H+

b) H

1. The first two electrons in a valence shell are s electrons, which are paired.

3. The second column of the periodic table

5.

a)

b)

7.

a)

b)

9.

a) Mg2+

b)

11.

a) Fe2+

b)

### One of the outermost electrons in a strontium atom in the ground state can be described by what sets of four quantum numbers?

#### Explanation:

Your starting point for this problem will be the electron configuration for a neutral strontium atom in its ground state, which looks like this

$' S r : ' 1 {s}^{2} 2 {s}^{2} 2 {p}^{6} 3 {s}^{2} 3 {p}^{6} 3 {d}^{10} 4 {s}^{2} 4 {p}^{6} \textcolor{red}{5} {s}^{2}$

As you can see, strontium's outermost electrons are located on the fifth energy level, in the 5s-orbital.

In order to completely describe these two electrons, you need to list the values of the four quantum numbers used to characterize their location and spin.

So, you know that the two electrons are located on the fifth energy level, which means that the principal quantum number, $n$, will be equal to $\textcolor{red}{5}$.

The angular momentum quantum number, $l$, which describes the subshell in which the electrons reside, will take the value specific for the s-subshell, i.e. $l$ will be equal to $0$.

For the s-subshell, the magnetic quantum number, ${m}_{l}$, can only take one possible value, and that is ${m}_{l} = 0$. Here ${m}_{l} = 0$ describes the s-orbital found in the s-subshell.

Finally, the spin quantum number, ${m}_{s}$, can take two possible values, ${m}_{2} = \pm \frac{1}{2}$.

### Strontium Ion Valence Electrons

Since two electrons occupy the 5s-orbital, one will have ${m}_{s} = \frac{1}{2}$, which describes spin-up, and the other ${m}_{s} = - \frac{1}{2}$, which describes spin-down.

Therefore, the quantum sets that describe the two outermost electrons electrons in a strontium atom will be

$n = 5 , l = 0 , {m}_{l} = 0 , {m}_{s} = + \frac{1}{2}$

and

### Ground State Electron Configuration Strontium

$n = 5 , l = 0 , {m}_{l} = 0 , {m}_{s} = - \frac{1}{2}$


text:
Trigonometric Substitution

1. Sep 20, 2007

Lanza52

$$\int\sqrt{16-(2x)^{4}}xdx$$

Hint says you may like to use the identity sin(theta)cos(theta)= sin(2theta)/2

However, I think I found a way to use 1-sin^2(theta)=cos^2(theta)

First, (2x)^4 = 16x^4

So make it 16(1-x^2)^2.

Take the 16 out of the root and the integral and you have:

$$4\int\sqrt{1-(x^{2})^{2}}xdx$$

Sub in sin(theta) for x^2. Sub in cos^2(theta) for 1-sin^2(theta). Sub in $$\sqrt{sin(\vartheta)$$ for the x outside the root and $$\frac{1}{2\sqrt{sin(\vartheta)}}$$ for dx.

Cancel out the sqrt{sin} on the top and bottom and drag the 1/2 outside the integral. Anti-derivative of cos is sin and you have 2sin(theta). Sin(theta) = x^2.

And so the final answer is; $$2x^{2}+c$$

Correct?

2. Sep 20, 2007

Dick

Is the derivative of 2x^2+c the same as the original integrand? I don't think so.

3. Sep 20, 2007

Lanza52

Hrmm....could you point me in the right direction?

4. Sep 20, 2007

Dick

Your substitution sin(theta)=x^2 is a great start. So cos(theta)*dtheta=2*x*dx. See where that takes you. I promise that's heading in the right direction.

5. Sep 20, 2007

Lanza52

Still a little stumped.

Realized I missed the chain rule in 2 steps in my original post...damn rustiness.

Got to $$\frac{-sin\vartheta(cos\vartheta)^{3}}{3}$$

First and foremost, am I correct to this point? Then, I can't figure out the back substitutions here.

6. Sep 20, 2007

Dick

I get to integral of cos(theta)^2.

7. Sep 20, 2007

Lanza52

Yea, I posted the anti-derivative of cos(theta)^2. I'm at the point where you need to substitute in x's for the trig functions.

8. Sep 20, 2007

Dick

That is NOT the antiderivative of cos(theta)^2. To do the antiderivative of cos(theta)^2 use that cos(2*theta)=cos(theta)^2-sin(theta)^2 (from the addition relation for cos) and that sin(theta)^2=1-cos(theta)^2. Solve for cos(theta)^2 in terms of cos(2*theta). Then it's easy to integrate cos(2*theta). After the integration is done you'll find that the hint actually does come in handy for expressing the result in terms of x.

9. Sep 20, 2007

Dick

Hint: When you think something is the antiderivative of something else then try differentiating it to check. Could save you some time, since you seem to be getting antiderivatives wrong a lot.

10. Sep 20, 2007

Dick

You could also try checking an integral table or an online site that computes integrals to check yourself. You can't do that on an exam, but it could be handy to avoid wasting time while practicing with homework problems.

11. Sep 20, 2007

Lanza52

Definitely having problems with the little/old things. That year off killed me. Just have to do all the little things a few times before it comes back to me.

Anyways. Came up with 2cos^2(theta)=cos2theta + 1

So integral cos2theta + 1

Comes to Sintheta * costheta + theta

Don't know what to do with that theta at the end. So I'm guessing I'm wrong.

12. Sep 20, 2007

Dick

You aren't that wrong this time. theta=arcsin(x^2). And what does sin(theta)*cos(theta) become? Hammer the multiplicative constants down and I think you've got it.

13. Sep 20, 2007

Lanza52

Just checking to make sure I'm not aiming for the wrong thing;

I'm currently at sin(2theta)/2 + arcsin(x^2)+c

An integration website tells me that I should have $$\sqrt{1-x^{4}}x^{2}$$+ arcsin(x^2)+c

Was the site right?

14. Sep 20, 2007

Dick

It's time to use the hint for sin(2theta)! You and the site are both right.

Last edited: Sep 20, 2007
15. Sep 20, 2007

Lanza52

Okay, so I got to the answer that the site gave me. But question, wasn't that 4 brought outside the integral used while converting cos^2theta to cos2theta and when solving for dx?

16. Sep 20, 2007

Dick

Probably. Sorry, I'm fading a bit and didn't want to check it myself. Just wanted to make sure you did.

17. Sep 20, 2007

Lanza52

Alright, thanks. =P

I owe ya, but I imagine there isn't much math I could help ya with.

18. Sep 20, 2007

Dick

You help me by making me feel that my time was not wasted on this internet busyness. The quicker you catch on, the better I feel. You're getting better, right?